ELEMENTS OF
TOPOLOGY
K13301_FM.indd 1
4/12/13 1:25 PM
K13301_FM.indd 2
4/12/13 1:25 PM
ELEMENTS OF
TOPOLOGY
Tej Bahadur Singh
K13301_FM.indd 3
4/12/13 1:25 PM
CRC Press
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Dedicated to my mother
and grandchildren
Amishi, Amil and Pradyumn
Contents
Author Bio
xi
Preface
xiii
Suggested Course Outlines
xvii
Acknowledgements
xix
List of Symbols
xxi
1 TOPOLOGICAL SPACES
1.1
1.2
1.3
1.4
1.5
Metric Spaces .
Topologies . . .
Derived Concepts
Bases . . . . . .
Subspaces . . . .
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2 CONTINUITY AND PRODUCTS
2.1
2.2
35
Continuity . . . . . . . . . . . . . . . . . . . . . . . . .
Product Topology . . . . . . . . . . . . . . . . . . . . .
3 CONNECTEDNESS
3.1
3.2
3.3
3.4
Connected Spaces . . . . . .
Components . . . . . . . . .
Path-connected Spaces . . .
Local Connectivity . . . . . .
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4 CONVERGENCE
4.1
1
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12
19
30
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . .
63
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93
vii
viii
4.2
4.3
4.4
Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . 106
5 COUNTABILITY AXIOMS
5.1
5.2
113
1st and 2nd Countable Spaces
Separable and Lindelöf Spaces
. . . . . . . . . . . . . . 113
. . . . . . . . . . . . . . 119
6 COMPACTNESS
6.1
6.2
6.3
6.4
6.5
Compact Spaces . . . . . .
Countably Compact Spaces
Compact Metric Spaces . .
Locally Compact Spaces .
Proper Maps . . . . . . . .
125
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7 TOPOLOGICAL CONSTRUCTIONS
7.1
7.2
7.3
7.4
7.5
7.6
Quotient Spaces . . . . . . . . . . .
Identification Maps . . . . . . . . .
Cones, Suspensions and Joins . . . .
Wedge Sums and Smash Products .
Adjunction Spaces . . . . . . . . . .
Coinduced and Coherent Topologies
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159
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8 SEPARATION AXIOMS
8.1
8.2
8.3
8.4
Regular Spaces . . . . . . . .
Normal Spaces . . . . . . . .
Completely Regular Spaces .
Stone–Čech Compactification
159
173
180
188
195
202
211
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9 PARACOMPACTNESS AND METRISABILITY
9.1
9.2
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155
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211
216
229
235
241
Paracompact Spaces . . . . . . . . . . . . . . . . . . . . 241
A Metrisation Theorem . . . . . . . . . . . . . . . . . . 252
ix
10 COMPLETENESS
257
10.1 Complete Spaces . . . . . . . . . . . . . . . . . . . . . . 257
10.2 Completion . . . . . . . . . . . . . . . . . . . . . . . . . 265
10.3 Baire Spaces . . . . . . . . . . . . . . . . . . . . . . . . 269
11 FUNCTION SPACES
275
11.1 Topology of Pointwise Convergence . . . . . . . . . . . 275
11.2 Compact-Open Topology . . . . . . . . . . . . . . . . . 283
11.3 Topology of Compact Convergence . . . . . . . . . . . . 301
12 TOPOLOGICAL GROUPS
12.1
12.2
12.3
12.4
Examples and Basic Properties
Subgroups . . . . . . . . . . .
Isomorphisms . . . . . . . . . .
Direct Products . . . . . . . .
313
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13 TRANSFORMATION GROUPS
313
324
331
341
347
13.1 Group Actions . . . . . . . . . . . . . . . . . . . . . . . 347
13.2 Orbit Spaces . . . . . . . . . . . . . . . . . . . . . . . . 365
14 THE FUNDAMENTAL GROUP
14.1
14.2
14.3
14.4
14.5
371
Homotopic Maps . . . . . . . . . .
The Fundamental Group . . . . .
Fundamental Groups of Spheres .
Some Group Theory . . . . . . . .
The Seifert–van Kampen Theorem
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15 COVERING SPACES
15.1
15.2
15.3
15.4
15.5
Covering Maps . . . . . . . . . .
The Lifting Problem . . . . . . .
The Universal Covering Space .
Deck Transformations . . . . . .
The Existence of Covering Spaces
371
383
397
408
424
439
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439
448
459
468
480
x
A Set Theory
A.1
A.2
A.3
A.4
A.5
A.6
A.7
A.8
Sets . . . . . . . . . . . .
Functions . . . . . . . . .
Cartesian Products . . .
Equivalence Relations . .
Finite and Countable Sets
Orderings . . . . . . . . .
Ordinal Numbers . . . .
Cardinal Numbers . . . .
B Fields R, C and H
483
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483
485
488
491
492
500
507
512
519
B.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . 519
B.2 The Complex Numbers . . . . . . . . . . . . . . . . . . 521
B.3 The Quaternions . . . . . . . . . . . . . . . . . . . . . . 523
Bibliography
525
Index
527
Author Bio
Dr. Tej Bahadur Singh is Professor in the Department of Mathematics,
University of Delhi, Delhi. He received his master’s degree from the
University of Delhi, Delhi, and doctorate degree from the University
of Allahabad, Allhabad. Since 1989 he has been at the University of
Delhi; prior to this, he has served the University of Allahabad and
Atarra P.G. College, Banda. He has written several research articles
on the cohomological theory of compact transformation groups. He has
taught graduate courses on General Topology and Algebraic Topology
many times during his teaching career.
xi
Preface
Topology is a branch of mathematics that studies the properties of
geometric objects which remain unaltered under “deformation.” The
idea of deformation involves strongly the notion of continuity. The notions of continuity of functions and convergence of sequences are the
two most fundamental concepts in analysis. Both concepts are based
on the abstraction of our intuitive sense of closeness of points of a set.
“Closeness” of elements of a set can be measured most conveniently
as distance between the elements. In any set endowed with a suitable
notion of distance, one can define convergence of sequences and talk
about continuity of functions between two such sets. Probably motivated by this observation, Maurice Fréchet (1906) introduced “metric
spaces.” In metric spaces, most of the important notions, for example,
limits, continuity, connectedness, compactness, etc. may be described,
and many important theorems of analysis can be proved solely in terms
of open sets. So, it is useful to abstract the basic properties of open
sets, and introduce a notion that is suitable for talking about these
concepts and is also independent of the idea of metrics. This led Felix Hausdorff (1914) to give the definition of “topological spaces” by
abstracting the basic properties of open sets. Topological spaces provide the most general setting for studying the notions of convergence
and continuity. The study of the properties of topological spaces which
are preserved by “homeomorphisms” (invertible continuous maps with
continuous inverses) is the subject matter of the (point-set) topology.
Historically, topology has roots scattered in nineteenth century
works on analysis and geometry. The term “topology” (actually
“topologie” in German) was coined by J.B. Listing in 1836. The seminal work of Henri Poincaré (“Analysis Situs” and its five compliments
published during 1895 - 1909) marks the beginning of the subject of
(combinatorial) topology. By the late twenties, topology has evolved as
a separate discipline, and it is now a large subject with many branches,
broadly categorised as algebraic topology, general topology (or pointset topology) and geometric topology (or the theory of manifolds). In
xiii
xiv
Elements of Topology
fact, point-set topology is today the main language for a broad variety
of mathematical disciplines, while algebraic topology serves as a powerful tool for studying the problems in geometry and many other areas
of mathematics.
The rapid growth of topology has led to the publication of many
excellent textbooks and treatises on the subject. However, most of the
existing books on the subject require a level of maturity and sophistication on the part of the reader which is rather beyond what is achieved
in mathematics undergraduate courses at many universities. The objective of the present work is to encourage average students to study
topology by providing “stepping stones” to help them into the subject.
It is intended to impart the important and useful ideas of present-day
mathematics to the reader. Accordingly, the book introduces the rudiments of general topology and algebraic topology as a part of the basic
vocabulary of mathematics for higher studies. Of course, no claim of
originality can be made in writing a book at this level. My contribution,
if any, is one of presentation and selection of the material. Certainly,
any selection of material is governed by one’s personal taste. It was felt
that even a short introduction to topological groups and transformation groups would make the subject more interesting. For, these partly
geometric objects form a rich territory of interesting examples in topology and geometry, and play an increasingly important role in modern
mathematics and physics. To keep the volume within reasonable limits, some topics such as “uniform spaces” and “the general metrisation
theorem” have been omitted.
This book is based on my experience in teaching the courses in
topology to Undergraduate and Graduate students at the University of
Delhi and elsewhere over the years. All the material presented here has
been found quite accessible by the students who have an elementary
knowledge of analysis, linear algebra and some group theory. With
these exceptions, we have collected enough material in the appendices
to make the book self contained. Some important properties of the
real numbers, complex numbers and quaternionic numbers are briefly
described in Appendix B. At certain points familiarity with cardinal
numbers and ordinal numbers are also assumed; necessary background
knowledge about these can be had from Appendix A.
The book can be organised into four main parts. The first part
comprises Chapters 1 through 7, and can be considered as the core of
the book. It deals with the notions of topological spaces, continuous
functions, connectedness, convergence, compactness and countability
Preface
xv
axioms. The discussion of quotient spaces has been postponed until
Chapter 6, for many of these spaces can be easily realised by the students who are familiar with certain topological tools. The material of
this part is now used in several branches of mathematics, and is suitable for a one-semester first course in general topology for advanced
undergraduates. The second part consisting of Chapters 8 through 11
is devoted to some more topics of point-set topology, specifically, separation axioms, paracompactness, metrisability, completeness and function spaces. The results of Chapters 10 and 11 are important to analysts. In the next part of the book, we present pretty basic information
on topological groups and some elementary facts about the actions of
these on topological spaces. It contains a study of classical groups, and
thus concretises the theory discussed in the preceding two parts. Based
on the direct observation of a rotating rigid body, a geometric meaning
has been given to the technical term ‘rotation’, and it is justified that
the rotation of an euclidean space at a particular time is an element of
the special orthogonal group. An understanding of topological groups
is useful to the students of several branches of mathematics. The last
part of the book (Chapters 14-15) introduces the reader to the realm of
algebraic topology. We discuss here fundamental groups and covering
spaces in some detail.
An effort has been made to sustain the reader’s interest in the subject. To give students an insight into the abstract concept, nearly every
new notion is followed by examples and counter-examples with clear
expositions, and the proofs are given in considerable detail. Numerous
figures have been included in an attempt to aid easier understanding
of the arguments presented in the text. Also, we have included a large
number of exercises of varying degree of difficulty at the end of each
section. These provide ample opportunity to consolidate the results
in the body of text, and in some exercises, a line of development related, but peripheral, to the work of the section is explored. A few
exercises are needed for the main development in the book, and these
are marked with a symbol ‘•’. All these things make the book suitable
for self-study too. It is hoped that a reader who completes this book
will feel inspired and encouraged to turn to a more advanced book on
topology and geometry.
Tej B. Singh
Delhi, India
Suggested Course Outlines
Obviously, the book has been arranged according to the author’s liking.
Also, several topics are independent of one another, so it is profitable
to advise the reader what should be read before a particular chapter.
The dependencies of chapters are roughly as follows:
Chapters
1→2
Chapters
4→5
-
?
?
-
Chapter 3
-
Chapter 6
Chapter 10
+
Chapter 7
?
Chapter 14
Q
Chapter 9
Q
QQ
s
3
?
-
Chapter 8
Chapter 11
?
Chapter 12
?
-
Chapter 15
Chapter 13
xvii
xviii
Elements of Topology
Undoubtedly, there is more material in the book than can be covered in a one-year course. But there is a considerable flexibility for an
individual course design. Chapters 1 through 11 are suitable for a fullyear course in general topology at the advanced undergraduate level.
For a one-year graduate course, we suggest Chapters 1 through 7, and
Chapters 12 through 15. The subject matter of Chapters 14 and 15
can be studied just after finishing the core part with the adaptability
of turning to materials of Chapters 12 and 13 as and when needed.
Acknowledgements
Anyone who writes a book at this level merely acts as selector of the
material and owes a great deal to other people. I acknowledge my
debt to previous authors of books on general topology and algebraic
topology, especially those listed in the Bibliography. I would also like to
acknowledge the facilities and support provided by the Harish Chandra
Institute, Allahabad, and the University of Delhi, Delhi, without which
this book would not have completed. I am deeply grateful to two of my
teachers Professor Shiv Kumar Gupta, West Chester University, PA,
and Professor Satya Deo Tripathi for their assistance and guidance
over the years. I have had many fruitful discussions with my colleagues
Professor Ramji Lal (University of Allahabad), Dr. Ratikanta Panda
and Dr. Kanchan Joshi about the material presented in this book.
I wish to express my great appreciation for their valuable comments
and suggestions. I have also been helped by several research students
one way or another. Among them, Mr. Sumit Nagpal deserves special
thanks. Finally, I thank my family, especially my wife, who has endured
and supported me during all these years.
xix
List of Symbols
The quantifier “there exists” is denoted by ∃, and the quantifier “for
all” is denoted by ∀; this is also read as “for each.” If p and q are
propositions, then the logical symbol p ⇒ q means p implies q, p ⇐ q
means p is implied by q, and p ⇔ q means “(p ⇒ q) and (q ⇒ p)”, or
p if and only if q.
A few particular sets frequently occur in this book; the following
special symbols will be used for them.
∅
N
Z
Q
R
C
H
Rn
I
In
Dn
Sn
ω
Ω
∏
Xα
ω
X
∑
Xα
emptyset
the set of natural (or positive) integers
the set of all (positive, negative, and zero) integers
the set of all rational numbers
the set (also field) of all real numbers
the set (also field) of all complex numbers
the set (also field) of all quaternions
the set of all n-tuples (x1 , . . . , xn ) of real numbers
the (closed) unit interval [0, 1]
the n-cube I × · · · × I (n factors)
the unit n-disc {x ∈ Rn |∥x∥ ≤ 1}
∑n+1
the unit n-sphere {(x1 , . . . , xn+1 ) ∈ Rn+1 : 1 x2i = 1}
the ordinal number isomorphic to the well-ordered set of all
nonnegative integers in its natural order
the first (or least) uncountable ordinal number
the cartesian product of the indexed family of sets Xα , α ∈ A
the cartesian product of countably infinite copies of X
the disjoint union of the indexed family of sets Xα , α ∈ A
xxi
Chapter 1
TOPOLOGICAL SPACES
1.1
1.2
1.3
1.4
1.5
Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Derived Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
Metric Spaces
1
7
12
19
30
Topology is a branch of mathematics that studies the properties of
geometric objects which remain unaltered under “deformation.” The
idea of deformation involves strongly the notion of continuity. The
notions of continuity of functions and convergence of sequences are
the two most fundamental concepts in analysis. Both the concepts are
based on the abstraction of our intuitive sense of closeness of points
of a set. For example, the usual ϵ-δ definition for continuity of real or
complex valued functions on the real line R1 (or the complex plane
C) and the definition of convergence of sequences in these spaces are
based on this idea. “Closeness” of elements of a set can be measured
most conveniently as distance between the elements. In any set endowed with a suitable notion of distance, one can define convergence
of sequences and talk about continuity of functions between such sets.
Maurice Fréchet (1906), perhaps motivated by this observation, introduced “metric spaces.”
In this section, we collect some basic facts about metric spaces.
Definition 1.1.1 Let X be a (nonempty) set. A metric on X is a
function
d:X ×X →R
such that the following conditions are satisfied for all x, y, z ∈ X:
(a) (positivity) d(x, y) ≥ 0 with equality if and only if x = y,
1
2
Elements of Topology
(b) (symmetry) d(x, y) = d(y, x), and
(c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z).
The set X together with a metric d is called a metric space; the
elements of X are called points. The value d(x, y) on a pair of points
x, y ∈ X is called the distance between x and y.
Example 1.1.1 A fundamental example of a metric space is the euclidean n-space Rn . Its points are the n-tuples x = (x1 , . . . , xn ) of real
numbers and the metric on this set is defined by
(∑n
)
2 1/2
d(x, y) =
.
i=1 (xi − yi )
To see that d actually satisfies the conditions of Definition 1.1.1,
we recall the definition of the “inner product” (or “scalar product”) in
n
n
n
R
∑n. This is a function (x, y) 7→ ⟨x, y⟩ of R × R into R, where ⟨x, y⟩ =
i=1 xi yi . It is linear in one coordinate when the other coordinate is
held fixed (that is, it is a bilinear function). The norm of x ∈ Rn is
defined by
√
∑
∥x∥ = ⟨x, x⟩ = ( x2i )1/2 .
The following properties of ∥x∥ can be easily verified:
(a) ∥x∥ > 0 for x ̸= 0;
(b) ∥ax∥ = |a| ∥x∥;
(c) |⟨x, y⟩| ≤ ∥x∥ ∥y∥
(Cauchy–Schwarz inequality); and
(d) ∥x + y∥ ≤ ∥x∥ + ∥y∥
for all x, y ∈ Rn and a ∈ R.
It is now easy to check that d is a metric on Rn , for d(x, y) =
∥x − y∥. The euclidean space R1 will usually be denoted by R.
Example 1.1.2 Let C be the field of complex numbers and H be the
(skew) field of quaternions. Let F denote one of these fields. Then the
set F n of ordered n-tuples (x1 , . . . , xn ), xi ∈ F , is a vector space under
the coordinatewise addition and scalar multiplication. For technical
reasons (to be observed later), we will consider F n as a right vector
space over F . Let ā denote the complex (resp. quaternionic) conjugate
of a in C∑(resp. H). The standard inner product on F n is defined by
n
⟨x, y⟩ = 1 xi yi for x = (x1 , . . . , xn ), y = (y1 , . . . , yn ), and it has the
following properties:
TOPOLOGICAL SPACES
3
(a) ⟨x, y + z⟩ = ⟨x, y⟩ + ⟨x, z⟩;
(b) ⟨x, ya⟩ = ⟨x, y⟩ a, ⟨xa, y⟩ = ā ⟨x, y⟩;
(c) ⟨x, y⟩ = ⟨y, x⟩;
(d) x ̸= 0 ⇒ ⟨x, x⟩ > 0.
√
We define a function ∥·∥ : F n → R by setting ∥x∥ = ⟨x, x⟩, and refer
to it as the euclidean norm on F n . It is easily seen that the function ∥·∥
has the properties analogous to those of the euclidean norm on Rn , and
hence there is a metric d on F n given by d(x, y) = ∥x − y∥, x, y ∈ F n .
We call the metric space Cn the n-dimensional complex (or unitary)
space, and the metric space Hn the n-dimensional quaternionic (or
symplectic) space.
Example 1.1.3 In the standard Hilbert
∑ 2space ℓ2 , the points are infinite
real sequences x = (xi ) satisfying
xi < ∞ and its metric is defined
by
)1/2
(∑
2
.
d(x, y) =
i (xi − yi )
We observe that d(x, y) is always finite. For each positive integer n, we
have
(∑
)1/2
(∑n 2 )1/2 (∑n 2 )1/2
n
2
+
≤
1 yi
1 xi
1 (xi − yi )
≤
(∑
i
x2i
)1/2
+
(∑
i
yi2
)1/2
< ∞.
The partial sums being bounded, this monotone, nondecreasing sequence must converge and one obtains
(∑
2
i (xi − yi )
)1/2
≤
(∑
i
x2i
)1/2
+
(∑
i
yi2
)1/2
.
This implies that d(x, y) is finite and it is indeed a metric.
Example 1.1.4 The set C(I) of all continuous real-valued functions on
I = [0, 1] with the metric
d(f, g) =
is an interesting metric space.
∫1
0
|f (t) − g(t)|dt
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Elements of Topology
Example 1.1.5 For any nonempty set X, let B (X) denote the set of
bounded functions X → R. The supremum metric ρ on B (X) is given
by
ρ(f, g) = sup {|f (x) − g(x)| : x ∈ X}.
Definition 1.1.2 Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is continuous at x ∈ X if, given ϵ > 0, there exists a
δ > 0 such that dX (x, x′ ) < δ ⇒ dY (f (x), f (x′ )) < ϵ. The function f
is called continuous if it is continuous at each x ∈ X.
Definition 1.1.3 In a metric space (X, d), the open r-ball with centre
x ∈ X and radius r > 0 is the set B (x; r) = {y ∈ X|d(x, y) < r}.
In the real line R, an open r-ball is just the open interval (x − r, x +
r), and an open r-ball in the plane R2 is a disk without its rim (see
Figure 1.2(a)).
In the terminology of open ball, a function f : X → Y between two
metric spaces X and Y is continuous if and only if for each open ball
B (f (x); ϵ) centred at f (x), there is an open ball B (x; δ) centred at x
such that f (B(x; δ)) ⊆ B (f (x); ϵ).
Example 1.1.6 Consider the function f : Rn → Rn given by f (x) =
(∑ 2 )1/2
x/ (1 + ∥x∥), where ∥x∥ =
. We have
xi
∥f (y) − f (x)∥ =
≤
∥(y − x) + x(∥x∥ − ∥y∥) + ∥x∥(y − x)∥
(1 + ∥x∥) (1 + ∥y∥)
(1 + 2∥x∥)
∥y − x∥.
(1 + ∥x∥) (1 + ∥y∥)
Consequently, for each ϵ > 0, f maps the open ball B (x; δ) into
the open ball B (f (x); ϵ), where δ = ϵ if x = 0, and δ =
min {1, ϵ (1 + ∥x∥) / (1 + 2∥x∥)} if x ̸= 0. So f is continuous.
Definition 1.1.4 A subset U of the metric space X is called open if,
for each point x ∈ U , there is an open r-ball with centre x contained
in U .
If y ∈ B(x; r), then B(y; r′ ) ⊆ B(x; r), where r′ = r − d(x, y). This
shows that all open balls are actually open sets.
TOPOLOGICAL SPACES
5
Theorem 1.1.5 Let X be a metric space. Then the union of any family of open sets is open and the intersection of any finite family of open
sets is also open.
Proof. The empty set ∅ and the full space X are obviously∪open. If
{Gα } is a nonempty collection of open subsets at X, then α Gα is
clearly open. It remains to show that the intersection of two open
subsets G1 and G2 is open. If G1 ∩G2 = ∅, we are through. So consider
a point x ∈ G1 ∩ G2 . We find positive numbers r1 and r2 such that
B (x; ri ) ⊆ Gi , i = 1, 2, and put r = min {r1 , r2 }. Then B(x; r) ⊆
G1 ∩ G2 and G1 ∩ G2 is open.
♢
It turns out that the continuity of functions between metric spaces
can be described completely in terms of open sets.
Theorem 1.1.6 Let (X, dX ) and (Y, dY ) be metric spaces. A function
f : X → Y is continuous ⇔ f −1 (G) is open in X for each open subset
G of Y .
Proof. Suppose that f is continuous and G ⊆ Y is open. If f −1 (G) = ∅,
then it is open in X. Let x ∈ f −1 (G) be arbitrary. Then f (x) ∈ G
and therefore there is an ϵ > 0 such that B (f (x); ϵ) ⊆ G. Since f is
continuous at x, there exists δ > 0 such that f (B(x; δ)) ⊆ B(f (x); ϵ).
This implies that B (x; δ) ⊆ f −1 (G); so f −1 (G) is open.
Conversely, suppose that x ∈ X and ϵ > 0 is given. Then
f −1 (B(f (x); ϵ)) is an open subset of X containing x, by our hypothesis. Consequently, there exists a δ > 0 such that B(x; δ) ⊆
f −1 (B(f (x); ϵ)) ⇒ f (B(f (x); δ)) ⊆ B (f (x); ϵ). This implies that f
is continuous at x.
♢
We recall some more terminologies used in a metric space (X, d). If
A and B are two nonempty subsets of X, the distance between them
is defined by
dist (A, B) = inf {d(a, b)|a ∈ A, b ∈ B}.
If A ∩ B ̸= ∅, then dist (A, B) = 0. However, there exist disjoint sets
with zero distance between them. When A or B is empty, we define
dist (A, B) = ∞. In particular, if x ∈ X and A ⊆ X, the distance of
x from A is dist (x, A) = dist ({x}, A). The diameter of A, denoted
by diam(A), is sup {d(a, a′ ) : a, a′ ∈ A}. By convention, the diameter
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Elements of Topology
of the empty set is 0. A set is called bounded if its diameter is finite; d
is a bounded metric if diam(X) is finite.
If X is a metric space and Y ⊆ X, then the restriction of the
distance function to Y × Y is clearly a metric on Y . The set Y , with
this metric, is referred to as a subspace of X. Thus any subset of a
metric space is itself a metric space in an obvious way. This construction
increases the supply of examples of metric spaces: We can now include
all subsets of Rn and ℓ2 . In particular, the closed unit n-disc
Dn = {x ∈ Rn : ∥x∥ ≤ 1}
and the unit (n − 1)-dimensional sphere
Sn−1 = {x ∈ Rn : ∥x∥ = 1}
(∑n 2 )1/2
for x = (x1 , . . . , xn ). Note
are metric spaces, where ∥x∥ =
1 xi
0
that S = {−1, 1} is a discrete two-point space and D0 is just a point.
The unit n-cube I n is the space
{(x1 , . . . , xn ) ∈ Rn |0 ≤ xi ≤ 1, i = 1, . . . , n} .
I 1 will be denoted by I.
Another technique of constructing a new metric space from old ones
involves definition of a metric in their cartesian product and this will
be discussed in §2 of Chapter 2.
Exercises
1. Given a set X, define d(x, y) = 0 if x = y, and d(x, y) = 1 if x ̸= y.
Check that d is a metric on X.
2. Let (X, d) be a metric space. Show that
(a) d′ (x, y) = d(x, y)/ (1 + d(x, y)), and
(b) d1 (x, y) = min {1, d(x, y)}
are bounded metrics on X.
3. • Let F = R, C or H, and given x ∈ F n , define ∥x∥ = max |xi |. Show
1≤i≤n
that the function ∥·∥ satisfies the conditions (a), (b) and (d) described
in Ex. 1.1.1, and hence defines a norm on F n . This is called the cartesian
norm on F n .
TOPOLOGICAL SPACES
7
4. • Prove that each of the following functions defines a metric on Rn .
(a) ρ ((xi ), (yi )) = max |xi − yi | (cartesian metric).
1≤i≤n
∑n
+
(b) ρ ((xi ), (yi )) = 1 |(xi − yi )| (taxi-cab metric).
5. For n = 2 and n = 3, describe geometrically the open r-balls in (Rn , ρ),
(Rn , ρ+ ) and (Rn , d), where d is the euclidean metric.
6. Verify that the functions d in Ex. 1.1.4, and ρ in Ex. 1.1.5 define metrics
for C(I) and B(X), respectively.
7. • Let (Y, d) be a metric space and X a set. Call a function f : X → Y
bounded if f (X) is a bounded subset of Y. Let B (X, Y ) be the set of
all bounded functions from X into Y . Show that d∗ defined by
d∗ (f, g) = sup {d (f (x), g(x)) |x ∈ X}
is a metric on B (X, Y ). (This is called the sup metric on B (X, Y )).
8. Show that (a) the translation function Rn → Rn , x 7→ x + a, where
a ∈ Rn is fixed, and (b) the dilatation function Rn → Rn , x 7→ rx,
where r ∈ R is fixed, are continuous.
9. If X is a metric space and A ⊆ X is nonempty, show that the function
f : X → R given by f (x) = dist (x, A) is continuous.
10. Show that a subset A of a metric space (X, d) is bounded if there exists
a point x ∈ X and a real number K such that d(x, a) ≤ K for every
a ∈ A.
11. Let A, B be bounded subsets of a metric space X. Show that
(a) diam (A ∪ B) ≤ diam (A) + diam (B), if A ∩ B ̸= ∅, and
(b) diam (A ∪ B) ≤ diam (A) + diam (B) + dist(A, B), if they don’t
meet.
1.2
Topologies
In metric spaces, most of the important notions, for example, limits,
continuity, connectedness and compactness, etc., may be described and
many important theorems of analysis can be proved solely in terms of
open sets. So, it is useful to abstract the basic properties of open sets,
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Elements of Topology
and introduce a notion that is suitable for talking about these concepts
and is also independent of the idea of metrics. This led Felix Hausdorff
(1914) to give the definition of “topological spaces.”
Definition 1.2.1 A topological structure or, simply, a topology on a
set X is a collection T of subsets of X such that
(a) the intersection of two members of T is in T;
(b) the union of any collection of members of T is in T; and
(c) the empty set ∅ and the entire set X are in T.
A set X endowed with a topological structure T on it is called a
topological space. The elements of X are called points and the members
of T are called the open sets. A topological space should, in general, be
denoted as a pair (X, T). But, it is customary to use the expression “X
is a topological space” or, more briefly, “X is a space” to mean (X, T)
without mentioning the topology T for X each time.
Example 1.2.1 Let X be any set. The family D of all subsets of X is
a topology on X, called the discrete topology; the pair (X, D) is called
the discrete space. On the other extreme, the family I = {∅, X} is
also a topology on X, called the indiscrete or trivial topology; the pair
(X, I) is called the indiscrete or trivial space.
Example 1.2.2 If X = {a, b}, then there are two topologies {∅, {a} , X}
and {∅, {b} , X} on X aside from the discrete and trivial ones. The set
X with one of these topologies is called the Sierpinski space.
Example 1.2.3 By Theorem 1.1.5, the collection of sets declared “open”
in a metric space (X, d) is a topology on X; this is called the topology
induced by the metric d or simply the metric topology. In future when
a metric space is mentioned, it will be understood that the space is a
topological space with the metric topology. In particular, the metric
topology generated by the euclidean metric on any subset of Rn will
be referred to as the usual topology. Unless otherwise stated, a subset
of Rn is assumed to have the usual topology. Similarly, the topologies
on Cn and Hn induced by the metrics in Ex. 1.1.2 are referred to as
the usual topologies.
TOPOLOGICAL SPACES
9
Example 1.2.4 Given any set X, the family of all those subsets of X
whose complements are finite together with the empty set forms a
topology Tf on X, called the cofinite (or finite complement) topology.
We call (X, Tf ) a cofinite space. Similarly, the family of all those subsets
of X whose complements are countable together with the empty set
is a topology Tc on X, called the cocountable topology (or countable
complement) topology.
We will encounter more serious examples later. It is obvious that
one can assign several topological structures to a given set and these can
be partially ordered by inclusion relation. If T and T ′ are the topologies
on the same set X, we call T ′ finer (or larger) than T if T ⊂ T ′ . In this
case, we also say that T is coarser (or smaller) than T ′ . The terms
“stronger” and “weaker” are also used in the literature to describe the
above situation. But there is no agreement on their meaning, so we will
not use these terms. It may happen that T is neither larger nor smaller
than T ′ ; in this case it is said that T and T ′ are not comparable. Clearly,
the trivial topology for a set X is the smallest possible topology on X,
while the discrete topology is the largest possible topology. Also, the
following proposition can be easily verified.
Proposition 1.2.2 The intersection of any (nonempty) collection of
topologies for a set X is a topology.
It follows that if S is any collection of subsets of X, then there is a
smallest topology (viz. the intersection of all topologies containing S)
on X such that all of the sets in S are open.
Definition 1.2.3 A subset F of a topological space X is closed if X−F
is open.
The following duality properties for closed sets hold in any space.
Proposition 1.2.4 Let X be a space. Then,
(a) the union of two closed sets is a closed set;
(b) the intersection of any family of closed sets is a closed set; and
(c) the entire set X and the empty set ∅ are closed sets.
Example 1.2.5 In R, any closed interval [a, b] is closed according to the
above definition, for R − [a, b] is the union of open sets (−∞, a) and
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Elements of Topology
(b, ∞). The set Z of integers is closed, but the set Q of rationals is not
closed.
The property (a) in Proposition 1.2.4, by iteration, implies that
the union of any finite number of closed sets is closed.
∪∞ But it does not
extend to infinite unions; for example, the union n=1 [1/n, 2] is not
closed in R.
Example 1.2.6 In a discrete space, every set is both open and closed.
Example 1.2.7 Consider the cofinite space Z of integers. In this topology, a finite subset of Z is closed but not open, Z − {0} is open but not
closed, and the set N of positive integers is neither open nor closed.
Example 1.2.8 In the euclidean space R2 , S1 and D2 are closed sets.
The set
{(x, y) : x ≥ 0 and y > 0}
is not closed (why?).
These examples suggest that a subset can be both closed and open
(called clopen) or it may not be either open or closed.
We observe that a topology for a set X can also be described by
specifying a family F of subsets of X satisfying the conditions in 1.2.4.
In fact, the family of complements of the members of F is a topology
for X such that F consists of precisely the closed subsets of X. Thus
the concept of closed set can be taken as the primitive notion to define
a topology.
Definition 1.2.5 If X is a topological space and x ∈ X, then a set
N ⊆ X is called a neighbourhood (written nbd) of x in X if there is an
open set U with x ∈ U ⊆ N .
We note that a nbd is not necessarily an open set, while an open
set is a nbd of each of its points. In particular, the entire space X is a
nbd of its every point. This suggests that a nbd need not be “small”
as one might think. If N itself is open, we will call it an “open nbd.”
This is standard practice, though some mathematicians use the term
“nbd.”
Proposition 1.2.6 For each point x of the topological space X, the
family Nx of all nbds of x satisfies the following properties:
TOPOLOGICAL SPACES
11
(a) x belongs to each N in Nx .
(b) The intersection of two members of Nx is again in Nx .
(c) If N ∈ Nx and N ⊆ M ⊆ X, then M ∈ Nx .
◦
(d) If N ∈ Nx , then N = {y ∈ N |N ∈ Ny } is also a member of Nx .
Conversely, if we are given a nonempty family Nx of subsets of X,
satisfying (a), (b) and (c), for each x ∈ X, then the collection
T = {U ⊆ X|U ∈ Nx for all x ∈ U }
is a topology on X. If (d) is also satisfied, then Nx is precisely the
collection of all nbds of x relative to T.
Proof. The verification of the axioms of topology is routine; we prove
the last statement only. If N is a nbd of the point x, then x ∈ U ⊆ N
for some open set U . Since x ∈ U ∈ T, we have U ∈ Nx . By (c),
N ∈ Nx . Conversely, let N ∈ Nx . We define U = {y ∈ X|N ∈ Ny }.
Then x ∈ U ⊆ N, clearly. We assert that U is open. If y ∈ U, then
◦
◦
N ∈ Ny . By (d), N ∈ Ny . For any y ′ ∈ N , N ∈ Ny′ so that y ′ ∈ U , by
◦
the definition of U . Thus N ⊆ U, and hence U ∈ Ny , as required. This
completes the proof.
♢
The preceding proposition shows that the concept of a neighbourhood of a point may be used as the primitive notion to define a topology.
Exercises
1.
(a) Find all possible topologies on the set X = {a, b, c}.
(b) Let T1 = {∅, X, {a} , {a, b}} , and T2 = {∅, X, {c} , {b, c}} on X.
Is the union of T1 and T2 a topology for X?
(c) Find the smallest topology containing T1 and T2 , and the largest
topology contained in T1 and T2 .
2.
(a) What is the topology determined by the metric d on X given by
d(x, y) = 1 if x ̸= y and d(x, x) = 0?
(b) Let X be a set containing more than one element. Can you define
a metric on X so that the associated metric topology is trivial?
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Elements of Topology
3. • Let X be an infinite set, x0 ∈ X a fixed point. Show that
T = {G|either X − G is finite or x0 ∈
/ G}
is a topology on X in which every point, except x0 , is both open and
closed. ((X, T) is called a Fort space.)
4. • Decide the openness and closedness of the following subsets in R:
(a) {x : 1/2 < |x| ≤ 1}, (b) {x : 1/2 ≤ |x| < 1},
(c) {x : 1/2 ≤ |x| ≤ 1},
(d) {x : 0 < |x| < 1 and (1/x) ∈
/ N}.
5. Find a topology on R, different from the trivial topology and the discrete
topology, so that every open set is closed and vice versa.
6. In R2 , show:
{
}
(a) The first quadrant A = (x, y) ∈ R2 |x, y ≥ 0 is closed.
(b) {(x, 0)| − 1 < x < 1} is neither open nor closed.
(c) {(x, 0)| − 1 ≤ x ≤ 1} is closed.
7. Show that Rn × {0} ⊂ Rn+m is closed in the euclidean metric on Rn+m .
8. Show that C (I) is closed in the space B (I) with the supremum metric
(see Ex. 1.1.5).
9. Find two disjoint closed subsets of R2 which are zero distance apart.
10. In a metric space (X, d), for any real number r ≥ 0, the closed r-ball at
x ∈ X is the set {y ∈ X : d(x, y) ≤ r} .
Show that a closed ball is always closed in the metric topology.
11. If every countable subset of a space is closed, is the topology necessarily
discrete?
1.3
Derived Concepts
In this section, we will study some derived concepts such as “interior”, “closure”, “boundary” and “limit points” of subsets of a topological space.
TOPOLOGICAL SPACES
13
Definition 1.3.1 Let X be a space and A ⊆ X. The set
∪
A◦ =
{G|G is open in X and G ⊆ A}
is the largest open set contained in A; it is called the interior of A in
X. The notation int(A) is also used for A◦ .
Example 1.3.1 In the real line R, [a, b]
◦
(R − Q) .
◦
= (a, b), and Q◦ = ∅ =
( )
( )
Example 1.3.2 In the space R2 , int S1 = ∅, int D2 = B(0; 1).
If X is a space and A ⊆ X, then a point of A◦ is called an interior
point of A. It is easily seen that a point x ∈ X is an interior point of
A if and only if A is a nbd of x, and A is open if and only if A = A◦ .
Proposition 1.3.2 Let X be a space. Then, for A, B ⊆ X, we have
◦
(a) (A◦ ) = A◦ ,
(b) A ⊆ B ⇒ A◦ ⊆ B ◦ ,
◦
(c) A◦ ∩ B ◦ = (A ∩ B) , and
◦
(d) A◦ ∪ B ◦ ⊆ (A ∪ B) .
We leave the simple proofs to the reader. Notice that the reverse inclusion in (d) fails in general; this is shown by Ex. 1.3.1.
Definition 1.3.3 Let X be a space and A ⊆ X. The set
∩
A=
{F |F is closed in X and A ⊂ F }
is the smallest closed set containing A. This is called the closure of
A, sometimes denoted by cl (A). A point x ∈ A is referred to as an
adherent point of A.
Example 1.3.3 In the space R, (a, b) = [a, b] and Q = R = R − Q.
Example 1.3.4 In the space R2 , B(0; 1) = D2 .
Example 1.3.5 In a cofinite space X, A = X for every infinite set
A ⊆ X.
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It is readily seen that a subset A of a space X is closed if and only
if A = A. We also leave the straightforward proofs of the following
proposition to the reader.
Proposition 1.3.4 Let X be a space and A, B ⊆ X. Then
(a) A = A,
(b) A ⊆ B ⇒ A ⊆ B,
(c) A ∪ B = A ∪ B, and
(d) A ∩ B ⊆ A ∩ B.
Note that the equality in (d) may fail, as is seen by taking A =
(−1, 0) and B = (0, 1) in the real line R.
Theorem 1.3.5 Let A be subset of a space X. Then x ∈ A ⇔ U ∩A ̸=
∅ for every (open) nbd U of X.
Proof. If there exists an open set U such that x ∈ U and U ∩ A = ∅,
then F = X − U is a closed set which contains A but not x. Thus
x ∈
/ A. Conversely, if x ∈
/ A, then U = X − A is an open nbd of x
disjoint from A.
♢
Definition 1.3.6 Let A be subset of a space X. A point x ∈ X is a
limit point (or accumulation point or cluster point) if every nbd of x
contains at least one point of A − {x}. The set A′ of all limit points of
A is called the derived set of A.
Example 1.3.6 In R, every point of [0, 1] is a limit of (0, 1), whereas
the set Z of integers has no limit points.
Example 1.3.7 In a discrete space, no point is a limit point of a given
subset.
Example 1.3.8 Every point of R3 is a limit point of the subset A of
those points all of whose co-ordinates are rational and, at the other
extreme, the subset B of points which have integer co-ordinates does
not have any limit points.
Theorem 1.3.7 Let A be a subset of a space X. Then A = A ∪ A′ .
TOPOLOGICAL SPACES
15
Proof. If x is neither a point nor a limit point of A, then there is an
open nbd U of x such that U ∩ A = ∅. Since U is a nbd of each of its
points, none of these is in A′ . So U is contained in the complement of
A ∪ A′ , and hence A ∪ A′ is closed. It follows that A ⊆ A ∪ A′ . On the
other hand, A′ ⊆ A, by Theorem 1.3.5. As A ⊆ A always, we find that
A ∪ A′ ⊆ A, completing the proof.
♢
Corollary 1.3.8 A set is closed if and only if it contains all its limit
points.
Proof. A is closed ⇔ A = A = A ∪ A′ ⇔ A′ ⊆ A.
♢
Definition 1.3.9 Let A be subset of X. The boundary (or frontier) of
A is defined to be the set ∂A = A ∩ X − A. The notation bd(A) is also
used for ∂A. A point x ∈ ∂A is called a boundary point of A.
Obviously, ∂A is identical with ∂ (X − A). Also, it is clear that a
point x ∈ X is a boundary point of A if and only if each (open) nbd
of x intersects both A and X − A.
Example 1.3.9 In R, ∂[0, 1] = {0, 1} and ∂Q = R.
Example 1.3.10 In R2 , ∂D2 = S1 and ∂S1 = S1 .
Example 1.3.11 Let A be the set of all points of R3 which have rational
coordinates. Then ∂A = R3 .
Theorem 1.3.10 Let A be a subset of space X. Then A = A ∪ ∂A.
Proof. By definition, A contains both A and ∂A, and hence their union.
Conversely, if x ∈ A − A, then x ∈ A ∩ (X − A) ⊆ ∂A and the reverse
inclusion follows.
♢
As an immediate consequence of this theorem, we have
Corollary 1.3.11 A set is closed if and only if it contains its boundary.
We have seen in the previous section that either of the notions of
closed set and neighbourhood of a point may be used as the primitive
notion for introducing a topology on a set X. The same is true of each
of the concepts of interior, closure, boundary and derived set.
The following notions will be needed later.
16
Elements of Topology
Definition 1.3.12 A subset A of a topological space X is called dense
(or everywhere dense) if A = X.
Example 1.3.12 In the real line R, both the set of rational numbers
and the set of irrational numbers are dense.
Example 1.3.13 If X is an infinite set with the cofinite topology, then
the dense subsets of X are its infinite subsets.
Definition 1.3.13 A subset A of a space(X )is called nowhere dense if
its closure has an empty interior (i.e., int A = ∅).
Clearly, A is nowhere dense in X if and only if no nonempty open
subset of X is contained in A.
Example 1.3.14 The set Z of integers is nowhere dense in the real line
R.
Example 1.3.15 Let I be the closed unit interval
) 1] with the
( 1 2sub)
( 1 2[0,
,
J
=
space
topology
induced
from
R.
Let
J
=
,
2
1
3 3
9, 9 ∪
(7 8)
n−1
intervals
9 , 9 , . . . . (In general, )let Jn , n > 1, be the union∪of 2
n−1
1+3k 2+3k
of the form 3n , 3n which∪are contained in I − i=1 Ji . The Can∞
tor set is defined by C = I − 1 Jn . This is the set of all points in I
whose at least one triadic expansion (base 3) contains no 1’s. Since C
is closed, and each open interval in I intersects some Jn , it follows that
C is nowhere dense in I.
Definition 1.3.14 Let A be a subset of a space X. A point a ∈ A is
called isolated if a ∈
/ A′ . The set A is called perfect if it is closed and
has no isolated points.
Example 1.3.16 The Cantor set is perfect. It is clear that C is closed.
To see that it is perfect, let x ∈ C be arbitrary and U be an open
interval containing x. Choose a sufficiently large integer n so that U
contains a closed interval [x − 1/3n , x + 1/3n ]. Now, find an integer k ≥
0 such that x belongs to a closed interval of the form [k/3n , (k + 1)/3n ] .
Obviously, one end point of this interval is different from x. Thus U
contains a point of C other than x, and x is a limit point of C.
TOPOLOGICAL SPACES
17
Exercises
1. Describe the boundary, closure, interior and derived set of each of the
following subsets of the real line R:
(a) {(1/n)|n = 1, 2, . . .};
(b) (−1, 0) ∪ (0, 1);
(c) {(1/m) + (1/n)|m, n ∈ N};
(d) {(1/n) sin n|n ∈ N}.
Observe that the interior operator and the closure operator do not generally commute.
2. Specify the boundary, closure, interior, and derived set of each of the
following subsets of R2 :
(a) {(x, 0)|x ∈ R};
(b) {(x, 0)|0 < x < 1};
(c) {(x, y)|x ∈ Q};
{
}
(e) (x, y)|1 < x2 + y 2 ≤ 2 ;
(d) {(x, y)|x, y ∈ Q};
(f) {(x, y)|x ≥ 0, y > 0};
{
}
(g) {(x, y)|x ̸= 0 and y ≤ 1/x}; (h) (x, y)|x ≥ y 2 ;
(i) R2 − {(x, sin(1/x)) |x > 0}.
3. Let {Aα } be an infinite family of subsets of a space X.
(a) Prove:
∩
∩
∪
∪
◦
◦
(i) ( Aα ) ⊆ A◦α ; (ii) A◦α ⊆ ( Aα ) ;
∩
∩
∪
∪
(iv) Aα ⊆ Aα .
(iii) Aα ⊆ Aα ;
(b) Give examples to show that the reverse inclusions in (a) fail in
general.
∪
(c) Prove that the equality in (a)(iv) holds if Aα is closed.
4. Let X be a space and A ⊆ X. Prove:
◦
(a) X − A = (X − A) ; (b) X − A◦ = X − A;
(c) ∂A = A − A◦ ;
(d) A◦ ∪ ∂A = A;
(e) A◦ ∩ ∂A = ∅;
(f) A◦ = A − ∂A.
5. Let X be a space and A ⊆ X. Prove that A is clopen ⇔ ∂A = ∅.
′
6. Let X be a space, and A, B ⊆ X. Prove that (A ∪ B) = A′ ∪ B ′ . How
does ∂ (A ∪ B) relate to ∂A and ∂B?
7. Let U be an open subset of a space X. Show:
( )
(a) U = int U .
(b) ∂U = U − U .
( )
(c) Is U = int U ? (d) U ∩ A ⊆ U ∩ A for every A ⊆ X.
18
Elements of Topology
8. Prove that G is open in a space X ⇔ G ∩ A = G ∩ A for every subset
A of X.
9. Let X be an infinite set with the cofinite topology and A ⊆ X. Prove
that if A is infinite, then every point of X is a limit point of A and if
A is finite then it has no limit points.
10. In a metric space (X, d), show:
(a) x is an interior point of a subset A of X ⇔ there exists an open
ball B(x; r) contained in A.
(b) x is a limit point of a set A ⊆ X ⇔ each open ball B(x; r) contains
at least one point of A − {x}.
(c) x ∈ A ⇔ dist(x, A) = 0.
11.
(a) In the euclidean space Rn , show that B(x; r) is the closed r-ball
B[x; r] = {y ∈ X|d(y, x) ≤ r}.
(b) Give an example of a metric space (X, d) in which B(x; r) is not
the closed r-ball at x, and ∂B(x; r) ̸= {y|d(y, x) = r} for some
point x ∈ X and some real r > 0.
(c) What is the relation between ∂B(x; r) and {y|d(y, x) = r}?
12.
(a) Prove that every nonempty subset of a trivial space is dense, while
no proper subset of a discrete space is dense.
(b) If no proper subset of the topological space X is dense, is the
topology necessarily discrete?
(c) What is the boundary of a subset of a discrete space? a trivial
space?
13. Let D be a subset of a space X.
(a) Prove that D is dense in X ⇔ X is the only closed superset of
D ⇔ X − D has an empty interior.
(b) If D is dense in X, prove that D ∩ G = G for every open subset
G of X.
(c) If G and H are open subsets of a space X such that G = X = H,
show that G ∩ H = X.
14. Let X be a space and A ⊆ X. Show that A is nowhere dense ⇔ A ⊆
)
(
X −A .
15. Prove that a closed set is nowhere dense ⇔ its complement is everywhere
dense. Is this true for an arbitrary set?
16. Show that the boundary of a closed (or open) set is nowhere dense. Is
this true for an arbitrary set?
17. Prove that the union of two nowhere dense sets is nowhere dense.
TOPOLOGICAL SPACES
18.
19
(a) If A has no isolated points, show that A is perfect.
(b) If a space X has no isolated points, prove that every open subset
of X also has no isolated points.
1.4
Bases
The specification of a topology by describing all of the open sets is
usually a difficult task. This can often be done more simply by using
the notion of a “generating family” for the topology. In this section,
we will study two such concepts.
Given a set X and a family S of subsets of X, we have already seen
in Section 2 that the intersection of the collection of all topologies on X
which contains S (certainly nonempty, for the discrete topology on X
is one such topology) is a topology, denoted by T (S). Clearly, T (S) is
the coarsest topology on X containing S. It consists of ∅, X, all finite
intersections of members of S and all unions of these finite intersections.
This can easily be ascertained by verifying that the collection of these
sets is a topology for X, which contains S and is coarser than T (S). It
follows that the topology T (S) is completely determined by the family
S.
Definition 1.4.1 Let X be a space with the topology T. A subbasis
(or subbase) for T is a family S of subsets of X such that T = T (S).
If S is a subbasis for the topology T on X, the members of S are open
in X, and referred to as subbasic open sets. As we have seen above,
any family S of subsets of X serves as a subbasis for some topology for
X, namely, T (S). Thus, to define a topology on a set X, it suffices to
specify a family S of subsets of X as a subbasis. The resulting topology
is said to be generated by the subbasis S.
We illustrate this by introducing a topology on an ordered set
(X, ≺). For each pair of elements a, b ∈ X, define
(a, b) = {x ∈ X|a ≺ x ≺ b}
(open interval),
[a, b] = {x ∈ X|a ≼ x ≼ b}
(closed interval),
20
Elements of Topology
[a, b) = {x ∈ X|a ≼ x ≺ b}
}
(a, b] = {x ∈ X|a ≺ x ≼ b}
(half-open or halfclosed intervals).
And, for each a ∈ X, define
(−∞, a) = {x ∈ X|x ≺ a}
}
(a, +∞) = {x ∈ X|a ≺ x}
(−∞, a] = {x ∈ X|x ≼ a}
[a, +∞) = {x ∈ X|a ≼ x}
}
(open rays or one-sided
open intervals),
(closed rays or one-sided
closed intervals).
It is obvious that [a0 , a) = (−∞, a) if a0 is the smallest element, and
(a, b0 ] = (a, +∞) if b0 is the largest element.
We ought to find a topology on X which justifies the use of adjectives closed and open here. Notice that an open interval (a, b) can
be obtained as an intersection of the rays (a, +∞) and (−∞, b) so
that a topology which contains these rays certainly contains (a, b). If
x ∈ X is not the largest element, then there exists a b ∈ X such that
x ∈ (−∞, b), and if x is not the smallest element, then x ∈ (a, +∞) for
some a ∈ X.
Definition 1.4.2 Let X be an ordered set. The order topology (or
the interval topology) on X is the topology generated by the subbasis
consisting of the “open rays” (−∞, a) and (a, +∞), where a ∈ X.
In the order topology on X, an open interval (a, b) is obviously open
and a closed interval [a, b], being the complement of (−∞, a)∪(b, +∞),
is closed.
Example 1.4.1 Consider the set R of real numbers with the usual order
relation on it. Since the open rays (−∞, a) and (a, +∞) (a ∈ R) are
open in the euclidean topology on R, the order topology for R is coarser
than the euclidean topology. On the other hand, every open ball, being
an open interval, is open in the order topology. Therefore every open
subset of the real line R is open in the order topology, and the two
topologies for R coincide. Thus we see that the family of all open rays
in R is a subbase for the topology of the real line R.
Since the operations of union and intersection both are involved in
TOPOLOGICAL SPACES
21
the construction of a topology from a subbasis, an important simplification occurs if the open sets are constructed only by taking unions
of members of S. This is possible, for example, if S is closed under the
formation of finite intersections; in that case S is termed without the
prefix “sub.”
Definition 1.4.3 Let (X, T) be a space. A basis (or base) for T is a
family B ⊆ T such that every member of T is a union of members of
B.
If B is a basis for a topology T on X, then T is the coarsest topology
on X containing B. For, if T ′ is a topology with B ⊂ T ′ , then all unions
of members of B are cetainly in T ′ and so T ⊆ T ′ . We say that the
topology T is generated by the basis B. The members of B are referred
to as the basic open sets in X, and B is also called a basis for the space
X. There is a simple characterization of bases, which some authors use
as a definition.
Theorem 1.4.4 A collection B of open subsets of a space X is a basis
if and only if for each open subset U of X and each point x ∈ U , there
exists a B ∈ B such that x ∈ B ⊆ U .
The straightforward proof is left to the reader.
By the preceding theorem, we have a useful way to describe the
open subsets of a space X when its topology is given by specifying a
basis B: A set G ⊆ X is open if and only if for each x ∈ G, there is a
B ∈ B such that x ∈ B ⊆ G. It follows that the topology of a space is
completely determined by a basis.
Example 1.4.2 In a discrete space X, the family of all singleton sets
{x} is a basis.
Example 1.4.3 In a metric space X, the collection
{B(x; r)|x ∈ X, and real number r > 0}
of open balls is a basis for the metric topology on X.
We note that a family S of subsets of a space X is a subbasis if and
only if the family of all finite intersections of members of S is a basis
for X. This basis is said to be generated by S. (We remark that some
authors elude the convention that X is the intersection of the empty
22
Elements of Topology
subfamily of S, and put the condition X =
to be a subbasis for X.)
∪
{S ∈ S} on the family S
Example 1.4.4 Let X be an ordered space. The basis generated by the
subbasis of X consists of all open rays, all open intervals (a, b), the
empty set ∅, and the full space X. If X has no largest element, then
(a, +∞) is a union of open intervals, and if X has no smallest element,
then (−∞, a) is a union of open intervals. Also, if x ∈ X is not the
largest or smallest element, then x obviously belongs to an open interval
in X. It follows that there is a basis for the topology of X consisting
of the open intervals (a, b), half open intervals [a0 , a) = (−∞, a) (if
a0 is the smallest element) and (a, b0 ] = (a, +∞) (if b0 is the largest
element). In particular, a basis for the order topology on R consists of
open intervals (a, b) alone, since there is no smallest or largest number
in R. This also follows from the fact that the order topology for R
coincides with the usual topology (see Ex. 1.4.1), and the (bounded)
open intervals are precisely the open balls in the euclidean metric on
R.
Example 1.4.5 Consider the “dictionary ordering” on the set X = R ×
R: For x = (x1 , x2 ) and y = (y1 , y2 ) in X, x ≺ y if and only if either
x1 < y1 or (x1 = y1 and x2 < y2 ). Let X have the order topology for
this ordering. Obviously, there is no largest or smallest element in X.
So the open intervals (x, y), x ≺ y in X, form a basis for the topology
^
_
(a;c)
x
_
(a;b)
_
y
FIGURE 1.1: Basic open sets in R2 with the dictionary order topology.
TOPOLOGICAL SPACES
23
in X. It is easily seen that an interval (x, y) with x1 < y1 is a union
of intervals of the form ((a, b), (a, c)). So the intervals of the form
((a, b), (a, c)) alone can generate the topology of X, and thus form
a basis.
Example 1.4.6 Let Ω be the least (first) uncountable ordinal number
and let [0, Ω] denote the set of ordinal numbers ≤ Ω. The order topology
for [0, Ω] is generated by the subbasis composed of sets [0, y) and (x, Ω]
for x, y ≤ Ω. Accordingly, these sets together with intervals (x, y) form
a basis for this topology. Since each ordinal number y < Ω has an
immediate successor, the topology of the space [0, Ω] is also generated
by the sets {0}, (x, y], where 0 ≤ x < y ≤ Ω, as a basis. Notice that
if x = 0 or has an immediate predecessor in [0, Ω], then {x} is open.
Moreover, observe that every basic open nbd of Ω intersects [0, Ω).
For, if (x, Ω] is disjoint from [0, Ω), then we have [0, Ω) = [0, x], which
contradicts the fact that [0, x] is countable. Therefore Ω is a limit point
of the set [0, Ω) (ref. Exercise 13).
In the above examples, we have described a base for a given topology. Conversely, it is desirable to introduce a topology on a set X by
specifying a basis for it. A natural question arises whether a given
family of subsets of X would be a base for some topology on X. The
answer to this question is not always positive. For example, the family
B = {{a, b}, {a, c}} of subsets of X = {a, b, c} cannot serve as a basis
for a topology on X, since any topology having B as a basis must contain {a} which cannot be expressed as the union of members of B. So
we ought to know when a given collection B of subsets of X can serve
as a basis for some topology on X. ∪
Assume that B is a basis for some
topology on X. Then we have X = {B|B ∈ B}, for X is open. And,
for every pair of sets B1 , B2 ∈ B and for each x ∈ B1 ∩ B2 , there exists
B3 ∈ B with x ∈ B3 ⊆ B1 ∩ B2 , since B1 ∩ B2 is open. In fact, these
conditions are also sufficient, as we see below.
Theorem∪1.4.5 Let B be a collection of subsets of the set X such
that X = {B|B ∈ B} , and for every two members B1 , B2 of B and
for each point x ∈ B1 ∩ B2 , there exists B3 ∈ B with x ∈ B3 ⊆ B1 ∩ B2 .
Then there is a topology on X for which B is a basis.
Proof. Let T (B) be the family of all sets U ⊆ X such that for each
x ∈ U there exists B ∈ B with x ∈ B ⊆ U . Then ∅, X ∈ T (B),
obviously. A union of members of T (B) is itself a union of members of
24
Elements of Topology
B, and is therefore in T (B). If U1 , U2 are in T (B) and x ∈ U1 ∩U2 , then
we can find Bi ∈ B such that x ∈ Bi ⊆ Ui , i = 1, 2. As x ∈ B1 ∩ B2 ,
we may choose a B3 ∈ B such that x ∈ B3 ⊆ B1 ∩ B2 ⊆ U1 ∩ U2 .
This implies that U1 ∩ U2 ∈ T (B), and thus T (B) is a topology on
X. Clearly, B is a basis for the topology T (B), for T (B) consists of
precisely the unions of members of B.
♢
We illustrate the use of bases to define some topologies on the set
R of real numbers. We have already seen that the open intervals (a, b)
constitute a basis for the usual topology on R. The family of “closed
intervals” [a, b] is also a basis for a topology on R, as is easily verified.
Since this family contains the singletons {a}, a ∈ R, the topology
generated by this basis is the discrete topology. Again, we fail to obtain
a new topology. Here is an interesting case.
Example 1.4.7 Consider the family of “right-half open intervals” [a, b),
where a, b ∈ R and a < b. One can readily verify that this family satisfies the conditions for a basis. The topology generated by this basis is
called the lower limit (or right half-open interval) topology for R. The
set R with this topology is denoted by Rℓ , and is referred to as the
“Sorgenfrey line.” In this space, all the intervals (−∞, a) and [a, +∞)
are both open and closed, and so is each basis ∪
element. The sets of the
form (a, b) ∪
or (a, +∞) are open, for (a, b) = {[x, b)|a < x < b} and
(a, +∞) = {[x, x + 1)|a < x}, but not closed.
Similarly, the family {(a, b]|a, b ∈ R and a < b} generates a topology on R, called the upper limit (or the left half-open) topology. These
topologies on R are useful for construction of counter examples.
Returning to the general case, we note that different bases (or subbases) may generate the same topology; for example, the collection of
open intervals with rational end points is also a basis for the usual
topology on R. We say that two bases (subbases) are equivalent if they
generate the same topology. To determine whether two bases are equivalent, we need to compare the topologies generated by them, and there
is a very simple criterion to carry this out.
Proposition 1.4.6 Let T and T ′ be topologies on a set X generated
by the bases B and B′ , respectively. Then T is coarser than T ′ ⇔ for
each B ∈ B and each x ∈ B, there exists B ′ ∈ B′ such that x ∈ B ′ ⊆ B.
Proof. ⇒: Given B ∈ B, we have B ∈ T ′ , for T ⊆ T ′ . If x ∈ B, then
there exists B ′ ∈ B′ with x ∈ B ′ ⊆ B, since B′ is a basis for T ′ .
TOPOLOGICAL SPACES
25
⇐: Let U be T-open. If x ∈ U , then there exists B ∈ B with
x ∈ B ⊆ U , since B is a basis for T. By our hypothesis, there exists
B ′ ∈ B′ such that x ∈ B ′ ⊆ B. So x ∈ B ′ ⊆ U . It follows that U is
T ′ -open. Thus T ⊆ T ′ .
♢
Example 1.4.8 The topology of the real line R is strictly smaller than
that of Rℓ (and the upper limit topology). We already know that the
open intervals (a, b) form a basis for the euclidean topology and the
half-open intervals [a, b) form a basis for the lower limit topology. Also,
for each x ∈ (a, b), we have x ∈ [x, b) ⊂ (a, b). So, by the preceding proposition, the euclidean topology is coarser than the lower limit
topology. But, the basis element [a, b) for the lower limit topology is not
open in the euclidean topology, since there is no open interval which
contains a and is contained in [a, b).
Example 1.4.9 As another example, we consider the topologies on Rn
induced by the euclidean metric d and the cartesian metric ρ given by
d ((xi ), (yi )) =
(∑
|xi − yi |2
)1/2
and
ρ ((xi ), (yi )) = max1≤i≤n |xi − yi |.
Figure 1.2 illustrates the open balls in each metric when n = 2. It is
•
•
x
x
B d (x;r)
(a)
B½ (x;r)
(b)
FIGURE 1.2: Open balls in (a) the euclidean metric and (b) the cartesian
metric on R2 .
26
Elements of Topology
obvious that, given a point in a disk, there is a square in the disk
centered at the point. Also, inside a square, we can find a disk centered
at a given point of the square. Specifically, let Bd and Bρ denote the
open balls in the metrics d and ρ, respectively. Then it is easily verify
that if y ∈ Bd (x; r), then Bρ (y; r′ ) ⊂ Bd (x; r) for 0 < r′ < (r −
√
d(y, x))/ n. And, if y ∈ Bρ (x; r), then Bd (y, r′ ) ⊂ Bρ (x; r) for r′ =
r − ρ(y, x). By Proposition 1.4.6, the topology induced by d coincides
with that induced by ρ.
Although the metric topology depends on the choice of metric, the
preceding example suggests that different metrics may determine the
same topology. This necessitates the following.
Definition 1.4.7 Two metrics d and d′ on a set X are called equivalent
if they induce the same topology on X.
There is a simple criterion to test the equivalence of two metrics.
Theorem 1.4.8 Two metrics d and d′ on the set X are equivalent if
and only if for each x ∈ X and for each ϵ > 0, there exists δ > 0 such
that Bd (x; δ) ⊆ Bd′ (x; ϵ), and Bd′ (x; δ) ⊆ Bd (x; ϵ).
The proof is an easy application of Proposition 1.4.6, and we leave this
to the reader.
We observe that any metric is equivalent to a bounded one. Let
(X, d) be a metric space. Given any real λ > 0, define
dλ (x, y) = min {λ, d(x, y)}.
It is easily verified that dλ is a metric on X such that diam(X) ≤
λ (with respect to dλ ). We have the inclusions Bd (x; ϵ) ⊆ Bdλ (x; ϵ)
and Bdλ (x; δ) ⊆ Bd (x; ϵ) for δ = min {λ, ϵ}. By Theorem 1.4.8, dλ is
equivalent to d. We will refer to this fact later as
Corollary 1.4.9 Let (X, d) be a metric space. Then, for each real
λ > 0, there is a metric dλ equivalent to d such that the diameter of
X in dλ is less than λ.
We conclude this section with a theorem which shows that the task
of giving a basis for a topology on a set X is generally accomplished
by specifying for each x ∈ X a “basis at x” in the following sense.
TOPOLOGICAL SPACES
27
Definition 1.4.10 If X is a topological space and x ∈ X, then a
collection Bx of subsets of X containing x is called a neighbourhood
basis (or a local basis) at x if each nbd of x in X contains some element
of Bx and every element of Bx is a nbd of x.
Example 1.4.10 In a topological space, the open nbds form a nbd basis.
Example 1.4.11 In a metric space, the open r-balls about a given point
x ∈ X for all real r > 0 form a nbd basis.
Example 1.4.12 In a discrete space, there is a neighbourhood basis at
a point x consisting of just one set, viz., {x}.
Theorem 1.4.11 Let X be a set and for each x ∈ X, let Bx be a
collection of subsets of X satisfying:
(a) Bx is a nonempty set for each x ∈ X;
(b) x ∈ B for every B ∈ Bx ;
(c) for every B1 , B2 ∈ Bx , there exists a B3 ∈ Bx such that B3 ⊆
B1 ∩ B2 ; and
(d) for each B ∈ Bx , there exists a B ′ ∈ Bx such that B contains a
member of By for every y ∈ B ′ .
Then there is a unique topology on X such that Bx is a nbd basis
at x for each x ∈ X.
Proof. The collection {Nx |x ∈ X}, where Nx is the family of all supersets of members Bx , satisfies the conditions (a)–(d) of Proposition
1.2.6, and hence the theorem.
♢
Note that a set G ⊆ X is open in the topology determined by the
local bases Bx , x ∈ X, if for each x ∈ G, there exists B ∈ Bx with
B ⊆ G.
Exercises
1.
(a) What is the order topology on the set N with the usual order
relation?
(b) Is the order topology on {1, 2} × N in the dictionary order relation
discrete?
28
Elements of Topology
2. Show that the topology on R generated by the subbasis
{[a, b)|a, b ∈ R} ∪ {(a, b]|a, b ∈ R}
coincides with the discrete topology.
3. Describe the topology on the plane for which the family of all straight
lines is a subbase.
4. Show that the sets {x ∈ R|x > r}, {x ∈ R|x < s}, where r, s ∈ Q, form
a subbasis for the euclidean topology of R. Is this still true if r, s are
restricted to the numbers of the form k/2n , where n and k are arbitrary
integers?
5. Show that the sets of form {x|x ≥ a} and {x|x < b} , a, b ∈ R, constitute
a subbasis for the topology of the Sorgenfrey line Rℓ .
6. • Let A = {1, 1/2, 1/3, . . .}. Show that the collection of open intervals
(a, b) and the sets (a, b) − A is a basis for a topology on R. Describe the
topology generated by this base (this topology is sometimes called the
Smirnov topology for R), and compare it with the different topologies
on R discussed in this section.
7. Find the boundary, closure and interior of the set {1/n|n ∈ N} in the
topology (on R) generated by the basis {(a, +∞)|a ∈ R}. (This topology
on R is referred to as the right order topology; a left order topology is
defined similarly.)
8.
(a) Determine
the
√
√ boundary, closure and interior of the subsets
(0, 2) and ( 3, 4) of R in the topology generated by the basis
{[a, b)|a, b ∈ Q} .
(b) Show that the topology in (a) is strictly coarser than the lower
limit topology.
9. The collection of all open intervals (a, b) together with the singletons
{n} , n ∈ Z, is a base for a topology on R. Describe the interior operation
in the resulting space.
10. Let B be a basis for the space X. Show that a subset D ⊆ X is dense
if and only if every nonempty member of B intersects D nontrivially.
11. Show that the rationals are dense in the Sorgenfrey line Rℓ .
12. Let S be a subbasis for the topology of a space X. If D ⊆ X and
U ∩ D ̸= ∅ for each U ∈ S, is D dense in X?
13. • Let B be a basis for the topological space X and A ⊆ X. Show that
x ∈ A ⇔ B ∩ A ̸= ∅ for every B in B with x ∈ B.
14. Let X be an ordered set with the order topology. Show that (a, b) ⊆
[a, b]. Find the conditions for equality.
TOPOLOGICAL SPACES
29
15. • Define an order relation ≼ on I × I by
(x, y) ≼ (x′ , y ′ ) ⇔ y < y ′ or (y = y ′ and x ≤ x′ ).
The order topology on I × I is called the television topology (the name
is due to E.C. Zeeman). Determine the closures of the following subsets
of I × I:
{
}
A = {(0, y)|0 < y < 1},
B = (0, n−1 )|n ∈ N ,
{
}
{
}
C = (x, 2−1 )|0 < x < 1 ,
D = (2−1 , y)|0 < y < 1 , and
)
}
{(
E = 2−1 , 1 − n−1 |n ∈ N .
16. Show that the taxi-cab metric ρ+ on Rn , ρ+ ((xi ), (yi )) =
is equivalent to the euclidean metric.
∑
|xi − yi |,
17. Let (X, d) be a metric space. Show that the metric d′ , defined by d′ =
d/(1 + d), is equivalent to d.
18. Show that the metrics d and ρ defined on C(I) by
d(f, g) =
ρ(f, g) =
∫1
0
|f (t) − g(t)|dt,
and
sup {|f (t) − g(t)| : 0 ≤ t ≤ 1}
are not equivalent.
19. Let X be a space with a basis B. Show that for each x ∈ X, the family
Bx = {B ∈ B|x ∈ B} is nbd basis at x.
20. In the real line R, show that the collection of open intervals (x−r, x+r),
r ranging over the set of all positive rational numbers, is a neighbourhood base at x.
21. Let (a, b) be a particular point of R2 . Show that the set of all squares
with sides parallel to the axes and centered at (a, b) is a neighbourhood
basis at (a, b).
22. Let X be a topological space, and Bx be a nbd basis at x ∈ X. For
A ⊆ X, show:
(a) x is an interior point of A ⇔ A contains some member of Bx .
(b) x is an adherence point of A ⇔ A intersects every member of Bx .
30
1.5
Elements of Topology
Subspaces
A subset of a space inherits a topology from its parent space, in an
obvious way. This is the simplest method of constructing a new space
from a given one.
Definition 1.5.1 Let (X, T) be a space and Y ⊆ X. The relative topology or the subspace topology TY on Y is the collection of all intersections
of Y with open sets of X, and Y equipped with this topology is called
a subspace of X.
A routine verification shows that TY = {Y ∩ U |U ∈ T} is, indeed,
a topology on Y . Each member H of TY is said to be open in Y and
its relative complement Y − H is closed in Y . We have
Proposition 1.5.2 Let Y be a subspace of a space X. A subset K ⊆ Y
is closed in Y if and only if K = Y ∩ F , where F is closed in X.
Proof. The proposition follows immediately from the equality Y − (Y ∩
G) = Y ∩ (X − G) for any G ⊆ X.
♢
Example 1.5.1 Let (X, d) be a metric space and Y ⊆ X. Then the
relative topology on Y induced by the metric topology on X coincides
with the metric topology determined by the restriction of d to Y . This
follows from the observation that an open ball about y of radius r in
Y is the intersection of Y with the open ball B(y; r) in X. Thus, the
metric spaces Dn (the unit disc), I n (the unit cube) and Sn−1 (the
unit sphere) are subspaces of Rn . The set Z, as a subspace of R, has
the discrete topology and so does the set {1/n|n ∈ N}. Note that Z is
closed in R, while {1/n|n ∈ N} is not closed.
If Y is a subspace of X, then any subset of Y which is open (or
closed) in X has the same property in Y . But, an open (or closed)
subset of Y need not be open (or closed) in X, as shown by the following
example.
Example 1.5.2 In the subspace Y = (0, 1] ∪ [2, 3] of R, the set (0, 1] is
open as well as closed. But this is not open or closed in R.
A simple condition on Y turns the above statements into affirmative.
TOPOLOGICAL SPACES
31
Proposition 1.5.3 Let Y be a subspace of a space X. If Y is closed
(or open) in X and A is closed (resp. open) in Y , then A is closed (resp.
open) in X.
Proof. By Proposition 1.5.2, A = Y ∩ F for some closed subset F of X.
Since the intersection of two closed subsets is closed, A is closed in X.
A similar argument applies to the “open” case.
♢
As a direct consequence of the definition of the relative topology,
we obtain
Proposition 1.5.4 If Y is a subspace of X, and Z is a subspace of Y ,
then Z is a subspace of X.
This property of relativisation is often used without explicit mention. The proof of the next proposition is also trivial, and left to the
reader.
Proposition 1.5.5 Let X be a space and Y ⊆ X.
(a) If B is a basis (resp. subbasis) of X, then {Y ∩ B|B ∈ B} is a
basis (resp. subbasis) for the relative topology of Y .
(b) If Bx is a nbd base at x ∈ X and x ∈ Y , then {B ∩ Y |B ∈ Bx }
is a nbd basis at x in Y .
Let Y be a subspace of a space X. For any A ⊆ Y , we can form
the boundary, closure, derived set, and interior of A using the topology
of Y or X. In such situations, we need to specify the space in which
the closure (boundary or interior or derived set) is taken. We shall use
the notations ∂AX , AX , A′X , A◦X , to indicate that these operations
are performed in X. The following proposition determines the various
relations.
Proposition 1.5.6 Let Y be a subspace of a space X, and A ⊆ Y .
Then AY = AX ∩ Y, A′Y = A′X ∩ Y, A◦Y ⊇ A◦X ∩ Y = A◦X and ∂AY ⊆
∂AX ∩ Y .
Proof. Since AX is closed in X, AX ∩Y is a closed subset of Y containing
A. So AY ⊆ AX ∩Y . On the other hand, AY is closed in Y , and therefore
AY = Y ∩ F for some closed subset F of X. It follows that A ⊆ F
whence AX ∩ Y ⊆ F ∩ Y = AY .
The other statements follow readily from the definitions.
♢
32
Elements of Topology
Example 1.5.3 Let Y = {(0, y)|y ∈ R} have the relative topology induced by the usual topology of the euclidean space R2 = X. For
A = {0} × [−1, +1], A◦Y = {0} × (−1, +1) while A◦X = ∅, and
∂AY = {0} × {−1, +1} while ∂AX = A. This example shows that
the inclusions in Proposition 1.5.6 may be strict.
Finally, in this section, we consider the behavior of relativisation
with order topology. Unfortunately, this is not so nice as we would like
it to be. If X is an ordered set and Y ⊆ X, then the restriction of
the simple order relation on X is a simple order relation on Y . Thus
Y receives two topologies: one induced by the restricted order and the
other one – the relative topology – inherited from the order topology
on X. These two topologies for Y don’t generally agree, as is observed
below.
Example 1.5.4 Consider the real line with its usual order and topology.
Let Y = {0} ∪ (1, 2). Then {0} is open in the subspace Y ; but, in the
order topology for Y , each basic open set containing 0 contains some
points greater than 1.
Example 1.5.5 Let X be an ordered space, and J be an interval in X.
By Proposition 1.5.5, the sets (−∞, a)∩J and (a, +∞)∩J, a ∈ X, form
a subbasis for the relative topology on J. One observes that if a ∈
/ J,
then these sets are ∅ or J, and if a ∈ J, then these are obviously open
rays in J. Consequently, the relative topology is coarser than the order
topology of J. On the other hand, an open ray in J is (−∞, a) ∩ J or
(a, +∞) ∩ J, which is open in the relative topology for J. It follows
from the definition of order topology that the relative topology is finer
than the order topology on J and the two topologies for J agree.
Exercises
1. A subset Y of a space X is called discrete if the relative topology for Y
is discrete.
Prove that every subspace of a discrete space is discrete and every
subspace of an indiscrete space is indiscrete.
2. Show that subset {0} ∪ {1/n|n ∈ N} of real line R is not discrete.
√ }
√
{
3. Verify that the set x ∈ Q| − 2 ≤ x ≤ 2 is both open and closed
in the subspace Q ⊂ R. (Notice that this is true for any open interval
with irrational end points.)
TOPOLOGICAL SPACES
33
4. Consider the sets Z and Q with the usual order relations. What is the
order topology on Z? Show that the order topology on Q is the relative
topology induced from the real line R.
5. Consider the subspace X = [−1, 1] of the real line R. Decide the openness and closedness in X of the sets in Exercise 1.2.4.
6. Let T and U be topologies for set X such that U is strictly finer than
T. For Y ⊆ X, what can be said about TY and UY ?
7. Let X be a space in which every finite subspace has the trivial topology. Show that X itself has the trivial topology. Is the corresponding
assertion for the discrete topology true?
8. Let F ⊆ X be closed and U ⊆ F open in F . Let V be any open subset
of X with U ⊆ V . Prove that U ∪ (V − F ) is open in X.
9. Let Y be a subspace of X. If A is dense in Y , show that A is dense in
Y.
10. Give an example of a space X which has a dense subset D and a subset
Y such that D ∩ Y is not dense in Y .
11. Consider the dictionary order on R2 and its restriction to I 2 , I = [0, 1].
Is the order topology for I 2 the same as the relative topology induced
from the order topology on R2 ?
12. Call a subset Y of an ordered set (X, ≼) convex if the interval (a, b) ⊆ Y
for every a ≺ b in Y .
(a) Verify that an interval in X, including a ray, is convex.
(b) Is a proper convex subset of an ordered set X an interval or a ray?
(c) Prove that the relative topology on a convex subset Y of an ordered space X agrees with the order topology for Y .
Chapter 2
CONTINUITY AND PRODUCTS
2.1
2.2
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Continuity
35
45
The continuity of functions between topological spaces is the central
notion in topology. The conditions of a topological structure have been
so formulated that the definition of a continuous function can be borrowed word for word from analysis. The first section is devoted to the
discussion of this concept. We will also study here the notion of equivalence for topological spaces, and related concepts. The second section of
this chapter concerns the construction of new topological spaces out of
old ones. We have already studied such a method, namely, Relativization. Given an indexed family of topological spaces, we can construct
their “cartesian product.” We shall go into a method of topologising
this set, in a natural and useful way.
Definition 2.1.1 Let X and Y be spaces and f : X → Y be a function. Then f is called continuous if f −1 (U ) is open in X for each open
set U ⊆ Y .
Example 2.1.1 A constant function c : X → Y is obviously continuous:
c−1 (U ) is either ∅ or X for every U ⊆ Y .
Example 2.1.2 Every function on a discrete space is continuous.
The following theorem provides some other ways of formulating the
continuity condition.
Theorem 2.1.2 Let X and Y be spaces and f : X → Y a function.
The following conditions are equivalent:
35
36
Elements of Topology
(a) f is continuous.
(b) f −1 (F ) is closed in X for every closed set F ⊆ Y .
( )
(c) f A ⊆ f (A) for every set A ⊆ X.
( )
(d) f −1 (B) ⊆ f −1 B for every set B ⊆ Y .
Proof. (a) ⇔ (b): This is immediate from the equality f −1 (Y − B) =
X − f −1 (B).
Now, we prove that (b) ⇒ (c) ⇒ (d) ⇒ (b).
(
)
(b) ⇒ (c): Since f (A) is closed, f −1 f (A) is closed, by (b). Obvi)
(
)
(
ously, A ⊆ f −1 f (A) ; so A ⊆ f −1 f (A) and (c) holds.
(
)
(c) ⇒ (d): Taking A = f −1 (B) in (c), we have f f −1 (B) ⊆
f (f −1 (B)) ⊆ B, which implies (d).
(d) ⇒ (b): If F is a closed subset of Y , then (d) implies f −1 (F ) ⊆
f (F ). But f −1 (F ) ⊆ f −1 (F ) always, so the equality holds and
f −1 (F ) is closed.
♢
−1
If a basis (or subbasis) for the range space of a function is known,
then the next theorem makes easier the task of proving that the given
function is continuous.
Theorem 2.1.3 A function f : X → Y between spaces is continuous
if and only if the inverse image of every set in a basis (or subbasis) of
Y is open.
We leave the proof as an exercise.
Example 2.1.3 The function f : [0, 2π) → S1 defined by f (t) = eıt is
continuous. The collection of all open segments of the circle is a basis
for the topology on S1 . If G is such a segment not containing 1 ∈ S1
then f −1 (G) is an open interval of the form (a, b), 0 < a < b < 2π.
And, if G contains 1, then f −1 (G) has the form [0, a) ∪ (b, 2π), a < b.
This is an open subset of [0, 2π), and the continuity of f follows from
the preceding theorem.
The proof the following theorem is also straightforward and left to
the reader.
CONTINUITY AND PRODUCTS
37
Theorem 2.1.4 (a) Given a space X, the identity map 1X : X → X
is continuous.
(b) If f : X → Y and g : Y → Z are continuous functions between
spaces, then the composition gf : X → Z is also continuous.
As in the case of metric spaces, there is also a localized form of
continuity.
Definition 2.1.5 (Cauchy) A function f : X → Y between spaces is
said to be continuous at x ∈ X, if given any nbd N of f (x) in Y, there
exists a nbd M of x in X such that f (M ) ⊆ N .
(
)
Since f f −1 (N ) ⊆ N , this is the same as saying that f −1 (N ) is
a nbd of x for each nbd N of f (x), and for this it is sufficient that the
condition holds for all members of a nbd basis of f (x). When X and
Y are metric spaces, this reduces to the ϵ - δ formulation.
Theorem 2.1.6 A function f : X → Y is continuous if and only if it
is continuous at each point of X.
Proof. Suppose that f is continuous, and let x ∈ X. If N is a nbd of
f (x), then there exists an open subset U of Y such that f (x) ∈ U ⊆ N .
We have x ∈ f −1 (U ) ⊆ f −1 (N ). By our hypothesis, f −1 (U ) is open in
X. So f −1 (N ) is a neighbourhood of x, and the continuity of f at x
follows.
Conversely, suppose that f is continuous at each point of X. Let V
be any open subset of Y . If x ∈ f −1 (V ), then V is a nbd of f (x). By our
assumption, there exists an open set U in X such that x ∈ U ⊆ f −1 (V ).
This implies that f −1 (V ) is open. Since V was an arbitrary open set
in Y , f is continuous.
♢
Let (X, T) be a topological space and A ⊆ X. If j : A ,→ X is the
inclusion map, then j −1 (U ) = U ∩A for every U ⊆ X. By the definition
of the relative topology, the function j is continuous in the relative
topology TA , and it is also clear that any topology on A which makes
the function j continuous must contain TA . So TA can be characterized
as the smallest topology on A for which j is continuous.
Let f : X → Y be a continuous function. If A is a subspace of X,
j
f
then the restriction f |A : A → Y is just the composite A ,→ X → Y ,
where j : A ,→ X is the inclusion map. Therefore f |A is continuous.
38
Elements of Topology
Also, if f (X) is given the relative topology, then the function f : X →
f (X) is continuous, since f −1 (O ∩ f (X)) = f −1 (O) for every O ⊆ Y .
We turn now to the discussion of piecewise construction of continuous functions. Let
∪ X be a set and {Ai } be a family of subsets of X
such that X = Ai . Suppose that, for each index i, fi : Ai → Y is
a function satisfying fi | (Ai ∩ Aj ) = fj | (Ai ∩ Aj ) for every i and j.
Then there is a function f : X → Y such that f |Ai = fi for every
i. For, given x ∈ X, there is an index i such that x ∈ Ai . We set
f (x) = fi (x). If x ∈ Aj also, then fi (x) = fj (x), by the hypothesis. So
f (x) is uniquely defined and x 7→ f (x) is a mapping extending each
fi . It may be noted that the function f thus defined is unique. The
following result, known as the Gluing (or Pasting) lemma, leads to a
useful process of constructing continuous functions.
Lemma 2.1.7 Let X and Y be topological spaces, and Ai , 1 ≤ i∪≤ n,
be a (finite) family of closed (or open) subsets of X such that X = Ai .
If, for each index i, fi : Ai → Y is a continuous function satisfying
fi | (Ai ∩ Aj ) = fj | (Ai ∩ Aj ) for every i and j, then the function f
defined by f |Ai = fi for every i is continuous.
Proof. Suppose that each Ai is closed in X, and let F ⊆ Y be any
closed set. Then, by the continuity of fi , fi−1 (F ) is closed in Ai . Since
Ai is closed in X for every
i, we see that each fi−1 (F ) is closed in X.
∪
n
Accordingly, f −1 (F ) = i=1 fi−1 (F ) is closed, and f is continuous.
In the case every Ai is open, the proof is similar.
♢
Note that the preceding lemma can be easily extended for any family of open sets {Ai }. However, this is not possible in the case involving
infinite families of closed sets Ai , as can be easily seen by taking onepoint sets {x} in an interval [a, b] ⊂ R. To remedy the situation, we
need to impose some restriction on the position of the (closed) sets Ai
in X.
Definition 2.1.8 A family {Ai } of subsets of a space X is called locally
finite (or nbd-finite) if each point of X has a nbd U such that U ∩Ai ̸= ∅
for at most finitely many indices i.
Proposition 2.1.9
∪ Let {Ai } be a locally finite family of subsets of a
space X. Then Ai is closed in X.
Proof. For x ∈ X, there exists an open nbd U of x such that U ∩Ai ̸= ∅
for at most finitely many indices i. The same is true of the sets U ∩ Ai ,
CONTINUITY AND PRODUCTS
39
∪
since U ∩ Ai = ∅ ⇒ U ∩ Ai = ∅. Now, if x ∈
/( Ai and U meets
the
))
∩n (
sets Ai1 , . . . , Ain only, then the nbd V = U ∩
of x
j=1 X − Aij
∪
does not meet any of the Ai . Consequently, the complement of Ai is
open, and the proposition follows.
♢
By the preceding proposition, it follows that the union of a locally
finite family of closed sets is closed.
Corollary 2.1.10 If a space X is the union of a family {Ai } such
that each Ai is closed in X and the family {Ai } is locally finite, then
a function f from X to a space Y is continuous if and only if the
restriction of f to each Ai is continuous.
Proof. Suppose that f |Ai is continuous for every index i. If F is a
closed subset of Y, then, for every index i, f −1 (F ) ∩ Ai is closed in
Ai and, therefore, in X. Since {A
so is the family
)
∪i }( is locally finite,
{f −1 (F ) ∩ Ai }. Hence f −1 (F ) = i f −1 (F ) ∩ Ai is closed in X, and
f is continuous. The converse is obvious.
♢
We shall often set about defining a continuous function f : X → Y
by cutting up X into closed (or open) subsets Ai and defining f on
each Ai separately in such a way that f |Ai is obviously continuous and
the different definitions agree on the overlaps.
We now introduce the concept of “topological equivalence” between
spaces.
Definition 2.1.11 A homeomorphism between spaces X and Y is a
bijective function f : X → Y such that both f and f −1 are continuous.
Two spaces X and Y are said to be homeomorphic, denoted by X ≈ Y ,
if there is a homeomorphism X → Y .
Clearly, a continuous function f : X → Y is a homeomorphism
if and only if there is a continuous function g : Y → X such that
gf = 1X and f g = 1Y . We must point out that a continuous bijection
need not be a homeomorphism; for example, consider the identity map
of a set with the discrete topology onto the same set, but equipped
with a different topology.
Example 2.1.4 The euclidean space Rn is homeomorphic to the open
ball B(0; 1) = {x ∈ Rn | ∥x∥ < 1}. The map x 7→ x/(1 + ||x||) is a
homeomorphism of Rn onto B(0; 1) with x 7→ x/(1−||x||) as its inverse.
40
Elements of Topology
Example 2.1.5 The punctured sphere Sn (i.e., Sn with one point removed) is homeomorphic to Rn . A homeomorphism Sn −{en+1 } → Rn ,
where en+1 = (0, . . . , 0, 1), is obtained by the stereographic projection (x0 , . . . , xn ) 7→ (x0 , . . . , xn−1 ) / (1 − xn ). The inverse of this
map
given by)(y0 , . . . , yn−1 ) 7→ (ty0 , . . . , tyn−1 , 1 − t), where t =
( is ∑
n−1
2/ 1 + 0 yi2 .
e3
º
•x
• y
R2
FIGURE 2.1: Stereographic projection.
Example 2.1.6 The cube I n is homeomorphic to the unit disc Dn . For
1 ≤ i ≤ n, let ei denote the vector of Rn whose all of coordinates are
zero except the ith coordinate, which is 1. The standard homeomor∑n
phism Dn → I n is realized by the translation with the vector 1 ei /2,
followed by the central projection.
Example 2.1.7 Note that the correspondence a + ıb ↔ (a, b) between
C and R2 defines a bijection Cn ↔ R2n which is clearly an isometry
(that is, a distance-preserving map). It follows that Cn with the usual
topology is homeomorphic to R2n . Similarly, the division ring H of
quaternions is identified with R4 and the bijection Hn ↔ R4n is a
homeomorphism between Hn with the usual topology and the euclidean
space R4n .
These examples illustrate that to prove two spaces are homeomorphic, one simply constructs a homeomorphism from one space to the
other. Unfortunately, the construction of a homeomorphism is usually
difficult, and it is even more difficult to prove that two given spaces are
CONTINUITY AND PRODUCTS
41
not homeomorphic. One of the main problems in topology is to find
methods of deciding whether two spaces are homeomorphic or not. We
give below some characterizations of homeomorphisms, but these do
not make our task any easier. For this, we need the following terminologies.
Definition 2.1.12 A function f : X → Y between spaces is called
open (resp. closed) if the image of each open (resp. closed) set in X is
open (resp. closed) in Y .
The following theorem provides alternative descriptions of homeomorphism.
Theorem 2.1.13 Let f : X → Y be a bijective continuous function.
Then f is a homeomorphism ⇔ f is open ⇔ f is closed.
(
)−1
Proof. For any set U ⊆ X, f −1
(U ) = f (U ). So f −1 is continuous
⇔ f is open.
Also, f (X − U ) = Y − f (U ) because f is bijective. This implies
that f is open ⇔ f is closed.
♢
By Theorem 2.1.13, a homeomorphism h : X ≈ Y not only establishes a one-to-one correspondence between the points of X and Y , but
also sets up a one-to-one correspondence between the open sets (resp.
closed sets) in X and the open sets (resp. closed sets) in Y . Thus any
assertion about X as a topological space is also valid for each homeomorph of X. For this reason, homeomorphic spaces are called topologically equivalent.We call a property of spaces a topological invariant
(or a topological property) if it is possessed by every homeomorph of a
space which has this property. Topology is often characterized as the
study of topological invariants. The problem of proving that two spaces
are not homeomorphic is generally solved by computing some suitable
invariant and showing that we obtain different answers. It is interesting
to notice that not all properties in a metric space are topological invariant, for example, the property of boundedness is not a topological
invariant.
The following examples suggest that the concepts “continuous functions,” “closed functions” and “open functions” are independent.
Example 2.1.8 If Y is a discrete space, then every function f : X → Y
is obviously closed and open.
42
Elements of Topology
Example 2.1.9 The inclusion map (0, 2) → R is open but it is not
closed, and the inclusion map of I into R is closed but not open.
Example 2.1.10 The continuous function t 7→ eıt of [0, 2π) onto S1 is
neither closed nor open. The interval [0, π) is open in [0, 2π), but its
image under the exponential map is not open; the interval [π, 2π) is
closed in [0, 2π), but its image under the above map is not closed.
A characterization of closed functions is given by
Theorem 2.1.14 Let X and(Y )be spaces and f : X → Y a function.
Then f is closed ⇔ f (A) ⊆ f A for each set A ⊆ X.
( )
( )
Proof. If f is closed, then f A is closed. Since f (A) ⊆ f A , we have
( )
( )
f (A) ⊆ f A = f A .
Conversely, suppose that the given condition holds,
( ) and let A be
any closed subset of X. Then f (A) ⊆ f (A) ⊆ f A = f (A), which
implies that f (A) = f (A). Thus f (A) is closed.
♢
Corollary 2.1.15 Let f : X → Y be a bijective function
( ) of a space
X into a space Y . Then f is a homeomorphism ⇔ f A = f (A) for
each A subset of X.
The proof is obvious.
The next theorem characterizes the open functions.
Theorem 2.1.16 The following properties of a function f : X → Y
are equivalent:
(a) f is an open map.
(b) f (A◦ ) ⊆ f (A)◦ for each set A ⊂ X.
(c) For each nbd U of x, x ∈ X, there is a nbd W of f (x) in Y with
W ⊆ f (U ).
(d) The image of each member of a basis for X is open in Y .
The proof is routine and left to the reader.
We close this section with the following notion.
CONTINUITY AND PRODUCTS
43
Definition 2.1.17 An embedding (or, more specifically, a topological
embedding) of a space X into a space Y is a function f : X → Y which
maps X homeomorphically onto the subspace f (X) of Y .
The inclusion of a subspace in its ambient space is an embedding,
and so is a closed (or open) continuous injection.
Example 2.1.11 The function f : [0, 2π) → R2 defined by f (t) = eıt
is a continuous injection. The image of f is S1 , the unit circle. The
function f fails to map the open neighbourhoods of 0 onto the open
neighbourhoods of 1; accordingly, f is not an embedding.
Exercises
1. Let T1 and T2 be two topologies on a set X. Show that
(a) the identity function i : (X, T1 ) → (X, T2 ) is continuous ⇔ T1 is
finer than T2 , and
(b) i is a homeomorphism ⇔ T1 = T2 .
2. Consider the set R of real numbers with the various topologies described
in the previous chapter except the trivial topology and the discrete
topology. Discuss the continuity of the identity function between each
pair of spaces.
3.
(a) Let a ̸= 0 and b be fixed real numbers. Show that the affine transformation x 7→ ax + b is a homeomorphism of R onto itself.
(b) Prove that any two open intervals (including the unbounded ones),
any two bounded closed intervals with more than one point, and
any two half-open intervals (including one-sided closed intervals)
in R are homeomorphic.
4. Prove that the inversion function x 7→ x−1 on R − {0} is continuous
and open.
5. Let X be an uncountable set with the cofinite topology. Show that every
continuous function X → R is constant.
6. Give an example of a function f : X → Y between spaces and a subspace
A ⊆ X such that f |A is continuous, although f is not continuous at
any point of A.
7. Let f : X → Y be continuous and A ⊆ X. If x is a limit point of A, is
f (x) a limit point of f (A)?
44
Elements of Topology
8. Prove that (a function
)◦ f : X → Y between
( spaces
) is continuous ⇔
f −1 (B ◦ ) ⊆ f −1 (B) for all B ⊆ Y ⇔ ∂ f −1 (B) ⊆ f −1 (∂B) for all
B ⊆Y.
∪
9. • Let {Uα } be a family of open subsets of a space X with X = Uα .
Show that a function f from X into a space Y is continuous if and only
if f |Uα is continuous for each index α. (This shows that continuity of a
function is a “local” property.)
10. Let f : X → Y be a function between spaces, and assume that X =
A ∪ B, where A − B ⊆ A◦ , and B − A ⊆ B ◦ . If f |A and f |B are
continuous, show that f is continuous.
11. Let X be a partially ordered set. Declare U ⊆ X to be open if it
satisfies the condition: if y ≼ x and x ∈ U , then y ∈ U . Show that
(a) {U |U is open} is a topology, and (b) a function f : X → X is
continuous ⇔ it is order preserving (i.e., x ≼ x′ ⇒ f (x) ≼ f (x′ )).
12. Let f, g : X → Y be continuous functions, where Y has an order topology. Show that (a) the set {x ∈ X|f (x) ≼ g(x)} is closed, and (b) the
function x 7→ min {f (x), g(x)} is continuous.
13. Prove that homeomorphism is an equivalence relation between spaces.
(equivalence classes are called homeomorphism types).
14. Let f : X → Y be a homeomorphism and A ⊆ X. Show that f |A is a
homeomorphism between A and f (A), and f |(X − A) is a homeomorphism between X − A and Y − f (A).
15. Let X and Y be two ordered sets with the order topology. If f : X → Y
is bijective and order preserving, show that it is a homeomorphism.
16.
(a) Let Y be the two point discrete space {0, 1}. Show that the function f : I → Y given by
{
0 for t ≤ 1/2, and
f (t) =
1 for t ≥ 1/2
is closed, open and surjective, but not continuous.
(b) Let X = [0, 1] ∪ [2, 3] and Y = [0, 2] have the subspace topology
induced from the real line R. Define a function f : X → Y by
{
x
for x ∈ [0, 1], and
f (x) =
x − 1 for x ∈ [2, 3].
Show that f is a closed continuous surjection which is not open.
17.
(a) Discuss the continuity, closedness and openness of the mappings
(i) t 7→ t2 and (ii) t 7→ 1/(1 + t2 ) on R.
CONTINUITY AND PRODUCTS
45
(b) Let p(x) be a polynomial over R. Show that x 7→ p(x) is a closed
mapping on R.
18.
(a) Show that the function R → [−1, 1], t 7→ sin t, is open.
(b) Show that the mapping t 7→ tan t is a homeomorphism of
(−π/2, π/2) onto R.
(c) • Show that the exponential map R1 → S1 , t 7→ eıt , is an open
map.
19. Give an example to show that a continuous open map need not send
the interior of a set onto the interior of the image.
20. • Let f : X → Y be open (closed).
(a) If A ⊆ X is open (closed), show that g : A → f (A) defined by f
is also open (closed).
(b) If A = f −1 (B) for B ⊆ Y , show that g : A → B defined by f is
open (closed).
21. Let f : X → Y and g : Y → Z be functions between spaces. Prove:
(a) If f, g are open (closed), so is gf .
(b) If gf is open (closed), and f is continuous, surjective, then g is
open (closed).
(c) If gf is open (closed), and g is continuous, injective, then f is open
(closed).
22. Let f : X → Y be a function between spaces. Show:
(
)
(a) f is open if and only if f −1 (∂B) ⊆ ∂ f −1 (B) for every B ⊆ Y ,
and
(b) f is closed if and only if for each y ∈ Y and each open nbd U of
f −1 (y), there exists an open nbd V of y such that f −1 (V ) ⊆ U .
2.2
Product Topology
There are three main methods of producing new sets out of old
ones, namely, forming subsets, cartesian products and quotients by an
equivalence relation. When these sets are constructed from topological
spaces, it is natural to look at the question of topologising them. We
46
Elements of Topology
have already addressed this question in the case of subsets of topological spaces. Here, we shall consider the problem of topologizing the
cartesian products of topological spaces in some natural and useful
way, and postpone the discussion about quotients of spaces to Chapter
7.
For simplicity, we first treat the case of finite products. Let X and
Y be spaces. An obvious way of defining a reasonable topology on the
set X × Y is to declare the sets U × V open for all open sets U ⊆ X
and V ⊆ Y . The family of all such subsets of X × Y obviously contains
∅ and X × Y , and is closed under the formation of finite intersections,
for (U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 ). This family is not
a topology on X × Y yet, since it is not closed under the formation
of unions. However, it can be used as a basis to define a topology for
X × Y , by Theorem 1.4.5.
Definition 2.2.1 Let X and Y be spaces, and let B be the collection
of sets U × V , where U ⊆ X and V ⊆ Y are open. The topology
generated by B as a basis is called the product topology for X × Y , and
the set X × Y , when equipped with the product topology, is called the
product space. The spaces X and Y are referred to as the coordinate
(or factor) space.
Forming the product of spaces is an associative and commutative
operation, in the sense that there are obvious canonical homeomorphisms (X × Y ) × Z ≈ X × (Y × Z), ((x, y), z) ↔ (x, (y, z)), and
X × Y ≈ Y × X, (x, y) ↔ (y, x).
The above definition easily extends to the cartesian product of a
finite number of spaces. If X1 , X2 , . . . , Xn are spaces, then a basis for
the product topology on X1 × · · · × Xn is the family of all products
U1 × · · · × Un , where each Ui is open in Xi . There is a canonical homeomorphism (X1 × · · · × Xn−1 ) × Xn → X1 × · · · × Xn ; accordingly,
the results for finite products follow by induction from those for the
product of two spaces.
There are projections pX : X × Y → X, (x, y) 7→ x, and pY :
X × Y → Y, (x, y) 7→ y, onto coordinate spaces X and Y, respectively.
For U ⊆ X and V ⊆ Y , we have p−1
(U ) = U ×Y and p−1
(V ) = X ×V .
X
Y
If X×Y is endowed with the product topology, then p−1
(U
) and p−1
(V )
X
Y
are open whenever U ⊆ X and V ⊆ Y are open. This shows that pX
and pY are continuous functions. Now, suppose that we have some
topology T on X × Y such that both projections are continuous. Then,
CONTINUITY AND PRODUCTS
47
for open sets U ⊆ X and V ⊆ Y , U × V = p−1
(U ) ∩ p−1
(V ) is open in
X
Y
T. Thus T contains all the basic open sets of the product topology, and
is therefore finer than the product topology. We summarize the above
as
Proposition 2.2.2 If X × Y has the product topology, then the projections pX : (x, y) 7→ x and pY : (x, y) 7→ y are continuous, and the
product topology is the smallest topology for which this is true. Also,
both maps pX and pY are open.
The last statement follows from the fact that the basic open sets of
the product topology are sent by each projection to open sets. However,
the projection maps need not be closed, as shown by Example 2.2.2
below.
We also notice that the collection of sets p−1
(U ) and p−1
(V ), where
X
Y
U ⊆ X and V ⊆ Y are open, is a subbase for the product topology on
X × Y , since U × V = p−1
(U ) ∩ p−1
(V ). Furthermore, if the bases for
X
Y
the coordinate spaces are known, then a basis for the product topology
consisting of fewer sets can be described.
Proposition 2.2.3 Let C and D be bases for the spaces X and Y ,
respectively. Then the collection B = {C × D|C ∈ C and D ∈ D} is a
basis for the product space X × Y .
Proof. Obviously, each set C × D in B is open in X × Y . If G is open
in X × Y and (x, y) ∈ G, then there are open nbds U of x and V of y
such that U × V ⊆ G. We find basis elements C ∈ C and D ∈ D with
x ∈ C ⊆ U and y ∈ D ⊆ V . Now, (x, y) ∈ C × D ⊆ U × V ⊆ G. This
implies that G is a union of members of B, and the proposition follows.
♢
Example 2.2.1 The product topology on R2 agrees with its usual topology. We have already seen that the euclidean metric d on R2 is equivalent to the cartesian metric ρ given by
ρ(x, y) = max {|x1 − y1 | , |x2 − y2 |}
(see. Ex. 1.4.9). In the metric ρ, an open ball Bρ (x; r) is the open
square (x1 − r, x1 + r) × (x2 − r, x2 + r), where x = (x1 , x2 ). Since
the open intervals (a, b) form a basis of R, the preceding proposition
shows that the collection B of open rectangles (a1 , b1 ) × (a2 , b2 ) is a
basis for the product topology on R2 = R1 × R1 . Obviously, every
48
Elements of Topology
open ball Bρ (x; r) is a member of the base B. On the other hand,
if x = (x1 , x2 ) ∈ (a1 , b1 ) × (a2 , b2 ), then we find a real r > 0 such
that Bρ (x; r) ⊆ (a1 , b1 ) × (a2 , b2 ). By Proposition 1.4.6, the topology
generated by the basis B agrees with the topology induced by ρ.
Similarly, the metric defined by the cartesian norm on F n (F =
C or H) (ref. Exercise 1.1.3) induces the product topology.
We use the above fact to prove the continuity of the addition function α : (x, y) 7→ x + y, and the multiplication function µ : (x, y) 7→ xy
on the euclidean space R2 . Given a point (a, b) ∈ R2 and a real ϵ > 0,
put U = (a − ϵ/2, a + ϵ/2) and V = (b − ϵ/2, b + ϵ/2). Then U × V is an
open nbd of (a, b) in R2 and α maps (U × V ) into (a + b − ϵ, a + b + ϵ).
So α is continuous. For the continuity of the multiplication function
µ, we note that a nbd base at a point r ∈ R is formed by the
open intervals (r − ϵ, r + ϵ), where 0 < ϵ < 1. Obviously, we have
|xy − ab| ≤ |x − a||y − b| + |x − a||b| + |y − b||a|. So |x − a| < δ and
|y − b| < δ implies that |xy − ab| < ϵ for δ < ϵ/(1 + |a| + |b|). Thus µ
sends the nbd (a − δ, a + δ) × (b − δ, b + δ) of (a, b) into (ab − ϵ, ab + ϵ),
and the continuity of µ at (a, b) follows.
Example 2.2.2 Consider R2 with the euclidean topology, and let p1 :
2
1
R
{ → R 2 be the }projection onto2 the first factor. The set F =
(x, y) ∈ R |xy = 1 is closed in R , since the multiplication of real
numbers is continuous and points in R are closed. But p1 (F ) = R − {0}
is not closed in R.
The product topology is compatible with the relative topology. Let
A and B be subspaces of the spaces X and Y , respectively. Then the
sets (A × B) ∩ (U × V ), where U ⊆ X and V ⊆ Y are open, form a
basis for the relative topology on A × B induced from X × Y, and the
collection of products (A ∩ U ) × (B ∩ V ) is a basis for the product space
A × B. The equality (A × B) ∩ (U × V ) = (A ∩ U ) × (B ∩ V ) implies
that the two bases for A × B are the same. Hence the relative topology
on A × B coincides with the product topology. This fact increases the
supply of examples of product spaces.
Example 2.2.3 The unit n-cube I n with its usual topology is the product of n copies of the unit interval I.
Example 2.2.4 The cylinder
{
}
C = (x, y, z) ∈ R3 |x2 + y 2 = 1 and 0 ≤ z ≤ 1
CONTINUITY AND PRODUCTS
49
with the usual topology is the product space S1 × I.
Example 2.2.5 The product space S1 × S1 is called the torus. This can
be considered as the surface of a solid ring or a doughnut. Technically,
it is the surface of revolution obtained by rotating the unit circle in the
xz-plane centered at (2, 0, 0) about the z-axis, the plane containing the
circle being always perpendicular to the xy-plane. If P = (x, y, z) is a
point of the circle, then the origin, the centre and the point Q = (x, y, 0)
are collinear. It follows that this surface is the set
{
}
(√
)2
3
2
x2 + y 2 − 2 + z = 1 .
T = (x, y, z) ∈ R |
Z
P
O
Y
1
z
X
2 C
Q
T
FIGURE 2.2: Torus as a surface of revolution.
To see a homeomorphism between the subspace T ⊆ R3 and the torus,
consider two circles
{
}
{
}
C1 = (x, y, 1)|x2 + y 2 = 4
and C2 = (x, 0, z)|(x − 2)2 + z 2 = 1
in R3 . Then there is a homeomorphism h : T → C1 × C2 given by
((
)
)
(√
)
2x
2y
√
h(x, y, z) =
,√
,1 ,
x2 + y 2 , 0, z
.
x2 + y 2
x2 + y 2
As both spaces C1 and C2 are homeomorphic to S1 , it is immediate
that T ≈ S1 × S1 .
Now, we turn to discuss the continuity of functions into product
50
Elements of Topology
spaces. For any function f : Z → X × Y , we have the coordinate
functions pX ◦ f : Z → X and pY ◦ f : Z → Y , where pX : X × Y → X
and pY : X × Y → Y are projections. When X, Y and Z are spaces,
there is a very useful characterisation of the continuity of f .
Theorem 2.2.4 A function f : Z → X × Y is continuous if and only
if the coordinate functions pX ◦ f : Z → X and pY ◦ f : Z → Y are
continuous.
Proof. If f is continuous, then pX ◦ f and pY ◦ f are continuous, since
pX and pY are continuous. Conversely, suppose that both pX ◦ f and
pY ◦ f are continuous. To establish the continuity of f , we need to show
that f −1 (U × V ) is open in Z for each basic open set U × V of X × Y .
−1
−1
It is easily seen that f −1 (U × V ) = (pX ◦ f ) (U ) ∩ (pY ◦ f ) (V ).
−1
−1
By our assumption, (pX ◦ f ) (U ) and (pY ◦ f ) (V ) are open in Z,
and so f −1 (U × V ) is open.
♢
Corollary 2.2.5 (a) Given continuous functions f : X ′ → X, g :
Y ′ → Y , the product function f × g : X ′ × Y ′ → X × Y defined by
(f × g) (x′ , y ′ ) = (f (x′ ), g(y ′ )) is continuous. Moreover, f × g is open
whenever f and g are open.
(b) Given continuous functions f : Z → X and g : Z → Y , there
is a unique continuous function ϕ : Z → X × Y defined by ϕ(z) =
(f (z), g(z)). (The function ϕ is often denoted by (f, g).)
The simple proofs are omitted.
As an application of this result, we see that the scalar multiplication
σ : R1 × Rn → Rn , σ(a, x) = ax is continuous. If µ : R2 → R1 is the
multiplication function and pj : Rn → R1 is the jth projection, then
µ ◦ (1 × pj ) = pj ◦ σ, where 1 is the identity map on R1 . So pj ◦ σ
is continuous for every j, and hence
σ is )continuous. Furthermore,
(
composing the restriction of σ to R1 − {0} × Rn with the product of
1
the inverse function a 7→ a−1 on
− {0}
and the identity map on Rn ,
( R
)
we obtain the scalar division R1 − {0} × Rn → Rn , (a, x) 7→ a−1 x.
So the scalar division is also continuous. This implies that the function
Rn → Rn , x 7→ x/(1+∥x∥), is continuous, since it is the
of
( composition
)
the scalar division with the continuous function Rn → R1 − {0} ×Rn ,
x 7→ (1 + ∥x∥, x) (see. Ex. 1.1.6).
From the preceding corollary, we see that a function f : X → Y is
continuous if and only if the function ϕ : X → X × Y, x →
7 (x, f (x)),
CONTINUITY AND PRODUCTS
51
is continuous. The restriction of pX to im(ϕ) is obviously the inverse
of ϕ. So ϕ is an embedding whenever f is continuous. The im(ϕ) is
usually referred to as the graph of f . In particular, the diagonal map
x 7→ (x, x) is an embedding of X into X × X for every space X.
Example 2.2.6 Let X be the subspace of the real line R consisting
of all non-zero real numbers, and f : X → R be the continuous
function defined by f (x) = sin 1/x. We determine the closure of the
graph G(f ) of f in R2 . Let p1 : X × R → X and p2 : X × R → R
be the projection maps. Then G(f ) = {(x, f (x)) |x ̸= 0 a real} is the
inverse of {0} under the function q = p2 − f ◦ p1 . Since the func[tion q is continuous,
] [ G(f ) is closed in] X × R. So,
[ we have G(f ) =
]
G(f ) ∩ (X × R) ∪ G(f ) ∩ ({0} × R) = G(f ) ∪ G(f ) ∩ ({0} × R) ,
where G(f ) is the closure of G(f ) in R2 . This implies that the points of
G(f ) − G(f ) belong to the y-axis. Notice that the function f assumes
-
•
-
FIGURE 2.3: Closure of the graph of the function sin 1/x.
]
[
1
1
for every positive integer
all values between −1 and 1 on (n+2)π
, nπ
n. Now, given a point (0, y0 ) with −1 ≤ y0 ≤ 1 and an open ball B of
radius r centered at (0, y0 ), we choose an integer n such that 1 < nrπ.
Then (x, y0 ) ∈ B for all x ≤ 1/nπ. It follows that B intersects G(f ),
52
Elements of Topology
and so (0, y0 ) ∈ G(f ). On the other hand, it is clear that a point (0, y),
|y| > 1, has a nbd which does not meet G(f ), since | sin 1/x| ≤ 1. Thus
we have G(f ) = G(f ) ∪ {(0, y) : |y| ≤ 1}.
INFINITE PRODUCTS
Now, we extend the concept of product topology to infinite cartesian products.
Let Xα , α ∈ A, be a family of sets.∪Their cartesian
∏
product
Xα is the set of all functions x : A → α Xα such that
x(α) ∈ Xα for every α ∈ A. We generally
write x(α) = xα and denote
∏
x by (xα ). If any Xα = ∅, then
Xα is empty. So we shall assume
that Xα ̸= ∅ for every α ∈ A. Note
∏ that, in case each of the sets Xα
is the same set X, the product
Xα is just the set of all functions
A → X, denoted by X A . This set may be referred to as the cartesian
product of A copies of X.
∏ If Xα , α ∈ A, is a family of spaces, then we should like to topologise
Xα so that important theorems for finite products remain true when
the indexing set A is infinite. A natural way to introduce a topology on
∏
Xα is to adopt the method used to topologize the product
∏of finitely
many spaces; in fact, this procedure gives a topology for
Xα . The
collection
{∏
}
Uα |Uα is open in Xα for every α ∈ A
∏
is a basis for a topology on Xα . The topology
generated by this basis
∏
is called the box topology. The map pβ : Xα → Xβ , (xα ) 7→ xβ , is
termed
∏ projection onto βth factor. Notice that each projection map
pβ : Xα → Xβ is continuous in the box topology. And, if the family
{Xα } is finite, and the indexing set A = {1, . . . , n}, then the canonical
∏
map η : x 7→ (x(1), . . . , x(n)) is a homeomorphism between
Xα
(with the box topology) and the product space X1 × · · · × Xn . But,
some results about finite products fail to hold good for infinite products
when this topology is used.
Example 2.2.7∏For each integer n ≥ 0, let Xn denote the
∏real line R
and consider Xn with the box topology. Let f : R → ∏ Xn be the
diagonal function given by f (t) = (t, t, . . .), t ∈ R. If pm : Xn → Xm
is the mth projection map, then∏pm ◦ f is the identity map on R.
Take Un = (−1/n, 1/n) and B = Un . The set B is open in the box
CONTINUITY AND PRODUCTS
53
∩
topology and f −1 (B) = Un = {0}, which is not open in R. So f is
not continuous, while all its compositions with the projections pn are
continuous.
This example shows that the analogue of Theorem 2.2.4 fails for
infinite products with the box topology. Therefore
we need to consider
∏
another topology for the cartesian product Xα of an arbitrary family
of spaces Xα , α ∈ A. Proposition 2.2.2 shows that the product topology
in the finite case could have been equivalently defined as the smallest
topology which makes all projections continuous.
This is the basis for
∏
the definition of the “product topology”
on
X
.
α Note that the con∏
tinuity of the projection map pβ : Xα → Xβ requires p−1
β (Uβ ) to be
open whenever Uβ is open in Xβ .
∏
Definition 2.2.6 The topology on Xα generated by the subbasis
{ −1
}
pα (Uα ) |Uα is open in Xα , and α ∈ A
is called the product topology (or Tychonoff topology).
x
y
z
ı
ı U®
ı
x
y
z
A
X®
FIGURE 2.4: Three elements in p−1
α (Uα ).
The following theorem characterizes the product topology in terms
of projections.
Theorem 2.2.7
∏ Given a family Xα , α ∈ A, of spaces, the product
topology
on
Xα is the smallest topology for which all projections
∏
pβ : Xα → Xβ are continuous.
54
Elements of Topology
∏
Xα
Observe that p−1
β (Uβ ), Uβ ⊆ Xβ , is the set of all elements of
−1
whose βth coordinates are restricted to lie in Uβ , that is, pβ (Uβ ) =
∏
−1
{x∏
∈ Xα |xβ ∈ Uβ }. So a finite intersection p−1
α1 (Uα1 )∩· · ·∩pαn (Uαn )
is Uα , where Uα = Xα for α ̸= α1 , . . . , αn . Thus the basis
∏ generated
by the subbasis in Definition 2.2.6 consists of the sets
Uα , where
Uα ⊆ Xα are open and Uα = Xα∏for all but a finite number of α’s. It
follows that each projection∏
pβ : Xα → Xβ is open. Moreover, if G is
a nonempty open subset of Xα , then pα (G) ∏
= Xα for all but finitely
many α ∈ A. Consequently, a set of the form Uα , where each Uα is
a nonempty proper open subset of Xα , is never open in the product
topology, unless the indexing set A is finite.
any
∏
∏ In contrast,
∏
∩ product
Fα of closed sets Fα ⊆ Xα is closed in Xα , for Fα = p−1
(Fα ).
α ∏
Note that each basic open set in the product topology on
Xα
is open in the box topology. So the box topology is finer than the
product topology, in general.
∏ If A is finite, then the box topology and
the product topology on Xα are identical. Hence, ∏
if the family {Xα }
is finite, and the indexing set A = {1, . . . , n}, then Xα just defined
is homeomorphic to X1 × · · · × Xn by the map η : x 7→ (x(1), . . . , x(n)).
If the topology
Sα for every α ∈ A,
{ of Xα is given by a subbasis
}
∏
−1
then the family pβ (Vβ ) |Vβ ∈ Sβ , β ∈ A , where pβ : Xα → Xβ
∏
is the projection map, is a subbasis for the product
topology on Xα .
∏
For, each set p−1
Xα . Moreover, if Uβ ⊆
α (Vα ) is obviously open in
Xβ is open and pβ (x) ∈ Uβ , then there
exist
( ) finitely many sets, say,
∩
n
−1
Vβ1 , . . . , Vβn in Sβ such that x ∈ 1 pβ Vβi ⊆ p−1
β (Uβ ). It follows that
each subbasic open set of the product space is open in the topology
generated by the given family, and hence the assertion.
If the topology of each Xα is given by specifying a base, then the
∏ following proposition describes a base for the product topology on Xα .
Proposition 2.2.8 Let Xα , α ∈ A, be a family of spaces. If Bα is a
basis for the ∏
topology of Xα for each α ∈ A, then the collection of sets
of the form
Bα , where Bα = Xα for all but finitely many indices
α and Bα ∈ ∏
Bα for the remaining indices, is a basis for the product
topology for Xα .
∏
Proof. Clearly, every set of the form Bα , where Bα = Xα for all but
finitely many indices α and B
∏α ∈ Bα for the remaining indices, is open
in
∏ the product topology on Xα . If G is a nonempty open subset of
Xα and x ∈ G, then there exist finitely many open sets Uαi ⊆ Xαi ,
CONTINUITY AND PRODUCTS
55
∩n
∏
i = 1, . . . , n, say, such that x ∈ 1 p−1
Xα →
αi (Uαi ) ⊆ G, where pαi :
Xαi are projections. Now, for each i = 1, . . . , n, xαi ∈ Uαi and so we
find sets Bαi ∈ Bαi such that xαi ∏
∈ Bαi ⊆∩Uαi . For α ̸= α1 , . . . , αn ,
n
put Bα = Xα . Then we have x ∈ Bα = 1 p−1
αi (Bαi ) ⊆ G and the
proposition follows.
♢
The proof of the following proposition is easy and left to the reader.
Proposition 2.2.9 Let Xα , α ∈ A, ∏
be a family of spaces and Yα ⊆
Xα . Then the∏relative topology on
Yα induced from the product
topology
for
Xα coincides with the topology of the product space
∏
Yα , where each Yα has the induced topology.
As we would hope, the analogue of Theorem 2.2.4 for arbitrary
products holds good.
Theorem 2.2.10 Let Xα , α ∈ A, be a∏
family of spaces. A function f
from a space Y into the product space Xα is continuous if and only
if the composition pα ◦ f is continuous for every α ∈ A.
Proof. If f is continuous, then each pα ◦ f , being the composition of
two continuous maps, is continuous. Conversely, suppose that pα ◦ f
is continuous for every α ∈ A. Then, for any open set Uα ⊆ Xα ,
−1
f −1 (p−1
(Uα ) is open in Y . Because∏the sets p−1
α (Uα ) = (pα ◦ f )
α (Uα ),
Uα ⊆ Xα open, constitute a subbasis for the space Xα , f is continuous.
♢
We will soon see that the preceding theorem characterizes∏the product topology (ref. Theorem 2.2.12). For a function f : Y → Xα , the
map pα ◦ f : Y → Xα is referred to as the αth coordinate function of f .
The preceding theorem is useful in the construction of continuous functions into cartesian products. In fact, if a continuous map fα : Y →
∏ Xα
for each α ∈ A is given, then the function y 7→ (fα (y)) of Y into Xα
is continuous. As another application, we see that if fα : Xα → Yα ,
α
is a family
∏∈ A,∏
∏ of continuous functions, then the product function
fα : Xα → Yα , which takes (xα ) into (fα (xα )), is continuous.
INDUCED TOPOLOGIES
The following notion unifies the studies of relative topology and
product topology.
56
Elements of Topology
Definition 2.2.11 Let X be a set and Yα , α ∈ A, a family of spaces
with a function fα : X → Yα for each α. The topology induced on X
by the functions fα is the smallest topology on X such that each fα is
continuous.
Some authors use the term “initial topology” or “weak topology”
to mean what we call induced topology. Observe that a subbasis for the
induced topology consists of the sets fα−1 (Uα ), where Uα runs through
the open subsets of Yα and α ranges over A.
Notice that the members of the topology induced on X by a function f of X into a space Y are precisely the inverse images of the open
subsets of Y under f. It is clear that if X ⊆ Y , then the topology on X
induced by the inclusion map X ,→ Y is simply the relative topology.
More generally, if f : X → Y is an embedding, then X has the topology induced by f , and conversely, if f : X → Y is an injection (or a
bijection), then the induced topology on X makes f an embedding (or
a homeomorphism). For example, the usual topology for Cn is induced
by the bijection Cn → R2n defined by a + ıb → (a, b). Similarly, the
usual topology for Hn is induced by the bijection Hn → R4n defined
by a + ıb + ȷc + kd → (a, b, c, d).
∏
For another instance, we see that if X =
Yα is the cartesian
product of the spaces Yα , and pα : X → Yα are the projection maps,
then the topology induced on X by the functions pα is just the product
topology for X.
We can restate Proposition 2.2.9 as: The relative topology on a
subset of a product space agrees with the topology induced by the
restrictions of the projections to that subset. Also, Theorem 2.2.10
carries over to the induced topology. In fact, we have
Theorem 2.2.12 If X has the topology induced by the function fα :
X → Yα , α ∈ A, then a function g of a space Z into X is continuous
⇔ fα ◦ g : Z → Yα is continuous for every α. Moreover, this property
characterizes the induced topology for X.
Proof. The first part follows as in Theorem 2.2.10. To establish the last
statement, suppose that X has the topology induced by the functions
fα , and let X ′ denote the set X with a topology such that a function
g of a space Z into X ′ is continuous if and only if fα ◦ g : Z →
Yα is continuous for every α. Then the functions fα : X ′ → Yα are
continuous, since the identity map on X ′ is continuous. If i : X → X ′
is the identity function, then the continuity of the compositions fα ◦i =
CONTINUITY AND PRODUCTS
57
fα : X → Yα implies that i is continuous, and the continuity of the
compositions fα ◦ i−1 = fα : X ′ → Yα implies that i−1 : X ′ → X is
continuous. Thus i is a homeomorphism, and the topologies of X and
X ′ coincide.
♢
Turning to the product of metric spaces, it would clearly be desirable to find a metric which induces the product topology. We have
already seen that the topology of a finite product of copies of the real
line R (cf. Ex. 2.2.1) is induced by a metric. More generally, given a
finite family of metric spaces (Xi , di ), 1 ≤ i ≤ n, there is a metric ρ on
∏
Xi defined by
ρ ((xi ), (yi )) = max {di (xi , yi ) |1 ≤ i ≤ n}.
Observe that an open ball B(x; r) about x = (xi ) in the metric ρ is
the product of the open balls B(xi ; r) in Xi , and the product of open
balls B(xi , ri ), 1 ≤ i ≤ n, contains B(x; r), where r = min {r1 , . . .∏
, rn }.
Consequently, the metric ρ induces the product topology on
Xi .
Thus a finite product of metric spaces is metrisable in the following
sense.
Definition 2.2.13 A space X is metrisable if there is a metric on X
which induces its topology.
An indiscrete space with more than one point and the Sierpinski
space are clearly not metrisable, while a discrete space is metrisable.
The above definition of metric for a finite product of metric spaces
is accessible to an arbitrary product also, provided that the diameters
of coordinate spaces are bounded by a fixed number. If (Xα , dα ), α ∈ A,
is a family of metric spaces each of diameter at most k (a constant),
then
ρ ((xα ), (yα )) = sup {dα (xα , yα )|α ∈ A}
∏
defines a metric on the set Xα . The topology induced by ρ is finer
than the product topology, if the indexing set A is infinite. It should be
noted that the product of uncountably many nontrivial metric spaces is
not metrisable (refer to Corollary 5.1.5). However, a slight modification
in the definition of ρ metrises a countable product of metric spaces. We
establish the following.
Theorem 2.2.14 Let Xn , n = 1, 2, . . ., ∏
be a countable family of
metrisable spaces. Then the product space Xn is metrisable.
58
Elements of Topology
Proof. By Corollary 1.4.9, there is a metric dn on Xn which metrises
its topology and∏
has the property: diam (Xn , dn ) → 0. So, for x = (xn )
and y = (yn ) in Xn , ρ(x, y) = sup {dn (xn , yn )|n = 1, 2, . . .} is a nonnegative real∏
number. We leave to the reader the verification that ρ is
a metric on Xn , and show that it induces the product topology.
∏ A
basic open set U in the product topology can be written as U = Gn ,
where Gn is open in Xn for n = n1 , . . . , nk (say) and Gn = Xn for
all other indices n. Given x ∈ U, choose open balls B(xni ; ri ) ⊆ Gni
for i = 1, 2, . . . , k, and put r = min {r1 , . . . , rk }. Then r > 0 and if
ρ(x, y) < r, then yni ∈ B(xni ; ri ) for each i = 1, . . . , k. It follows
that B(x; r) ⊆ U , and hence the metric topology induced by ρ is finer
than the product topology.
Conversely, let B(x; r) be any open ball
∏
in the metric space
Xn . Since diam(Xn ) → 0, there is an integer
n0 such that diam(Xn ) < r/2, for all n > n0 . Let Gn be the open
ball B(xn ; ∏
r/2) for n = 1, 2, . . . , n0 , and Gn = Xn , for all n > n0 .
Then U = Gn is a basic open nbd of x in the product topology and
U ⊆ B(x; r). This implies that the metric topology is coarser than the
product topology and the two topologies are the same.
♢
As an immediate application of the preceding theorem, we see that
if X is a metric space and A is a countably infinite set, then the product
of A copies of X is metrisable. We usually write X ω for the product
of countably many copies of X, since only the number of elements in
A is important in determining X A . In particular, the space Rω , the
countable product of copies of the real line, and its subspace I ω , which
is the product of countably many unit intervals, are metrisable. There
is another metric on Rω which also induces the product topology. This
is given by
∑
ρ(x, y) = 2−i |xi − yi |/ (1 + |xi − yi |),
where xi is the ith coordinate of x ∈ Rω . To see this, recall that
d(a, b) = |a − b|/(1 + |a − b|)
defines a metric d on R equivalent to the euclidean metric. So the real
line R can be considered equipped with this metric. Then we have
diam(R) = 1. It is easily verified that ρ is a metric for Rω . Now let U
be a subbasic open set of the product topology. Then there exists an
open set O ⊆ R and an integer i > 0 such that U = {x ∈ Rω |xi ∈ O}.
If x ∈ U(, then we
) can find a real ϵ > 0 such that B(xi ; ϵ) ⊆ O. We
have Bρ x; 2−i ϵ ⊆ U , for ρ(x, y) < 2−i ϵ ⇒ yi ∈ O. It follows that U is
CONTINUITY AND PRODUCTS
59
open relative to the metric ρ and the product topology is weaker than
the metric topology for Rω . For the converse, it suffices to show that
for each x ∈ Rω and any real number r > 0, the open ball Bρ (x; r)
contains a nbd of x in the product topology. Let n be
∏∞so large that
r−1 < 2n . Then V = B(x1 ; r/2) × · · · × B(xn+1 ; r/2)× n+2 R is a nbd
of x = (xi ) in the product topology. For y ∈ V , we have
∑n+1 −i
∑∞
ρ(x, y) =
2 d(xi , yi ) + n+2 2−i d(xi , yi )
1
∑n+1 −i−1
∑∞
<
2
r + n+2 2−i
1
< r/2 + r/2 = r
so that y ∈ Bρ (x; r). Thus x ∈ V ⊆ Bρ (x; r), and the topology induced
by ρ on Rω coincides with the product topology. The metric space
(Rω , ρ) is sometime referred to as Fréchet space.
Exercises
1. Let X and Y be spaces, and X × Y have the product topology. If
A ⊆ X and B ⊆ Y , show that (A × B)◦ = A◦ ×( B ◦ , A ×) B(= A × B,
)
(A × B)′ = (A′ × B) ∪ (A × B ′ ) and ∂(A × B) = ∂A × B ∪ A × ∂B .
2. Prove that (a) [0, 1) × [0, 1) ≈ [0, 1] × [0, 1), and (b) Dm × Dn ≈ Dm+n .
3. Let R be the real line and Rd denote the set of all real numbers with
the discrete topology.
(a) Compare the product topologies on R × R and Rd × R, and the
dictionary order topology on R × R.
(b) Show that R × R in the dictionary order topology is metrisable.
4. Compare the product topologies on I × I and I × Id , and the television
topology on I×I (see Exercise 1.4.15), where Id denotes the unit interval
with the discrete topology.
5. Let X be a straight line in the plane R2 (e.g., (x, y) ∈ X ⇔ x + y = 1).
Describe the topologies induced on X from Rℓ × R and Rℓ × Rℓ .
6. Let x0 ∈ X and y0 ∈ Y . Prove that the functions x 7→ (x, y0 ) and
y 7→ (x0 , y) are embeddings of X and Y into X × Y , respectively.
7. Let (X, d) be a metric space. Show that the metric function d : X ×
X → R is continuous in the product topology. Also, prove that the
metric topology is the smallest topology on X such that each function
fy : X → R, fy (x) = d(x, y), is continuous.
60
Elements of Topology
8. Prove that both the addition function and the multiplication function
on R2 are open but neither of them is closed.
{
}
9. (a) Let Y be the subspace (x, y) ∈ R2 |xy = 0 of R2 and p : R2 → R1
be the map defined by p(x, y) = x. Show that p|Y is closed but
not open.
(b) Let X be the subspace
of R2 consisting
of the points of the closed
{
}
right half plane (x, y) ∈ R2 |x ≥ 0 and the x-axis {(x, 0)|x ∈ R}.
Show that the restriction of p to X is neither open nor closed.
10. Let X be a space and f, g : X → R continuous functions. Define the
sum f + g, the difference f − g, the product f.g, the multiple af , a ∈ R,
and the quotient f /g, provided g(x) ̸= 0 for all x ∈ X, by
(f ± g)(x) = f (x) ± g(x),
(f · g)(x) = f (x)g(x),
(af )(x) = af (x),
(f /g) (x) = f (x)/g(x).
Show that each of these functions is continuous.
11. Let X be a space and f, g : X → R continuous functions. Prove that
(a) the graph Gf of f is closed in X × R, and (b) the set of points on
which f and g agree is closed in X.
12. A function f : Rm × Rn → Rk is called bilinear if f (x + x′ , y) =
f (x, y) + f (x′ , y), f (x, y + y ′ ) = f (x, y) + f (x, y ′ ) and f (ax, y) =
af (xy) = f (x, ay), for all x, x′ ∈ Rm , y, y ′ ∈ Rn , and a ∈ R.
Show that a bilinear function Rm × Rn → Rk is continuous.
13. A function f from the product of two spaces X and Y to another space
Z is continuous in x if, for each y ∈ Y , the function x 7→ f (x, y)
is continuous. Similarly, f is continuous in y if for each x ∈ X, the
function y 7→ f (x, y) is continuous.
(a) If f : X × Y → Z is continuous, prove that f is continuous in each
variable.
(b) Consider the function f : R × R → R given by
{
(
)
2xy/ x2 + y 2
for x =
̸ 0 ̸= y, and
f (x, y) =
f (0, 0) = 0.
Show that f is continuous in each variable, but it is not continuous
on R × R.
14. Prove that the subspace {(xn ) ∈ ℓ2 |0 ≤ xn ≤ 1/n} of the Hilbert space
ℓ2 is homeomorphic to the product I ω . (For this reason, it is called the
Hilbert cube.)
CONTINUITY AND PRODUCTS
15. Let Xα , α ∈ A, be a family of discrete spaces. Show that
discrete if and only if A is finite.
61
∏
Xα is
16. Let Xα , α ∈ A, be a family of topological spaces and Eα ⊆ Xα . Show
that
∏
∏
(a)
Eα = Eα .
∏
∏
(b)
Eα is dense in Xα ⇔ each Eα is dense in Xα , and
∏
◦
(c) ( Eα∏
) ̸= ∅ only
∏ if all but finitely many factors Eα = Xα , and
◦
then ( Eα ) = Eα◦ .
17. • Let Xα , α ∈ A,
∏ be a family of spaces, and y = (yα ) be a fixed point
in the product Xα . Show that
(a) for each β ∈ A, (i) Xβ is homeomorphic to the subspace
∏
{x ∈ Xα |xα = yα for every α ̸= β and xβ ∈ Xβ }, and
∏
∏
(ii) {x ∈ Xα |xβ = yβ } is homeomorphic to α̸=β Xα .
∏
(b) The set D = {x ∈ Xα |xα = yα for all but finitely many α ∈ A}
is dense.
18. Let fα : Xα → Yα , α ∈ A, be a family of open
∏ maps
∏ such that
∏ all
but finitely many fα are surjective. Show that fα : Xα → Yα ,
(xα ) 7→ (fα (xα )) , is open.
19. Let (Xi , di ), 1 ≤ i ≤ n, be metric
∏ spaces. Show that each of the following
functions defines a metric on Xi which induces the product topology:
[∑
]1/2
(a) d ((xi ), (yi )) =
di (xi , yi )2
, and
∑
+
(b) d ((xi ), (yi )) = di (xi , yi ).
20. • Let ρ be the metric on the set Rω defined by
ρ ((xn ), (yn )) = sup{d1 (xn , yn )|n = 1, 2, . . .},
where d1 is the standard bounded metric for the set R of real numbers:
d1 (a, b) = min{1, |a − b|}. (This is referred to as sup metric or the
uniform metric for Rω .)
(a) Show that the topology for Rω induced by the metric ρ is strictly
finer than the product topology and coarser than the box topology.
(b) Verify that the function ρ∗ on Rω × Rω given by
ρ ∗ ((xn ), (yn )) = sup{d1 (xn , yn )/n|n = 1, 2, . . .}
is a metric and it induces the product topology on Rω .
62
Elements of Topology
21. Let R∞ = {x ∈ Rω |xn = 0 for almost all n}. Find the closure of R∞ in
the box, product and uniform topologies for Rω .
22. Let Xα ,∏α ∈ A, be an indexed family of topological spaces and consider
the set Xα with the box topology. Show:
∏
(a) If Yα ⊆ Xα are
∏ subspaces, then Yα , with the box topology, is a
subspace of Xα .
∏
(b) If Bα is a ∏
basis of Xα , then the family of sets Bα , Bα ∈ Bα , is
a basis of Xα .
23. Let f : X → Y be a mapping of a set X onto a space Y . If X has the
topology induced by f , prove that f is closed and open.
24. Determine the topology induced by the function f : R → R, where
(a) f (x) = 0 if x ≤ 0, and f (x) = 1 if x > 0,
(b) f (x) = −1 if x < 0, f (0) = 0, and f (x) = 1 if x > 0.
25. Let F be a collection of real-valued functions on R. Describe the topology on R induced by F, if
(a) F consists of all constant functions,
(b) F consists of all functions which are continuous in the usual topology on R,
(c) F consists of all bounded functions which are continuous in the
usual topology for R.
26. Determine the topology on X such that every real-valued function on
X is continuous.
27. Let F be a family of real-valued functions on a set X. Show that a basis
for the topology on X induced by F is given by the sets of the form
{y ∈ X : ∃ x ∈ X, a real ϵ > 0 and finitely many functions f1 , . . . , fn in
F such that |fi (y) − fi (x)| < ϵ}.
28. • Suppose that a space X has the topology induced by a collection
of functions fα : X → Yα , α ∈ A. If Z ⊆ X, show that the relative
topology on Z is induced by the restrictions of the fα to Z.
29. Let X be a set, and Yα , α ∈ A, a family
of spaces. Given functions fα :
∏
X → Yα for every α, let f : X → Yα be defined by f (x) = (fα (x)).
Prove:
(a) The topology induced by f is the same as that induced by the
family {fα } .
(b) If for each pair of distinct points x and x′ in X, there is an index
α ∈ A such that fα (x) ̸= fα (x′ ), then f is an embedding.
Chapter 3
CONNECTEDNESS
3.1
3.2
3.3
3.4
Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Path-connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Local Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1
Connected Spaces
63
72
77
82
The notion of connectedness corresponds roughly to the everyday
idea of being in one piece. One condition when we intuitively think
a space to be all in one piece is if it does not have disjoint “parts.”
Another condition when we would like to call a space one piece is that
one can move in the space from any one point to any other point.
These simple ideas have had important consequences in topology and
its applications to analysis and geometry. In this section, we make
the former intuitive notion precise and establish the basic properties
of this concept. The formulation of the other intuitive notion will be
considered in an other section.
Definition 3.1.1 A space X is called disconnected if it can be expressed as the union of two disjoint, nonempty, open subsets. A space
which is not disconnected is called connected.
If X = A ∪ B, where A and B are disjoint, nonempty and open,
then the pair {A, B} is called a separation or disconnection of X. Notice that, in this case, A and B are also closed. Intuitively, X has a
disconnection if it falls into two parts.
Example 3.1.1 The Sierpinski space and an infinite cofinite space are
connected.
Example 3.1.2 An indiscrete space is connected while a discrete space
having more than one point is disconnected.
63
64
Elements of Topology
Example 3.1.3 The Sorgenfrey line Rℓ is not connected, since the open
sets {x|x < a} and {x|x ≥ a} form a disconnection.
We now determine some convenient characterizations of connectedness.
Theorem 3.1.2 Let X be a space. The following conditions are equivalent:
(a) X is connected.
(b) The empty set ∅ and X are the only clopen subsets of X.
(c) Every continuous function of X into a discrete space is constant.
Proof. (a) ⇔ (b): Trivial.
(b) ⇒ (c): Let f : X → D be continuous, where D is discrete.
Then f −1 (p) is clopen in X for every p ∈ D. If f is not constant and
p ∈ im(f ), then f −1 (p) is a nonempty proper subset of X.
(c) ⇒ (b): Suppose that X contains a proper nonempty set Y which
is clopen in X. Let D = {0, 1} be the two-point discrete space. Then
f : X → D defined by f (Y ) = 0 and f (X − Y ) = 1 is a non-constant
continuous map of X onto the discrete space D.
♢
Next, we see that a subspace of a connected space need not be
connected; for example, a finite subspace of the cofinite space R having
more than one point is not connected. By a connected subset of a space
X, we mean a connected subspace of X. Thus a subset Y ⊆ X is
connected if there do not exist two open sets U and V in X such that
Y ⊂ U ∪ V , U ∩ Y ̸= ∅ ̸= V ∩ Y and U ∩ V ∩ Y = ∅. Note that if
Y is a subspace of X, then a subset Z ⊆ Y is connected if and only
if Z is connected as a subset of X. Thus connectedness is an absolute
property of sets. The empty set and singletons are obviously connected
sets in any space.
The following theorem increases the supply of connected subsets.
Theorem 3.1.3 A nonempty subset X of the real line R is connected
if and only if it is an interval. In particular, R is connected.
Proof. If X is not an interval, then there exist real numbers x, y ∈ X,
and a real number z outside X such that x < z < y. Since z ∈
/ X,
CONNECTEDNESS
65
X = [(−∞, z) ∩ X] ∪ [(z, +∞) ∩ X] is a disconnection of X. So X
connected implies that it is an interval.
Conversely, suppose that X is an interval. We show that X is connected. Assume otherwise, and let X = A ∪ B be a disconnection of
X. Choose a ∈ A and b ∈ B. By relabeling, we can assume that a < b.
Then [a, b] ⊆ X, for X is an interval. Consider S = {x|[a, x) ⊂ A}.
Since A is open, there exists a δ > 0 such that (a − δ, a + δ) ∩ X ⊂ A.
Since X contains [a, b], for sufficiently small δ, [a, a + δ) ⊂ A so that
a + δ ∈ S and S ̸= ∅. Clearly, x ≤ b for all x ∈ S, for A ∩ B = ∅.
Therefore sup(S) exists; put c = sup(S). We have a < c ≤ b, so c ∈ X.
For any δ > 0, c − δ < c. Consequently, there exists an x ∈ S such that
c−δ < x. Since [a, x) ⊂ A, we have (c−δ, c+δ)∩A ⊃ (c−δ, x)∩A ̸= ∅.
It follows that c ∈ A ∩ X = AX , the closure of A in X. As A is closed
in X, AX = A. Thus c ∈ A, and therefore c < b. Since A is open, there
exists a δ > 0 such that (c − δ, c + δ) ∩ X ⊂ A. Then, for δ satisfying
c + δ < b, we have [a, c + δ) ⊂ A, which implies that c + δ ∈ S. This
contradicts the definition of c.
♢
We call two subsets A and B of X separated if A ∩ B = ∅ = A ∩ B.
We can describe the connectedness of a subset of a space in terms of
separated sets.
Proposition 3.1.4 A subset Y of a space X is connected if and only
if Y is not the union of two nonempty separated sets in X.
Proof. Recall that, for any A ⊂ Y , AY = AX ∩ Y . If Y is disconnected,
then Y = A ∪ B, where A ∩ B = ∅, and A and B are both nonempty
closed subsets of Y . We have AX ∩ B = Y ∩ AX ∩ B = AY ∩ B =
A ∩ B = ∅; similarly, A ∩ B X = ∅. Thus A and B are separated sets
in X.
Conversely, suppose that Y = A ∪ B, where A and B are( nonempty
)
and
in X. Then AY = Y ∩AX = (A∪B)∩AX = A ∩ AX ∪
( separated
)
B ∩ AX = A, and therefore A is closed in Y . Similarly, B is closed
in Y , and Y is disconnected.
♢
Clearly, connectedness is a topological invariant; in fact, a continuous image of a connected space is connected.
Theorem 3.1.5 If f : X → Y is continuous and X is connected, then
f (X) is connected.
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Elements of Topology
Proof. Let D be a discrete space, and g : f (X) → D be a continuous
map. If f ′ : X → f (X) is the map defined by f , then the composition
gf ′ : X → D is continuous. Since X is connected, gf ′ is constant and
so g is constant. By Theorem 3.1.2, f (X) is connected.
♢
As a consequence of the above theorem, we obtain a generalization
of the Weierstrass intermediate value theorem.
Corollary 3.1.6 A continuous real-valued function on a connected
space X assumes all values between any two given values.
Proof. Let f : X → R be a continuous function. By Theorem 3.1.5,
f (X) is connected, and therefore it is an interval. So, if x, y ∈ X and
f (x) < c < f (y), there exists a z ∈ X with c = f (z).
♢
We give two simple applications of the preceding corollary.
Corollary 3.1.7 Let f : I → I be a continuous function. Then there
exists a point t ∈ I with f (t) = t.
Proof. If f (0) = 0 or f (1) = 1, then we are through. So assume that
f (0) > 0 and f (1) < 1. Consequently, the continuous function g : I →
R defined by g(t) = f (t) − t satisfies g(1) < 0 < g(0). By Corollary
3.1.6, there exists a t ∈ I with g(t) = 0 which implies that f (t) = t. ♢
Corollary 3.1.8 Let p : R → R be a polynomial function of odd
degree. Then p(x) = 0 for some x ∈ R.
Proof. Clearly, we can assume that p(x) is monic. Then find a real a > 0
such that p(x) < 0 for x ≤ −a, and p(x) > 0 for x ≥ a. By Corollary
3.1.6, there is a point x ∈ (−a, a) with p(x) = 0.
♢
Theorem 3.1.9 Let {Yα } be a family of connected subsets of ∪
a space
X such that Yα ∩ Yβ ̸= ∅ for every pair of indices α, β. Then Yα is
also connected.
∪
Proof. Let f : Yα → D be a continuous function, where D is discrete.
Since Yα is connected, f |Yα is constant for each α. Because any two
members
∪ of the family {Yα } intersect, we see that f is constant, and
hence Yα is connected.
♢
It is clear that the union of a family of connected subsets of a space
having at least one point in common is also connected. So each open
or closed ball in Rn is connected, for it is the union of its diameters
CONNECTEDNESS
67
which are homeomorphic to an interval. It is also immediate that a
space in which any two points are contained in some connected subset
is connected.
Corollary 3.1.10 Let C be a connected subset of a space X, and let
{Yα } be a family
∪ of connected subsets of X such that C ∩ Yα ̸= ∅ for
each α. Then ( α Yα ) ∪ C is connected.
Proof. Set Zα = C ∪ Yα . Then Zα is ∪
connected, by Theorem 3.1.9.
Again, the above theorem implies that Zα is connected.
♢
Corollary 3.1.11
∏ Let X1 , . . . , Xn be connected spaces. Then the
product space Xi is connected.
Proof. By induction, it suffices to prove the corollary for the case n = 2.
Choose a fixed point x1 ∈ X1 . Then C = {x1 } × X2 is homeomorphic
to X2 , so C is connected. Similarly, X1 × {x2 } is connected for each x2
in X2 . It is ∪
obvious that C ∩ (X1 × {x2 }) ̸= ∅ for every x2 ∈ X2 . So
X1 × X2 = x2 ∈X2 (X1 × {x2 }) ∪ C is connected.
♢
Example 3.1.4 The unit n-cube I n and the euclidean space Rn are
connected. This is immediate from Theorem 3.1.3 and Corollary 3.1.11.
Example 3.1.5 The n-sphere Sn is connected for n ≥ 1. Choose two
distinct points x and y of Sn , and put U = Sn − {x} and V = Sn − {y}.
Then both U and V are homeomorphic to Rn and, therefore, connected.
Obviously, we have U ∩ V ̸= ∅, for n ≥ 1. Hence Sn = U ∪ V is
connected.
Example 3.1.6 The subspace Rn −{0} of the euclidean space Rn is connected for n ≥ 2. The function x 7→ (x/ ∥x∥ , ∥x∥) is a homeomorphism
of Rn − {0} onto Sn−1 × (0, ∞), and the latter space is connected, by
Corollary 3.1.11.
The last example enables us to distinguish between R1 and Rn for
n > 1. For, if there were a homeomorphism h : Rn → R1 , then h|(Rn −
{0}) would be a homeomorphism between Rn −{0} and R1 −{h(0)}. But
Rn − {0} is connected and R1 − {h(0)} is disconnected, so they cannot
be homeomorphic. More generally, Rm is not homeomorphic to Rn for
m ̸= n, but the proof of this fact requires tools from algebraic topology.
An argument similar to the above one shows that I and I n , n ≥ 2, are
not homeomorphic. Accordingly, there cannot be a continuous bijection
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Elements of Topology
I ↔ I n . However, we will see later that there is a continuous surjection
I → I n.
Applying the same technique, we see that there is no embedding
of
( )
S1 in R. For, if f : S1 → R1 is a continuous injection, then f S1 , being
a connected subset of R, is an interval. If we remove a point from S1 ,
then the resulting space is connected. But the removal of an interior
point from an interval results in a disconnected space.
( 1 ) Consequently,
1
f can not be a homeomorphism between S and f S .
The following simple result is quite useful, as we shall see in a
moment.
Theorem 3.1.12 Let A be a connected subset of a space X. Then
any set B satisfying A ⊂ B ⊂ A is also connected; in particular, A is
connected.
Proof. Let D be a discrete space, and f : B → D be a continuous
function. Since A
f |A is constant. We have AB = A∩B =
( is connected,
)
B, so f (B) = f AB ⊂ f (A) = f (A), which implies that f is constant.
Hence B is connected.
♢
Example 3.1.7 Consider the portion S = {(x, sin(1/x)) |0 < x ≤ 1}
of the graph of the curve y = sin(1/x) and the set {0} × J, where
J = {y| − 1 ≤ y ≤ 1}. Then S, being a continuous image of an interval,
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S
FIGURE 3.1: The closed topologist’s sine curve.
CONNECTEDNESS
69
is connected. As we have observed in Ex. 2.2.6, each point of the
set{0} × J is a limit point of S; in fact, S ∪ ({0} × J) is the closure of S
in R2 (Figure 3.1(a)). By the preceding theorem, the subspace S ⊂ R2
is connected. This is usually referred to as “the closed topologist’s sine
curve.” The subspace S ∪{(0, 0)} is called “the topologist’s sine curve.”
As another application of Theorem 3.1.12, we generalize Corollary
3.1.11 to arbitrary products.
∏
Theorem 3.1.13 Let Xα , α ∈ A, be a family of spaces. Then Xα
is connected if and only if each Xα is connected.
∏
Proof. Since the projection
pβ : Xα → Xβ is a continuous surjection,
∏
Xβ is connected if Xα is so, by Theorem 3.1.5.
Conversely, suppose
that Xα is connected for every α. Let z be
∏
a
fixed
point
in
X
and
Y be the union of all connected subsets of
α
∏
Xα which contain z. By Theorem 3.1.9, Y is connected
∏ and Theorem
3.1.12 shows that Y is connected. We claim that Y∏= Xα . Let B =
−1
p−1
Xα . We choose a
α1 (Uα1 ) ∩ · · · ∩ pαn (Uαn ) be a basic open set in
point u ∈ B such that uα = zα for α ̸= α1 , . . . , αn . Now, define sets
∏
E1 = {x ∈ ∏ Xα |xα = zα for α ̸= α1 },
E2 = {x ∈ Xα |xα1 = uα1 , xα = zα for α ̸= α1 , α2 },
..
..
..
.
.
.∏
En = {x ∈ Xα |xαi = uαi for i = 1, . . . , n − 1 and
xα = zα for α ̸= α1 , . . . , αn }.
Then clearly z ∈ E1 , u ∈ En and Ei ∩Ei+1 ̸= ∅ for every i = 1, . . . , n−
1. Also, Ei ≈ Xαi for
∪n each i = 1, . . . , n (ref. Exercise 2.2.17), and is
thus connected.
So i Ei is connected and, by the definition of Y, we
∪n
have 1 Ei ⊂ Y . It follows that Y ∩ B ̸= ∅, and hence our claim. This
completes the proof.
♢
Exercises
1. Let (X, T) be a connected space. If T ′ is a topology on X coarser than
T, show that (X, T ′ ) is connected.
2. • Let A and B be separated subsets of a space and C be a connected
subset of A ∪ B. Show that C ⊂ A or C ⊂ B. Give an alternative proof
of Theorem 3.1.12.
3.
(a) Prove that every nonempty proper subset of a connected space
has a nonempty boundary. Is the converse also true?
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Elements of Topology
(b) Let X be a connected metric space with an unbounded metric d.
Prove that the set {y|d(x, y) = r} is nonempty for each x ∈ X. Is
this also true when X is disconnected?
4. Let A be a connected subset of a space X. Are A◦ and ∂A connected?
Does the converse hold?
5. Let An , n = 1, 2, . . ., be a countable family of connected subsets
of
∪
a space X such that An ∩ An+1 ̸= ∅ for each n. Prove that An is
connected.
6. • Let
A
= {(x, y)|x is irrational and 0 ≤ y ≤ 1}, and
B
= {(x, y)|x is rational and −1 ≤ y ≤ 0}.
Prove that A ∪ B is a connected subset of the euclidean space R2 .
7. • For n = 1, 2,∪
. . . , let Xn = {(1/n, y)| − n ≤ y ≤ n}. Show that the
subspace R2 − n Xn is connected.
8. Show that Rn+1 − Sn is the union of two disjoint open connected sets.
9. Discuss the connectedness of the following subsets of Rn , n ≥ 2:
(a) Rn − A, where A is countable.
(b) {(xi ) ∈ Rn |at least one xi is irrational}.
(c) {(xi ) ∈ Rn |xi is irrational for 1 ≤ i ≤ n}.
10. Prove that Rω is disconnected in both the uniform metric topology (ref.
Exercise 2.2.20) and the box topology.
11. • An ordered set X is order complete if each nonempty subset of X
having an upper bound has a supremum. A linear continuum is a complete ordered set X which has no gaps, that is, if x < y in X, then there
exists a z ∈ X such that x < z < y (e.g., R with the usual order relation
is a linear continuum, but the complete ordered set [0, Ω] is not.)
(a) Prove that an ordered space X is connected ⇔ it is a linear continuum.
(b) Show that I × I with the order relation in Exercise 1.4.15 is a
linear continuum, and therefore I ×I is connected in the television
topology.
12. Let A and B be closed subsets of a connected space X such that X =
A ∪ B. Show:
(a) A and B are connected if A ∩ B is connected, and
(b) A or B is connected if A ∩ B contains at most two points.
CONNECTEDNESS
71
13. Give an example of two connected subsets of a space, whose intersection
is disconnected.
14. Let C be a connected subset of a space X, and let A ⊂ X. If C ∩ A ̸=
∅ ̸= C ∩ (X − A), show that C ∩ ∂A ̸= ∅.
15. Let X be a metric space.
(a) If A, B ⊆ X are separated, prove that there exist open sets U, V ⊆
X such that A ⊆ U , B ⊆ V and U ∩ V = ∅.
(b) Prove that a subset C of X is connected ⇔ there do not exist
open sets U, V ⊆ X such that C ⊂ U ∪ V , C ∩ U and C ∩ V are
nonempty, and U ∩ V = ∅.
(c) Give an example to show that (b) is not true for arbitrary topological spaces.
16. Let X be a connected space. Suppose that C ⊂ X is connected and D
is clopen in the subspace X − C ⊆ X. Show that C ∪ D is connected.
17. Prove that an open interval is not homeomorphic to a closed or halfopen interval, and a closed interval is not homeomorphic to a half-open
interval.
18. Prove:
(a) Every continuous open mapping R → R is monotonic.
(b) A monotonic bijection R → R is a homeomorphism.
(c) Let X be the subspace (0, ∞) ⊂ R. Given integer n > 0, prove
that the function f : X → X, defined by f (x) = xn√
, is a homeomorphism. (This implies that the function x 7→ n x on X is
continuous.)
19. Let f : S1 → R1 be a continuous map. Show that there exists x ∈ S1
such that f (x) = f (−x).
20. Let f : X → Y be a continuous open surjection such that f −1 (y) is
connected for each y ∈ Y . Show that C ⊂ Y is connected ⇔ f −1 (C) is
connected.
21. Prove that a connected subset of Rn having more than one point is
uncountable.
22. Let X and Y be connected spaces, and A ⊂ X, B ⊂ Y be proper
subsets. Prove that (X × Y ) − (A × B) is connected.
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3.2
Elements of Topology
Components
If a space X is not connected, then the knowledge about its maximal connected subspaces becomes indispensable in any description of
the complete structure of the topology of X. We now consider these
“pieces” of the space X.
Definition 3.2.1 Let X be a space, and x ∈ X. The component C(x)
of x in X is the union of all connected subsets of X which contain x.
By Theorem 3.1.9, the component C(x) is connected. And, it is
evident from its very definition that C(x) is not properly contained in
any connected subset of X. Thus C(x) is a maximal connected subset
of X; we shall see in a moment that the components of different points
of X are either equal or disjoint. So we refer to them as the components
of X.
A connected space X has only one component, viz., X itself, and
at the opposite extreme, the components of a discrete space are the
one-point subsets.
Proposition 3.2.2 Let X be a space. Then:
(a) Each component of X is closed.
(b) Each connected subset of X is contained in a component of X.
(c) The set of all components of X forms a partition of X.
Proof. (a): If C(x) is a component of x ∈ X, then C(x) is connected.
So C(x) is also connected. By the maximality of C(x), we have C(x) =
C(x), for C(x) ⊂ C(x). Hence C(x) is closed in X.
(b): If A is a nonempty connected subset of X, then A ⊂ C(a) for
any a ∈ A; this follows from the definition of a component.
(c): By definition, each point x ∈ X belongs to a component of X,
viz., C(x). If C(x) and C(x′ ) intersect, then C(x) ∪ C(x′ ) is connected.
The maximality of C(x) and C(x′ ) implies that C(x) = C(x)∪C(x′ ) =
C(x′ ). Thus any two components of X are either equal or disjoint. ♢
Proposition 3.2.2 provides a decomposition of the space X into
connected pieces which are also closed. However, the components of this
CONNECTEDNESS
73
decomposition do not have to be open. If the number of components is
finite, then obviously each component is open as well.
Example 3.2.1 Let Q be the space of rational numbers with the usual
topology. If X ⊂ Q has two distinct points x and y, then, for an
irrational number c between x and y, X = [X ∩(−∞, c)]∪[X ∩(c, +∞)]
is a separation of X, and therefore X is disconnected. It follows that
the components of Q are its one-point subsets. But the points in Q are
not open, for every nonempty open subset of Q is infinite.
We next determine the components of a product space in terms of
those of its factor spaces.
Proposition 3.2.3 Let Xα , α ∈ A,
of spaces. Then the
∏ be a family
∏
component of a point x = (xα ) in Xα is C(xα ), where C(xα ) is
the component of xα in Xα .
∏
Proof.
Let E be the component of
Xα . By Theorem 3.1.13,
∏ x in
∏
C(xα ) is connected, so we have C(x
)
⊆
E. To prove the reverse
∏α
inclusion, assume that y ∈ E. If pβ :
Xα → Xβ is the projection
map, then pβ (E) is a connected subset of Xβ containing
both xβ and
∏
yβ . Consequently, yβ ∈ C(xβ ), and we have y ∈ C(xα ).
♢
It is of some interest to observe that the components of ∏
the product
of discrete spaces Xα , α ∈ A, are one-point sets, although Xα is not
discrete when A is infinite.
If f : X → Y is continuous, and C(x) is the component of x in
X, then f (C(x)) ⊆ C (f (x)), the component of f (x) in Y . If f is a
homeomorphism, then we also have f −1 (C(f (x))) ⊆ C(x), and therefore f (C(x)) = C (f (x)). This shows that the number of components
of a space and structure of each component are topological invariants.
So the knowledge about components can be used for distinguishing
between spaces.
We now apply Proposition 3.2.3 to find the components of the Cantor set.
Example 3.2.2 The components of the Cantor set C (with the subspace
topology induced
∏∞ from R) are singleton sets. To establish this, we show
that C ≈ n=1 Xn , where each Xn is the discrete space {0, 2}. Note
that there is at least one triadic expansion of each member in C that
does not require the use of “1” (e.g. 1/3 = 0.0222 · · · ). Observe that
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Elements of Topology
such a representation of each element of C is unique and take this
∏ type
of representation of every∑
element of C. Then the function f : Xn →
∞
C, defined by f ((xn )) = 1 xn /3n , is a bijection. We claim that f is
a homeomorphism.
To prove the continuity of f , consider an arbitrary
∏
point x ∈ Xn . Given a real number ϵ > 0, find an integer m > 0 so
large that 3−m < ϵ. Then the set
∏
U = {(yn ) ∈ Xn |yi = xi for 1 ≤ i ≤ m}
is a basic open neighbourhood of x, and |f (y) − f (x)| < ϵ for all y ∈ U .
This implies that f is continuous at x. Finally, to see that f is open,
suppose that B is a basic open nbd of x. Then, for some integer k,
∏
B = {y ∈ Xn |xni = yni for 1 ≤ i ≤ k}.
∏
Let m = max{n1 , . . . , nk } and y ∈ Xn be an arbitrary point with
|f (y) − f (x)| < 1/3m+1 . Then yn = xn for every 1 ≤ n ≤ m.
For,
∑ if l ≤ m is the least integer such that yl ̸= xl , then we have
| (yn − xn ) /3n | ≥ 1/3l . Thus y ∈ B and it follows that f (B) is a
neighbourhood of f (x) in C. Therefore f (B) is open, and hence f is
open.
There an important class of spaces typified by the Cantor set. We
introduce them in the following.
Definition 3.2.4 A space X is called totally disconnected if its components are singleton sets.
Obviously, a discrete space is totally disconnected. More interesting
examples are the Cantor set, the set Q of rationals, and the set R − Q
of irrationals with the relative topologies induced from R. Note that
a one-point space is both connected and totally disconnected. If X is
totally disconnected and Y is a subspace of X, then Y is also totally
disconnected, for a component of Y must be contained in some component of X. Also, we see from Proposition 3.2.3 that the product of
a family of totally disconnected spaces is totally disconnected.
Returning to the general discussion, let X = A ∪ B be a separation
of X. If x and y belong to a component of X, then both points are
either in A or in B (ref. Exercise 3.1.2). However, there are spaces X
with distinct points x and y, which do not lie together in a connected
set, and yet they cannot be separated by any separation of X.
CONNECTEDNESS
75
Example 3.2.3 Let X be the subspace of R2 consisting of the line segments {1/n} × I, n = 1, 2, . . . , together with the set K = {0} × I − {p},
where p = (0, 1/2). The sets {(0, t)|0 ≤ t < 1/2} and {(0, t)|1/2 <
t ≤ 1} are components of X containing the points (0, 0) and (0, 1),
respectively, but there is no separation of X which separates them.
K
)
)
p
K
FIGURE 3.2: The space X in Example 3.2.3.
The above situation leads us to the following.
Definition 3.2.5 A subset K of a space X is a quasi-component of
X if for any separation {A, B} of X, K is contained in either A or B,
and K is maximal with respect to this property (i.e., K is not a proper
subset of another set J ⊆ X with the same property).
In Example 3.2.3, K is a quasi-component that is not a component.
If a space has just one quasi-component, then it is clearly connected.
In future, we shall see two different conditions under which components and quasi-components are identical (refer to Exercise 3.4.7 and
Theorem 6.1.18)
We close this section by describing some properties of quasicomponents.
Proposition 3.2.6 Let X be a space. Then:
(a) Each point of X belongs to a unique quasi-component.
(b) The quasi-component of X containing x is the intersection of all
clopen subsets of X which contain x, and hence is closed.
(c) Each component of X is contained in a quasi-component.
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Elements of Topology
Proof. (a): Given x ∈ X, let K(x) be the set of all y in X such that
there is no separation X = A ∪ B with x ∈ A, and y ∈ B. Then
K(x) is clearly a quasi-component of X with x ∈ K(x). Now, we show
that two quasi-components of X are either equal or disjoint. Suppose
that K and K ′ are two quasi-components of X with x ∈ K ∩ K ′ . If
X = A ∪ B is a separation of X, then K and K ′ , both, are contained
in either A or B, according as x ∈ A or x ∈ B. Thus K ∪ K ′ is also
contained in either A or B. By the maximality of a quasi-component,
we have K = K ∪ K ′ = K ′ . This proves the uniqueness of K(x).
(b): Let A be a clopen subset of X with x ∈ A. Then X = A ∪
(X − A) is a separation of X. If K(x) is the quasi-component
∩ of X
containing x, then K(x) ⊂ A. This implies that K(x) ⊂ {A|x ∈
A and A is clopen} = L, say. If X = A ∪ B is a separation, then L ⊂ A
or L ⊂ B according as x ∈ A or x ∈ B. Therefore K(x) = L, by the
maximality of K(x).
(c): If C(x) is the component of x, and X = A ∪ B is a separation
with x ∈ A, then C(x) ⊆ A, since C(x) is connected. So C(x) ∪ K(x)
is contained in A. By the maximality of K(x), we have C(x) ∪ K(x) =
K(x), which implies that C(x) ⊆ K(x).
♢
Exercises
1. • In a space X, show that the relation x ∼ y if both x and y belong to a
connected subset C of X is an equivalence relation, and its equivalence
classes are the components.
2. Prove that a connected clopen subset of a space X is a component of
X.
3. • Determine the components of
(a) the Sorgenfrey line Rℓ and
(b) the subspace {(x, y)|y = 0} ∪ {(x, 1/x)|x > 0} of R2 .
4. Let U be a connected open subset of a space X. Prove that U is a
component of X − ∂U .
5. Let A be a connected subset of a connected space X. If C is a component
of X − A, show that X − C is connected.
6. Let f : X → Y be a continuous map, and C be a component of Y . Show
that f −1 (C) is the union of components of X.
7. Find spaces X and Y such that X ̸≈ Y , but X can be embedded in
Y and Y can be embedded in X. (This shows that the analogue of the
Schroeder–Bernstein theorem for topological spaces does not hold.)
CONNECTEDNESS
77
∪∞
8. Let X = n=0 [(3n, 3n + 1) ∪ {3n + 2}] and Y = (X − {2}) ∪ {1}. Give
X and Y the relative topology induced from R, and define f : X → Y
by f (x) = x for x ̸= 2, and f (2) = 1, and g : Y → X by

for 0 < x ≤ 1,

 x/2
(x − 2)/2 for 3 < x < 4, and
g(x) =


x−3
for x ≥ 5.
Show that both f and g are continuous bijections, but X ̸≈ Y .
9. Show that the subspace A = {0} ∪ {1/n|n = 1, 2, . . .} of the real line is
totally disconnected.
10. Prove that an open subset of Rn can have at most countably many
components. Is this true for closed subsets also?
11. Let Cn denote the circle with centre at the origin and radius 1 − 1/n for
n = 2, 3, . . . , and let T be the subspace of R2 consisting of the Cn , and
the two lines y = ±1. Determine the components and quasi-components
of T .
12. Let X be a space, and consider the relation ∼ in X defined by x ∼ y
if f (x) = f (y) for every continuous map f of X into a discrete space.
Show that ∼ is an equivalence relation and the equivalence classes are
the quasi-components of X.
13. Let X and Y be spaces, and x1 , x2 belong to a quasi-component of X,
and y1 , y2 belong to a quasi-component of Y . Show that (x1 , y1 ) and
(x2 , y2 ) belong to a quasi-component of X × Y .
3.3
Path-connected Spaces
Path-connectedness corresponds to our intuitive notion of considering a space one piece if one can move in the space from any point to
any other point. This kind of property has led to many sophisticated
algebraic techniques for the study of geometry of the spaces.
Let X be a space. A path in X is a continuous function f : I → X,
where I is the unit interval with the usual subspace topology. The point
f (0) is referred to as the origin (or initial point) and the point f (1) as
the end (or terminal point) of f . We usually say that f is a path in
78
Elements of Topology
X from f (0) to f (1) (or joining them). It is important to note that a
path is a function, not the image of the function.
Definition 3.3.1 A space X is called path-connected if for each pair
of points x0 , x1 ∈ X, there is a path in X joining x0 to x1 .
Example 3.3.1 An indiscrete space and the Sierpinski space are obviously path-connected.
Example 3.3.2 Given x, y ∈ Rn , the set of points (1 − t)x + ty ∈ X,
0 ≤ t ≤ 1, is called a (closed) line segment between x and y. A set
X ⊆ Rn is called convex if it contains the line segment joining each
pair of its points. Thus, if X is a convex subspace of Rn , then, given
x, y ∈ X, we have (1 − t)x + ty ∈ X for every t ∈ I. So there is a
path in X joining x to y, namely, t 7→ (1 − t)x + ty. In particular, the
euclidean space Rn , the n-disc Dn , and the open balls in Rn are all
path-connected.
Lemma 3.3.2 A space X is path-connected if and only if there is a
point x0 ∈ X such that each x ∈ X can be joined to x0 by a path in
X.
Proof. The necessity is obvious. To establish the sufficiency, let x, y ∈
X. Suppose that f and g are paths in X joining x and y to x0 , respectively. Consider the function h : I → X defined by
{
f (2t)
0 ≤ t ≤ 1/2,
h(t) =
g(2 − 2t)
1/2 ≤ t ≤ 1.
For t = 1/2, we have f (2t) = x0 = g(2 − 2t). By the Gluing lemma,
h is continuous, and thus defines a path in X joining x to y. So X is
path-connected.
♢
As a consequence of Lemma 3.3.2, we have
Corollary 3.3.3 The union of a family {Xα } of path-connected subspaces of X having a common point is path-connected.
∩
∪
Proof. Let x0 ∈ Xα , and Y = Xα . Then each point y ∈ Y belongs
to some Xα . Since Xα is path-connected, there is a path f : I → Xα
joining y to x0 . If i : Xα → Y is the inclusion map, then the composition if is a path in Y joining y to x0 . Hence Y is path-connected.
♢
CONNECTEDNESS
79
Example 3.3.3 Consider the subspace A of R2 consisting of the points
(x, y), where (x = 1/n, n ∈ N, and 0 ≤ y ≤ 1) or (0 ≤ x ≤ 1 and
y = 0) (see Figure 3.3). For each (x, y) ∈ A, the function f : I → X
defined by
{
(2xt, 0)
0 ≤ t ≤ 1/2,
f (t) =
(x, (2t − 1)y)
1/2 ≤ t ≤ 1
is a path from (0, 0) to (x, y) in X. From the preceding lemma, it is
immediate that A is path-connected. Its closure A = A ∪ ({0} × I) is
also path-connected, since A and {0} × I have a common point. The
subspace A ⊂ R2 is called the Comb space.
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
FIGURE 3.3: Comb space
Example 3.3.4 The n-sphere Sn , n > 0, is path-connected. For, if H1 =
{x ∈ Sn |xn+1 ≥ 0} is the upper hemisphere and y ∈ H1 is the north
pole, then
t → ((1 − t)x + ty) / ∥(1 − t)x + ty∥
is a path in Sn joining x to y. So H1 is path-connected. Similarly,
the lower hemisphere H2 = {x ∈ Sn |xn+1 ≤ 0} is also path-connected.
Obviously, H1 ∩ H2 ̸= ∅, and hence Sn = H1 ∪ H2 is path-connected.
Theorem 3.3.4 Every path-connected space is connected.
Proof. Suppose that X is path-connected, and choose a point x0 ∈ X.
For each x ∈ X, there is a path fx in X from x0 to x. Since I is
connected, so is im(fx ). Clearly, X is the union of the connected subsets
im(fx ), x ∈ X. By Theorem 3.1.9, X is connected.
♢
80
Elements of Topology
In general, the converse of the above theorem is false, although it
is true for all subspaces of R. Thus path-connectedness is a stronger
form of connectedness.
Example 3.3.5 We have already seen that the subspace S =
{(x, sin 1/x) |0 < x ≤ 1} ⊂ R2 is connected (refer to Ex. 3.1.7), and
therefore the closed topologist’s sine curve S is connected. But, it is
not path-connected. To see this, consider the points z0 = (0, 0) and
z1 = (1/π, 0) in S and suppose that f : I → S is a path with origin
z0 and end z1 . Let pi : R2 → R1 , i = 1, 2, be the projection maps.
Then the compositions p1 f and p2 f must be continuous. Let t0 be the
supremum of the points t ∈ I such that p1 f (t) = 0. We assert that p2 f
is discontinuous at t0 , a contradiction. By the continuity of p1 f , it is
easily seen that p1 f (t0 ) = 0; accordingly, t0 < 1. Since f (t) ∈ S for
t > t0 , the image of each interval [t0 , t1 ) under p1 f contains numbers
of the form 2/nπ for all large integers n. Thus p2 f takes on both values +1 and −1 in [t0 , t1 ). Hence the open interval (y0 − 1/3, y0 + 1/3),
where y0 = p2 f (t0 ), cannot contain the image of [t0 , t1 ) under p2 f , and
this proves our assertion.
The invariance properties of path-connectedness are similar to those
of connectedness. The simple proof of the folowing theorem is left to
the reader.
Theorem 3.3.5 If f : X → Y is continuous and X is path-connected,
then f (X) is path-connected.
The preceding theorem shows that path-connectedness is a topological invariant. Like connectedness, this property is also productive,
but not hereditary.
Theorem 3.3.6 The product of a family of path-connected spaces is
path-connected.
Proof. Let Xα , α ∈ ∏
A, be a family of path-connected spaces, and
suppose that x, y ∈
Xα . Then, for each α∏ ∈ A, there is a path
fα : I → Xα ∏
joining xα to yα . Define ϕ : I → Xα by ϕ(t) = (fα (t)),
t ∈ I. If pβ : Xα → Xβ is the projection map, then pβ ϕ = fβ is continuous. So ϕ is continuous, by Theorem 2.2.10. Also, we have ϕ(0)
∏= x
and ϕ(1) = y, obviously. Thus x and y are joined by a path in Xα ,
and the theorem follows.
♢
CONNECTEDNESS
81
Note that there is no analogue of Theorem 3.1.12 for pathconnectedness, as shown by the subset S in Example 3.3.5.
By Corollary 3.3.3, for each point x in a given space X, there exists
a maximal path-connected subset of X containing x, viz., the union of
all path-connected subsets of X which contain x.
Definition 3.3.7 Let X be a space, and x ∈ X. The path component
of x in X is the maximal path-connected subset of X containing x.
Clearly, the path component of a point x in a space X is the set of all
points y ∈ X which can be joined to x by a path in X. If P (x) and P (y)
are path components of two points x and y in a space X, and z ∈ P (x)∩
P (y), then P (x) ∪ P (y) is path-connected. By the maximality of P (x),
we have P (x) = P (x) ∪ P (y). So P (x) ⊆ P (y). Similarly, P (y) ⊆ P (x)
and the eqality holds. It follows that the path components of the space
X partition the set X, and X is path-connected if and only if it has no
more than one path component. Since a path component is connected,
it is contained in a component. Accordingly, each component of X is a
disjoint union of its path components.
Exercises
1. Show that a space having at most three open sets is path-connected.
2. Show that a subset of R is path-connected ⇔ it is an interval.
3. Show that the space in Exercise 1.4.6 is connected, but not pathconnected.
4. Is the space in Exercise 3.1.7 path-connected?
5. If A ⊂ Rn , n > 1, is countable, show that Rn − A is path-connected.
6. For each integer n > 0, let Ln be the line segment in R2 joining the origin
∪
0 to the point (1, 1/n). Is the subspace A ∪ B ⊂ R2 , where A = Ln
and B = [1/2, 2] × {0}, connected, path-connected?
7. Is the union of a closed disc and an open disc in R2 , which are externally
tangent to each other, connected, path-connected?
8. Determine the components and the path components of I × I in the
television topology (refer to Exercise 3.1.11(b)).
9. Let∏Xα , α ∈ A, be a family of spaces. Prove that the path components
of Xα are the direct product of path components of the factors.
82
Elements of Topology
10. Let X be a space and consider the relation ∼ on X defined by x ∼ y if
there exists a path in X from x to y. Prove:
(a) ∼ is an equivalence relation.
(b) The equivalence classes determined by ∼ are the path components
of X.
11. Let π0 (X) be the set of all path components of a space X. If f :
X → Y is continuous, show that there is a well-defined function
π0 (f ) : π0 (X) → π0 (Y ) given by π0 (f )P (x) = Q (f (x)), P (x) denotes
the path component of x in X and Q(f (x)) denotes the path component
of f (x) in Y . If f is a homeomorphism, prove that π0 (f ) is a bijection.
12. An arc in a space X is an embedding f : I → X. The space X is
called arcwise connected if any two points in X are the end points of
an arc in X. (Some authors use this terminology in the sense of pathconnectedness.)
Give an example of a path-connected space that is not arcwise connected.
3.4
Local Connectivity
The path components of a space X are not necessarily closed subsets of X, nor are they necessarily open, as shown by the path components S and {0} × [−1, 1] of the closed sine curve (refer to Ex.
3.1.7). Also, the components of a space need not be open (cf. Ex.
3.2.1). We will study here the localized version of connectedness and
path-connectedness under which components and path components,
respectively, are open and hence closed.
LOCAL CONNECTEDNESS
Definition 3.4.1 A space X is called locally connected at x ∈ X if,
for each open neighbourhood U of x, there is a connected open set V
such that x ∈ V ⊂ U . The space X is locally connected if it is locally
connected at each of its points.
CONNECTEDNESS
83
It is clear that the family of all connected open sets in a locally
connected space forms a basis. Conversely, if a space X has a basis
consisting of connected sets, then it is obviously locally connected.
Example 3.4.1 A discrete space is locally connected, and so is an indiscrete space.
Example 3.4.2 The euclidean space Rn is locally connected, since its
base of open balls consists of connected sets. Similarly, Sn is also locally
connected.
Example 3.4.3 The subspace X = {0} ∪ {1/n|n = 1, 2, . . .} of the real
line is not locally connected. Because the singleton set {1/n} are clopen
in X, and any neighbourhood of 0 contains them for large values of n.
Thus there is no connected neighbourhood of the point 0 in X.
A locally connected space need not be connected, as shown by the
subspace [0, 1/2) ∪ (1/2, 1] of I. On the other hand, the following examples show that a connected space need not be locally connected.
Example 3.4.4 For each integer n > 0, let Ln be the line segment in
R2 joining the origin 0 to the point (1, 1/n), and L0 be the segment
{(s, 0)|0 ≤ s ≤ 1} (Figure 3.4). Then
∪ each Ln , n ≥ 0, is connected2 and
contains the point 0. Hence X = Ln is a connected subset of R . We
L1
L2
O
x•
¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦ ¦ ¦
¦
¦
¦
¦¦
¦
¦¦
¦¦¦
¦ ¦ ¦
¦
L3
L0
FIGURE 3.4: A connected space which is not locally connected.
84
Elements of Topology
show that X is not locally connected at any point x of L0 other than
0. Consider a small ball B about x such that 0 ∈
/ B. For 1/n less than
the radius of B, Ln ∩ B ̸= ∅, and this intersection is a component of
U = X ∩ B, for it is connected, closed and open in U . Now, if V is open
and x ∈ V ⊂ U , then clearly V ∩ Ln ̸= ∅ for all large n. Therefore V
cannot be connected, and we see that X is not locally connected at x.
Example 3.4.5 The closed topologist’s sine curve (ref. Ex. 3.1.5) is not
locally connected at any point p = (0, y), −1 ≤ y ≤ 1. Consider the
open nbd U = S ∩ B of p, where B is an open ball of radius less than
1/2 and centered at p (see Figure 3.5). Clearly, the boundary of B
divides the wiggly part of S into infinitely many arcs, and U = S ∩ B
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B
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S
FIGURE 3.5: The closed topologist’s sine curve is not locally connected.
consists of a segment C of {(0, y)| − 1 ≤ y ≤ 1}, and some of the arcs of
S. Each arc of S lying in B is closed and connected, being a continuous
image of a closed interval. Since C is separated from every arc contained
in U , it follows that C is the component of p in U . Obviously, every
neighbourhood of p intersects S; so there is no connected open set V
in S containing p and contained in U . Thus S is not locally connected
at p.
Similarly, one sees that the sine curve S ∪ {0} is also not locally
connected at 0, while the subspace S is.
CONNECTEDNESS
85
Now, we turn to see a characterisation of locally connected spaces
in terms of the components of open subsets of the space. By the components of a subset U ⊂ X, we mean the components of the subspace
U of X. Thus, the component of a point x in U is a subset of U .
Theorem 3.4.2 A space X is locally connected if and only if the
components of each open subset of X are open.
Proof. Suppose that X is locally connected. Let U be an open subset
of X, and C be a component of U . If x ∈ C, then there is a connected
open set V ⊆ X such that x ∈ V ⊆ U , by our hypothesis. Since C is a
component of x in U and V is a connected subset of U containing x,
we have V ⊆ C. Thus C is a neighbourhood of each of its points, and
therefore open.
To prove the converse, let U ⊂ X be open, and x ∈ U . By our hypothesis, the component V of x in U is open. So X is locally connected
at x. This is true for every x ∈ X, and X is locally connected.
♢
As a particular case of the above theorem, we see that each component of a locally connected space is open.
Next, we come to the usual questions involving continuous functions, products and subspaces of locally connected spaces. Although,
local connectedness is not hereditary, every open subset of a locally
connected space is locally connected. It is noteworthy that every connected subspace of the real line R is locally connected. It is also clear
from the definition that a continuous open image of a locally connected
space is locally connected. Thus local connectedness is a topological invariant. Moreover, we have
Theorem 3.4.3 Let f : X → Y be a continuous closed surjection. If
X is locally connected, then so is Y .
Proof. Suppose that X is locally connected. By Theorem 3.4.2, we need
to show that the components of each open set U ⊂ Y are open. Let C be
a component of U . We assert that f −1 (C) is open. If x ∈ f −1 (C), then
there exists a connected open set V in X such that x ∈ V ⊆ f −1 (U ),
since X is locally connected and f −1 (U ) is open in X. It follows that
f (x) ∈ f (V ) ⊆ C, for f (V ) is connected. So x ∈ V ⊆ f −1 (C), and
f −1 (C) is a neighbourhood
This) proves our assertion. Now, since
( of x.
−1
f is closed, Y − C = f X − f (C) is closed. So C is open and the
theorem follows.
♢
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Elements of Topology
However, local connectedness is not preserved by a continuous map,
as shown by the following.
Example 3.4.6 Let Y = N ∪ {0} with the discrete topology, and X be
the space in Ex. 3.4.3. It is known that Y is locally connected, and X
is not. Obviously, the function f : Y → X, defined by f (n) = 1/n,
n ∈ N, and f (0) = 0, is a continuous bijection.
The property of local connectedness is not transmitted to arbitrary
products, as shown by an infinite product of discrete spaces. In this
regard, we have the following.
Theorem
∏ 3.4.4 Let Xα , α ∈ A, be a collection of spaces. Then the
product
Xα is locally connected if and only if each Xα is locally
connected and all but finitely many spaces Xα are also connected.
∏
Proof.∏Suppose that Xα is locally
connected. Then each component
∏
C of Xα is open. Let pβ : Xα → Xβ denote the ∩
projection map
n
for every index β ∈ A. Find a basic open set B = i=1 p−1
αi (Uαi ),
say, contained in C. Then, for α ̸= α1 , . . . , αn , Xα = pα (B) = pα (C)
is connected. To see the local connectedness of Xβ , let x ∈ Xβ be
arbitrary, ∏
and Uβ be an open neighbourhood of x in Xβ . Then p−1
β (Uβ )
is open in Xα and contains a point∏ξ with x = ξβ . By our assumption,
there is a connected open set V in Xα such that ξ ∈ V ⊆ p−1
β (Uβ ).
We have x ∈ pβ (V ) ⊆ Uβ . The set pβ (V ) is connected and open, for
pβ is continuous and open. Thus, Xβ is locally connected
at x.
∏
To
prove
the
converse,
let
ξ
=
(x
)
∈
X
,
and
let B =
α
α
∩n −1
1 pαi (Uαi ) be a basic neighbourhood of ξ. By our hypothesis, there
are at most finitely many indices α ̸= α1 , . . . , αn such that Xα is not
connected. Assume that these indices are αn+1 , . . . , αn+m . By local
connectedness of the Xα , there exists a connected open neighbourhood
m, such that Vαi ⊂ Uαi for 1 ≤ i ≤ n. By
Vαi of xαi , 1 ≤ i ≤ n +
∩n+m
Theorem 3.1.13,
p−1
αi (Vαi ) is a connected open set, and
∏C = 1
ξ ∈ C ⊆ B. So Xα is locally connected at ξ, and this completes the
proof.
♢
CONNECTEDNESS
87
LOCAL PATH-CONNECTEDNESS
Definition 3.4.5 A space X is said to be locally path-connected at a
point x ∈ X if each open neighbourhood of x contains a path-connected
neighbourhood of x. The space X is called locally path-connected if it
is locally path-connected at each of its points.
A discrete space is obviously locally path-connected. The euclidean
space Rn is locally path-connected, since the open balls are pathconnected. The n-sphere Sn is locally path-connected, because each
point of Sn has a neighbourhood which is homeomorphic to an open
subset of Rn . It should be noted that a path-connected space need not
be locally path-connected. This can be seen by the union of the closed
topologist’s sine curve with an arc connecting the points (1, sin 1) and
(0, 1). (see Figure 3.6.)
+1
•
•
0
−1
FIGURE 3.6: A path-connected space which is not locally path-connected.
Clearly, a space X is locally path-connected at x ∈ X if and only
if there exists a neighbourhood basis at x consisting of path-connected
sets. A frequently used criterion for local path-connectedness is described by
Proposition 3.4.6 A space X is locally path-connected if and only if
the path components of open subsets of X are open.
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Proof. ⇒: Suppose that X is locally path-connected, and let U ⊂ X
be open and P a path component of U . If x ∈ P , then there exists
a path-connected neighbourhood V of x with V ⊂ U . As x ∈ P ∩ V ,
P ∪ V is path-connected and contained in U . Since P is a maximal
path-connected subset of U , we have P ∪ V = P , which implies that
V ⊂ P . Thus P is a neighbourhood of x, and P is open.
⇐: Obvious.
♢
From the preceding proposition, it is clear that a space X is locally
path-connected if and only if it has a basis of path-connected open sets.
Another equivalent formulation of local path-connectedness is given in
Exercise 14.
Corollary 3.4.7 Let X be a locally path-connected space. Then each
path component P (x) of X is clopen, and therefore coincides with the
component C(x) of X.
Proof. By Proposition 3.4.6, P (x) is open. The complement of P (x) in
X is the union of all path components of X which are different from
P (x), and is therefore open. Thus P (x) is closed, too. Finally, P (x) is
connected, by Theorem 3.3.4, and hence a component of X.
♢
As a consequence of the preceding corollary, we obtain
Corollary 3.4.8 A connected, locally path-connected space is pathconnected.
A discrete space with at least two points shows that the condition
of connectedness in this corollary is essential.
It is clear from the definition that an open subspace of a locally
path-connected space is locally path-connected. Hence every connected
open subspace of Rn and of Sn is path-connected. It is also obvious that
a locally path-connected space is locally connected, but the converse is
not true (see Exercise 5).
The invariance properties of local path-connectedness are similar
to those of local connectedness. The proof of the following theorems is
similar to that of Theorem 3.4.3, and left to the reader.
Theorem 3.4.9 Let f : X → Y be a continuous closed or open surjection. If X is locally path-connected, then so is Y .
By the preceding theorem, the local path-connectedness is a topological invariant. However, the property is not continuous invariant.
CONNECTEDNESS
89
Example 3.4.7 Consider the subspaces X = S ∪ {(−1, 0)} and Y =
S∪{(0, 0)} of the closed topologist’s sine curve (Ex. 3.1.7). Clearly, X is
locally path-connected, but Y is not (see Ex. 3.4.5). Define f : X → Y
by setting f (x) = x for every x ∈ S, and f (−1, 0) = (0, 0). Then f is
a continuous surjection.
Theorem
∏ 3.4.10 Let Xα , α ∈ A, be a collection of spaces. Then the
product α Xα is locally path-connected if and only if each Xα is
locally path-connected, and all but finitely many Xα are also pathconnected.
The proof is similar to that of Theorem 3.4.4 and is left to the reader.
The principal applications of local connectedness lie in the theory
of (continuous) curves. By a curve in a space X is meant the range of
a path f : I → X. The following result, which shows the existence of a
curve filling up the square I 2 , is somewhat startling because it violates
one’s naive geometric intuition.
Theorem 3.4.11 There exists a continuous surjection I → I 2 .
Proof. Let C be the Cantor set, and Xn denote the two-point discrete
space
∏ {0, 2} for every n = 1, 2, . . .. Then there is a homeomorphism
f : ∏ Xn → C (ref. Ex. 3.2.2). Similarly,
there is a continuous function
∑
g : Xn → I given by g ((xn )) = xn /2n+1 . Using dyadic expansion
of the numbers in I, we see that
Also,
∏ g is surjective.
∏
∏it is not difficult
to see that the function h :
Xn → ( Xn ) × ( Xn ) defined by
h ((xn )) = ((x2n−1 ), (x2n )) is a homeomorphism. Accordingly, we have
a continuous surjection ϕ : C → I 2 , where ϕ is the composition (g ×
g) ◦ h ◦ f −1 . Now, let (a, b) denote an open interval removed from I
in the construction of C. Since I 2 is convex, there is a line segment L
in I 2 with ends ϕ(a) and ϕ(b). Let ψ : [a, b] → L be the continuous
map defined by ψ(t) = [(b − t)ϕ(a) + (t − a)ϕ(b)]/(b − a), a ≤ t ≤ b.
Obviously, ψ maps a into ϕ(a), and b into ϕ(b). By the Gluing lemma,
ϕ and ψ can be combined to give a continuous map C ∪ (a, b) → I 2 .
We can clearly repeat this process with other open intervals in I − C.
Thus we obtain a continuous map ξ : I → I 2 which is an extension of
ϕ, and therefore surjective.
♢
There is an interesting construction of a curve, due to Hilbert, which
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fills up the square (with its interior). For each integer n > 0, a continuous functions fn : I → I 2 is defined as suggested by the following
figures.
•
•
f1
f2
•
f3
FIGURE 3.7: Three stages of filling up the square I 2 .
It is easily checked that the sequence ⟨fn ⟩ converges uniformly to a
continuous function on I. The range of the limit function fills I 2 . A
curve in I 2 , which fills up the entire space, is referred to as a “Peano
space-filling curve.”
Exercises
1. Prove that the space in Exercise 3.1.6 is not locally connected.
2. Which of the spaces in Exercise 3.2.3 are locally connected?
3. • Discuss local connectedness and local path-connectedness of the following:
(a) The subspace A of the comb space (see Ex. 3.3.3).
(b) The deleted comb space A ∪ {(0, 1)}.
(c) The comb space.
4. Let X be the union of all line segments in R2 joining the point (1, 1)
to each point (r, 0), where r ∈ I is a rational. Show that X is locally
connected only at the point (1, 1). Find a path-connected subspace of
R2 which is not locally connected at any of its points.
5. • Show that the space I × I with the television topology (see Exercise
1.4.15) is locally connected but not locally path-connected.
6.
(a) If X = A ∪ B is a separation of the space X in Ex. 3.2.3, show
that K is contained in either A or B.
CONNECTEDNESS
91
(b) Prove that the space X in (a) is not locally connected at (0, 0).
7. Let X be a locally connected space and x ̸= y be points lying in different
components of X. Show that there exists a separation X = A ∪ B with
x ∈ A and y ∈ B (that is, quasi-components of X equal its components).
8. Prove: A space X is locally connected ⇔ for each component C of a
subspace Y ⊂ X, ∂C ⊂ ∂Y .
9. Let U be an open subset of a locally connected space X. Show:
(a) If C is a component of U , then U ∩ ∂C = ∅.
(b) If C − C ̸= ∅, then X − C ̸= ∅.
(c) Even if U is connected, ∂U may fail to be connected or locally
connected.
10. Let X be locally connected, and Y ⊂ X. If U is a connected, open
subset of Y , show that U = Y ∩ G for some connected open subset G
of X.
11. Let X be a locally connected space, and A ⊂ X. If A is closed, and C
is a component of A, prove that ∂C = C ∩ ∂A.
12. Let X be a locally connected space. If A ⊂ X with ∂A locally connected,
prove that A is locally connected.
13. Let X be a locally connected space, and A and B closed subsets of X
such that X = A ∪ B. If A ∩ B is locally connected, show that A and
B are locally connected.
14. • Prove that a space X is locally path-connected ⇔ for each point
x ∈ X, and each open neighbourhood U of x, there is a neighbourhood
V of x such that V ⊂ U and any point of V can be joined to x by a
path in U .
15. Let X be a space, and the set Y = X have the topology generated by
the path components of open sets in X. Prove:
(a) Y is locally path-connected.
(b) The identity function i : Y → X is continuous.
(c) A function f of a locally path-connected space Z into Y is continuous ⇔ the composition if : Z → X is continuous.
Chapter 4
CONVERGENCE
4.1
4.2
4.3
4.4
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
Sequences
93
96
103
106
We are familiar with the notion of convergence of sequences of numbers (real or complex) and its useful role in analysis. This notion can
also be introduced in topological spaces. But, as we will see, the basic
theorems describing the topology of spaces and continuity of functions
in terms of the convergence of sequences fail in the general setting.
This problem can be dealt with by means of the “nets” or the “filters”
which will be treated in the next two sections.
Definition 4.1.1 A sequence in a set X is a function ϕ : N → X,
where N denotes the ordered set of natural numbers.
For n ∈ N, the image ϕ(n) of n under ϕ is referred to as the nth
term of the sequence, and is usually denoted by xn . The sequence ϕ
is written in the form ⟨xn ⟩ or {x1 , x2 , . . .}. If Y ⊆ X and xn ∈ Y for
every n, then ⟨xn ⟩ is said to be in Y . It is easy enough to generalize
the notion of convergence of sequences of real numbers to sequences in
any space, as in
Definition 4.1.2 A sequence ⟨xn ⟩ in a space X converges to a point
x ∈ X if, for each nbd U of x, there exists a positive integer n0 such
that xn ∈ U for every n ≥ n0 .
If ⟨xn ⟩ converges to x, we write xn → x and call x a limit of ⟨xn ⟩.
If ⟨xn ⟩ is a sequence in X and n1 < n2 < · · · is a strictly increasing
sequence of positive integers, then ⟨xnk ⟩ is a subsequence of ⟨xn ⟩.
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By weakening the condition of convergence, we obtain
Definition 4.1.3 Let ⟨xn ⟩ be a sequence in a space X. A point x ∈ X
is called a cluster (or accumulation) point of ⟨xn ⟩ if each nbd U of x
contains infinitely many terms of the sequence.
In elementary analysis, we have seen a very close relation between
the limit points of (sub)sets and the limits of convergent sequences.
Specifically speaking, in a metric space X ⊇ A, x ∈ A ⇔ there exists a sequence in A which converges to x. This enables us to express
continuity of functions between metric spaces in terms of the convergence of sequences: A function f : X → Y is continuous ⇔ xn → x
in X implies that f (xn ) → f (x) in Y . It is also known that cluster
points of a sequence in a metric space are precisely the limits of the
convergent subsequences of the given sequence. Unfortunately, all these
statements, when considered in a general topological space, are false,
as the following examples show.
Example 4.1.1 Let Rc denote the space of real numbers with the cocountable topology. Then the complement of the range of a sequence
is open in this space. Consequently, no rational number is a limit of
a sequence in R − Q. On the other hand, every rational number is
an adherent point of R − Q, since it intersects every nonempty open
set. Next, suppose that a sequence ⟨xn ⟩ in Rc converges to a point
x. Then the complement of {xn |xn ̸= x, and n = 1, 2, . . .} is a nbd of
x. Accordingly, there exists an integer n0 such that xn = x for all
n ≥ n0 . Thus a convergent sequence in Rc must be constant from some
place on. It follows that the identity function Rc → R (the reals with
the usual topology) preserves convergent sequences, although it is not
continuous.
Example 4.1.2 (R. Arens) Let Y be the set of all ordered pairs of positive integers and X = Y ∪ {(0, 0)}. Consider the topology on X
in which every point of Y is open, and a subset U of X containing (0, 0) is open if for all except a finite number of integers m, the
sets {n|(m, n) ∈
/ U } are each finite. The bijective mapping Y → N,
1
(m, n) 7→ n + 2 (m + n − 1)(m + n − 2), defines a sequence with (0, 0)
as a cluster point, since given an integer k > 0, for each m, there is
an n such that n + 21 (m + n − 1)(m + n − 2) > k. We observe that
no sequence in Y can converge to (0, 0). If a sequence ⟨yn ⟩ contains
points from finitely many columns only, then it is clear that it cannot
CONVERGENCE
95
converge to (0, 0). Otherwise, we obtain an infinite subsequence ⟨zn ⟩
of ⟨yn ⟩, which contains at most one point from each column. Since the
complement of ⟨zn ⟩ is a nbd of (0, 0), it cannot converge to (0, 0), and
therefore the sequence ⟨yn ⟩ does not converge to (0, 0).
Exercises
1.
(a) What are limits of ⟨1/n⟩ when R is assigned the cofinite topology?
(b) Find limit(s) of a convergent sequence in R if it has the lower limit
topology.
2. Let X be space with the cofinite topology. Prove that a sequence ⟨xn ⟩
in X converges to a point x iff, for each y ̸= x, the set {n|xn = y} is
finite.
3. Let ⟨xn ⟩ be a sequence in R with the range Q, the set of all rational
numbers. Show that every real number is a cluster point of ⟨xn ⟩.
⟨
⟩
4. If a sequence x(n) in the Hilbert space ℓ2 converges to x, show that
⟨
⟩
(n)
(n)
xi → xi for every i. Find a sequence x(n) in ℓ2 such that xi → 0 for
⟨ (n) ⟩
every i but x
fails to converge in ℓ2 . (This shows that the topology
of ℓ2 is different from the topology it would inherit as a subspace of
Rω .)
∏
5. • Let RI denote the product space i∈I Ri , where each Ri is a copy of
the real line R and I denotes the unit interval [0,1]. Prove:
(a) The point c0 which takes value 0 everywhere is an adherent point
of the set A = {f : I → R|f (r) = 0 for finitely many points r
and f (r) = 1 elsewhere} , but there is no sequence in A which
converges to c0 .
(b) The sequence ⟨fn ⟩ in RI given by fn (t) = tn converges to the
function which is zero for all points of I except 1, where it takes
value 1 (accordingly, the subset C(I) is not closed in RI ).
6. Let (Y, d) be a metric space and X a set. A sequence of functions fn :
X → Y is said to converge uniformly to a function f : X → Y if for
each ϵ > 0, there exists an integer m such that d (fn (x), f (x)) < ϵ for
all n ≥ m and all x ∈ X.
Prove that a sequence ⟨fn ⟩ in the set B(X, Y ) of all bounded functions
X → Y converges to g with respect to the sup metric d∗ (ref. Exercise
1.1.7) if and only if ⟨fn ⟩ converges uniformly to g.
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7. • Let X be a topological space, and Y be a metric space. If a sequence
of continuous functions ⟨fn : X → Y ⟩ converges uniformly to a function
f : X → Y , show that f is continuous. Give an example to show that
this need not be true if the convergence is not uniform.
4.2
Nets
It is now evident that sequences alone are generally incapable of expressing all topological concepts and do not carry enough information
about the topology of a space. This difficulty is overcome by introducing a natural generalization of sequences – “nets” – in which the set of
natural numbers is replaced by ordered sets. We shall use this concept
to prove some important results in subsequent chapters.
Definition 4.2.1 A directed set A is a nonempty set together with
a reflexive and transitive relation ≼ such that, for any two elements
α, β ∈ A, there is a γ ∈ A satisfying α ≼ γ and β ≼ γ. We say that the
relation ≼ directs A, and sometimes write β ≽ α for α ≼ β.
Some authors require the relation ≼ to be antisymmetric, too.
Example 4.2.1 The set N of natural numbers with its usual ordering is
a directed set.
Example 4.2.2 The family of all finite subsets of a given set X is directed by inclusion ⊆ (i.e., Y ≼ Z if Y ⊆ Z). Similarly, the reverse
inclusion ⊇ directs any family of subsets of X which is closed under
finite intersections.
Definition 4.2.2 A net in a space X is a function ϕ from a directed
set A into X.
A net with domain the set N of natural numbers is a sequence. We
remark that nets are often called Moore–Smith sequences or generalized
sequences.
CONVERGENCE
97
Definition 4.2.3 Let ϕ : A → X be a net in the space X and Y ⊆ X.
We say that
(a) ϕ is in Y if ϕ(α) ∈ Y for every α ∈ A;
(b) ϕ is eventually in Y if there is an α ∈ A such that ϕ(β) ∈ Y for
all β ≽ α; and
(c) ϕ converges to a point x ∈ X (written ϕ → x) if it is eventually
in every nbd of x.
Notice that, for sequences, Definition 4.2.3 (c) reduces to that in
4.1.2. It is common to write xα for ϕ(α), α ∈ A, and denote the net ϕ
by ⟨xα ⟩ or {xα , α ∈ A}. If xα → x, we occasionally write lim xα = x.
Limits, when they exist, are not necessarily unique. For example, every
net in the space X with the trivial topology converges to every point of
X. A nontrivial example is a sequence with distinct terms in a cofinite
(infinite) space. It is clear that a net cannot be eventually in each of
two disjoint sets. Accordingly, limits of a convergent net are unique in
a space in which every pair of distinct points can be separated by open
sets in the sense that if x ̸= y, then there exist disjoint open sets U
and V containing x and y, respectively. A space with this property is
called a Hausdorff space. Interestingly, this condition is equivalent to
the uniqueness of limits of nets in a topological space.
Theorem 4.2.4 A space X is Hausdorff if and only if each net in X
converges to at most one point in X.
Proof. As seen above, a net in a Hausdorff space can not have more
than one limit. To see the converse, suppose that X is not Hausdorff.
Then there are two points x ̸= y in X which do not have disjoint
nbds. Let P = {(U, V )|U be a nbd of x, and V a nbd of y}. We direct
P by ≼, where (U, V ) ≼ (U ′ , V ′ ) if U ⊇ U ′ and V ⊇ V ′ . For each
(U, V ) in P, we choose a point ϕ(U, V ) in U ∩ V and consider the net
ϕ : P → X. We assert that ϕ converges to both x and y, contrary to our
hypothesis. To see this, let N be any nbd of x. Then there is an open
set U ⊆ X such that x ∈ U ⊆ N . Take any open nbd V of y. Then
(U, V ) ∈ P. Now, if (U ′ , V ′ ) ≽ (U, V ), then U ′ ⊆ U and V ′ ⊆ V so
that ϕ(U ′ , V ′ ) ∈ U ′ ∩ V ′ ⊆ U ∩ V ⊆ N . This shows that ϕ is eventually
in N , and therefore it converges to x. Similarly, ϕ converges to y, and
hence our assertion.
♢
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This theorem enables us to speak of “the” limit of a convergent
net in a Hausdorff space. It is remarkable that the “if” part of Theorem 4.2.4 is not valid for sequences. The space Rc of reals with the
cocountable topology is not Hausdorff, although each sequence in this
space has at most one limit.
The next two theorems suggest that nets are adequate to describe
all the basic topological notions, without much difficulty.
Theorem 4.2.5 Let X be a space and A ⊆ X. Then x ∈ A ⇔ there
is a net in A which converges to x.
Proof. If x ∈ A, then each open nbd U of x intersects A. So we can
pick a point xU ∈ U ∩ A. The family Ux of all open nbds of x is
directed by the reverse inclusion (that is, U1 ≼ U2 if U1 ⊇ U2 ). The
net {xU , U ∈ Ux } obviously converges to x.
Conversely, if {aλ , λ ∈ Λ} is any net in A which converges to a point
x ∈ X, then each nbd U of x intersects A, and hence x ∈ A.
♢
By Theorem 4.2.5, x ∈ A if and only if there exists a net in A which
converges to x. It follows that the closed sets and the open sets can be
described in terms of convergence of nets; in other words, the topology
of the space can be completely determined.
Theorem 4.2.6 A function f : X → Y between spaces is continuous
if and only if for every net ϕ in X converging to a point x ∈ X, the
net f ◦ ϕ in Y converges to f (x).
Proof. The direct part is easy. To prove the converse, suppose that f
is not continuous at some point x ∈ X. Then there is a nbd V of f (x)
such that f (U ) ̸⊂ V for every nbd U of x. Choose a point xU in U
such that f (xU ) ∈
/ V . Direct the set of all nbds U of x by the reverse
inclusion and define a net ϕ by putting ϕ(U ) = xU . If U0 is any nbd
of x and U ≽ U0 , then xU ∈ U0 so that ϕ is eventually in U0 . Thus ϕ
converges to x. But (f ϕ)(U ) ∈
/ V for any U ; so f ◦ ϕ does not converge
to f (x).
♢
This theorem can be rephrased as:
f : X → Y is continuous ⇔ f (lim xα ) = lim f (xα ) for every convergent net ⟨xα ⟩ in X.
The following theorem describes convergence in a product space.
CONVERGENCE
99
Theorem
4.2.7 Let Xα , α ∈ A, be a family of topological spaces, and
∏
pβ : ∏ Xα → Xβ be the projection onto the βth factor. Then a net
ϕ in Xα converges to x = (xα ) if and only if pα ◦ ϕ → xα for each
α ∈ A.
Proof. If ϕ → x, then pα ◦ ϕ → pα (x) = xα , since each pα is continuous.
Conversely, suppose that p∏
α ◦ ϕ → xα in Xα for every α ∈ A. Let B
be a basic nbd of x = (xα ) in Xα . Then there are∩finitely many open
sets Uαi ⊆ Xαi , i = 1, . . . , n, say, such that B = p−1
αi (Uαi ). By our
−1
hypothesis, the net ϕ is eventually in pαi (Uαi ) for every i = 1, . . . , n.
It is now plain that ϕ is eventually in B, and hence ϕ → x.
♢
The sequential version
∏ of Theorem 4.2.7( is )the familiar statement:
A sequence ⟨(xnα )⟩ in α Xα converges to x0α ⇔ xnα → x0α for every
α.
We next introduce the notion of “subnets” of a net – a generalization of subsequences. Call a subset B of a directed set (A, ≼) cofinal if
for each α ∈ A, there is β ∈ B such that α ≼ β. Each cofinal subset
of A is also directed by the ordering of A. Accordingly, a definition of
“subnet” can be made by restricting the net to a cofinal subset of the
indexing set. But this simple definition of subnet turns out to be inadequate for many purposes (ref. Ex. 4.1.2). Recall that subsequences
are defined by precomposing the sequence with an order preserving injection, and this injection is strictly monotone increasing. Analogously,
a “subnet” of a net ϕ : A → X can be defined to be the composition ϕθ, where θ is a function of a directed set B into A such that
β ≼ β ′ ⇒ θ(β) ≼ θ(β ′ ) and θ(B) is cofinal in A. This definition includes the previous case and is good for almost all purposes, yet the
condition on θ is relaxed to give a more general notion of “subnet.”
Definition 4.2.8 A subnet of a net ϕ : A → X is a net ψ : B → X
together with a function θ : B → A such that
(a) ψ = ϕθ, and
(b) for each α ∈ A, there exists β ∈ B satisfying α ≼ θ(β ′ ) for all
β ′ ≽ β.
It is immediate that this definition includes the earlier two plausible
definitions of a subnet. Thus, every subsequence of a sequence ⟨xn ⟩ is a
subnet of ⟨xn ⟩. It should be noticed that a sequence may have subnets
which are not subsequences.
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Definition 4.2.9 Let X be a space and Y ⊆ X. A net ϕ : A → X is
frequently in Y if for each α ∈ A, there is a β ∈ A such that α ≼ β and
ϕ(β) ∈ Y . A point x ∈ X is called a cluster (or an accumulation) point
of ϕ if it is frequently in every nbd of x.
When restricted to sequences this definition gives our earlier notion of cluster point. As we would hope, the pathological behavior of
sequences shown in Ex. 4.1.2 is not found in nets. This follows from
Theorem 4.2.10 Let ϕ : A → X be a net in a space X. A point
x ∈ X is a cluster point of ϕ ⇔ there is a subnet of ϕ which converges
to x.
Proof. The “if” part is trivial. To prove the converse, suppose that x
is a cluster point of ϕ. Let Ux be the family of all open nbds of x and
define
D = {(α, U )|α ∈ A, U ∈ Ux , and ϕ(α) ∈ U } .
Consider the relation ≼ in D given by (α, U ) ≼ (β, V ) ⇔ α ≼ β
and V ⊆ U. Then D is directed by ≼ because the intersection of two
members of Ux is again a member of Ux , and ϕ is frequently in each
member of Ux . Define θ : D → A by setting θ(α, U ) = α. Now, for any
α ∈ A, we have (α, X) ∈ D and if (β, U ) ≽ (α, X), then β ≽ α. Thus θ
satisfies the condition (b) of Definition 4.2.8, and therefore determines
a subnet of ϕ. We observe that ϕθ converges to x. Given any open nbd
U of x, find α ∈ A such that ϕ(α) ∈ U . Then (α, U ) ∈ D and, for
(β, V ) ≽ (α, U ), we have ϕθ ((β, V )) = ϕ(β) ∈ V ⊆ U . This implies
that ϕθ converges to x.
♢
It is clear that if a net ϕ converges to x, then x is a cluster point
of ϕ, and every subnet of ϕ converges to x. But, there are nets which
have a single cluster point and yet fail to converge, for example, the
sequence ϕ(n) = n + (−1)n n in R (0 is the only cluster point of ϕ). We
describe here the nets for which the converse holds.
Definition 4.2.11 A net ϕ in a set X is called a universal net (or an
ultranet) if, for every S ⊆ X, ϕ is eventually in either S or X − S.
As an example, a constant net is a universal net. Note that a net
ϕ is frequently in U ⇔ ϕ is not eventually in X − U . From this, it
is immediate that a universal net in a topological space converges to
each of its cluster points. The next theorem guarantees the existence
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101
of nontrivial universal nets (of course, assuming the axiom of choice
A.3.3).
Theorem 4.2.12 Every net has a universal subnet.
Proof. Let ϕ : A → X be a net. Consider the collection C of all families
E of subsets of X such that ϕ is frequently in each member of E and
the intersection of any two members of E is also in E. The family {X}
belongs to C so that it is nonempty. We partially order the set C by the
inclusion relation. If B is a chain in C, then the union of the families in
B is an upper bound (for B). So Zorn’s lemma applies and we obtain
a maximal family E0 in C. The set D = {(α, E) ∈ (A × E0 ) |ϕ(α) ∈ E}
is directed by the ordering ≼ defined by
(α, E) ≼ (α′ , E ′ ) ⇔ E ′ ⊆ E and α ≤ α′ .
The mapping θ : D → A, which takes (α, E) to α, clearly satisfies the
condition (b) of Definition 3.1.11 and, therefore, determines a subnet
of ϕ. We show that the subnet ϕθ is universal. Let S ⊆ X be any set.
If ϕθ is not frequently in S, then it is eventually in X − S, and we are
through. So assume that ϕθ is frequently in S. We claim that S ∈ E0 .
It is obvious that ϕ is frequently in S. Given α ∈ A and E ∈ E0 , we find
β ≽ α such that ϕ(β) ∈ E. Then (β, E) ∈ D. By our assumption, there
is (γ, F ) ≽ (β, E) such that ϕθ ((γ, F )) ∈ S. Accordingly, ϕ(γ) ∈ E ∩S,
where γ ≽ α. Thus ϕ is frequently in E ∩ S for every E ∈ E0 . It follows
that the family consisting of the sets in E0 , S, and their intersections
is in C. By the maximality of E0 , we have S ∈ E0 , and hence our claim.
Now, if ϕθ is also frequently in X − S, then we have X − S ∈ E0 ,
as above. Consequently, E0 contains the empty set, contrary to the
definition of C. Hence ϕθ is eventually in S, and this completes the
proof.
♢
We remark that there is no analogue of this theorem for sequences.
Exercises
1. Prove that a net in a discrete space is convergent if and only if it is
constant from some point on.
2. • Let S be a subbasis for a space X. If a net ⟨xα ⟩ in X is eventually in
each member of S containing x, prove that xα → x.
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3. Let (X, d) be a metric space and x0 ∈ X be a limit point of X − {x0 }.
Direct the set X − {x0 } by the relation x ≼ x′ if d(x′ , x0 ) ≤ d(x, x0 ).
Show that a net ϕ : X −{x0 } → Y , where Y is a metric space, converges
to y0 ∈ Y if and only if limx→x0 ϕ(x) = y0 in the sense of elementary
analysis.
4. Let T1 and T2 be topologies on a set X. If each net in X which converges
relative to T1 also converges to the same point with respect to T2 , prove
that T2 ⊆ T1 .
5. Let (X, ≼) be a linearly ordered set. A net ϕ : A → X is called monotone
increasing (resp. decreasing) if α ≼ β ⇒ ϕ(α) ≼ ϕ(β) (resp. ϕ(β) ≼
ϕ(α)).
Let (X, ≼) be an order complete linearly ordered set. Show that each
monotone increasing net in X whose range has an upper bound converges in the order topology to the supremum of its range.
6. Let X be a metric space and ϕ : [0, Ω) → X be a net. Show that ϕ → x
in X if and only if ϕ(α) equals x eventually.
7.
(a) Show that there is a net in [0, Ω) which converges to Ω in the
ordinal space [0, Ω], but there is no sequence in [0, Ω) converging
to Ω.
(b) Show that the real valued function f defined on [0, Ω] by f (α) = 0
if α < Ω and f (Ω) = 1 is not continuous at Ω, even though it does
preserve convergent sequences.
8. Let X be any subset of R and F be the family of all∑finite subsets
of X ordered by inclusion. For F ∈ F, put ϕ(F ) =
x∈F x. Prove
that the net ϕ has a limit if and
only
if
X
is
a
countable
set
and,
∑
∑∞ for a
enumeration {x1 , x2 , . . .} of X, |xn | < ∞. In this case, ϕ → n=1 xn .
9.
(a) If x is a cluster point of a net which is eventually in a closed set
F , show that x ∈ F .
(b) Prove that a subset F of a space X is closed if and only if no net
in F converges to a point of X − F .
10. Show that a cofinal subset of a directed set (A, ≼) is also directed by
≼.
11. Find a subnet of a sequence which fails to be a subsequence.
12. Let ϕ : A → X be a net and θ : B → A be a function of a directed set
B to A which is increasing (β1 ≼ β2 ⇒ θ(β1 ) ≼ θ(β2 )) and cofinal (i.e.,
θ(B) is cofinal in A). Show that ϕθ is a subnet of ϕ.
13. If every subnet of a net ϕ : A → X has a subnet converging to x, show
that ϕ converges to x. (It remains true if we replace “nets” by sequences
in this statement.)
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14. Let ϕ : A → X be a net in the space X and, for each α ∈ A, let
Yα = {ϕ(β)|β
≽ α in A} . Show that x is a cluster point of ϕ if and only
∩
if x ∈ α Y α .
∏
15. Let Xα , α ∈ A, be a family of topological spaces, and
∏ pβ : Xα → Xβ
be the projection onto the βth factor. Let ϕ : Λ → Xα be a net with
x as a cluster point. Show that for every α, pα ◦ ϕ has pα (x) for a cluster
point. Give an example to show that the converse fails.
16. Show that a sequence is a universal net if and only if it is eventually
constant.
17. Prove that a subnet of a universal net is universal.
18. Let f : X → Y be a function, where X and Y are sets. If ϕ is a universal
net in X, prove that the composition f ϕ is a universal net in Y .
4.3
Filters
In the previous section, we have developed a natural generalisation
of the theory of sequential convergence. However, the relation between
convergence and topologies is best understood by means of the “filters.”
As this concept is not essential for our future discussions, we set forth
the subject on a rather fast pace and relegate most properties of the
filters to exercises.
A filter F on a set X is a nonempty family of nonempty subsets of
X with the properties:
(a) if F ∈ F and F ⊂ F ′ , then F ′ ∈ F, and
(b) if F1 , F2 ∈ F, then F1 ∩ F2 ∈ F.
The family of all subsets of X which contains a given nonempty
set E ⊆ X is a filter. If X is a topological space and x ∈ X, then the
family Nx of all nbds of x in X is a filter on X.
For an other example, consider a net {xα , α ∈ A} in X. Then the
family F consisting of subsets F of X for which there exists a α ∈ A
(depending on F ) such that xβ ∈ F for all β ≽ α is a filter on X. We
call F the filter generated by the net ⟨xα ⟩. On the other hand, given a
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filter F on X, let D = {(x, F )|x ∈ F ∈ F}. Direct D by the ordering
(x, F ) ≼ (x′ , F ′ ) ⇔ F ′ ⊆ F. Then the function ϕ : D → X defined by
ϕ(x, F ) = x is a net, called the net based on F.
If F and F′ are two filters on X and F ⊆ F′ , then we say that F′ is
finer than F. A filter F on a topological space X converges to x ∈ X
(written as F → x) if F is finer than the nbd filter Nx . In this case,
we call x a limit of F. The filter F is said to accumulate at x ∈ X if
x ∈ F for every F ∈ F. If F accumulates at x, then we say that x is an
accumulation point or cluster point of F.
A filter base in X is a family B of nonempty subsets of X such that
for every pair of sets B1 , B2 in B, there exists a set B3 ∈ B such that
B3 ⊆ B1 ∩ B2 . Obviously, the family Ux of all open nbds of a point x
in a space X is a filter base in X. If B is a filter base in X, then the
family of supersets of members of B is a filter on X. This is referred to
as the filter generated by B. If F and G are two filters on X such that
F ∩G ̸= ∅ for all F ∈ F and G ∈ G, then F∩G = {F ∩G|F ∈ F, G ∈ G}
is a filter base, and the filter generated by F ∩ G is finer than both F
and G.
A filter base B in a space X is said to converge to a point x ∈ X if
the filter generated by B converges to x. When B converges to x, we
write B → x. We say that B accumulates at x if the filter generated
by B accumulates at x.
A filter F on a set X is called an ultrafilter if there is no filter on X
finer than F. In other words, an ultrafilter on X is a maximal member of
the collection of filters on X partially ordered by the inclusion relation.
Proposition 4.3.1 Every filter on a set X is contained in an ultrafilter.
Proof. Let F be a filter on X, and consider the collection C of all filters
on X which contain F. Partially order C by the inclusion relation. If D
is a chain in C, then it is easily checked that the union of members of D
is a filter on X. Thus every chain in C has an upper bound. By Zorn’s
lemma, C has a maximal member G, say. Clearly, G is an ultrafilter on
X containing F.
♢
Given a point x ∈ X, the family of all subsets of X which contain
x is an ultrafilter (called the principal filter generated by x). It follows
that an ultrafilter on X is not unique.
(A detailed discussion about this can be found in the text General
Topology by Bourbaki (see also James [5] and Willard [16]).)
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Exercises
1. Which filters F converge to x in the trivial space X? Answer the same
question in a discrete space.
2. Let X be an infinite set with cofinite topology.
(a) Let F be a filter on X such that every member of F is infinite.
Find its accumulation points.
(b) If F consists of all cofinite sets, what are its limits?
3. If a filter F on a space X converges to x, show that x is an accumulation
point of F.
4. Prove that a filter F on a space X accumulates at x ⇔ some filter finer
than F converges to x.
5. Let X be a space and E ⊆ X. Show that x ∈ E ⇔ there exists a filter
F on X such that E ∈ F and F → x.
6. Prove that a space X is Hausdorff if and only if F → x in X implies
that each accumulation point of F coincides with x. (Thus limits of
convergent filters on a Hausdorff space are unique.)
7. Let F be the filter generated by a net ϕ in X. Verify:
(a) ϕ → x ⇔ F → x.
(b) ϕ accumulates at x ⇔ F accumulates at x.
8. Let F be a filter on X, and ϕ be the net based on F. Prove
(a) F → x in X if and only if ϕ → x.
(b) F accumulates at x if and only if ϕ accumulates at x.
9. If ψ is a subnet of a net ϕ in X, show that the filter generated by ψ is
finer than the filter generated by ϕ.
10. Let ϕ : A → X be a net. Show that the family of subsets Bα =
{ϕ(β)|β ≽ α}, α ∈ A, is a filter base.
11. Let B be a filter base in a space X. Prove that
(a) B → x ⇔ for each nbd U of x, there exists B ∈ B such that
B ⊆ U.
(b) B
( accumulates at x∩⇔ for every )B ∈ B and every nbd U of x,
U ∩ B ̸= ∅ ⇔ x ∈ {B : B ∈ B} .
12. Let f : X → Y be a function.
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Elements of Topology
(a) If F is a filter on X, show that {f (F ) : F ∈ F} is a filter base in
Y . (The filter on Y generated by this filter base will be denoted
by f∗ (F).)
(b) If G is a filter on Y and f −1 (G) ̸= ∅ for all G ∈ G, show that
{f −1 (G) : G ∈ G} is a filter base in X. (The filter on X generated
by this filter base will be denoted by f ∗ (G).)
13. Let X and Y be spaces and f : X → Y be a function. Prove that f
is continuous at x0 ∈ X if and only if for every filter F → x0 in X,
f∗ (F) → f (x0 ) in Y .
14. Let∏
Xλ , λ ∈ Λ, be a family of topological spaces. Show that a filter F
on ∏Xλ converges to (x0λ ) ⇔ (pµ )∗ (F) → x0µ for each µ ∈ Λ, where
pµ : Xλ → Xµ is the canonical projection map.
15. Let f : X → Y be a function.
(a) If B is a filter base in X, prove that f (B) = {f (B)|B ∈ B} is a
filter base in Y .
(b) If B is a filter base in Y and f −1 (B) ̸= ∅ for every B ∈ B, prove
that f −1 (B) = {f −1 (B)|B ∈ B} is a filter base in X.
16. Let F be a filter on X. Show that F is an ultrafilter if and only if for
each set S ⊆ X, either S ∈ F or X − S ∈ F.
17. Prove: A filter F on a set X is an ultrafilter if and only if for every pair
of subsets S, T ⊂ X, S ∪ T ∈ F implies that S ∈ F or T ∈ F.
18. Prove: The net based on an ultrafilter is an ultranet, and the filter
generated by an ultranet is an ultrafilter.
19. If a filter F on a set X is contained in a unique ultrafilter F0 , show that
F = F0 .
20. If f : X → Y is a surjection and F is an ultrafilter on X, prove that
f∗ (F) is an ultrafilter on Y .
4.4
Hausdorff Spaces
As seen in Section 2, the separation of points by disjoint open sets
in a topological space is needed to ensure the uniqueness of “limits”
(i.e., no net converges to more than one point). There are more than ten
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107
“separation axioms,” traditionally denoted by T0 , T1 , . . .. These stipulate the degree to which distinct points or closed sets may be separated
by open sets. The axiom T2 is quite an important separation axiom and
will be treated here in more detail. We will also deal with the weaker
axioms T0 and T1 , and postpone the discussion about some other important separation axioms until Chapter 7.
Definition 4.4.1 A space X is called Hausdorff (or T2 ) if for every
pair of distinct points x, y of X there exist disjoint open sets U and V
with x ∈ U and y ∈ V .
Example 4.4.1 A discrete space is obviously a T2 -space.
Example 4.4.2 Any metric space is Hausdorff. For, if x ̸= y and 2r is
the distance between x and y, then the open balls B(x; r) and B(y; r)
are disjoint nbds of x and y.
Example 4.4.3 An ordered space is Hausdorff: Suppose that x ≺ y
are two points of the ordered space (X, ≼). If (x, y) = ∅, then the
subbasic open sets (−∞, y) and (x, +∞) are disjoint nbds of x and
y, respectively. And, if there is a point z ∈ (x, y), then (−∞, z) and
(z, +∞) satisfy the requirement.
By Theorem 4.2.4, the Hausdorff condition essentially means that
“limits” are unique. Here are other useful characterizations of this property.
Proposition 4.4.2 The following properties of a space X are equivalent:
(a) X is Hausdorff.
(b) For each point x ∈ X, the intersection of the closed nbds of x is
{x}.
(c) The diagonal ∆ = {(x, x)|x ∈ X} is closed in X × X.
Proof. (a) ⇔ (b): If y ̸= x, then there is an open nbd U of x and an
open nbd V of y such that U ∩ V = ∅. So X − V is a closed nbd of
x, for U ⊆ X − V . Obviously y ∈
/ X − V and (b) holds. Conversely,
if y ̸= x, then there is a closed nbd K of x such that y ∈
/ K. Choose
an open set U with x ∈ U ⊆ K. Then U and X − K are disjoint open
nbds of x and y, respectively.
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(a) ⇔ (c): For any U, V ⊆ X, we have U ∩ V = ∅ if and only if
(U × V ) ∩ ∆ = ∅, and the result follows immediately.
♢
Corollary 4.4.3 Let Y be a Hausdorff space.
(a) If f : X → Y is a continuous function, then the graph of f is
closed in X × Y .
(b) If f, g : X → Y are continuous functions, then the set E =
{x ∈ X|f (x) = g(x)} is closed in X.
Proof. (a) Clearly, the graph of f is the inverse image of the diagonal
∆ under the continuous map
f ×1
X × Y −−−→ Y × Y,
(x, y) 7→ (f (x), y).
Since Y is T2 , ∆ is closed in Y × Y , and the conclusion holds.
(b) Suppose that f, g : X → Y are continuous. Then the map
h : X → Y × Y , defined by h(x) = (f (x), g(x)), is continuous. As ∆ is
closed in Y × Y , E = h−1 (∆) is closed in X, and the result follows. ♢
By the preceding corollary, it is immediate that if two continuous
functions f, g : X → Y agree on a dense subset of X, and Y is a
T2 -space, then f = g.
The Hausdorff property is clearly a topological invariant, but it
is not preserved by continuous maps or even by continuous open or
continuous closed maps.
Example 4.4.4 The space Rf of real numbers with the cofinite topology
is not Hausdorff, for there are no nonempty disjoint open sets. The
identity map of the real line R onto Rf is, of course, continuous.
For the remaining cases, the reader is referred to Ex. 7.1.11 and
7.1.12.
A given property of a topological space is said to be hereditary
if every subspace of a space with the property also has the property.
Hausdorffness is obviously hereditary, and the next result shows that
it is also acquired by the product topology.
Theorem 4.4.4 Let
∏ Xα , α ∈ A, be a family of Hausdorff spaces. Then
the product space Xα is Hausdorff.
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∏
Proof. Let x = (xα ) and y = (yα ) be distinct points of
Xα . Then
xβ ̸= yβ for some index β ∈ A. Since Xβ is Hausdorff, there exist
disjoint
If
∏ open nbds Uβ and Vβ of xβ and yβ in Xβ , respectively.
−1
pβ : Xα → Xβ is the projection map, then p−1
(U
)
and
p
(V
)
are
β
β
β
β
clearly disjoint open nbds of x and y, respectively.
♢
It is remarkable that the converse of this theorem is also true (provided Xα ̸=∏∅ for all α), because each Xα is homeomorphic to a
subspace of Xα .
We now discuss two weaker conditions, viz., T0 and T1 which are
sufficient for some purposes, and will be talked about occasionally.
Definition 4.4.5 A space X is called T1 if for each pair of distinct
points x, y ∈ X, there is an open set containing x but not y, and
another open set containing y but not x.
Definition 4.4.6 A space X is called T0 if for any two distinct points
of X, there is an open set which contains one point but not the other.
The axiom T1 is also referred to as the Fréchet condition, and axiom
T0 as the Kolmogorov condition. Evidently, a T2 -space is T1 , and a T1 space is T0 . But the converse is false in either case.
Example 4.4.5 The Sierpinski space satisfies T0 -axiom but not the axiom T1 .
Example 4.4.6 The space Rf in Ex. 4.4.4 satisfies T1 -axiom but not
T2 .
The assertion in the preceding example follows at once from the
following.
Proposition 4.4.7 A space X is T1 ⇔ {x} is closed for every x ∈
X ⇔ the intersection of all nbds of x ∈ X is {x}.
Proof. Suppose that X is a T1 -space. Then, for each y ∈ X − {x},
there
is an open nbd Uy of y such that x ∈
/ Uy . So X − {x} =
∪
{Uy |y ∈ X and y ̸= x} is open. Conversely, if every one-point set in
X is closed and x ̸= y, then the open sets X − {y} and X − {x} are
nbds of x and y, respectively, not containing the other point.
To see the second equivalence, let x0 be a fixed element of X. If {x}
is closed for every x ∈ X and Nx denotes the family of all nbds of x in
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∩
X, then X − {x} ∈ N
∩x0 for all x ̸= x0 . So we have {N |N ∈ Nx0 } =
{x0 }. Conversely, if {N |N ∈ Nx } = {x} for every x ∈ X, then X −
{x0 } contains a N ∈ Nx for every x ̸= x0 . It follows that {x0 } is closed
in X.
♢
It is immediate from the preceding proposition that every finite
set in a T1 -space is closed. As another consequence, we see that a
closed image of a T1 -space is T1 . But a continuous function or even
a continuous open function fails to preserve this property. The first
assertion can be seen by considering the identity map of a two-point
discrete space onto the Sierpinski space, and the second will be justified
later (see Ex. 7.1.11). It is also clear that subspaces and products of
T1 -spaces are T1 .
Proposition 4.4.8 Let X be a T1 -space and A ⊆ X. Then a point
x ∈ X is a limit point of A if and only if each open nbd of x contains
infinitely many points of A.
Proof. The sufficiency is obvious. To prove the other half, assume that
there is an open nbd U of x such that A ∩ U is finite. Then the set
F = A ∩ U − {x} is closed in X, since X is T1 . Thus U − F is an open
nbd of x, which contains no point of A except possibly x itself. So x is
not a limit point of A, and the proposition follows.
♢
It follows that if x is a limit point of a subset A of a T1 -space, then
there are infinitely many points of A which are arbitrarily close to x.
As an application of the above proposition, we show that an infinite
Hausdorff space contains a large number of open sets.
Theorem 4.4.9 An infinite Hausdorff space has infinitely many disjoint open sets.
Proof. Let X be an infinite T2 -space. If there are no limit points in
X, then X is a discrete space and the theorem is obvious. So assume
that y is a limit point of X. Choose a point x1 ∈ X different from y.
Then there exist disjoint open sets U1 and V1 in X such that x1 ∈ U1 ,
y ∈ V1 . As y is a limit point, V1 − {y} is nonempty. So we can find a
point x2 ∈ V1 different from y. By our hypothesis, there are disjoint
open sets U2 and V2 contained in V1 such that x2 ∈ U2 , y ∈ V2 . Notice
that U1 ∩ U2 = ∅ for U2 ⊆ V1 . Clearly, the above argument is inductive
because every nbd of y contains infinitely many points. Thus we obtain
a sequence of open sets Un and Vn such that Un ∩ Vn = ∅, and both
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Un and Vn are contained in Vn−1 . Since V1 ⊇ V2 ⊇ · · · , Un is disjoint
from every Ui for 1 ≤ i < n.
♢
Exercises
1. Prove that space X in Example 4.1.2 is T2 .
2. Let X be any uncountable set, and Tc be the cocountable topology for
X. Show that (X, Tc ) is a T1 -space which is not T2 .
3. On the set R of real numbers, consider the topology
{
}
T = {∅} ∪ R − F |F is closed and bounded in R1 .
Show that (R, T) is T1 but not T2 .
4. Let X be a T2 -space and f : X → Y be a closed bijection. Show that
Y is T2 .
5. Let Y be a T2 -space.
(a) If f : X → Y is a continuous injection, show that X is T2 .
(This implies that a topology finer than a Hausdorff topology is
Hausdorff.)
(b) Let f : X → Y and g : Y → X be continuous with gf = 1X . Show
that X is T2 and f (X) is closed in Y .
6. Let f : X → Y be a continuous open surjection. Show that Y is Hausdorff if and only if the set {(x, x′ )|f (x) = f (x′ )} is closed in X × X.
7. Let F be a family of real-valued functions on a set X and assign X the
topology induced by F. Show that X is T2 ⇔ for each pair of distinct
points x and y in X, there exists an f ∈ F with f (x) ̸= f (y).
8. Let X be an infinite Hausdorff space. Prove:
(a) X contains a countably infinite discrete subspace.
(b) X contains a strictly decreasing sequence of closed sets.
9. Show that any finite T1 -space is discrete.
10. For a set X, show that cofinite topology is the smallest topology for X
satisfying T1 -axiom.
11. If X is a T1 -space with more than one point, show that a base which
contains X as an element remains a base if X is dropped.
12. Let X be a T1 -space and A ⊆ X. Prove that A′ is a closed set.
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13. If A is a subset of the T1 -space X, show that the intersection of all the
nbds of A in X is A itself.
14. Suppose that X is a T1 -space such that the intersection of every family
of open sets is open. Show that X is discrete.
15. If a T1 -space X has no isolated points, show that every dense set in X
also has no isolated points. Is the condition of T1 -ness on X necessary?
16. Prove: A T1 -space in which each point has a local base consisting of
clopen sets is totally disconnected.
17. Show that a connected subset of a T1 -space having more than one point
is infinite.
18. Give an example of a countable connected Hausdorff space.
Chapter 5
COUNTABILITY AXIOMS
5.1
5.2
1st and 2nd Countable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Separable and Lindelöf Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
1st and 2nd Countable Spaces
113
119
This section concerns itself with topological spaces which have
countable bases or countable local bases. These spaces have many pleasant properties. Besides, it will be seen that the theory of sequential
convergence is adequate to express most topological concepts in such
spaces.
We recall that a nbd basis at a point x of a space X is a family Bx
of nbds of x (in X) such that each nbd of x contains some member of
Bx .
Definition 5.1.1 A topological space is said to satisfy the first axiom
of countability (or be first countable) if it has a countable neighbourhood basis at each point.
Definition 5.1.2 A topological space is said to satisfy the second axiom of countability (or be second countable) if its topology has a countable basis.
Since a family S of subsets of X is countable if and only if the collection of finite intersections of members of S is countable (ref. A.5.12),
a space is second countable if it admits a countable subbasis. It is clear
that a second countable space is first countable, but the converse is not
true.
Example 5.1.1 A metric space is obviously first countable, for the open
balls of radii 1/n, n = 1, 2, . . ., about a given point of the space constitute a countable nbd basis. But, an uncountable set X with the discrete
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Elements of Topology
metric, [d(x, y) = 1 if x ̸= y, and d(x, x) = 0], is not second countable,
since the topology determined by this metric is discrete.
Example 5.1.2 The open intervals with rational end points form a
countable basis for the real line R. More generally, the euclidean space
Rn is second countable for every n ≥ 1. The cubes (a1 , b1 ) × · · · ×
(an , bn ), where ai and bi are rationals for all i, form a countable base.
Example 5.1.3 The Sorgenfrey line Rℓ satisfies the first axiom of countability, since the intervals [x, x + 1/n), n = 1, 2, . . ., form a nbd basis
at x, for each x ∈ R. But it is not second countable. Given a basis B
of Rℓ , choose Bx ∈ B for each x ∈ R such that x ∈ Bx ⊆ [x, x + 1).
Then x 7→ Bx is an injective mapping R → B; consequently, B is
uncountable. Thus, Rℓ fails to satisfy the second axiom of countability.
Example 5.1.4 An uncountable set X with cofinite topology is not first
countable. If Bx is a local basis at x ∈ X,∩then, for every y ̸= x, there
is a member of Bx not containing
y. So {B : B ∈ Bx } = {x} which
∪
implies that X − {x} = {X − B : B ∈ Bx }. If Bx were countable,
then X−{x}, being a countable union of finite sets, would be countable,
a contradiction. Thus there is no countable nbd basis at x.
A similar analysis applies when X is assigned the cocountable topology.
The following proposition shows that both axioms of countability
are topological invariants.
Proposition 5.1.3 The continuous open image of a first countable
(resp. second countable space) is first countable (resp. second countable).
The proof is simple and we leave this to the reader.
The above proposition does not hold for continuous or continuous
closed maps (cf. Example 7.1.13).
By Proposition 1.5.5, we see that both axioms of first countability
and second countability are hereditary. The following theorem answers
the questions regarding inheritance of countability properties by the
product space.
Theorem 5.1.4 A product of topological spaces is first countable
(resp. second countable) if and only if each factor is first countable
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115
(resp. second countable) and all but a countable number of the factors
are indiscrete.
∏
Proof. Let Xα , α ∈ A, be a family of
Xα is first
∏ spaces such that
countable. Since the projections pβ : Xα → Xβ are continuous open
surjections, each Xα is first countable, by Proposition 5.1.3. Let B =
{α ∈ A|Xα is not indiscrete}. For each β ∈ B, we choose a ∏
point cβ ∈
Xβ which has a proper nbd in Xβ and consider a point x ∈ Xα with
xβ = cβ . Let M be a countable nbd basis at x. By the definition of
product topology, for every nbd N of x, pα (N ) = Xα for all but finitely
indices α. So, for each M ∈ M, there exists a finite subset AM ⊂ A
such ∪
that pα (M ) = Xα for all α ∈ A − AM . Since M is countable,
Γ = {AM |M ∈ M} is also countable. We assert that B ⊆ Γ which
implies that B is countable. By the choice of x, for each index β ∈ B,
there exists a proper open subset Uβ ⊂ Xβ such that pβ (x) ∈ Uβ .
Therefore there is a member M in M such that pβ (M ) ̸= Xβ , and
β ∈ Γ. This proves our assertion.
Conversely, let Xα , α ∈ A, be a family of first countable spaces,
and B be a countable subset of A such
∏ that Xα is indiscrete for every
α ∈ A−B. Given a point x = (xα ) ∈ Xα , let Mα be a countable local
base at xα .{Note that Mα = {Xα } for} every α ∈ A − B. Let V denote
the family p−1
β (M )|M ∈ Mβ , β ∈ B . Then V is clearly a countable
family of nbds of x, and so is the family N of finite intersections of
members of V. We claim that N is a local base at x. For a proper nbd
N of x, we have pα (N ) = Xα for all but finitely many indices α1 , . . . αn ,
say. Obviously, the indices α1 , . . . , αn are in B. For each
i = 1, . . . , n,
∩n −1
there exists an open subset Ui ⊆ Xαi such that x ∈ 1 pαi (Ui ) ⊆ N .
Now, we
a set Mi ∈ Mαi such that Mi ⊂ Ui for every = 1, . . . , n.
∩n find
−1
Then 1 pαi (Mi ) ⊆ N , and hence our claim.
∏
Now, we turn to the case of second countability. If Xα is second
countable, then
∏ each Xα is second countable, again by Proposition
5.1.3. Since Xα is first countable, there are at most countably many
nontrivial factors.
If Uα is a countable basis for Xα for every α ∈ A, then, by our
hypothesis, there is a countable set B ⊆ A such that Uα = {Xα }
for
that the family
{ all α ∈ A − B. It
} is not difficult to show ∏
−1
pβ (U )|U ∈ Uβ , β ∈ B is a countable subbasis of Xα , and hence
it is second countable.
♢
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∏
Corollary 5.1.5 If {Xα |α ∈ A} is a family of spaces such that Xα
is metrisable, then each Xα is metrisable, and all but countably many
coordinate spaces are one-point spaces.
∏
Proof. Since Xβ , β ∈ A, is homeomorphic to a subspace of
Xα , it
is
metrisable.
The
second
statement
follows
from
Theorem
5.1.4,
since
∏
Xα is first countable.
♢
It is clear from the preceding corollary that the product of an uncountable family of nontrivial metric spaces is not metrisable (cf. Theorem 2.2.14).
Next, we discuss the results which establish that the essential features of the topology of a first countable space (and, a fortiori, of a
second countable space) can be expressed satisfactorily in terms of sequences. It is worth noticing that there is a countable nbd basis {Un }
at each point of a first countable space X such that Un ⊇ Un+1 for
every n = 1, 2, . . .. In fact, given a countable nbd basis {Bn } at a point
x in X, one needs to define Un = B1 ∩ · · · ∩ Bn .
Theorem 5.1.6 Let X be a first countable space and A ⊆ X. Then
x ∈ A ⇔ there is sequence in A which converges to x.
Proof. Suppose that x ∈ A. Since X satisfies the first axiom of
countability, there is a nbd basis {Un |n = 1, 2, . . .} at x such that
U1 ⊇ U2 ⊇ . . .. For each n, we can select a point an ∈ Un ∩ A for
x ∈ A. It is clear that an → x.
Conversely, if there is a sequence in A converging to x, then every
nbd of x intersects A, and x ∈ A, by Theorem 1.3.5.
♢
Theorem 5.1.7 Let X be a first countable space and Y a space. A
function f : X → Y is continuous ⇔ xn → x in X implies f (xn ) →
f (x).
Proof. Suppose that f : X → Y is continuous, and xn → x in X. For
any nbd U of f (x), f −1 (U ) is a nbd of x. So there is an integer n0 such
that xn ∈ f −1 (U ) for all n ≥ n0 . This implies that f (xn ) ∈ U for all
n ≥ n0 , and f (xn ) → f (x).
Conversely, assume that f (xn ) → f (x) in Y whenever
xn → x in X.
( )
We show that f is continuous by showing that f A ⊆ f (A) for every
A ⊆ X. If x ∈ A, then there is a sequence ⟨an ⟩ in A which converges to
COUNTABILITY AXIOMS
117
x, by Theorem 5.1.6. Our hypothesis implies that f (an ) → f (x), and
hence f (x) ∈ f (A). This completes the proof.
♢
We have already seen that the Hausdorff condition implies that a
limit of a sequence, when it exists, is unique. However, the converse
is not true, in general. For the spaces satisfying the first axiom of
countability we have
Proposition 5.1.8 A first countable space X is Hausdorff if and only
if each convergent sequence in X has a unique limit.
Proof. The necessity is clear, for a sequence cannot be eventually in each
of two disjoint sets. To prove the sufficiency, assume that the Hausdorff
condition fails in X for two points x and y. Since X is first countable,
there are monotonically decreasing countable nbd bases {Un } and {Vn }
at x and y, respectively. By our assumption, Un ∩ Vn ̸= ∅ for every
n = 1, 2, . . .. So we can choose a point zn in Un ∩Vn for every n. Clearly,
zn → x as well as zn → y.
♢
Exercises
1. Prove that a cofinite space X is second countable if and only if X is
countable.
2. Show that each of the following spaces is second countable.
(a) The Hilbert space ℓ2 .
(b) The set R with the Smirnov topology (see Exercise 1.4.6).
3. Show that none of the following spaces is first countable.
(a) An uncountable space with the cocountable topology.
(b) An uncountable Fort space (see Exercise 1.2.3).
4. Is Sorgenfrey line Rℓ metrisable?
5. Show that Rω with the uniform metric topology is first countable, but
not second countable.
6. • Let X be a first countable space. Prove:
(a) A point x ∈ X is a cluster point of a sequence ⟨xn ⟩ in X if and
only if ⟨xn ⟩ has a subsequence which converges to x.
(b) For A ⊆ X, a point x ∈ A′ if and only if there is a sequence in
A − {x} which converges to x.
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(c) A subset F of X is closed if and only if the cluster points of each
sequence in F belong to F .
(d) A subset G of X is open if and only if each sequence which converges to a point of G is eventually in G.
(A space with this property is called sequential.)
(e) A subset G of X is open if and only if G ∩ A is open in A for every
countable subset A of X.
7. If Y is an uncountable subset of a second countable space X, show that
Y contains uncountably many of its limit points. Does the converse hold
good?
8. Prove that every basis of a second countable space contains a countable
subfamily which is also a basis.
9. A subset A of a space X is called Gδ (resp. Fσ ) if it is the intersection
(resp. union) of at most countably many open (resp. closed) sets.
Prove:
(a) A closed interval [a, b] in R is a Gδ -set as well as an Fσ -set.
(b) The set Q of rationals is an Fσ -set in R, while the set of irrationals
is Gδ .
10. Show that every closed subset of metric space X is a Gδ -set. Give an
example of a topological space X and a closed set A ⊂ X which is not
a Gδ -set.
11. Prove:
(a) The complement of an Fσ -set is a Gδ -set, and conversely.
(b) A countable intersection and a finite union of Gδ -sets are Gδ .
(c) A countable union and a finite intersection of Fσ -sets are Fσ .
12. Let X be a first countable T1 -space. Show that every one-point set is a
Gδ -set. What about the converse?
13. If X is a second countable T1 -space, prove that |X| ≤ c.
COUNTABILITY AXIOMS
5.2
119
Separable and Lindelöf Spaces
In this section, we continue our discussion about the spaces satisfying some sort of countability conditions. In fact, we shall study two
conditions, both weaker than the second axiom of countability. In metric spaces, however, each of these conditions turns out to be equivalent
to the axiom of second countability.
SEPARABILITY
Definition 5.2.1 A topological space is called separable if it has a
countable dense subset.
Example 5.2.1 The euclidean space Rn is separable, since the set
{(qi ) ∈ Rn |qi is rational for all i}
is countable and dense in Rn .
Example 5.2.2 The space ℓ2 is separable, since the set of all points (qi )
with qi ’s rational numbers and having only a finite number of nonzero
terms is countable and dense in ℓ2 .
Example 5.2.3 An uncountable set with the cocountable topology is
not separable.
We note that any second countable space is separable, since a
countable dense set can be obtained by taking one point from each
(nonempty) member of a countable basis for the space. Example 5.1.4
shows that the converse is not true. Even a separable, first countable
space need not be second countable; the Sorgenfrey line Rℓ provides a
counter example. But there is a partial converse.
Proposition 5.2.2 A separable metric space is second countable.
Proof. Let X be a separable metric space and A ⊆ X be a countable
dense set. It is easy to see that each open ball B(x; r) in X contains an
open ball B(a; 1/n) containing x, where a ∈ A and n > 2/r. Therefore,
the family {B(a; 1/n)|a ∈ A, n ∈ N} is a countable basis for X.
♢
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Elements of Topology
Corollary 5.2.3 A metric space is second countable if and only if it
is separable.
It is easily seen that the separability is invariant under continuous
mappings. We restate this as
Proposition 5.2.4 If f is a continuous mapping of a separable space
X onto a space Y , then Y is separable.
The property of separability does not behave well with relativisation. The following example shows that even a closed subspace of a
separable space need not be separable.
Example 5.2.4 The Sorgenfrey plane Rℓ × Rℓ is separable, since the
countable set Q × Q is dense in this space. The antidiagonal subspace
L = {(x, −x)|x ∈ R} is closed in Rℓ × Rℓ , for the open set [x, x + r) ×
[y, y + r), where y ̸= −x, and r = −(x + y)/3 if y < −x, does not meet
L. Also, L is discrete, since L ∩ ([x, ∞) × [−x, ∞)) is exactly the point
(x, −x). Being uncountable and discrete, L cannot be separable.
However, an open subspace Y of a separable space X is separable,
for if A is a countable dense subset of X, then A ∩ Y is clearly a
countable dense subset of Y .
Regarding the product topology, we content ourselves with the following simple result, though a stronger statement is true (see Dugundji
[3], p.175).
Proposition 5.2.5 A countable product of separable spaces is separable.
Proof. For every n = 1, 2, . . ., let Xn be a separable space and let An be
a countable dense subset of Xn . Choose a ∏
fixed element cn from each
An . Then the set S of all elements (an ) ∈ An such that an = cn for
all but finitely
many indices n is clearly countable. We assert that
∏
∏ it is
dense in Xn . To see this, consider a basic open subset B of Xn .
∩k
We have B∏= 1 p−1
ni (Uni ), where Uni is a proper open subset of Xni ,
and pni :
Xn → Xni is the projection map for every i = 1, . . . , k.
Since Ani is dense in Xni , we can find an∏element ani ∈ Uni ∩ Ani
for i = 1, . . . , k. Now, the point (xn ) ∈
Xn , where xn = cn for
n ̸= n1 , . . . , nk and xni = a∏
ni for i = 1, . . . , k, belongs to B ∩ S. Thus
every
basic
open
subset
of
Xn intersects S, and hence S is dense in
∏
Xn .
♢
COUNTABILITY AXIOMS
121
LINDELÖF SPACES
The other countability property of spaces involves a preliminary
definition. Let X be
∪ a set. A covering of X is a family G of subsets of
X such that X = {G : G ∈ G}. If this is the case, we also say that
G covers X. A subcovering of G is a subfamily H of G such that H
covers X. If X is a topological space and every member of G is open
(resp. closed), then G is called an open (resp. closed) covering of X.
For a subset Y of X, we sometimes consider a family G of subsets of
X whose union contains Y . In this case, we say that G is a covering of
Y by subsets of X.
Example 5.2.5 Let r > 0 be a real number, and let G be the family of
intervals (x − r, x + r) for every x ∈ R. Then G is an open cover of R.
Similarly, in any metric space X, the family of all open balls B(x; r),
x ∈ X, is an open cover of X.
Example 5.2.6 Let X be a metric space and x0 ∈ X be a fixed point.
The family of open balls B(x0 ; n), n ∈ N, is an open cover of X.
Example 5.2.7 The family of intervals (1/n, 1], n a positive integer, is
an open cover of (0, 1].
The main property of second countable spaces is
Theorem 5.2.6 (Lindelöf ) Let X be a second countable space.
Then each open covering of X has a countable subcovering.
Proof. Let {Bn |n = 1, 2, . . .} be a countable basis for X. Given an open
cover G of X, let M be the set of all integers n such that Bn is contained
in some member of G. For each m ∈ M , choose a member Gm ∈ G such
that Bm ⊆ Gm . Then the family {Gm |m ∈ M } is obviously countable.
We observe that this family covers X. For x ∈ X, there exists G ∈ G
with x ∈ G. Since G is open, there is a set Bn such that x ∈ Bn ⊆
G. Accordingly, n ∈ M and x ∈ Bn ⊆ Gn . Thus {Gm |m ∈ M } is a
countable subcovering of G.
♢
This theorem leads to the following abstraction.
Definition 5.2.7 A space X is Lindelöf if each open covering of X
admits a countable subcovering.
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Elements of Topology
With this terminology, the preceding theorem can be restated as:
A second countable space is Lindelöf. However, the converse of this is
in general not true.
Example 5.2.8 Consider the Sorgenfrey line Rℓ . As seen in Ex. 5.1.3,
it is not second countable. To see that it is a Lindelöf
space, let
∪
◦
{Gα |α ∈ A} be an open covering of Rℓ . Put E = α Gα , where G◦α
denotes the interior of Gα in the euclidean space R. Since R is second
countable, so is E. Because a second countable
space
{
} is Lindelöf, there
◦
is a countable set B ⊆ A such that Gβ |β ∈ B covers E. Next, we
observe that F = R − E is countable. Given x ∈ F , there is an index
α ∈ A such that x ∈ Gα . As Gα is open in Rℓ , there exists a rational
number rx such that [x, rx ) ⊆ Gα . Obviously, (x, rx ) ⊆ G◦α ⊆ E. So
(x, rx ) ∩ F = ∅, and rx ≤ y < ry for every x < y in F . It follows
that the mapping x → rx is an injection from F into Q, and hence
F is countable. Accordingly, there is a countable set Γ ⊆ A such that
{Gγ |γ ∈ Γ} covers F . It is now clear that the family {Gα |α ∈ B ∪ Γ}
is a countable subcovering of {Gα |α ∈ A}, and thus Rℓ is Lindelöf.
In this direction, we have
Proposition 5.2.8 A Lindelöf metric space is second countable.
Proof. Let X be a Lindelöf metric space. Then, for each integer n > 0,
there is a countable set An ⊆ X such that the open balls B(a; 1/n),
a ∈ An , cover X. If G ⊆ X is open and x ∈ G, then there exists an
open ball B(x; r) ⊆ G. For each n = 1, 2, . . . , x ∈ B(ax ; 1/n) for some
point ax ∈ An . Choose an integer n such that 2 < nr. Then we have
x ∈ B(ax ; 1/n) ⊆ G. Therefore the family {B(a; 1/n)|a ∈ An , n ∈ N}
is a countable basis of X, and X is second countable.
♢
Thus, it follows that the concepts of being a Lindelöf space, second
countability, and separability are equivalent in a metric space. It should
be noted that, in general, there is no relation between separability and
the property of being Lindelöf.
Example 5.2.9 A separable space, which is not Lindelöf. It has been
shown in Ex. 5.2.4 that Rℓ × Rℓ is separable and its subset L =
{(x, −x)|x ∈ R} is closed. If Rℓ ×Rℓ were Lindelöf, then the open covering which consists of Rℓ × Rℓ − L and the sets [x, x + r) × [−x, −x + r),
x ∈ R, would have a countable subcovering. But this is impossible,
COUNTABILITY AXIOMS
123
since L is uncountable and each set [x, x + r) × [−x, −x + r) meets L
in exactly one point, viz., (x, −x).
Example 5.2.10 A Lindelöf space, which is not separable. Consider the
closed ordinal space [0, Ω]. Given any open covering G of this space,
choose a member G of G with Ω ∈ G. Then there is an ordinal number β < Ω such that (β, Ω] ⊆ G. The interval [0, β] is countable,
and therefore it can be covered by countably many members of G.
Thus G has a countable subcovering, and [0, Ω] is Lindelöf. To see
that it is not separable, let E be a countable subset of [0, Ω]. Then
α = sup (E − {Ω}) < Ω. Obviously, the open set (α, Ω) does not meet
E, and therefore E cannot be dense in [0, Ω].
By Ex. 5.2.9, the product of two Lindelöf spaces need not be Lindelöf. The property of being a Lindelöf space is not hereditary either.
The subspace [0, Ω) of the Lindelöf space [0, Ω] is not Lindelöf, for the
open covering {[0, α)|α < Ω} does not have a countable subcovering.
However, we have the following.
Proposition 5.2.9 A closed subspace of a Lindelöf space is Lindelöf.
Proof. Let X be a Lindelöf space and Y a closed subspace of X. Let
{Hα }α∈A be any open covering of Y. Then for each index α, there is
an open subset Gα of X such that Hα = Y ∩ Gα . Since Y is closed in
X, X − Y is open. So the family {X − Y } ∪ {Gα |α ∈ A} is an open
covering of X. Since X is Lindelöf, this open covering has a countable
subcovering.
Accordingly, we find a countable
∪
∪subset B of A such that
Y ⊆ {Gβ |β ∈ B}. It follows that Y = {Hβ |β ∈ B}, and thus
{Hβ }β∈B is a countable subcovering of {Hα }α∈A .
♢
The straightforward proof of the following proposition is left to the
reader.
Proposition 5.2.10 The continuous image of a Lindelöf space is Lindelöf.
Thus Lindelöfness is a topological invariant.
Exercises
1. Show that a discrete space is separable if and only if it is countable.
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Elements of Topology
2. Show that a cofinite space is separable.
3. Let X be a metric space in which every infinite subset has a limit point.
Prove that X is separable.
4. Prove that the space B(I) of all bounded functions I → R with the
supremum metric (ref. Ex. 1.1.5) is not separable but the subspace C(I)
is.
5. Prove that the space in Ex. 1.1.4 is separable.
6. Find a countable dense subset of the space R − Q (of all irrationals).
7. Let X be an uncountable set, and let x0 ∈ X be a fixed point. Let T be
the family of sets G ⊆ X such that G = ∅ or x0 ∈ G. Show that (X, T)
is a separable space which has a non-separable subspace.
8. Prove that every subspace of a separable metric space is separable.
9. Let X be a second countable or separable space. Show that every family
of pairwise disjoint open subsets of X is countable. (Thus, a second
countable or separable space satisfies the countable chain condition.)
10. Let X be an uncountable set, and let x0 ∈ X be a fixed point. Let U
be the family of sets G ⊆ X such that G = X or x0 ∈
/ G. Show that
(X, U) is a Lindelöf space which has a subspace that is not Lindelöf.
11. Let X be a Lindelöf space. Show that each uncountable subset of X has
a limit point.
12. (a) Show that the ordered space I×I in Exercise 1.4.15 is first countable
and Lindelöf but not second countable or separable.
(b) Show that the subspace (0, 1) × I of the ordered space I × I is not
Lindelöf.
13. Show that the ordinal space [0, Ω) satisfies the first axiom of countability, but no other countability properties.
14. Show that the ordinal space [0, Ω] does not satisfy any countability
condition except the Lindelöf property. Deduce that it cannot be given
a metric consistent with its topology.
15. If X is a separable T2 -space, show that |X| ≤ 2c . If X is also first
countable, show that |X| ≤ c.
16. Let C(X) be the family of all continuous real-valued functions on a space
X. If X is separable, prove that |C(X)| ≤ c.
Chapter 6
COMPACTNESS
6.1
6.2
6.3
6.4
6.5
Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Countably Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Compact Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proper Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Compact Spaces
125
136
140
148
155
The term “compact” was initially coined to describe the property
of a metric space in which every infinite subset has a limit point. However, this sense of the term failed to give some desirable theorems for
topological spaces, especially its invariance under the product topology.
On the other hand, it was found that, in metric spaces, this property is
equivalent to the Borel–Lebesgue property: Each open cover has a finite
subcover. After Tychonoff proved the invariance of this last property
under the formation of topological products, the following definition of
compactness was universally adopted.
Definition 6.1.1 A space X is said to be compact if every open covering of X has a finite subcover.
Thus, the notion of compactness is a strong form of the Lindelöf
condition. In the literature, the term “bicompact” has also been used
by some mathematicians to describe the property of present-day compactness. We will study here some characterisations and the invariance
properties of compactness.
Example 6.1.1 A cofinite space is compact.
Example 6.1.2 The space in Exercise 1.2.3 is compact.
Example 6.1.3 An infinite discrete space is not compact.
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Example 6.1.4 The euclidean space Rn is not compact, since the open
covering {B(x; 1)|every coordinate of x is an integer} does not have a
finite subcovering.
It is obvious that a compact space is Lindelöf but it need not satisfy
other countability axioms; this can be seen from the preceding examples. However, it is easily seen that a compact metrisable space satisfies
all the countability axioms.
There is a useful formulation of compactness in terms of closed sets.
Theorem 6.1.2 A space X is compact if and
∩ only if for every family
{Fα }α∈A of closed subsets∩of X satisfying α∈A Fα = ∅, there is a
finite set B ⊆ A such that β∈B Fβ = ∅.
The proof follows by De Morgan’s rules.
Definition 6.1.3 A collection C of subsets of a set has the finite intersection property if the intersection of any finite subcollection of C is
nonempty.
With this terminology, Theorem 6.1.2 can be restated as
Theorem 6.1.4 A space X is compact if and only if each family of
closed subsets of X with the finite intersection property has a nonempty
intersection.
There are several other characterisations of compactness which will
be discussed soon. We return to see some more illustrations.
Definition 6.1.5 A subset Y of a space X is said to be compact if
every covering of Y by sets open in X has a finite subcovering.
Recall that a subset U of Y is open in the relative topology if and
only if U = Y ∩ G for some open set G ⊆ X. Therefore Y is a compact
subset of X if and only if the subspace Y (of X) is compact. Note that
if K ⊆ Y ⊆ X, then K is a compact subset of X if and only if K is a
compact subset of the subspace Y . Contrast this with the property of
being open or closed.
Example 6.1.5 If ⟨xn ⟩ is a convergent sequence in a space X with a
limit x, then {x} ∪ {xn |n = 1, 2, . . .} is a compact subset of X.
COMPACTNESS
127
Theorem 6.1.6 (Heine–Borel) A closed interval J = [a, b] in the
real line R is compact.
Proof. Let G be a covering of J by open subsets of R, and consider the
set X = {x ∈ J|[a, x] is covered by a finite subfamily of G}. Obviously,
X is nonempty (a ∈ X) and bounded above by b. So c = sup X exists.
As a < c ≤ b, there exists a set G in G such that c ∈ G. Since G is
open in R, there is a real number r > 0 such that (c − r, c + r) ⊆ G. By
the definition of c, there exists an element x ∈ X such that c − r < x.
Now, by the definition of X, we find a finite subfamily {G1 , . . . , Gn }
of G which covers [a, x]. Then {G, G1 , . . . , Gn } obviously covers [a, c],
and therefore c ∈ X. Further, if c were less than b, there would be a
number d ∈ J such that (c, d] ⊂ (c, c + r) ⊆ G. This forces d in X,
contradicting the choice of c. Therefore b = c, and J is covered by a
finite subfamily of G. Thus J is compact.
♢
The proof of the preceding theorem shows that every closed interval
in an ordered space having the least upper bound property is compact.
Example 6.1.6 The interval (0, 1] is not compact, for the open cover
(1/n, 1], n a positive integer, has no finite subcover.
As shown by the preceding example, a subspace of a compact space
need not be compact. However, we have
Theorem 6.1.7 A closed subset of a compact space is compact.
Proof. Let X be a compact space, and Y ⊆ X be closed. Let {Fα |α ∈ A}
be any family of closed subsets of Y with the finite intersection property. Since Y is closed in
∩ X, the Fα are closed in X, too. By the direct
part of Theorem 6.1.4, Fα ̸= ∅, and its converse part implies that Y
is compact.
♢
The converse holds in Hausdorff spaces. To see this, we first prove
Proposition 6.1.8 If A is a compact subset of a Hausdorff space X
and x ∈ X − A, then there are disjoint open sets U and V such that
x ∈ U and A ⊂ V .
Proof. Let X be a Hausdorff and A ⊆ X be compact. Suppose that
x ∈ X − A. Then, for each a ∈ A, there exist open sets Ua and Va such
that x ∈ Ua , a ∈ Va and Ua ∩Va = ∅. Obviously, the family {Va |a ∈ A}
covers A. Since A is compact, there are finitely many points a1 , . . . , an
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∪n
∩n
in A such that A ⊆ 1 Vai = V . If U = 1 Uai , then U does not
intersect V , since Uai ∩ Vai = ∅. Thus U and V are disjoint nbds of x
and A, respectively. This completes the proof.
♢
Theorem 6.1.9 Let X be a Hausdorff space.
(a) A compact subset of X is closed.
(b) Any two disjoint compact subsets of X have disjoint nbds.
Proof. (a): Let A be a compact subset of X. We show that X − A is
open. If x ∈ X − A, then, by Proposition 6.1.8, there exist open sets U
and V in X such that x ∈ U , A ⊂ V and U ∩ V = ∅. So we have the
inclusions x ∈ U ⊆ X − V ⊆ X − A which imply that x is an interior
point of X − A. Since x ∈ X − A is arbitrary, (a) follows.
(b): Suppose A and B are disjoint compact subsets of X. By Proposition 6.1.8, for each b ∈ B, there are open sets Ub and Vb such that
b ∈ Ub , A ⊂ Vb and Ub ∩Vb = ∅. The family {Ub |b ∈ B} covers the compact set
∪n B. So there are finitely many
∩n points b1 , . . . , bn in B such that
B ⊆ 1 Ubi = U . The set V = 1 Vbi is open and contains A. Since
Ubi ∩ Vbi = ∅ for every index i, we have U ∩ V = ∅. This completes
the proof.
♢
We remark that the Hausdorff condition in the preceding theorem
cannot be relaxed, for an infinite proper subset of a cofinite space is
compact but not closed. From Theorems 6.1.7 and 6.1.9, it follows that
a subset of a compact Hausdorff space is compact if and only if it is
closed.
Compactness is preserved by continuous maps, as shown by the
following theorem; thus it is a topological property.
Theorem 6.1.10 Let f : X → Y be a continuous map. If X is compact, then so is f (X).
Proof. Let U be a covering of f (X) by open subsets of Y . Then
the sets f −1 (U ), U ∈ U, form an open covering of X. Since X
is
compact, this open
{ −1
} covering of X has a finite subcovering, say,
f (Ui )|i = 1, . . . , n . Then U1 , . . . , Un cover f (X), and the theorem
follows.
♢
Corollary 6.1.11 Every continuous map of a compact space into a
Hausdorff space is closed.
COMPACTNESS
Proof. This follows from Theorems 6.1.7, 6.1.9 and 6.1.10.
129
♢
As an immediate consequence of the preceding theorem, we have
Corollary 6.1.12 A one-to-one continuous map of a compact space
onto (resp. into) a Hausdorff space is a homeomorphism (resp. embedding).
The next theorem is crucial for the adoption of Tychonoff topology
on product spaces, and shows the invariance of compactness under the
formation of products.
Theorem 6.1.13 (Tychonoff ) Let ∏
Xλ , λ ∈ Λ, be a family of compact spaces. Then the product space Xλ is compact.
We postpone the proof of this theorem for a while, and discuss its
finite version, which can be proved rather easily. The proof in this case
depends upon the following simple lemma which is useful in proving
several other propositions as well.
6.1.14 (Tube Lemma) Let Y be compact and X be any space. Let
N be an open nbd of the slice {x} × Y in X × Y . Then there is an open
nbd U of x such that U × Y ⊆ N . (The set U × Y is referred to as a
tube about {x} × Y .)
Proof. For each y ∈ Y, there exist open sets Uy ⊆ X and Vy ⊆ Y such
that (x, y) ∈ Uy × Vy ⊆ N . Since Y is compact, the open covering
{Vy |y ∈ Y } has a finite subcover. Accordingly,
there are finitely
∪n
∩n many
points y1 , . . . , yn in Y such that Y = 1 Vyi . Then U = 1 Uyi is a
nbd of x and U × Y ⊆ N . (See Figure 6.1 below).
♢
Theorem 6.1.15 If X and Y are compact spaces, then so is X × Y .
Proof. Let G be an open cover of X × Y . For each x ∈ X, the slice
{x}×Y is homeomorphic to Y , and hence compact. So there are finitely
n(x)
in G, depending on the point x, which cover
many sets G1x , . . . , Gx
{x} × Y . By Lemma 6.1.14, there exists an open nbd Ux of x such that
∪n(x)
Ux ×Y ⊆ i=1 Gix . Since X is compact, the open covering {Ux |x ∈ X}
has a finite subcover.
∪m So there are finitely many points
∪m ( x1 , . . . ,)xm in
X such that X = 1 Uxj ;{accordingly, X × Y = 1 Uxj × }
Y . It is
now clear that the family Gixj |1 ≤ i ≤ n(xj ) and 1 ≤ j ≤ m
X × Y, and the theorem follows.
covers
♢
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Elements of Topology
x×Y
Y
Uy × V y
N
•
g
X
U
FIGURE 6.1: Proof of the Tube Lemma.
As another application of the Tube Lemma, we obtain an interesting
characterisation of compact spaces.
Theorem 6.1.16 A space X is compact if and only if the projection
p : X × Y → Y, (x, y) 7→ y, is closed for all spaces Y .
Proof. Suppose first that X is compact, and let F ⊆ X × Y be closed.
Let y ∈ Y − p(F ). Then (X × {y}) ∩ F = ∅ so that (X × Y ) − F is nbd
of X × {y}. By Lemma 6.1.14, there exists an open nbd U of y such
that X × U ⊆ (X × Y ) − F . Consequently, U ∩ p(F ) = ∅ which implies
that U ⊆ Y − p(F ). Thus Y − p(F ) is a nbd of y. Since y ∈ Y − p(F ) is
arbitrary, it follows that Y − p(F ) is open, and therefore p(F ) is closed.
Conversely, assume that the mapping p : X × Y → Y, p(x, y) = y,
is closed for every space Y . Let U be an open cover of X. Choose an
element ∞ ∈
/ X, and let Y = X ∪ {∞}. Give Y the topology generated
by the subbase which consists of one-point sets {x} , x ∈ X, and sets
Y −U, U ∈ U. Consider the subset D = {(x, x)|x ∈ X} of X ×Y . By our
assumption, the projection p : X × Y → Y is closed; so pD = pD = X.
For each x ∈ X, there exists U ∈ U with x ∈ U . Then U × (Y − U ) is a
nbd of (x, ∞) and [U × (Y − U )] ∩ D = ∅. So (x, ∞) is not an adherent
point of D, and D = D. It follows that X = X and the singleton set
{∞} is open in Y∩. So there exist finitely many sets U∪
1 , . . . , Un in U
n
n
such that {∞} = 1 (Y − Ui ), which implies that X = 1 Ui . Thus U
has a finite subcovering, and X is compact.
♢
COMPACTNESS
131
We turn now to prove the Tychonoff theorem in full generality.
Unfortunately, there is no way to generalise the proof of the theorem
in the finite case to arbitrary products. However, our task is made
easier considerably by the following characterisations of compactness.
Theorem 6.1.17 In any space X, the following statements are equivalent:
(a) X is compact.
(b) Every universal net in X converges.
(c) Every net in X has a convergent subnet.
Proof. (a)⇒(b): Let ϕ : A → X be a universal net, and assume that
ϕ fails to converge in X. Then, for each x ∈ X, there exists an open
nbd Ux of x such that ϕ is eventually in X − Ux . Accordingly, there is
an element αx ∈ A such that ϕ(α) ∈ X − Ux for all α ≽ αx . Since X
is compact,
there exist finitely many points x1 , . . . , xn in X such that
∪n
X = i=1 Uxi . As A is directed, ∩
there is an α ∈ A such∪that α ≽ αxi
n
n
for every i = 1, . . . , n. So ϕ(α) ∈ i=1 (X − Uxi ) = X − i=1 Uxi = ∅,
a contradiction.
(b)⇒(c): This follows from Theorem 4.2.12.
(c)⇒(a): Let F be a family of closed subsets of X with the finite intersection property. Consider the family E which consists of all
members of F and their finite intersections. Direct E by the reverse
inclusion, that is, E ≼ E ′ ⇔ E ⊇ E ′ . Notice that every member of E is
nonempty, so we can choose a point xE ∈ E for every {E in E. By our
}
hypothesis, the net ⟨xE ⟩ has a convergent subnet, say, xf (α) , α ∈ D .
Then D is a directed set and f : D → E is a function with the property that, for each E ∈ E, there exists a β ∈⟨D such
⟩ that f (γ) ⊆ E
for every γ ≽ β. It follows that the subnet xf (α) is eventually in
E. Suppose
∩ that xf (α)∩→ x in X. Since E is closed, we have x ∈ E.
So x ∈ {E ∈ E} ⊆ {F ∈ F}, for F ⊆ E. By Theorem 6.1.4, X is
compact, and this completes the proof.
♢
∏
Proof∏of Theorem 6.1.13: Let ϕ : A →
Xλ be
net.
∏ a universal
If pµ :
Xλ → Xµ is the projection map, then
Xλ − p−1
(S)
=
µ
−1
−1
pµ (Xµ − S) for every S ⊆ Xµ . So ϕ is eventually in either pµ (S) or
p−1
µ (Xµ − S); accordingly, pµ ◦ ϕ is eventually in either S or Xµ − S.
Thus for each µ ∈ Λ, the composition pµ ◦ ϕ is a universal net in Xµ .
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Elements of Topology
By
∏ Theorem 6.1.17, pµ ◦ ϕ → xµ in Xµ , say. Then ϕ → (xλ ), and hence
Xλ is compact.
♢
The converse of the preceding theorem is also true; this is immediate
from Theorem 6.1.10.
The Tychonoff theorem is unquestionably the most useful result
in point-set topology; we will see some applications of this theorem
as we proceed in the book. We remark that most of the proofs of
the Tychonoff theorem are difficult; here the entire difficulty has been
subsumed in the result about universal nets. In fact, it is known that
the Tychonoff theorem is equivalent to the axiom of choice. Using an
equivalent axiom, viz. the Hausdorff Maximal Principle, we prove the
following.
Theorem 6.1.18 The components of a compact Hausdorff space coincide with its quasi-components.
Proof. Let X be a compact Hausdorff space, and x ∈ X. Let C(x)
be the component, and K(x) be the quasi-component of x ∈ X. We
have already seen that C(x) ⊆ K(x). So we need to prove the reverse
inclusion. Given y ∈ K(x), it suffices to produce a connected subset of
X which contains x and y, both. Let {Fν |ν ∈ N } be the family of all
closed subsets of X such that x, y ∈ Fν , and Fν is not the union of two
separated sets, one containing x and the other containing y. Then K(x)
is a member of {Fν |ν ∈ N } and this family is nonempty. We partially
order {Fν |ν ∈ N } by the reverse inclusion. By the maximal principle,
{Fν |ν ∈ N∩
} contains a maximal simply ordered subfamily {Fµ |µ ∈ M }.
Put F0 = {Fµ |µ ∈ M }. Then both x and y are in F0 , obviously. We
claim that F0 is connected. To prove this, we first show that F0 = Fν for
some ν ∈ N . Notice that F0 is closed in X. If F0 ̸= Fν for every ν ∈ N ,
then there exists a separation F0 = A ∪ B of F0 such that x ∈ A and
y ∈ B. Both A and B are compact, being closed subsets of X. Hence
there exist disjoint open sets U and V in X such that A ⊂ U , and
B ⊂ V . If Fµ ⊂ U ∪ V , then Fµ = (Fµ ∩ U ) ∪ (Fµ ∩ V ) is a separation
of Fµ with x ∈ Fµ ∩ U and y ∈ Fµ ∩ V , contrary to our assumption. So
each Eµ = Fµ −(U ∪V ) is nonempty. Clearly, the family {Eµ |µ ∈ M } is
simply ordered,∩
and therefore has the finite intersection property. Since
X is compact, {Eµ |µ ∈ M } ̸= ∅; consequently, F0 − (U ∪ V ) ̸= ∅.
This contradicts the fact that F0 ⊆ U ∪ V . Therefore F0 = Fν for some
ν ∈ N . Now, if F0 = A ∪ B is a separation, then either A or B must
contain both x and y. To be specific, suppose that x, y ∈ A. Then
COMPACTNESS
133
A = Fν ′ for some ν ′ ∈ N because, otherwise, we have a separation
of F0 by two disjoint closed subsets, one containing x and the other
containing y. It is obvious that Fν ′ is properly contained in Fµ for
every µ ∈ M . Thus {Fν ′ } ∪ {Fµ |µ ∈ M } is a chain in {Fν |ν ∈ N }
which contains the chain {Fµ |µ ∈ M } properly. This contradicts the
maximality of {Fµ |µ ∈ M }, and hence our claim. This completes the
proof.
♢
By the preceding theorem and Proposition 3.2.6, it follows that the
component of a compact Hausdorff space X containing a point x is the
intersection of all clopen nbds of x in X.
Exercises
1. Show that the interval [0,1] is not compact in either of the topologies
induced by the cocountable topology, the lower limit topology and the
topology described in Exercise 1.4.6 for R.
2. Prove that a space X is compact ⇔ every open covering consisting of
members of a base for X has a finite subcovering. (The result also holds
if the condition is satisfied by subbasic open coverings.)
3. Prove that a space X is compact if and only if each open covering of X
has a nbd-finite subcovering.
4.
(a) Prove that a finite union of compact subsets of a space is compact.
(b) Give an example of two compact subsets K1 and K2 of a space
such that K1 ∩ K2 is not compact.
5. Give an example of a compact set whose closure is not compact.
6. Give an example of a T1 -space which contains a compact set that is not
closed.
7. If every subset of a T2 -space X is compact, show that X is discrete.
8.
(a) If X is order-complete, prove that a closed subset of X with both
lower and upper bounds is compact.
(b) Prove that I × I with the television topology is compact.
9. Show that the ordinal space [0, Ω] is compact.
10. Let X be an ordered space. If every closed interval of X is compact,
show that X has the least upper bound property.
11. Let T be a compact Hausdorff topology on a set X. Show that a topology
strictly larger than T is not compact, and the one strictly smaller than
T is not Hausdorff.
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Elements of Topology
12. Let X be a compact Hausdorff space, and f : X → Y be a continuous
surjection. Show that Y is Hausdorff ⇔ f is closed.
13. If a compact Hausdorff space X has no isolated points, prove that X is
uncountable. Deduce that the Cantor set is uncountable.
14. Let X and Y be Hausdorff spaces and f : X → Y a continuous function.
If {F
∩ n } is a∩decreasing sequence of compact subsets of X, show that
f ( Fn ) = f (Fn ).
15. Let X be a compact Hausdorff space and f : X → X a continuous
function. Prove that there exists a nonempty closed set A ⊆ X such
that A = f (A).
16. Give an example of a continuous injection of a compact space into a
Hausdorff space that is not open.
17. Prove (without using universal nets) that every net in a compact space
has a cluster point (and hence a convergent subnet).
18. Prove: A space X is compact ⇔ each filter on X accumulates at some
point x ∈ X ⇔ each ultrafilter on X is convergent.
19. Let F be a filter on a compact space X. Prove:
(a) If E is the set of accumulation points of F, then each nbd of E is
a member of F.
(b) If there is only one accumulation point of F in X, then F is convergent.
20. Let X be a compact space, and {fn } be a sequence of continuous realvalued functions on X such that fn (x) ≤ fn+1 (x) (or fn (x) ≥ fn+1 (x))
for all n and for all x ∈ X. If fn (x) → f (x) for each x ∈ X and f is
continuous, show that fn → f uniformly.
21. • Let {Kα } be a family of compact subsets of a Hausdorff space X.
Prove:
(a) If the∩intersection of every finite subfamily of {Kα } is nonempty,
then Kα is nonempty and compact.
∩
(b) Every open nbd U of Kα contains a finite intersection of members of {Kα }.
22. Prove that a compact Hausdorff space X is locally connected ⇔ for each
open covering {Uα } of X, there exists
∪ a finite family {Ci } of connected
open subsets of X such that X = Ci and each Ci is contained in some
Uα .
23. Prove that a compact locally connected space has a finite number of
components.
COMPACTNESS
135
24. Let X be a compact Hausdorff space.
(a) Suppose that C is a nonempty collection of closed connected
sub∩
sets that is simply ordered by inclusion. Prove that {C : C ∈ C}
is connected.
(b) Let B be a filter base in X consisting of closed connected subsets
of X. Show that the intersection of members of B is a closed
connected set.
25. A compact connected set in a space is called a continuum. Show that a
Hausdorff continuum having more than one point is uncountable.
26.
(a) Prove that the intersection of a (nonempty) family of continua in
a Hausdorff space X directed by the reverse inclusion is a continuum.
(b) Let {Fα } be a family of continua in a T2 -space ∩
X such that finite
intersections of Fα ’s are connected. Show that Fα is connected.
27. Let X be a continuum, and A ⊂ X be closed. Show that there exists
a closed connected subset B ⊃ A such that no proper closed connected
subset of B contains A.
28. Let X be a compact Hausdorff space, C a component of X. Show that
for each open set U ⊂ X with C ⊂ U , there exists an open set V such
that ∂V = ∅ and C ⊂ V ⊂ U .
29. Let X be a Hausdorff continuum. If C is a component of an open subset
U ⊆ X, show that C has a limit point in U − U .
30. Let X be a compact Hausdorff space. Consider the relation ∼ on X
defined by x ∼ y if for every continuous function f : X → R with
f (x) = 0, f (y) = 1, there is a z ∈ X such that f (z) = 1/2. Show that ∼
is an equivalence relation and its equivalence classes are the components
of X.
31. Prove: A compact Hausdorff space X is totally disconnected if and only
if for every pair x, y of distinct points, there exists a clopen set A ⊂ X
such that x ∈ A and y ∈
/ A.
32. Let f be a function from a space X into a compact Hausdorff space Y .
Show that f is continuous if and only if its graph Gf is closed in X × Y .
33. If X is compact and Y is Lindelöf, show that X × Y is Lindelöf.
34. • Let X and Y be spaces, and A ⊆ X, B ⊆ Y be compact subsets. If
W is a nbd of A × B in X × Y, prove that there exist nbds U of A and
V of B such that U × V ⊆ W .
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Elements of Topology
6.2
Countably Compact Spaces
For a long time, a space X was said to be compact if it had the
Bolzano–Weierstrass property (briefly, B-W property): Every infinite subset of X has a limit point. Some authors call this property
“countable compactness”; others term it the “limit point compactness.”
In this section, we study this and other related notions. The following
theorem shows that this property is a generalisation of the compactness.
Theorem 6.2.1 A compact space has the B-W property.
Proof. Suppose that X is a compact space and A ⊆ X is infinite. If A
has no limit point, then each point x ∈ X has an open nbd Gx such
that A ∩ Gx − {x} = ∅. The family {Gx } forms an open covering of X.
Since X is compact,
∪n xi ∈ X, 1 ≤ i ≤ n,
∪n there exist finitely many points
such that X = 1 Gxi . This implies that A = 1 (A ∩ Gxi ) is finite, a
contradiction.
♢
The converse of the theorem is not true, as shown by
Example 6.2.1 On the set X = R − Z, consider the topology generated
by the open intervals (n − 1, n). If a subset A of X contains a point
x ∈ (n − 1, n), then every point of (n − 1, n) except possibly x is a limit
point of A. But it is not compact, since the open covering consisting
of the basis elements is not reducible to a finite subcovering.
Clearly, the B-W property is a topological invariant but, unlike
compactness, it is not invariant under continuous mappings (ref. Ex.
6.2.2 below). However, the property is preserved by continuous maps
with domains also satisfying the T1 -axiom. To see this, suppose that
a T1 -space X has the B-W property, and f : X → Y is a continuous
surjection. Given an infinite set B ⊆ Y , choose a point x ∈ f −1 (y) for
every y ∈ B. The set A of all such points x is an infinite subset of X,
and therefore has a limit point p, say. If G is an open nbd of f (p) in
Y , then f −1 (G) is an open nbd of p. Since X is T1 , f −1 (G) contains
infinitely many points of A; consequently, G also contains infinitely
many points of B. So f (p) is a limit point of B.
Example 6.2.2 Let Z have the discrete topology, and X be the space of
COMPACTNESS
137
Ex 6.2.1. The function f : X → Z, given by f (x) = n, for n−1 < x < n,
is continuous and surjective. But Z does not have the B-W property.
There is a compactness condition which agrees with the B-W property on a very wide class of topological spaces.
Definition 6.2.2 A space X is called countably compact if every
countable open covering of X has a finite subcovering.
Theorem 6.2.3 A T1 -space X is countably compact if and only if it
has the Bolzano–Weierstrass property.
Proof. Assume that some infinite subset A ⊆ X has no limit point.
Then we find a countably infinite set B ⊆ A. Since no point of X is a
limit point of B, for each x ∈ X, there exists an open nbd Ux of x such
that Ux ∩ B = ∅ for x ∈
/ B, and Ux ∩ B = {x} for x ∈ B. The family
{X − B} ∪ {Ux : x ∈ B} is a countable open covering of X which has
no finite subcover.
Conversely, if X is not countably compact, then there exists a
countable open covering {Un |n = 1, 2, . . .} of X which contains no finite subcovering.
Now, for each n, we choose a point xn ∈ X such that
∪n
xn ∈
/ ∪ 1 Ui and xn ̸= xi for every 1 ≤ i ≤ n − 1. This is possible, since
n
X − 1 Ui is infinite for every n. Since each x ∈ X belongs to some
Un and xi ∈
/ Un for all i > n, the infinite set {xn |n = 1, 2, . . .} has no
limit point in X.
♢
Obviously, every compact space is countably compact. It is also
immediate that countable compactness is equivalent to compactness in
Lindelöf spaces. For, given an open covering of a Lindelöf space, we
find a countable subcovering, and then countable compactness gives a
finite subcovering. The invariance properties of countable compactness
are given in the exercises. Countable compactness is characterised in
terms of convergence of subnets of sequences.
Theorem 6.2.4 A space X is countably compact ⇐⇒ each sequence
in X has a cluster point (equivalently, every sequence has a convergent
subnet).
Proof.
in X and put Tn = {xn , xn+1 , . . .}.
}
∩ ⇒: Let ⟨xn ⟩{be a sequence
If Tn = ∅, then X − Tn is a countable open covering of X. Since
X is countably compact, this open cover has a finite subcovering.
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Elements of Topology
Consequently,
there exist finitely many integers n1 , . . . , nk such that
∩k
T
=
∅.
Obviously,
this is not true and, therefore, we have a point
ni
1 ∩
x ∈ Tn . Then, given a nbd U of x in X and an integer n > 0, we
have U ∩ Tn ̸= ∅. So there is an m ≥ n such that xm ∈ U . It follows
that x is a cluster point of the sequence ⟨xn ⟩.
⇐: If there is a countable open covering {Un } of X, which contains
no finite subcovering, then, as in the proof of Theorem 6.2.3, we obtain
a sequence ⟨xn ⟩ in X such that xn ∈
/ Ui for every 1 ≤ i ≤ n. As each
x ∈ X belongs to some Un and xi ∈
/ Un for all i ≥ n, x is not a cluster
point of ⟨xn ⟩.
♢
There is another type of compactness defined in terms of convergence of sequences.
Definition 6.2.5 A space X is called sequentially compact if every
sequence in X contains a convergent subsequence.
It is clear that if X is a sequentially compact space, then every sequence in X has at least one cluster point, and therefore X is countably
compact. However, there are compact spaces which are not sequentially
compact. Because a limit point of a subset of a first countable space is
the limit of a sequence of points of the set, the converse statement can
be expected to hold in such spaces.
Theorem 6.2.6 A countably compact space satisfying the first countability axiom is sequentially compact.
Proof. Let ⟨xn ⟩ be a sequence in X. Since X is countably compact, the
sequence ⟨xn ⟩ has a cluster point x ∈ X, by Theorem 6.2.4. We show
that ⟨xn ⟩ has a subsequence which converges to x. By the first axiom
of countability, we have a monotonically decreasing countable nbd base
{Bi } at x. Now, given an integer k > 0, each Bi contains a term xn
with n ≥ k. Accordingly, we obtain a subsequence ⟨xni ⟩ of < xn >
♢
such that xni ∈ Bi . Clearly, xni → x.
It follows that the conditions of compactness and sequential compactness are equivalent to the B-W property for second countable T1 spaces. This holds good for metrisable spaces without the condition of
second countability, as we shall see in the next section. We remark that
sequential compactness is neither stronger nor weaker than compactness (ref. Exercise 14). The invariance properties of sequential compactness are discussed in the exercises.
COMPACTNESS
139
Exercises
1. Find a space in which every uncountable subset has a limit point but
no countable subset has a limit point.
2. Prove that the following are equivalent:
(a) A space X is countably compact.
(b) Each countable family of closed sets in X having the finite intersection property has a nonempty intersection.
(c) Every decreasing sequence C1 ⊃ C2 ⊃ · · · of nonempty closed sets
in X has a nonempty intersection.
3. Let X be a T1 -space. Prove that a space X is countably compact ⇔
every infinite open cover of X has a proper subcover.
4. Is the subspace [0, 1] of the Sorgenfrey line Rℓ countably compact?
5.
(a) Prove that a closed subset of a countably compact space is countably compact.
(b) Prove that the subspace (0, 1) of the real line R is not countably
compact. (Thus an open subset of a countably compact space need
not be countably compact.)
6. Let X be a T2 -space. Show that X is countably compact ⇔ each discrete
closed subset of X is finite.
7. If A is a countably compact subset of a first countable T2 -space, show
that A is closed.
8. Let X be a countably compact space, and Y a first countable T2 -space.
Show that a continuous bijection f : X → Y is a homeomorphism.
9. Show that the continuous image of a countably compact space is countably compact.
10. Let X be compact, and Y be a countably compact space. Show that
X × Y is countably compact.
11. Prove that the sequential compactness is preserved by continuous functions.
12. Prove that a closed subset of a sequentially compact space is sequentially
compact.
13. Prove that a countable product of sequentially compact spaces is sequentially compact.
14. • Show: (a) An uncountable product of copies of the unit interval I is
not sequentially compact.
(b) The ordinal space [0, Ω) is sequentially compact but not compact.
140
15.
Elements of Topology
(a) Let X be a Lindelöf space. If every sequence in X has a cluster
point, show that X is compact.
(b) Give an example of a non-compact space X such that every sequence in X has a cluster point.
16. Let X be a countably compact T2 -space satisfying the first axiom of
countability. Show that a sequence in X converges ⇔ it has a single
cluster point.
17. Let X be a first countable space, Y a countably compact space. If f :
X → Y is a function such that the mapping X → X × Y, x 7→ (x, f (x)),
is closed, show that f is continuous.
6.3
Compact Metric Spaces
The word “compact” has been used to describe many (related)
properties of topological spaces before its current definition was accepted as the most satisfactory, in particular, the ones discussed in
the preceding section. We shall see here that these notions coincide in
metric spaces, which were then the most widely studied objects.
If (X, d) is a compact metric space, then X is bounded. For, the
covering of X by the open balls B(x0 ; n), where x0 ∈ X is fixed point
and n ranges over N, has a finite subcover so that X = B(x0 ; n) for
some n. In fact, a compact metric space satisfies a condition stronger
than boundedness.
Definition 6.3.1 A metric space X is called totally bounded or precompact if for each ϵ > 0, there is a finite set F ⊆ X such that the
family {B(x; ϵ)|x ∈ F } covers X. The subset F is called an ϵ-net for
X.
If a metric space X is totally bounded, then it is also bounded.
For, if F is an ϵ-net for X, then diam(X) ≤ diam(F ) + 2ϵ. However, a
bounded metric space is not necessarily totally bounded; for example,
the set R of reals with the bounded metric d(x, y) = min {1, |x − y|} is
not totally bounded.
It is clear that a compact metric space is totally bounded. This
COMPACTNESS
141
is true even for countably compact spaces, as shown by the following
proposition.
Proposition 6.3.2 Every countably compact metric space is totally
bounded.
Proof. Let (X, d) be a countably compact metric space. If X is not
totally bounded, then there is a real number ϵ > 0 for which X has
no ϵ-net. Choose a point x1 ∈ X. By our assumption, X ̸= B(x1 ; ϵ).
So we can choose a point x2 outside B(x1 ; ϵ). Then d(x∪
1 , x2 ) ≥ ϵ.
2
Since {x1 , x2 } is not an ϵ-net for X, we find x3 outside 1 B(xi ; ϵ).
Then d(xi , x3 ) ≥ ϵ, for i=1,2. Proceeding inductively, suppose that we
have chosen points x1 , . . . , xn in X such that d(xi , xj ) ≥ ϵ for i ̸= j.
Again,
∪nsince {x1 , . . . , xn } is not an ϵ-net for X, we can choose xn+1 ∈
X − i=1 B(xi ; ϵ). Then d(xi , xn+1 ) ≥ ϵ for i = 1, . . . , n. By induction,
we have a sequence of points xn ∈ X such that d(xi , xj ) ≥ ϵ for i ̸= j.
Since X is countably compact, the infinite set E = {xn |n = 1, 2, . . .}
must have a limit point x ∈ X. Then the open ball B(x; ϵ/2) contains
an infinite number of points of E, for X is T1 . On the other hand,
B(x; ϵ/2) does not contain more than one point of E, since d(xi , xj ) ≥ ϵ
for i ̸= j. Therefore X must be totally bounded.
♢
Furthermore, we have
Proposition 6.3.3 A totally bounded metric space is separable.
Proof. Let X be a totally bounded metric
for each
}
{ kspace. Then,
k
,
.
.
.
,
x
for
X. Let
positive
integer
k,
there
exists
an
1/k-net
x
nk
1
}
∪ { k
k
E = k x1 , . . . , xnk . Then E is a countable subset of X. We observe
that X = E. Let x ∈ X, and ϵ > 0 be arbitrary.
a positive
)
( k Choose
integer k > 1/ϵ. Then, for some j ≤ nk , x ∈ B xj ; 1/k . This implies
that xkj ∈ B(x; ϵ) ∩ E, as desired.
♢
By the above two propositions, we see that a countably compact
metric space is separable. Since a separable metric space is second
countable, a countably compact metric space is sequentially compact
(Theorem 6.2.6). The converse also holds good, by Theorem 6.2.4. Thus
we have established the following.
Theorem 6.3.4 In a metric space X, the following conditions are
equivalent:
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Elements of Topology
(a) X is compact;
(b) X is countably compact; and
(c) X is sequentially compact.
We now come to see compact subsets of metric spaces. If A is
a compact subset of a metric space X, then it is closed, by Theorem 6.1.9. Also, the open covering {B(a; 1)|a ∈ A} reduces to a finite subcovering {B(ai ; 1)|1 ≤ i ≤ n}. So diam(A) ≤ M + 2, where
M = max{d(ai , aj )|1 ≤ i < j ≤ n}. It follows that every compact
subset of a metric space is closed and bounded. However, the converse
is not true, in general. For, the unit sphere ∑
S in the Hilbert space ℓ2 ,
which consists of all points x ∈ ℓ2 such that
x2n = 1, is clearly closed
and bounded. But it is not sequentially
⟨
⟩ compact. This is immediate
from the fact that the sequence x(n) in S, where x(n) is the element
of ℓ2 with its nth coordinate 1 and all other coordinates 0, has no
convergent subsequence.
In this direction, we have the following.
Theorem 6.3.5 (Generalized Heine–Borel theorem) A subset A
of Rn is compact if and only if it is closed and bounded.
Proof. If A is compact, then we have already seen that it is closed
and bounded. Conversely, suppose that A is closed and bounded. Let
pi : Rn → R1 be the projection map onto the ith factor. Then pi (A) is a
bounded subset of R1 . So, for each i, we can find a closed interval [ai , bi ]
such that pi (A) ⊆ [ai , bi ]; consequently, A ⊆ [a1 , b1 ] × · · · × [an , bn ] =
X ⊆ Rn . By Theorems 6.1.6 and 6.1.13, X is compact. Since A is closed
in Rn , A is closed in X, and hence compact.
♢
It follows from the preceding theorem that all the cubes, the cylinders and the discs in the euclidean spaces are compact sets. We will
soon see that these sets and many other familiar objects in a euclidean
space are homeomorphic to each other.
Lemma 6.3.6 Let A ⊆ Rn be a compact, convex set. If the interior
A◦ of A contains the origin 0, then each ray from 0 intersects ∂A (the
boundary of A) in exactly one point.
Proof. Suppose that R is a ray from the origin, and let u ∈ Rn be a
unit vector in the direction of R. Then R = {tu|t ≥ 0 real}. Since A is
bounded, the set T = {t > 0|tu ∈
/ A} is nonempty and bounded below.
COMPACTNESS
143
Put t0 = inf T . We observe that t0 u ∈ ∂A. By the definition of t0 ,
tu ∈ A for every 0 ≤ t < t0 . Since A is closed, we have t0 u ∈ A. The
convexity of A forces that tu ∈
/ A for all t > t0 ; consequently, t0 u ∈
/ A◦ .
◦
Thus t0 u ∈ A − A = ∂A. To establish the uniqueness of t0 u, suppose
that there is t1 > t0 with t1 u ∈ ∂A. Let B(0; r) be a ball centered at
0 and contained in A. Let V be the geometric cone over B(0; r) with
the vertex t1 u (that is, V is the union of all line segments joining t1 u
to the points of B(0; r)) (Figure 6.2). Then V ⊆ A, since A is convex.
If r1 = r(t1 − t0 )/t1 , then B(t0 u; r1 ) ⊆ V ⊆ A so that t0 u ∈ A◦ , a
contradiction. Similarly, there is no t1 < t0 with t1 u ∈ ∂A, and the
proof is complete.
t1u
t0u
•
A
•
0
FIGURE 6.2: Proof of Lemma 6.3.6.
♢
Theorem 6.3.7 Let K be a compact, convex subset of Rn with K ◦ ̸=
∅. Then there is a homeomorphism of K with Dn carrying ∂K onto
Sn−1 .
Proof. Let w ∈ K ◦ , and τ : Rn → Rn be the translation x 7→ x − w.
Then τ is a homeomorphism and preserves convex combinations. So
A = τ (K) is also compact and convex, and the origin 0 ∈ A◦ . We
observe that A ≈ Dn by a homeomorphism which carries ∂A onto Sn−1 .
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Elements of Topology
Consider the mapping h : ∂A → Sn−1 defined by h(x) = x/∥x∥. By
the preceding lemma, h is a bijection. The function h is the restriction
of the continuous map Rn − {0} → Sn−1 , x 7→ x/∥x∥, and therefore
continuous. Since ∂A is compact and Sn−1 is T2 , h is a homeomorphism.
Let f : Sn−1 → ∂A denote the inverse of h. We extend f to a function
ϕ : Dn → A by setting
{
∥x∥f (x/∥x∥)
for x ̸= 0, and
ϕ(x) =
0
for x = 0.
We show that ϕ is a homeomorphism. If x ̸= 0, then ϕ is clearly continuous at x. The continuity of ϕ at x = 0 follows from the inequality
∥ϕ(x)∥ ≤ b∥x∥, where b is a bound for {∥x∥ : x ∈ A}. To see the injectivity of ϕ, suppose ϕ(x) = ϕ(y). Since 0 ∈ A◦ , and f (x/∥x∥) ∈ ∂A,
ϕ(x) = 0 implies x = 0. Accordingly, ϕ(x) = ϕ(y) = 0 ⇒ x = y = 0. If
x ̸= 0 ̸= y, then f (y/∥y∥) = ∥x∥
∥y∥ f (x/∥x∥) implies that both f (x/∥x∥)
and f (y/∥y∥) are on the same ray from 0. Because these points belong to ∂A, we must have f (x/∥x∥) = f (y/∥y∥), which implies that
x/∥x∥ = y/∥y∥. Our assumption now implies that ∥x∥ = ∥y∥, and
therefore x = y. Next, to see that ϕ is surjective, let 0 ̸= v ∈ A. If
v ∈ ∂A, then there is an x ∈ Sn−1 ⊆ Dn such that v = f (x) = ϕ(x).
If v ∈ A◦ , then w = f (v/∥v∥) ∈ ∂A satisfies v/∥v∥ = w/∥w∥. So
x = v/∥w∥ ∈ Dn , and ϕ(x) = v. Thus ϕ is a continuous bijection between Dn and A, and hence a homeomorphism. Now, it is clear that
the composition ϕ−1 ◦ (τ |K) is a homeomorphism of K onto Dn having
the desired property.
♢
Notice that every one of a cube, a (solid) cylinder and a disc in
Rn satisfies the hypotheses of the preceding theorem, and hence each
of these sets is homeomorphic to the unit n-disc Dn . We remark that
there are compact subsets of Rn which are not convex, for example,
the (n − 1)-sphere Sn−1 .
Next, we give an important property of open coverings of compact
metric spaces.
Definition 6.3.8 Let G be an open covering of a metric space X. A
real number r > 0 is called a Lebesgue number of G if each subset of X
of diameter less than r is contained in some member of G.
Obviously, a Lebesgue number of an open covering G of a metric
COMPACTNESS
145
space X is not unique. For, if a number r > 0 is a Lebesgue number
of G, then every positive real number less than r is also a Lebesgue
number of G. It is also clear that G has a Lebesgue number if and only
if there exists a real r > 0 such that, for each point x ∈ X, the open
ball B(x; r) is contained in a member of G.
Example 6.3.1 Any number 0 < r ≤ 1/2 is a Lebesgue number of the
open covering {(n, n + 2)|n ∈ Z} of the real line R.
Example 6.3.2 The open covering {(1/n, 1)|n = 1, 2, . . .} of the subspace (0,1) of R has no Lebesgue number.
Theorem 6.3.9 (Lebesgue Covering Lemma) Every open covering of a compact metric space has a Lebesgue number.
Proof. Let G be an open covering of the compact metric space (X, d).
Given x ∈ X, there is a set G ∈ G with x ∈ G. Choose rx > 0
such that the open ball B(x; rx ) ⊆ G. Consider the open covering
{B (x; rx /2) |x ∈ X} of X. Since X is compact, this open covering
has a finite subcovering. Accordingly, there exist finitely many points
x1 , . . . , xn such that the family {B (xi ; rxi /2) |i = 1, . . . , n} covers X.
We observe that λ = (1/2) min {rx1 , . . . , rxn } > 0 is a Lebesgue number. Let A be any subset of X with diam(A) < λ. Choose a point
a ∈ A. Then a ∈ B (xi ; rxi /2) for some i; consequently, d(a′ , xi ) ≤
d(a′ , a) + d(a, xi ) < λ + rxi /2 ≤ rxi for every a′ ∈ A. This implies that
A ⊆ B (xi ; rxi ). Since the latter set is contained in some member of G,
the result follows.
♢
Finally, in this section, we generalise a well-known result in analysis
by way of an application of the preceding lemma: A continuous realvalued function on a closed interval is uniformly continuous.
Definition 6.3.10 Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is uniformly continuous if for each ϵ > 0, there
exists a δ > 0 (depending only on ϵ) such that dX (x, x′ ) < δ ⇒
dY (f (x), f (x′ )) < ϵ for all x, x′ ∈ X.
Theorem 6.3.11 Let f be a continuous function of a compact metric
space (X, dX ) into a metric space (Y, dY ). Then f is uniformly continuous.
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Elements of Topology
Proof. Given ϵ > 0, consider the open
{ ball B(y; ϵ/2) for each
} point y ∈
Y . Since f is continuous, the family f −1 (B(y; ϵ/2)) |y ∈ Y is an open
covering of X. Since X is compact, this open covering has a Lebesgue
number δ. If x, x′ ∈ X and dX (x, x′ ) < δ, then there exists a y ∈ Y
such that both f (x) and f (x′ ) belong to B(y; ϵ/2). So dY (f (x), f (x′ )) ≤
dY (f (x), y) + dY (y, f (x′ )) < ϵ, and f is uniformly continuous.
♢
The condition of compactness on X in the preceding theorem is
essential, as shown by the function f : R → R given by f (x) = x2 .
Exercises
1. Let X be a metric space. If the projection q : X × R → R is a closed
map, show that X is bounded.
2. Show that a continuous function of a compact space into a metric space
is bounded.
3.
(a) Show that a real-valued continuous function f on a compact space
X attains its bounds, that is, there exist points x0 , y0 in X such
that f (x0 ) = sup f (X), and f (y0 ) = inf f (X).
(b) If f (x) > 0 for every x ∈ X, show that there exists an r > 0 such
that f (x) > r for all x ∈ X.
4. Do as in Exercise 3 for a countably compact space.
5. Let X, Y be a spaces and f : X × Y → R be a continuous function. If
X is compact, prove that the function g, h : Y → R defined by g(y) =
sup {f (x, y)|x ∈ X} and g(y) = inf {f (x, y)|x ∈ X} are continuous.
6. Let A be a (nonempty) compact subset of a metric space (X, d). Show
that there exist points a, b in A such that diam(A) = d(a, b).
7. Let A and B be nonempty disjoint closed subsets of a metric space
(X, d). Show:
(a) If B is compact, then there exists b ∈ B such that dist(A, B) =
dist(A, b) > 0.
(b) If both A and B are compact, then there exist a ∈ A, b ∈ B such
that dist(A, B) = d(a, b).
(c) The conclusion fails if A and B are not compact.
8. Let X be a metric space and A ⊆ X. Prove:
(a) A is bounded if and only if A is bounded.
(b) A is totally bounded if and only if A is totally bounded.
COMPACTNESS
147
(c) A is totally bounded if X is so.
9. Give an example of a bounded metric space which is not totally
bounded.
10. Prove that a bounded subspace of Rn is totally bounded.
11. Give an example to show that total boundedness is a property of metrics
(i.e., this is not a topological property).
12.
(a) Prove that the Hilbert cube {(xn ) ∈ ℓ2 |0 ≤ xn ≤ 1/n for every n}
is sequentially compact (and hence compact).
(b) Is every closed bounded subspace of ℓ2 sequentially compact?
13. Give an example of a separable metric space that is not countably compact.
14. • Prove that a totally bounded (and therefore a compact) metric space
is second countable. (see also Exercise 8.2.26).
15. Prove that a countably compact metrisable is second countable.
16. If every closed ball in a metric space X is compact, show that X is
separable.
17. Show that I × I with the dictionary order topology is not metrisable.
18. Show that the exponential map R → R is not uniformly continuous.
19. Let (X, d) be a compact metric space and f : X → X be a mapping
such that d(x, y) = d(f (x), f (y)). Prove that f is onto.
20. A space X is pseudocompact if and only if every continuous real-valued
function on X is bounded.
(a) Prove that every countably compact space is pseudocompact.
(b) Give an example of a pseudocompact space which is not countably
compact.
21. If A is a compact subset of a metric space, prove that its derived set A′
is compact. Is it true in any topological space?
148
6.4
Elements of Topology
Locally Compact Spaces
There are many spaces, the most important being the euclidean
spaces, which are not compact, but have instead a localized version of
the property. These spaces and their embeddings in compact spaces as
dense subsets will be discussed here.
Following an established practice, we adopt
Definition 6.4.1 A space X is locally compact at x ∈ X if it has a
compact neighbourhood of x. X is called locally compact if it is locally
compact at each of its points.
Since a space is a nbd of each of its points, a compact space is locally
compact. The converse of this is not true, as shown by the following.
Example 6.4.1 An infinite discrete space is locally compact that is not
compact.
Example 6.4.2 A euclidean space Rn is locally compact, for the closed
unit ball centered at x ∈ Rn is a compact neighbourhood of x.
Example 6.4.3 The Hilbert space ℓ2 is not locally compact. To see this,
consider the point x0 ∈ ℓ2 all of whose coordinates are zero. If N is a
nbd of x0 in ℓ2 , then there exists a closed ball {x ∈ ℓ2 : ∥x∥ ≤ r} ⊆ N .
(n)
For each positive integer n, let x(n) ∈ ℓ2 be the point given by xi = 0
(n)
i ⟩̸= n and xn = r. Then x(n) ∈ N for every n, and( the sequence
⟨for(n)
)
x
does not have a convergent subsequence, for d x(n) , x(m) =
√
2r when n ̸= m. Hence N is not compact.
Although the above formulation of local compactness is contrary to
the general spirit of local properties, this can be observed in Hausdorff
spaces.
Theorem 6.4.2 Let X be a Hausdorff space. The following conditions
are equivalent:
(a) X is locally compact.
(b) For each x ∈ X, there is an open nbd W of x such that W is
compact.
COMPACTNESS
149
(c) For each x ∈ X, and each nbd U of x, there is an open set V
such that x ∈ V ⊆ V ⊆ U and V is compact (that is, the closed
compact nbds of x form a nbd basis at x).
(d) X has a basis consisting of open sets with compact closures.
Proof. (a)⇒(b): Let N be a compact nbd of x in X. Since X is Hausdorff, N is closed in X. Let W be the interior of N . Then x ∈ W and
W ⊆ N is compact.
(b)⇒(c): Let x ∈ X be arbitrary and U be any nbd of x. Obviously,
it suffices to prove (c) when U is open. By our hypothesis, there exists
an open nbd W of x with W compact. Then F = W − U is a closed set
with x ∈
/ F . Since W is compact Hausdorff, there exist disjoint sets G
and H open in W such that x ∈ H and F ⊆ G (see Theorem 6.1.8). So
H ⊆ W − G ⊆ W − F ⊆ U and we have H W ⊆ U . By definition, there
exists an open subset of O ⊆ X such that H = W ∩O. Put V = O ∩W .
Then V is an open nbd of x in X, and V ⊆ H ∩ W , obviously. Now,
we have V ⊆ H ∩ W = H W ⊆ U . As W is compact, so is V . Thus V
is the desired nbd of x.
The implications (c) ⇒ (d) ⇒ (a) are trivial.
♢
For the invariant properties, we first see that local compactness is
invariant under continuous open mappings, and therefore is a topological invariant. For, if f : X → Y is a continuous open surjection
and U is a compact nbd of x, then f (U ) is a compact nbd of f (x). It
follows that Y is locally compact if X is locally compact. However, the
condition is not invariant under continuous mappings, as shown by the
identity map ℓ2 → ℓ2 , where the domain is assigned the discrete topology. The property is not preserved by even closed continuous maps
(refer to Exercise 7.1.14).
As regards the hereditary property of local compactness, we have
the following counter example.
Example 6.4.4 The subspace Q ⊂ R is not locally compact. For, if K is
a compact subset of Q, then it is a compact subset of R, and therefore
K must be closed in R, by the Heine–Borel theorem. If K also contains
an open subset of Q, then some irrational numbers must be its limit
points, a contradiction.
However, we see that every closed subset of a locally compact space
X is locally compact. Suppose that F is a closed subset of X. If x ∈ F
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has an open nbd W contained in a compact subset K ⊆ X, then
F ∩ W is an open nbd of x in F , and the set F ∩ K, being closed in
K, is compact. It follows that F is locally compact at x. If X is also
Hausdorff, then every open subset of X is locally compact, by Theorem
6.4.2. Furthermore, if A = F ∩ G, where F ⊆ X is closed and G ⊆ X
is open, then A is locally compact, since F is locally compact and A is
open in F . We show below that locally compact subspaces of a locally
compact Hausdorff space are precisely the intersections of its closed
and open sets.
Definition 6.4.3 A subspace A of a space X is called locally closed if
each point of A is contained in an open subset G of X such that G ∩ A
is closed in G.
Lemma 6.4.4 A subspace A of a space X is locally closed if and only
if A = F ∩ G for a closed subset F of X and an open subset G of X.
Proof. Suppose that A is locally closed. Then for each a ∈ A, there
exists an open nbd Oa of a such that Oa ∪
∩ A is closed in Oa . Then
∩
A.
So
G
=
Oa ∩ A =∪Oa ∩(A ∩ Oa )= O∪
a
a∈A Oa is open in X and
G ∩ A = a∈A Oa ∩ A = a∈A (Oa ∩ A) = G ∩ A = A. The converse
is obvious.
♢
Theorem 6.4.5 A locally compact subspace of a Hausdorff space is
locally closed.
Proof. Suppose that A is a locally compact subspace of a Hausdorff
space X. For a ∈ A, let U be a nbd of a in A such that U A is compact.
Then there exists a nbd V of a in X with U = A ∩ V . We have
U A = U ∩ A = A ∩ V ∩ A. Thus A ∩ V ∩ A is compact and hence closed
in X. This implies that A ∩ V ⊆ A ∩ V ⊆ A. It follows that A contains
a nbd of a in A, so A is open in A. Accordingly, there exists an open
set G ⊆ X such that A = G ∩ A, and the proof is complete.
♢
It follows that a subspace of a locally compact Hausdorff space
is locally compact if and only if it is locally closed. Concerning the
product of locally compact spaces, we have the following.
∏
Theorem 6.4.6 The product Xα of a family of spaces Xα , α ∈ A,
is locally compact if and only if each Xα is locally compact and all but
finitely many Xα are compact.
COMPACTNESS
151
∏
Proof. If
Xα is locally compact, then each Xα is locally compact
because the projections are continuous and open. To∏see the second
statement, choose a compact nbd K of some point x ∈ ∩
Xα . Let V be
n
a basic nbd of x contained in K, and suppose that V = 1 p−1
αi (Uαi ) ,
where Uαi is open in Xαi . Then, for α ̸= α1 , . . . , αn , we have pα (K) =
Xα , so Xα is compact.
Conversely, assume that the condition holds, and let
∏ α1 , . . . , αn be
the indices such that Xαi is not compact. Given x ∈ Xα , choose a
compact
∏ nbd Vαi of xαi in Xαi , i = 1, . . . , n. Then W = Vα1 × · · · ×
Vαn × {Xα |α ̸= α1 , . . . , αn } is a compact nbd of x.
♢
We now turn to see the construction of a compact space which
contains a given noncompact space as a subspace. This method is quite
similar to the construction of a Riemann sphere (identified with the
extended complex plane) which the reader might have seen in the study
of functions of a complex variable. In the construction of the extended
complex plane, an ideal point ∞ is adjoined to the complex plane R2
and the complements of the closed and bounded subsets of R2 (which
are, in fact, compact subsets of R2 ) are taken to be the neighbourhoods
of ∞. We shall generalise this construction to an arbitrary topological
space.
Definition 6.4.7 Let X be a space. The (Alexandroff) one-point compactification of X is the space X ∗ = X ∪ {∞}, where ∞ is an element
that is not in X, and its topology T ∗ consists of all open subsets of
X and all subsets U of X ∗ such that X ∗ − U is a closed and compact
subset of X.
Of course, we must check that the above specification does give a
topology for X ∗ . We do this verification now. Since the empty set ∅ is
trivially compact, X ∗ and ∅ are members of T ∗ . Suppose U, V ∈ T ∗ . If
both U and V are contained in X, then U ∩ V is open in X. If U ⊆ X
and ∞ ∈ V , then U ∩V is open in X, for V ∩X = X −(X ∗ −V ) is open
in X. And, if ∞ ∈ U ∩ V , then X ∗ − (U ∩ V ) = (X ∗ − U ) ∪ (X ∗ − V ) is
∗
closed and compact in X. So U ∩ V ∈ T∪
. Next, let {Uα } be a family
of members of T ∗ . If each Uα ⊆ X, then Uα is ∪
open in X.
∩ And, if ∞
∗
belongs to some member Uβ of {Uα }, then X − Uα = (X − Uα ) is
closed in X. This∪is also compact, being a closed subset of the compact
set X ∗ − Uβ . So Uα ∈ T ∗ , and T ∗ is a topology for X ∗ .
The point ∞ is often referred to as the point at infinity in X ∗ . If X
is already compact, then ∞ is an isolated point of X ∗ ; so, in this case,
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the one-point compactification is uninteresting. If X is not compact,
then ∞ is a limit point of X so that X is dense in X ∗ . Note that this
point may have just the trivial nbd in X ∗ . Of course, any space contains
compact subsets (e.g., singletons), but they need not be closed.
Theorem 6.4.8 Let X be a space and X ∗ be its one-point compactification. Then the inclusion X ,→ X ∗ is an embedding, and X ∗ is
compact. Furthermore, X ∗ is Hausdorff if and only if X is Hausdorff
and locally compact, and in this case, any homeomorphism between X
and the complement of a single point of a compact Hausdorff space Y
extends to a homeomorphism between X ∗ and Y .
Proof. We first show that the relative topology on X induced from X ∗
is the same as the given topology on X. If U ⊆ X is open, then U is
open in X ∗ . Conversely, for any U ⊆ X ∗ , U ∩ X = X − (X ∗ − U ) or U,
according as U contains or does not contain ∞. Thus U ∩ X is open in
X for every open U ⊆ X ∗ .
To see that X ∗ is compact, let U be any open cover of X ∗ . Then
there is a member U∞ of U which contains ∞. By definition, X ∗ − U∞
is a compact subset of X. Hence
∪n there is a finite subfamily {U1 , . . . , Un }
∗
of U such that X − U∞ ⊆ 1 Ui . It follows that {U1 , . . . , Un , U∞ } is
a finite subcovering of U, and X ∗ is compact.
Next, suppose that X is locally compact and T2 . Let x, y be distinct
points of X ∗ . If x and y both belong to X, then there are disjoint open
sets U and V in X such that x ∈ U and y ∈ V . By definition, U and
V are also open in X ∗ . If y = ∞, then x ∈ X and there is an open set
U ⊂ X such that x ∈ U and U (in X) is compact. So V = X ∗ − U is an
open nbd of ∞, which is obviously disjoint from U . Conversely, suppose
that X ∗ is T2 . Then X is T2 , for X is a subspace of X ∗ . To see that
X is locally compact, let x ∈ X be arbitrary. By our hypothesis, there
are disjoint open sets U and V in X ∗ such that x ∈ U and ∞ ∈ V .
Then X ∗ − V is a compact nbd of x, for x ∈ U ⊆ X ∗ − V . Thus X is
locally compact at x.
To establish the last statement, suppose that X ∗ is T2 , and let h be
a homeomorphism of X into a compact Hausdorff space Y such that
Y −h(X) is the singleton {y0 }. Define h∗ : X ∗ → Y by h∗ (x) = h(x) for
all x ∈ X and h∗ (∞) = y0 . Then h∗ is obviously bijective. We observe
that h∗ is open. Let U be any open subset of X ∗ . If ∞ ∈
/ U , then U
∗
is open in X so that h (U ) = h(U ) is open in Y , for {y0 } is closed.
And, if ∞ ∈ U , then X ∗ − U is a compact subset of X. So h(X ∗ − U )
COMPACTNESS
153
is compact, and hence closed in Y . As h(X ∗ − U ) = Y − h∗ (U ), h∗ (U )
is open in Y . Thus h∗ is open. By symmetry, h∗−1 : Y → X ∗ is also
open, and therefore h∗ is a homeomorphism.
♢
The last statement of the theorem implies that the topology of the
one-point compactification X ∗ of a locally compact Hausdorff space
X is the only topology which makes X ∗ a compact Hausdorff space
such that X ⊆ X ∗ as a subspace. Also, it enables us to determine the
one-point compactification of spaces.
Example 6.4.5 The one-point compactification of N (with the discrete
topology) is the subspace {1/n|n ∈ N} ∪ {0} of R.
Example 6.4.6 The one-point compactification of the euclidean space
Rn is the n-sphere Sn , since Sn is compact Hausdorff and the stereographic projection Sn − {north pole} → Rn is a homeomorphism.
As an illustration of the usefulness of one-point compactification,
we prove the following.
Theorem 6.4.9 The one-point compactification of a locally compact
Hausdorff space X is second countable if and only if X is second countable.
Proof. Let X be a locally compact Hausdorff space and X ∗ be its
one-point compactification. Suppose first that X is second countable.
Then, by Theorem 6.4.2, there is a countable
(∪n )basis {Un } of X such
that U n is compact. Put Vn = X ∗ −
1 U i for n = 1, 2, . . .. We
assert that {Vn } is a countable local base at ∞, the point at infinity in
X ∗ . Obviously, each Vn is an open nbd of ∞. If O is an open set in X ∗
containing ∞, then X ∗ − O is a compact subset of X. As {Un } covers
X ∗ − O, there are finitely many sets, say, Un1 , . . . , Unk in {Un } such
∪k
that X ∗ − O ⊆ 1 Uni . If m = max {n1 , . . . , nk }, then Vm ⊆ O, and
our assertion follows. Since X is open in X ∗ , the family {Un } ∪ {Vn }
is a countable basis for X ∗ .
The converse is immediate from the fact that the property of being
second countable is hereditary.
♢
One might guess that if X is a locally compact metric space, then
the one-point compactification X ∗ could be metrisable. This is not true
unless X is second countable. For, if X ∗ were a compact metric space,
then it would be second countable (ref. Exercise 6.3.14). So X, being
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Elements of Topology
a subspace of X ∗ , must be second countable. We shall see later that
X ∗ is metrisable for second countable, locally compact metric spaces
X (see Theorem 8.2.8).
Exercises
1. Prove: Every open subset of a compact metric space is locally compact
and second countable.
2. Is a locally compact metric space second countable?
3. Prove that a locally compact dense subset of a compact Hausdorff space
is open.
4. Let X be a locally compact Hausdorff space and K be a compact subset
of X. Show:
(a) If U is an open set in X with K ⊆ U , there exists an open set V
such that K ⊆ V ⊆ V ⊆ U and V is compact.
(b) If K1 , K2 are disjoint compact subsets of X, then they have disjoint nbds with compact closures.
5. Prove that a second countable, locally compact Hausdorff space has a
countable basis consisting of open sets with compact closures.
6. Prove that a separable metric space X is locally compact if and only if
X is the union of a countable family of open subsets Un such that U n
is compact and U n ⊆ Un+1 for every n.
7. Let X be a connected, locally connected, locally compact Hausdorff
space and x, y ∈ X. Show that there is a compact connected set C in
X which contains both x and y.
8. Prove that a connected, locally compact, Hausdorff space X is locally
connected ⇔ for each compact set K ⊂ X and each open set U ⊃ K,
all but a finite number of components of X − K lie in U .
9. Prove that in a locally compact Hausdorff space, every compact quasicomponent is a component, and every compact component is a quasicomponent.
10. Determine the one-point compactification of the following:
(a) The subspace (0, 1] of R.
(b) The ordinal space [0, Ω).
(c) The Sorgenfrey line Rℓ .
(d) The metric spaces B(I) and C(I) with the sup metric ρ.
COMPACTNESS
11.
155
(a) If locally compact Hausdorff spaces X and Y are homeomorphic,
show that their one-point compactifications are also homeomorphic.
(b) Find two spaces X and Y such that X ̸≈ Y but X ∗ ≈ Y ∗ .
12. Prove that one-point compactification of a T1 -space is a T1 -space.
13. Let f be a continuous open mapping of a locally compact T2 -space X
onto a T2 -space Y . Given a compact subset B ⊆ Y , show that there
exists a compact subset A ⊆ X such that f (A) = B.
14. If X is a compact space, show that {∞} is a component of the one-point
compactification X ∗ (of X).
15.
(a) If X is connected, is X ∗ necessarily connected?
(b) If the one-point compactification X ∗ of a space X is connected, is
X connected?
16. Give an example of a countable connected Hausdorff space.
6.5
Proper Maps
In this section, we study special functions which generally preserve
topological properties.
Definition 6.5.1 A continuous map f : X → Y is called proper (or
perfect) if, for all spaces Z, the product f × 1 : X × Z → Y × Z is a
closed map.
f
If f : X → Y is proper, then, for any B ⊆ Y, the function g :
(B) → B defined by f is also proper. This follows from the equality
((
)
)
(B × Z) ∩ (f × 1)(K) = (f × 1) f −1 (B) × Z ∩ K
−1
for all K ⊆ X × Z. In particular, we see that the constant map
f −1 (y) → y is proper. Therefore, by Theorem 6.1.16, f −1 (y) is compact. Also, by taking Z to be a one-point space, we see that f is closed.
The following theorem shows that these conditions are sufficient, too.
Theorem 6.5.2 Let f : X → Y be continuous. Then f is proper if
and only if f is closed and f −1 (y) is compact for every y ∈ Y .
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Proof. (Sufficiency) Let Z be any space and consider the product f ×
1 : X × Z → Y × Z. Given a closed set F ⊆ X × Z, suppose that
(y, z) ∈
/ (f × 1)(F ). If y ∈
/ f (X), then (Y − f (X)) × Z is a nbd of
(y, z) disjoint from (f × 1)(F ). So assume that y = f (x) for some
x ∈ X. Then f −1 (y) × {z} ⊆ (X × Z) − F . Since F is closed and
f −1 (y) is compact, there exists an open nbd U of f −1 (y) and V of
z such that U × V ⊆ (X × Z) − F (ref. Exercise 6.1.34). So F ⊆
(X × Z) − (U × V ) = X × (Z − V ) ∪ (X − U ) × Z = E, say. It is obvious
that (y, z) ∈
/ (f × 1)(E) and (f × 1)(F ) ⊆ (f × 1)(E). Since f is closed,
(f × 1)(E) = [f (X) × (Z − V )] ∪ [f (X − U ) × Z] is closed. Therefore
Y × Z − (f × 1)(F ) is a nbd of (y, z), and it follows that (f × 1)(F ) is
closed. Thus f × 1 is a closed map, and the proof is complete.
♢
Corollary 6.5.3 Let X be a compact space.
(a) If Y is a Hausdorff space, then any continuous map f : X → Y
is proper.
(b) For any space Y, the projection X × Y → Y is proper.
The simple proofs are omitted.
The most topological properties are preserved by proper mappings.
By way of illustration, we prove
Proposition 6.5.4 Let f : X → Y be a proper surjection.
(a) If X is a second countable, then so is Y .
(b) If X is Hausdorff, then so is Y.
Proof. (a): Let {Bn } be a countable basis for X. Then the family G
of all finite unions of Bn is countable. So G can be indexed by the
positive integers; let G = {Gn }. Since f is closed, Un = Y − f (X − Gn )
is open for every n. We observe that {Un } is a countable basis for
Y . Let V ⊆ Y be open, and y ∈ V be arbitrary. Then f −1 (V ) is
open in X and contains f −1 (y). For each x ∈ f −1 (y), there is an
integer n such that x ∈ Bn ⊆ f −1 (V ). Varying x over f −1 (y), we
obtain a covering of f −1 (y) consisting of basis elements each of which
is contained in f −1 (V ). By Theorem 6.5.2, f −1 (y) is compact, so this
open cover reduces to a finite subcover. Consequently, we find a set Gn
in G such that f −1 (y) ⊆ Gn ⊆ f −1 (V ). Then X − f −1 (V ) ⊆ X − Gn ⊆
COMPACTNESS
157
X − f −1 (y) = f −1 (Y − y), which implies that y ∈ Un ⊆ f (Gn ) ⊆ V .
Thus V is a union of sets in {Un } and (a) holds.
(b): Let y1 , y2 be two distinct points of Y. By Theorem 6.5.2, each
f (yi ), i = 1, 2, is a compact subset of X. Since X is Hausdorff, there
exist disjoint open nbds U1 and U2 of f −1 (y1 ) and f −1 (y2 ), respectively,
by Theorem 6.1.9. Set Vi = Y − f (X − Ui ), i = 1, 2. Then each Vi is
open in Y, since f is closed, and yi ∈ Vi . Also, we have f −1 (Vi ) ⊆ Ui .
Since U1 ∩ U2 = ∅ and f is surjective, V1 ∩ V2 = ∅. Thus V1 and V2
are disjoint nbds of y1 and y2 , respectively, and Y is Hausdorff.
♢
−1
In the other direction, there are some properties of the range of
a proper mapping, especially those defined by the behavior of open
coverings, which are acquired by its domain.
Proposition 6.5.5 Let f : X → Y be a proper map. If Y is compact,
then so is X.
Proof. Let {Gα |α ∈ A} be an open cover of X. By Theorem 6.5.2,
f −1 (y) is compact for every y ∈ Y and f is closed.
∪ Consequently, there
−1
exists a finite set By ⊆ A such that f (y) ⊆ {Gα |α ∈ By } = Ny ,
say. As f (X − Ny ) is closed in Y , we find an open nbd Vy of y such that
f −1 (Vy ) ⊆ Ny . If Y is compact,∪then there are finitely many∪points
n
n
y1 , . . . , yn in Y such that Y = 1 Vyi . It follows that X = 1 Nyi
and the family {Gα |α ∈ Byi , i = 1, . . . , n} is a finite subcovering of
{Gα |α ∈ A}. Thus X is compact.
♢
If f : X → Y is a proper map and C ⊆ Y , then the map g :
f −1 (C) → C, defined by f , is clearly proper. By the preceding theorem,
we see that f −1 (C) is compact for every compact C. The converse holds
under some mild conditions on the spaces involved.
Theorem 6.5.6 Let X be a Hausdorff space and Y be a locally compact Hausdorff space. Then a continuous map f : X → Y is proper if
and only if f −1 (C) is compact for every compact C ⊆ Y .
Proof. The necessity of the condition follows from the preceding remarks. For sufficiency, suppose that f −1 (C) is compact for every compact subset C of Y . Then, in particular, f −1 (y) is compact for each
y ∈ Y . By Theorem 6.5.2, it suffices to prove that f is closed. Note
also that X is locally compact, by our hypothesis. Denote the one-point
compactifications of X and Y by X ∗ and Y ∗ , respectively, and define
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a function f ∗ : X ∗ → Y ∗ by setting f ∗ (x) = f (x) for every x ∈ X and
f ∗ (∞X ) = ∞Y . We assert that f ∗ is continuous. Let U be an open set
subset of Y ∗ . If U ⊆ Y , then U is open in Y . So f ∗−1 (U ) = f −1 (U ) is
open in X, and hence in X ∗ . If ∞Y ∈ U , then C = Y ∗ − U is compact,
and so f −1 (C) is compact, by our assumption. Since X is T2 , f −1 (C)
is closed in X. Thus f ∗−1 (U ) = X ∗ − f −1 (C) is open in X ∗ , and our
assertion follows. Now, if F is a closed subset of X, then F ∪ {∞X }
is closed in X ∗ and, therefore, compact. It follows that f ∗ (F ∪ {∞X })
is compact, and hence closed in Y ∗ . So f (F ) = Y ∩ f ∗ (F ∪ {∞X }) is
closed in Y , and the proof is complete.
♢
Exercises
1. Let f : X → Y be a continuous injection. Show that f is closed ⇔ f is
proper ⇔ f is an embedding and im(f ) is closed in Y .
2. Let f : X → Y, g : Y → Z be continuous surjections. Prove:
(a) If f is proper and A ⊆ X is closed, then f |A is proper.
(b) If f and g are proper, then gf is proper.
(c) If gf is proper, then g is proper.
(d) If gf is proper and g is injective or Y is T2 , then f is proper.
3. Let f : X → Y be a proper surjection. Prove:
(a) If Y is Lindelöf, then so is X.
(b) If Y is countably compact, then so is X.
4. Let f be a proper map of a Hausdorff space X onto a space Y . Prove
that X is locally compact ⇔ Y is locally compact (and Hausdorff).
5. Let X and Y be locally compact Hausdorff spaces, and f : X → Y
be a continuous surjection. Show that f is proper ⇔ f extends to a
continuous function f ∗ : X ∗ → Y ∗ with f ∗ (∞X ) = ∞Y , where X ∗
denotes the one-point compactification of X, etc.
Chapter 7
TOPOLOGICAL
CONSTRUCTIONS
7.1
7.2
7.3
7.4
7.5
7.6
Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Identification Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Cones, Suspensions and Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wedge Sums and Smash Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Adjunction Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Coinduced and Coherent Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
k-spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1
Quotient Spaces
159
173
180
188
195
202
206
Until now, we have studied two methods of producing new spaces
from old ones, namely, forming subspaces and cartesian products. For
the same purpose, we treat here another technique which is akin to the
physical process of gluing things together. This makes precise the intuitive idea of identifying points in a space, and leads to many interesting
topological constructions such as “collapsing subsets of a topological
space to points,” “attaching spaces via a map,” “joining two spaces,”
etc. Thus the new technique plays an important role in topology.
Let X be a space with an equivalence relation ∼, and X/ ∼ be the
set of equivalence classes. Let π : X → X/ ∼ be the natural projection
which takes x ∈ X to its equivalence class [x]. Then the family of
subsets U ⊆ X/ ∼ such that π −1 (U ) is open in X is a topology on
X/ ∼, since π −1 behaves well with the operations of taking intersection
and union.
Definition 7.1.1 The set X/ ∼ together with the topology
{
}
U |π −1 (U ) is open in X
is called the quotient space of X by the relation ∼, and this topology
is referred to as the quotient topology for X/ ∼.
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Elements of Topology
We think of X/ ∼ as the space obtained from X by identifying each
of the equivalence classes [x], x ∈ X, to a single point. Notice that a
set F ⊆ X/ ∼ is closed if and only if π −1 (F ) is closed in X. Observe
that the quotient topology for X/ ∼ is the largest topology such that
π is continuous. However, the mapping π need not be open or closed
(refer to Exercise 1). It is clear that π is closed (resp. open) if and only
if π −1 (π(F )) is closed (resp. open) in X for every closed (resp. open)
subset F ⊆ X.
For a subset A ⊆ X, the set
π −1 (π(A)) = {x ∈ X|x ∼ a for some a ∈ A}
is called the saturation of A. We say that the set A is saturated if
A = π −1 (π(A)). With this terminology, we have the following.
Proposition 7.1.2 A necessary and sufficient condition for the projection π : X → X/ ∼ to be a closed (resp. open) mapping is that the
saturation of every closed (resp. open) set is closed (resp. open).
To see a quotient space, we usually need to find a homeomorphism
between this space and a known space. With this end in view, we first
describe a method of constructing continuous functions on quotient
spaces.
Let X/ ∼ be a quotient space of X and π : X → X/ ∼ be the natural
projection. We know that if a function g : X/ ∼ → Y is continuous, then
the composition gπ
( : X →) Y is continuous. Conversely, if gπ is continuous, then π −1 g −1 (U ) = (gπ)−1 (U ) is open in X for every open
set U ⊆ Y. Thus, by the definition of the quotient topology, g −1 (U ) is
open in X/ ∼ for every open subset U of Y, and g is continuous. Next, if
f : X → Y is a function which is constant on each equivalence class [x],
x ∈ X, then we can define a function f¯ : X/ ∼ → Y by f¯ ([x]) = f (x),
x ∈ X. The composition
π
f¯
X −−→ (X/ ∼) −−→ Y
is obviously the function f , that is, f¯◦π = f. So, if f is also continuous,
then f¯ is continuous. We summarise this in the following.
Proposition 7.1.3 Let X be a space with an equivalence relation ∼.
If f : X → Y is a continuous function which is constant on each
equivalence class, then the induced function f¯ : X/ ∼ → Y , given by
TOPOLOGICAL CONSTRUCTIONS
f¯ ([x]) = f (x), is continuous. Furthermore,
and only if f (U ) is open for each saturated
lar statement obtained by replacing “open”
holds good.
161
f¯ : X/ ∼ → Y is open if
open set U ⊆ X. A simiwith “closed” everywhere
Proof. To prove the second statement, suppose that f (U ) is open for ev−1
ery saturated open set U ⊆ X. For each open set
⊆ X/
( V
) ∼, π (V ) is
−1
an open saturated set. Consequently, f¯(V ) = f π (V ) is open. Thus
f¯ is open. Conversely, if f¯ is open, and U ⊆ X is an open saturated
set, then f (U ) = f¯π(U ) is open, since π(U ) is open.
A similar proof applies to the “closed” case.
♢
Example 7.1.1 Intuitively, a circle is obtained from an interval by identifying the end points. Let I be the unit interval with its usual topology
and ∼ the equivalence relation: 0 ∼ 1 and x ∼ x for x ̸= 0, 1. The quotient space I/ ∼ is homeomorphic to the unit circle S1 by the mapping
[x] 7→ e2πıx .
Example 7.1.2 Intuitively, a cylinder is obtained from a rectangle
(which is homeomorphic to the square I × I) by gluing a pair of its
opposite sides together (Figure 7.1 below). Accordingly, the quotient
space (I × I)/ ∼, where ∼ is the equivalence relation (s, 0) ∼ (s, 1),
0 ≤ t ≤ 1, is referred to as the cylinder. This( is homeomorphic
to
)
1
2πıs
S × I; a homeomorphism is given by [(s, t)] 7→ e
,t .
(s;1)
^
^
I£I
^
(s;0)
Cylinder
FIGURE 7.1: Construction of a cylinder.
Example 7.1.3 A Möbius band (or strip) is a surface in R3 generated
by moving a line segment of finite length in such a way that its middle
point glides along a circle, and the segment remains normal to the circle
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Elements of Topology
and turns uniformly through a total angle of 180◦ . A model of a Möbius
band can be constructed by gluing the ends of a rectangular strip of
paper after giving it a half twist (of 180◦ ). (See Figure 7.2 below). The
quotient space (I × I)/ ∼, where the equivalence relation ∼ is given by
(0, t) ∼ (1, 1 − t), 0 ≤ t ≤ 1 is homeomorphic to a Möbius band.
• (1,1-t)
^
I£I
^
(0;t) •
Möbius Band
FIGURE 7.2: Construction of a Möbius band.
Example 7.1.4 We consider another equivalence relation on I × I given
by (s, 0) ∼ (s, 1) and (0, t) ∼ (1, t) for all s, t ∈ I. The quotient space
1
1
(I × I)/ ∼ is homeomorphic
( 2πıs 2πıtto
) S × S , the torus (see. Ex. 2.2.5); the
function [(s, t)] 7→ e
,e
provides a homeomorphism. Intuitively,
we first identify a pair of opposite edges of a rectangle to get a cylinder
and then the circular ends are glued together, giving a torus (Figure
7.3 below).
^
^
I£I
^
^
Torus
FIGURE 7.3: Construction of a torus.
Example 7.1.5 The quotient space (I × I)/ ∼, where ∼ is the equivalence relation (0, t) ∼ (1, t) and (s, 0) ∼ (1 − s, 1) for all s, t ∈ I, is
TOPOLOGICAL CONSTRUCTIONS
163
^
^
called a Klein bottle. We can think of a Klein bottle as the space obtained from a (finite) cylinder by identifying the opposite ends with the
orientation of two circles reversed. This space cannot be represented in
3-dimensional space without self-intersection.
^
^
I£I
^
^
FIGURE 7.4: Construction of a Klein bottle.
Example 7.1.6 For any integer n ≥ 0, the quotient space of the nsphere Sn by the equivalence relation ∼, where x ∼ y if x = −y, is called
the n-dimensional real projective space, denoted by RPn . There are
other equivalent descriptions of the space RPn , given in the exercises.
The quotient spaces discussed in the preceding examples will be
further investigated in later chapters. Next, we study the particular
construction which involves the idea of collapsing a subset of a space
to a point.
Definition 7.1.4 Let X be a space and A ⊆ X. The quotient space
of X by the equivalence relation ∼ given by a ∼ a′ for every a, a′ ∈ A
is usually denoted by X/A, and called the quotient of X by A.
We think of X/A as the space X with the subspace A identified
(collapsed) to a point. If A is either open or closed, then the projection
map π : X → X/A is open (resp. closed). In this case, it is also easily
seen that X − A is homeomorphic to X/A − [A], where [A] is the
equivalence class of any point of A.
Example 7.1.7 As an illustration of the above notion, we show that the
2
quotient space D2 /S1 , where S1 = ∂D2 , is homeomorphic
( 2 )◦ to S . 2The
map h : x → x/(1 − ∥x∥) is a homeomorphism of D
onto R . If
p = (0, 0, 1) ∈ S2 , then the map g : S2 − {p} → R2 , (x0 , x1 , x2 ) 7→
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Elements of Topology
(x0 , x1 ), is a homeomorphism. So g −1 h is a homeomorphism be( )◦
tween D2 and S2 − {p}. We define a function f : D2 → S2 by
{ −1
g h(x)
for x ∈ D2 − S1 and
f (x) =
p
for x ∈ S1.
1
1−x2
(The function f maps a concentric circle of radius r in D2 homeomorphically onto a parallel at the latitude (2r − 1)/(1 − 2r + 2r2 ) in S2 and
takes the radii in D2 onto the meridians running from the south pole
to the north pole (see Figure 7.5).)
•
^
D2
^
•
S2
•
FIGURE 7.5: Construction of a sphere from a disc.
Clearly, f is continuous and induces a bijection f¯ : D2 /S1 → S2 such
that f = f¯π, where π : D2 → D2 /S1 is the quotient map. We observe
that f¯ is a homeomorphism. By Proposition 7.1.3, it suffices to show
that f is continuous and closed. To see the continuity of f, consider a
point x ∈ D2 , and let U ⊂ S2 be an open nbd of f (x). If x ∈ D2 − S1 ,
then V = f −1 (U − {p}) is a nbd of x such that f (V ) ⊆ U. And, if
x ∈ S1 , then we choose a small open ball B(p; ϵ) ⊆ U . It is easily
2
checked that V = f −1
) annulus {x ∈ D |r < ∥x∥ ≤ 1},2
( (B(p;√ϵ)) is the
2
2
2
where r = (4−ϵ )/ 4 + 2ϵ 4 − ϵ . Thus V is an open nbd of x in D
with f (V ) ⊆ U, and the continuity of f follows. Finally, the closedness
of f is immediate from Corollary 6.1.11, since D2 is compact and S2 is
Hausdorff.
Similarly, we see that the quotient space Dn /Sn−1 is homeomorphic
to Sn .
Turning to the general discussion, we remark that an equivalence
TOPOLOGICAL CONSTRUCTIONS
165
relation ∼ on X is often regarded as its graph
G(∼) = {(x, x′ ) |x, x′ ∈ X and x ∼ x′ }.
We call ∼ open (resp. closed) if G(∼) is open (resp. closed) in the
product space X × X. If ∼ is open (resp. closed), then every equivalence class [x] is open (resp. closed) in X, because [x] is the image of
({x} × X) ∩ G(∼) under the homeomorphism {x} × X ≈ X. Consequently, the quotient topology for X/ ∼ is discrete and π is open, if ∼
is open. But, the projection map defined by a closed relation need not
be closed (see to Ex. 7.1.8). Incidentally, we would like to mention that
the terms open relation and closed relation are used in the literature
in a different sense too.
Example 7.1.8 Consider the equivalence relation ∼ on R defined by
x ∼ y if x = y or x ̸= 0 and y = 1/x. This relation is obviously
closed in R2 , but the projection map π : R → R/ ∼ is not closed (the
saturation of the closed set [1, ∞) is (0, ∞)).
Example 7.1.9 Let X be the Sierpinski space with the identity relation.
Then the projection map X → X/ = is obviously closed, while G(=)
is not closed in X × X.
Example 7.1.10 In the real line R, consider the equivalence relation
x ∼ y if x − y is rational. The projection π : R → R/ ∼ maps each open
interval (a, b) onto R/ ∼ so that π is open. But the relation G(∼) is
not open in R2 , since it does not contain any rectangle (a, b) × (c, d).
(cf. Exercise 8.1.16).
The last two examples show that the relation ∼ is not necessarily
closed (open) in X × X if the projection map π : X → X/ ∼ is closed
(open).
Now, we deal with the behaviour of the quotient topology with
respect to various topological properties (discussed so far). It is clear
that a quotient space X/ ∼ is T1 if and only if each equivalence class
of the space X is closed. Said differently, if a quotient map is closed,
Axiom T1 certainly passes down to the quotient space. In particular, if
X satisfies the T1 -axiom and A ⊂ X is closed, then X/A is a T1 -space.
However, a quotient space of a T1 -space need not be T1 in general, even
if the quotient map is open.
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Elements of Topology
Example 7.1.11 Let X = [0, 2] be the subspace of the real line R and
A = (1, 2]. It is easy to see that the projection map π : X → X/A
is open. But X/A fails to satisfy T1 -axiom. For, every saturated open
nbd of the point 1 contains the set A; so there is no nbd of the point
π(1), which excludes the point π(2).
It is obvious from the preceding example that even open quotient
maps need not transmit Hausdorffness. This is also true for closed
quotient maps, as shown by
Example 7.1.12 Let A = {1/n|n = 1, 2, . . .} and X be the space of real
numbers with the topology
T = {U − B|U is open in the real line R and B ⊆ A}
(ref. Exercise 1.4.6). Then A is closed in X, and X is Hausdorff, since
the topology of X is finer than the usual topology of R. It follows that
the projection map X → X/A is closed, but the points [0] and [A] in
X/A cannot be separated by open sets, since every open set containing
A intersects every nbd of 0 in X. The quotient space X/A satisfies the
T1 -axiom, however.
A positive case in point is the following.
Proposition 7.1.5 If X is a compact Hausdorff space and A ⊂ X is
closed, then X/A is (compact and) Hausdorff.
Proof. Suppose that X is a compact Hausdorff space and A ⊂ X is
closed. Let π : X → X/A be the quotient map. If x ̸= y are in X − A,
then there exist disjoint open sets U and V in X with x ∈ U and y ∈ V .
Since A is closed in X, we can assume that U ∩ A = ∅ = V ∩ A. Then
π(U ) and π(V ) are disjoint nbds of π(x) and π(y) in X/A. If x ∈
/ A and
y ∈ A, then we find disjoint open sets U and V in X with x ∈ U and
A ⊂ V , by Proposition 6.1.8. Now, π(U ) and π(V ) are disjoint nbds of
π(x) and π(y) in X/A. Thus X/A is a Hausdorff space.
♢
For a generalisation of the preceding result, the reader is referred
to Exercise 8.1.13. It is clear that a quotient space of a space X is
Hausdorff if and only if any two distinct equivalence classes of X have
disjoint saturated open nbds. The following proposition may sometimes
come in useful for checking the Hausdorff property of quotient spaces.
TOPOLOGICAL CONSTRUCTIONS
167
Proposition 7.1.6 Let X be a space with an equivalence relation ∼
and π : X → X/ ∼ be the natural projection. If the quotient space
X/ ∼ is Hausdorff, then G(∼) is closed in X × X. Conversely, if G(∼)
is closed in X × X and π is open, then X/ ∼ is Hausdorff.
Proof. Assume that X/ ∼ is T2 and x ̸∼ y. Then there exist disjoint open
nbds U of π(x) and V of π(y). We have (x, y) ∈ π −1 (U ) × π −1 (V ) ⊆
(X × X) − G(∼), where G(∼) = {(x, x′ ) ∈ X × X|x ∼ x′ }. This shows
that (X × X) − G(∼) is open; so G(∼) is closed.
Conversely, suppose that π is open and G(∼) is closed in X × X.
If π(x) ̸= π(y) for x, y ∈ X, then (x, y) ∈ (X × X) − G(∼). Since
G(∼) is closed in X × X, there exist nbds N of x and M of y such
that N × M ⊆ (X × X) − G(∼). This implies that π(N ) ∩ π(M ) = ∅.
Since π is open, π(N ) and π(M ) are disjoint nbds of π(x) and π(y),
respectively.
♢
It should be noted that if the projection π : X → X/ ∼ is open, the
graph of ∼ is not necessarily closed (cf. Exercise 9).
Since a quotient map is a continuous surjection, separability and
Lindelöfness are obviously transmitted to quotient spaces. As regards
the other countability axioms, we have counter-examples.
Example 7.1.13 Consider the real line R with the subspace Z of all
integers pinched to a point. Let {Un } be a countable family of open
nbds of [Z] in the quotient space Y = R/Z. Then each set π −1 (Un ) is
a nbd of Z, where π : R → Y is the quotient map. So, for each integer
n, we can find a real rn such that 1/2∪> rn > 0 and (n − rn , n + rn ) ⊆
π −1 (Un ). The image of the open set (n − rn /2, n + rn /2) under π is
an open nbd G of [Z], and it is clear that G contains no Un . Therefore
Y does not satisfy the first axiom of countability.
By Proposition 5.1.3, the quotient space X/ ∼ of a first countable
(resp. second countable) space X is first countable (resp. second countable) if the saturation of every open set in X is open. However, it is
readily seen that X/ ∼ is second countable if and only if there is a
countable family O of open saturated sets in X such that any open
saturated set of X can be expressed as a union of members of O.
The quotient topology behaves rather well with the connectivity
properties. About connectedness and path-connectedness, this is immediate from the fact that a quotient map is continuous; about local
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Elements of Topology
connectedness and local path-connectedness, this is justified by the
following.
Theorem 7.1.7 Let ∼ be an equivalence relation on a space X. If X
is locally connected (resp. locally path-connected), then so is X/ ∼.
Proof. Suppose that X is locally connected. By Theorem 3.4.2, it suffices to show that the components of each open set U ⊂ X/ ∼ are open.
Let π : X → X/ ∼ be the projection map and C be a component of
U . We assert that π −1 (C) is open. Let x ∈ π −1 (C) be an arbitrary
element and V be the component of π −1 (U ) containing x. Then V is
open, since X is locally connected and π −1 (U ) is open in X. It is obvious that π(x) ∈ π(V ) ⊆ C, for π(V ) is connected. So x ∈ V ⊆ π −1 (C),
and π −1 (C) is a neighbourhood of x. This proves our assertion. It is
now immediate from the definition of quotient topology that C is open.
A similar argument applies in the case of local path-connectedness.
♢
Clearly, the quotient space of a compact space is compact; however,
the same is not true about local compactness (cf. Exercise 14).
Now, we discuss the behaviour of quotient topology vis-à-vis the
subspace topology and the product topology. Let X be a space and
A ⊆ X. An equivalence relation ∼ on X determines, by restriction,
an equivalence relation ∼A on A. There is a canonical map λ : A/ ∼A
→ X/ ∼ which maps the equivalence class of a under ∼A into its
equivalence class under ∼. The mapping λ is obviously injective with
the image π(A), where π is the projection X → X/ ∼. So we can
identify A/ ∼A with π(A), and regard A/ ∼A as a subset of X/ ∼.
Then A/ ∼A also receives the relative topology induced by the quotient
topology for X/ ∼ . This topology for A/ ∼A is, in general, coarser than
its quotient topology (see Example 7.1.14 below). Notice that if A is
saturated, then A/ ∼A is actually a subset of X/ ∼. If A is also open (or
closed), then the quotient space A/ ∼A is clearly a subspace of X/ ∼
(cf. Proposition 7.2.7).
Example 7.1.14 In Example 7.1.1, consider the subspace A = [0, 1/2)∪
(2/3, 1). The relation ∼A on A is the identity relation. So the quotient
space A/ ∼A can be identified with A itself. Thus [0, 1/4) is open in the
quotient topology for A/ ∼A , while a nbd of 0 in the relative topology
induced from I/ ∼ must contain a set [0, ϵ) ∪ (1 − ϵ, 1) for some ϵ > 0.
TOPOLOGICAL CONSTRUCTIONS
169
The behaviour of quotient topology in relation to product topology
is also not satisfactory. To see this, consider two spaces X1 and X2 with
the equivalence relations ∼1 and ∼2 , respectively. Then an equivalence
relation R on X1 × X2 can be defined by (x1 , x2 ) R (y1 , y2 ) iff x1 ∼1 y1
and x2 ∼2 y2 . The relation R on X1 ×X2 is usually denoted by ∼1 × ∼2 .
There is a canonical map
ϕ : (X1 × X2 ) /R → (X1 / ∼1 ) × (X2 / ∼2 )
such that the diagram
X1 × X2
Q
ν +
(X1 × X2 ) /R
ϕ
Q π1 × π2
Q
Q
Q
Q
s
- (X1 / ∼1 ) × (X2 / ∼2 )
commutes, where ν, π1 and π2 are natural projections. The mapping ϕ
is a continuous bijection, but need not be a homeomorphism (see Ex.
7.1.15 below). The following proposition describes a simple sufficient
condition when the product of two quotient spaces can be regarded as
a quotient space.
Proposition 7.1.8 With the above notations, if the projection maps
π1 : X1 → X1 / ∼1 and π2 : X2 → X2 / ∼2 are open, then the canonical
map
ϕ : (X1 × X2 ) /R → (X1 / ∼1 ) × (X2 / ∼2 )
is a homeomorphism.
Proof. This follows from the fact that the map
π1 × π2 : X1 × X2 → (X1 / ∼1 ) × (X2 / ∼2 )
is open under the condition on π1 and π2 .
♢
In the next section (Corollary 7.2.9), we will see that the preceding
proposition remains valid if only one of the mappings π1 and π2 is open,
provided both spaces X1 and X2 are locally compact Hausdorff.
Example 7.1.15 Let ∼ be the equivalence relation on Q which identifies
Z to a point and = denote the identity relation on Q. The relation P
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Elements of Topology
on Q × Q, given by (x, y)P (x′ , y ′ ) iff x, x′ ∈ Z and y = y ′ , is obviously
the relation ∼ × =. We show that the canonical mapping
ϕ : (Q × Q)/P → (Q/ ∼) × Q,
ϕ ([(x, y)]) = (π(x), y),
where π : Q → Q/ ∼ is the projection map, is not a homeomorphism.
For each positive integer n, choose an irrational number αn such that
αn → 0. Let Gn ⊆ [n, n + 1] × R be an open set such that Gn meets
{n, n + 1} × R in the points (n, αn ) and (n + 1, αn+1 ) only (see Figure
7.6).
R
—
Gn
(n;®n )
(n+1;®n+1)
Q
………………………………………………………………………………………
(n;y)
•
•
(n+1;y)
•
(n+2;y)
FIGURE 7.6: Proof of Example 7.1.15.
Set Fn = Gn ∩ (Q × Q) for every n. Then {F
∪n } is a locally finite family
of closed subsets of Q × Q, and hence A = Fn is closed. Since A does
not meet Z × Q, it can be regarded as a closed subset of (Q × Q)/P.
But, B = ϕ(A) is not closed in (Q/ ∼) × Q. The point p = ([0], 0) does
not belong to B, for its preimage [(0, 0)] is not in A. We contend that
p is a limit point of B. The sets of the form∪N = O × ((−δ, δ) ∩ Q),
where the open set O ⊆ Q/ ∼ is the image of n ((n − rn , n + rn ) ∩ Q)
(rn small) under π, constitute a nbd basis of p. For the integers n
satisfying αn < δ, the intersection of (n − rn , n + rn ) × (−δ, δ) with Gn
is a nonempty open set so that ((n − rn , n + rn ) × (−δ, δ)) ∩ Fn ̸= ∅.
The image of this set under π × 1 is contained in N ∩ B, and p ∈ B.
We end this section by describing a useful construction of continuous functions between quotient spaces. Let X and Y be spaces
TOPOLOGICAL CONSTRUCTIONS
171
with the equivalence relations ∼ and ≃, respectively. Let f : X → Y
be a relation-preserving function, that is, it satisfies the condition:
f (x) ≃ f (x′ ) whenever x ∼ x′ . Then the equivalence class of f (x)
is independent of the choice of the representative x of [x], for [x] = [x′ ]
in X/ ∼ implies that x ∼ x′ which, in turn, implies that f (x) ∼ f (x′ ).
So there is a function f¯ : (X/ ∼) → (Y / ≃) defined by f¯ ([x]) = [f (x)].
If
( π ): X → X/ ∼ and ϖ : Y → Y / ≃ are the natural projections, then
f¯π (x) = (ϖf ) (x) for all x ∈ X, by the definition of f¯. Accordingly,
we have the commutative diagram
X
f
π
- Y
ϖ
?
X/ ∼
f¯
?
- Y/≃
that is, f¯π = ϖf . Since π is surjective, f¯ is uniquely determined by
this equation. Moreover, if f is continuous, then so also is f¯. Thus, we
have established the following.
Proposition 7.1.9 If f : X → Y is a relation-preserving continuous
map, then the mapping f¯ : (X/ ∼) → (Y / ≃), [x] 7→ [f (x)], is continuous, where X/ ∼ and Y / ≃ have the quotient topologies.
Exercises
1. • In the unit interval I, consider the equivalence relation such that
x ∼ y if both are less than 1/2 or both are ≥ 1/2. Show that I/ ∼
is homeomorphic to the Sierpinski space. Also, prove that the quotient
map I → I/ ∼ is neither open nor closed.
2.
(a) Does there exist an equivalence relation on I so that the quotient
space is homeomorphic to a two-point discrete space?
(b) Consider the equivalence relation ∼ on I defined by x ∼ y if both
x and y are either rational or irrational. Determine the quotient
space I/ ∼.
3. Find the cardinality and topology of the quotient space R/ ∼ in Ex.
7.1.10
4. Show that (a) RP0 is the one-point space, and (b) RP1 ≈ S1 .
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Elements of Topology
5. Is the quotient space in Ex. 7.1.13 T2 ?
6. Let X be the euclidean plane and A ⊆ X be the x-axis. Show that the
projection map X → X/A is closed, but X/A is not first countable.
7. Give an example of a sequential space which does not satisfy the axiom
of first countability.
8. Let A = {1/n : n ∈ N} and π : I → I/A be the projection map. Let
G = [0, 1/2] ∪ {1}. Show that G is a saturated nbd of 0, but π(G) is not
a nbd of π(0).
9. • On the interval J = [−1, 1] ⊂ R, consider the equivalence relation ∼
which identifies x with −x for −1 < x < 1. Show:
(a) The projection map π : J → J/ ∼ is open, but the graph G(∼) is
not closed in J × J.
(b) The quotient space J/ ∼ is T1 but not T2 .
10. Let X be a space with an equivalence relation ∼. Prove that the projection map X → X/ ∼ is closed if and only if for each element [x] of
X/ ∼ and each open set U ⊆ X containing the equivalence class [x],
there is a saturated open set V ⊆ X such that [x] ⊆ V ⊆ U . (Such an
equivalence relation is called upper semicontinuous.)
11. For each integer n = 1, 2, . . ., let U be the subspace of R2 consisting of
circles (x − n)2 + y 2 = n2 , X be the subspace of R2 which is the union
of circles (x − 1/n)2 + y 2 = (1/n)2 , and Y denote the quotient space
R/Z. Show that no two of U, X and Y are homeomorphic.
12. Let X = I × N ⊂ R2 and A = {0} × N. Find a continuous bijection from
the quotient space X/A on to the subspace Y ⊂ R2 which consists of
the closed line segments joining the origin to the points (1, 1/n), n ∈ N.
Are they homeomorphic?
13. Let ∼ be an equivalence relation on a locally compact space X such
that the quotient map π : X → X/ ∼ is closed and each equivalence
class is compact. Prove that X/ ∼ is also locally compact.
14. • Show that the quotient space R/Z is not locally compact, although
the quotient map R → R/Z is closed.
15. Show that the quotient space X/A in Example 7.1.12 is connected but
not path-connected.
16. Find an example showing that if A ⊂ X is not open or closed, then
X − A need not be homeomorphic X/A − [A].
17. Show that (I × I)/∂(I × I) is homeomorphic to S2 .
18. Describe the space obtained by identifying the boundary (edge) of the
Möbius band to a point.
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173
19. Let X be a compact Hausdorff space. If A is closed in X, prove that
X/A is homeomorphic to the one-point compactification of X − A.
20. Prove that the one-point compactification of the open Möbius band
(i.e., without its boundary) is homeomorphic to RP2 .
21. Let A ⊆ B be subsets of a space X and A closed in X. Prove that B/A
is a subspace of X/A.
22. Let π : X → X/ ∼ be open or closed. If A ⊆ X meets every equivalence
class of X nontrivially, show that the induced map A/ ∼ → X/ ∼ is a
homeomorphism.
23. Show that each of the quotient spaces
(
)
(a) Rn+1 − {0} / ∼, where x ∼ y if there exists 0 ̸= λ ∈ R with
y = λx, and
(b) Dn / ∼, where x ∼ y if x = −y for x, y ∈ Sn−1 = ∂Dn
is homeomorphic to RPn .
24. Show that RP2 is homeomorphic to (I × I)/ ∼, where (s, 0) ∼ (1 − s, 1)
and (0, t) ∼ (1, 1 − t), for all s, t ∈ I.
7.2
Identification Maps
The quotient spaces can often be described more conveniently by
means of “identification maps.”
Definition 7.2.1 Let X and Y be spaces. A function f : X → Y is
called an identification map or a quotient map if it is surjective and Y
has the largest topology such that f is continuous.
Clearly, a surjection f : X → Y is an identification map if and only
if (a set U is open in Y ⇔ f −1 (U ) is open in X). If we substitute closed
sets for open sets, we obtain another equivalent definition. When f is an
identification map, the space Y is referred to as an identification space
of f. By the definition of the quotient topology, the natural projection
of a space onto its quotient space is an identification map.
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Elements of Topology
Given a function f of a space X onto a set Y , the procedure of
topologising the quotient space of an equivalence relation can be used
to construct a topology on Y so that f is an identification map. The
family T(f ) of all subsets U of Y such that f −1 (U ) is open in X
is a topology on Y because the inverse of an intersection (or union)
of members of T(f ) is the intersection (resp. union) of the inverses.
Obviously, f is an identification map when Y is assigned the topology
T(f ). This topology for Y is called the identification topology or the
quotient topology determined by f .
Note that not every continuous surjection is an identification map,
for example, the identity map of R where the domain is assigned the
discrete topology and the range has the usual topology. The following
result provides a useful class of identification maps.
Proposition 7.2.2 A closed (or open) continuous surjection is an
identification map.
We leave the straightforward proof to the reader.
Corollary 7.2.3 Let f : X → Y be a continuous surjection. If X is
compact and Y is Hausdorff, then f is an identification.
Proof. This is immediate from the preceding proposition and Corollary
6.1.11.
♢
Given a function f : X → Y , a subset A ⊆ X is called f -saturated
if A = f −1 f (A). The following result is essentially Proposition 7.1.3.
Theorem 7.2.4 Let f : X → Y be an identification map and, for
some space Z, let h : X → Z be a continuous map which is constant
on each set f −1 (y), y ∈ Y . Then there is a continuous map g : Y → Z
such that gf = h. Also, g is an open (closed) map if and only if h(U )
is open (closed) for every f -saturated open (closed) set U ⊆ X.
Proof. Given y ∈ Y , there is x ∈ X such that f (x) = y. We set
g(y) = h(x) ⇔ f (x) = y. If f (x) = f (x′ ), our hypothesis implies that
h(x) = h(x′ ). Thus g is a single-valued function satisfying gf (x) =
h(x), for all x ∈ X. By the proof of Proposition 7.1.3, it is immediate
that g is continuous. For the last statement, we note that g(V ), for each
open (resp. closed) set V ⊆ Y , is the image under h of an f -saturated
open (resp. closed) subset of X.
♢
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175
Now, we show that an identification map f differs from a quotient map by a homeomorphism; this justifies the use of the terminology in Definition 7.2.1. Suppose that f : X → Y is a function
between topological spaces. Then there is an equivalence relation on
X: x ∼ x′ if f (x) = f (x′ ). The equivalence classes of X under this relation are nonempty sets f −1 (y), y ∈ Y , and the associated quotient
e will be referred to as the decomposition space
space, denoted by X,
e → Y , which
of f . The function f induces an injective function f˜ : X
maps the equivalence class [x] into f (x), and there is a factorisation
f = f˜ ◦ π that is, the following triangle
f
X
π
f˜ ? e - Y
3
X
commutes, where π is the natural projection. If f is continuous, then it
is clear from Theorem 7.2.4 that f˜ is a continuous injection. A necessary
and sufficient condition for f˜ to be a homeomorphism is given by
Theorem 7.2.5 Let X and Y be spaces and f : X → Y be a function.
e → Y is a homeomorphism if and only if f is
Then the mapping f˜ : X
an identification.
Proof. Assume that f is an identification. Then, as seen above, f˜ is a
e is open, then f −1 f˜(O) = π −1 (O) is
continuous bijection. Also, if O ⊆ X
˜
open in X. By our assumption, f (O) is open in Y , and f˜ is a homeomorphism. Conversely, assume that f˜ is a homeomorphism. Then f = f˜◦ π
is obviously a continuous surjection. Now suppose that for( U ⊆ Y,
)
e so U = f˜ f˜−1 (U )
f −1 (U ) is open in X. Then f˜−1 (U ) is open in X,
is open in Y. Thus f is an identification.
♢
It follows from the preceding theorem that the identification space
Y of a map f can always be regarded as a quotient space of its
domain via the equivalence relation whose equivalence classes are
f −1 (y), y ∈ Y. The preceding theorem is often used to compare different descriptions of the same space.
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Elements of Topology
It is easily seen that the composition of two identification maps is
an identification map. In particular, we see that a quotient space of a
quotient space of X is a quotient space of X; in other words, taking
quotient is a transitive operation:
Let ∼ be an equivalence relation on a space X and ≃ be an equivalence relation on Y = X/ ∼. If π : X → X/ ∼ and ϖ : Y → Y / ≃ are the
π
ϖ
projection maps, then the composition X −−→ Y −−→ Y / ≃ is an identification map. Therefore Y / ≃ is homeomorphic to the decomposition
space of the composition ϖ ◦ π.
Next, we see a characteristic of the identification maps. If f : X →
Y is an identification map and g : Y → Z is a function such that
gf : X → Z is continuous, then g is continuous, by the proof of
Theorem 7.2.4. Conversely, suppose that f : X → Y is a continuous surjection such that for any space Z the continuity of a function
g : Y → Z follows from that of the composition gf : X → Z. Let
e of f, π : X → X
e be the projection
Z be the decomposition space X
˜
e
map, and f : X → Y be the continuous bijection, [x] 7→ f (x). Then
f˜−1 ◦f = π is continuous. By our hypothesis, f˜−1 is continuous. Theree → Y is a homeomorphism, and hence an identification map.
fore f˜ : X
Being a composition of two identification maps, f is an identification
map. Thus, we have established
Theorem 7.2.6 Let f : X → Y be a continuous surjection. Then f
is an identification if and only if, for any space Z, the continuity of a
function g : Y → Z follows from that of gf : X → Z.
Let f : X → Y be an identification map, and A be a subspace of
X. Then, as shown by Example 7.1.14, the map g : A → f (A) defined
by f is not necessarily an identification map. But the desired result
holds good in certain situations.
Proposition 7.2.7 Let f : X → Y be an identification map, and
A ⊆ X be an f -saturated set. Then each of the following conditions
implies that g : A → f (A), a →
7 f (a), is an identification map.
(a) f is open (or closed).
(b) A is open (or closed).
Proof. If V ⊆ Y and U = f (A) ∩ V, then g −1 (U ) = A ∩ f −1 (V ). This
shows that g −1 (U ) is open in A whenever U is open in f (A).
TOPOLOGICAL CONSTRUCTIONS
177
Conversely, let U be a subset of f (A) such that g −1 (U ) is open in
A. We need to prove that U is open in f (A). Observe that g −1 (U ) =
f −1 (U ), since A is f -saturated.
Case (a): Suppose that f is open. We have f −1 (U ) = A ∩ G for
some open set G ⊆ X. We observe that U = f (A) ∩ f (G), which
implies that U is open in f (A), since f is open. If f (x) ∈ f (A), then
x ∈ f −1 f (A) = A; consequently, f (A) ∩ f (G) ⊆ f (A ∩ G). The reverse
inclusion is obvious, and the equality holds. The case f is closed is
proved similarly.
Case (b): If A is open in X, then f −1 (U ) is open in X. Since f is
an identification, U is open in Y and, therefore, in f (A). The proof for
closed A is similar.
♢
More generally, it can be seen that g is an identification map if each
g-saturated set which is open (resp. closed) in A is the intersection of
A with an f -saturated set open (resp. closed) in X. The converse also
holds.
Finally, in this section, we establish a useful theorem, due to
J.H.C. Whitehead, for quotient topologies in the cartesian products.
If f : X → Y and g : Z → T are identification maps, then we have
already seen that the mapping f × g : X × Z → Y × T may not
be an identification. The fundamental theorem in this direction is the
following.
Theorem 7.2.8 Let f : X → Y be an identification map, and let Z
be locally compact Hausdorff. Then f × 1 : X × Z → Y × Z is an
identification map.
Proof. By Theorem 7.2.6, it suffices to prove that every function g
from Y × Z into a space T is continuous for which the composition
f ×1
g
X × Z −−−→ Y × Z −→ T is continuous. Write h = g ◦ (f × 1), and let
O ⊆ T be open. Choose a point (y0 , z0 ) in g −1 (O), and find x0 ∈ X so
that f (x0 ) = y0 . Then h (x0 , z0 ) ∈ O. By the continuity of h, there exist
open sets U ⊆ X and V ⊆ Z such that (x0 , z0 ) ∈ U × V ⊆ h−1 (O).
Since Z is locally compact Hausdorff, V contains a compact nbd N
of z0 . Note that N is closed in Z and U × N ⊆ h−1 (O). Let W =
{y ∈ Y |g(y, z) ∈ O for all z ∈ N }. Then y0 ∈ W and W ×N ⊆ g −1 (O).
We show that f −1 (W ) is open in X, which implies that W is open in
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Elements of Topology
Y . Clearly,
f −1 (W ) = {x ∈ X|h(x, z) = g(f (x), z) ∈ O for all z ∈ N }
= {x ∈ X|h ({x} × N ) ⊆ O} .
So we have
X − f −1 (W ) = {x ∈ X|h ({x} × N ) ∩ (T − O) ̸= ∅}
(
)
= p h−1 (T − O) ∩ (X × N ) ,
where p : X × N → X is the projection map. Since N is compact, p is
closed and, therefore, X − f −1 (W ) is closed. This completes the proof.
♢
Corollary 7.2.9 Let f : X → Y and g : Z → T be identification
maps. If X and T are locally compact Hausdorff, then f × g is an
identification.
Proof. Obviously, the mapping f ×g : X ×Z → Y ×T is the composition
1×g
f ×1
X × Z −−−→ X × T −−−→ Y × T . By the preceding theorem, both
1X ×g and f ×1T are identification maps if X and T are locally compact
Hausdorff. Hence f × g is an identification.
♢
Exercises
1. Prove that an injective identification map is a homeomorphism.
2.
(a) • Let f : X → Y be a continuous map. If f admits a continuous
right inverse, prove that it is an identification map.
(b) • Let f : X → Y and g : Y → Z be continuous surjections such
that gf is an identification. Show that g is an identification.
3. Prove that the function f¯ in Proposition 7.1.9 is an identification map
if f is so.
4. Prove that an identification map f : X → Y is closed (resp. open) if
and only if the saturation of every closed (resp. open) subset of X is
closed (resp. open).
5. Let χ : [0, 2] → {0, 1} be the characteristic function of the set [1, 2].
Determine the identification topology on {0, 1} induced by χ.
6. Determine the quotient space R2 / ∼, where the equivalence relation ∼
is given by
TOPOLOGICAL CONSTRUCTIONS
179
(a) (x, y) ∼ (x′ , y ′ ) if y = y ′ ;
(b) (x, y) ∼ (x′ , y ′ ) if x + y 2 = x′ + y ′2 ; and
(c) (x, y) ∼ (x′ , y ′ ) if x2 + y 2 = x′2 + y ′2 .
7. Describe each of the following spaces:
(a) The cylinder with each of its boundary circles identified to a point.
(b) The torus with the subset consisting of a meridianal and a longitudinal circle identified to a point.
8. Let X = [0, 1] ∪ [2, 3] and Y = [0, 2] have the subspace topology from
the real line R. Show that the mapping f : X → Y defined by
{
x
for 0 ≤ x ≤ 1, and
f (x) =
x − 1 for 2 ≤ x ≤ 3
is an identification map. Is it open? Is the restriction of f to the subspace
A = [0, 1] ∪ (2, 3] an identification map?
9. Let X be the closed topologist’s sine curve (see Ex. 3.1.5) and f : X → I
be the projection map (x, y) 7→ x. Show that f is an identification map
which is not open.
10. Let f : R → [−1, 1] be the function f (x) = sin 1/x for x ̸= 0 and
f (0) = 0. Give [−1, 1] the identification topology induced by f . Prove
that the subspace [−1, 0)∪(0, 1] has its usual topology but the only nbd
of 0 is [−1, 1].
11. Show that the quotient space of a space X by the equivalence relation
in Exercise 3.2.1 is totally disconnected.
12. Let f : X → Y be an identification such that f −1 (y) is connected for
each y ∈ Y . Prove that X is connected ⇔ Y is connected.
13. Let f : X → Y be an identification map. If Y is T2 , show that the
e of f is also T2 .
decomposition space X
14. Let X be a compact Hausdorff space, and f : X → Y be an identification. Show that Y is Hausdorff ⇔ the set {(x1 , x2 )|f (x1 ) = f (x2 )} is
closed in X × X.
180
7.3
Elements of Topology
Cones, Suspensions and Joins
In this and the subsequent sections, we will employ the operations
of forming cartesian products and taking quotients to develop several
interesting techniques of generating new spaces from old ones. These
methods of generating new spaces are of great importance in the study
of algebraic topology.
Definition 7.3.1 For a space X, the quotient space (X × I) / ∼,
where the equivalence relation ∼ is given by (x, 0) ∼ (x′ , 0) for all
x, x′ ∈ X, is called the cone on X and is usually denoted by CX.
Notice that the cone on X can also be described as the quotient
space (X × I) / (X × {0}). The identified point [X × {0}] is called the
apex or vertex of the cone. The equivalence class of (x, t) ∈ X × I will
be denoted by [x, t]. Intuitively, the cone on a space X is obtained from
the cylinder X × I (over X) by pinching the bottom X ×{0} to a single
point.
X
•
X£I
vertex
FIGURE 7.7: Cone on X.
The mapping x 7→ [x, 1] is an embedding of X into CX; so X can be
regarded as a subspace of CX. We refer to the subspace X ⊂ CX as
the base of the cone. If X ⊆ Rn ⊂ Rn+1 and v ∈ Rn+1 − Rn , then the
subspace
T X = {tx + (1 − t) v|x ∈ X, and 0 ≤ t ≤ 1}
is referred to as the geometric cone on X. Clearly, T X is obtained by
joining each point of X to v by a line segment. In general, CX has
TOPOLOGICAL CONSTRUCTIONS
181
more open sets than the geometric cone T X. Yet, the cone CX over
a space X is intuitively considered as the union of all line segments
joining points of X to a point outside X.
If f : X → Y is a continuous function, then the function (x, t) 7→
(f (x) , t) of X × I into Y × I is continuous and relation preserving.
Hence, there is an induced continuous map Cf : CX → CY taking a
point [x, t] into [f (x) , t].
Example 7.3.1 The cone CSn over the n-sphere Sn is homeomorphic
to the disc Dn+1 . To see this, consider the function f : Sn × I →
Dn+1 defined by f (x, t) = tx. This is clearly a continuous surjection,
which is constant on each equivalence class. Therefore, f induces a
continuous bijection f˜ : CSn → Dn+1 . Since Sn × I is compact and
Dn+1 is Hausdorff, the induced mapping f˜ is a homeomorphism.
Definition 7.3.2 Given a space X, the quotient space (X × I) / ∼,
where the equivalence relation ∼ is defined by (x, 0) ∼ (x′ , 0) , (x, 1) ∼
(x′ , 1) for all x, x′ ∈ X, is called the suspension of X. We denote the
the suspension of X by ΣX.
The identified points [X × {0}] and [X × {1}] are the vertices of
ΣX. The subspace [X × { 21 }] is homeomorphic to X and is referred to
as the base of the suspension.
X£I
X
FIGURE 7.8: Suspension of X.
The space ΣX can be alternatively described as CX/X. Intuitively,
the suspension of a space X is obtained from the cylinder X × I by
pinching the bottom X ×{0} to a point and the top X ×{1} to another
point. As in the case of the cone over a space, a continuous function
f : X → Y induces a continuous function Σf : ΣX → ΣY , since it is
constant on each equivalence class.
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Elements of Topology
Example 7.3.2 ΣSn is homeomorphic to Sn+1 . The mapping f : Sn ×
I → Sn+1 defined by f (x, t) = (x sin πt, cos πt) is a continuous surjection. Hence it induces a continuous bijection f˜ : ΣSn → Sn+1 . Since
Sn × I is compact and Sn+1 is Hausdorff, the induced mapping f˜ is a
homeomorphism.
Definition 7.3.3 Let X and Y be spaces. Their join X ∗ Y is the
quotient space of X × Y × I by the equivalence relation (x, y, 0) ∼
(x, y ′ , 0) for all y, y ′ ∈ Y and (x, y, 1) ∼ (x′ , y, 1) for all x, x′ ∈ X.
The relation ∼ identifies each of the sets {x} × Y × {0}, and X ×
{y} × {1}, where x ∈ X and y ∈ Y , to a point. The mappings
x 7→ {x} × Y × {0}
and
y 7→ X × {y} × {1}
are embeddings of X and Y into X ∗ Y , respectively. Their images are
usually identified with X and Y , and called the bases of the join. Then,
intuitively speaking, X ∗ Y can be regarded as the space consisting of
X, Y and all line segments joining each x ∈ X to every y ∈ Y , where
no two of the segments have interior points in common (see Figure 7.9).
Y
X×Y×t
I
X
Y
X×Y×I
X
FIGURE 7.9: Join of X and Y .
Example 7.3.3 X ∗ {p} ≈ CX, where {p} is a one-point space. Obviously, the mapping (x, p, t) 7→ (x, t) is a homeomorphism of X ×{p}×I
with X × I. By passing to the quotients, we obtain X ∗ {p} ≈ CX.
Example 7.3.4 For any space X, X ∗ S0 ≈ ΣX. The mapping f : X ×
S0 ×I → X ×I defined by f (x, −1, t) = (x, (1 − t) /2) and f (x, +1, t) =
(x, (1 + t) /2), x ∈ X and t ∈ I, is a continuous closed surjection. So,
TOPOLOGICAL CONSTRUCTIONS
183
the composition of f with the natural projection X × I → ΣX is an
identification ϕ : X × S0 × I → ΣX. The decomposition space of ϕ is
X ∗ S0 , and hence X ∗ S0 ≈ ΣX.
The join operation ∗ is commutative, in the sense that X ∗ Y ≈ Y ∗
X. The correspondence X ×Y ×I ↔ Y ×X ×I, (x, y, t) ↔ (y, x, 1 − t),
induces the required homeomorphism. However, the join operation is
not associative with this topology. Also, this topology for the join is
not convenient for maps into the join. Therefore, we introduce another
topology for the join, which enjoys these properties and agrees with
the aforesaid topology for a wide class of spaces.
Let ∆ be the subspace {(1 − t, t) |0 ≤ t ≤ 1} of R2 . Then I ≈ ∆ under the homeomorphism t 7→ (1 − t, t). Consider the equivalence relation ∼ on X ×Y ×∆ generated by (x, y, 1, 0) ∼ (x, y ′ , 1, 0), (x, y, 0, 1) ∼
(x′ , y, 0, 1). The canonical homeomorphism (x, y, t) ↔ (x, y, 1 − t, t) between X × Y × I and X × Y × ∆ clearly preserves the equivalence relations. Hence there is a homeomorphism of X ∗ Y onto (X × Y × ∆) / ∼
given by the mapping [x, y, t] 7→ [x, y, 1 − t, t]. For notational convenience, we will denote the element [x, y, 1 − t, t] by (1 − t) x + ty, and
write the same symbol X ∗ Y for the space (X × Y × ∆) / ∼. It is obvious that (1 − t′ ) x′ + t′ y ′ = (1 − t) x + ty if t = t′ , x = x′ for t ̸= 1
and y = y ′ for t ̸= 0. In particular, note that 1x + 0y = 1x + 0y ′ even
if y ̸= y ′ , and 0x + 1y = 0x′ + 1y even if x ̸= x′ . With this definition of
X ∗ Y , we have four functions
ρ : X ∗ Y → I,
(1 − t) x + ty 7→ 1 − t;
σ : X ∗ Y → I,
(1 − t) x + ty 7→ t;
ξ : ρ−1 (0, 1] → X,
(1 − t) x + ty 7→ x;
ψ : σ −1 (0, 1] → Y,
(1 − t) x + ty 7→ y.
These functions are referred to as the coordinate functions.
Definition 7.3.4 The coarse topology for X ∗ Y is the smallest topology which makes the four coordinate functions ρ, σ, ξ and ψ continuous,
where the domains of ξ and ψ have the relative topologies.
In other words, the coarse topology for X ∗ Y is generated by the
subbase consisting of subsets: the inverse images under ρ and σ of the
open subsets of I, the inverse images under ξ of open subsets of X,
and the inverse images under ψ of the open subsets of Y . It follows
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Elements of Topology
from Theorem 2.1.3 that a function f : Z → X ∗ Y is continuous
in the coarse topology for X ∗ Y if and only if its composition with
the coordinate functions are continuous. In particular, we see that the
natural projection X × Y × ∆ → X ∗ Y is continuous in this topology
for X ∗ Y . Therefore, the coarse topology for X ∗ Y is indeed coarser
than the quotient topology. This justifies the choice of the terminology
for the new topology on X ∗ Y . By contrast, the quotient topology on
X ∗ Y is referred to as the “fine topology”. Observe that x 7→ 1x + 0y,
and y 7→ 0x + 1y are embeddings of X and Y , respectively, into X ∗ Y
with the coarse topology, too. Accordingly, X and Y can be identified
with the corresponding subspaces of X ∗ Y .
The fine topology and the coarse topology for X ∗ Y are identical
for many nice spaces, for example, if X and Y are compact Hausdorff.
To see this, we first prove the following.
Proposition 7.3.5 If X and Y are Hausdorff spaces, then so is X ∗ Y
(in both the topologies).
Proof. Suppose that X and Y are Hausdorff spaces, and let ζ =
(1 − t) x + ty and ζ ′ = (1 − t′ ) x′ + t′ y ′ be distinct elements of X ∗ Y .
Clearly, it suffices to produce disjoint nbds of ζ and ζ ′ in the coarse
topology for X ∗ Y . If t ̸= t′ , then we find disjoint open sets V and V ′
in I with t ∈ V and t′ ∈ V ′ . By the definition of the coarse topology for
X ∗Y, σ −1 (V ) and σ −1 (V ′ ) are open subsets of X ∗Y. It is obvious that
σ −1 (V ) and σ −1 (V ′ ) are disjoint and contain ζ and ζ ′ , respectively. If
t = t′ ̸= 0, then y ̸= y ′ . Since Y is T2 , there exist disjoint open nbds U
and U ′ of y and y ′ , respectively, in Y. Then ψ −1 (U ) and ψ −1 (U ′ ) are
disjoint open nbds of ζ and ζ ′ , respectively. A similar argument applies
in the case t = t′ = 0, for x ̸= x′ .
♢
Now, suppose that X and Y are compact Hausdorff spaces. Then
the product X × Y × ∆ is also compact Hausdorff, since ∆ ≈ I. By
Corollary 7.2.3, the projection map ν : X × Y × ∆ → X ∗ Y (with the
coarse topology), (x, y, 1 − t, t) 7→ (1 − t) x + ty, is an identification.
It follows from Theorem 7.2.5 that the decomposition space of ν is
homeomorphic to its range space. Therefore, the fine topology and the
coarse topology for X ∗ Y coincide.
The commutativity of the join operation remains valid in the coarse
topology also. Moreover, we have the following.
Proposition 7.3.6 The join operation ∗ is associative in the coarse
TOPOLOGICAL CONSTRUCTIONS
185
topology, that is, there is a natural homeomorphism (X ∗ Y ) ∗ Z ≈
X ∗ (Y ∗ Z) for all spaces X, Y and Z.
Proof. Consider X ∗ Y with the coarse topology. The mapping ηXY :
X ∗ Y → CX × CY , defined by ηXY ((1 − t) x + ty) = ([x, 1 − t], [y, t]),
−1
is a continuous injection. The compositions of ηXY
: im (ηXY ) → X ∗ Y
−1
with the coordinate functions are obviously continuous. Therefore ηXY
is continuous, and hence ηXY is an embedding. For [x, r] ∈ CX, [y, s] ∈
CY and t ∈ I, define t ([x, r], [y, s]) = ([x, tr], [y, ts]). Then we have a
continuous injection
η(XY )Z : (X ∗ Y ) ∗ Z → CX × CY × CZ
given by η(XY )Z ((1 − t) u + tz) = ((1 − t) ηXY (u) , [z, t]), where u ∈ X∗
(
)
Y . Obviously, im η(XY )Z = {([x, r], [y, s], [z, t]) |r + s + t = 1}. The
function
(
)
−1
η(XY
: im η(XY )Z → (X ∗ Y ) ∗ Z
)Z
takes the element ([x, r], [y, s], [z, t]) into (1 − t) u + tz, where u =
r
s
1−t x + 1−t y for t ̸= 1, and u ∈ X ∗ Y is arbitrary for t = 1. The com−1
with the coordinate functions on (X ∗ Y ) ∗ Z are
positions of η(XY
)Z
−1
is continuous. It follows that η(XY )Z
continuous, and therefore η(XY
)Z
is an embedding. Similarly, the function
η(Y Z)X : (Y ∗ Z) ∗ X → CY × CZ × CX
is an embedding. The canonical
(
) homeomorphism
(
) CX × CY × CZ ≈
CY ×CZ×CX maps im η(XY )Z onto im η(Y Z)X . Hence (X ∗ Y )∗Z ≈
(Y ∗ Z) ∗ X ≈ X ∗ (Y ∗ Z).
♢
The preceding proposition enables us to discuss the multiple join
of a finite number of spaces, with the coarse topology, without using
parentheses. Note that if the spaces X, Y and Z are compact Hausdorff,
then (X ∗ Y ) ∗ Z ≈ X ∗ (Y ∗ Z) in the fine topology, too.
As an immediate consequence of this result, we see that X ∗ Sn ≈
X ∗ ΣSn−1 ≈ X ∗ Sn−1 ∗ S0 ≈ X ∗ S0 ∗ Sn−1 ≈ ΣX ∗ Sn−1 , since
X ∗ S0 ≈ ΣX (ref. Ex. 7.3.4). By induction, we deduce that X ∗ Sn is
homeomorphic to the (n + 1)-fold suspension of X. In particular, we
have Sm ∗ Sn ≈ Sm+n+1 .
The next simple result helps us to find the join in some cases.
Proposition 7.3.7 With the fine topology for X ∗ Y , C (X ∗ Y ) ≈
CX × CY for any two spaces X and Y.
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Elements of Topology
Proof. Clearly, there is a homeomorphism h : ∆ × I → I × I defined by
{
(s, st/(1 − t))
for 0 ≤ t ≤ 1/2,
h ((1 − t, t), s) =
(s(1 − t)/t, s)
for 1/2 ≤ t ≤ 1.
Z
c
b
a
Y
d
h
X
c
1
d
b
a
0
Δ£I
1
I£I
FIGURE 7.10: The homeomorphism h in the proof of Proposition 7.3.7.
Consequently, the composition
1X ×g×1Y
1 ×1 ×h
X × Y × ∆ × I −−X−−−Y−−→ X × Y × I × I −−−−−−−→ X × I × Y × I
is a homeomorphism, where g is the homeomorphism
Y × I ≈ I × Y,
(y, u) 7→ (u, y).
Denote this composite by η. Then we have
{
(x, s, y, st/(1 − t))
η(x, y, 1 − t, t, s) =
(x, s(1 − t)/t, y, s)
for 0 ≤ t ≤ 1/2,
for 1/2 ≤ t ≤ 1.
If p : X × I → CX and q : Y × I → CY are the natural projections,
then the composite
η
p×q
θ :X ×Y ×∆×I −
→ X × Y × I × I −−→ CX × CY
is an identification. It is easily checked that the decomposition space
of θ is C(X ∗ Y ). Hence C (X ∗ Y ) ≈ CX × CY .
♢
TOPOLOGICAL CONSTRUCTIONS
187
Observe that θ maps X ∗ Y × {1} ⊂ C(X ∗ Y ) homeomorphically
onto (CX × Y × {1}) ∪ (X × {1} × CY ). Thus, with the fine topology,
X ∗ Y ≈ (CX × Y ) ∪ (X × CY ). Of course, this result and the preceding proposition are valid in the coarse topology also when the spaces
involved are compact and Hausdorff.
Example 7.3.5 Dm ∗ Dn is homeomorphic to Dm+n+1 . By Example
the preceding
result,
D)m ∗ Dn ≈
( m−1 7.3.3 ) and( n−1
)
( m−1 we haven−1
S(
∗ {pt} ∗ S ) ∗ {pt} ≈ S
∗ {pt} ∗ S
∗ {pt} ≈
C Sm−1 ∗ {pt} ∗ Sn−1 ≈ CSm−1 × C{pt} × CSn−1 ≈ Dm × I × Dn ≈
Dm+n+1 .
Exercises
{
}
1. Let X be the subspace (n, 0) ∈ R2 |n = 1, 2, . . . and T X be the geometric cone on X with vertex v = (0, 1). Show that there is a continuous
bijection CX → T X, but CX ̸≈ T X.
2. Let X ⊆ Rn be compact. Prove that the cone CX is homeomorphic to
the geometric cone T X on X.
3. If X is T2 , show that CX is also T2 .
4. Let X be a compact Hausdorff space. Show that the one-point compactification of X × (0, 1] is homeomorphic to the cone CX on X.
5. Show that the cone CX and the suspension ΣX of any space X are
path-connected.
6. Prove that CX can be regarded as a closed subspace of ΣX.
7. If the cone CX over a space X is locally compact Hausdorff, show that
X must be compact.
8. Prove that X ∗ Y is path-connected for any two spaces X and Y .
9. Show that (X ∗ Y ) ∗ Z ≈ X ∗ (Y ∗ Z) in the fine topology if (a) If X
and Y , or (b) Y and Z) are compact Hausdorff, or (c) X and Z are
locally compact Hausdorff.
10. If f : X → X ′ and g : Y → Y ′ are continuous, show that (a) there is a
continuous map f ∗ g : X ∗ Y → X ′ ∗ Y ′ in either topology for joins, (b)
if f and g are homeomorphisms, then so is f ∗ g.
11. If X ≈ X ′ and Y ≈ Y ′ , show that X ∗ Y ≈ X ′ ∗ Y ′ .
12. Let A ⊆ X and B ⊆ Y be subspaces.
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Elements of Topology
(a) Show that A ∗ B can be identified with a subspace of X ∗ Y in the
coarse topology.
(b) If both A ⊆ X and B ⊆ Y are closed, then show that A ∗ B can
be identified with a closed subspace of X ∗ Y in the fine topology.
13. Let Y be a compact T2 -space, and Ai , i = 1, 2, . . . , n, be compact subsets
of a Hausdorff space X. Prove:
∪n
∪n
(a) ( 1 Ai ) ∗ Y ≈ 1 (Ai ∗ Y ).
∩n
∩n
(b) ( 1 Ai ) ∗ Y ≈ 1 (Ai ∗ Y ).
14. Show that the quotient space of X ∗ Y (with the fine topology) obtained
by identifying its bases to points is homeomorphic to Σ (X × Y ).
7.4
Wedge Sums and Smash Products
There is a simple way of topologising the disjoint union of a family
of topological spaces. This is dual to the formation of cartesian product
and, together with the previously discussed constructions, leads to several interesting techniques of forming new spaces from old ones. In this
section, we discuss the notion of “topological sum” and the technique
of gluing two or more spaces together at prespecified points. We will
also study the concept of “Smash Product.”
Suppose that a space Z is the union of disjoint subsets X and Y .
Then, for every U ⊆ Z, U = (U ∩ X) ∪ (U ∩ Y ). So, if both X and Y
are open in Z, then the topology of Z is determined by the topologies
of X and Y , for a set U is open in Z if and only if U ∩ X is open in X
and U ∩ Y is open in Y . This situation is generalised in
Definition 7.4.1 Let Y be a space,
and {Xα } be a family of disjoint
∪
subspaces of Y such that Y = Xα . ∑
We say that Y is the topological
sum of the spaces Xα , and write Y =
Xα , if a subset U ⊆ Y is open
if and only if U ∩ Xα is open in Xα (equivalently, a subset F of Y is
closed if and only if F ∩ Xα is closed in Xα ) for each index α.
TOPOLOGICAL CONSTRUCTIONS
189
This notion can be easily extended to an indexed family of spaces
{Xα }, where the spaces Xα are not necessarily disjoint and inherit
their topologies from one and the same space. In this case, we construct
the sets Xα × {α} which are pairwise disjoint. For each index α, the
bijective function (x, α) 7→ x from Xα × {α} to Xα induces a topology
on Xα × {α} so that it is a homeomorphic copy of Xα . The topological
∑
sum
(or
disjoint
union
or
free
union)
of
the
spaces
X
is
the
set
Xα =
α
∪
(Xα∑× {α}) equipped with the topology whose members are the sets
U ⊆
Xα such that U ∩ (Xα × {α}) is open in Xα × {α} for every
α ∈ A.
∑
For each β ∈ A, there is an injection iβ : Xβ →
Xα , x
∑7→ (x, β).
The set Xβ ×
{β}
is
the
range
of
the
embedding
i
:
X
→
Xα , and
β
β
∑
is clopen in Xα . In practice,∑
we generally identify Xβ with X∑
β ×{β},
and treat Xβ as a subspace of
Xα . With this identification,
Xα is
a disjoint union of the clopen subsets Xα . In case the indexing set A is
finite and consists of the numbers 1, . . . , n, then we also use the notation
X1 + ·∑
· · + Xn for the topological sum. To define a continuous function
from
Xα into a space Y , we simply
∑ need a family {fα : Xα → Y }
of continuous functions. For, if x ∈
X
α , then there is a unique Xα
∑
containing x. So there is a function ϕ :
Xα → Y defined by ϕ(x) =
fα (x). Since ϕ|Xα = fα is continuous for every α, it is easily seen that
ϕ is continuous.
Next, we discuss the technique of forming sums in the category of
“pointed spaces.”
Definition 7.4.2 A pointed space is a pair (X, x0 ), where X is a topological space and x0 is a point of X; the distinguished point x0 is called
the base point.
(
)
Definition 7.4.3 Let Xα , x0α ,∑
α ∈ A, be a family of pointed spaces.
The quotient space of the sum
Xα obtained by identifying all the
base points x0α is called the wedge (or the one-point
union or the bou∨
quet) of the spaces Xα , and is denoted
by Xα . The identified point
∨
[x0α ] is taken as the base point of Xα .
iβ ∑
π ∨
Clearly, for each β ∈ A, the composition Xβ −−→ Xα −−→ ∨Xα ,
denoted by jβ , is an embedding. Also, note that each point x of Xα
other than the base point lies in a unique Xα . So there are natural
190
maps qβ :
Elements of Topology
∨
Xα → Xβ defined by
{
x
if x ∈ Xβ − {x0β }, and
qβ (x) =
x0β otherwise.
It is obvious that qβ ◦ jβ = 1Xβ and qβ ◦ jβ ′ is the constant map if
′
β ̸= β ′ . It is easily seen that each qβ is continuous.
( 0)
∏Let Xα denote
the slice parallel
Xα . Then the
∏to Xα through the point xα of
′
subspace Xα ⊂ Xα is homeomorphic∨to Xα , and
∪the′ correspondence
x
↔
(q
(x))
is
a
bijection
between
X
and
Xα . The mapping
α
α
∨
∏
Xα → Xα , x 7→ (qα (x)), is clearly continuous but it need not be
a homeomorphism, even if the family {Xα } is finite. If the indexing
set A = {1, 2, . . . , n}, then the wedge of the Xα is also written as
X1 ∨ · · · ∨ Xn . If, for every α = 1, . ∏
. . , n, the base point x0α is closed in
′
Xα , then each slice Xα is closed in Xα , and the continuous bijection
x ↔ (qα (x)) is a homeomorphism between X1 ∨ · · · ∨ Xn and the
× · · · × Xn . Thus, for
subspace X1′ ∪ · · · ∪ Xn′ of the product∪space X1∏
n
n
T1 spaces X1 , . . . , Xn , the subspace 1 Xα′ ⊂ 1 Xα is regarded as
the wedge X1 ∨ · · · ∨ Xn(. We emphasise
case )n = 2; X1 ∨ X2 is
) ( the
0
0
essentially the subspace X1 × {x2 } ∪ {x1 } × X2 ⊂ X1 × X2 , where
x01 , x02 are the base points of X1 and X2 , respectively.
Example 7.4.1 The wedge S1 ∨ S1 is homeomorphic to the subspace
{
}
(z1 , z2 ) ∈ S1 × S1 |z1 = 1 or z2 = 1
of the torus. This space can also be viewed as the subspace of the plane
R2 , which is the union of two unit circles centered at (-1,0) and (1,0),
or the quotient space of the unit circle S1 obtained by identifying the
points -1 and 1, that is, S1 / {−1, 1}. We usually call it the figure 8
space.
•
-1
•
1
FIGURE 7.11: The “figure 8” space.
TOPOLOGICAL CONSTRUCTIONS
191
Example 7.4.2 For each positive integer n, consider the circle Cn in
the plane R2 having radius 1/n and centre (1/n, 0). Notice
∪ that each2
Cn touches the y-axis at the origin. The subspace X = Cn of R
is usually referred
∨ to as the Hawaiian earring (Figure 7.12(a)). Let
Y be the wedge Cn with the equivalence class of the origin as the
base point (Figure 7.12(b)). Then there exists a continuous bijection
between X and Y , but they are not homeomorphic. For, the set F =
{(2/n, 0) |n = 1, 2, . . .} is closed in Y (since F ∩ Cn is closed in Cn for
every n), while it is not closed in X (since the origin is its limit point).
C3
C2
0
C2
C1
C1
C3
X
(a)
Y
(b)
FIGURE 7.12: (a) Hawaiian earring; (b) Wedge of circles.
The Hausdorff property remains invariant under the formation of
one-point union.
Proposition
7.4.4 If (Xα , xα ), α ∈ A, are Hausdorff pointed spaces,
∨
then Xα is also Hausdorff.
∨
Proof. Let∨
x0 = [xα ] denote the base point of Xα . Then
it is obviously
∨
closed in Xα . If x, y are ∨
two distinct points in Xα − {x0 }, then
they have disjoint
nbds
in
Xα , since each Xα − {xα } is Hausdorff
∨
and open in Xα . If x = x0 ̸= y, then there exists a unique index β
such that y ∈ Xβ − {xβ }. So we have disjoint open sets Uβ and Vβ in
Xβ such that xβ ∈ Uβ and y ∈ V . ∑
For each index α ̸= β, find an open
nbd Uα of xα in ∑
Xα and put
Uα and V = Vβ . Then π(U ) and
∨ U =
π(V ), where π :
Xα → Xα is the identification map, are disjoint
open nbds of x and y, respectively.
♢
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Elements of Topology
Moreover, the wedge sum of a family of connected (resp. pathconnected) spaces is connected (resp. path-connected).
The following concept interwines the product and wedge sum of
two spaces.
Definition 7.4.5 Let (X, x0 ) and (Y, y0 ) be pointed spaces. The quotient space (X × Y ) / (X ∨ Y ) is called the smash product of X and Y ,
and is denoted by X ∧ Y (some authors write X#Y ).
Notice that X ∧ Y depends on the base points x0 and y0 . For example, consider the pointed space (I, 0) . It is easily seen that I ∧ I is
homeomorphic to D2 . On the other hand, if we choose 1/2 as the base
point of I, then I ∧ I is homeomorphic to the wedge of four copies of
D2 .
Obviously, we have X∨S0 = (X × {−1})∪{(x0 , 1)} so that X∧S0 ≈
X × {1} ≈ X. To see X ∧ S1 , we regard S1 as the quotient space I/∂I
with the base point z0 = [∂I]. Since the projection p : I → I/∂I is a
1×p
proper mapping, the mapping(X × I) −−−→ X ×S1 is closed. Therefore
the composition π ◦ (1 × p) : X × I → X ∧ S1 is an identification, where
π : X × S1 → X ∧ S1 is the quotient
map.) Clearly, the inverse of the
(
1
subspace X ∨S = (X × {z0 })∪ {x0 } × S1 ⊂ X ×S1 under π ◦(1 × p)
is the subspace (X × {0}) ∪ (X × {1}) ∪ ({x0 } × I) of X × I. Therefore
(
) [
(
)]
X ∧ S1 = X × S1 / (X × {z0 }) ∪ {x0 } × S1
≈ (X × I) / [(X × ∂I) ∪ ({x0 } × I)].
The quotient space (X × I) /[(X × {0}) ∪ (X × {1}) ∪ ({x0 } × I)] is
called the reduced suspension of the pointed space (X, x0 ) , and is denoted by SX. Thus, X ∧ S1 ≈ SX.
Example 7.4.3 The reduced suspension SSn−1 is homeomorphic to Sn .
n−1
Denote
X = SSn−1 − {[x0 , 0]} ≈
( n−1 the base
) point of Sn−1 by x0 . Then
S
− {x0 } × (0, 1) ≈ R
× R ≈ Rn . So their one-point compact∗
n ∗
ifications X and (R ) are homeomorphic. Since SSn−1 is compact
∗
Hausdorff, X ∗ ≈ SSn−1 . So SSn−1 ≈ (Rn ) ≈ Sn .
It is clear that the operation of taking a smash product is commutative: X ∧ Y ≈ Y ∧ X. The associativity of smash product can also
be seen in certain cases. First, we prove the following.
TOPOLOGICAL CONSTRUCTIONS
193
Proposition 7.4.6 If X and Y are Hausdorff spaces, then so also is
X ∧ Y.
Proof. Let x0 and y0 be base points of X and Y, respectively, and ∗
denote the base point of X ∧ Y. It is clear that X ∨ Y is closed in X × Y.
So X ∧ Y − {∗} is an open subspace of X × Y ; consequently, every pair
of distinct points in X ∧ Y − {∗} has disjoint open nbds in X ∧ Y. To
complete the proof, we need to separate ∗ and another point (x, y) in
X ∧ Y. Since X is T2 , there exist disjoint open sets Ux0 and Ux in X
with x0 ∈ Ux0 and x ∈ Ux . Similarly, there are disjoint open sets Vy0
and Vy in Y with y0 ∈ Uy0 and y ∈ Uy . Then (Ux0 × Y ) ∪ (X × Vy0 )
and U × V are disjoint nbds of ∗ and (x, y) .
♢
Theorem 7.4.7 Let X, Y and Z be Hausdorff pointed spaces. If X
and Z are locally compact or two of X, Y, Z are compact Hausdorff,
then X ∧ (Y ∧ Z) ≈ (X ∧ Y ) ∧ Z.
Proof. Let ∗ denote all base points, and π denote the canonical projection of the product of two spaces onto their smash product. If X is
1×π
locally compact Hausdorff, then X × Y × Z −−−→ X × (Y ∧ Z) is an
identification, by Theorem 7.2.8. If Y and Z are compact Hausdorff,
then π : Y × Z → Y ∧ Z is a proper map, by Corollary 6.5.3. Thus
1 × π is a closed map, and hence an identification. Let ϕ denote the
composition
1×π
π
X × Y × Z −−−→ X × (Y ∧ Z) −−→ X ∧ (Y ∧ Z).
Then ϕ, being a composition of two identifications, is an identification.
Similarly,
π×1
π
ψ : X × Y × Z −−−→ (X ∧ Y ) × Z −−→ (X ∧ Y ) ∧ Z
is an identification, if Z is locally compact Hausdorff or X and
Y are compact Hausdorff. The mappings ϕ and ψ each have the
decomposition space (X × Y × Z) /A, where A = ({∗} × Y × Z) ∪
(X × {∗} × Z) ∪ (X × Y × {∗}). By Theorem 7.2.5, (X ∧ Y ) ∧ Z ≈
(X × Y × Z) /A ≈ X ∧ (Y ∧ Z).
♢
We apply the above result to compute Sm ∧ Sn . Clearly,
(
) (
)
X ∧ S2 ≈ X ∧ S1 ∧ S1 ≈ X ∧ S1 ∧ S1 ≈ SX ∧ S1 ≈ S(SX),
194
Elements of Topology
etc. In particular, we have Sm ∧ Sn ≈ Sm+n .
Exercises
1. Let X and Y be disjoint subspaces of a space Z such that Z = X ∪ Y .
Show that the following conditions are equivalent:
(a) Z = X + Y ,
(b) X, Y are both open in Z,
(c) X is clopen in Z, and
(d) X ∩ Y = ∅ = Y ∩ X.
2. Prove that a space X is connected if and only if X is not the sum of
two nonempty disjoint subspaces.
3. Prove that the formation of a topological sum is associative and commutative:
(a) (X + Y ) + Z ≈ X + (Y + Z),
(b) X + Y ≈ Y + X.
4. Prove that there is a canonical homeomorphism: (X + Y ) × Z ≈ (X ×
Z) + (Y × Z).
5. Prove that the sum X + Y of two spaces X and Y is first countable
(resp. separable) if and only if both X and Y are first countable (resp.
separable).
6. If X and Y are metrisable spaces, prove that X + Y is metrisable.
∑
7. Let {Xα } be a family of T1 -spaces. Show that
Xα is also T1 .
∑
8. Let {Xα } be a family of T2 -spaces. Show that
Xα is also T2 .
∑
∑
9. Prove that Y ×
Xα ≈
(Y × Xα ) for any family of spaces Xα ,
α ∈ A.
10. Let Xα , α ∈ A, be a family
∑of spaces each of which is homeomorphic
to a space X. Prove that
Xα ≈ X × A, where A has the discrete
topology.
11. Let {Xα } , α ∈ A, be a collection of spaces. Prove:
(a) If all spaces Xα satisfy the
∑ first axiom of countability, then the
same holds for their sum
Xα .
(b) If the index set A is countable,
and each Xα is second countable
∑
(resp. separable), then
Xα is also second countable (resp. separable).
TOPOLOGICAL CONSTRUCTIONS
195
12. Prove that any sum of locally compact spaces is locally compact.
13. Let Xα , α ∈ A, be a family of metrisable spaces and let dα be the
standard bounded metric on Xα . Verify that δ given by
{
dα (x, y)
if x, y ∈ Xα and
δ(x, y) =
1
if x ∈ Xα , y ∈ Xβ and α ̸= β
∑
is
Xα and the topology induced by δ is the topology of
∑a metric on
Xα .
14. Show that the space Y in Ex. 7.4.2 is homeomorphic to the quotient
space of R obtained by collapsing Z to a point.
15. Describe each of the following spaces:
(a) S2 with the equator identified to a point;
(b) R2 with each of the circles centred at the origin and of integer
radius identified to a point.
16. Show that the wedge sum is associative, that is, (X ∨ Y ) ∨ Z ≈ X ∨
(Y ∨ Z) for any three pointed spaces X, Y and Z all with closed base
points.
17. If Xα , α ∈ A, is a family of connected
(resp. path-connected) pointed
∨
spaces, prove that the wedge sum Xα is also so.
18. Prove that S (X ∨ Y ) ≈ SX ∨ SY for any two pointed T1 -spaces X, and
Y.
19. If (X, x0 ) and (Y, y0 ) are compact Hausdorff pointed spaces, prove that
X ∧ Y is the one-point compactification of (X − {x0 }) × (Y − {y0 }).
7.5
Adjunction Spaces
In this section, we shall study the process of “attaching two spaces
by a continuous map.” These constructs find many applications in algebraic topology.
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Elements of Topology
Definition 7.5.1 Let X and Y be spaces, A a closed set, and f :
A → Y a continuous map. The quotient space (X + Y ) / ∼, where the
equivalence relation ∼ on X + Y is generated by the relation a ∼ f (a),
a ∈ A, is denoted by X ∪f Y . The space X ∪f Y is said to be obtained
by attaching X to Y by f or the adjunction space determined by f .
The map f is called the attaching map.
A
[
X
]
f
Y
[
]
f(A)
X [f Y
FIGURE 7.13: Adjunction space of X to Y by f .
Example 7.5.1 Let A ⊆ X be closed and Y a one-point space {p}.
Then X ∪f Y ≈ X/A, where f is the obvious map. Since a ∼ p in
X + Y for all a ∈ A, the inclusion map λ : X → X + Y is relation
preserving. So it induces a continuous bijection λ̄ : X/A → X ∪f Y. If
π : X + Y → X ∪f Y and ϕ : X → X/A are the natural projections,
then we have π −1 λ̄(F ) = ϕ−1 (F )+{p} or ϕ−1 (F ), according to [A] ∈ F
or [A] ∈
/ F . It follows that λ̄ is a closed mapping, and therefore a
homeomorphism.
Example 7.5.2 A torus can be considered as an adjunction space of a
map from the boundary of a 2-disk to the figure 8 space. As seen in
Ex. 7.1.4, the torus S1 × S1 can be considered as the quotient space
of the square I × I obtained by identifying opposite sides. Under this
identification all four vertices of the square are identified, and the two
pairs of identified sides result in the figure 8 space, essentially S1 ∨S1 . If
π : I × I → (I × I) / ∼ is the quotient map, then the image of ∂ (I × I)
under π is clearly S1 ∨S1 . Let h : D2 → I×I be a homeomorphism. Then
h sends S1 = ∂D2 onto ∂ (I × I). So we have a continuous function
f : S1 → S1 ∨ S1 defined by the composition πh. The mapping g : D2 +
2
S1 ∨ S1 → (I × I) / ∼ given by g(x) = πh(x) for
( x1 ∈ D1 ), and g(x) = x
1
1
2
for x ∈ S ∨S is an identification. Hence D ∪f S ∨ S ≈ (I × I) / ∼,
the torus. See Figure 7.14 below.
TOPOLOGICAL CONSTRUCTIONS
197
^
^
b
a
b
^
b
^
^
^
a
a
FIGURE 7.14: Torus as an adjunction space of a 2-disk to the figure 8 space.
Proposition 7.5.2 Let X be attached to Y via a map f : A → Y and
let π : X + Y → X ∪f Y be the quotient map. Then
(a) π|Y is an embedding of Y onto a closed subspace of X ∪f Y , and
(b) π| (X − A) is an embedding of X − A onto an open subspace of
X ∪f Y .
Proof. (a): If F ⊆ Y is closed, then f −1 (F ) is closed in A and hence
in X. So π −1 (π (F )) = f −1 (F ) ∪ F is closed in X + Y which, in turn,
implies that π (F ) is closed in X ∪f Y . Thus π|Y is a closed mapping;
in particular, π (Y ) is closed in X ∪f Y . Also, π|Y is a continuous
injection, and therefore Y ≈ π (Y ).
(b): If G ⊆ X − A is open, then π −1 (π (G)) = G is open in X + Y .
This implies that π (G) is open in X ∪f Y . It follows that π| (X − A) is
an open mapping; in particular, π (X − A) is open in X∪f Y . Obviously,
π| (X − A) is a continuous injection of X − A into X ∪f Y . So it is a
homeomorphism between X − A and its image under π.
♢
By identifying Y with π (Y ), it is usually regarded as a closed subspace of X ∪f Y . Similarly, X − A may be considered as an open
subspace of X ∪f Y . It is clear that π (X − A) and π (Y ) are disjoint,
and X ∪f Y equals their union. Therefore X ∪f Y can be considered as
the disjoint union of X − A and Y glued together by a topology defined
by the map f . When A is also open in X, then X ∪f Y is a topological
sum of X − A and Y .
Next, we describe a method of constructing continuous maps from
adjunction spaces. Let A be a closed subset of a space X and f : A → Y
be a continuous map. Then we have the following commutative diagram
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Elements of Topology
f
A
- Y
ρY
j
?
ρX
X
?
- X ∪f Y
of spaces and continuous maps, where j is the inclusion, and ρX , ρY
are the restrictions of the quotient map π : X + Y → X ∪f Y to X,
Y , respectively. A function ϕ : X ∪f Y → Z induces two functions
k : X → Z and g : Y → Z making the diagram
A
j
f
3
Z
k
?
ρX
X
- Y
g ρ
+
Y
Q
k ϕ
Q
Q
Q ?
- X ∪f Y
commutative, that is, k (x) = ϕ ◦ ρX (x) and g (y) = ϕ ◦ ρY (y). By the
definition of the topology of X ∪f Y , it is clear that ϕ is continuous
if and only if both k and g are continuous. The other way round, we
have the following.
Proposition 7.5.3 Suppose that g : Y → Z and k : X → Z are
continuous maps such that kj = gf , where the other notations have the
above meaning. Then there is a unique continuous map ϕ : X ∪f Y → Z
such that ϕ ◦ ρY = g and ϕ ◦ ρX = k.
Proof. Consider the map F : X + Y → Z defined by
{
k(x) for x ∈ X,
F (x) =
g(y) for y ∈ Y.
If x ∈ X, y ∈ Y and x ∼ y, then x ∈ A and y = f (x). So g (y) =
gf (x) = kj (x) = k (x). Therefore F induces a function ϕ : X ∪f Y → Z
such that ϕπ = F (Figure 7.5). We have ϕ ◦ ρX = F |X = k, ϕ ◦ ρY =
F |Y = g, and hence ϕ is continuous. If a mapping ψ : X ∪f Y → Z
satisfies ψ ◦ρX = k and ψ ◦ρY = g, then (ψπ) |X = F |X and (ψπ) |Y =
F |Y . Thus ψπ = F whence ψ = ϕ.
♢
TOPOLOGICAL CONSTRUCTIONS
199
Now, we turn to see the invariance of topological properties under
this construction. If X and Y are T1 -spaces and f is a continuous map
of a closed subset of X into Y , then X ∪f Y is clearly a T1 -space. The
Hausdorff property is not inherited by adjunction spaces, in general.
But, under a mild condition on the domain of the attaching map, we
have a positive statement.
Theorem 7.5.4 Suppose that X, Y are Hausdorff spaces, A a compact
subspace of X, and f : A → Y is a continuous map. Then X ∪f Y is
Hausdorff.
Proof. Let π : X + Y → X ∪f Y be the quotient map. If F ⊆ X + Y
is closed, then F ∩ A is closed in A, and hence compact. Therefore f (F ∩ A) is closed in Y ; thus π −1 π(F ) = F ∪ f (F ∩ A) ∪
f −1 ((F ∩ Y ) ∪ f (F ∩ A)) is closed. It follows that π (F ) is closed, and
thus π is a closed map. For a point z in X + Y , the equivalence class
of z is a singleton set if z ∈ (X − A) ∪ (Y − f (A)) , and {z} ∪ f −1 (z)
if z ∈ f (A). It follows that π −1 π (z) is compact. Therefore X ∪f Y is
Hausdorff (cf. Exercise 6.1.11).
♢
The behaviour of adjunction spaces relative to the other topological
properties is discussed in the exercises and in the following chapters.
In the end of this section, we deal with the notion of the “mapping
cylinder” and “mapping cone” of a map f . Let f : X → Y be a
continuous map. Note the set X × {0} is closed in X × I, where I is as
usual the unit interval. Let ϕ : X × {0} → Y be the map defined by f ,
that is, ϕ ((x, 0)) = f (x). The adjunction space (X × I) ∪ϕ Y is called
the mapping cylinder of f , and is denoted by Mf (see Figure 7.15).
X£1
Y
FIGURE 7.15: Mapping cylinder of f .
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Elements of Topology
Let π : (X × I) + Y → Mf be the quotient map. By Proposition
7.5.2, π|Y is a homeomorphism of Y onto a closed subspace of Mf . By
means of this embedding, Y is considered as a closed subspace of Mf .
Moreover, the map i : X → Mf given by i (x) = π ((x, 1)),x ∈ X, is
a closed embedding. Thus X is also embedded as a closed subspace of
Mf .
The quotient space Mf / (X × {1}) is called the mapping cone of f ,
and denoted by Cf . Observe that if Y is a one-point space, then Mf is
CX (the cone over X), and Cf is ΣX (the suspension of X).
Exercises
1. Let X = [0, 1], Y = [2, 3] be the subspaces of the real line R and
A = {0, 1}. Let f : A → Y be the map given by f (0) = 3, f (1) = 2.
Show that X ∪f Y is homeomorphic to S1 .
2. Let A ⊆ X be a closed set and f : A → Y be a continuous map. Let
π : X + Y → X ∪f Y denote the natural projection.
(a) For a closed set F ⊆ X, show that π (F ) is closed in X ∪f Y if
and only if f (F ∩ A) is closed in Y .
(b) If U ⊆ X and V ⊆ Y are open sets such that f −1 (V ) = A ∩ U ,
show that π (U + V ) is open in X ∪f Y .
3. Show that RP2 ≈ D2 ∪f S1 for a continuous map f : S1 → S1 .
4. Suppose that X and Y are closed subsets of a space Z such that Z =
X ∪ Y . Let A = X ∩ Y and f : A → Y be the inclusion map. Show that
Z = X ∪f Y .
5. Give an example to show that the adjunction space of two Hausdorff
spaces need not be Hausdorff.
6. Let A be a nonempty subset of a space X, and f : A → Y be a continuous map. If X and Y are connected (resp. path-connected), prove
that X ∪f Y is also connected (resp. path-connected).
7. Let A be a nonempty subset of a space X, and f : A → Y be a continuous map. If A and X ∪f Y are connected, show that Y is also connected.
8. Prove that the adjunction space of two compact spaces via any continuous function is compact.
9. Let X and Y be Lindelöf spaces, and A ⊆ X be closed. If f : A → Y is
continuous, show that X ∪f Y is Lindelöf.
TOPOLOGICAL CONSTRUCTIONS
201
10. Let X, Y be spaces and A ⊂ X be closed. Let f : A → Y be an
identification. Prove that the composition X → X + Y → X ∪f Y is
also an identification.
11. Let A be a closed subset of a space X, and f : A → Y be a continuous
function. Suppose that S ⊆ X, T ⊆ Y are subspaces with A ⊆ S. If
g : A → T is the map defined by f , show that S ∪g T is a subspace of
X ∪f Y .
12. Let A be a closed subspace of X, and let f : A → Y, g : Y → Z be
continuous maps. Prove that (X ∪f Y ) ∪g Z ≈ X ∪gf Z.
13. Let X and Y be spaces, and A ⊆ B be closed subspaces of X. If
f : A → Y is a continuous map, prove that X ∪f Y ≈ X ∪g (B ∪f Y ),
where g is the restriction of the quotient map B + Y → B ∪f Y to B.
14. If X, Y are Hausdorff spaces, prove that for any continuous map f :
X → Y Mf and Cf are Hausdorff.
15. Let A ⊆ X be closed and f : A → Y be a continuous map. Let T be a
locally compact Hausdorff space, and g : X ×T → Z and h : Y ×T → Z
be continuous maps such that g(x, t) = h(f (x), t) for every x ∈ A, t ∈ T .
Prove that there exists a continuous map ϕ : (X ∪f Y ) × T → Z such
that ϕ ◦ (π × 1) = (g, h), where π : X + Y → X ∪f Y is the quotient
map.
16. Prove:
(
)
(a) S1 × S1 ≈ D2 ∪f S1 ∨ S1 for a continuous map f : S1 → S1 ∨ S1 .
(
)
(b) The Klein bottle K is homeomorphic to D2 ∪f S1 ∨ S1 for a
continuous map f : S1 → S1 ∨ S1 .
(
) (
)
17. Prove that Sm+n+1 ≈ Dm+1 × Sn ∪ Sm × Dn+1 . Use it to show that
Sm × Sn is homeomorphic to Dm+n ∪f (Sm ∨ Sn ) for a continuous map
f : Sm+n−1 → Sm ∨ Sn .
18. Show that X ∗ Y with the fine topology is homeomorphic to the space
obtained by attaching X × Y × I to the topological sum X + Y by the
mapping ϕ : X × Y × {0, 1} → X + Y given by ϕ (x, y, 0) = x and
ϕ (x, y, 1) = y, for all x ∈ X, y ∈ Y .
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7.6
Elements of Topology
Coinduced and Coherent Topologies
The notion of coinduced topology is a generalisation of the identification topology, and it dualizes the notion of induced topology. This
leads to the concept of “coherent topology” which makes the intuitive
idea of “pasting” spaces together along prespecified subsets precise.
Let Xα , α ∈ A, be a family of spaces, and Y a set. Given functions
fα : Xα → Y , one for each α, we wish to find a topology for Y such that
each fα is continuous. Obviously, the trivial topology on Y satisfies this
requirement, but it is not interesting. If T is any topology on Y which
makes each fα continuous, then fα−1 (U ) is open in Xα for all U ∈ T
and α ∈ A. As we would like, the family
{
}
U = U ⊆ Y |fα−1 (U ) is open in Xα for each α ∈ A
is a topology on Y having the desired property. Obviously, U is finer
than any topology T for Y with this property. Thus U is the largest
topology on Y such that each fα is continuous.
Definition 7.6.1 Let Xα , α ∈ A, be a family of spaces, and Y a set.
Given functions fα : Xα → Y , one for each α, the largest (or finest)
topology on Y which makes each fα continuous is called the topology
coinduced by the collection {fα }. This is also referred to as the final
topology with respect to {fα }.
Since fα−1 (Y − U ) = Xα − fα−1 (U ), it is clear that a set F ⊆ Y
is closed in the topology coinduced by {fα } if and only if fα−1 (F ) is
closed in Xα for every α ∈ A. The following theorem characterises the
topology on Y coinduced by the functions fα : Xα → Y .
The topology coinduced by a function f of a space X onto a set Y is
the identification topology. It is obvious that the topology of the ∑
sum of
a family of spaces Xα is coinduced by the injections iβ : Xβ →
Xα .
The topology of the adjunction space X ∪f Y is clearly coinduced by
the canonical mappings ρX :X→X∪f Y and ρY : Y → X ∪f Y .
Theorem 7.6.2 Let Xα , α ∈ A, be a family of spaces for each index
α. Then the topology of a space Y is coinduced by the functions fα :
Xα → Y if and only if, for any space Z, a function g : Y → Z is
continuous ⇔ g ◦ fα : Xα → Z is continuous for every α ∈ A.
TOPOLOGICAL CONSTRUCTIONS
203
Proof. Suppose that Y has the topology coinduced by {fα }. If g is continuous, then g ◦ fα is continuous for every α, since each fα( is continu)
ous. Conversely, if g ◦ fα is continuous for all α, then fα−1 g −1 (U ) =
(g ◦ fα )−1 (U ) is open in Xα for every open U ⊆ Z. So g −1 (U ) is open
in Y and g is continuous. This proves the necessity of the condition.
To prove the sufficiency, suppose that Y satisfies the condition.
Then each fα : Xα → Y is continuous, since the identity map on Y is
continuous. If Y ′ is the set Y together with the topology coinduced by
{fα }, then the functions fα : Xα → Y ′ are continuous. If i : Y → Y ′
fα
i
is the identity function, then the composition Xα −−→ Y → Y ′ is
continuous for every α. By our hypothesis, i is continuous. Also, each
fα
i−1
composition Xα −−→ Y ′ −−−→ Y is continuous. Since the topology
of Y ′ is coinduced by {fα } , i−1 is continuous. Thus i : Y → Y ′ is a
homeomorphism, and Y = Y ′ as spaces.
♢
The behavior of coinduced topology with respect to subspaces,
products and quotients will be studied in the exercises.
The notion of coinduced topology is especially useful when all the
spaces Xα are subsets of (a set) Y. Assume that the set Y is already
given a topology, and each Xα is a subspace of Y. Consider the topology
on Y coinduced by the inclusion maps Xα ,→ Y . Then the sets open
(or closed) in the original topology of Y remain open (or closed) in the
new topology. So the new topology on Y is, in general, finer than the
given topology of Y . Also, the relative topology on each Xα induced
by the new topology of Y coincides with its original topology. This
suggests the following.
Definition 7.6.3 Let Y be a space, and {Xα } , α ∈ A, be a collection
of subspaces of Y . The topology of Y is said to be coherent with {Xα } if
it is coinduced from the subspaces Xα by the inclusion maps Xα ,→ Y ,
α ∈ A.
The coherent topology for Y relative to the collection {Xα } is also
called the weak topology. (Notice that this term is also used in the literature to mean something quite different (refer to Definition 2.2.11).)
Of course, there is only one topology on Y coherent with a given collection {Xα } of subspaces Xα ⊆ Y , and this is the largest topology
on Y which induces the initial topology back on each Xα . Thus, the
topology of a space Y coherent with a given family of subsets of Y
preserves the predetermined topologies of the members.
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Elements of Topology
The simple proof of the following proposition is left to the reader.
Proposition 7.6.4 A necessary and sufficient condition that a space
Y has a topology coherent with a collection of its subspaces Xα is that
a set U ⊆ Y be open (or closed) if and only if U ∩ Xα is open (or
closed) in Xα for every index α.
From the preceding proposition, it is immediate that if a space Y
is the union of a collection {Xα } of open sets, then the topology of
Y is coherent with {Xα }. Also, by Proposition 2.1.9, if a space Y is
the union of a locally finite collection {Xα } of closed sets, then the
topology of Y is coherent with {Xα }. ∨
Note that the topology
of∨the wedge Xα is coinduced by the in∨
jections jβ : Xβ → Xα . So Xα is a coherent union of the subspaces
im (jα ) ≈ Xα .
Example 7.6.1 The canonical embedding
Rn → Rn+1 , (x1 , . . . , xn ) 7→ (x1 , . . . , xn , 0),
permits us to identify Rn with the subspace of Rn+1 consisting of
all points having their last∪coordinate zero. Thus, we have inclusions
R1 ⊂ R2 ⊂ · · · . Set R∞ = n≥1 Rn , and topologize this set by the metric d(x, y) = max {|xn − yn | : n ∈ N} . The metric space R∞ is called a
generalised euclidean space, and contains Rn as a subspace. The topology coherent with the subspaces Rn is called the inductive topology for
R∞ . This topology is distinct from the metric∑
topology of R∞ because
∞
∞
1
the function f : R → R defined by f (x) = 1 nxn is continuous in
the inductive topology, but discontinuous in the metric topology.
If the topology of a∪space Y is coherent with a collection of subspaces Xα , and Y = Xα , then we often say that Y is a coherent
union of the Xα . In this case, intuitively speaking, we can think of the
space Y as obtained by the “pasting on” of the subspaces Xα . This is
evident from the following.
Proposition 7.6.5 Let Y be
Xα , α ∈ A, a family of
∪ a space and ∑
subspaces of Y with Y =
Xα . Let f :
Xα → Y be the map
defined by f (x, α) = x, where x ∈ Xα , α ∈ A. Then the topology of Y
is coherent with the collection {Xα } if and only if f is an identification
∑
map. In this case, Y ≈ ^
Xα , the decomposition space of f .
TOPOLOGICAL CONSTRUCTIONS
205
Proof. Obviously, f∑
is surjective. The composition of f with each injection iβ : Xβ →
X
∑α , x 7→ (x, β), is the inclusion map Xβ ⊆ Y .
Since the topology of
Xα is coinduced by the injections iβ , β ∈ A,
f is continuous, by Theorem 7.6.2. For any set U ⊆ Y , we obtain
f −1 (U ) ∩ (Xβ × {β}) = iβ (U ∩ Xβ ). Since i∑
β is a homeomorphism be−1
tween Xβ and Xβ × {β}, f (U ) is open in
Xα if and only if U ∩ Xβ
is open in Xβ for every β ∈ A. It follows that if f −1 (U ) is open, and
Y has a topology coherent with {Xα }, then U is open in Y. So f is an
identification. Conversely, if f is an identification, and U ∩ Xα is open
in Xα for every α, then U is open in Y, and Y has a topology coherent
with {Xα }.
The last statement follows from Theorem 7.2.5.
♢
It should be noted that if {Xα } is a family of spaces with each
Xα ⊆ Y , then there is in general no topology on Y such that each Xα
is a subspace of Y . However, there is a simple (sufficient) condition
which guarantees the existence of a topology on Y which induces the
preassigned topology back on each Xα .
Theorem 7.6.6 Let {Xα } be a family of subsets of a set Y . Suppose
that each Xα has a topology such that (a) the relative topologies on
Xα ∩ Xβ induced from Xα and Xβ agree, and (b) Xα ∩ Xβ is closed (or
open) in Xα and in Xβ for each pair of indices α and β. Then there is
a topology on Y in which each Xα is a closed (resp. open) subspace of
Y and which is coherent with {Xα }.
Proof. We show that the topology on Y coinduced by the inclusion maps
Xα ,→ Y has the required properties. Clearly, it suffices to establish the
first statement, that is, each Xα , as a subspace of Y , retains its original
topology and is closed in Y . By the definition of coinduced topology,
F ∩ Xα is closed in Xα for every closed set F ⊆ Y . Conversely, suppose
that E ⊆ Xα is closed. Then, for any index β, E ∩ Xβ = E ∩ Xα ∩ Xβ
is closed in Xα ∩ Xβ . By our hypothesis, Xα ∩ Xβ is a closed subspace
of Xβ . So E ∩ Xβ is closed in Xβ , and therefore E is closed in Y.
It follows that the relative topology on Xα induced from Y coincides
with its original topology. Also, taking E = Xα , we see that Xα itself
is closed in Y .
The other case is proved similarly.
♢
The following simple result enables us to test the continuity of
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Elements of Topology
functions on a space having a topology coherent with a given collection
of its subsets.
Proposition 7.6.7 Suppose that a space Y has the topology coherent
with {Xα }. Then, for any space Z, a function f : Y → Z is continuous
if and only if each restriction f |Xα is continuous.
Proof. This is immediate from Theorem 7.6.2.
♢
k-spaces
In this vein, we also study the spaces whose topologies are determined by the families of all their compact subsets. These spaces are
important to our discussion about function spaces because continuous
functions on them are precisely those which behave well on compact
subsets.
Definition 7.6.8 A Hausdorff space X is called a k-space (or a compactly generated space) if it is the coherent union of its compact sets.
It follows that a subset F of a k-space X is closed if and only if the
intersection of F with each compact subset is closed. By Proposition
7.6.7, a mapping of a k-space X into a space Y is continuous if and
only if it is continuous over every compact subset of X.
There is a very wide class of k-spaces, which contains all Hausdorff
spaces satisfying the first axiom of countability, as is shown by the
following proposition.
Proposition 7.6.9 Every Hausdorff space satisfying the first axiom
of countability is a k-space.
Proof. Let X be a first countable T2 -space. If A ⊆ X is not closed, then
there is a point x ∈ A−A. Consequently, we can find a sequence ⟨an ⟩ in
A such that an → x. The set C = {x} ∪ {an |n = 1, 2, . . .} is obviously
compact, but A ∩ C is not closed. Again, the converse is obvious and
X is a k-space.
♢
From the preceding proposition, it is immediate that every metric
space is a k-space. Another class of k-spaces is given by the following.
TOPOLOGICAL CONSTRUCTIONS
207
Proposition 7.6.10 Every locally compact Hausdorff space is a kspace.
Proof. Let X be a locally compact T2 -space. Assume that A ∩ C is
closed in C for each compact set C ⊆ X. We show that X − A is open.
Let x ∈ X − A be arbitrary. Let N be a compact nbd of x. By our
hypothesis, (X −A)∩N is open in N ; accordingly, (X −A)∩N ◦ = G is
open in N ◦ . Since N ◦ is open in X, G is an open nbd of x contained in
X − A. If follows that X − A is open, and so A is closed. The converse
is obvious, and thus X is a k-space.
The relation between k-spaces and local compactness is given by
Theorem 7.6.11 (D.E. Cohen) A Hausdorff space X is a k-space if
and only if it is homeomorphic to a quotient space of a locally compact
space.
Proof. Let X be a k-space, and {C∑
α } the family of all compact subsets
of X. Then the topological sum
Cα∑is obviously locally compact.
By Proposition 7.6.5, the mapping ϕ :
Cα → X, where ϕ|Cα is the
inclusion map Cα ,→ X, is a quotient map. Conversely, suppose that
X is a quotient space of a locally compact space Y and let π : Y → X
be the quotient map. Let F be a subset of X such that F ∩ C is closed
in C for every compact C ⊆ X. We need to show that F is closed.
Since π is a quotient map, F is closed in X ⇔ π −1 (F ) is closed in
Y . Assume, on the contrary, that π −1 (F ) is not closed. Then there is
a point y ∈ π −1 (F ) − π −1 (F ). Let N be a compact nbd of y. Since
π(y) ∈ F , we see that π(y) ∈ π(N ) ∩ F ; consequently, π(N ) ∩ F is
not closed. This contradicts the property for F , for π(N ) is compact.
Therefore π −1 (F ) is closed, and X is a k-space. This completes the
proof.
♢
As an immediate consequence of the above theorem, we have
Corollary 7.6.12 A Hausdorff image of a k-space under an identification map is a k-space.
It is a fact that a subspace of a k-space is not necessarily a k-space,
and the product of two k-spaces may fail to be a k-space.
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Elements of Topology
Exercises
1. Let Y have the topology coinduced by f : X → Y . Prove that f : X →
f (X) is an identification map.
2. Show that the topology on R coinduced by the inclusion maps i : Q ,→ R
and j : R − Q ,→ R, where Q and R − Q have the subspace topologies
induced from the usual topology on R, is strictly finer than the topology
of R.
3. Let {Xα } be a family of spaces each having the same underlying set Y .
What is the topology coinduced by the identity functions Xα → Y ?
4. Suppose that {Xα } is a family of subspaces of a space Y, U, and Z ⊆ Y .
Let T be the topology on Y coinduced by the inclusions Xα ,→ Y , and
let T Z be the topology on Z coinduced
( by the
) inclusions Z ∩ Xα ,→ Z.
Show that (a) the identity map from Z, T Z into (Z, TZ ) is continuous,
and (b) if Z and each Xα are closed in Y , then T Z = TZ , and Z is a
closed subspace of (Y, T).
5. Let Xα , α ∈ A, be a family of spaces, and Y a set.
∑Given a function
fα : Xα → Y for each α ∈ A, there is a function f : Xα → Y defined
by f ((x, α)) = fα (x), (x, α) ∈ Xα × {α}. Prove that the topology on Y
coinduced by the functions fα coincides with that coinduced by f .
6. Prove that coinduced topologies are transitive: Suppose that Yλ , λ ∈ Λ,
is a family of spaces and a space Z has the topology coinduced by the
functions gλ : Yλ → Z. For each λ ∈ Λ, let {Xλα } be a collection of
spaces with functions fλα : Xλα → Yλ , α ∈ Aλ , such that the topology of Yλ is coinduced by the functions fλα . Then the topology of Z
coincides with the topology coinduced by the composites {gλ ◦ fλα }.
7. Give an example to show that without condition (b) in Theorem 7.6.6,
there may not be a topology on Y such that each Xα , as a subspace of
Y , retains its original topology.
8. Let Y be the coherent union of the subspaces {Xα }. Show that if Z is
a closed (or open) subspace of Y , then Z is the coherent union of its
subspaces {Z ∩ Xα }.
′
9. Suppose that
{ Y}and Y have topologies coherent with the collections
′
{Xα } and Xβ , respectively. Let T be the topology on Y × Y ′ coin-
duced by the inclusion Xα × Xβ′ ,→ Y × Y ′ and P be the product
topology on Y × Y ′ . Show:
(a) The identity map ı : (Y × Y ′ , T) → (Y × Y ′ , P) is continuous.
∪
∪
(b) If Y = Xα , Y ′ = Xβ′ , where each Xα is open in Y and each
Xβ′ is open in Y ′ , then the mapping ı in (a) is a homeomorphism.
TOPOLOGICAL CONSTRUCTIONS
209
(c) The conclusion in (b) holds if each y ∈ Y lies in the interior of
some Xα , and each y ′ ∈ Y ′ lies in the interior of some Xβ′ .
{ }
(d) The conclusion in (b) also holds if {Xα } and Xβ′ are locally
finite collections
of closed
of Y and Y ′ , respectively, such
∪
∪ subsets
′
′
that Y = Xα , Y = Xβ .
10. Prove that a coherent union of T1 -spaces is a T1 -space.
11. Let X be a k-space and let ⟨fn ⟩ be a sequence of continuous functions
from X into metric space Y . Suppose that ⟨fn ⟩ converges to f : X → Y
uniformly on each compact K ⊆ X. Show that f is continuous.
12. Prove that a closed or open subspace of a k-space is a k-space.
13. If the topology of X is coherent with the subspaces Xα and if Y is a
locally compact Hausdorff, prove that the topology of X ×Y is coherent
with the subspaces Xα × Y .
14. Let X be a k-space and Y be a locally compact Hausdorff. Show that
X × Y is also a k-space.
15. Let X be a T2 -space. Show that there exists a k-space Y and a continuous bijection f : Y → X such that f is a homeomorphism on every
compact set of either Y or X.
16. Prove that a sequential T2 -space is a k-space.
17. Let f : X → Y be a continuous surjection. If Y is a k-space and f −1 (K)
is compact for every compact K ⊆ Y , show that f is proper.
Chapter 8
SEPARATION AXIOMS
8.1
8.2
8.3
8.4
Regular Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Completely Regular Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Stone–Čech Compactification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1
Regular Spaces
211
216
229
235
The properties of topological spaces are in general quite different
from those of metric spaces, so some additional restrictions are often
imposed on a topology in order to bring the properties of the corresponding space closer to those of metric spaces. You might have seen
that distinct points or closed sets in metric spaces can be separated
by open sets. In Chapter 4, we have already studied the conditions
which separate distinct points by open sets. Here, we shall study the
condition which separates points and closed sets.
Definition 8.1.1 A space X is T3 if for each point x ∈ X and each
closed set F ⊆ X with x ∈
/ F, there are disjoint open sets U and V
such that x ∈ U and F ⊆ V . The space X is called regular if it satisfies
both T1 and T3 axioms.
(Caution: The opposite terminology is used by many authors for this
and the other separation properties to be discussed later in this chapter.)
Example 8.1.1 A metric space is regular. In particular, a euclidean
space is regular.
Example 8.1.2 A trivial space with more than one point is T3 (vacuously), but not T1 .
Example 8.1.3 An infinite cofinite space is T1 but not T3 .
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Elements of Topology
A regular space is Hausdorff, since points are closed. But the axiom
T2 is not a consequence of T3 .
Example 8.1.4 A Hausdorff space which is not regular. In the set R of
real numbers, consider the topology T generated by the set Q of rational
numbers and the (open) intervals (a, b), a < b reals, as subbasis. This
topology is finer than the usual topology for R, so (R, T) is Hausdorff.
The set R − Q is obviously closed in T. If q ∈ Q and U is a nbd of q,
then there is an open interval (a, b) such that q ∈ (a, b) ∩ Q ⊆ U . Let
x be an irrational number in (a, b). Then every nbd of x intersects U
so that there is no nbd of R − Q which is disjoint from U . Thus (R, T)
is not regular.
This example also shows that a topology finer than a regular topology is not necessarily regular.
By a nbd of a subset A of a space X is meant a subset N ⊆ X
such that there is an open set U ⊆ X with A ⊆ U ⊆ N . The condition
in Definition 8.1.1 can be rephrased as: Each point x ∈ X and each
closed subset F of X not containing x have disjoint nbds. An equivalent
formulation is given by
Proposition 8.1.2 A space X is T3 if and only if for each point x ∈ X
and each nbd U of x, there exists a nbd V of x such that V ⊆ U (that
is, the closed nbds of x form a nbd basis at x).
Proof. Suppose that X satisfies the T3 axiom. Let U be an open nbd
of x. Then there exist disjoint open sets V and W such that x ∈ V
and X − U ⊆ W . Since V ∩ W = ∅ and X − W is closed, we have
V ⊆ X − W ⊆ U.
Conversely, let F be a closed subset of X and x ∈ X − F . By our
hypothesis, there exists an open set V such that x ∈ V ⊆ V ⊆ X − F .
♢
Put U = X − V . Then U is a nbd of F and U ∩ V = ∅.
By Proposition 6.1.8, every compact Hausdorff is regular. It is also
clear from the preceding proposition and Theorem 6.4.2 that every
locally compact Hausdorff space is regular.
Proposition 8.1.3 A subspace of a regular space is regular.
Proof. This follows immediately from the definition of regularity and
the fact that a closed subset of a subspace is the intersection of a closed
subset of the space with the subspace.
♢
SEPARATION AXIOMS
213
Theorem 8.1.4 Let
∏ Xα , α ∈ A, be a family of regular spaces. Then
the product space Xα is regular.
∏
Proof. Since each Xα satisfies the∏
T1 -axiom, so does Xα . To see that
it is a T3 -space, let x = (x∏
X
of x. Then
α) ∈
∏α and O be any nbd∏
there is a basic open set
Uα in
Xα such that x ∈
Uα ⊆ O.
Assume that Uα = Xα for all α ̸= α1 , . . . , αn . For each i = 1, . . . , n,
Uαi , since Xαi is regular.
there exists a nbd Vαi of xαi such that V αi ⊆∏
Write∏Vα = X∏
V∏
α for α ̸= α1 , . . . , αn . Then
α is a nbd of x such
that V α ⊆ Uα ⊆ O. By Proposition 8.1.2, Xα is T3 , and thus
regular.
♢
∏
Note that if Xα is regular, then so is each Xα because regularity
is hereditary. Clearly, regularity is a topological invariant. But, like the
Hausdorff property, it is also not preserved under continuous closed or
open maps, as shown below.
Example 8.1.5 Consider the the subspace X = I × {0, 1} of R2 and
let ∼ be the equivalence relation (t, 0) ∼ (t, 1), t ∈ [0, 1). Clearly, the
projection map π : X → X/ ∼ is open, and the quotient space X/ ∼
is T1 . Also, the points x0 = (1, 0) and x1 = (1, 1) do not have disjoint
saturated open nbds in X, so the points π(x0 ) and π(x1 ) cannot be
separated by open sets. Thus X/ ∼ fails to satisfy the T3 -axiom.
{
}
Example 8.1.6 Let H = (x, y) ∈ R2 |y > 0 and L = {(x, 0)|x ∈ R}.
Let T be the topology on Z = H ∪ L generated by the basis consisting
of the open balls contained in H, and the sets {(x, 0)}∪B((x, r); r); the
latter are the open balls in the upper half plane which are tangent to
the x-axis together with the points of tangency (see Figure 8.1 below).
H
L
r
•
x
FIGURE 8.1: Basic open sets in Z.
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Elements of Topology
It is easy to see that T is finer than the euclidean subspace topology
for Z so that it is T1 . To see that Z is regular, let U be an open nbd
of p ∈ Z. If p ∈ H, then there is an open ball B(p; r) ⊆ U , and the
set V = B(p; r/2) satisfies p ∈ V ⊆ V ⊆ U . If p = (x, 0) ∈ L, then
there is an open ball B((x, r); r) such that {p} ∪ B((x, r); r) ⊆ U . The
set {p} ∪ B((x, r/2); r/2) = V is a nbd of p with V ⊆ U . So (Z, T) is
regular.
Next, we observe
that every subset F ⊆ L is closed in Z. It is ob∩
vious that F = p∈F
/ (L − {p}). Also, for each p ∈ L, H ∪ {p} is open
in Z so that L − {p} is closed in Z. This F is closed. In particular, the
disjoint sets A = {(q, 0)|q is rational} and B = {(r, 0)|r is irrational}
are closed. Now, consider the equivalence relation ∼ on Z, which identifies A to a point. Obviously, the quotient map π : Z → Z/ ∼ is closed
and continuous. We claim that Z/ ∼ does not satisfy the T3 -axiom.
If there are disjoint nbds of the point [A] and the closed set π(B) in
X/ ∼, then we have disjoint open sets G and H in Z containing A and
B, respectively. So, for each rational q, there exists a real δq > 0 such
that Uq = {(q, 0)}∪B ((q, δq ); δq ) ⊆ G. Also, for each irrational r, there
exists a real δr > 0 such that Vr = {(r, 0)} ∪ B ((r, δr ); δr ) ⊆ H. For
each n = 1, 2, . . . , let Sn = {(r, 0) ∈ B|δr > 1/n}. Then the family of
2
all Sn and A covers L. As
( a )subspace of the euclidean space R , L is
complete, and hence int Sn0 ̸= ∅ for some n0 . (If the reader is not
already familiar with this result, (s)he may refer to Theorem 10.3.4.)
So there exist reals s < t such that N = {(x, 0) |s < x < t} ⊆ Sn0 .
Since N is open in L, N ∩ Sn0 is dense in N . So we can find a rational
q ∈ (s, t) and a real ϵ such that 0 < ϵ < min{q − s, t − q, δq + 1/n0 }.
Then there exists a point (r, 0) ∈ Sn0 such that |r −q| < ϵ. This implies
that Uq ∩ Vr ̸= ∅ and, therefore, G ∩ H ̸= ∅, a contradiction. Hence
our claim.
In Chapter 7, we have already seen that the product of two quotient
maps is not necessarily a quotient map. Here is another example justifying this statement. Consider the space Z in Ex. 8.1.6, and let ∼ be the
equivalence relation on Z with the graph G(∼) = ∆∪(A×A)∪(B ×B),
where A, B are defined as in the above example and ∆ denotes the
diagonal of Z × Z. This relation identifies A and B to points. If
π : Z → Z/ ∼ is the natural projection, then the image of the saturated open set (Z × Z) − [∆ ∪ (A × A) ∪ (B × B)] under the map
π × π : Z × Z → (Z/ ∼) × (Z/ ∼) is not open, and so π × π is not a
quotient map.
SEPARATION AXIOMS
215
Note that the quotient space X/ ∼ of a space X is T3 if and only if
for any equivalence class [x] and any saturated closed set F ⊆ X with
F ∩ [x] = ∅, there exist disjoint saturated open nbds of [x] and F in
X.
Exercises
1. On the set {a, b, c}, find a topology which is T3 but not T2 .
2. Is the space X in Example 7.1.12 regular?
3. Let H, L and Z be as in Ex. 8.1.6. Show that:
(a) The family B of open balls contained in H, and the sets {(x, 0)} ∪
B((x, 0); r) ∩ H is a basis for a topology on Z.
(b) The topology generated by B is T2 but not T3 .
4. Let X be a T2 -space such that each point of X admits a closed nbd
which is T3 . Show that X is T3 .
5. Let A be a closed subspace of the regular space X. Show that
(a) for each point x ∈
/ A, there are open nbds U of A and V of x such
that U ∩ V = ∅, and
(b) A coincides with the intersection of its closed nbds.
6. Let X be a space such that for each x ∈ X and each subbasic nbd U
of x, there exists a nbd V of x with V ⊆ U . Prove that X satisfies the
T3 -axiom.
7. Let f : X → Y be a proper surjection. If X is regular, show that Y is
also regular.
8. Prove that a countably compact, first countable T2 -space is regular.
9. Let X be a regular space and A ⊆ X infinite. Show that there exists a
countable family {Un } of open sets such that A ∩ Un ̸= ∅ for all n and
U n ∩ U m = ∅ for n ̸= m.
10. Let X be a regular locally connected space, and U ⊂ X be connected
and open. If x, y ∈ U , prove that there is a closed connected set C ⊂ X
such that x, y ∈ C ⊂ U .
11. Let X be a T3 -space. Show that
(a) the relation x ∼ y if every nbd of x contains y is an equivalence
relation on X;
(b) the natural projection π : X → X/ ∼ is both open and closed;
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Elements of Topology
(c) X/ ∼ is regular; and
(d) if Y is T2 and ϕ : X → Y is continuous, then there exists a
continuous map ψ : X/ ∼ → Y such that ϕ = ψπ.
12. Prove that the sum of a family of T3 -space is T3 .
13. If X is regular and A is closed in X, show that X/A is Hausdorff.
14. Let A be a closed nbd retract of a regular space X (that is, A is closed
in X, and has a nbd N in X such that there exists a continuous map
r : N → A fixing every element of A). If Y is Hausdorff, prove that
X ∪f Y is Hausdorff for every continuous map f : A → Y .
15. Give an example to show that an adjunction space of two regular spaces
need not be regular.
16.
(a) Let ∼ be an equivalence relation on a regular space X. If the
natural projection X → X/ ∼ is closed, show that G(∼) is closed
in X × X. (cf. Ex. 7.1.10).
(b) Give an example of an equivalence relation ∼ on a regular space X
such that G(∼) is closed in X × X, but the projection X → X/ ∼
is not closed.
8.2
Normal Spaces
In this section, we shall discuss the condition which separates closed
sets by open sets. Topological spaces satisfying this condition are quite
close to metric spaces; in fact, it gives a characterisation of second
countable metric spaces. We will also see that a continuous function
defined on a closed subspace of a topological space with this property
admits a continuous extension to the whole space.
Definition 8.2.1 A space X is T4 if each pair of disjoint closed subsets
of X have disjoint nbds. X is normal if it satisfies both the T1 and T4
axioms.
Example 8.2.1 Every compact Hausdorff space is normal, by Theorem
6.1.9.
SEPARATION AXIOMS
217
Example 8.2.2 Every metric space is normal. Let X be a metric space
and A, B be disjoint closed subsets of X. For each
∪ a ∈ A, there is an
open ball B(a; ra ) disjoint from B. Put U = a∈A B(a; ra /2). Then
U
∪ is a nbd of A. Similarly, construct a nbd V of B, where V =
b∈B B(b; rb /2) and B(b; rb ) ∩ A = ∅ for all b ∈ B. If U ∩ V ̸= ∅, then
there exist points a ∈ A and b ∈ B such that d(a, b) < max{ra , rb }.
This forces a ∈ B(b; rb ) or b ∈ B(a; ra ), a contradiction in either case.
Therefore the nbds U and V are disjoint.
In particular, each euclidean space Rn is normal.
Axiom-T4 is neither stronger nor weaker than axiom-T3 .
Example 8.2.3 The right-hand topology on R, which consists of ∅, R
and all open right rays (a, +∞) is T4 , since there are no disjoint closed
sets. But it is not T3 , for there are no disjoint open sets.
Example 8.2.4 The regular space Z in Ex. 8.1.6 is not T4 , since
the disjoint closed sets A = {(q, 0)|q is rational} and B =
{(r, 0)|r is irrational} do not have disjoint nbds in Z.
A normal space is obviously regular but, as seen in the preceding
example, a regular space is not normal in general. For Lindelöf spaces,
regularity forces normality.
Theorem 8.2.2 A regular Lindelöf space is normal.
Proof. Let X be a Lindelöf regular space and A, B be disjoint closed
sets in X. Then, for each a ∈ A, there exists an open nbd Ua of a
such that U a ⊆ X − B. Similarly, for each b ∈ B, we find an open
nbd Vb of b such that V b ⊆ X − A. Since X is Lindelöf, the open
covering {Ua |a ∈ A} ∪ {Vb |b ∈ B} ∪ {X − (A ∪ B)} has a countable
subcovering. So there is a countable subfamily {Uan } of {Ua |a ∈ A}
which covers A, and a countable subfamily {Vbn } of
{ b |b ∈ B} }which
∪ {V
V bi |i ≤ n , and
covers B. For
each
integer
n
≥
1,
let
G
=
U
−
n
an
}
∪{
Hn = Vbn −
U ai |i ≤ n . Then the sets Gn and Hn are open for each
n and Gn ∩ Hm = ∅ for all n and m. Since {Uan } covers A and no V bi
meets A, {Gn } is an open
of A. Similarly, {Hn } is an open cover
∪ cover ∪
of B. It follows that Gn and Hn are disjoint open nbds of A and
B, respectively, and X is normal.
♢
Corollary 8.2.3 Every second countable regular space is normal.
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Elements of Topology
Regarding the invariance properties of normal spaces, we have
Proposition 8.2.4 Let X be normal space. Then
(a) every closed subspace of X is normal, and
(b) each closed continuous image of X is normal.
Proof. (a): This follows from the fact that a closed subset of a closed
subspace is a closed subset of the space.
(b): Let f : X → Y be a continuous closed surjection. Then Y is
obviously T1 . Suppose that A, B are disjoint closed sets in Y . Then
f −1 (A) and f −1 (B) are disjoint closed subsets of X. So there exist
disjoint open sets U ⊇ f −1 (A) and V ⊇ f −1 (B). Since f is closed,
U ′ = Y − f (X − U ) and V ′ = Y − f (X − V ) are open subsets of Y with
A ⊆ U ′ and B ⊆ V ′ . Since f −1 (U ′ ) ⊆ U, f −1 (V ′ ) ⊆ V and U ∩ V = ∅,
we have U ′ ∩ V ′ = ∅. This proves (b).
♢
We remark that nonclosed subspaces of normal spaces may fail to be
normal; however, an example exhibiting this situation is not convenient
to give at this stage. Moreover, the product of two normal spaces need
not be normal.
Example 8.2.5 The Sorgenfrey line Rℓ is normal. To see this, let A, B
be disjoint closed subsets of Rℓ . Then for each a ∈ A, there exists xa > a
such that [a, xa ) ⊆ R − B, and for each b ∈ B, there exists xb > b such
that [b, xb ) ⊆ R − A. It is easily
∪ checked that [a, xa ) ∩ [b,
∪xb ) = ∅ for all
a ∈ A and b ∈ B. So U = {[a, xa )|a ∈ A} and V = {[b, xb )|b ∈ B}
2
are disjoint nbds of A and B, respectively.
} R2ℓ is
{ But the2 product space
not normal. For, the antidiagonal L = (x, y) ∈ Rℓ |x + y = 0 of Rℓ is
closed and discrete (see Ex. 5.2.4), and so a proof similar to Ex. 8.1.6
can be given to show that the disjoint closed sets {(x, y) ∈ L|x, y ∈ Q}
and {(x, y) ∈ L|x, y ∈ R − Q} do not have disjoint nbds in R2ℓ .
Alternatively, consider the subsets
{
}
D = (x, y) ∈ R2ℓ |x and y are rationals ,
{
}
L = (x, y) ∈ R2ℓ |x + y = 0
and
of R2ℓ . Clearly, D is countable and dense, and L is uncountable discrete
and closed in R2ℓ . If R2ℓ were normal, then for every subset F ⊆ L,
there would be disjoint open sets U (F ) and V (F ) such that F ⊆ U (F )
SEPARATION AXIOMS
219
and L − F ⊆ V (F ). Accordingly, for any two subsets F1 ̸⊆ F2 of L,
U (F1 ) ∩ V (F2 ) is a nonempty open subset of R2ℓ . Since the set D is
dense in R2ℓ , D ∩ U (F1 ) ∩ V (F2 ) ̸= ∅. Also, it is obvious that D ∩ U (F2 )
is disjoint from D ∩ V (F2 ). Consequently, D ∩ U (F1 ) and D ∩ U (F2 ) are
distinct subsets of D,and there is an injection from P(L) (the power
set of L) into P(D). So we have c = |L| < |P(L)| ≤ |P(D)| = c (see
Propositions A.8.1, A.8.5 and A.8.6), a contradiction.
In view of the fact that products of normal spaces need not be
normal, the following consequence of the Tychonoff theorem is interesting: The product of any family of closed unit intervals is compact
Hausdorff, and therefore normal.
It is evident from Ex. 8.1.5 that not every quotient of a normal
space is normal (or T4 ). Clearly, the quotient space X/ ∼ of a space X
is T4 if and only if for any saturated, closed subsets of A and B of X
with A ∩ B = ∅, there exist disjoint saturated open nbds of A and B
in X. Also, if the saturation of every closed subset of a normal space
X under an equivalence relation ∼ is closed, then the quotient space
X/ ∼ is normal, by Proposition 8.2.4(b). In particular, if X is normal
and A ⊆ X is closed, then X/A is normal. Moreover, this property
behaves well with the construction of adjunction spaces.
Theorem 8.2.5 If X and Y are normal spaces and A ⊆ X is closed,
then X ∪f Y is normal for any map f : A → Y .
Proof. Since each of the equivalence classes of X + Y is closed, X ∪f Y
is T1 . Now, let B1 and B2 be disjoint closed subsets of X ∪f Y , and let
π : X +Y → X ∪f Y be the natural projection. Then Fi = π −1 (Bi )∩Y,
i = 1, 2, are disjoint closed sets in Y . As Y is normal, there exist open
sets Gi ⊆ Y such that Fi ⊆ Gi for i = 1, 2, (and) G1 ∩ G2 = ∅. Since
π|Y is a closed injection, the sets Ci = Bi ∪π Gi , i = 1, 2, are disjoint
and closed in X ∪f Y . So π −1 (C1 ) ∩ X and π −1 (C2 ) ∩ X are disjoint
closed subsets of X. By the normality of X, there exist disjoint open
sets O1 , O2 in X such that π −1 (Ci ) ∩ X ⊆ Oi . It is clear that Ui =
π ((Oi − A) ∪ Gi ) contains Bi , and U1 ∩ U2 = ∅. We show that U1 and
U2 are open. We have π −1 (Ui )∩Y = Gi and π −1 (Ui )∩X = (Oi − A)∪
f −1 (Gi ). Since f −1 (Gi ) is open in A, there exist open sets Vi ⊆ X,
i = 1, 2, with f −1 (Gi ) = Vi ∩ A. Obviously, f −1 (Gi ) ⊆ π −1 (π (Gi )) ∩
X ⊆ Oi so that (Oi − A) ∪ f −1 (Gi ) = (Oi − A) ∪ (Oi ∩ Vi ∩ A) =
Oi ∩ ((X − A) ∪ (Vi ∩ A)) = Oi ∩ ((X − A) ∪ Vi ), which is open. This
completes the proof.
♢
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Elements of Topology
We now come to the first major result of the book, the proof of
which consists of a different idea not encountered so far. It guarantees the existence of non-constant continuous real valued functions on
normal spaces – a purely topological assumption. This fact is used to
prove many important theorems for normal spaces; we shall content
ourselves with two of them – the Urysohn metrisation theorem and the
Tietze extension theorem. These theorems justify the use of the adjective “normal” for this class of the spaces despite the abnormalities
shown by them when forming subspaces and products.
A dyadic rational number is a number of the form m/2n , where m
and n > 0 are integers. The set of all dyadic rational numbers is dense
in the real line R, for if r < r′ are two real numbers, then we can find
integers m and n > 0 such that 2−n < r′ − r and m − 1 < 2n r < m.
Theorem 8.2.6 (Urysohn Lemma) If X is a normal space, then for
each pair of nonempty disjoint closed subsets A and B of X, there is a
continuous function f : X → I such that f (A) = {0} and f (B) = {1}.
Proof. Let D be the set of dyadic rational numbers of I. For each t ∈ D,
we associate an open nbd Ut of A such that
(a) B ∩ Ut = ∅, and
(b) U t ⊆ Ut′ for t < t′ .
Set U1 = X − B. Then U1 is an open nbd of A. Since X is T4 , there
exists an open set U0 ⊆ X such that A ⊆ U0 ⊆ U 0 ⊆ U1 . For the same
reason, we find an open set U1/2 such that U 0 ⊆ U1/2 ⊆ U 1/2 ⊆ U1 .
Suppose that for some n the open sets Um/2n satisfying conditions (a)
and (b) are already defined. Then,
( using the normality of
) X again,
we can find open sets Um/2n+1 m = 1, 3, 5, . . . , 2n+1 − 1 such that
U (m−1)/2n+1 ⊆ Um/2n+1 ⊆ U m/2n+1 ⊆ U(m+1)/2n+1 . By induction, for
each t ∈ D, there exists an open set Ut such that the conditions (a)
and (b) hold.
Define a function f : X → I by
{
inf {t|x ∈ Ut } for x ∈ U1 , and
f (x) =
1
for x ∈ X − U1 .
It is evident that f (x) = 0 if x ∈ A, and f (x) = 1 if x ∈ B. To
establish the continuity of f , we note that the intervals [0, r) and (r, 1],
0 < r < 1, constitute a subbase of I. Therefore it suffices to prove
SEPARATION AXIOMS
221
−1
that the sets f −1 ([0, r)) and
are open for every 0 < r <
∪ f ((r, 1])−1
−1
1. Clearly, f ([0, r)) = t<r U∩t , and f ∩((r, 1]) = X − f −1 ([0, r]).
We observe that f −1 ([0, r]) = t>r Ut = t>r U t . Consider the first
equality. Assume that f (x) ≤ r. If t ∈ D and r < t, then f (x) < t,
so there an s ∈ D such that s < t and x ∈ Us . By condition (b),
x ∈ Ut . Conversely, suppose that x ∈ Ut for all t > r and t ∈ D.
If r < f (x), then there is a t ∈ D such that r < t < f (x). By our
assumption, x ∈ Ut , which implies that f (x) ≤ t, a contradiction.
It
∩
follows that f (x) ≤∩r and the ∩equality f −1 ([0, r]) = t>r Ut holds.
Next, the inclusion t>r Ut ⊆ t>r U t is obvious. To see the reverse
inclusion, assume that x ∈
/ Ut for some t > r. Then
∩ there is an
∩s ∈ D
such that t > s > r and, by (b), x ∈
/ U s . Hence t>r
U
⊆
t>r)Ut ,
∪ t(
−1
and the second equality holds. Thus f ((r, 1]) = t>r X − U t is
also open, and this completes the proof.
♢
Remarks 8.2.7 (a) The proof of the above theorem does not use
axiom T1 ; accordingly the theorem is true for any T4 -space.
(b) A continuous function f : X → I such that f (A) = 0 and f (B) =
1 is referred to as a Urysohn function for the pair A, B. Notice
that a Urysohn function f for a pair A, B obviously satisfies A ⊆
f −1 (0); the theorem does not assert that A = f −1 (0). Thus, f
may take the value 0 outside of A.
(c) It is clear that if each pair of disjoint closed subsets of X admits
a Urysohn function, then X is T4 . So, the converse holds for T1 spaces.
(d) The unit interval I can be replaced by any closed interval [a, b].
In fact, the composition of a Urysohn function for the pair A, B
with the homeomorphism t → a + (b − a)t between I and [a, b]
is a continuous function from X into [a, b] which maps A into a
and B into b.
We now come to some very useful applications of the Urysohn
Lemma. Recall that a topological space X is metrisable if there exists a metric on the set X that induces the topology of X.
Theorem 8.2.8 (Urysohn Metrisation Theorem) Every regular
second countable space is metrisable.
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Proof. Let X be a regular space with a countable basis B. Since a space
homeomorphic to a subspace of a metric space is metrisable, it suffices
to prove that X can be embedded into a metric space. Each point
x ∈ X belongs to some member B of B and, by regularity of X, we find
another
B ′ of B such that x}∈ B ′ ⊆ B ′ ⊆ B. So the collection
{ member
C = (B, B ′ )|B, B ′ ∈ B and B ′ ⊆ B is nonempty and countable. We
index the pairs (B, B ′ ) in C by positive integers. By Corollary 8.2.3,
X is normal and, therefore, for each integer n > 0, there is a Urysohn
function fn : X → I such that fn (x) = 0 if x ∈ Bn′ , and fn (x) = 1 if
x∈
/ Bn . For x ∈ X, write ϕ(x) = (f1 (x), f2 (x)/2, f3 (x)/3, . . .). Then
ϕ(x) ∈ ℓ2 , the Hilbert space. Thus, there is a function ϕ : X → ℓ2 . We
assert that ϕ is an embedding. First, if x ̸= y in X, then there is a
pair (Bn , Bn′ ) in C such that x ∈ Bn′ and y ∈
/ Bn , by regularity of X.
So fn (x) = 0 and fn (y) = 1. This implies that ϕ(x) ̸= ϕ(y), and ϕ is
injective. Next, to see the continuity ∑
of ϕ, let x0 ∈ X be a fixed point
and ϵ > 0 be given. Since the∑series
n−2 converges, we can choose
−2
an integer m > 0 such that n>m n < ϵ2 /2. By the continuity of
√
fn , there is a nbd Vn ∩
of x0 such that |fn (x) − fn (x0 )|∑< nϵ/ 2m for
m
m
all x ∈ Vn . Then U = 1 Vn is a nbd of x0 such that 1 n−2 |fn (x) −
fn (x0 )|2 < ϵ2 /2 for every x ∈ U . It follows that ∥ϕ(x) − ϕ(x0 )∥ < ϵ
for all x ∈ U , and ϕ is continuous at x0 . Finally, we show that the
function ψ : ϕ(X) → X, the inverse of ϕ, is continuous.( Given a )point
y0 ∈ ϕ(X) and a nbd N of ψ(y0 ) = x0 , we find a pair Bn0 , Bn′ 0 in C
′
such that x0 ∈ B
and Bn0 ⊆ N . Then, for ϵ = 1/n0 and ϕ(x) = y,
∑n0 −2
∥y − y0 ∥ < ϵ ⇒
n |fn (x)−fn (x0 )|2 < ϵ2 ⇒ |fn0 (x) − fn0 (x0 )| < 1.
Since fn0 vanishes on Bn′ 0 , we have |fn0 (x)| < 1, which forces x ∈ Bn0 .
Thus, if y ∈ ϕ(X) and ∥y − y0 ∥ < ϵ, then ψ(y) ∈ N. This establishes
the continuity of ψ at y0 , and our assertion follows.
♢
Notice that we have proved a little more than the theorem actually
requires; in fact, we have established the following.
Theorem 8.2.9 (Urysohn Embedding Theorem) Every regular
second countable space can be embedded in ℓ2 .
Since a metric space is certainly regular, the Urysohn Metrisation
theorem can be restated as
Theorem 8.2.10 A second countable space is metrisable if and only
if it is regular.
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The last theorem is a characterisation of second countable metric spaces. But there are metric spaces which are not separable and,
therefore, second countable. So this is not a characterisation of general
metric spaces. In Chapter 9, we shall see a complete characterisation
of metric spaces, separable or not.
Another fundamental problem in topology is the following “extension problem”: Whether or not a continuous function defined on a
subspace of a topological space admits a continuous extension to the
whole space. For example, the Urysohn lemma implies that the function g : A∪B → I defined by g(A) = 0, g(B) = 1 has an extension over
X, if A and B are closed subsets of the normal space X. On the other
hand, the identity function on the unit circle S1 cannot be extended to
a continuous function on the unit disc D2 . It is difficult to justify the
statement at this stage, but we shall do it in Chapter 14.
Theorem 8.2.11 (Tietze Extension Theorem) If A is a closed
subset of the normal space X, then every continuous function f : A →
R extends to a continuous function g : X → R. Moreover, g can be chosen so that inf x∈X g(x) = inf a∈A f (a) and supx∈X g(x) = supa∈A f (a).
Proof. To begin with, we shall assume that f is bounded, and let α =
inf a∈A f (a) and β = supa∈A f (a). Then f maps A into [α, β]. If f is
constant, the theorem is obviously true. So we may assume that α < β.
Then the mapping h : x → (2x − β − α)/(β − α) is a homeomorphism
between [α, β] and [−1, 1]. It is clear that if g : X → [−1, 1] is a
continuous extension of hf , then h−1 g : X → [α, β] is a continuous
extension of f . So we can assume further that α = −1 and β = 1.
Now, define two subsets E1 and F1 of A by E1 = f −1 ([−1, −1/3]) and
F1 = f −1 ([1/3, 1]). Since A is closed in X, E1 and F1 are nonempty,
disjoint, closed subsets of X. By Theorem 8.2.6, there is a continuous
function g1 : X → [−1/3, 1/3] which takes the value −1/3 on E1 ,
and 1/3 on F1 . Observe that |f (x) − g1 (x)| ≤ 2/3 for x ∈ A, and
|g1 (x)| ≤ 1/3 for x ∈ X. Assume, by induction, that we have already
constructed continuous functions gi : X → R, i = 1, 2, . . . , n, such that
|f (x) −
∑n
1
gi (x)| ≤ (2/3)n
|gi (x)| ≤ 2
i−1
/3
if x ∈ A, and
i
(∗)
if x ∈ X.
Then, to construct gn+1 , we repeat the above process with f −
∑n
1
gi
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in place of f . Specifically, we consider the closed sets
}
{
∑n
En+1 = x ∈ A|f (x) − 1 gi (x) ≤ −2n /3n+1 , and
{
}
∑n
Fn+1 = x ∈ A|f (x) − 1 gi (x) ≥ 2n /3n+1 ,
and apply Theorem 8.2.6 again to find a continuous function
gn+1 : X → [−2n /3n+1 , 2n /3n+1 ]
such that gn+1 (E
) = −2n /3n+1 and gn+1 (Fn+1 ) = 2n /3n+1 .
∑n+1
n+1
Clearly, |f (x) − 1 gi (x)| ≤ (2/3)n+1 for x ∈ A, and |gn+1 (x)| ≤
2n /3n+1 for x ∈ X.
We thus have a sequence of∑functions gn : X → R, n = 1, 2, . . . ,
satisfying (∗). Since the series
2n−1 /3n converges
to 1, we see, by
∑
the second inequality in (∗), that the series
gn (x) converges to a
number in [−1, 1] for every
x
∈
X.
We
may
therefore
define a function
∑
g : X → R by g(x) =
gn (x). Obviously, |g(x)| ≤ 1 for all x ∈ X
and, by the first inequality in (∗), g(x) = f (x) for every x ∈ A. It
remains to show that g is continuous. Let x0 ∈ X be an arbitrary but
fixed point. Then, given ϵ > 0, we can find an integer m such that
∑
i−1 i
/3 < ϵ/4. Since the function gi is continuous, there exists
i>m 2
an open nbd Ui of x0 such that |gi (x) − gi (x0 )| < ϵ/2m for x ∈ Ui . If
x ∈ U1 ∩ · · · ∩ Um , then we have (by the second inequality in (∗))
∑m
∑
|g(x) − g(x0 )| ≤ 1 |gi (x) − gi (x0 )| + 2 i>m 2i−1 /3i < ϵ,
and so g is continuous at x0 .
Finally, suppose that f is not bounded. We choose an appropriate
homeomorphism h
(a) (−∞, ∞) ≈ (−1, 1)
in case f is unbounded in both directions,
(b)
[α, ∞) ≈ [−1, 1)
in case f is bounded below by α,
(c)
(−∞, β] ≈ (−1, 1]
in case f is bounded above by β,
and consider the composition hf . This is bounded by −1, 1 and
has a continuous extension ϕ : X → [−1, 1]. We put B =
{x ∈ X||ϕ(x)| = 1} in case (a), B = {x ∈ X|ϕ(x) = 1} in case (b),
and B = {x ∈ X|ϕ(x) = −1} in case (c). Then B is closed in X
and B ∩ A = ∅. By Theorem 8.2.6, there is a Urysohn function
ψ : X → [0, 1] such that ψ(B) = 0 and ψ(A) = 1. We define g : X → R
SEPARATION AXIOMS
225
by g(x) = ϕ(x)ψ(x). Then g is a continuous extension of hf and maps
X into (-1,1) in case (a), into [-1,1) in case (b) and into (-1,1] in case
(c). It is now immediate that h−1 g is the desired extension of f .
♢
Remarks 8.2.12 (a) If f maps A into (α, β), then an extension g
of f can be found so that g maps X into (α, β) (see the proof of
the unbounded case).
(b) We have proved the Tietze theorem by invoking the Urysohn
lemma. Conversely, the Urysohn lemma can be derived from the
Tietze theorem. For, given disjoint closed subsets A, B of X, the
function f : A ∪ B → R defined by f (a) = 0 for all a ∈ A, and
f (b) = 1 for all b ∈ B is continuous. Then an extension g of f is
a Urysohn function for the pair A and B.
(c) The theorem fails if we omit the assumption that the subspace A
is closed. For example, the continuous function x → log (x/1 − x)
defined on (0, 1) does not extend to a continuous function on [0, 1]
because any continuous function on [0, 1] must be bounded.
(d) The theorem can be generalised to maps into Rn . If A is a closed
subset of the normal space X, then every continuous function
A → Rn extends to continuous function X → Rn . To see this,
one needs to apply the above theorem to the coordinate functions
of the given map A → Rn . The claim remains true if one takes
the cube I n instead of Rn .
Corollary 8.2.13 Let A be a closed subset of a normal space X, and
f be a continuous function of A into the n-sphere Sn , n ≥ 0. Then
there is an open set U ⊆ X such that A ⊆ U , and f has an extension
U → Sn.
Proof. By the preceding theorem, Sn ≈ ∂I n+1 . Let h : Sn → ∂I n+1 be
a homeomorphism. Then the composition
f
h
A−
→ Sn −
→ I n+1
is continuous. If pi : I n+1 → I is the projection map onto the ith
factor, 1 ≤ i ≤ n + 1, then each composition pi hf can be extended
to a continuous mapping gi : X → I, by the Tietze theorem. Define
g : X → I n+1 so that pi g = gi for every i. Then g is a continuous
extension of hf . Now we choose the point z = (1/2, . . . , 1/2) ∈ I n+1 .
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Then for each point y ̸= z in I n+1 , the ray from z through y meets
∂I n+1 in exactly one point k(y), say. The function k : I n+1 − {z} →
∂I n+1 taking y into (k(y) is clearly
) continuous and fixes the points in
∂I n+1 . Set U = g −1 I n+1 − {z} . Then U is open in X and A ⊆ U =
X − g −1 (z). It is obvious that g maps U into I n+1 − {z}, and the
composition kg : U → ∂I n+1 is defined and continuous. If x ∈ A, then
kg(x) = khf (x) = hf (x) ⇒ h−1 kg(x) = f (x). Thus h−1 kg : U → Sn
is a continuous extension of f .
♢
Exercises
1. On the set {a, b, c}, find a topology which is T4 but not T3 .
2. Let I be the unit interval with the euclidean subspace topology, and let
J ⊆ I be the set of all irrationals. Consider the topology T generated
by the subbasis consisting of subsets of J and the open subsets of I.
Prove that (I, T) is normal.
3. Prove that a countable regular space is normal.
4. In the space R2ℓ , consider the sets A = {(x, −x)|x ∈ Q}, and B =
{(x, −x)|x ∈ R − Q}. Show that A and B are disjoint closed subsets of
R2ℓ which fail to have disjoint nbds.
5. Show that the space Z in Ex. 8.1.6 is separable, but not Lindelöf. (Thus,
a regular separable space need not be normal.)
6. Prove that the ordinal space [0, Ω) is normal. Generalise this to an
ordered space with a well-ordering.
7. Let X be a T3 -space, K a compact subset of X and U an open nbd of
K in X. Show that there exists an open set V ⊆ X such that K ⊆ V ⊆
V ⊆ U . Deduce that a compact T3 -space is T4 .
8. Give an example of a proper map f : X → Y such that Y is normal,
but X is not.
9. Prove that the sum of a family of T4 -spaces is T4 .
10. Prove that the wedge sum of a family of normal pointed spaces is normal.
11. Suppose that a space Y is a coherent union of a countable family of
closed subsets Xn . If each Xn is normal, show that Y is normal.
12. Suppose that X is a binormal space (i.e., X × I is normal) and Y is
a normal space. Prove that Mf is normal for any continuous function
f :X →Y.
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227
13. Show that the condition of closedness on the sets A and B in the
Urysohn lemma is essential.
14. Let D be a dense subset of the non-negative reals. Suppose that for each
r ∈ D, there
∪ is an open subset Ur of a space X such that r < s ⇒ U r ⊆
Us , and r∈D Ur = X. Show that the function f : X → R defined by
f (x) = inf {r|x ∈ Ur } is continuous.
15.
(a) Prove that a T1 -space is normal ⇔ for each closed A ⊆ X and
for every open U ⊇ A, there exists an open set V ⊆ X such that
A ⊆ V ⊆ V ⊆ U.
(b) If A, B are disjoint closed subsets of a normal space X, show that
there are open sets U ⊇ A and V ⊇ B such that U ∩ V = ∅.
∩n
(c) Prove (b) for finitely many closed sets A1 , . . . , An with 1 Ai = ∅.
16. Prove that a T1 -space X is normal if and only if for each finite covering {U1 , . . . , Un } of X by open sets, there exist continuous functions
f1 , . . . , fn of X into I such that fi (x) = 0 for x ∈
/ Ui and
∑
fi (x) = 1 for all x ∈ X.
17. Let A be a subset of a space X.
(a) If there exists a continuous function f : X → R such that A =
f −1 (0), show that A is a closed Gδ -set.
(b) If X is normal and A is a closed Gδ -set, prove that there exists a
continuous real-valued function f such that A = f −1 (0).
18. If A and B are disjoint closed Gδ -sets in a normal space X, show that
there exists a continuous map f : X → I such that A = f −1 (0) and
B = f −1 (1).
19. Let K be a compact subset of a locally compact Hausdorff space X and
U be an open set in X with K ⊆ U . Show that there exists a continuous
function f : X → I such that f (x) = 0 for x ∈
/ U and f (x) = 1 for
x ∈ K.
20. Does there exist a countable connected regular space?
21. Is a second countable Hausdorff space metrisable?
22. Prove that every separable metric space is homeomorphic to a subspace
of a Fréchet space.
23. Prove that the continuous image of a compact metric space in a Hausdorff space is metrisable.
24. Suppose that X is a compact Hausdorff space. If there exists a continuous function f : X × X → R such that f (x, y) = 0 ⇔ x = y, prove
that the diagonal ∆ in X × X is a Gδ -set, and hence deduce that X is
second countable. (Thus X is metrisable.)
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25. Prove that a countably compact Hausdorff space is metrisable if and
only if it is second countable.
26. Prove that a metrisable space has a totally bounded metric ⇐⇒ it is
second countable.
27. Prove that every infinite subset of a pseudocompact normal space has
a limit point.
28. Let X be a metrisable space. Prove that the following conditions are
equivalent:
(a) X is compact.
(b) X is bounded in every metric that gives the topology of X.
(c) X is pseudocompact.
29.
(a) If a Hausdorff space X is the union of two compact metrisable
spaces, show that X is metrisable. (This is known as the addition
theorem for compacta).
(b) Give an example of a non-metrisable space that is the union of two
metrisable subsets.
30. Show that the one-point compactification of a second countable, locally compact Hausdorff space is metrisable. (This implies that a second
countable, locally compact Hausdorff space is metrisable.)
31. A space X is called T5 if for every pair of separated sets A, B (i.e., with
A ∩ B = ∅ = A ∩ B) in X, there exist disjoint open sets containing
them. X is said to be completely normal if it is both T1 and T5 .
Prove: (a) A metric space is completely normal.
(b) Rℓ is completely normal.
(c) R2ℓ is not completely normal.
(d) Complete normality is hereditary.
(e) X is completely normal if and only if every subspace of X is
normal.
(f) A regular second countable space is completely normal.
(g) An ordered space is completely normal.
32. A normal space in which each closed set is Gδ is called perfectly normal.
(a) Prove that a metric space is perfectly normal.
(b) Show that a perfectly normal space is completely normal.
(c) Give an example of a completely normal space which is not perfectly normal.
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8.3
229
Completely Regular Spaces
Let A, B be a disjoint pair of subsets of the space X. The previous
section is concerned with the question whether A and B can be separated by open sets: Do there exist disjoint open nbds U, V of A, B, respectively? The present section is concerned with the question whether
A and B can be separated by a continuous function: Does there exist
a Urysohn function f for the pair A, B? The latter condition implies
the former, since we can take U = f −1 [0, 1/2), V = f −1 (1/2, 1]. The
Urysohn lemma says that in a normal space we can separate each pair
of disjoint closed sets by a continuous function, and thus the two notions are equivalent. Naturally, one would be interested to know if an
analogous statement holds good in a regular space, where points can
be separated from closed sets by open sets. The answer is no, because
there is a regular space, due to E. Hewitt, on which every continuous real-valued function is constant. This suggests a new separation
condition intermediate between regularity and normality.
Definition 8.3.1 A space X is T3 21 if for each closed set F ⊆ X and
each point x ∈ X − F , there exists a continuous function f : X → I
such that f (F ) = 0 and f (x) = 1. A space which satisfies both T1 and
T3 21 axioms is called completely regular or Tychonoff.
Example 8.3.1 Every metric space is completely regular. For if X is a
metric space, F is a closed subset of X and p ∈
/ F , then f : X → R
defined by f (x) = dist(x, F ) is continuous. If r = f (p) > 0, then the
mapping x → | sin (f (x)π/2r) | is a Urysohn function for F and {p}.
It is obvious that a completely regular space is regular. Since points
are closed in normal spaces, it follows from the Urysohn lemma that
normal spaces are completely regular. There are, however, completely
regular spaces which fail to be normal.
Example 8.3.2 The space Z in Ex. 8.1.6 is completely regular. To see
this, consider a closed subset F ⊂ Z and a point p in Z − F . Let U
be a basic nbd of p contained in Z − F . If p ∈ H (the open upper half
plane), then U is a disc B centered at p, and if p ∈ L (the x-axis), then
U = B ∪ {p}, where B is a disc tangent to L at p. We define a function
f : Z → R by sending each straight line segment from p to a point on
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the boundary of B linearly onto I, and z ∈
/ U to 1. More precisely, we
put in the first case
{
1
for z ∈
/ U,
f (z) =
|z − p|/r for z ∈ U ,
where r is the radius of U, and in the second case

for z ∈
/ U,

 1
0
for z = p,
f (z) =


2
|z − p| /2ry for z ∈ B,
where y is the ordinate of z and r is the radius of B. To see the
continuity of f in the first case, we observe that the function z →
|z − p|/r is defined on ∂U and assumes value 1. Therefore the Gluing
lemma applies, and f is continuous. Similarly, the continuity of f in
the latter case is established, once we check that it is continuous at
p = (x, 0). But this is clear because the nbd B((x, δ); δ) ∪ {p} of p is
mapped by f into [0, ϵ), where δ = rϵ. It follows that f is a Urysohn
function for F and p, and Z is completely regular. We have already
seen (in Ex. 8.2.4) that it is not normal.
Example 8.3.3 Every subspace of a normal space is completely regular.
Suppose that Y is a subspace of a normal space X. Let A be a closed
subset of Y and y ∈
/ A. We have A = Y ∩ F for some closed subset
F of X. Since X is normal and y ∈
/ F , there is a Urysohn function
f : X → I for F and y. Clearly, the restriction of f to Y is the desired
function.
The argument given in the preceding example shows that every
subspace of a completely regular space is completely regular.
Theorem 8.3.2 Every locally compact Hausdorff space is completely
regular.
Proof. Let X be a locally compact Hausdorff space, and let X ∗ be
its one-point compactification. Then X ∗ , being compact Hausdorff, is
normal and, therefore, completely regular. Since every subspace of a
completely regular space is completely regular, X is completely regular.
♢
The next theorem exhibits invariance of this property under the
formation of products.
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231
Theorem 8.3.3 The product of a family of completely regular spaces
is completely regular.
Proof. Let X be the product of a family of completely regular spaces
Xα , α ∈ A. Since each Xα is T1 , so is X. Given a closed ∏
set F ⊂ X and
a point x = (xα ) in X − F , we choose a basic open nbd Uα of x that
does not meet F . Then Uα = Xα for all but finitely many indices α, say,
α = α1 , . . . , αn . Since Xαi is completely regular, there is a continuous
function fi : Xαi → I, for each i = 1, . . . , n, such that fi (xαi ) = 1 and
pα
fi
i
fi (Xαi − Uαi ) = 0. The composition X −−→
Xαi −→ I takes x into
−1
1 and vanishes outside pαi (Uαi ), where pαi is the natural projection
map. We define g : X → I by g(y) = min {fi pαi (y)|i = 1, . . . , n} for
every y ∈ ∏
X. Then∪g is continuous
and )g(x) = 1, and g = 0 throughout
n (
F ⊆ X − Uα = i=1 X − p−1
(U
♢
αi ) .
αi
It is easy to see that complete regularity is a topological invariant.
Since this property
is hereditary, it follows that each Xα is completely
∏
regular when Xα is so. Thus the converse of Theorem 8.3.3 is also
true. As has been the case with previously stated separation properties,
this property too does not behave well with the quotient topology.
Example 8.3.4 The quotient space of the completely regular space
Z in Ex. 8.3.2 obtained by identifying the closed set A =
{(x, 0)|x is a rational} to a point is not even regular because the point
[A] and the closed set F = {(x, 0)|x is an irrational} cannot be separated by open sets. Hence Z/A is not completely regular, although the
natural map Z → Z/A is closed.
We now establish an analogue of the Urysohn Embedding Theorem
for completely regular spaces.
Definition 8.3.4 Let Yα , α ∈ A, be a family of spaces, and let Φ be
a family of functions fα : X → Yα , each defined on a fixed space X.
We say that
(a) Φ separates points of X if for each pair of points x ̸= x′ in X,
there is an α ∈ A such that fα (x) ̸= fα (x′ ); and
(b) Φ separates points and closed sets if for each closed set F ⊂ X and
each point x ∈ X −F , there is an α ∈ A such that fα (x) ∈
/ fα (F ).
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∏
The function e : X → Yα defined by e(x)α = fα (x), x ∈ X, is
referred to ∏
as the evaluation map induced by Φ. Note that pβ ◦ e = fβ ,
where pβ : Yα → Yβ is the projection map.
In the above notations and terminology, we have
Proposition 8.3.5 (a) If Φ separates points, then e is an injection,
and conversely.
(b) If Φ separates points and closed sets, then e is an open map of
X onto e(X).
(c) If each fα is continuous, then e is continuous.
Proof. (a): This is obvious.
(b): Let U ⊆ X be open, and x ∈ U . Then, by our hypothesis,
there is an index β ∈ A such that fβ (x) does not belong to fβ (X − U ).
−1
If
∏ O = Yβ − fβ (X − U ), then pβ ∏(O) is a subbasic open subset of
Yα containing e(x), where pβ :
Yα → Yβ is the projection map.
−1
So G = e(X) ∩ pβ (O) is an open nbd of e(x) in subspace e(X). It is
evident that G ⊆ e(U ); consequently, e(U ) is open in e(X).
(c) Suppose that each fα , α ∈ A, is continuous.
Since the compo∏
sition of e with the projection map pβ :
Yβ → Yβ is fβ for every
β ∈ A, e is continuous.
♢
Observe that if X is a T1 -space and a family of continuous functions
X → I separates points from closed sets, then it certainly separates
points. Thus the family Φ of all continuous functions from a completely
regular space X to the closed unit interval I satisfies the hypotheses of
this proposition. It follows from the preceding proposition that a completely∏
regular space X is homeomorphic to a subspace of the product
space
If , where If = I and f ∈ Φ. For any indexing set A, the
cartesian product of A copies of I with the product topology is called
a cube and denoted by I A . With this terminology, the space X can be
embedded in the cube I Φ . Conversely, if X can be embedded in a cube,
then it must be a completely regular space, since the property of being
a completely regular space is hereditary and every cube is completely
regular, by Theorem 8.3.3. Thus we have the following characterisation
of a completely regular space.
8.3.6 Tychonoff Embedding Theorem A space is completely regular if and only if it can be embedded in a cube.
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It follows from the preceding theorem that a space is completely
regular if and only if it is homeomorphic to a subspace of a normal
space.
We end this section with an alternative proof of the Urysohn Metrisation Theorem.
Theorem 8.3.7 A completely regular, second countable space X is
metrisable.
Proof. Let B be a countable basis for X, and let C be the family of
all pairs (B, B ′ ) of members of B such that B ⊇ B ′ . Clearly, C is
countable. For each pair (B, B ′ ) in C, we select a continuous function
f : X → I such that f ≡ 0 on X − B and f ≡ 1 on B ′ , provided
such a function exists. Since C is countable, so is the family Φ of these
functions f. We observe that Φ satisfies the hypotheses of Proposition
8.3.5. If F ⊆ X is closed and x ∈ X − F , then there exists B ∈ B
such that x ∈ B ⊆ X − F . Since X is completely regular, there is a
continuous function g : X → I such that g ≡ 0 on X − B and g(x) = 1.
The function h : I → I given by h(t) = 2t for 0 ≤ t ≤ 1/2 and h(t) = 1
for 1/2 ≤ t ≤ 1 is continuous, so the composition hg = f : X → I is
continuous. Obviously, f ≡ 1 on the open set g −1 ((3/4, 1]) and f ≡ 0 on
X−B. Choose a basic open nbd B ′ of x contained in g −1 ((3/4, 1]). Then
f ≡ 1 on B ′ so that B ′ ⊂ B. Accordingly, we have f ∈ Φ, f (F ) = 0
and f (x) = 1. It follows that Φ separates points and closed sets. Since
X is a T1 -space, Φ also separates points. By Proposition 8.3.5, the
evaluation map e : X → I Φ is an embedding. Since Φ is countable, I Φ
is metrisable (by Theorem 2.2.14), and hence X is metrisable. This
completes the proof.
♢
Exercises
1. Show that R2ℓ is completely regular.
2. If a space X satisfies the T3 and T4 axioms, show that it satisfies T3 12 .
3. Let X be a completely regular space and F ⊆ X be closed. Show that
for each point x ∈ X − F , there is a continuous function f : X → I
such that f ≡ 0 on F and f ≡ 1 on a nbd of x.
4. Prove that a T1 -space X is completely regular if and only if the topology
of X is generated by the cozero-sets X − f −1 (0) of continuous functions
f : X → R (as a base).
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Elements of Topology
5. Let X be a completely regular space. Show that the sets {y ∈ X :
|f (x) − f (y)| < ϵ}, where x ∈ X, f ranges over all bounded continuous
functions X → R, and ϵ ranges over positive reals, form a basis for the
topology of X.
6. If a T1 -space X has the topology induced by the bounded continuous
functions f : X → R, prove that X is completely regular.
7. Let X be a completely regular space. Let F be a closed subset and
K a compact subset of X with K ∩ F = ∅. Show that there exists a
continuous function f : X → I such that f (x) = 0 for all x ∈ K and
f (x) = 1 for all x ∈ F .
8. Show that a continuous open image of completely regular space need
not be completely regular.
9. Let X be completely regular, and V be an open nbd of x ∈ X. Prove
that a necessary and sufficient condition that there be a continuous
function f : X → I such that f −1 (1) = x, f (X − V ) = 0, is that {x}
be a Gδ -set.
10. Let X be a space, and consider the relation x ∼ y iff {x} and {y} have
the same closure (equivalently, x and y have the same nbd base). Show
that ∼ is an equivalence relation on X, and if X satisfies the Ti -axiom,
i = 3 21 or 4, so does X/ ∼ (cf. Exercise 8.1.11).
11. Give an example to show that the adjunction space of two completely
regular spaces need not be completely regular.
12. Let fα : X → Yα , α ∈ A, be a family of continuous functions which
separates points from closed sets. Show that X has the topology induced
by the functions fα .
13. Give an example of a space which has topology induced by a collection
of functions which separates points, but does not separate points from
closed sets.
14. Let f : X → Y be a proper, open surjection. If X is completely regular,
show that Y is also completely regular.
15. If X is a connected, completely regular space having more than one
point, then every nonempty open subset of X is uncountable.
16. Let f be a proper mapping of a completely regular space X onto a
k-space Y . Show that X is a also a k-space.
SEPARATION AXIOMS
8.4
235
Stone–Čech Compactification
The study of a non-compact space X is often made easier by constructing a compact space which contains X as a subspace. We have
already discussed such a construction, viz. the one-point compactification. The ambient space obtained by this method is somewhat unf
satisfactory because, for example, the function x → sin 1/x defined on
(0, 1] cannot be extended continuously to [0, 1], which is the one-point
compactification of (0, 1]. However, it is possible to find a compact
space K which contains a homeomorphic copy X ′ of (0, 1] and admits
a continuous extension of f carried over to X ′ via the homeomorphism
X ′ ≈ (0, 1]. Let X ′ be the graph of f . Then (0, 1] ≈ X ′ . Since X ′ is
contained in the product [0, 1]×[−1, 1], K = X ′ is compact. Obviously,
the function K → R, (x, y) → y, is an extension of the composition
f
X ′ ≈ (0, 1] → R.
The one-point compactification is only one of many ways of compactifying a space. In general, a compactification of a space X is a
e together with an embedding η : X → X
e such that
compact space X
e
η(X) is dense in X. We will study here a very special type of compactification – one in which X is embedded in such a way that every
bounded real-valued continuous function on X extends continuously to
the compactification of X. Such a compactification of X is called the
Stone–Čech compactification, and is usually denoted by β(X). In the
year 1937, M.H. Stone and E. Čech each published important papers
which provided independent proofs of the existence of β(X).
Since a compact Hausdorff space is completely regular and the latter property is hereditary, only completely regular spaces can have
Hausdorff compactifications.
Theorem 8.4.1 (Stone–Čech) For every completely regular space
X, there exists a compact Hausdorff space β(X) such that
(a) X is homeomorphic to a dense subspace X ′ of β(X), and
(b) every bounded real-valued continuous function on X ′ can be extended uniquely to a continuous function β(X) → R.
Proof. Since X is completely regular, the hypotheses of Proposition 8.3.5 are satisfied by the family Φ of all continuous functions f :
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Elements of Topology
X → I (the unit interval). Therefore the evaluation map e : X → I Φ ,
x 7→ (f (x)), is an embedding. If X ′ denotes the range of e, then X is
homeomorphic to the subspace X ′ ⊆ I Φ . We write β(X) = X ′ (the
closure of X ′ in I Φ ). Since I Φ is compact and Hausdorff, so is the
subspace β(X) ⊆ I Φ .
To establish the last statement, assume that g : X ′ → R is a
bounded continuous function. We choose appropriate positive numbers b and c so that 0 ≤ b(ge(x) + c) ≤ 1 holds for every x ∈ X.
Then b(ge + c) = f for some f ∈ Φ. By the definition of e, we have
pf e(x) = f (x) for all x ∈ X, where pf is the projection map I Φ → I.
Accordingly, we define F : β(X) → I to be the restriction of pf to
β(X). Then, for x ∈ X, F e(x) = f (x), while ge(x) = b−1 f (x) − c. It
follows that b−1 F − c = G is a continuous extension of g to β(X). The
map G is unique because X ′ is dense in β(X).
♢
Remarks 8.4.2 (a) We usually identify the space X with the subspace X ′ ⊆ β(X), and consider X as a subspace of β(X).
(b) The proof of the preceding theorem also establishes that if X is
a compact T2 -space, then β(X) ≈ X.
The next theorem shows that not only every real-valued bounded
continuous function on X extends to β(X), but also any continuous
function of X into a compact T2 -space can be extended to β(X).
Theorem 8.4.3 (Stone) Let f be a continuous function of a completely regular space X into a compact T2 -space Y . Then there exists
a unique continuous extension to β(X) of f .
Proof. Let Φ be the family of all continuous functions of Y into R. For
each ϕ ∈ Φ, denote the range of ϕ by Rϕ . Note that Rϕ is a closed
bounded subset of R. By Theorem 8.4.1, there exists a continuous map
Fϕ : β(X) → R such that Fϕ (x) = (ϕf )(x) for every x ∈ X. Since
into Rϕ .
β(X) = X and Rϕ is closed, it follows that Fϕ maps β(X)
∏
Consequently, there is a continuous function k : β(X)
→
R
ϕ given by
∏
pψ (k(x)) = Fψ (x) for all x ∈ β(X), where pψ : Rϕ → Rψ is the ψth
projection
∏ map. Next, for y ∈ Y , let j(y) be the element of the product
space Rϕ , whose ϕth coordinate is ϕ(y). Then the composition
j
Y −→
∏
pψ
Rϕ −→ Rψ
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237
is ψ for every ψ ∈ Φ, and hence j is continuous. We have the following
commutative diagram of spaces and continuous functions
X
f
i
Q
Q
ψf QQ
s
j
Y
k
Rψ
3
ψ ? - β(X)
+ Fψ
Q
k
Q pψ
Q
Q∏?
Rϕ
that is, pψ ◦ j ◦ f = ψ ◦ f = Fψ ◦ i = pψ ◦ k ◦ i, where i : X →
β(X) is the inclusion map. Therefore k ◦ i = j ◦ f , which implies that
k (β(X)) ⊆ k(X) = (jf )(X) ⊆ j(Y ). Since Y is compact Hausdorff,
j is a closed mapping so that k (β(X)) ⊆ j(Y ). It is obvious that the
family Φ
∏ separates points in Y , so y 7→ j(y) is a closed embedding of
Y into Rϕ . Thus the composition
k
j −1
β(X) −→ j(Y ) −−→ Y
is defined. The mapping H = j −1 ◦ k is obviously continuous and
satisfies H ◦ i = f . The uniqueness of H is clear from Corollary 4.4.3
because X is dense in β(X), and Y is a T2 -space.
♢
The above property of β(X) enables us to show that the Stone–
Čech compactification of a space is essentially unique.
e of X to which
Corollary 8.4.4 Any Hausdorff compactification X
every continuous function of X into a compact space has an extension
e ≈ β(X)
is homeomorphic to β(X); indeed there is a homeomorphism X
which leaves points of X fixed.
e be the inclusions. By
Proof. Let i : X → β(X) and j : X → X
e → β(X) such
our hypothesis, there exists a continuous map ei : X
that ei ◦ j = i. And, by Theorem 8.4.3, there exists a continuous map
e such that e
e
j : β(X) → X
j ◦ i = j. So ei ◦ e
j ◦ i = i, that is, ei ◦ e
j is the
identity map on X. Since X is dense in β(X), ei ◦ e
j is the identity map
e and therefore e
on β(X). Similarly, e
j ◦ ei is the identity map on X,
j is
−1
e
e
a homeomorphism with i = j . Also, it is immediate that e
j fixes the
points of X.
♢
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Elements of Topology
Corollary 8.4.5 Any compactification of X is a continuous image of
β(X) under a mapping which leaves points of X fixed.
e is a compactification of X. By Theorem 8.4.3,
Proof. Suppose that X
e extending the inclusion map
there is a continuous map e
j : β(X) → X
e Clearly, the image of e
j : X → X.
j is a closed set containing X. Since
e
e
X is dense in X, we have X = im(e
j), that is, e
j is surjective.
♢
e1 and X
e2 are two compactifications of a space X, we define X
e1 ≼
If X
e
e
e
X2 , if there exists a continuous mapping F of X1 onto X2 such that
F (x) = x for all x ∈ X. Then ≼ is a partial ordering on the collection of
compactifications of X. It follows from the above corollary that β(X)
is a maximal element in the collection of Hausdorff compactifications
of X, while one-point compactification of a non-compact space is a
minimal element. In general, β(X) is fairly complicated, even for simple
spaces X. The following example is rather exceptional.
Example 8.4.1 The Stone–Čech compactification of [0, Ω) is the space
[0, Ω]. First, note that [0, Ω) is a dense subset of the Hausdorff space
[0, Ω]. Next, we show that the ordinal space [0, Ω] is compact. Let U be
an open covering of [0, Ω] . For each x ∈ (0, Ω], the sets (y, x] form a
nbd basis at x, since every point in [0, Ω] has an immediate successor.
So we can select a nbd V (x) = (y, x] such that V (x) is contained in
some U ∈ U. Consider the sequence of open nbds V (Ω) = (x1 , Ω],
V (x1 ) = (x2 , x1 ], V (x2 ) = (x3 , x2 ], . . .. Since x1 > x2 > · · · , and the
subset {x1 , x2 , . . .} has a least element in [0, Ω), we must have xn = 0
for some integer n > 0. Now, find sets Ui ∈ U, i = 0, 1, . . . , n + 1, such
that U0 ⊇ V (Ω), Ui ⊇ V (xi ) for 1 ≤ i ≤ n, and Un+1 ⊇ {0}. Then the
family {Ui |0 ≤ i ≤ n + 1} is a subcovering of U, and therefore [0, Ω] is
compact.
Now, by the proof of the uniqueness of Stone–Čech compactification, it clearly suffices to establish that every bounded continuous function f : [0, Ω) → R can be extended continuously to [0, Ω]. To this end,
we assert that for each integer n > 0, there exists an xn < Ω such that
|f (xn ) − f (y)| < 1/n for all y ≥ xn in [0, Ω). Assume otherwise. Then
there is an integer m > 0 such that for each ordinal number x < Ω,
there exists y ∈ (x, Ω) with |f (y) − f (x)| ≥ 1/m. Let y1 be the least
element of (0, Ω) such that |f (y1 ) − f (0)| ≥ 1/m. Next, let y2 be the
least element of (y1 , Ω) such that |f (y2 ) − f (y1 )| ≥ 1/m. Continuing in
this way, we obtain a sequence of ordinals 0 = y0 < y1 < y2 < · · · such
that |f (yn ) − f (yn−1 )| ≥ 1/m. If y is the supremum of {y0 , y1 , . . .},
SEPARATION AXIOMS
239
then 0 < y < Ω, and each basic open nbd (x, y] of y in [0, Ω) contains all but finitely many yn . So f fails to map (x, y] into the open
nbd (f (y) − 1/3m, f (y) + 1/3m) of f (y). This contradicts the continuity of f at y, and hence our assertion. If b < Ω is an upper bound
of {xn |n = 1, 2, . . .}, then, for any x ≥ b, we have |f (x) − f (b)| ≤
|f (x) − f (xn )| + |f (xn ) − f (b)| < 2/n for all n. So f (x) = f (b) for
all x ≥ b in [0, Ω). Finally, a desired continuous extension F of f is
obtained by defining F (Ω) = f (b) and F (x) = f (x) for x < Ω. Thus
β ([0, Ω)) = [0, Ω].
We close this chapter with the following theorem which characterizes the topology of a completely regular space.
Theorem 8.4.6 Let X and Y be completely regular, first countable
spaces. Then X ≈ Y ⇔ β(X) ≈ β(Y ).
Proof. Suppose first that h : X → Y is a homeomorphism, and let
i : X → β(X) and j : Y → β(Y ) be the inclusions. By Theorem 8.4.3,
there exist unique continuous functions ϕ : β(X) → β(Y ) and ψ :
β(Y ) → β(X) such that ϕi = jh and ψj = ih−1 . So, we have ψϕ(x) = x
for all x ∈ X, and ϕψ(y) = y for all y ∈ Y . Since X = β(X) and
Y = β(Y ), ψϕ = 1X and ϕψ = 1Y . Thus ϕ is a homeomorphism with
ψ as its inverse.
Conversely, suppose that ϕ : β(X) → β(Y ) is a homeomorphism.
We show that ϕ maps β(X) − X onto β(Y ) − Y, which implies that it
maps X onto Y . To establish this, we prove that there is no countable
open basis at any point of both β(X) − X or β(Y ) − Y . Assume the
contrary, and let {Un } be a countable open basis at a point p ∈ β(X) −
X. By regularity, we may assume that U n+1 ⊆ Un for every n. Since
X is dense in β(X), each Un contains infinitely many points of X. We
construct two sequences ⟨sn ⟩ and ⟨tn ⟩ of distinct points of X such that
sn , tn ∈ U∩
n and sn∩̸= tm for all n and m. Clearly, no point of X belongs
to {p} = Un = U n+1 . So, for each integer i > 0, there is an integer
ni such that si ∈
/ U ni . Note that U ni contains all but finitely many
sn ’s and tn ’s. So we can find an open nbd Vi of si in X such that
Vi ⊆ Ui − U ni and Vi does not contain any tj and sj ̸= si . Since X
is completely regular, there exists a continuous map fi : X → I such
that fi (si ) = 1 and fi (x) = 0 for all x ∈
/ Vi . We define a function
g : X → R by setting g(x) = supi f∪i (x). We assert that g is continuous.
It is obvious that g −1 (a, ∞) = i fi−1 (a, ∞), which is open for all
a ∈ R. To see that g −1 (−∞, a) is also open, consider a point x with
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Elements of Topology
0 ≤ g(x) < a. We have an integer k such that x ̸∈ U(k . As the function f)i
(
) ∩k−1 −1
vanishes on X −U k for all i ≥ k, W = X − U k ∩
i=1 fi (−∞, a)
is a nbd of x. It is easily checked that g maps W into (−∞, a), and
our assertion follows. Now, we have g(sn ) = 1 and g(tn ) = 0 for all n.
Since lim sn = p = lim tn , g cannot have a continuous extension over
β(X), a contradiction. Therefore β(X) has no countable open basis at
p. By the same argument, β(Y ) does not have a countable open basis
at any point of β(Y ) − Y, and ϕ maps β(X) − X onto β(Y ) − Y. This
completes the proof.
♢
Exercises
1. Justify the following:
(a) [0, 1] is not β((0, 1]).
(b) [−1, +1] is not β((−1, 1)).
(c) S1 is not β(R1 ).
2. Let X be a completely regular space, Y a compact Hausdorff space, and
f a homeomorphism of X into Y . Show that the Stone–Čech extension
F : β(X) → Y sends β(X) − X into Y − f (X).
3. Prove that a completely regular space X is connected ⇔ β(X) is connected.
4. Let X be a discrete space. Prove that the closure of every open subset
of β(X) is open and hence deduce that β(X) is totally disconnected.
5. Show that a sequence in N converges in β(N) if and only if it converges in
N. (Thus the sequence ⟨1, 2, . . .⟩ in β(N) has no convergent subsequence,
although it has a convergent subnet.) Conclude that β(N) is not second
countable or metrisable. (The space β(N) is one of the most widely
studied topological spaces. The interested reader may see the text by
Walker [15].)
6. If a sequence ⟨xn ⟩ in a normal space X converges to some point y in
β(X), show that y ∈ X.
7. Suppose that the Stone–Čech compactification of a completely regular
X is metrisable. Show that X is a compact metric space.
Chapter 9
PARACOMPACTNESS AND
METRISABILITY
9.1
9.2
Paracompact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A Metrisation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1
Paracompact Spaces
241
252
The concept of paracompactness was introduced in 1944 by J.
Dieudonné as a generalisation of compact spaces. Paracompact Hausdorff spaces are very close to metrisable spaces, and have proved quite
useful in differential geometry and topology. Here, we will study the
basic properties of these spaces.
Definition 9.1.1 Let {Uα |α ∈ A} and {Vβ |β ∈ B} be two coverings
of a space X. We say that {Uα } is a refinement of {Vβ } if for each
α ∈ A there is a β ∈ B such that {Uα ⊆ Vβ }. In this case, we also say
that {Uα } refines {Vβ }. If each member of {Uα } is open (resp. closed),
we call {Uα } an open (resp. closed) refinement of {Vβ }.
Recall that a family {Uα |α ∈ A} of subsets of a space X is locally
finite if each point x ∈ X has a nbd which meets at most finitely many
members (ref. Definition 2.1.8).
Definition 9.1.2 A space X is called paracompact if each open covering of X has a locally finite open refinement.
A discrete space is paracompact, since the open covering by singleton sets is locally finite and refines every open covering of the space. It
is also obvious that a finite covering is locally finite; so every compact
space is paracompact. As we proceed further, we will see some more
examples of paracompact spaces. Here is a negative one.
Example 9.1.1 The ordinal space [0, Ω) is not paracompact. To prove
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Elements of Topology
this, we observe that the open covering G = {[0, x)|0 < x < Ω} of [0, Ω)
does not have a locally finite open refinement. Assume otherwise, and
let {Uj |j ∈ J} be a locally finite open refinement G. Since the sets
(y, x], 0 < x < Ω, and {0} form a basis of [0, Ω), for each nonzero
ordinal number x < Ω, we choose an ordinal number f (x) < x such
that (f (x), x] ⊆ Uj for some j. Also, put f (0) = 0. By our assumption,
there exists a finite set K ⊂ J (depending on x) such that x ∈
/ Uj for
j ∈
/ K. Then there is an ordinal number y < Ω such that Uk ⊆ [0, y)
for every k ∈ K. It follows that x ≤ f (z) for all z ≥ y, and the set
{y|f (z) ≥ x for all z ≥ y} is nonempty. Denote its least element by
g(x). Thus we have a function g from [0, Ω) to itself. By the principle
of recursive definition (Appendix A.5.6), there exists a function h :
[0, ω) → [0, Ω) such that h(n + 1) = g(h(n)), where h(0) = 1. Then
s = sup{h(n)} exists and s < Ω. Obviously, h(n) < s + 1 for every
n < ω. So we find an ordinal yn ≤ s such that f (z) ≥ h(n) for all
z ≥ yn . In particular, we have f (s) ≥ h(n) for every n < ω. This
implies that f (s) ≥ s, a contradiction.
In a T3 -space X, the existence of a locally finite refinement (not
necessarily open) for each open cover ensures that it is paracompact.
This is established by the following.
Theorem 9.1.3 In a T3 -space X, the following conditions are equivalent:
(a) X is paracompact.
(b) Each open covering of X has a locally finite refinement (consisting
of sets not necessarily either open or closed).
(c) Each open covering of X has a locally finite closed refinement.
Proof. (a) ⇒ (b): Obvious.
(b) ⇒ (c): Let U be an open covering of X. Since X satisfies the
T3 -axiom, there exists an open covering V of X such that the closure of
each member of V is contained in some member of U. By our hypothesis,
V has a locally finite refinement W, say. Then the family {W |W ∈ W}
is also locally finite and refines U.
(c) ⇒ (a): Suppose that X is a space with the property that every
open covering of X has a locally finite closed refinement. Let U be an
open covering of X, and let F = {Fα |α ∈ A} be a locally finite closed
PARACOMPACTNESS AND METRISABILITY
243
refinement of U. Then for each x ∈ X, there exists an open nbd Vx of
x such that {α ∈ A|Vx ∩ Fα ̸= ∅} is finite. Next, find a locally finite
closed refinement {Eβ |β ∈ B} of the open cover {Vx |x ∈ X} of X, and
set
∪
Wα = X − {Eβ |Eβ ∩ Fα = ∅}
for every α ∈ A. Since the family {Eβ } is locally finite, Wα is open.
Clearly, Fα ⊆ Wα and, therefore, {Wα |α ∈ A} is an open covering
of X. We assert that it is locally finite, too. Let x ∈ X be arbitrary.
Then there exists a nbd G of x and
∪ a finite subset Γ ⊂ B such that
G ∩ Eβ = ∅ for all β ∈
/ Γ. So G ⊂ γ∈Γ Eγ . By definition,
Eβ ∩ Wα ̸= ∅ ⇐⇒ Eβ ∩ Fα ̸= ∅.
Since the set {α ∈ A|Eβ ∩ Fα ̸= ∅} is finite for every β ∈ B, it follows
that {α ∈ A|G ∩ Wα ̸= ∅} is finite, and hence our assertion. Now,
for each index α, choose a member Uα of U such that Fα ⊂ Uα . Then
{Uα ∩ Wα |α ∈ A} is a locally finite open refinement of U, and this
completes the proof.
♢
Though there are several other characterisations of paracompactness, we are interested in the one that is useful in proving the paracompactness of a metric space. For this purpose, we need the following
terminology.
Definition 9.1.4 A∪family {Uα |α ∈ A} of subsets of a space X is σlocally finite if A = n∈N An and for each n, the family {Uα |α ∈ An }
is locally finite.
The following specification of a σ-locally finite family is quite useful.
Suppose that {Uα |α ∈ A} is a σ-locally finite family of subsets of a
space X. If we put Un,α = Uα for α ∈ An , and Un,α = ∅ for α ∈
/
An , then we obtain a family {Un,α |n ∈ N and α ∈ A} such that for
each fixed∪n ∈ N, the family {Un,α |α ∈ A} is locally finite. Note that
∪
α Uα =
n,α Un,α .
Theorem 9.1.5 A T3 -space X is paracompact if and only if each open
covering of X has a σ-locally finite open refinement.
Proof. Sufficiency: Let X be a T3 -space with the property that its
every open covering has a σ-locally finite open refinement. Let U be
any open covering of X. By our hypothesis, there is an open refinement
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Elements of Topology
V = {Vn,α |n ∈ N and α ∈ A} of U such that for each fixed
n, the family
∪
{Vn,α |α ∈ A} is locally finite.
For
each
n,
put
V
=
V
n
α n,α . Also, set
∪
W1 = V1 and Wn = Vn − i<n Vi for n > 1. Given x ∈ X, observe
that there is a least integer n(x) such that x ∈ Vn(x) . Then we have
x ∈ Wn(x) , so the family {Wn |n ∈ N} covers X. Now, consider the
family
{Wn ∩ Vn,α |n ∈ N and α ∈ A}.
We show that it is a locally finite refinement of U. Clearly, this family
covers X, and is a refinement of U. To see that it is locally finite, let
x ∈ X be arbitrary. Find the least integer n(x) such that x ∈ Vn(x),β
for some β. Then Vn(x),β does not intersect Wn for n > n(x). Since the
family {Vn,α |α ∈ A} is locally finite, we can find, for each i ≤ n(x),
an open nbd Gi of x such that {α ∈ A|Gi ∩ Vi,α ̸= ∅} is finite. Then
)
∩n(x) (
i=1 Gi ∩ Vn(x),β is a nbd of x, which can meet only finitely many
sets Wn ∩ Vn,α . It follows that the family {Wn ∩ Vn,α |n ∈ N and α ∈
A} is a locally finite refinement of U and, by Theorem 9.1.3, X is
paracompact.
♢
Corollary 9.1.6 Every Lindelöf T3 -space is paracompact.
Proof. Let {Uα |α ∈ A} be an open covering of a Lindelöf T3 -space
X. Then there exists a countable subcovering {Uαn |n ∈ N} of {Uα }.
Obviously, {Uαn |n ∈ N} is an open refinement of {Uα } and decomposes
into countably many locally finite families, each consisting of just one
♢
set Uαn .
We shall see soon that a paracompact Hausdorff space is regular,
but it need not be Lindelöf.
Example 9.1.2 An uncountable discrete space is not Lindelöf, although
it is paracompact Hausdorff.
The following examples are immediate from the preceding corollary.
Example 9.1.3 The euclidean space Rn is paracompact and so are any
of its subspaces.
Example 9.1.4 The Sorgenfrey line Rℓ is paracompact.
An important class of paracompact Hausdorff spaces is given by
the next theorem.
PARACOMPACTNESS AND METRISABILITY
245
Theorem 9.1.7 Every metric space is paracompact.
Proof. Let (X, d) be a metric space and let U = {Uα |α ∈ A} be an
open covering
∪ of X. For any subset E ⊂ X and any real r > 0, define
Br (E) = {B(x; r)|x ∈ E}, the open ball about E. Now, for each
integer n ∈ N and each α ∈ A, let En,α = X − B1/n (X − Uα ). Notice
E1β
E2β
E2α E3α
E3β
Uβ
Uα
E1α
FIGURE 9.1: Proof of Theorem 9.1.7.
that En,α = {x ∈ Uα |dist(x, X − Uα ) ≥ 1/n}∪
and may be empty. Next,
well order A by ≺, and put Fn,α = En,α − α′ ≺α Uα′ . We show that
the family
V = {B1/3n (Fn,α )|n ∈ N and α ∈ A}
is a σ-locally finite open refinement of U. Obviously, each Vn,α =
B1/3n (Fn,α ) is open. To see that V covers X, let x be an arbitrary
element of X. Then there exists an index α in A such that x ∈ Uα
and x ∈
/ Uα′ for every α′ ≺ α. Since dist(x, X − Uα ) > 0, x ∈ En,α
for some n. Then x ∈ Fn,α ⊂ Vn,α . We next verify that Vn,α ⊂ Uα
for all n. Suppose that x ∈ Vn,α . Then there exists y ∈ Fn,α such
that d(x, y) < 1/3n. If x ∈
/ Uα , then we have dist(y, X − Uα ) < 1/3n,
a contradiction. So x ∈ Uα , and V is an open refinement of U. It
remains to show that the family {Vn,α |α ∈ A} is locally finite for every n. To this end, we observe that dist(Vn,α , Vn,β ) ≥ 1/3n for every
α ̸= β. Assume that x ∈ Vn,α and y ∈ Vn,β . Then there exist p ∈ Fn,α
and q ∈ Fn,β such that d(x, p) < 1/3n and d(y, q) < 1/3n. Consequently, d(p, q) ≤ d(p, x) + d(x, y) + d(y, q) < d(x, y) + 2/3n. We have
d(p, q) ≥ 1/n, for if β ≺ α, then Fn,α ⊂ X − Uβ . So d(x, y) > 1/3n
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Elements of Topology
and dist(Vn,α , Vn,β ) ≥ 1/3n. Now, given x ∈ X, consider the open
ball B(x; 1/6n). If this contains points p ∈ Vn,α and q ∈ Vn,β , where
α ̸= β, then d(p, q) ≤ d(p, x) + d(x, q) < 1/3n, contrary to the fact that
dist(Vn,α , Vn,β ) ≥ 1/3n. This completes the proof.
♢
Theorem 9.1.8 A paracompact Hausdorff space is normal.
Proof. Let X be a paracompact Hausdorff space. We first show that
X is regular. Let F be a closed subset of X and x0 ∈ X − F. Then
for each x ∈ F, there exist disjoint open sets Ux and Vx such that
x0 ∈ Ux and x ∈ Vx . Since X is paracompact, the open covering
{Vx |x ∈ F }∪{X −F } has a locally finite open refinement {Wα |α ∈ A},
say. Let
B = {α ∈ A|Wα ⊆ Vx for some x ∈ F } .
Then for each β ∈ B, there
∪ is an x ∈ F such∪that Ux ∩ Wβ = ∅. So
x0 ∈
/ Wβ . Put U = X − β∈B Wβ and V = β∈B Wβ . Then x0 ∈ U,
F ⊆ V and U ∩ V = ∅. It is obvious that V is open, and U is also
open by Proposition 2.1.9. Thus X is regular.
Now, let F1 and F2 be two disjoint closed subsets of X. For each
x ∈ F1 , we find disjoint open nbds Ux of x and Vx of F2 as above. The
open covering of X which consists of sets X − F1 and Ux , x ∈ F1 , has
a locally finite open refinement {Gλ |λ ∈ Λ}, say. Let
M = {λ ∈ Λ|Gλ ⊆ Ux for some x ∈ F1 } .
Then for each µ ∈ M, there
∪ exists x ∈ F1 such∪that Gµ ∩ Vx = ∅. So
F2 ∩ Gµ = ∅. Put G = µ Gµ and H = X − µ Gµ . Then G and H
are disjoint open nbds of F1 and F2 , respectively.
♢
It is clear that paracompact Hausdorff spaces are close to being
metric spaces because all that is needed is Lindelöfness. Also, a paracompact T3 -space is T4 , by the proof of the preceding theorem.
Now, we turn to see the invariance properties of paracompactness.
Observe that it is a topological invariant; we shall show later that it
is preserved by continuous closed maps, too. But this is not hereditary, as shown by the ordinal space [0, Ω] (see Example 8.4.1 for its
compactness). However, we can prove the following.
Theorem 9.1.9 A closed subspace of a paracompact space is paracompact.
PARACOMPACTNESS AND METRISABILITY
247
Proof. Let F be a closed set in a paracompact space X, and let
V = {Vα |α ∈ A} be an open covering of the subspace F. For each
α ∈ A, there exists an open set Uα ⊆ X such that Vα = F ∩ Uα . Since
X is paracompact, the open covering {Uα |α ∈ A} ∪ {X − F } has a locally finite open refinement {Wλ |λ ∈ Λ} , say. Then the open covering
{F ∩ Wλ } is locally finite and refines V.
♢
More generally, we have
Proposition 9.1.10 Every Fσ -set in a paracompact Hausdorff space
is paracompact. (Recall that a subset A of a space X is Fσ if it is the
union of at most countably many closed sets.)
Proof. Let∪Y be an Fσ -set in a paracompact space X and suppose
that Y = n Fn , where each Fn is closed in X. Let V = {Vα |α ∈ A}
be an open covering of the subspace Y. Then for each α ∈ A, we
find an open set Uα ⊆ X such that Vα = Y ∩ Uα . For each fixed n,
{Uα |α ∈ A} ∪ {X − Fn } is an open covering of X, and therefore has a
locally finite open refinement {Wn,λ |λ ∈ Λn }. Set
Gn = {Y ∩ Wn,λ |Fn ∩ Wn,λ ̸= ∅}
for every n. Then each Gn is a locally
∪ finite family and consists of open
subsets of Y. It is easily seen that n Gn is an open covering of Y and
refines V. By the preceding theorem, X is regular and so is the subspace
Y ⊂ X. From Theorem 9.1.5, we see that Y is paracompact.
♢
Unlike compactness, the property of paracompactness is not productive. Even the product of two paracompact spaces need not be
paracompact. For example, the Sorgenfrey line Rℓ is paracompact, but
the product space Rℓ × Rℓ is not paracompact, since it is Hausdorff but
not normal (ref. Exercise 8.4.4). However, we can prove the following.
Proposition 9.1.11 The product of a paracompact space and a compact space is paracompact.
Proof. Let X be a paracompact space and Y be a compact space. Let
U be an open cover of X × Y . Let x ∈ X be fixed. Given y ∈ Y, choose
a member U in U with (x, y) ∈ U. Then there exist open nbds Gy
of x and Hy of y such that (x, y) ∈ Gy × Hy ⊆ U. The open covering
{Hy |y ∈ Y } of compact space Y has a finite subcovering. So there
∪nx exist
finitely many points y1 , . . . , ynx (say) in Y such that Y = j=1 Hyj .
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Elements of Topology
∩nx
Gyj . Then each Gx × Hyj is contained in some member
Put Gx = j=1
of U. Since X is paracompact, the open covering {Gx |x ∈ X} has a
locally finite open refinement {Vλ |λ ∈ Λ} , say. For each λ ∈ Λ, choose
x ∈ X such that Vλ ⊆ Gx and write Hj,λ = Hyj , nλ = nx . Then
{Vλ × Hj,λ |j = 1, . . . , nλ and λ ∈ Λ} is an open refinement of U. This
refinement of U is obviously locally finite, and X × Y is paracompact.
♢
The following lemma is used in establishing several properties of
paracompact Hausdorff spaces.
Lemma 9.1.12 Let X be a paracompact Hausdorff space and
{Uα |α ∈ A} be an open covering of X. Then there is a locally finite
open refinement {Vα |α ∈ A} of {Uα } such that V α ⊆ Uα for every
α ∈ A.
Proof. Since X is paracompact and Hausdorff, it is regular. So there is
an open covering {Wβ |β ∈ B} of X such that each W β is contained in
some Uα . Now, we find a locally finite open refinement {Gγ |γ ∈ Γ} of
{Wβ }. Then each Gγ ⊆ Uα for some α ∈ A. For each γ ∈ Γ, choose
f (γ) ∈ A such that Gγ ⊆ Uf (γ) . Thus we obtain a function f : Γ → A.
Put Γα = f −1 (α) for every α ∈ A. Note that
∪ some Γα may be empty
′
and Γα ∩ Γα′ = ∅ for α ̸= α . Put Vα = γ∈Γα Gγ . Then each Vα is
an open
∪ subset (possibly empty) of X and Vα ⊆ Uα . In fact, we have
V α = γ∈Γα Gγ ⊆ Uα for all α ∈ A, since the family {Gγ } is locally
finite. Also, it is clear that a subset N ⊆ X meets Vα (nontrivially)
if and only if it meets some Gγ , where f (γ) = α. As Γα ∩ Γα′ = ∅
for α ̸= α′ and the family {Gγ } is locally finite, so is the family {Vα }.
Thus {Vα } is a desired refinement of {Uα }.
♢
Theorem 9.1.13 Let X be a paracompact Hausdorff space and f :
X → Y be a continuous closed surjection. Then Y is also paracompact.
Proof. By Proposition 8.2.4, Y is normal and, therefore, regular. So it
suffices to show that every open covering of Y has a σ-locally finite open
refinement,
Let {Uα |α ∈ A} be an open covering of
{ by Theorem 9.1.5.
}
Y. Then f −1 (Uα ) |α ∈ A is an open covering of X. Consider a wellordering ≼ on A. For each n ∈ N, we construct by induction a locally
finite open covering {Vn,α |α ∈ A} of X such that
(
)
(a) f V n,α ⊆ Uα for every α ∈ A, and
PARACOMPACTNESS AND METRISABILITY
(b) f
(∪
α≺β
249
)
(
)
V n−1,α ∩ f V n,β = ∅
(n > 1).
By Lemma
exists a precise locally finite open refinement
{ 9.1.12, there
}
{V1,α } of f −1 (Uα ) such that V 1,α ⊆ f −1 (Uα ). Assume that n > 1
and that the family
) been defined for i ≤ n. For β ∈ A, let
(∪ {Vi,α } has
Wn,β = Uβ − f
α≺β V n,α . Then Wn,β is open, since f is closed
}
{
and V n,α is locally finite. We assert that {Wn,α |α ∈ A} covers Y.
Given y ∈ (Y, let) β0 be the first element of A such that y ∈ Uβ0 .
Then y ∈
/ f V n,α for α ≺ β0 , since V{n,α ⊆ f −1 (Uα ). It }follows that
y ∈ Wn,β0 and hence our assertion. So f −1 (Wn,α ) |α ∈ A is an open
covering of X. We invoke the preceding lemma
precise
{ −1again to find a }
locally finite open refinement {Vn+1,α } of f ( (Wn,α )) |α ∈(A such
)
−1
that V n+1,α ⊆ f −1 (W
( n,α ) ⊆ )f (Uα ). Then f V n+1,β ∩ f V(n,α =)
∅ for α ≺ β, since f V n+1,β ⊆ Wn,β , which does not meet f V n,α
when α ≺ β. Thus conditions
(a) and (b)) are satisfied by the Vn+1,α .
(∪
Now, put Gn,α = Y − f
β̸=α V n,β for every n ∈ N and α ∈ A.
(
)
Then it is clear that each Gn,α is open, Gn,α ⊆ f V n,α ⊆ Uα and
Gn,α ∩ Gn,β = ∅ if α ̸= β. Moreover, {Gn,α |n ∈ N, α ∈ A} covers
Y. To see this, suppose (that y) ∈ Y and n ∈ N. Then there is an
α ∈ A such that
( y )∈ f V n,α . Denote the first element of the set
{α ∈ A|y ∈ f V n,α } by λn , and let λm be the first element
of) the
(
set {λn |n ∈ N}. Then, by the definition of λ(m , y ̸∈ )f V m+1,β for
every β ≺ λm . Also, if λm ≺ β, then y ̸∈ f V m+1,β , by (b). Thus
y ∈ Gm+1,λ
∪ m and {Gn,α |n ∈ N, α ∈ A} is a covering of Y. Next, set
{Gn |n ∈ N} is an open covering of Y whence
Gn = α Gn,α . Then
{
}
f −1 (Gn ) |n ∈ N is an open covering of X. Invoking Lemma 9.1.12
once
finite open refinement {Hn |n ∈ N}
{ again, we
} find a precise
( locally
)
of f −1 (Gn ) such that f H n ⊆ (Gn . )Since Y is normal, there exists
an open set On ⊆ Y such that f ∪H n ⊆( On )⊆ On ⊆ Gn . For each
n ∈ N and α ∈ A, write Ln = On − k<n f H k and Ln,α = Ln ∩Gn,α .
Then {Ln } is an open covering of Y. For, if y ∈ Y, and n is the least
integer such that y ∈ On , then y ∈ Ln . Thus {Ln,α |n ∈ N and α ∈ A}
is an open refinement of {Uα }. We observe that it is σ-locally finite.
Clearly, the family {Gn,α |α ∈ A} ∪ {Y − On(} is an open
covering of
)
Y . Also, Ln,α ∩ Gn,β for α ̸= β and Ln,α ∩ Y − On = ∅ for all α.
Thus the family {Ln,α |α ∈ A} is locally finite for every n ∈ N, and this
completes the proof.
♢
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Elements of Topology
Notice that while proving local finiteness of the family {Ln,α |α ∈ A}
in the proof of Theorem 9.1.13, we have, in fact, shown that it satisfies
a stronger condition rather than what is required. This leads to
Definition 9.1.14 A family F of subsets of a topological space X is
called discrete if each point of X has a nbd which meets at most one
member of F. The family F is called σ-discrete if it can be decomposed
as the union of countably many discrete families of subsets of X.
With this terminology, we have another characterisation of paracompactness.
Theorem 9.1.15 A regular space X is paracompact ⇔ every open
covering of X has a σ-discrete open refinement.
Proof. The sufficiency is immediate from Theorem 9.1.5, since a discrete
family is obviously locally finite, so a σ-discrete family is σ-locally
finite.
The proof of the necessity is very similar to that of Theorem 9.1.13,
and is left to the reader.
♢
In the end of this section we describe, perhaps, the most important
property of paracompact Hausdorff spaces. For any continuous map
f : X → R, the support of f is defined to be the closure of the set
{x ∈ X|f (x) ̸= 0}, and denoted by suppf . Notice that if x lies outside
the support of f, then f vanishes on a nbd of x, and conversely.
Definition 9.1.16 Let X be a space. A partition of unity on X is a
family {fα |α ∈ A} of continuous maps fα : X → I such that
(a) the family {suppfα } is a locally finite closed covering of X, and
∑
(b)
α fα (x) = 1 for each x ∈ X.
If {Uα |α ∈ A} is an open covering of X, we say that a partition of
unity {fα }α∈A is subordinate to (or dominated by) {Uα } if the support
of each fα is contained in the corresponding Uα .
Theorem 9.1.17 Every open covering of a paracompact Hausdorff
space has a partition of unity subordinate to it.
PARACOMPACTNESS AND METRISABILITY
251
Proof. Let X be a paracompact Hausdorff space and let {Uα |α ∈ A}
be an open covering of X. By Lemma 9.1.12, we find a precise locally
finite open refinement {Vα |α ∈ A} of {Uα } such that Vα ⊆ Uα for every
α ∈ A. We further shrink {Vα } to get a precise locally finite open
refinement {Wα |α ∈ A} of {Vα } such that Wα ⊆ Vα for every α. By
Urysohn’s
there exists a continuous function fα : X → I such
( lemma,
)
that fα Wα = 1 and fα (X − Vα ) = 0. Then Sα = {x ∈ X|fα (x) >
0} ⊆ Vα whence suppfα ⊆ Vα ⊆ Uα . Since {Vα |α ∈ A}
∑ is locally finite,
so is {Sα }. It follows that for each x ∈ X, f (x) = α fα (x) is a real
number, and we obtain a function f : X → R. We observe that f is
continuous. Suppose that N (x) is an open nbd of x in X which meets
only finitely
∑n many sets Sα1 , . . . , Sαn , say. Then for every y ∈ N (x),
f (y) = 1 fαi (y) and f is continuous on N (x). Since {N (x)|x ∈ X}
covers X, f is continuous. We also note that f (x) ≥ 1 for every x ∈ X
because there is an α ∈ A such that fα (x) = 1. Now, for each α ∈ A, we
define a continuous function gα : X → I by setting gα (x) = fα (x)/f (x),
x ∈ X. It is easily seen that the family {gα } is a desired partition of
unity subordinate to {Uα }.
♢
We remark that the converse (due to C.H. Dowker) of the preceding
theorem is also true. The proof of the preceding theorem also shows
that a partition of unity subordinate to a locally finite open covering
of a normal (or T4 ) space always exists.
Exercises
1. Prove that a space X is paracompact if and only if each open covering
of X has a precise locally finite open refinement.
2. Prove:
(a) A second countable locally compact Hausdorff space is paracompact.
(b) A locally compact Hausdorff space that is a countable union of
compact sets is paracompact.
3. If each open subset of a paracompact space is paracompact, show that
every subspace is paracompact.
4. Suppose that a Hausdorff space Y is the union of countably many compact sets. If X is a paracompact Hausdorff space, show that X × Y is
paracompact.
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Elements of Topology
5. Let p : X → Y be a proper map. If Y is paracompact, show that X is
also paracompact.
6. Give an example of a normal space which is not paracompact.
7. Prove that a space X is T4 ⇔ each finite open covering of X has a
locally finite closed refinement.
8. Prove that a space X is T4 ⇔ each point-finite open covering
{Uα |α ∈ A} of X has an open refinement {Vα |α ∈ A} such that Vα ⊆ Uα
for every α. (This is known as the shrinking lemma for normal spaces.)
9.2
A Metrisation Theorem
Given a topological space X, there are several known conditions
which guarantee that X is metrisable. The Urysohn metrisation theorem provides a simple condition for metrisability of regular spaces.
Recall that a metric space is certainly regular, and the most important
criterion for metrisability of regular spaces is given by the following.
Theorem 9.2.1 (Nagata–Smirnov) A space is metrisable if and
only if it is regular and has a basis that is the union of at most countably many locally finite families of open sets.
This theorem was established independently by J. Nagata (1950)
and Yu M. Smirnov (1951). Before proving it, we consider a concept
which is slightly more general than that of a metric space.
A pseudo-metric on a set X is a function d : X × X → R such that
the following conditions are satisfied for all x, y, z ∈ X:
(a) d(x, x) = 0,
(b) d(x, y) = d(y, x), and
(c) d(x, z) ≤ d(x, y) + d(y, z).
If d is a pseudo-metric on a set X, then we have d(x, y) ≥ 0 for all
x, y in X. The pair (X, d) is called a pseudo-metric space. The value
d(x, y) on a pair of points x, y ∈ X is called the distance between x
PARACOMPACTNESS AND METRISABILITY
253
and y. As in the case of metric spaces, we have the notions of “an open
ball,” “an open set,” “diameter of a set,” etc., in a pseudo-metric space.
Notice that a pseudo-metric space satisfying the T0 -axiom is actually a
metric space. A topological space X is called pseudo-metrisable if there
is a pseudo-metric d on X such that the topology induced by d is the
same as that of X.
The following proposition will be required to establish the above
theorem.
Proposition 9.2.2 Let (Xn , dn ) , n = 1, 2, . . ., be∏a sequence of
pseudo-metric spaces. Then the product space P =
Xn is pseudometrisable.
Proof. If d is a pseudo-metric on a set X, then d1 (x, y) = min{1, d(x, y)}
is also a pseudo-metric on X, and the topology induced by d1 is the
topology of (X, d). So we may assume that diam(Xn ) ≤ 1 for every n.
Then the function δ : P × P → R, defined by
∑ −n
δ ((xn ), (yn )) =
2 dn (xn , yn ),
is easily checked to be a pseudo-metric on P . We show that δ induces
the product topology of P . Let O ⊆ P be a subbasic open set of the
product topology. Then there exists an integer i > 0 and an open set
U ⊆ Xi such that O = {x ∈ P |xi ∈ U }. If x ∈ O, then we can find a real
ϵ > 0 such
Bdi (xi ; ϵ) ⊆ U . Obviously, δ(x, y) < 2−i ϵ ⇒ yi ∈ U so
( that
)
that Bδ x; 2−i ϵ ⊆ O. Thus O is open relative to the metric δ and the
product topology is weaker than the metric topology for P . To see the
converse, we observe that for each x ∈ P and any real number r > 0,
the open ball Bδ (x; r) contains a nbd of x in the product topology.
−1
n
Let n be so large (that
∏∞ r ) < 2 . Then V = Bd1 (x1 ; r/2) × · · · ×
Bdn+1 (xn+1 ; r/2)× n+2 Xi is a nbd of x in the product topology.
For any y ∈ V , we have
∑n+1 −i
∑∞
δ(x, y) =
2 di (xi , yi ) + n+2 2−i di (xi , yi )
1
<
∑n+1
1
2−i−1 r +
∑∞
n+2
2−i
< r/2 + r/2 = r.
So y ∈ Bδ (x; r) and V ⊆ Bδ (x; r). It follows that the topology induced
by δ on P is weaker than the product topology.
♢
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Elements of Topology
Theorem 9.2.3 If a normal space X has a σ-locally finite base, then
X can be embedded into a pseudo-metrisable space, and therefore X
is metrisable.
∪
Proof. Suppose that X is normal and B = Bn is a basis of X, where
each Bn is locally finite.
} fixed integer n > 0 and any
{ Then, for each
V ⊆ X, the
∪ { family B ⊆ V |B ∈ B}n is locally finite, and therefore
F (V ) =
B|B ∈ Bn and B ⊆ V is closed in X, by Proposition
2.1.9. Thus, for an open subset V ⊆ X, we have disjoint closed sets
X − V and F (V ). Since X is normal, there exists a continuous function
ϕn,V : X → I such that ϕn,V (X − V ) = 0 and ϕn,V (F (V )) = 1. Since
each point of X belongs
to at most finitely many members of Bm , the
∑
sum dm,n (x, y) = V ∈Bm |ϕn,V (x) − ϕn,V (y)| is finite for every integer
m > 0 and all ordered pairs of points (x, y) in X. It is easily checked
that dm,n is a pseudo-metric on X. Denote the pseudo-metric
space
∏
(X, dm,n ) by Xm,n . Then the product space P = (m,n)∈N×N Xm,n
is pseudo-metrisable. We show that there is an embedding of X into
P. By Proposition 8.3.5, it suffices to construct a continuous function
fm,n : X → Xm,n for every m, n ∈ N such that the family {fm,n }
separates points and closed sets in X. Let fm,n be the identity map
on X. To see the continuity of fm,n , consider a point x0 ∈ X and a
real ϵ > 0. Then find a nbd U0 of x0 in X which meets at most finitely
many members of Bm . If U0 does not meet any member of Bm , then
dm,n (x0 , x) = 0 for all x ∈ U0 , and we are through. So assume that U0
meets the sets V1 , . . . , Vk in Bm and is disjoint from the other members.
Now, by the continuity of the functions ϕn,Vi , 1 ≤ i ≤ k, there exists
an nbd Ui of x0 in X such that |ϕn,Vi (x0 ) − ϕn,Vi (x)| < ϵ/k for every
∩k
x ∈ Ui . Then W = 0 Ui is a nbd of x0 and dm,n (x0 , x) < ϵ for all
x ∈ W . This implies that fm,n is continuous at x0 . Now, suppose that
E ⊆ X is closed and x ∈ X − E. Then there exists an integer m ∈ N
and a set V ∈ Bm such that x ∈ V ⊂ X − E. Since X is regular, we
find a basic open set W ∈ B such that x ∈ W ⊆ W ⊆ V . If W ∈ Bn ,
then ϕn,V (x) = 1 and ϕn,V (E) = 0, for x ∈ F (V ) and E ⊂ X − V .
It follows that dm,n (x, y) ≥ |ϕn,V (x) − ϕn,V (y)| = 1 for all y ∈ E.
Accordingly, x0 does not belong to the closure of E in Xm,n , and the
family {fm,n |m, n ∈ N} separates points from closed sets in X. Since
the points in X are closed, it also separates points of X (this is also
obvious otherwise), and this completes the proof.
♢
Now, we come to see
PARACOMPACTNESS AND METRISABILITY
255
Proof of Theorem 9.2.1: If X is a metrisable space, then it is regular
and paracompact, by Theorem 9.1.7. So, for each integer n > 0, the
covering of X by open balls B(x; 1/n), x ∈∪X, has a locally finite open
refinement Vn , say. We assert that B = n Vn is a base for X. Let
U ⊆ X be a nbd of x. Then there exists a sufficiently large integer
n > 0 so that B(x; 1/n) ⊆ U . Now, we have some V ∈ V2n with x ∈ V .
Since V ⊆ B(y; 1/2n) for some y ∈ X, we see that V ⊆ B(x; 1/n) and
hence our assertion.
Conversely, suppose that B is a σ-locally finite base of a regular
space X. Then every open covering of X has a refinement which consists
of members of B. It follows from Theorems 9.1.5 and 9.1.8 that X is
normal. Now, by invoking the preceding theorem, we see that X is
metrisable.
♢
We end this section with another metrisation theorem proved independently at the same time by R. H. Bing (1951).
Theorem 9.2.4 (Bing) A space X is metrisable if and only if it is
regular and has a σ-discrete base.
Proof. The sufficiency of the condition is immediate from Theorem
9.2.1, since a σ-discrete family is σ-locally finite.
To see the necessity, we consider the covering of X by the open
balls B(x; 1/n), x ∈ X and n > 0 a fixed integer. Then, by the proof of
Theorem 9.1.7, there is a σ-discrete refinement Vn of this
∪ open cover.
As in the proof of Theorem 9.2.1, it is easily seen that n Vn is a basis
for X.
♢
Exercises
1. A space X is called locally metrisable if each point of X has a nbd which
is metrisable in the relative topology.
Show that a compact Hausdorff space X is metrisable if it is locally
metrisable. Generalise this for regular Lindelöf spaces.
2. Prove Exercise 1 for paracompact Hausdorff spaces.
3. If a completely regular space X is the union of a locally finite family of
closed, metrisable subspaces, show that X is metrisable.
Chapter 10
COMPLETENESS
10.1
10.2
10.3
10.1
Complete Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Baire Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
257
265
269
Complete Spaces
Unlike most other concepts, the notion of completeness is not a
topological invariant. But it is closely related to some important topological properties. Our considerations here are limited to some useful
theorems in topology which find frequent applications in analysis.
Definition 10.1.1 Let X be a metric space with metric d. A sequence
⟨xn ⟩ in X is called a Cauchy sequence if for every real number ϵ > 0,
there exists an integer k (depending on ϵ) such that d (xm , xn ) < ϵ
whenever m, n > k.
It is easy to see that every convergent sequence in a metric space is
a Cauchy sequence. But the converse is not true in general; for example, the Cauchy sequence ⟨1/n⟩ in the space (0, ∞) with the euclidean
metric fails to converge.
Definition 10.1.2 A metric space X is called complete if every
Cauchy sequence in X is convergent.
It should be noted that completeness is a property of metrics; it is
not a topological invariant.
Example 10.1.1 It is well known that the real line R is complete in the
usual metric (we shall also see this shortly). Consider another metric
d′ given by d′ (x, y) = |x/(1 + |x|) − y/(1 + |y|)|. The sequence ⟨n⟩ in the
space (R, d′ ) is a Cauchy sequence and does not converge to any point
of R. Thus the space (R, d′ ) is not complete, although the associated
topology is euclidean.
257
258
Elements of Topology
Accordingly, we introduce
Definition 10.1.3 A metrisable space X is called topologically complete if X has a metric d such that the metric space (X, d) is complete,
and d metrises the topology of X.
The following result will lead up to some interesting examples of
complete spaces.
Theorem 10.1.4 A compact metrisable space X is complete in every
metric which induces its topology.
Proof. This is immediate from Theorem 6.2.4 and Exercise 1 (below);
however, we give a direct simple proof. Let ⟨xn ⟩ be a Cauchy sequence
in X. It suffices to show that ⟨xn ⟩ has a convergent subsequence. If ⟨xn ⟩
does not have a convergent subsequence, then no point of X can be a
cluster point of ⟨xn ⟩. Therefore each x ∈ X has an open nbd Ux which
contains at most finitely many terms of ⟨xn ⟩. Since X is compact,
we find finitely many such open sets, say Ux1 , . . . , Uxn , covering X.
It now follows that xn is defined for only finitely many indices n, a
contradiction. Therefore ⟨xn ⟩ has a convergent subsequence, and the
theorem follows.
♢
Example 10.1.2 The real line R is complete. If ⟨xn ⟩ is a Cauchy sequence in R, then the set {xn } is bounded, and therefore it is contained
in a closed interval K, say. Since K is compact, ⟨xn ⟩ converges to a
point of K ⊆ R.
⟨
⟩
Example 10.1.3 The euclidean space Rk is complete. Let x(n) be a
(n)
Cauchy sequence in Rk . If xi denote the ith coordinate of⟨ x(n)⟩
, then
(n)
(m)
(n)
|xi − xi | ≤ ∥x(n) − x(m) ∥ for every i = 1, . . . , k. So xi
is a
Cauchy sequence in R. Because R is complete, there is a real number
(n)
xi , say, such that limn→∞ x⟨i =⟩xi . Clearly, the point x = (xi ) in Rk
is the limit of the sequence x(n) .
⟨
⟩
Example 10.1.4 The Hilbert space ℓ2 is complete. Let x(n) be a
Cauchy sequence in ℓ2 . Given
( ϵ > 0,) there exists a positive integer
k such that m, n > k ⇒ d x(m) , x(n) < ϵ, where d is the metric on
(n)
(m)
(n)
(n)
ℓ2 . If xi denote the
⟨ i th⟩ coordinate of x , then |xi − xi | < ϵ,
(n)
for m, n > k. Thus xi
is a Cauchy sequence in R. By Ex. 10.1.2,
COMPLETENESS
259
(n)
xi → xi for some xi ∈ R. Put x = (xi ). We observe that x ∈ ℓ2 , and
x(n) → x. For a fixed integer j > 0 and integers m, n > k, we have
j (
j (
)2
)2
∑
∑
(m)
(n)
(n)
xi − xi
< ϵ2 . Letting m → ∞, we get
xi − xi
≤
)2
∑(
(n)
ϵ . As this is true for every integer j, we have
xi − xi
≤ ϵ2 .
i=1
i=1
2
Also, by using the inequality 2ab ≤ a2 + b2 , which holds for any two
reals a, b, we have
(
)2
(
)2
(n)
(n)
x2i ≤ 2 xi − xi
+ 2 xi
.
So
∑
(
)
x2i < ∞ and d x, x(n) < ϵ for n > k.
Example 10.1.5 Let X be a set. Then the metric space B(X) of all
bounded functions X → R with the supremum metric ρ (refer to
Ex. 1.1.5) is complete. To see this, let ⟨fn ⟩ be a Cauchy sequence
in B(X). Then, for each x ∈ X, ⟨fn (x)⟩ is a Cauchy sequence in
R, since |fn (x) − fm (x)| ≤ ρ (fn , fm ). Therefore ⟨fn (x)⟩ converges
to a real number g(x), say. Thus we have a function g : X → R
defined by g(x) = limn→∞ fn (x) for all x ∈ X. We show that g is
bounded, and fn → g. Given ϵ > 0, there exists an integer k such that
ρ (fm , fn ) < ϵ for m, n > k. Consequently, |fn (x) − fm (x)| < ϵ for all
x ∈ X, and m, n > k. While keeping n and x fixed, let m → ∞. Since
fm (x) → g(x) as m → ∞, we obtain |fn (x) − g(x)| ≤ ϵ for all n > k
and all x ∈ X. In particular, this implies that there is an integer n
such that |fn (x) − g(x)| ≤ 1 for all x ∈ X. Since fn is bounded, g is
also bounded. Thus g ∈ B(X), and ρ (fn , g) ≤ ϵ for all n > k. It follows
that fn → g, and hence our assertion.
Returning to the general discussion, we prove a characterisation of
completeness, which is analogous to the characterisation of compactness in terms of closed sets.
Theorem 10.1.5 (G. Cantor) A metric space (X, d) is complete if
and only if the intersection of every family {Fα } of closed subsets of X
with the finite intersection property and inf {diam(Fα )} = 0 contains
exactly one point.
Proof. ⇒: Suppose that {Fα } is a family of closed subsets of X
such that the intersection of every finite subfamily is nonempty, and
inf {diam(Fα )} = 0. For each integer n ≥ 1, there is a member of {Fα },
260
Elements of Topology
∩n
say Fαn , such that diam (Fαn ) < 1/n. Since 1 Fαi ̸= ∅, we choose
a point xn which lies in each Fαi , 1 ≤ i ≤ n. Then ⟨xn ⟩ is a Cauchy
sequence in the complete space
∩ X, and therefore it converges to a point
x ∈ X. We claim that x ∈ Fα . If x ∈
/ Fα for an index α, then there
exists an open ball B(x; r) ⊂ X − Fα . Now, we find an integer n so
large that 1/n
∩n< r/2 and xn ∈ B(x; r/2). Then Fαn ⊂ B(x; r), which
forces Fα ∩ ( 1 Fαi ) = ∅, a contradiction. Hence our claim.∩To see the
uniqueness of x, assume that there is another point y ∈ Fα . Then
we have diam(Fα ) ≥ d(y, x) > 0 for all α, contrary to our hypothesis.
⇐: Let ⟨xn ⟩ be a Cauchy sequence in X. For {each} n, let En =
our
{xm |m ≥ n} . Then diam (En ) → 0, so the family E n satisfies
∩
hypothesis. Consequently, there exists a unique point x ∈ E n . We
assert( that
) xn → x. Given ϵ > 0, there exists an integer nϵ such that
diam E n < ϵ, for all n ≥ nϵ . As x ∈ E n , we have d(xn , x) < ϵ for all
n ≥ nϵ , and hence our assertion.
♢
The condition inf{diam (Fα )} = 0 in the preceding theorem is essential: The family Fn = [n, ∞), n ∈ N, in the real line R has an empty
intersection.
The next theorem generalises the Bolzano–Weierstrass theorem for
the real line and gives another characterisation of complete metric
spaces.
Theorem 10.1.6 A metric space X is complete if and only if every
infinite totally bounded subset of X has a limit point in X.
Proof. Suppose that X is complete, and let A ⊆ X be infinite totally
bounded. Choose a sequence ⟨xn ⟩ of distinct points from A. Since A
is totally bounded, it is covered by a finite number of open balls, each
of radius 1/2. Accordingly, at least one of these balls, say B1 , contains
an infinite number of terms of ⟨xn ⟩. Thus there exists an infinite set
M1 ⊂ N such that xn ∈ B1 for all n ∈ M1 . Let n1 ∈ M1 be the least
integer. Again, consider a finite covering of X by the open balls each
of radius 1/4. Then some member of this covering, say B2 , contains
xn for infinitely many indices n ∈ M1 . Accordingly, there is an infinite
set M2 ⊂ M1 such that xn ∈ B2 for all n ∈ M2 . Let n2 ∈ M2 be
the least integer greater than n1 . Then xn2 ∈ B1 ∩ B2 . Proceeding by
induction, we obtain a subsequence ⟨xnk ⟩ of ⟨xn ⟩, and a sequence of
∩k
open balls Bk of radii 1/2k such that xnk ∈ j=1 Bj . The sequence
⟨xnk ⟩ satisfies the Cauchy condition, since the points xnk and xnl lie in
COMPLETENESS
261
Bj for k, l > j, and diam(Bj ) < 1/j. Since X is complete, the sequence
⟨xnk ⟩ converges to a point x ∈ X. It is clear that x is a limit point of
A because ⟨xnk ⟩ is composed of distinct points of A.
To prove the converse, suppose that every infinite totally bounded
subset of X has a limit point. Let ⟨xn ⟩ be Cauchy sequence in X. Then
the range {xn } of the sequence ⟨xn ⟩ is totally bounded. If {xn } is finite,
then one of the terms of ⟨xn ⟩ must be repeated infinitely many times,
and ⟨xn ⟩ obviously converges to that point. In the other case, it has a
limit point x ∈ X. Consequently, we find a subsequence ⟨xnk ⟩ of ⟨xn ⟩,
which converges to x. Since ⟨xn ⟩ is a Cauchy sequence, xn → x and X
is complete.
♢
We note that the property of completeness is not hereditary; however, it is inherited by the closed sets. This is shown by the following.
Theorem 10.1.7 Every closed subspace of a complete metric space is
complete; the converse is also true.
Proof. Let X be a complete metric space, and Y ⊆ X be closed. If
⟨yn ⟩ is a Cauchy sequence in Y (with the induced metric), then there
is a point x ∈ X such that yn → x. Therefore x ∈ Y = Y , and Y is
complete.
The simple proof of the converse is left to the reader.
♢
Open sets in a complete metric space are in general not complete
with respect to the induced metrics; however, these are topologically
complete. More generally, we have
Theorem 10.1.8 If X is a complete metric space, then every Gδ -set
in X is topologically complete.
Proof. Let (X, d) be ∩
a complete metric space and Y ⊂ X be a Gδ -set.
∞
Suppose that Y = 1 Un , where each Un is open in X. Obviously,
we may assume that Un ⊇ Un+1 for all n. Note that if x ∈ Un , then
dist(x, X − Un ) > 0. Put
ϕn (x)
= 1/dist(x, X − Un ), and
ψn (x, y) = |ϕn (x) − ϕn (y)|/ (1 + |ϕn (x) − ϕn (y)|)
for x, y ∈ Un . Now,
R given by
∑∞consider the function ρ : Y × Y∑→
∞
ρ(x, y) = d(x, y)+ 1 2−n ψn (x, y). Clearly, the series 1 2−n ψn (x, y)
is convergent so that ρ is well defined. Since
|ϕn (x) − ϕn (z)| ≤ |ϕn (x) − ϕn (y)| + |ϕn (y) − ϕn (z)|,
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Elements of Topology
we have ψn (x, z) ≤ ψn (x, y) + ψn (y, z), and it is easily checked that ρ
is a metric on Y .
Next, we show that the metric space (Y, ρ) is complete. Let ⟨yi ⟩
be a ρ-Cauchy sequence in Y. Then it is clearly a d-Cauchy sequence.
Since X is d-complete, yi → x for some x ∈ X. We assert that x ∈ Y.
If x ∈
/ Y, then there exists an integer n0 > 0 such that x ∈
/ Un for all
n > n0 . Let i be a fixed positive integer and n > n0 . Then
dist(yi+j , X − Un ) ≤ d(x, yi+j ) → 0 as j → ∞,
since yi+j → x. Consequently,
ψn (yi , yi+j ) → 1 and it follows that
∑∞
−n
limj→∞ ρ(yi , yi+j ) ≥
2
. This contradicts our hypothesis
n>n0
about the sequence ⟨yi ⟩, and hence our assertion. Now, we observe
that ∑
yi → x in the metric ρ. Given ϵ > 0, choose an integer n0 > 0 so
∞
that n>n0 2−n < ϵ/3. Since yi → x in the metric d and the distance
function x → dist(x, X − Un ) is continuous, we can find an integer
i0 > 0 such that d(yi , x) < ϵ/3 and ψn (x, yi ) < ϵ/3n0 for all i > i0 and
n = 1, . . . , n0 . Then, for i > i0 , we have ρ(x, yi ) < ϵ, and yi → x. Thus
(y, ρ) is complete.
Finally, notice that a sequence in Y converges in the metric ρ if and
only if it converges in the metric d, so a subset F ⊆ Y is d-closed if and
only if it is ρ-closed. Thus ρ is equivalent to dY , and this completes the
proof.
♢
Example 10.1.6 The subspace Q ⊂ R is not topologically complete (see
Ex.
∩ 10.3.1), but the subspace of irrational numbers is. For, R − Q =
q∈Q (R − {q}) and each {q} is closed in R.
The property of topological completeness is invariant under countable products.
Theorem 10.1.9 Let (Xi , di ) , i = 1, 2, . . . , be a∏countable family of
complete metric spaces. Then the product space Xi is topologically
complete.
Proof. Choose a sequence ⟨λi ⟩ of positive real numbers such that
λi → 0. For each i, define a new metric d¯i on Xi by d¯i (xi , yi ) =
¯
min {λi , di (x∏
i , yi )}. Then Xi is complete in the{metric di also, and the
}
¯
metric ρ on Xi given by ρ ((xi ), (yi )) = sup di (xi , yi ) |i = 1, 2, . . .
metrises
the product topology (ref. Theorem 2.2.14). We assert that
∏
Xi is complete in the metric ρ. Let ⟨xn ⟩ be a Cauchy sequence in
COMPLETENESS
263
∏
Xi . If xni denote the i th coordinate of xn , then each coordinate sequence ⟨xni ⟩ is a Cauchy sequence in Xi . So there exists a point xi ∈ Xi
such that xni → xi in the metric d¯i . By Theorem
∏ 4.2.7, we see that the
sequence ⟨xn ⟩ converges to the point (xi ) in Xi .
♢
One can apply the preceding theorem to give an alternative proof
of Theorem 10.1.8 (refer to Exercise 17).
We end this section by establishing a close relationship between
completeness and compactness.
Theorem 10.1.10 A space is compact if and only if it is totally
bounded, complete metric space.
Proof. A compact metric space is obviously totally bounded and, by
Theorem 10.1.4, is complete.
Conversely, let X be a complete and totally bounded metric space.
Then, by Theorem 6.3.4, it is enough to prove that X is sequentially
compact. Let ⟨xn ⟩ be a sequence in X. Applying the method used in
Theorem 10.1.6, we can extract a Cauchy subsequence ⟨xnk ⟩ of ⟨xn ⟩.
Since X is complete, the sequence ⟨xnk ⟩ converges to a point of X, and
this completes the proof.
♢
Exercises
1. • Prove that a Cauchy sequence in a metric space X is convergent if it
has a convergent subsequence.
2. • Let (X, d) be a metric space and d1 be the standard bounded metric
corresponding to d : d1 (x, y) = min {1, d(x, y)}. Show that
(a) d1 is a metric equivalent to d.
(b) d1 does not alter Cauchy sequences in X.
(c) X is d-complete ⇔ X is d1 -complete.
3. Prove that every discrete space is topologically complete.
4. Prove that the interval [0, 1) with the metric d(x, y) = |(1 − x)−1 − (1 −
y)−1 | is complete.
5. Let z be an irrational number, and define a metric d on the set Q of
all rational numbers by d(x, y) = |(x − z)−1 − (y − z)−1 |. What are the
Cauchy sequences in Q?
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Elements of Topology
6. Is the sequence ⟨fn ⟩ defined by fn (x) = xn a Cauchy sequence in the
space B(I) with the supremum metric?
7. For each integer n > 0, define fn

1



1 + 3n−1 − 3n t
fn (t) =



0
: I → R by
for 0 ≤ t ≤ 1/3,
(
)
for 1/3 < t ≤ 1 + 3n−1 /3n ,
)
(
for 1 + 3n−1 /3n < t ≤ 1.
Show that ⟨fn ⟩ is a Cauchy sequence in in the metric space in Ex. 1.1.4
which fails to converge.
8. • Let X be a space, and C ∗ (X) be the set of all real-valued bounded
continuous functions on X with the metric
ρ(f, g) = sup {|f (x) − g(x)| : x ∈ X} .
Show that the metric space (C ∗ (X), ρ) is complete. (Notice that if X is
compact, then C ∗ (X) consists of all real-valued continuous functions on
X; in particular, C(I) is complete in the supremum metric.)
9. Prove that the Fréchet space Rω is complete.
10. Show that a complete subspace of a metric space X is closed.
( )
11. Let X be a metric space. Prove that diam (A) = diam A for every
A ⊆ X.
12. Let X be a metric space, and suppose that there is a real r > 0 such
that B(x; r) is compact for every x ∈ X. Prove that X is complete.
13. If every closed and bounded subset of a metric space X is compact,
show that X is complete.
14. Prove that a metric space X is totally bounded if and only if every
sequence in X has a Cauchy subsequence.
15. Let X be a complete metric space. Show that a subset A ⊆ X has
compact closure if and only if A is totally bounded.
16. Let X be a complete metric space. If {Fn } is a sequence of closed and
bounded
∩ subsets of X such that Fn ⊇ Fn+1 and diam (Fn ) → 0, show
that Fn consists of exactly one point. Also, prove the converse.
17. • In a metric space X, prove:
(a) The graph of the function x 7→ 1/dist(x, X − U ) of an open set
U ⊂ X is closed in X × R.
(b) If X is complete, and U ⊆ X is open, then U is topologically
complete.
COMPLETENESS
265
(c) If U1 , U2 , . . . is a sequence ∩
of subspaces
∏ of X, then the image of
the embedding x → (x) of Un into Un is closed.
(d) If X is complete, then every Gδ -set in X is topologically complete.
18. Let (X, d) be a complete metric space, and f : X → X be a function
for which there exists a real α < 1 such that d (f (x), f (y)) ≤ αd(x, y)
∀ x, y ∈ X. Prove:
(a) f is continuous.
(b) For every x ∈ X, the sequence ⟨f n (x)⟩ is a Cauchy sequence in
X.
(c) If x0 = lim f n (x) (which exists), then f (x0 ) = x0 . (The point x0
is called is called a fixed point of f .)
19. Prove that a complete, connected and locally connected metric space is
arcwise connected.
10.2
Completion
In this section, we will see that an incomplete metric space can
always be enlarged to become complete. To make this statement precise, we recall that an isometry (or an isometric embedding) of a metric
space (X, dX ) into a metric space (Y, dY ) is a distance-preserving map
f : X → Y , that is, one which satisfies dY (f (x), f (x′ )) = dX (x, x′ )
for all x, x′ ∈ X. If f : X → Y is a surjective isometry, then we say
that X and Y are isometric. It is clear that two isometric spaces are
homeomorphic, and therefore an isometry of X into Y is an embedding
(topological). With this terminology, we have
Theorem 10.2.1 Any metric space X can be isometrically embedded
into a complete metric space.
Proof. Let C ∗ (X) be the space of all bounded continuous real-valued
functions on X with the supremum metric ρ. If ⟨fn ⟩ is a Cauchy
sequence in C ∗ (X), then there exists a function g : X → R such
that fn → g uniformly on X (cf. Ex. 10.1.5). Since fn is continuous for all n, g is continuous. Thus the sequence ⟨fn ⟩ converges to
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the function g in C ∗ (X), and C ∗ (X) is complete. Let us fix a point
x0 ∈ X. For each x ∈ X, consider the function fx : X → R defined
by fx (p) = d(p, x) − d(p, x0 ) for all p ∈ X, where d is the metric on
X. Since d is continuous, so is fx . Moreover, by the triangle inequality, |fx (p)| ≤ d(x, x0 ) for all p ∈ X, and therefore fx ∈ C ∗ (X). For
x, y ∈ X, we have
ρ (fx , fy ) = supp∈X {|fx (p) − fy (p)|} = supp∈X {|d(p, x) − d(p, y)|}.
Since |d(p, x) − d(p, y)| ≤ d(x, y) for every p ∈ X and the equality is
attained for p = y, ρ (fx , fy ) = d(x, y). Thus the mapping x 7→ fx is
an isometric embedding of X into C ∗ (X), and the proof is complete.♢
By Theorem 10.1.7, the closure of the image of the isometry x 7→ fx
b
in C ∗ (X) is also complete. Thus there is a complete metric space X
b
which contains a dense subset isometric to X. We call X a completion of
b depends on the choice of the fixed
X. Observe that this definition of X
point x0 . However, we will soon establish that any two completions of
b is unique upto an isometry, and we
X are isometric; accordingly, X
may call it the completion of X. This is a simple consequence of the
following extension theorem for uniformly continuous functions into
complete metric spaces.
Theorem 10.2.2 Let A be a dense subset of a metric space X, and
Y be a complete metric space. Then a uniformly continuous function
f : A → Y has a unique uniformly continuous extension to X.
Proof. We define a function g : X → Y as follows: For x ∈ A, we put
g(x) = f (x). If x ∈ X − A, then there is a sequence ⟨an ⟩ in A which
converges to x. Since f is uniformly continuous, it is easily checked that
⟨f (an )⟩ is a Cauchy sequence in Y. So ⟨f (an )⟩ converges to a point of
Y , since Y is complete . Put g(x) = lim f (an ). We observe that g(x)
is independent of the sequence ⟨an ⟩ used in its definition. Let ⟨bn ⟩ be
another sequence in A such that bn → x. Then dX (an , bn ) → 0 which
implies that dY (f (an ), f (bn )) → 0. Therefore f (bn ) → g(x). Notice
that g extends the function f to X, by its definition.
Next, we show that g is uniformly continuous on X. Given
ϵ > 0, choose δ > 0 (by the uniform continuity of f ) such that
dY (f (a), f (b)) < ϵ whenever dX (a, b) < δ. Let x, x′ ∈ X satisfy
dX (x, x′ ) < δ/3. Find sequences ⟨an ⟩ and ⟨a′n ⟩ in A such that an → x
and a′n → x′ . (If x ∈ A, we may take an = x for every n.) Then there
COMPLETENESS
267
exists an integer n0 such that dX (an , x) < δ/3 and dX (a′n , x′ ) < δ/3 for
all n > n0 . Now, for n > n0 , dX (an , a′n ) < δ ⇒ dY (f (an ), f (a′n )) < ϵ.
Because f (an ) → g(x) and f (a′n ) → g(x′ ), we obtain dY (g(x), g(x′ )) =
lim dY (f (an ), f (a′n )) ≤ ϵ (by Exercise 1). This shows that g is uniformly continuous. By Corollary 4.4.3, g is unique, and this completes
the proof.
♢
We remark that the hypotheses about f and Y in the preceding
theorem are essential. This can be seen by considering the functions
R − {0} → R, x 7→ x/|x|, and the identity map on the subspace Q ⊂ R.
b Yb be complete metric spaces, and X ⊆ X
b
Corollary 10.2.3 Let X,
b
b
b
and Y ⊆ Y be dense. If X and Y are isometric, then so are X and Y .
Proof. Let f be an isometry of X onto Y . It is immediate that f is
b and j : Y → Yb be the inclusion
uniformly continuous. Let i : X → X
mappings. By Theorem 10.2.2, j ◦ f extends to a uniformly continuous
b → Yb , and i ◦ f −1 extends to a uniformly continuous
function g : X
b Thus, we have the following commutative diagram
function h : Yb → X.
X
f
i
?
b
X
- Y
j
h
g
?
b
- Y
of continuous maps. It is clear that the composition hg is an extension
b is also an extension of i, we have hg =
of i. Since the identity map on X
1Xc , by uniqueness. Similarly, gh = 1Yb , and g is a homeomorphism.
b is a limit point of a
Since j ◦ f is an isometry, and each point of X
sequence in X, it is easily seen that g is an isometry.
♢
The uniqueness property for completion of a metric space can now
be readily observed. We remark that there is a different method of
constructing the completion, which is an emulation of the Cantor process for constructing the real numbers from the rationals by means of
Cauchy sequences. This construction is outlined in Exercise 5 below.
A homeomorphism f : X → Y between metric spaces is called a
uniform homeomorphism if both f and f −1 are uniformly continuous.
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Elements of Topology
A surjective isometry is a uniform homeomorphism, but the converse
is not true, in general. A property which is preserved under uniform
homeomorphisms is called a uniform property; for example, completeness is a uniform property. Two metrics d1 and d2 on a set X are
called uniformly equivalent if the identity map iX : (X, d1 ) → (X, d2 )
is a uniform homeomorphism.
The simple proof of the following corollary is left to the reader.
Corollary 10.2.4 Let X and Y be complete metric spaces and A ⊆ X
and B ⊆ Y be dense. Then each uniform homeomorphism f : A → B
can be extended to a uniform homeomorphism g : X → Y .
Exercises
1. • Let (X, d) be a metric space. Suppose that ⟨xn ⟩ and ⟨x′n ⟩ are
two sequences in X converging to x and x′ , respectively. Show that
d(xn , x′n ) → d(x, x′ ).
2. • Let f : X → Y be a uniformly continuous function. If ⟨xn ⟩ is a Cauchy
sequence in X, show that ⟨f (xn )⟩ is a Cauchy sequence in Y . Give an
example of a continuous map which transforms Cauchy sequences into
Cauchy sequences, but fails to be uniformly continuous.
3. Give an example to show that the image of a complete space under a
uniformly continuous map is not necessarily complete.
4. Prove that the completion of a metric space X is separable if and only
if X is separable.
5. • Let (X, d) be a metric space, and S be the set of all Cauchy sequences in X. Consider the relation ∼ on S defined by ⟨xn ⟩ ∼ ⟨yn ⟩ ⇔
lim d(xn , yn ) = 0. Prove:
(a) ∼ is an equivalence relation.
(b) If ⟨xn ⟩ and ⟨yn ⟩ are Cauchy sequences in X, then d(xn , yn ) is a
Cauchy sequence in R, and hence converges.
(c) If x̂ and ŷ are equivalence classes of ⟨xn ⟩ and ⟨yn ⟩, respectively,
then dˆ(x̂, ŷ) = lim d(xn , yn ) is independent of the choice of these
b = S/ ∼.
representations and defines a metric on X
(
)
b dˆ is complete.
(d) The metric space X,
COMPLETENESS
269
(e) If f (x) denotes the equivalence class of the constant sequence ⟨x⟩ ,
x ∈ X, then dˆ(f (x), f (y)) = d(x, y). Thus the mapping x 7→ f (x)
b
is an isometry of X into X.
b and f (X) = X
b when X is complete.
(f) f (X) is dense in X,
6. (a) Give an example of a uniform homeomorphism which is not an
isometry.
(b) Show that the homeomorphism x → x3 of R onto itself is not a
uniform homeomorphism, although it preserves Cauchy sequences
in R.
7. (a) Show that the euclidean metric on Rn and the two metrics described in Exercise 1.1.4 are all uniformly equivalent.
(b) Find a metric on Rn which is equivalent but not uniformly equivalent to the euclidean metric.
8. Prove:
(a) If (X, d) is a metric space, then the metric d′ = d/(1 + d) is
uniformly equivalent to d.
(b) Boundedness is a metric but not a uniform property.
(c) Total boundedness is a uniform property, but not a topological
property.
10.3
Baire Spaces
In this section, we prove one of the most important theorems in
topology, which has extensive applications in analysis.
Theorem 10.3.1 Let (X, d) be a complete metric space. Then the
intersection of any countable family of open dense subsets of X is
nonempty; in fact, it is dense in X.
Proof. Let {Un } be a countable collection
of open sets such that U n =
∩
X for every n. We show that G ∩ ( Un ) ̸= ∅ for each nonempty open
set G ⊆ X. Since U1 is dense, G ∩ U1 ̸= ∅. Because G ∩ U1 is open,
there is an open ball B1 of radius r1 ≤ 1 and centered at x1 ∈ G ∩ U1
such that B 1 is contained in G ∩ U1 . Since U2 is dense, there is a point
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Elements of Topology
x2 ∈ B1 ∩ U2 . So we can find an open ball B2 of radius r2 ≤ 1/2 and
centered at x2 such that B 2 ⊆ U2 . For r2 < r1 − d(x1 , x2 ), we have
B 2 ⊆ B1 . Proceeding inductively, we obtain a sequence of open balls
Bn ⊆ Un such that diam(Bn ) ≤ 2/n and B n+1 ⊆ Bn ∩ Un+1 . Let xn
denote the centre of Bn . Then ⟨xn ⟩ is a Cauchy sequence in X, since
xn ∈ Bm for all n ≥ m, and rn → 0. By the completeness of X, xn → x
for some x ∈ X. It is clear that x ∈ B n+1 ⊆ B∩
n for every n.∩Since
B 1 ⊆ G and B n ⊆ Un for every n, we have x ∈ B n ⊆ G ∩ ( Un ).
This completes the proof.
♢
The preceding theorem can also be established for locally compact
Hausdorff spaces. This property of locally compact Hausdorff spaces
makes them interesting to analysts.
Theorem 10.3.2 The intersection of any countable family of open
dense sets in a locally compact Hausdorff space X is dense.
Proof. Let {Un } be a countable family of open sets, each Un being
dense
in X. Let O be any nonempty open subset of X. We show that
∩
( Un ) ∩ O ̸= ∅. Since X = U 1 , O ∩ U1 ̸= ∅. By Theorem 6.4.2,
there exists a nonempty open set V1 such that V 1 is compact and
V 1 ⊆ O ∩ U1 . For the same reason, we find a nonempty open set V2
such that V 2 is compact and V 2 ⊆ V1 ∩ U2 . Proceeding by induction,
we obtain for each positive integer n, a nonempty open set Vn such
that V n is compact and V n ⊆ Vn−1 ∩
. It is obvious that the sets
{ Un}
V n are closed∩in V 1 , and the family V n has the finite intersection
property. So V n ̸= ∅,
is compact. As V n ⊆ Un for every n,
∩ for V 1 ∩
and V 1 ⊆ O. We have V n ⊆ ( Un ) ∩ O. This completes the proof.
♢
We remark that the hypotheses “open” and “countable” in the
above theorems are essential, for the set Q and its complement are
dense in R, and the family {R − {x} |x ∈ R} has empty intersection,
although all its members are open and dense.
Definition 10.3.3 A space X in which the intersection of each countable family of dense open sets is dense is referred to as a Baire space.
By the preceding two theorems, we see that locally compact Hausdorff spaces and topologically complete spaces are two main classes of
Baire spaces.
COMPLETENESS
271
We recall that a subset E of a space X is nowhere dense if E has an
empty interior (equivalently, X − E is dense). With this terminology,
the above theorems can be unified into a more standard form.
Theorem 10.3.4 (Baire Category Theorem) A Baire space is not
the union of a countable number of nowhere dense sets.
Proof. Let {En } be a countable family of nowhere dense subsets of a
Baire space X. Then Un =∩X − E n is an open dense set for every n,
and so there is a point x ∈ Un . This implies that x ∈
/ En for all n. ♢
This theorem implies that a Baire space cannot be decomposed into
countably many closed sets, each of which has no interior.
A subset E of a space X is said to be of the first category if it is
the union of a countable family of nowhere dense sets. A set which
is not of the first category is said to be of the second category (or
nonmeager.) In these terms, a Baire space is of the second category.
The converse is also true. For, if X is a space of the second category
and Un , n = 1, 2, . . ., are open dense
∩ subsets of X,
∪ then each X − Un is
closed and nowhere
dense
so
that
U
=
X
−
(X − Un ) ̸= ∅. This
n
∩
implies that G ∩ ( Un ) ̸= ∅ for each nonempty open set G ⊆ X, since
G ∩ Un is also dense in X. Notice that a Hausdorff space of the second
category cannot be countable and perfect.
Example 10.3.1 In the real line R, the set Q of rationals is of the
first category, and its complement is of the second category. For,
{{r}|r ∈ Q} is a countable closed covering of Q, and no {r} is an
open set. If R − Q were also of first category, then the complete space
R would be a countable union of nowhere dense sets. This contradicts
Theorem 10.3.4.
By Theorem 10.3.1, it is also clear that Q is not topologically complete.
As an application of the Baire Category Theorem, we establish the
uniform boundedness principle.
Theorem 10.3.5 Let X be a complete metric space, and F a family
of real-valued continuous functions on X with the property that for
each x ∈ X, there is a number Mx such that |f (x)| ≤ Mx for all f ∈ F.
Then there exists a nonempty open set U ⊆ X and a number M such
that |f (x)| ≤ M for all f ∈ F and all x ∈ U .
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Elements of Topology
Proof. For each f ∈ F and each integer n > 0, let En,f =
{x ∈ X||f (x)| ≤ n}. Then En,f
∩ is closed, for it is the inverse image
of [−n,
∪n] under f . Set En = {En,f |f ∈ F}. Then En is closed and
X = En . By Theorem 10.3.4, some set En is not nowhere dense.
Since En is closed, it must contain a nonempty open set U , say. Then,
for every x ∈ U, |f (x)| ≤ n for all f ∈ F. This completes the proof. ♢
To see another application of Baire’s Theorem, we ask the reader
to prove Exercise 19. It goes to show the existence of a continuous
real-valued function on I which has no derivative at any point.
Exercises
1. Show that the Sorgenfrey line Rℓ is a Baire space.
2. Prove that the set Q of rational numbers is not a Gδ -set in R.
3. Give an example of an uncountable nowhere dense set.
4. Prove that the intersection of countably many open dense sets in a Baire
space is a set of the second category.
5. Prove that an open (or closed) subset of a topologically complete space
is of the second category.
6. Give an example of a second category set which contains an open subset
that is a set of the first category.
7. Prove that an open subset of a Baire space is also a Baire space.
8. Suppose that each point of a space X has a nbd that is a Baire space.
Show that X is a Baire space.
9. Show that Baire spaces are invariant under continuous open surjections.
10. Show that a space X is a set of second category ⇔ the intersection of
every countable family of open dense sets in X is nonempty.
11. Prove:
(a) A subset of a set of the first category is also of the first category.
(b) The countable union of sets of the first category is a first category
set.
(c) The notion of category is not topological.
12. Let X be a Baire space and A ⊆ X be a set of the first category. Show
that (a) A has no interior, and (b) X − A is dense in X, and is a set of
the second category.
COMPLETENESS
273
13. Give an example to show that the complement of a second category set
in a complete metric space need not be a first category set.
14. Prove that a compact Hausdorff space is a Baire space.
15. If X is a complete metric space which has no isolated points, prove that
X is uncountable.
16. Prove that a locally compact Hausdorff space cannot be countable and
perfect both.
17. Construct a function that is continuous at each irrational number and is
discontinuous at each rational number. Prove that there is no function
that is continuous at the rationals and discontinuous at the irrationals.
18. Prove that the set of points of discontinuity of a real-valued function on
a space is an Fσ -set. Deduce that there is no function from R to itself
which is continuous precisely on the rationals.
19. • Consider the metric space C(I) with the sup metric ρ. For each positive
integer n, let Fn denote the set of all f ∈ C(I) for which there exists
(x)
| ≤ n for all 0 < δ with x + δ ∈ I. Prove:
x ∈ I such that | f (x+δ)−f
δ
(a) If a sequence ⟨fp ⟩ in Fn converges to f ∈ C(I), then f ∈ Fn . (So
Fn is closed.)
(b) Given f ∈ Fn and r > 0, there exists a piecewise linear function
g such that ρ(f, g) < r/2.
(c) There exists a “sawtoothed” function h within r/2 of g such that
the absolute value of the gradient of each line segment of h is
greater than n. (Thus Fn is nowhere dense.)
∪
(d) A function in the complement Fn is not differentiable anywhere.
Chapter 11
FUNCTION SPACES
11.1
11.2
11.3
11.1
Topology of Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Topology of Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Compact-Open Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Topology of Compact Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
275
278
283
301
Topology of Pointwise Convergence
Let (X, dX ) and (Y, dY ) be metric spaces. Recall that a sequence
⟨fn ⟩ of functions of X into Y is said to
(a) converge simply (or pointwise) to a function f : X → Y if for each
x ∈ X and each ϵ > 0, there is an integer n0 > 0 (depending on
x and ϵ) such that dY (fn (x), f (x)) < ϵ for all n ≥ n0 .
(b) converge uniformly to a function f : X → Y if for each
ϵ > 0, there is an integer n0 > 0 (depending on ϵ) such that
dY (fn (x), f (x)) < ϵ for all n ≥ n0 (ϵ), and for every x ∈ X.
(c) converge continuously to a function f : X → Y if for each x ∈ X
and each sequence ⟨xn ⟩ in X converging to x, fn (xn ) → f (x) in
Y.
Note that the metric of X does not play any role in the first two
types of convergence. In this section, we shall generalise the first two
concepts in the realm of topological spaces and investigate the topologies describing them.
Given sets X and Y , the set F(X, Y ) of all functions f ∏
: X →
Y , also denoted by Y X , is essentially the cartesian product
Yx of
the family {Yx |Yx = Y and x ∈ X}. If Y is a topological space, then
F(X, Y ) can be given the product topology. The space F(X, Y ) with
the product topology is∏denoted by F(X, Y )p . When X is finite, we
have two definitions of
Yx , and we also have the product topology
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Elements of Topology
defined on Y × · · · × Y (|X| copies), albeit in a different way. However,
there is a canonical bijection
η : F(X, Y ) → Y × · · · × Y
(|X| copies) ,
f 7→ (f (x)), where f (x) is the xth coordinate of (f (x)), and our definitions of topologies for F(X, Y )p and Y × · · · × Y make the mapping
η : F(X, Y )p → Y × · · · × Y a homeomorphism (ref. §2 of Chapter
2) so that the two spaces can be identified. By Theorem 4.2.7, a net
⟨fα ⟩ in F(X, Y )p converges to a function f : X → Y if and only if
fα (x) → f (x) in Y for every x ∈ X. This kind of convergence for nets
(or sequences) of functions is called the pointwise convergence. Because
of this, the product topology for F(X, Y ) is appropriately called the
topology of pointwise convergence or simply the pointwise topology.
For each x ∈ X, the projection mapping F(X, Y ) → Y, f 7→ f (x),
is referred to as the evaluation mapping at x, and denoted by ex . A
subbasis for F(X, Y )p consists of the sets
e−1
x (U ) = {f : X → Y |f (x) ∈ U } ,
where x ∈ X and U ⊆ Y is open (Figure 11.1). Accordingly, the topology of F(X, Y )p is also called the point-open topology.
………
f3
U
f2
f1
x
Yx
FIGURE 11.1: Three elements of e−1
x (U ).
X
FUNCTION SPACES
277
From our knowledge about the product (topology, we )see that if
Y has compactness, connectedness or the Ti i = 1, 2, 3, 3 12 property,
then F(X, Y )p has that property. As normality and the countability
axioms are not product invariant, F(X, Y )p may fail to inherit these
properties of Y . If X is infinite, then F(X, Y )p is locally connected
(resp. locally compact) if and only if Y is locally connected and connected (resp. locally compact and compact), in which case F(X, Y ) is
also connected (resp. compact). This follows by Theorem 3.4.4 (resp.
Theorem 6.4.6).
It is quite common to be concerned with a subset Φ of F(X, Y ).
Then one natural way to topologise Φ would be to give it the relative
topology induced from F(X, Y )p . We denote this space by Φp , and
refer to its topology with any of the four adjectives used to describe the
topology of F(X, Y )p . Clearly, the pointwise topology on Φ is generated
by the sets {f ∈ Φ|f (x) ∈ U }, where x ∈ X, and U ⊆ Y is open, as a
subbasis. In functional analysis, it is often important to know whether
such a space Φp is compact. If Y is a compact Hausdorff space, then a
function space Φp is compact if and only if it is closed in F(X, Y )p . This
is evident from the Tychonoff theorem. A more general characterisation
of those function spaces which are compact in the pointwise topology
is given in
Theorem 11.1.1 Let X be a set, and Y be a T2 -space. A subspace
Φ of F(X, Y )p is compact if and only if Φ is closed in F(X, Y )p , and
each set {f (x)|f ∈ Φ}, x ∈ X, has compact closure in Y.
The simple proof is omitted.
Note that Hausdorffness of Y is not needed for “if” part of the
theorem.
A useful fact about the pointwise topology Tp for Φ ⊂ F (X, Y ) is
that any topology strictly finer than Tp is not compact, if Y is Hausdorff. For if a topology T on Φ is compact, and finer than Tp , then
the identity function (Φ, T) → (Φ, Tp ) is a homeomorphism, since Tp
is Hausdorff. Thus T = Tp . This simple remark provides the standard
method of checking the compactness of a function space (Φ, T), when
the range of the functions in Φ is T2 . One needs to prove that the space
(Φ, Tp ) is compact, and a net in Φ, which converges to f ∈ Φ in the
topology Tp , also converges to f in the topology T. Observe that if
either condition fails, then (Φ, T) is not compact.
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Elements of Topology
Topology of Uniform Convergence
Let Y be a metrisable space and d be a bounded metric on Y,
which induces the topology of Y. Such a metric always exists, by
Corollary 1.4.9. For any set X and any two functions f, g : X → Y,
sup {d (f (x), g(x)) |x ∈ X} is always a real number. We leave it to the
reader to verify that d∗ defined by
d∗ (f, g) = sup {d (f (x), g(x)) |x ∈ X}
is a metric on F(X, Y ).
Definition 11.1.2 The metric d∗ on F(X, Y ) is called the sup metric
or the uniform metric induced by d, and the topology it induces for
F(X, Y ) is called the topology of uniform convergence.
The metric space (F(X, Y ), d∗ ) will usually be denoted by
F(X, Y )u . Note that, for any real ϵ > 0, d∗ (f, g) ≤ ϵ ⇔ d (f (x), g(x)) ≤
ϵ for all x ∈ X. Therefore a sequence ⟨fn ⟩ in the metric space
(F(X, Y ), d∗ ) converges to f : X → Y if and only if, for each ϵ > 0,
there exists a positive integer n0 (ϵ) such that n ≥ n0 implies that
d (fn (x), f (x)) ≤ ϵ for all x ∈ X. Thus fn → f in F(X, Y )u if and only
if ⟨fn ⟩ converges uniformly to f. This justifies the use of the adjective
“uniform” with the metric d∗ . It should be emphasised that the topology of uniform convergence for F(X, Y ) generally depends not only on
the topology of Y, but also on the particular bounded metric used in
Y, as is shown by the following.
Example 11.1.1 Let X = Y = R, and d1 , d2 be the metrics on Y
defined by
d1 (y, y ′ ) = min {1, |y − y ′ |},
′
and
′
d2 (y, y ) = |y/(1 + |y|) − y /(1 + |y ′ |)|.
Both the metrics d1 and d2 are bounded and generate the usual topology for Y. But the induced metrics d∗1 and d∗2 on F(X, Y ) are not
equivalent. To see this, consider the functions fn : X → Y for each
n = 1, 2, . . . , given by
{
x if x ≤ n, and
fn (x) =
n if x > n.
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Let f : X → Y be the identity map. Then we have d∗1 (fn , f ) = 1, and
d∗2 (fn , f ) = 1/(1 + n). Accordingly, fn → f in the metric d∗2 , but fails
to do so in the metric d∗1 . It follows that the metrics d∗1 and d∗2 generate
different topology for F(X, Y ).
However, if Y is a compact metric space, then the topology of uniform convergence for F(X, Y ) is independent of the metric used to
give the topology of Y. To see this, let d1 , d2 be two metrics on Y, each
generating its topology. Then, by the compactness of Y , both metrics
d1 and d2 are bounded. Also, given ϵ > 0, there exists δ1 > 0 such that
d1 (y, y ′ ) < δ1 ⇒ d2 (y, y ′ ) < ϵ for all y, y ′ ∈ Y. So, for f, g in F(X, Y ),
d∗1 (f, g) < δ1 ⇒ d1 (f (x), g(x)) ≤ δ1 for all x ∈ X ⇒ d2 (f (x), g(x)) ≤ ϵ
for all x ∈ X ⇒ d∗2 (f, g) ≤ ϵ. By symmetry, we find a real δ2 > 0 such
that d∗2 (f, g) < δ2 ⇒ d∗1 (f, g) ≤ ϵ. Hence d∗1 ∼ d∗2 .
Returning to the general situation, we note that each evaluation
ex : (F (X, Y ), d∗ ) → (Y, d),
f 7→ f (x), is uniformly continuous. This is immediate from the definition of d∗ . Because the pointwise topology for F(X, Y ) is the coarsest
topology such that all evaluations ex are continuous,the topology of uniform convergence is finer than the topology of pointwise convergence; in
other words, every uniformly convergent sequence in F(X, Y ) is simply
convergent.
We now assume that X also has a topology, and denote the subspace
C(X, Y ) ⊆ F(X, Y )u of all continuous maps X → Y by C(X, Y )u . The
following proposition describes the main properties of this subspace.
Proposition 11.1.3 Let X be a topological space and (Y, d) be a
bounded metric space. Then C(X, Y ) is closed in the metric space
F(X, Y )u .
Proof. Let ⟨fn ⟩ be a sequence in C(X, Y ), and suppose that fn → f
in F(X, Y )u . We need to show that f is continuous. Let x0 ∈ X be
arbitrary but fixed. Given ϵ > 0, there is an integer n0 (ϵ) such that
d (fn0 (x), f (x)) < ϵ/3 for every x ∈ X. We have d (f (x), f (x0 )) ≤
d (f (x), fn0 (x)) + d (fn0 (x), fn0 (x0 )) + d (fn0 (x0 ), f (x0 )). Since fn0 is
continuous, we can find an open nbd G of x0 such that fn0 (G) ⊆
B (fn0 (x0 ) ; ϵ/3). Then d (f (x), f (x0 )) < ϵ for all x ∈ G, and f is
continuous at x0 . Thus f ∈ C(X, Y ), and (a) follows.
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Notice that while establishing the preceding proposition, we have
actually proved the classical result: The uniform limit of continuous
functions is continuous. On the other hand, the pointwise limit of continuous functions need not be continuous, that is, C(X, Y ) is not always
closed in the topology of pointwise convergence for F(X, Y ).
Example 11.1.2 Let X = I and Y = R. For each positive integer n,
let fn : I → R be the function defined by fn (t) = tn , t ∈ I. Then the
sequence ⟨fn ⟩ converges to f ∈ F(X, Y )p , where f is given by f (t) = 0
if t ̸= 1, and f (1) = 1. Notice that each fn is continuous, while f is
discontinuous.
Proposition 11.1.4 Let X be a topological space, and (Y, d) be a
bounded, complete metric space. Then the metric space F(X, Y )u is
complete, and hence the subspace C(X, Y )u is also complete.
Proof. Let ⟨fn ⟩ be a Cauchy sequence in F(X, Y )u . Given ϵ > 0, there is
an integer n0 (ϵ) such that d∗ (fn , fm ) < ϵ whenever n, m ≥ n0 (ϵ). So for
every x ∈ X and n, m ≥ n0 , we have d (fn (x), fm (x)) ≤ d∗ (fn , fm ) < ϵ.
This shows that ⟨fn (x)⟩ is a Cauchy sequence in Y for each x ∈ X.
Since Y is complete, there is a point in Y , say f (x), such that fn (x) →
f (x). If n ≥ n0 , then d (fn (x), f (x)) = d (fn (x), limm→∞ fm (x)) =
limm→∞ d (fn (x), fm (x)) ≤ ϵ for every x ∈ X. Thus d∗ (fn , f ) ≤ ϵ for
all n ≥ n0 whence fn → f in F(X, Y )u .
The last statement follows from Proposition 11.1.3.
♢
It should be noted that if an unbounded metric d is chosen to
generate the topology of Y, then the above discussion can be carried
on with the set of all d-bounded functions of X into Y (cf. Exercise
16). To see that d∗ (f, g) is a real number for every pair of d-bounded
functions f, g : X → Y , consider a point x0 ∈ X. Then
d (f (x), g(x)) ≤ d (f (x), f (x0 )) + d (f (x0 ), g(x0 )) + d (g(x0 ), g(x))
≤ diam (f (Y )) + d (f (x0 ), g(x0 )) + diam (g(Y ))
for every x ∈ X. So d∗ (f, g) < ∞ always.
In particular, if X is a compact space, then the sup metric d∗ is
defined on C(X, Y ), even if the metric d in Y is unbounded, since every
continuous function from X to Y is bounded. In this case, we will later
see that the topology on C(X, Y ) induced by d∗ (that is, the uniform
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convergence topology) is independent of the metric d on Y (refer to
Theorem 11.2.6).
Exercises
1. Let X = Y = R, and f : X → Y be a function. Given a real ϵ > 0, and
a finite set F ⊆ X, define
B(f, ϵ, F ) = {g ∈ F (X, Y )| |g(x) − f (x)| < ϵ for all x ∈ F }.
Show that the sets B(f, ϵ, F ), as F ranges over all finite subsets of
X, and ϵ ranges over all positive reals, form a nbd basis at f in the
pointwise topology on F(X, Y ).
2. Let X be a completely regular space. Prove that C(X) is dense in
F (X, R)p .
3. Suppose that f ∈ Φ ⊆ F(X, Y ), where X is a set and Y is a space. Given
an integer n > 0, let xi ∈ X and Ui ⊆ Y be a nbd of f (xi ), i = 1, . . . , n.
Prove that the set {g ∈ Φ|g(xi ) ∈ Ui for each i = 1, . . . , n} is a nbd of
f ∈ Φp , and the family of all such sets is a nbd basis at f.
4. Which of the following subspaces of F(I, I)p is compact?
(a) {f : I → I|f (0) = 0}.
(b) {f : I → I|f is continuous and f (0) = 0}.
5. Is the space RI = F(I, R)p separable?
6. Show that I I = F(I, I)p is separable, normal but not first countable
(and hence is not metrisable).
7. For each t ∈ I, let ft : I → I be the characteristic function of {t}. Suppose that A = {fr |r ∈ I is rational} and B = {fs |s ∈ I is irrational}.
Prove that A and B are separated sets in F(I, I)p , which can not be
separated by disjoint open sets. (Thus F(I, I)p is a normal space which
is not completely normal.)
8. If Y is Hausdorff (regular, completely regular), prove that C(X, Y )p is
Hausdorff (regular, completely regular) for any space X.
9.
(a) Is the sequence ⟨fn ⟩ given by fn (t) = tn , t ∈ I, convergent in the
topology of uniform convergence for F (I, I)?
(b) Are the topology of pointwise convergence and the topology of
uniform convergence for F (I, I) identical?
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(a) For each integer n = 1, 2 . . . , define a function fn : I → R
n
by fn (x) = nx (1 − x) . Show that the sequence ⟨fn ⟩ converges
pointwise to the constant function at 0, but fails to converge uniformly.
10.
(b) Do ( as in) (a) with the sequence ⟨gn ⟩, where gn (x)
n
for all x ∈ I.
nx 1 − x2
=
11. Compare the topology of pointwise convergence for C (I) with the topology of uniform convergence.
12. • Let d be a bounded metric in R. Prove that the topology of uniform
convergence for F (R, R) defined by d is strictly finer than the topology
of pointwise convergence.
13. Suppose that Xi and Yi , i = 1, 2, are topological spaces such that
X1 ≈ X2 and Y1 ≈ Y2 . If Y1 and Y2 are compact and metrisable, show
that C (X1 , Y1 )u ≈ C (X2 , Y2 )u for any metric di in Yi .
14. Determine whether the following spaces are compact. Separable.
(a) F (I, I)u , (b) C (I, I)u , (c) C (I)u .
15. If a Cauchy sequence ⟨fn ⟩ in C (I)u converges pointwise to f , show that
⟨fn ⟩ converges uniformly to f .
16. •
(a) Let (Y, d) be a metric space and Φ be a set of bounded functions
X → Y . If d1 be the corresponding standard bounded metric,
show that d∗1 = min {1, d∗ } on Φ.
(b) • Let X be a topological space and (Y, d) be a complete metric
space. Let B (X, Y ) be the set of all bounded functions X → Y ,
and C ∗ (X, Y ) be the set of all bounded continuous functions X →
Y. Show that B (X, Y ) and C ∗ (X, Y ), both, are closed in the metric
d∗1 on F (X, Y ) , where d1 = min {1, d} .
(c) • If (Y, d) is complete, show that the spaces B (X, Y ) and
C ∗ (X, Y ), both, are complete in the sup metric d∗ .
17. Let Y = R − {0} have the standard bounded metric. Discuss the completeness of the spaces
(a) F (R, Y )u ,
(b) the subspace of F (R, Y )u which consists of functions f with |f | ≥
1, and
(c) C (R, Y )u .
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11.2
283
Compact-Open Topology
In the previous section, we have studied two topologies for a given
set Φ of functions of one space into another, namely, the pointwise
topology and the uniform metric topology (when the range space of
the functions in Φ is metrisable). The first one is too small for many
purposes though it inherits many properties of the range space and is
easy to handle, while the second one is rather large. These remarks can
be seen by the following examples.
Example 11.2.1 Let X = R and Y = I, the unit interval. For each
finite set A ⊂ X, consider the characteristic function fA : X → Y
(that is, fA (x) = 1 if x ∈ A, and fA (x) = 0 otherwise). The family A
of the sets A is directed by the inclusion, and therefore we have a net
{fA , A ∈ A} in F(X, Y ). Let c : X → Y be the constant function at 1,
and U be any open set in Y with 1 ∈ U. Then, for each x ∈ X, e−1
x (U ) is
a subbasic open set in F(X, Y )p and contains c. Because each singleton
{x} is a member of A, and if {x} ⊆ A, then fA (x) = 1, the net ⟨fA ⟩
is eventually in e−1
x (U ). Hence fA → c in the pointwise topology on
F(X, Y ). This convergence does not conform to our intuition, since no
term of the net ⟨fA ⟩ seems close to c.
Example 11.2.2 Consider the space C ∗ (R) of all bounded continuous
functions with the uniform metric topology. For each positive integer
n, define the function fn : R → R by

0 for |x| ≤ n,



1 for |x| > n + 2 and
fn (x) =



(|x| − n)/2 for n ≤ |x| ≤ n + 2.
(See Figure 11.2). The sequence ⟨fn ⟩ clearly fails to converge in the
space C ∗ (R); however, it converges to the constant function f (x) = 0
on every bounded interval of R.
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……………… … … … … … ……
f3
f2
f1
f1
f2
f
f3
f
FIGURE 11.2: Illustration of Example 11.2.2.
One would like to see a topology for C ∗ (X) in which sequences such
as in Example 11.2.2 are convergent. With this end in view, we examine
the pointwise topology further. Suppose that X and Y are topological
spaces and a set Φ ⊂ F(X, Y ) is given the pointwise topology. Then
the evaluation mapping ex : f 7→ f (x) from Φ to Y is continuous for
each x ∈ X, since the inverse image of an open set U ⊂ Y under ex
is a subbasic open set of Φp . In fact, the pointwise topology on Φ is
the smallest topology for which each evaluation map ex is continuous.
And, if Φ consists of continuous functions, then f (x) is also continuous
in the variable x. Thus, in this case, the natural function e : Φp ×
X → Y , defined by e(f, x) = f (x), is separately continuous in each
of its variables. However, e may not be jointly continuous in both the
variables, f and x, if Φp × X is given the product topology. This is
shown by the following.
Example 11.2.3 For each integer n = 1, 2, 3, . . . , consider the function
2
fn : I → R given by fn (x) = nxe−nx . Then fn → 0 in C (I, R)p , and
fn (1/n) → 1. If the function e : C (I, R)p × I → R, (f, x) 7→ f (x), were
continuous, then the sequence ⟨fn (1/n)⟩ would have converged to 0.
So e is not continuous.
The natural function e : Φ × X → Y , (f, x) → f (x), plays an important role in the study of function spaces, and its continuity requires
that the functions in Φ be continuous. Accordingly, our main interest
centres around the set C(X, Y ) of all continuous functions X → Y .
Definition 11.2.1 A topology for Φ ⊆ C(X, Y ) is called admissible (or
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jointly continuous) if the natural function e : Φ×X → Y, (f, x) → f (x),
is continuous, and e is referred to as the evaluation map of Φ.
Because the pointwise topology is the smallest topology on Φ which
makes the evaluation map e : Φ × X → Y separately continuous, an
admissible topology for Φ must be finer than this topology. At the
other extreme, given any two spaces X and Y , the discrete topology
on C(X, Y ) is admissible. For, if (f, x) ∈ e−1 (U ), where U ⊆ Y, then
there is an open nbd O of x such that f (O) ⊆ U, since f is continuous.
Now, the set {f } × O is an open nbd of (f, x) ∈ C(X, Y ) × X, and is
mapped by the evaluation map e into U. So e−1 (U ) is a nbd of each
of its points, and hence open. Thus every set Φ ⊆ C(X, Y ) has an
admissible topology. So it is natural to seek the smallest admissible
topology, if it exists. However, there is no such smallest topology, in
general. In fact, R. Arens (1946) has shown that X must be locally
compact Hausdorff if there is an admissible topology on C(X, I), which
is coarsest. Thus, for example, there is no coarsest admissible topology
on C (Rω , I).
One might have noticed that the pointwise topology for Φ depends
only on Φ and the topology of Y, and the uniform metric topology for it
depends on the metric in Y . The topology of X does not play any role
in the definitions of these topologies and the results stated in §1. We
shall now discuss a method of topologizing Φ in which the topology of
domain space plays a significant role. This topology, viz. the compactopen topology for Φ is fairly satisfactory, and of particular importance
in Analysis.
Let Φ be a set of functions from X to Y. For each pair of sets
A ⊆ X, and B ⊆ Y, define
(A, B) = {f ∈ Φ|f (A) ⊆ B}.
With this notation, the subbase generating the pointwise topology Tp
on Φ is the family of the sets (x, U ), where x ∈ X and U ⊆ Y is open.
It is easily observed that the family
{(F, U )|F ⊆ X finite, and U ⊆ Y open}
also generates the topology Tp . But, if we use compact sets in place of
the one-point sets used in the defining subbase for Tp , we obtain a new
topology.
Definition 11.2.2 Let Φ be a set of functions from a space X to a
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space Y . The compact-open topology Tco on Φ is the topology generated
by the subbasis
{(K, U )|K ⊆ X compact, and U ⊆ Y open}.
It is also called the k-topology for Φ.
The set Φ endowed with the topology Tco will be denoted by Φco . It
is easy to verify that the compact-open topology for Φ coincides with
the relative topology it inherits from F(X, Y )co as a subspace.
Notice that if Y is indiscrete, then the compact-open topology on
F(X, Y ) is also indiscrete for all spaces X. And, if X is discrete, then
the compact-open topology and the point-open topology are identical for all Y , because the only compact sets in X are the finite sets.
In general, the two topologies are distinct (ref. Exercise 5). However,
the compact-open topology is always finer than the pointwise topology. This is immediate from the fact that the defining subbase for the
compact-open topology contains a subbasis for the pointwise topology,
since each one-point subset of {X} is compact. The following proposition shows that the pointwise topology Tp on C(X, Y ) is not admissible
if Tco is strictly finer than Tp .
Proposition 11.2.3 Let Φ be a set of continuous functions from X to
Y . The compact-open topology Tco for Φ is coarser than any admissible
topology.
Proof. Let Ta be an admissible topology for Φ and Φa denote the space
Φ with the topology Ta . It suffices to prove that each subbasis member
(K, U ) of Φco is open in Φa . Let e : Φ × X → Y be the evaluation map
(f, x) 7→ f (x). If f ∈ (K, U ), and x ∈ K, then e−1 (U ) contains (f, x).
By our hypothesis, e−1 (U ) is open in the product Φa ×X. So there exist
open sets Vx ⊂ X and Gx ⊂ Φa such that (f, x) ∈ Gx × Vx ⊂ e−1 (U ).
Since K is compact,
∩n many points x1 , . . . , xn , say, in K
∪n we find finitely
such that K ⊂ 1 Vxi . Put G = 1 Gxi . Then G is a nbd of f in Φa .
If g ∈ G, and x ∈ Vxi , then (g, x) ∈ Gxi × Vxi ⊂ e−1 (U ) which implies
that g(x) ∈ U . It follows that G ⊂ (K, U ) and (K, U ) is a nbd of f in
the topology Ta . Thus (K, U ) is a nbd of each of its points with respect
to Ta , and hence open in Φa .
♢
The next proposition gives a sufficient condition for the continuity
of the evaluation map.
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Proposition 11.2.4 Let Φ be a set of continuous functions of a space
X into a space Y . If X is locally compact Hausdorff, then the compactopen topology on Φ is the coarsest admissible topology.
Proof. Suppose that X is locally compact Hausdorff. By Proposition
11.2.3, we need to prove that e : Φco ×X → Y is continuous. Let V ⊂ Y
be open and (f, x) ∈ e−1 (V ). Since f is continuous, f −1 (V ) is an open
nbd of x. Since X is a locally compact Hausdorff space, there exists a
compact nbd K of x such that K ⊆ f −1 (V ) ⇒ f (K) ⊆ V . Thus (K, V )
is an open set of Φco , and contains f . We have e[(K, V ) × K ◦ ] ⊆ V so
that (f, x) ∈ (K, V ) × K ◦ ⊆ e−1 (V ). This implies that e−1 (V ) is open
in Φco × X, and the map e is continuous.
♢
The following example shows that the condition of local compactness on X is essential in the preceding proposition, and this can not
be generalized for k-spaces.
Example 11.2.4 By Ex. 6.4.4, the subspace Q ⊆ R of the rational
numbers is not locally compact. We show that the compact-open
topology for C(Q, I) is not admissible, that is, the evaluation map
e : C(Q, I) × Q → I, (f, x) 7→ f (x), is not continuous. Let q ∈ Q,
and c : Q → I be the constant
map at 1. We prove that e is not contin∩n
uous at (c, q). If B = 1 (Ki , Ui ) is a basic nbd of c in C(Q, I)co∪
, then
n
each Ki is compact and 1 = c(Ki ) ⊆ Ui for every i. Thus F = 1 Ki
is compact, and therefore closed in Q. If a nbd N of q in Q is contained in F , then its closure N in Q is compact and, consequently, N
would be closed in R. But, this is obviously false. So we can find a
point x ∈ N − F. Since Q is completely regular, there exists a continuous function f : Q → I such that f (x) = 0 and f (F ) = 1. We have
f (Ki ) = 1 ∈ Ui for every i so that f ∈ B. Thus e−1 ((0, 1]) does not
contain B × N. The set (0, 1] is obviously a nbd of c(q). Since the sets
B × N form a local base at (c, q), e is not continuous at this point.
To see a comparison between the compact-open topology and the
topology of uniform convergence for C(X, Y ), we prove the following.
Proposition 11.2.5 Let X be a topological space and (Y, d) be a
bounded metric space. Then the topology of C(X, Y )u is admissible.
Proof. Let e : C(X, Y )u × X → Y be the evaluation map. To see the
continuity of e, consider a point x0 ∈ X, and a continuous function
f : X → Y. Given ϵ > 0, let U be the open ball in C(X, Y )u with radius
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ϵ and centre f . The inverse image of the open ball B (f (x0 ); ϵ) in Y
under f is a nbd of x0 , for f is continuous. Put V = f −1 (B(f (x0 ); ϵ)).
Then U × V is an open nbd of (f, x0 ) in the domain space of e. Since
d (g(x), f (x0 )) ≤ d (g(x), f (x)) + d (f (x), f (x0 )) < 2ϵ for every g ∈ U
and x ∈ V, the function e maps U × V into the open ball B(f (x0 ); 2ϵ).
It follows that e is continuous at the point (f, x0 ), and this completes
the proof.
♢
By the preceding proposition and Proposition 11.2.3, it is clear that
the compact-open topology for C(X, Y ) is coarser than topology of uniform convergence for metrisable spaces Y . Moreover, the two topologies
are identical for compact spaces X, as shown by the following.
Theorem 11.2.6 Let X be a compact space, and (Y, d) be a metric
space. Then the topology of uniform convergence determined by d for
C(X, Y ) is identical with the compact-open topology.
Proof. We have already seen that the topology of C(X, Y )u is finer
than the topology of C(X, Y )co . Here is an alternative argument. Let
(K, U ) be a member of the subbase for C(X, Y )co , and f ∈ (K, U ).
Then K ⊆ X is compact, U ⊆ Y is open and f (K) ⊂ U . So, for each
y ∈ f (K), we can find a real ry > 0 such that B(y; 2ry ) ⊆ U . By the
compactness of f (K), there ∪
exist finitely many points y1 , . . . , yn , say,
n
in f (K) such that f (K) ⊂ 1 B (yi ; ryi ). Put ϵ = min {ry1 , . . . , ryn }.
Then ϵ > 0, and B(f ; ϵ) ⊆ (K, U ). For, if d∗ (g, f ) < ϵ and x ∈ K, then
there is an index i such that d (yi , f (x)) < ryi so that d (g(x), yi ) <
2ryi ⇒ g(x) ∈ U . Since f ∈ (K, U ) is arbitrary, (K, U ) is open in
C(X, Y )u .
The topology of C(X, Y )co is finer than the topology of C(X, Y )u :
Consider a d∗ -open ball B(f ; ϵ) in C(X, Y )u . Choose a real δ such
that 0 < 3δ < ϵ. By the the compactness of X,
∪n we find finitely many
points x1 , . . . , xn , say, in X(such that f (X)
) ⊂ 1 B (f (xi ); δ). For each
i = 1, . . . , n, let Ki = f −1 B(f (xi ); δ) and Vi = B (f (xi ); 2δ). Then
each Ki is compact, being a closed subset of the compact space X,
and each Vi is open in Y. Obviously,
∩nf (Ki ) ⊆ B (f (xi ); δ) ⊂ Vi so
that f ∈ (Ki , Vi ) for every i. Thus 1 (Ki , Vi ) is an open nbd of f
in C(X, Y )co . We assert that it is contained in B(f ; ϵ). Suppose that
g ∈ (Ki , Vi ) for every i = 1, . . . , n. If x ∈ X, then f (x) belongs to some
B (f (xi ); δ). Accordingly, x ∈ Ki and g(x) ∈ Vi . So d (g(x), f (x)) ≤
d (g(x), f (xi )) + d (f (xi ), f (x)) < 3δ, which implies that d∗ (g, f ) < ϵ.
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Thus we have g ∈ B(f ; ϵ), and hence our assertion. It follows that
B(f ; ϵ) is open in C(X, Y )co , and this completes the proof.
♢
From the preceding theorem, it is immediate that the compact-open
topology for C(X, Y ) is metrisable if X is compact and Y is metrisable.
Moreover, in this case, the topology of uniform convergence on C(X, Y )
is independent of the metric d on Y . By Proposition 11.1.4 and Exercise
10.1.2, we also see that the space C(X, Y )co is topologically complete
whenever Y is so. Recall that if X is a discrete space, then the compactopen topology for C(X, Y ) coincides with the topology of pointwise
convergence. So we can expect C(X, Y )co to inherit in general only
the productive properties of Y . Thus, there is no hope that C(X, Y )co
would inherit any of the countability properties or normality from Y .
The following theorem describes the separation properties which can
be inherited from Y.
Theorem 11.2.7 Let X be a space.
(a) If Y satisfies the Ti -axiom (i = 0, 1, 2), then so does F(X, Y )co .
(b) If Y satisfies the Ti -axiom (i = 0, 1, 2, 3, 3 21 ), then so does
C(X, Y )co .
Proof. (a): We prove this part of the theorem for i = 2 only; the other
cases are proved similarly. Suppose that Y is a T2 -space. Let f, g : X →
Y be distinct functions. Then there is an x ∈ X such that f (x) ̸= g(x).
Since Y is T2 , there exist disjoint open sets U and V in X containing
f (x) and g(x), respectively. Clearly, (x, U ) and (x, V ) are disjoint nbds
of f and g, respectively.
(b): The statements for i = 0, 1, 2 are true, by (a). To see the case
(i = 3), suppose that Y is a T3 -space. Let f : X → Y be a continuous
function, and let (K, U ) be a subbasic nbd of f . Then f (K) is compact
and contained in U . Since Y is T3 , for each y ∈ f (K), there exists
an open set Gy ⊆ Y such that y ∈ Gy ⊂ Gy ⊂ U . Since f (K) is
compact,∪we find finitely many points y1 , . . .∪
, yn in f (K) such that
n
n
f (K) ⊂ 1 Gyi = V , say. So we have V = 1 Gyi ⊂ U , obviously.
We observe that (K, V ) ⊂ (K, U ). Let g : X → Y be a continuous
function such
/ (K, U ). Then, for some( x ∈ K, g(x)
∈
/ U so
( that g) ∈
)
that g ∈ x, Y − V . It is clear that the set x, Y − V is open in
C(X, Y )co and is disjoint from (K, V ). This implies that g ∈
/ (K, V ).
Thus f ∈ (K, V ) ⊂ (K, V ) ⊂ (K, U ), where (K, V ) is a subbasic open
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set in C(X, Y )co . Now, if N is any nbd of f in C(X, Y )co , then there exist
finitely many compact
sets K1 , . . . , Km in X, and open sets U1 , . . . , Um
∩m
in Y such that f ∈ 1 (Ki , Ui ) ⊂ N. As shown above, we find open
sets V1 , . . . , Vm in Y such that f ∈ (K∩i , Vi ) ⊂ (Ki , V∩
i ) ⊂ (Ki , Ui ) for
m
m
every i = 1, . . . , m. Consequently, f ∈ 1 (Ki , Vi ) ⊆ 1 (Ki , Vi ) ⊂ N,
and C(X, Y )co satisfies the condition T3 .
Case (i = 3 12 ). Let Y be a T3 12 -space, and let f0 ∈ C(X, Y )co . We
first show the existence of a Urysohn function for the pair {f0 } and the
complement of a subbasic open nbd (K, U ) of f0 in C(X, Y )co . We have
f0 (K) ⊂ U . Since Y is T3 21 , for each y ∈ f0 (K), there is a continuous
function ϕy : Y → I such that ϕy (y) = 0 and ϕy (Y − U ) = 1. Choose a
real 0 < r < 1 and put Vy = ϕ−1
y [0, r). Then Vy is open and the family
{Vy |y ∈ f0 (K)} covers f0 (K). Since K is compact and f0 is continuous,
f0 (K) is compact. So there
∪nexist finitely many points y1 , . . . , yn in
f0 (K) such that f0 (K) ⊂ 1 Vyi . Let ξ = min {ϕy1 , . . . , ϕyn }. Then
ξ : Y → I is continuous and ξ(y) < r for every y ∈ f0 (K), and
ξ(Y −U ) = 1. Consider the function µ = max {ξ, χr }, where χr denotes
the constant map Y → I at r. Obviously, the function µ is continuous
and takes f0 (K) into r, Y − U into 1, and maps Y into [r, 1]. For any
continuous map f : X → Y , the real-valued function µf |K attains
its maximum; consequently, there exists a point x0 ∈ K (depending
upon f ) such that µf (x0 ) = sup {µf (x)|x ∈ K}. Define a mapping
ζ : C(X, Y ) → [r, 1] by ζ(f ) = µf (x0 ). Clearly, the number ζ(f ) is
independent of the choice of point x0 in K, and ζ(f0 ) = r, and ζ(f ) = 1
if f ∈
/ (K, U ). To verify the continuity of ζ at f , let ϵ > 0 be given. By
the continuity of µ, there exists an open nbd W1 of f (x0 ) = y0 in Y
such that |µ(y) − µ(y0 )| < ϵ for every y ∈ W1 . Put ζ(f ) = µ(y0 ) = s
and let W2 = µ−1 [r, s + ϵ). Then N = (x0 , W1 ) ∩ (K, W2 ) is an open
subset of C(X, Y )co . For x ∈ K, we have r ≤ µ (f (x)) ≤ µ (f (x0 )) =
s < s + ϵ; accordingly, f ∈ N. If g ∈ N, then g(x0 ) ∈ W1 ; consequently,
µg(x0 ) > s − ϵ. This implies that ζ(g) = sup {(µg)(x)|x ∈ K} > s − ϵ,
for x0 ∈ K. Since g(K) ⊂ W2 , we have µg(x) < s + ϵ for all x ∈
K; so ζ(g) < s + ϵ. Thus |ζ(g) − s| < ϵ, and ζ is continuous at f.
The composition of ζ with the homeomorphism [r, 1] ≈ [0, 1] yields a
continuous function ψ : C(X, Y )co → I such that ψ(f0 ) = 0 and ψ(f ) =
1 for f ∈
/ (K, U ). Now let G be any open nbd of f0 in C(X, Y )co . Then
there exist compact sets
∩n K1 , . . . , Kn in X, and open sets U1 , . . . , Un
in Y such that f0 ∈ 1 (Ki , Ui ) ⊂ G. As above, we find continuous
functions ψi : C(X, Y )co → I such that ψi (f0 ) = 0, and ψi (f ) = 1 for
FUNCTION SPACES
291
f∈
/ (Ki , Ui ) for i = 1, . . . , n. Then the function ψ = max {ψ1 , . . . , ψn }
satisfies ψ(f0 ) = 0, ψ(f ) = 1 for f ∈
/ G.
♢
We remark that the converse of the preceding theorem is also valid,
since we can embed Y in C(X, Y )co , and the property of being a Hausdorff, regular or completely regular space is hereditary. To see this,
consider the function λ : Y → C(X, Y ), which takes y ∈ Y to the
constant map X → Y at y. We have
Proposition 11.2.8 The mapping λ : Y → C(X, Y )co is an embedding. Moreover, if Y is Hausdorff, then λ(Y ) is closed in C(X, Y )co .
Proof. Obviously, λ is injective. The continuity of λ is also clear, for
if G = (K, U ) is a subbasic open set in C(X, Y )co , then λ−1 (G) = U
is open in Y. Next, let U ⊂ Y be open, and x ∈ X. Then we have
λ(U ) = (x, U ) ∩ λ(Y ). As (x, U ) is open in C(X, Y )co , λ(U ) is an open
subset of λ(Y ). It follows that λ is an embedding.
To prove the last statement, suppose that Y is T2 . We show that
the complement of λ(Y ) in C(X, Y )co is open. If f ∈
/ λ(Y ), then there
exist two distinct points x1 , x2 in X such that f (x1 ) ̸= f (x2 ). Since
Y is T2 , we find disjoint open nbds Ui of f (xi ), i = 1, 2, in Y. Set
O = (x1 , U1 ) ∩ (x2 , U2 ). Then O is open in C(X, Y )co and contains
f . If g ∈ O, then g(x1 ) ̸= g(x2 ), for U1 ∩ U2 = ∅; consequently,
g∈
/ λ(Y ). Therefore, O is contained in the complement of λ(Y ), and
this completes the proof.
♢
The map λ is referred to as the natural injection of Y into
C(X, Y )co .
The following facts are immediate from the definitions, and will be
useful for our discussion below. Let {Kα } and {Vα }, α ∈ A, be families
of subsets of X and Y , respectively. Then we have
∩
∪
for every V ⊆ Y ,
α (Kα , V ) = ( α Kα , V )
∩
∩
for for every K ⊆ X, and
α (K, Vα ) = (K, α Vα )
∩
∪
∪
α (Kα , Vα ) ⊆ ( α Kα , α Vα ).
The next theorem describes the conditions for the second axiom of
countability of the space C(X, Y ) with the compact-open topology.
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Theorem 11.2.9 If X is a second countable, locally compact, Hausdorff space and Y is a second countable space, then C(X, Y )co is also
second countable.
Proof. Suppose that both X and Y satisfy the second axiom of countability. Let A and B be countable bases of X and Y , respectively. If X is
locally compact Hausdorff, then the family A′ = {A ∈ A|A is compact}
is also a basis of X. For, if N is a nbd of a point x ∈ X, then there exists
an open set V ⊂ X such that x ∈ V ⊂ V ⊂ N and V is compact. Since
A is a base for X, we find a set A ∈ A such that x ∈ A ⊆ V . Clearly,
and A ⊆ N . Now, for each A ∈ A′ and B ∈ B, the set
(A is compact
)
A, B is open {(
in the )compact-open topology
} for C(X, Y ). Consider
the family Σ = A, B |A ∈ A′ and B ∈ B . We contend that it is a
subbasis for C(X, Y )co . Clearly, it suffices to prove that each subbasic
open subset (K, U ) of C(X, Y )co is open in the topology generated by
Σ. Suppose that f ∈ (K, U ). Then, for each x ∈ K, we find a Bx ∈ B
such that f (x) ∈ Bx ⊆ U . Since X is regular, there is an Ax ∈ A′
there are
such that x ∈ Ax ⊂ Ax ⊂ f −1 (Bx ). Since K is compact
∪n
finitely many points x1(, . . . ,)xn in K such that K ⊂ 1 Axi . For each
i = 1, . . . , n, we have f Axi ⊂ Bxi ⊂ U . So
) (∪
)
∩n (
∪
f ∈ 1 Axi , B xi ⊂
i Axi , i Bxi ⊂ (K, U ).
It follows that (K, U ) is a nbd of f in T (Σ) and hence our contention.
Obviously, Σ is countable, so the base generated by it is also countable.
Thus C(X, Y )co is second countable.
♢
We ask the reader to see for himself that the condition of local
compactness on X in the preceding theorem is essential (ref. Exercise
23). Since Y can be embedded in C(X, Y )co , there is a partial converse:
If C(X, Y )co is second countable, then so is Y.
The compactness of a subset of C(X, Y ) in the compact-open topology generally requires some strong conditions which will be discussed
at the end of the next section. In this regard, it is worth recalling that if
Y is Hausdorff, and the compact-open topology for F(X, Y ) is strictly
finer than the pointwise topology, then F(X, Y )co is not compact.
We now turn to see another important feature of the compact-open
topology. Suppose that X, Y and Z are three spaces. Given a function
f : X × Y → Z, write fx (y) = f (x, y) for all (x, y) ∈ X × Y. Then,
for a fixed x ∈ X, we have a function fx : Y → Z, y 7→ fx (y). If
FUNCTION SPACES
293
f is continuous, then each fx is continuous, since Y ≈ {x} × Y and
f | ({x} × Y ) is continuous. Thus a continuous function f : X × Y → Z
determines a one-parameter family of continuous functions fx : Y → Z
indexed by X. So we can define a function fˆ : X → C(Y, Z) by setting
fˆ(x) = fx , x ∈ X. With the compact-open topology for C(Y, Z), the
function fˆ is continuous.
Theorem 11.2.10 If f : X × Y → Z is continuous, then the function
fˆ : X → C(Y, Z)co is also continuous.
Proof. It is enough to prove that the inverse image of each subbasic
open set (K, U ) ⊆ C(Y, Z)co under fˆ is open in X. Let x0 be a point in
fˆ−1 (K, U ). Then fˆ(x0 ) ∈ (K, U ) so that f (x0 , y) ∈ U for every y ∈ K.
Thus x0 × K ⊂ f −1 (U ). Since f is continuous, and U ⊆ Z is open,
f −1 (U ) is open in X × Y . By Lemma 6.1.14, there exists a nbd N of
x0 such that N × K ⊂ f −1 (U ). Accordingly, f (N × K) ⊂ U , which
implies that fˆ maps N into (K, U ). Hence fˆ−1 (K, U ) is a nbd of each
of its points, and is therefore open.
♢
We remark that the converse of the preceding theorem is not true,
in general. However, by the preceding theorem, we can define a function
α : C(X × Y, Z) −→ C (X, C(Y, Z)co )
by α(f ) = fˆ. We refer to the function fˆ as the associate of f , and vice
versa; the function α is called the association function. This function is
injective, for if f and g are distinct members of C(X × Y, Z), then there
is a point (x, y) ∈ X ×Y such that f (x, y) ̸= g(x, y), which implies that
fˆ(x) ̸= ĝ(x). The other way round, given a function ϕ : X → C(Y, Z)co ,
we can define a function f : X × Y → Z by setting f (x, y) = ϕ(x)(y).
Then we have fˆ = ϕ so that each function fx : Y → Z defined by f is
continuous, but f itself need not be continuous, even if ϕ is. Clearly, f
is the composition
ϕ×1
e
X × Y −−−→ C(Y, Z)co × Y → Z,
where 1 denotes the identity map on Y and e is the evaluation map.
Hence f would be continuous if e and ϕ, both, are continuous. It follows
that α is bijective, provided the compact-open topology for C(Y, Z) is
admissible.
Thus, for locally compact Hausdorff spaces Y, we have the following.
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Theorem 11.2.11 (Exponential Correspondence) Let X, Y and
Z be topological spaces. If Y is locally compact Hausdorff, then a
function f : X × Y → Z is continuous if and only if each fx : Y → Z,
y 7→ f (x, y), is continuous, and the function fˆ : X → C(Y, Z)co , x 7→
fx , is also continuous.
The preceding theorem is often used to establish the continuity of
a function ϕ : X → C(Y, Z)co by showing that the associated function
e ◦ (ϕ × 1) : X × Y → Z is continuous. In the other direction, too, there
are some interesting applications of the theorem. As an illustration, we
give an alternative and somewhat easier proof of Theorem 7.2.8.
Theorem 11.2.12 Let f : X → Y be an identification, and suppose
that Z is a locally compact Hausdorff space. To show that f × 1 :
X × Z → Y × Z is also an identification, consider a space T and a
function g : Y × Z → T such that the composition
f ×1
g
X × Z −→ Y × Z −→ T
is continuous. Put h = g ◦ (f × 1). Then the associated function ĥ :
X → C(Z, T )co is continuous. Given y ∈ Y, there is an x ∈ X with
f (x) = y. We have g(y, z) = h(x, z) = ĥ(x)(z) for every z ∈ Z. This
shows that ĥ(x)(z) is independent of the choice of x in f −1 (y). So
we can define a function θ : Y → C(Z, T )co by θ(y) = ĥ(x), where
f (x) = y. It is obvious that the composite θf = ĥ is continuous. Since
f is an identification, we see that θ is also continuous. Since Z is locally
compact Hausdorff, the associated function g = e◦(θ×1) is continuous.
By Theorem 7.2.6, f × 1 is an identification.
We now see the conditions when the association function α is a
homeomorphism. Towards this end, we first prove the following.
Lemma 11.2.13 Let X be a Hausdorff space and Y be any
space. If S is a subbase for the topology of Y , then the family
{(K, V )|K ⊆ X compact and V ∈ S} is a subbase for the compactopen topology on C(X, Y ).
Proof. Let Σ be the family of all sets (K, V ) ⊆ C(X, Y ), where K ⊆ X
is compact, and V ∈ S, the subbasis for Y . It suffices to prove that each
subbasic open set (K, U ) of C(X, Y )co is open in the topology generated
by Σ. Let f ∈ (K, U ) be arbitrary. Then f : X → Y is a continuous
function such that f (K) ⊂ U . If B is the base for Y determined by
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the subbasis S, then U is the union of a subfamily {Wα } ⊂ B. Since
f (K) ⊂ U , the family f −1 (Wα ) covers K. Because K is compact
Hausdorff, it is regular. So, if x ∈ K ∩ f −1 (Wα ), then there is an
open set Hx in K such that x ∈ Hx ⊂ Hx ⊂ K ∩ f −1 (Wα ). The open
covering {Hx |x ∈ K}
} of the compact space K has a finite subcover
{
Hxj |j = 1, . . . , m , say. Now, for each
) 1, . . . , m, there
( is a set
)
( j =
Wαj ∈ {Wα } such that Hxj ⊂ f −1 Wαj . Then f ∈ Hxj , Wαj ,
where Hxj , being a closed subset of K, is compact. Also,
∩nj there are
=
in
S
such
that
W
finitely many
sets
V
,
.
.
.
,
V
α
j1
jn
j
(
) ∩nj (
) (
) i=1 Vji . We
∩j nj
have f ∈ Hxj , Wαj = Hxj , i=1
Vji = i=1
Hxj , Vji whence
f
)] ∩ (
)
∩m [∩nj (
∈
Hxj , Vji = j Hxj , Wαj
j=1
∪ i=1
⊆ (K, j Wαj ) ⊆ (K, U ).
It follows that (K, U ) is open in the topology generated by the subbase
Σ, and this completes the proof.
♢
Lemma 11.2.14 Let X, Y be Hausdorff spaces and Z be any space.
Then the sets (A × B, W ), where A ⊆ X, B ⊆ Y are compact and
W ⊆ Z is open, form a subbase for the compact-open topology for
C(X × Y, Z).
Proof. Let (K, W ) be a subbasic open set of C(X × Y, Z)co , and f ∈
(K, W ) be arbitrary. Then f : X × Y → Z is continuous with f (K) ⊂
W . So f −1 (W ) is a nbd of K. Let pX : X × Y → X, pY : X × Y → Y
be the projection maps, and put E = pX (K) and F = pY (K). Then
E ⊆ X and F ⊆ Y are compact, and K ⊆ E × F . The subspace
E × F ⊆ X × Y is regular, being compact Hausdorff. So, for each
k ∈ K, we can find an open nbd Nk of k in E × F such that Nk ⊂
(E × F ) ∩ f −1 (W ). Notice that the closures of Nk in E × F and X × Y
are identical. Obviously, we may assume that Nk = Uk × Vk , where
Uk is open in E, and Vk is open in F . Since K is compact,
there
∪n
∪n exist
finitely many (points k)1 , . . . , kn in K such that K ⊆ 1 Nki ⊆ 1 Nki .
We have f ∈ Nki , W for every i; consequently,
) ∩n (
) (∪n
)
∩n (
f ∈ 1 Uki × Vki , W = 1 Nki , W =
1 Nki , W ⊆ (K, W ).
Since E is a compact subspace of X, so is Uki for every i. Similarly, each
Vki is compact in Y . Because the compact-open topology for C(X ×
Y, Z) is generated by the sets (K, W ), the lemma follows.
♢
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Elements of Topology
Theorem 11.2.15 (The Exponential Law) Let X, Y be Hausdorff
spaces and Z be any space. Then the association map
α : C(X × Y, Z)co → C(X, C (Y, Z)co )co ,
f 7→ fˆ, is an embedding. If, in addition, Y is locally compact, then α
is a homeomorphism.
Proof. Let Φ and Ψ denote the domain and the codomain of α, respectively. We have already observed that α is injective. Since X is T2 , and
the collection of the sets (B, W ), where B ⊆ Y is compact and W ⊆ Z
is open, is a subbase for C(Y, Z)co , the topology of Ψ is generated by
the sets (A, (B, W )), A ⊆ X compact (by Lemma 11.2.13). Clearly,
f ∈ (A × B, W )
⇐⇒
fx (y) ∈ W for all x ∈ A and all y ∈ B
⇐⇒
fx ∈ (B, W ) for all x ∈ A
⇐⇒
α(f ) = fˆ ∈ (A, (B, W )).
So α−1 (A, (B, W )) = (A × B, W ) . Since the sets (A × B, W ) are open
in Φ, α is continuous.
Next, by Lemma 11.2.14, the sets (A × B, W ), where A ⊆ X, B ⊆
Y are compact, and W ⊆ Z is open, form a subbase for the topology
of Φ. Obviously, we have α (A × B, W ) = (A, (B, W )) ∩ α(Φ), so α
carries members of a subbase for Φ to open subsets of α(Φ). Since α is
injective, it preserves the intersections, and therefore α : Φ → α(Φ) is
a homeomorphism.
The final assertion follows from our discussion preceding Theorem
11.2.11.
♢
If we denote the set C(X, Y ) by Y X , etc., then the Exponential Law
can be expressed as Z X×Y ≈ (Z Y )X , provided that X and Y satisfy
the condition of the preceding theorem. We remark that this holds good
also under some other conditions on X and Y (Exercises 22 and 24).
This theorem is highly appreciated by the mathematicians working in
the field of Functional Analysis (duality theory) or Algebraic Topology
(homotopy theory).
In the end of this section, we discuss a notion of convergence which
describes the compact-open topology in some cases.
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Definition 11.2.16 Let X and Y be spaces. We say that a net
{fν , ν ∈ N } in C(X, Y ) converges continuously to f ∈ C(X, Y ) if for
each x ∈ X and each net {xν , ν ∈ N } in X converging to x, the net
⟨fν (xν )⟩ converges to f (x) in Y .
The condition of continuous convergence is equivalent to the following: For each x ∈ X and each nbd Of (x) of f (x) in Y , there exist
a nbd Ux of x in X and an index µ ∈ N such that fν (Ux ) ⊆ Of (x)
for all ν ≽ µ. To see this, suppose that there is a point x ∈ X and
a nbd Of (x) of f (x) such that for each nbd Ux of x and each ν ∈ N,
there exist
η (Ux , ν) ≽ ν and a point xη(Ux ,ν) in Ux with
( an index
)
fη(Ux ,ν) xη(Ux ,ν) ∈
/ Of (x) . Direct the set Nx × N by
(Ux , ν) ≼ (Vx , µ) ⇐⇒ Ux ⊇ Vx and ν ≼ µ.
Then, for every ν ∈ N, we ⟨have (X,
⟩ ν) in Nx × N and (X, ν) ≼
(Ux , µ) ⇒ ⟨η(Ux , µ) ⟩≽ ν. So xη(Ux ,ν) is a subnet
⟨ of ⟨xν ⟩.
( It is ob)⟩
vious that xη(Ux ,ν) converges to x while the net fη(Ux ,ν) xη(Ux ,ν)
fails to converge to f (x).
It should be noted that the concept of continuous convergence does
not require a topology for the set C(X, Y ). However, we have the following.
Theorem 11.2.17 In the smallest admissible topology for C(X, Y ), if
it exists, continuous convergence is equivalent to net convergence.
Proof. Let T be an admissible topology for C(X, Y ). If a net ⟨fν ⟩ in
C(X, Y ) converges to a function f ∈ C(X, Y ) relative to T, and a net
⟨xν ⟩ in X converges to x ∈ X, then (fν , xν ) → (f, x). By the continuity
of the evaluation map e : C(X, Y ) × X → Y , fν (xν ) → f (x). Thus
fν → f continuously.
Now, suppose that a net {fν , ν ∈ N } in C(X, Y ) converges continuously to f . Put Tν = {fµ |ν ≼ µ} and consider the family
U = {U ⊆ C(X, Y )|f ∈ U ⇒ Tν ⊆ U for some ν ∈ N }.
Then it is easily checked that U is a topology for C(X, Y ), and fν → f
with respect to U. We claim that U is admissible: Let (g, x) ∈ C(X, Y )×
X and Og(x) be an open nbd of g(x) in Y. If g ̸= f, then {g} ∈ U. By the
continuity of g, there exists a nbd Vx of x in X such that g(Vx ) ⊆ Og(x) .
Thus e({g} × Vx ) ⊆ Og(x) and e is continuous at (g, x). And, if g = f,
then there exists a nbd Vx of x in X such that f (Vx ) ⊆ Of (x) . Since
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fν → f continuously, there exist a nbd Wx of x and an index ν0 ∈ N
such that fν (Wx ) ⊆ Of (x) for all ν ≽ ν0 . Write Gx = Vx ∩ Wx . Then
U = {f } ∪ Tν0 ∈ U and f ∈ U. Clearly, we have e(U × G) ⊆ Of (x) .
This establishes the continuity of e at (f, x), and hence our claim. If
T is the smallest admissible topology for C(X, Y ), then T ⊆ U. Since
fν → f with respect to U, and therefore fν → f with respect to T. ♢
It follows from the preceding theorem and Proposition 11.2.3 that
if the compact-open topology for C(X, Y ) is admissible (e.g., the case
where X is locally compact Hausdorff), then a net in C(X, Y ) converges to a continuous function f : X → Y if and only if it converges
continuously to f . Another case where this holds good is described
below.
Proposition 11.2.18 Let X be a first countable space, and Y be any
space. Then a sequence of continuous functions fn : X → Y converges
to f relative to the compact-open topology on C(X, Y ) if and only if
⟨fn ⟩ converges continuously to f .
Proof. Suppose first that fn → f in the compact-open topology on
C(X, Y ). If xn → x in X, then f (xn ) → f (x), by the continuity of f . So,
for an open nbd U of f (x) in Y , there exists an integer n0 > 0 such that
f (xn ) ∈ U for every n ≥ n0 . The subset K = {x}∪{xn |n ≥ n0 } of X is
compact, and (K, U ) contains f. By our hypothesis, there is an integer
n1 such that fn ∈ (K, U ) for every n ≥ n1 . Now, for n ≥ max {n0 , n1 },
we have fn (xn ) ∈ U , which implies that fn (xn ) → f (x).
Conversely, suppose that a sequence ⟨fn ⟩ fails to converge to f
relative to the compact-open topology for C(X, Y ). Then there exist a
compact subset K ⊆ X, an open subset U ⊆ Y such that f (K) ⊆ U
and ⟨fn ⟩ is frequently in the complement of (K, U ) in C(X, Y ). So there
is a sequence of integers n1 < n2 < n3 < · · · such that fni (K) ̸⊆ U for
every i. For each integer i, put gi = fni and choose a point xi ∈ K such
that gi (xi ) ∈
/ U. Since K is compact, the sequence ⟨xi ⟩ has a cluster
point, say, x0 in K. Since X is first countable, the sequence ⟨xi ⟩ has a
subsequence ⟨xik ⟩ which converges to x0 . If ⟨fn ⟩ converges continuously
to f , then ⟨gik (xik )⟩ converges to f (x0 ). But this is clearly not true,
and the proposition follows.
♢
By the proof of the preceding theorem, we see that if both X and
C(X, Y )co satisfy the axiom of first countability, then the compact-open
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topology on C(X, Y ) is admissible, and therefore continuous convergence is equivalent to sequential convergence.
Exercises
1. Show that the compact-open topology for C (R) is different from the
the topology of uniform convergence defined by the standard bounded
metric.
2. Show that the family {(F, U )|F ⊆ X finite, and U ⊆ Y open} generates the pointwise topology on F(X, Y ). (For this reason, the pointwise
topology for F(X, Y ) is also referred to as the finite-open topology.)
3. Suppose that T is an admissible topology on a set Φ ⊆ C(X, Y ). Show
that every topology finer than T is also admissible.
4. If X is a discrete space with n elements, show that C(X, Y )co ≈ Y ×
· · · × Y, (n-factors).
5. • Prove that the pointwise topology for C (R) is different from the
compact-open topology.
6. Prove that the function ψ : Rn+1 → C (R, R)co defined by
ψ (a0 , . . . , an ) = a0 + a1 t + · · · + an tn is continuous.
(
)
7. (a) In the space C(X, Y )co , prove that (A, B) ⊆ A, B .
(b) Is C(I) closed in the space F (I, R)co ?
8. Let A be a closed, locally compact, Hausdorff subspace of a space X, and
y0 be an element of a space Y. Let Φ = {f ∈ C(X, Y )|f (X − A) = {y0 }}.
Prove that the evaluation map e : Φco × X → Y is continuous.
9. Let X be a discrete space of all positive integers, and Y be the discrete
two-point space. Show that the compact-open topology for C (X, Y )
satisfies the second axiom of countability. What is the uniform metric
topology for it? Are the two topologies identical?
10. Let X be a second countable, compact, Hausdorff space and Y be a
second countable space. Prove that C(X, Y )co is metrisable ⇔ Y is
regular.
11. For f, g ∈ C (R) and each integer n > 0, denote the supremum of
{|f (x) − g (x) | : |x| ≤ n} by cn . Show that ρ defined by ρ (f, g) =
∞
∑
cn
is a metric on C (R) and it induces the compact-open
n
2 (1 + cn )
1
topology.
12. Give an example of a regular space Y and a space X such that F(X, Y )co
is not regular.
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13. Give an example of a locally compact, second countable space X and a
second countable space Y such that C(X, Y )p is not second countable.
14. Let Z be a subspace of a space Y. Show that C(X, Z)co is homeomorphic
to a subspace of C(X, Y )co and the restriction C(Y, X)co → C(Z, X)co
is continuous for every space X.
15. Let X, Y, Z be spaces, and ϕ : Y → Z be a continuous function. Define
the function ϕ∗ : C(X, Y )co → C(X, Z)co by ϕ∗ (f ) = ϕf. Prove:
(a) ϕ∗ is continuous.
(b) If ψ : Z → T is another continuous function, then (ψϕ)∗ = ψ∗ ◦ϕ∗ .
(c) If ϕ is an embedding, then so is ϕ∗ .
16. Let X, Y, Z be spaces, and ϕ : X → Y be a continuous function. Define
the function ϕ∗ : C(Y, Z)co → C(X, Z)co by ϕ∗ (f ) = f ϕ. Prove:
(a) ϕ∗ is continuous.
(b) If ψ : T → X is another continuous function, then (ϕψ)∗ = ψ ∗ ◦ϕ∗ .
(c) If ϕ is surjective, then ϕ∗ is injective for any space Z.
(d) If X is Hausdorff, Y is locally compact Hausdorff and ϕ is a proper
surjection, then ϕ∗ is an embedding.
17. Let Y be a space, X a compact T2 space and A ⊆ X closed. If π : X →
X/A is the quotient map, show that the image of π ∗ : C (X/A, Y )co →
C(X, Y )co is the subspace consisting of continuous functions f : X → Y
such that f (A) is a singleton set.
18. Let X, Y, Z be Hausdorff spaces, and suppose that Y is locally compact
also. Show that the function θ : C(Y, Z) × C(X, Y ) → C(X, Z) defined
by θ(g, f ) = gf is continuous, when all function spaces are given the
compact-open topologies.
19.
(a) Let X1 , X2 and Y be spaces. Prove that C (X1 + X2 , Y )co ≈
C (X1 , Y )co + C (X2 , Y )co .
(b) Let Y1 , Y2 be spaces. If X is a Hausdorff space, show that
C (X, Y1 × Y2 )co ≈ C (X, Y1 )co × C (X, Y2 )co .
(c) If X1 , X2 are Hausdorff spaces, show that the mapping
C (X1 , Y1 )co × C (X2 , Y2 )co −→ C (X1 × X2 , Y1 × Y2 )co ,
(f1 , f2 ) 7→ f1 × f2 , is an embedding.
20. Let F be a closed subset of a space Y, and X be a locally compact
Hausdorff space. Prove that {(f, x)|f (x) ∈ F } is closed in C(X, Y )co ×X.
21. Let X be a compact Hausdorff space, and G ⊆ X be open. Prove:
{
}
(a) If F ⊆ Y is closed, then f |f −1 (F ) ⊆ G is open in C(X, Y )co .
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301
{
}
(b) (f, y)|f −1 (y) ⊆ G is open in C(X, Y )co × Y , if Y is Hausdorff.
22. • Prove:
(a) Let X and Y be first countable spaces, and Z be a space. If ψ :
X → C(Y, Z)co is continuous, then the associated function ψ̂ : X ×
Y → Z, ψ̂(x, y) = ψ(x)(y), is continuous. (Thus the association
map α : C(X × Y, Z) → C (X, C(Y, Z)co ) , α(f ) = fˆ, is onto.)
(b) If X and Y are also Hausdorff, then α is a homeomorphism in the
compact-open topology for both function spaces.
23. • Prove that the identity map on C(X, Y ) is the associate of the evaluation map e : C(X, Y ) × X → Y, and deduce that C(Q, I)co is not second
countable, where Q is the subspace of the rationals.
24. • Let X, Y and Z be spaces, ψ : X → C(Y, Z)co a continuous map and
ψ̂ : X × Y → Z be the associated function. Prove:
(a) ψ̂| (X × K) is continuous for every compact set K ⊆ Y.
(b) If X ×Y is a Hausdorff k-space, then the map α : C(X ×Y, Z)co →
C (X, C(Y, Z)co )co , f 7→ fˆ, is a homeomorphism.
25. Let Y and Z be spaces, and suppose that C(Y, Z) is given a topology. If
the association function C(X × Y, Z) → C (X, C(Y, Z)) is surjective for
every space X, show that the topology on C(Y, Z) is admissible.
26. Let X be a space and Y be a metric space. Suppose that ⟨fn ⟩ is a
sequence of continuous functions of X into Y. If the sequence ⟨fn ⟩ converges uniformly to f , prove that fn (xn ) → f (x) whenever xn → x in
X.
27. Let X, Y be spaces and Φ ⊆ C(X, Y ). Suppose that T is a topology on
Φ such that a net in Φ converges continuously to f ∈ Φ if it converges
to f in the topology T. Show that T is admissible.
11.3
Topology of Compact Convergence
Given a topological space X and a metric space Y , we study here
an alternative description of the compact-open topology for C(X, Y ) in
terms of a more familiar concept. We will also see a characterisation of
compact subsets of C(X, Y ) with this topology.
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Definition 11.3.1 Let X be a set, and (Y, d) be a bounded metric
space. Given a family A of subsets A ⊆ X, consider the functions
γA : F(X, Y ) → F (A, Y )u defined by γA (f ) = f |A, where F(A, Y )u
denotes the space F(A, Y ) with the uniform metric induced by the
metric d. The topology on F(X, Y ) induced by the family of functions
γA , A ∈ A, is called the topology of uniform convergence on members
of A, and will be denoted by T (d∗ |A).
It is useful to know a neighborhood basis at a point f ∈ F(X, Y )
in the topology T (d∗ |A). By the definition, it is immediate that a
neighbourhood N ∩of f relative to the topology T (d∗ |A) contains a
−1
finite intersection A γA
(B(f |A; rA )), where B(f |A; rA ) is the open
ball of radius rA centred at f |A in F(A, Y )u . Observe that the family
consisting of sets of the form
{g ∈ F(A, Y )|d (g(x), f (x)) < ϵ for all x ∈ A},
where ϵ varies over the set R+ of all positive real numbers, is a nbd
basis at f |A in F(A, Y )u . The inverse image of such a set under γA is
obviously
B(f, ϵ, A) = {g ∈ F(X, Y )|d (g(x), f (x)) < ϵ for all x ∈ A}.
It follows that N contains a finite intersection of sets B(f, ϵ, A), ϵ ∈
R+ and A ∈ A. Since γA is continuous, each set B(f, ϵ, A) is a nbd
of f relative to the topology T (d∗ |A). Thus the family of all finite
intersections of sets B(f, ϵ, A) is a nbd basis at f . Consequently, if A
is closed under formation of finite unions, then the family
{B(f, ϵ, A)|ϵ ∈ R+ and A ∈ A}
is a nbd base at f.
Definition 11.3.2 Let A ⊆ X be sets and (Y, d) be a metric space. A
net ⟨fν ⟩ in F(X, Y ) is said to converge to g : X → Y uniformly on A
if for each ϵ > 0, there is an index µ, depending on ϵ and A, such that
µ ≼ ν ⇒ d (fν (x), g(x)) < ϵ for all x ∈ A.
Let X be a set and (Y, d) be a bounded metric space. Suppose that
A is a covering of X. Let ⟨fν ⟩ be a net in F(X, Y ) such that the net
⟨fν |A⟩ converges to gA in the space F(A, Y )u for every A ∈ A. Since
A covers X, each point x ∈ X belongs to some set A ∈ A. Write
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g(x) = gA (x), if x ∈ A. Then g(x) = lim fν (x), and therefore g(x) is
well-defined. Clearly, ⟨fν ⟩ converges to g uniformly on each A ∈ A.
The topology of uniform convergence on members of a family of
subsets of a set X derives its name from the following.
Proposition 11.3.3 Let X be a set, and (Y, d) be a bounded metric
space. Suppose that A is a covering of X, and F(X, Y ) is given the
topology of uniform convergence on members of A. Then a net ⟨fν ⟩ in
the space F(X, Y ) converges if and only if the net ⟨fν |A⟩ converges in
the space F(A, Y )u for every A ∈ A.
Proof. Suppose that fν → g in the space F(X, Y ). By the definition
of the topology on F(X, Y ), the function γA : F(X, Y ) → F(A, Y )u ,
f 7→ f |A, is continuous for every A ∈ A. Therefore fν |A → g|A in
F(A, Y )u for all A ∈ A.
Conversely, suppose that fν |A → gA in the space F(A, Y )u for
every A ∈ A. Denote the metric of F(A, Y )u by d∗A . Then there exists
a function g : X → Y such that g|A = gA . We assert that fν → g
in the space F(X, Y ). To see this, consider a nbd of g of the form
B(g, ϵ, A). By our hypothesis, there exists an index µ such that µ ≼
ν ⇒ d∗A (fν |A, g|A) < ϵ ⇒ d (fν (x), g(x)) < ϵ for all x ∈ A ⇒ fν ∈
B(g, ϵ, A). Since the family of finite intersections of the sets B(g, ϵ, A)
is a nbd basis at g, we see that fν → g relative to the topology of
F(X, Y ).
♢
The following facts are easily verified.
Theorem 11.3.4 (a) The topology of uniform convergence for
F(X, Y ) is finer than the topology of uniform convergence on
members of A, and coincides with it if X ∈ A.
(b) If the family A covers X, then the topology of pointwise convergence for F(X, Y ) is coarser than the topology of uniform
convergence on members of A.
It is known from Real Analysis that a sequence of continuous functions fn : R → R, n = 1, 2, . . . , has a continuous limit function, if
it converges uniformly on every compact subset of R. This result can
also be established for nets of continuous functions of a k-space into a
metric space. With this end in view, we prove
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Elements of Topology
Proposition 11.3.5 Let X be a topological space, and (Y, d) be a
metric space. If a net ⟨fν ⟩ in C(X, Y ) converges to a function g : X → Y
uniformly on a set A ⊆ X, then g|A is continuous.
Proof. Let x0 ∈ A be arbitrary. Given ϵ > 0, there is an index µ
such that d (fµ (x), g(x)) < ϵ/3 for all x ∈ A. Since fµ is continuous,
Gx = fµ−1 (B(fµ (x); ϵ/3)) is open in X. Thus A ∩ Gx0 is a nbd of x0 in
A and, for every x ∈ A ∩ Gx0 , we have
d (g(x), g(x0 )) ≤ d (g(x), fµ (x))+d (fµ (x), fµ (x0 ))+d (fµ (x0 ), g(x0 )) < ϵ.
So g|A is continuous at x0 .
♢
From the preceding proposition, it follows immediately that the
limit of a net of continuous functions from a k-space X to a metric
space Y , which converges uniformly on every compact subset of X, is
continuous because a function g : X → Y is continuous if and only if
g|K : K → Y is continuous for every compact set K ⊆ X. A trivial,
but useful, reformulation of the foregoing discussion is given in
Corollary 11.3.6 If X is a k-space, and Y is a bounded metric space,
then C(X, Y ) is closed in the topology of uniform convergence on the
compact subsets of X for F(X, Y ).
Note the particular case that X is a locally compact Hausdorff
space or satisfies the first axiom of countability; in both cases X is a
k-space.
Definition 11.3.7 Let (Y, d) be a bounded metric space, X a space
and K be the family of all compact subsets of X. The topology of
uniform convergence on members of K for F(X, Y ) is called the topology of uniform convergence on compacta or the topology of compact
convergence.
We will denote the set F(X, Y ) endowed with the topology of
uniform convergence on compacta by F(X, Y )c , and the subspace
C(X, Y ) ⊆ F(X, Y )c by C(X, Y )c .
If Tu , Tp and Tc denote the topology of uniform convergence, pointwise convergence and compact convergence for F(X, Y ), respectively,
then Tu ⊇ Tc ⊇ Tp , by Theorem 11.3.4. Clearly, the first two are identical for compact X, and the last two are identical for discrete X. But,
in general, the three topologies are distinct from each other, as shown
by the following.
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Example 11.3.1 For each n = 1, 2, . . ., define fn : I → R by setting
fn (t) = tn (1 − tn ). The sequence ⟨fn ⟩ in C(I) clearly converges to
the constant function at 0 in the pointwise topology. But, given n,
fn (t) = 1/4 at t = 2−1/n so that this convergence is not uniform. This
shows that the topology of compact convergence on C(I) is strictly finer
than the topology of pointwise convergence.
Example 11.3.2 Let C ∗ (R) be the set of all bounded, continuous, realvalued functions. For each n = 1, 2, . . ., define a function fn : R → R
by fn (x) = |x|/n for |x| ≤ n, and 1 otherwise. (See Figure 11.3 below.)
It is clear that the sequence ⟨fn ⟩ converges to the constant function
f (x) = 0 in the pointwise topology for C ∗ (R). But it does not converge
to f in the topology of uniform convergence, since for each integer
n > 0, |fn (x) − f (x)| = 1 for x > n. If A is a covering of R consisting of
bounded sets, then it can be shown that the sequence ⟨fn ⟩ converges
to f in the topology of uniform convergence on members of A. To see
–––– ––––
f3
f2
f1
f1
f2
f3
f
FIGURE 11.3: Illustration of Example 11.3.2.
this, consider a subbasic nbd B(f, ϵ, A) of f. Since A is bounded, there
is a real number δ such that |x| ≤ δ for every x ∈ A. Now choose an
integer n0 > δ/ϵ. Then, for n > n0 , we have |fn (x) − f (x)| = |x|/n < ϵ
for all x ∈ A. This implies that fn ∈ B(f, ϵ, A) for all n > n0 , and hence
fn → f. In particular, ⟨fn ⟩ converges to f in the topology of compact
convergence. It follows that the topology of uniform convergence for
C ∗ (R) is strictly finer than the topology of compact convergence.
Observe that the topology of compact convergence on C(X, Y ) is
induced by the functions C(X, Y ) → C(K, Y )u , f 7→ f |K, where K ∈
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K, the family of all compact subsets of X (see Exercise 2.2.28). Also,
notice that each set
B(f, ϵ, K) = {g ∈ C(X, Y )|d (g(x), f (x)) < ϵ for all x ∈ K}
is open in the topology of C(X, Y )c . Since K is closed under formation
of finite unions, the sets B(f, ϵ, K) form a base for C(X, Y )c . Moreover,
the topology of C(X, Y )c is independent of the metric on Y , since the
topology of C(K, Y )u is so. In fact, we have
Theorem 11.3.8 Let X be a topological space, and Y be a metric
space. Then the topology of compact convergence and the compactopen topology for C(X, Y ) are identical.
Proof. Assume, first, that (K, U ) is a member of the subbase for
the compact-open topology for C(X, Y ) and f ∈ (K, U ). As in the
proof of Theorem 11.2.6, we obtain a real ϵ > 0 such that the
open ball B(f (x); ϵ) ⊆ U for every x ∈ K. Then B(f, ϵ, K) =
{g ∈ C(X, Y )|d (g(x), f (x)) < ϵ for all x ∈ K} is an open nbd of f in
C(X, Y )c , and K is mapped into U by all its members. Thus f ∈
B(f, ϵ, K) ⊆ (K, U ). This implies that (K, U ) is an open set in
C(X, Y )c , and hence the topology of C(X, Y )co is coarser than that
of C(X, Y )c .
To prove the converse, let f : X → Y be a continuous map. Since
the family of sets B(f, ϵ, K), where ϵ > 0 is a real and K ⊆ X is
compact, forms a nbd base at f in the topology of compact convergence,
it suffices to prove that each set B(f, ϵ, K) is a nbd of f in the compactopen topology. But this is easy, because a slight modification in the
second half of the proof of Theorem 11.2.6 gives finitely∩many closed
subsets Ki of K and open sets Ui ⊆ Y such that f ∈ i (Ki , Ui ) ⊆
B(f, ϵ, K).
♢
Corollary 11.3.9 Let X be a locally compact Hausdorff space, and
Y be a metric space. Then continuous convergence in C(X, Y ) is equivalent to uniform convergence on compacta.
This is evident from the preceding theorem and the remark following
Proposition 11.2.17.
We now investigate the conditions for compactness of function
spaces in the compact-open topology. With this end in view, we have
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Definition 11.3.10 Let X be a topological space, and Y a metric
space. A subset Φ ⊆ F(X, Y ) is equicontinuous at a point x ∈ X if for
each ϵ > 0, there is a nbd U of x such that f (U ) ⊆ B(f (x); ϵ) for every
f ∈ Φ. The family Φ is equicontinuous if it is equicontinuous at every
point of X.
Obviously, Φ is equicontinuous at x if and only if
∩{
}
f −1 (B(f (x); ϵ)) |f ∈ Φ
is a nbd of x for each ϵ > 0. Also, it is clear that the members of
an equicontinuous family Φ are continuous. Note that equicontinuity
depends on the metric used in Y.
Example 11.3.3 Any finite family Φ ⊆ C(X, Y ) is equicontinuous.
Example 11.3.4 For each real c > 0, the family
Φ = {f ∈ C(I) : |f ′ (t)| ≤ c for all t ∈ (0, 1)}
is equicontinuous on I, since |f (t) − f (t′ )| ≤ |t − t′ |c, by the mean value
theorem.
Example 11.3.5 For each n = 1, 2, . . . , define fn : I → R by
2
fn (x) = x2 /[x2 + (1 − nx) ].
The family {fn } is obviously not equicontinuous at 0 ∈ I, since
fn (1/n) = 1 for every n.
Lemma 11.3.11 If Φ is an equicontinuous family of functions of a
topological space X into a metric space (Y, d), then (a) the topology
of pointwise convergence for Φ coincides with the topology of compact
convergence, and (b) the closure of Φ in F(X, Y )p is equicontinuous.
Proof. (a): We have already seen that the topology of compact convergence is finer than the topology of pointwise convergence (see Theorem
11.3.4). To establish the reverse, we show that if a net ⟨fν ⟩ in Φ converges to f in Φp , then fν → f relative to the topology of compact
convergence. Consider a basic nbd B(f, ϵ, K) of f in Φc , where K ⊆ X
is compact, and ϵ > 0. Since Φ is equicontinuous, each point x ∈ X has
an open nbd Ux such that g(Ux ) ⊆ B (g(x); ϵ/4) for every g ∈ Φ. In particular, we have fν (Ux ) ⊆ B (fν (x); ϵ/4) for all ν. Since fν (x) → f (x)
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in Y, there exists an index µx such that fν (x) ∈ B (f (x); ϵ/4) whenever
ν ≥ µx . So
d (fν (u), f (x)) ≤ d (fν (u), fν (x)) + d (fν (x), f (x)) < ϵ/2
for all u ∈ Ux and all ν ≥ µx . This implies that d (f (u), f (x)) ≤ ϵ/2
for every u ∈ Ux , for fν (u) → f (u). By compactness of∪K, there exist
n
finitely many points x1 , . . . , xn in K such that K ⊆ 1 Uxi . Choose
now an index µ0 such that µ0 ≥ µxi for every i = 1, . . . , n. Then, for
u ∈ Uxi , and ν ≥ µ0 , we have
d (fν (u), f (u)) ≤ d (fν (u), f (xi )) + d (f (xi ), f (u)) < ϵ.
Thus fν ∈ B (f, ϵ, K) for ν ≥ µ0 , and (a) holds.
(b): Let Φ denote the closure of Φ in F(X, Y )p . Let ϵ > 0 and
x0 ∈ X be given. Since Φ is equicontinuous at x0 , there is a nbd U of
x0 such that f (U ) ⊆ B (f (x0 ); ϵ/3) for every f ∈ Φ. If g ∈ Φ, then there
is a net ⟨fν ⟩ in Φ such that fν (x) → g(x) for every x ∈ X. So there
exists an index µx0 such that d (fν (x0 ), g(x0 )) < ϵ/3 for all ν ≥ µx0 .
Consequently,
d (fν (x), g(x0 )) ≤ d (fν (x), fν (x0 )) + d (fν (x0 ), g(x0 )) < 2ϵ/3
for every ν ≥ µx0 and every x ∈ U. Since fν (x) → g(x), we have
d (g(x), g(x0 )) ≤ 2ϵ/3 < ϵ for every x ∈ U. Thus g(U ) ⊆ B (g(x0 ); ϵ) ,
and the lemma follows.
♢
Theorem 11.3.12 (Arzela–Ascoli) Let X be a locally compact
Hausdorff space and (Y, d) a metric space. If Φ is a compact subset
of C(X, Y )c , then it is a closed, equicontinuous family, and ex (Φ) is
compact for every x ∈ X. The converse holds for any topological space
X.
Proof. Suppose that Φ is a compact subset of C(X, Y )c . Since the topology of compact convergence is finer than the pointwise topology and
Y is T2 , C(X, Y )c is T2 . Hence Φ is closed in C(X, Y )c . Also, for the
same reason, each evaluation map ex : C(X, Y )c → Y , f 7→ f (x),
is continuous, and therefore ex (Φ) is compact. As Y is T2 , we have
ex (Φ) = ex (Φ). It remains to establish that Φ is equicontinuous. Let
x0 ∈ X and ϵ > 0 be arbitrary. By Theorem 11.3.8 and Proposition
11.2.4, the evaluation map e : C(X, Y )c × X → Y is continuous. So,
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309
for each f ∈ Φ, there exist open sets Of ⊆ C(X, Y )c and Uf ⊆ X
such that f ∈ Of , x0 ∈ Uf and g (Uf ) ⊆ B (f (x0 ); ϵ/2) for every
g ∈ Of . Since d (g(x), g(x0 )) ≤ d (g(x), f (x0 )) + d (f (x0 ), g(x0 )), we
have g (Uf ) ⊆ B (g(x0 ); ϵ) for every g ∈ Of . The family of open sets
Of , f ∈ Φ, covers the compact set Φ; accordingly, there are finitely
cover Φ. Let Uf1 , . . . , Ufn
many such sets Of1 , . . . , Ofn , say, which also ∩
n
be the corresponding nbds of x0 . Then U = 1 Ufi is a nbd of x0 and
f (U ) ⊆ B (f (x0 ); ϵ) for every f ∈ Φ. Thus Φ is equicontinuous at x0 ,
as desired.
Conversely, suppose that Φ is a family of functions X → Y having
the given properties. Let Φ be the closure of Φ in F(X, Y )p . By Lemma
11.3.11, Φ ⊆ C(X, Y ) and the topology of compact convergence for Φ is
identical with the∏
topology of pointwise convergence. Since∏
each ex (Φ)
is compact, so is x ex (Φ). Since F(X, Y )p is Hausdorff, x ex (Φ) is
∏
∏
closed in F(X, Y )p . Obviously, Φ ⊆ x ex (Φ) whence Φ ⊆ x ex (Φ).
Being a closed subset of a compact space, Φ is compact in the pointwise
topology, and therefore it is a compact subspace of the space C(X, Y )c .
Since Φ is closed in C(X, Y )c , it is certainly closed in the subspace
Φ ⊆ C(X, Y )c . Hence Φc is compact.
♢
In the foregoing theorem, the condition of equicontinuity on Φ can
be weakened for k-spaces. We introduce the following variation of this
notion.
Definition 11.3.13 Let X be a space and (Y, d) be a metric space.
A family Φ ⊆ C(X, Y ) is called equicontinuous on a set K ⊆ X if the
family {f |K : f ∈ Φ} is equicontinuous at each point of K.
Clearly, a family of continuous functions X → Y that is equicontinuous at every point of K ⊆ X is equicontinuous on K, but the converse
is not true, in general. With this terminology, we have
Theorem 11.3.14 (Arzela–Ascoli) Let X be a k-space, and Y be
a metric space. Then a set Φ ⊆ C(X, Y ) is compact in the topology of
compact convergence if and only if Φ is closed in C(X, Y )c , equicontinuous on each compact subset of X, and ex (Φ) is compact for every
x ∈ X.
Proof. The proof of the necessity of the conditions is entirely similar
to that of Theorem 11.3.12 because the evaluation map C(K, Y )u ×
K → Y is continuous, by Proposition 11.2.5. And, if Φ ⊆ C(X, Y )c is
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compact, then {f |K : f ∈ Φ} is a compact subset of C(K, Y )u , since
the restriction map
γK : C(X, Y )c → C(K, Y )u ,
γK (f ) = f |K, is continuous for every compact subset K ⊆ X.
To prove the sufficiency of the conditions, we first show that Φ, the
closure of Φ in F(X, Y )p , is contained in C (X, Y ). By our hypothesis,
γK (Φ) = {f |K : f ∈ Φ} is equicontinuous at each x ∈ K, and so is
γK (Φ) (the closure of γK (Φ) in F (K, Y )p ), by Lemma 11.3.11. Because the restriction map γK : F (X, Y )p → F (K, Y )p is continuous,
g|K ∈ γK (Φ) for every g ∈ Φ. Thus Φ is equicontinuous on K and, in
particular, g|K is continuous for all g ∈ Φ. It follows that each g ∈ Φ
is continuous on X, since X is a k-space. So Φ ⊆ C (X, Y ).
Next, we show that the topology of compact convergence for Φ
agrees with the pointwise topology. Recall that the sets
B (f, ϵ, K) = {g ∈ C (X, Y ) |d (g (x) , f (x)) < ϵ for all x ∈ K},
where ϵ > 0 real, f ∈ C (X, Y ), and K ⊆ X compact, form an open
base for the topology of C (X, Y )c . So the sets B (f, ϵ, K) ∩ Φ generate
the topology of compact convergence on Φ. It suffices to show that
these sets are open in the pointwise topology. Since K is compact, it
−1
follows that B (f, ϵ, K) = γK
(B (f |K; ϵ)) and B (f |K; ϵ) is open in
the topology of compact convergence
for C (K, Y ). By Lemma 11.3.11,
( )
topology for
we (see) that B (f |K; ϵ) ∩ γK Φ is open in the pointwise
( )
γK Φ . The continuity of the map γK : Φ → γK Φ in the pointwise
topology on both spaces implies that B (f, ϵ, K) ∩ Φ is open relative
to the pointwise topology. From this, we conclude that the pointwise
topology for Φ coincides with the topology of compact convergence.
Now, the proof proceeds in the same way as that of Theorem
11.3.12.
♢
Remarks 11.3.15 (a) The form of Arzela-Ascoli theorem given here
is quite general; we will develop in Exercise 14 (below) the classical
form of the theorem.
(b) The concept of uniform convergence and the results discussed
in this section can be generalised to a class of spaces which lie between
metric spaces and topological spaces. These spaces have a structure
which is a generalisation of the notion of metric, and are referred to
as the uniform spaces. A study of this notion is outside the scope of
FUNCTION SPACES
311
this book; the interested reader may refer to James [5], Kelley [6] or
Willard [16].
Exercises
1. Does the sequence in Ex. 11.3.5 converge to some function in the pointwise topology?
2.
(a) For each integer n = 1, 2 . . . , define a function fn : R → R by
fn (x) = (n + 1) x/n. Prove that the sequence ⟨fn ⟩ converges uniformly on compacta, but fails to converge uniformly.
(b) Do as in (a) with the functions fn (x) = x/n.
(c) What about the sequence ⟨fn ⟩, where fn (x) = n sin (x/n)?
3. Let X denote the subspace (−1, 1)∑
of R, and for each integer n = 1, 2 . . . ,
n
define sn : X → R by sn (x) = 1 ixi . Show that the sequence ⟨sn ⟩
converges to a continuous function uniformly on every compact subset
of X, but convergence is not uniform.
4. Let X be a space, and ⟨fν ⟩ be a monotonically decreasing (or increasing)
net of continuous functions fν : X → R. If ⟨fν ⟩ converges pointwise to a
continuous function f : X → R, show that ⟨fν ⟩ converges to f uniformly
on compacta.
5. Let X be a topological space and Y be a metric space. Without using
Theorem 11.3.8, prove that a net ⟨fν ⟩ in C (X, Y ) converges to an f ∈
C (X, Y ) relative to the compact-open topology ⇐⇒ ⟨fν ⟩ converges to
f uniformly on every compact subset of X.
6. Let X be a first countable space, Y a metric space and f : X → Y
a continuous map. Show that a sequence ⟨fν ⟩ in C (X, Y ) converges
continuously to f if and only if it converges uniformly to f on every
compact subset of X. Is this true for nets?
7. Let X and Y be topological spaces, and Φ be a family of functions from
X to Y. Let K be the family of all compact subsets of X. Prove:
(a) Any topology on Φ which makes the evaluation map eK : Φ×K →
Y , eK (f, x) = f (x), continuous for every K ∈ K, is finer than the
compact-open topology.
(b) If X is T2 or T3 , and each function in Φ is continuous on every
K ∈ K, then the evaluation map eK : Φco × K → Y is continuous
for every K ∈ K.
8. Let X be a topological space and (Y, d) be a bounded metric
∪ space. Let
A be a family of subsets of X, and suppose that X = {A|A ∈ A}.
Prove:
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(a) A net ⟨fν ⟩ in F (X, Y ) converges in the topology T (d∗ |A) ⇔ ⟨fν ⟩
converges pointwise and ⟨fν |A⟩ is a Cauchy net in F (A, Y )u for
every A ∈ A.
(b) If Φ is the family of all functions f : X → Y such that f |A : A →
Y is continuous for each A ∈ A, then Φ is closed in the space
F (X, Y ) with the topology T (d∗ |A).
(c) The evaluation map eA : Φ×A → Y is continuous for every A ∈ A,
where Φ has the relative topology induced by T (d∗ |A).
(d) If the family A consists of all compact subsets of X, then the
relative topology for Φ is finer than the compact-open topology.
9. Prove that the pointwise topology on an equicontinuous family of functions of a space X into a metric space Y is admissible.
10. Let X be a space and Y a metric space. If a sequence ⟨fn ⟩ in C (X, Y )
converges uniformly, show that the family {fn } is equicontinuous.
11. Let X be a space and Y a metric space. Suppose that an equicontinuous
sequence of functions fn ∈ C (X, Y ) converges pointwise to f : X → Y.
Prove that f is continuous and ⟨fn ⟩ converges uniformly to f on every
compact subset of X.
12. Let X be a space and (Y, d) be a compact metric space. If Φ ⊆ C (X, Y )u
is totally bounded, show that it is equicontinuous.
13. Let X be a compact space and (Y, d) a compact metric space. If a
set Φ ⊆ C (X, Y ) is equicontinuous, show that Φ is totally bounded
under d∗ . Deduce that the closure of Φ in the compact-open topology
on C (X, Y ) is compact.
14. • Let (X, d) be a compact metric space and Φ ⊆ C(X). Prove:
(a) If Φ is an equicontinuous family, then, for each real ϵ > 0, there
exists a real δ > 0 such that d(x′ , x′′ ) < δ ⇒ |f (x′ ) − f (x′′ )| < ϵ
for all f ∈ Φ.
(b) If Φ is equicontinuous and pointwise bounded (that is, {f (x)|f ∈
Φ} is bounded for every x ∈ X), then it is uniformly bounded (i.e.
there exists a real M such that |f (x)| ≤ M for all f ∈ Φ and all
x ∈ X).
(c) (Ascoli theorem) In case (b), the closure of Φ in C(X)u is compact.
Chapter 12
TOPOLOGICAL GROUPS
12.1
12.2
12.3
12.4
12.1
Examples and Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
313
324
331
341
Examples and Basic Properties
The study of the symmetries of geometric objects has interested
generations of mathematicians for hundreds of years. A symmetry of
a geometric object is merely a self-equivalence of the object, and the
groups (in the algebraic sense) are intended to analyse symmetries of
such objects. In most instances, groups arise as a family of continuous
transformations acting on a space. These groups of continuous transformations receive topological structures, in a natural way. The mathematical structures, in which the notions of group and topology are
blended, are called “Topological Groups.” Their origin can be traced
in Klein’s (1872) programme to study geometries through transformation groups associated with them, and in the work of Lie (1873)
on “continuous groups.” However, the abstract notion of a topological
group was introduced by O. Schreier (1925) and F. Leja (1927). These
partly geometric objects form a rich territory of interesting examples
in topology and geometry, due to the presence of the two basic interrelated mathematical structures in one and the same set. In this section,
we see some important examples and the basic properties of topological
groups.
Definition 12.1.1 A topological group consists of a topological space
G and a group structure on G such that the group operations (in the
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Elements of Topology
multiplicative notation)
µ : G × G → G, (x, y) 7→ xy,
and
ı : G → G, x 7→ x−1 ,
are continuous, where G × G is given the product topology.
If G is a topological group and{ A, B ⊆ G,
} then we write AB =
{ab|a ∈ A, and b ∈ B} and A−1 = a−1 |a ∈ A . Obviously, AB is the
direct image of A × B under the multiplication mapping µ, and A−1
is the image of A under the inversion mapping ı. It is easily seen that
the condition of continuity of the multiplication mapping µ and the inversion mapping ı is equivalent to the continuity of the single mapping
G × G → G, (x, y) 7→ xy −1 . Accordingly, if x and y are two elements
of G, then for every nbd W of xy −1 , there exist nbds U of x and V of
y such that U V −1 ⊆ W.
Example 12.1.1 A group with the discrete topology is a topological
group.
Example 12.1.2 The real line R with the addition of real numbers as
the group operation is a topological group. More generally, the euclidean space Rn is a topological group, the group multiplication being
the usual addition. In particular, the complex plane C is a topological
group, the group multiplication being the addition of complex numbers.
Example 12.1.3 The punctured real line R0 = R − {0} with the relative topology and the multiplication of reals as a group operation
is a topological group. As already seen (in §2.1), the multiplication
R × R → R is continuous, and so its restriction to R0 × R0 is continuous. Since the range of this function is R0 , the multiplication mapping
µ : R0 × R0 → R0 is continuous. To check the continuity of the inversion mapping ı : R0 → R0 , x 7→ x−1 , note that the intersection of the
open interval (a, b) with R0 is (a,( b) or (a,)0) ∪ (0, b). Furthermore, the
−1 −1
inverse of (a, b) ⊂ R0 under
( −1 ı)is b ( , a −1 ) when a ̸= 0 ̸= b, and those
of (a, 0) and (0, b) are a , 0 and 0, b
, respectively. Thus ı is also
continuous, and hence our assertion.
Similarly, one verifies that the punctured complex line C0 = C−{0}
forms a topological group under the multiplication of complex numbers
and the relative topology.
TOPOLOGICAL GROUPS
315
Example 12.1.4 Consider the set GL (n, R) of all invertible real n × n
matrices; it consists of all nonsingular matrices, that is,
GL (n, R) = {A ∈ M (n, R) |det(A) ̸= 0},
where det(A) denotes the determinant of A. This is obviously a group
under the matrix multiplication. We find a topology for GL (n, R)
which turns it into a topological group. To this end, consider the set
M (n, R) of all real n × n matrices. Each matrix (aij ) in M (n, R)
determines by stringing out its rows a unique ordered n2 -tuple
(a11 , . . . , a1n , a21 , . . . , ann )
of real numbers, and there is a bijective mapping of M (n, R) onto
2
Rn . So we can assign to the set M (n, R) the topology induced by
2
this mapping. With this topology, M (n, R) is homeomorphic to Rn ;
consequently, there is a continuous projection map
pij : M (n, R) → R,
(aij ) 7→ aij , for every 1 ≤ i, j ≤ n, and a mapping of a space into
M (n, R) is continuous if and only if all its compositions with the mappings pij are continuous. Now,
∑n if A = (aij ) and B = (bij ) are in
M (n, R), then pij (AB) = k=1 aik bkj . Since the multiplication and
the addition of real numbers are continuous operations, we see that the
composition of the matrix multiplication
µ : M (n, R) × M (n, R) → M (n, R)
with each pij is continuous. Therefore µ is continuous and it follows
that the group multiplication in the subspace GL (n, R) ⊂ M (n, R) is
continuous. To see the continuity of the inversion
function
A 7→ A−1
( −1
)
in this topology on GL (n, R), we note that pij A
is
(1/det(A)) × (ji)th cofactor of A.
Because det(A) is a sum of products of entries in A (with proper signs),
the determinant function det : M (n, R) → R is continuous. A similar
argument shows that the function of GL (n, R) into R, which takes
A into the (ji)th cofactor of A, is continuous.
( Since
) det(A) does not
vanish on GL (n, R), the function A 7→ pij A−1 is continuous for
every i and j. Hence the function A 7→ A−1 of GL (n, R) into itself is
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Elements of Topology
continuous, and GL (n, R) is a topological group. It is called the general
linear group over the real numbers.
Similarly, we have the general linear group GL (n, C) of invertible
complex n × n matrices, and the general linear group GL (n, H) over
the (skew) field H of quaternions, which also consists of invertible n×n
matrices. The group GL (n, C) is given the relative topology induced
2
from the complex n2 -space Cn , and the group GL (n, H) is given the
2
relative topology induced from the quaternionic n2 -space Hn .
Since GL (n, R) is the complement of the inverse image of {0} un2
der the continuous map det, it is an open subset of M (n, R) ≈ Rn .
Therefore GL (n, R) is locally compact and Hausdorff. Note that it is
not compact or connected, for R0 is its continuous image under the
determinant function and R0 is not compact or connected. It is wellknown that
GL (n, C) = {A ∈ M (n, C) |det(A) ̸= 0},
2
and therefore GL (n, C) is open in Cn . The group GL (n, H) is also
2
open in Hn , although the argument in this case is harder, since the
determinant function is not defined for quaternionic matrices.
Example 12.1.5 The group Z of integers with the cofinite topology is
not a topological group because addition is not continuous, although
it is continuous in each variable separately.
Now, we turn to the general discussion. Let G be a topological
group. For a fixed g ∈ G, the function ρg : G → G, x 7→ xg,
is continuous, for it is the composition of the continuous function
G → G×G, x 7→ (x, g) with the multiplication function µ : G×G → G.
This is referred to as the right translation by the element g. Clearly,
ρh ◦ ρg = ρgh for every g, h ∈ G, and therefore ρg is a homeomorphism
with ρg−1 as its inverse. Similarly, the left translation λg : x 7→ gx is
a homeomorphism. It is immediate that the conjugation (inner automorphism) determined by g is also a homeomorphism. Notice that if
x, y ∈ G, then the homeomorphism ρx−1 y carries x into y. So G is a
homogeneous space by which it is meant that for every pair of points
x, y ∈ G, there exists a homeomorphism h : G → G such that h (x) = y.
This fact can be used to show that a particular topological space, e.g.,
the unit interval I, cannot be made into a topological group under any
multiplication.
TOPOLOGICAL GROUPS
317
If A is closed (or open) in G, then so are A−1 , the translates Ag =
{ag|a ∈ A} and gA∪= {ga|a ∈ A}
∪ of A for every g ∈ G. For any A, B ⊆
G, we have AB = b∈B Ab = a∈A aB. So the products AB and BA
are open in G whenever A or B is open in G. On the other hand, the
product of two closed subsets, even if they are subgroups, need not be
closed. For example,
{ √ let A =} Z, the subgroup of integers in the real
line R, and B = n 2|n ∈ Z . The product AB is not a closed set (cf.
Ex. 12.3.5). However, we shall soon see that the product of a closed set
and a compact set in a topological group is necessarily closed.
If U is an open nbd of the identity element e of a topological group
G, then U x is obviously an open nbd of x ∈ G. Conversely, if O is an
open nbd of x ∈ G, then Ox−1 is an open nbd of e in G. Therefore a
subset H ⊆ G is open if and only if for each x ∈ H there is a nbd U
of e such that U x ⊆ H. In other words, H is open if and only if Hx−1
is a nbd of e for every x ∈ H. It is now evident that if Ve is a nbd
basis at the identity element e ∈ G, then {V x|V ∈ Ve } is a nbd basis
at the point x of G (so also is {xV |V ∈ Ve }). Therefore the topology
of G is completely determined by the nbd basis Ve . The family Ve has
the following properties:
Theorem 12.1.2 (a) If V ∈ Ve and x ∈ V , then there is a W ∈ Ve
such that xW ⊆ V .
(b) If V1 , V2 ∈ Ve , then there is a V3 ∈ Ve such that V3 ⊆ V1 ∩ V2 .
(c) If V ∈ Ve , then there is a W ∈ Ve such that W W −1 ⊆ V .
(d) If V ∈ Ve and x ∈ G, then there is a W ∈ Ve such that xW x−1 ⊆
V.
Proofs. (a) and (b) are trivial, and (d) is immediate from the continuity
of the function y → xyx−1 of G into itself.
To see (c), consider the function ϕ : G×G → G defined by ϕ(x, y) =
−1
xy . Since ϕ is continuous, given V ∈ Ve , there exist nbds M and N
of e such that ϕ(m, n) = mn−1 ∈ V for all m ∈ M and n ∈ N . Now,
we have a set W ∈ Ve such that W ⊆ M ∩ N . Then, for all x, y ∈ W ,
we have xy −1 ∈ V .
♢
Interestingly, we have the following.
Proposition 12.1.3 Let G be an abstract group with the identity e,
and Ve be a nonempty family of nonempty subsets of G satisfying the
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Elements of Topology
conditions (a)–(d) in 12.1.2. Then there is a unique topology on G
which turns G into a topological group with Ve a local base at e.
Proof. Let B = {xV |V ∈ Ve , x ∈ G}. We observe that B is a basis for
a topology on G. Since Ve is nonempty,
there exists a V ∈ Ve and,
∪
by (c), e ∈ V. So x ∈ xV and G = {B|B ∈ B}. Next, suppose that
Bi = xi Vi , i = 1, 2, are in B and z ∈ B1 ∩ B2 . Then there exist yi ∈ Vi ,
i = 1, 2, such that z = x1 y1 = x2 y2 . By (a), we find Wi ∈ Ve such that
x−1
i zWi ⊆ Vi ⇒ zWi ⊆ xi Vi for i = 1, 2. By (b), we have a W3 ∈ Ve
such that W3 ⊆ W1 ∩ W2 . So zW3 ⊆ B1 ∩ B2 and hence the family B
is a basis for a topology on G. Let T be the topology generated by the
base B. Then Ve is a local basis at e in the topology T.
Next, we show that (G, T) is a topological group. Consider the
mapping ϕ : G × G → G defined by ϕ(x, y) = xy −1 , and assume that
xy −1 ∈ zV . Then we have xy −1 = zv for some v ∈ V. By (a), there
exists a W ∈ Ve such that vW ⊆ V . And, by (d), there exists a U ∈ V
such that yU y −1 ⊆ W. Then we have xU y −1 ∈ zvW ⊆ zV. Now, by (c),
we find a T ∈ Ve such that T T −1 ⊆ U . So (xT )(yT )−1 = xT T −1 y −1 ⊆
zV . Thus ϕ(xT × yT ) ⊆ zV and ϕ is continuous at (x, y).
Finally, T is unique because if T ′ is a topology for G in which Ve is
a local basis at e, then B must be a base for T ′ .
♢
The following proposition will be required later.
Proposition 12.1.4 Let G be a topological group. If A ⊆ G is closed,
and B ⊆ G is compact, then AB is closed in G.
Proof. Suppose that A is closed in G, and B is a compact subset of G.
Let x ∈ G−AB be arbitrary. Then the identity element e of G does not
belong to x−1 AB. Since A is closed in G, so is x−1 Ab for every b ∈ B.
By Proposition 12.1.2, there exists a nbd Wb of e such that Wb Wb−1 ⊆
G − x−1 Ab. Since B is compact, the open covering {bWb |b ∈ B} has a
finite subfamily covering B. Accordingly,
we find ∩
finitely many points
∪n
n
b1 , . . . , bn in B such that B ⊆ 1 bi Wbi . Set W = 1 Wbi . Then xW is
obviously a nbd of x. We assert that it is contained in G − AB. Assume
otherwise. Then we∪have elements w ∈ W , a ∈ A, and b ∈ B satisfying
n
xw = ab. As B ⊆ 1 bi Wbi , there exists wi ∈ Wbi such that b = bi wi .
So xw = abi wi , which implies that wwi−1 = x−1 ab. This contradicts
the definition of Wbi , and hence our assertion. Thus G − AB is a nbd
of each of its points, and therefore open. This completes the proof. ♢
TOPOLOGICAL GROUPS
319
SEPARATION PROPERTIES
It is generally required that the identity element e of a topological
group G be closed. From the homogeneity of G, it follows that this is
equivalent to the requirement that each one-point set {x}, x ∈ G, be
closed. Thus a topological group with this property is a T1 -space. We
will soon see that a topological group having the weakest separation
property T0 actually satisfies the stronger separation condition T3 21 .
We call a subset A ⊆ G symmetric if A = A−1 . It is clear that A ∩
−1
A is symmetric, and so is AA−1 . Also, the intersection of symmetric
sets is symmetric. If U is a nbd of the identity e of G, then so is U −1 .
The nbd V = U ∩ U −1 of e is symmetric, and U contains V , obviously.
It follows that the symmetric nbds of e form a local base at e, and this
basis completely describes the topology of G.
Lemma 12.1.5 Let G be a topological group with the identity e. If
U is a nbd of e in G, then there is a symmetric nbd V of e such that
V V ⊂ U.
Proof. By the continuity of multiplication µ : G × G → G, there exist
nbds W1 and W2 of e such that W1 × W2 ⊂ µ−1 (U ). So U contains
the nbd W = W1 ∩ W2 of e, and also the product W W . It is now clear
that the set V = W ∩ W −1 satisfies the requirements of the lemma. ♢
Proposition 12.1.6 Let G be a topological group.
(a) If G is T0 , then it is T2 .
(b) G is T3 .
Proof. (a): Let e denote the identity element of G. We first prove that
{e} is closed in G (this implies that G is T1 ). Let x ̸= e be any point
of G. Since G satisfies the T0 -axiom, G − {e} is a nbd of x or G − {x}
is a nbd of e. In the latter case, G − {e} is a nbd of x−1 , since the
left translation by x−1 maps G − {x} onto G − {e} . So the image
of G − {e} under the inversion homeomorphism ı is a nbd of x. As
ı (G − {e}) = G − {e} , it follows that G − {e} is a nbd of x. Thus
G − {e} is open, and therefore {e} is closed in G. Now, if x, y ∈ G, and
x ̸= y, then xy −1 ̸= e. By Lemma
there exists a symmetric nbd
{ −112.1.5,
}
V of e such that V V ⊆ G − xy
. Obviously, V x and V y are nbds
of x and y, respectively. If vx = v ′ y, then xy −1 = v −1 v ′ ∈ V −1 V , a
contradiction. So V x ∩ V y = ∅, and G is Hausdorff.
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Elements of Topology
(b): Let G be a topological group with the identity element e. By
the homogeneity of G, it suffices to consider the nbds of e. Let U be a
nbd of e. Since ee−1 = e, there exits a nbd V of e such that V V −1 ⊆ U .
We assert that V ⊆ U. Let x ∈ V . Then every nbd of x intersects V.
So xV ∩ V ̸= ∅, and we can find points v, v ′ ∈ V such that xv = v ′ .
This implies that x = v ′ v −1 ∈ V V −1 ⊆ U , and hence our assertion. ♢
It follows that a topological group satisfying the T0 -axiom is regular.
In fact, we can establish the following.
Theorem 12.1.7 A topological group satisfying the T0 -axiom is completely regular.
Proof. Let G be a topological group with the identity element e. Since
the left translation λg : x 7→ gx is a homeomorphism of G onto itself for
every g ∈ G, it suffices to prove the existence of a continuous function
f : G → I such that f (e) = 0 and f (F ) = 1 for each closed set F ⊂ G
not containing e. We first show that, for each dyadic rational r = k/2n ,
where k = 1, . . . , 2n and n = 0, 1, 2, . . . , there is an open nbd Ur of e
such that
Uk/2n U1/2n ⊂ U(k+1)/2n for all k < 2n .
(∗)
We set V0 = G − F, and find a symmetric open nbd V1 of e such
that V12 ⊆ V0 . Then find a symmetric open nbd V2 of e such that
V22 ⊆ V1 , and so on. Thus we obtain a sequence of symmetric open
2
⊆ Vn . For each positive integer k, there
nbds Vn of e such that Vn+1
exists a unique integer m ≥ 0 such that 2m ≤ k < 2m+1 . So we can
write k = 2m a1 + 2m−1 a2 + · · · + am+1 , where the ai are uniquely
determined integers 0 or 1. For each dyadic rational k/2n between 0
a
a1
and 1, we define U1 = V0 and Uk/2n = Vn−m
· · · Vn m+1 , where Vi1 = Vi
0
and Vi = {e}. Observe that U1/2n = Vn , and U2k/2n = Uk/2n−1 so that
each Ur depends on the dyadic rational r, and not on the particular
representation by k/2n . To see that the sets Ur satisfy the inclusions
(∗), we apply induction on n. If n = 1, then k = 1 and we have
U1/2 U1/2 = V1 V1 ⊂ V0 = U1 . Thus (∗) holds for n = 1. Now, assume
that n > 1 and (∗) is true for n − 1. Then we have
U2k/2n U1/2n = Uk/2n−1 U1/2n
a
a1
m+1
= Vn−m−1
· · · Vn−1
Vn1
= U(2k+1)/2n
and
TOPOLOGICAL GROUPS
321
U(2k+1)/2n U1/2n = Uk/2n−1 U1/2n U1/2n = Uk/2n−1 Vn2
⊂ Uk/2n−1 U1/2n−1 ⊂ U(k+1)/2n−1
= U(2k+2)/2n ,
by our induction assumption. This completes the inductive argument,
and the family {Ur } has the property (∗).
We next observe that Ur ⊆ Ur′ for r < r′ ≤ 1. Suppose that
′
r = k/2n and r′ = k ′ /2n , where k = 2m a1 + 2m−1 a2 + · · · + am+1 and
′
′
k ′ = 2m a′1 + 2m −1 a′2 + · · · + a′m′ +1 . If n − m > n′ − m′ , then we have
a
a1
Ur = Vn−m
· · · Vn m+1 ⊆ Vn−m · · · Vn
⊆ Vn−m · · · Vn Vn ⊆ Vn−m · · · Vn−1 Vn−1
⊆ · · · ⊆ Vn−m−1 ⊆ Vn′ −m′
a′
a′
′
⊆ Vn′1−m′ · · · Vn′m +1 = Ur′ .
If n − m = n′ − m′ , then we find the least integer l such that al = 0 and
al
= {e}, we have
a′l = 1. Put W = Vn−m · · · Vn−m+l−2 . As Vn−m+l−1
a
a
l+1
Ur = W Vn−m+l
· · · Vn m+1 ⊆ W Vn−m+l · · · Vn
⊆ W Vn−m · · · Vn Vn ⊆ W Vn−m+l · · · Vn−1 Vn−1
a′
a′
⊆ · · · ⊆ W Vn−m+l−1 = Vn′1−m′ · · · Vn′l−m′ +l−1
a′
a′
′
⊆ Vn′1−m′ · · · Vn′m +1 = Ur′ .
Now, we define f : G → I by putting f (x) = 0 if x ∈ Ur for every
r, f (x) = 1 if x ∈
/ U1 and f (x) = sup{r|x ∈
/ Ur } otherwise. It is obvious
that f (F ) = 1, and f (e) = 0, since e ∈ Ur for all r. To see the continuity
of f , let x ∈ G and ϵ > 0 be given. Find a positive integer n such that
2−n < ϵ. If f (x) = 0, then each U1/2m is a nbd of x, and |f (x)−f (y)| ≤
1/2m for every y ∈ U1/2m . Consequently, we have |f (x) − f (y)| < ϵ for
all y ∈ U1/2m and m > n. If 0 < f (x) < 1, then there exists a positive
integer m such that m > n and 2−m < min{f (x), 1 − f (x)}. Also,
there is an integer k such that 1 < k < 2m and x ∈ Uk/2m − U(k−1)/2m ,
for x ∈ U(2m −1)/2m − U1/2m . Then (k − 1)/2m ≤ f (x) ≤ k/2m . Note
that xU1/2m is an open nbd of x. We verify that |f (x) − f (y)| < ϵ for
all y ∈ xU1/2m . If y ∈ xU1/2m , then y ∈ Uk/2m U1/2m ⊆ U(k+1)/2m .
Since x−1 y ∈ U1/2m which is symmetric, we have y −1 x ∈ U1/2m , and
so x ∈ yU1/2m . If y ∈ U(k−2)/2m , then we would have x in U(k−1)/2m ,
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Elements of Topology
a contradiction. Accordingly, y ∈
/ U(k−2)/2m and (k − 2)/2m ≤ f (y) ≤
m
(k + 1)/2 . It follows that |f (x) − f (y)| ≤ 2/2m ≤ 1/2n < ϵ. Finally,
suppose that f (x) = 1. Again, we choose m as above, and consider
the nbd xU1/2m of x. Let y ∈ xU1/2m be arbitrary. Then x ∈ yU1/2m ,
since U1/2m is symmetric. If y ∈ Uk/2m and k/2m ≤ 1 − 2/2m , then
we have x ∈ U(k+1)/2m which implies that f (x) ≤ (k + 1)/2m < 1, a
contradiction. Therefore y ∈
/ Uk/2m for k/2m ≤ 1 − 2/2m , and we have
1 ≥ f (y) ≥ 1 − 2/2m . It follows that |f (x) − f (y)| ≤ 2/2m ≤ 1/2n < ϵ,
and this completes the proof.
♢
Exercises
1. Verify that the set M (n, R) (resp. M (n, C)) of all n×n matrices over R
(resp. C) under addition with the usual topology is a topological group.
2. • Prove that the topology of uniform convergence defined by a bounded
metric in R for the group F (R, R) with pointwise addition and negation
(that is, (f + g)(x) = f (x) + g(x) and (−f )(x) = −f (x)) makes it a
topological group.
3. • Prove that the set C (I) of all real-valued continuous functions on I
with the topology induced by the supremum metric
ρ (f, g) = sup {|f (t) − g(t)| : 0 ≤ t ≤ 1}
is a topological group under the pointwise addition.
4. Let X be a compact Hausdorff space. Show that C (X)co is a topological
group under addition.
5. Let (X, d) be a compact metric space, and topologise the group
Homeo(X) of all homeomorphisms of X onto itself by the metric
ρ (f, g) = supx∈X d (f (x) , g (x)). Show that Homeo(X) is a topological
group.
6.
(a) Can the open interval (−1, 1) be given a topological group structure?
(b) Suppose that G is a topological group and H is a homeomorphic
copy of G. Show that H can be given a topological group structure.
7. • Let F = R, C or H. A map f : F n → F n is called an affine map if
there exists a linear map λ : F n → F n and an element v ∈ F n such
that f (x) = λ (x) + v for all x ∈ F n . Prove:
(a) The set of all invertible affine maps of F n form a group, denoted
by Af fn (F ), under the composition of mappings.
TOPOLOGICAL GROUPS
323
(b) There is a canonical bijection between Af fn (F ) and GL (n, F ) ×
F n.
(c) The group Af fn (F ) with the topology of the product space
GL (n, F ) × F n is a topological group.
(d) Is Af f1 (R) abelian?
8. Let G be a locally compact Hausdorff space, and suppose that it has a
group structure. If the group multiplication µ : G×G → G is continuous,
prove:
(a) For each compact K ⊆ G and open U ⊆ G, {g|gK ⊆ U } is open
in G.
(b) If G is also locally connected, then it is a topological group.
9. Prove that the subspace R × {0} ∪ {0} × R of R2 is not homogeneous.
10. Let p be a prime and V0 be a family of all subsets V ⊆ Z such that for
some integer n > 0, kpn ∈ V for all k ∈ Z. Show that V0 is a nbd basis
at 0 relative to a topology which makes the group (Z, +) a topological
group. (This is referred to as the p-adic topology for Z.) Prove that it
is totally disconnected.
11. Let p be a fixed prime. For each k ∈ Z, define sets Uk = {mpk /n :
m, n ∈ Z and (p, n) = 1}. Show that there is a topology on Q, the
additive group of rationals, which makes Q a topological group with
the family {Uk } a local basis at 0. (This is referred to as the p-adic
topology for Q.) Is it totally disconnected?
12. Let G be a topological group with the identity e and U be a nbd of e.
Show that for each integer n > 0, there is a symmetric nbd V of e such
that V n ⊆ U , where V n = V · · · V (n factors).
13. Let x1 , . . . , xn be n elements of a topological group G, and let y =
xk11 · · · xknn , where k1 , . . . , kn are integers. If W is a nbd of y, show that
there exist nbds Ui of xi , 1 ≤ i ≤ n, such that U1k1 · · · Unkn ⊆ W .
14. Let G be a topological group and U be an nbd of g in G. Show that
there exists a symmetric nbd V of the identity element e of G such that
V gV −1 ⊂ U .
15. In the topological group R2 , what is the product AB, where A =
{(0, y) |y ∈ R} and B = {(x, 1/x) |x > 0 real}? Is AB closed in R2 ?
16. Let G be a topological group, and A, B be compact subsets of G. Show
that AB is compact.
17. Prove that a topological group G is Hausdorff ⇔ the intersection of all
nbds of e is {e}.
324
12.2
Elements of Topology
Subgroups
Let G be a topological group and H ⊂ G be a subgroup. If H is
given the relative topology, then it becomes a topological group. This
follows from the fact that if f : X → Y is continuous, and A ⊂ X,
then the map g : A → f (A) defined by f is also continuous when
A and f (A), both, are given the relative topologies. A subgroup of a
topological group is always assumed to be given the relative topology
when it itself is being considered as a topological group.
Example 12.2.1 Let F = R (the field of real numbers), C (the field of
complex numbers) or H (the skew field of quaternions). Then the unit
sphere
S (F ) = {x ∈ F : |x| = 1}
in F is clearly a subgroup of the topological group F0 = F − {0} under
the multiplication in F . Note that S (F ) ≈ S0 , S1 or S3 , according as
F = R, C or H. It is a fact that these are the only spheres which are
also groups.
Example 12.2.2 Recall that a real matrix A is orthogonal if At A = I,
the identity matrix. The set O (n) of all orthogonal matrices forms a
subgroup of GL (n, R), and is called the orthogonal group. We observe
n2
that it is a closed and bounded subset
∑ of M (n, R) ≈ R . If A ∈
O (n) and A = (aij ), then we have k aik ajk = δij for 1 ≤ i, j ≤ n.
For every
∑ pair of indices i and j, the functions ϕij : M (n, R) → R,
A 7→ k aik ajk , are clearly continuous, and O (n) is the intersection
−1
of the sets ϕ−1
ii (1), and ϕij (0), i ̸= j. Therefore O (n), being a finite
intersection of closed subsets of M (n, R), is closed. The boundedness
of O (n) follows from the fact that each row of A ∈ O (n) has unit
length. Hence O (n) is a compact subgroup of GL (n, R).
t
A complex matrix A is unitary if A A = I, the identity matrix.
The set U (n) of all unitary matrices forms a subgroup of GL (n, C),
and is called the unitary group. Similarly, there is a subgroup Sp (n) ⊂
t
GL (n, H) consisting of quaternionic matrices A such that A A = I,
t
where A is the quaternionic conjugate transpose of A. It is called the
symplectic group. As above, we see that both topological groups U (n)
and Sp (n) are compact.
TOPOLOGICAL GROUPS
325
To see the geometric interpretations of the topological groups in
the preceding example, suppose that F is one of the three fields R,
C or H. Recall that a subset S ⊆ F n is orthonormal if ⟨x, y⟩ = 0
for every pair of distinct points x and y in S, and ∥x∥ = 1 for all
x ∈ S. The standard basis of F n , for example, is an orthonormal set.
Let f : F n → F n be a linear mapping, and let {vi } be an orthonormal
basis of F n . If y ∈ F n , then gy∑
: F n → F , x 7→ ⟨f (x) , y⟩, is a linear
n
map, and the element f ∗ (y) = 1 vi gy (vi ) is uniquely determined by
the condition gy (x) = ⟨x, f ∗ (y)⟩ for each x ∈ F n . So there is a mapping
f ∗ : F n → F n given by ⟨f (x) , y⟩ = ⟨x, f ∗ (y)⟩ for all x, y ∈ F n . It is
easily checked that f ∗ : F n → F n is also linear, called the adjoint of
f . For each j ∑
= 1, . . . , n, there exist scalars aij ∈ F, i = 1, . . . , n, such
n
that f (vj ) = 1 vi aij . Thus we obtain the matrix A = (aij ) (over F ),
called the matrix of f relative to the ordered basis {vi }. One readily
t
verifies that A is the matrix of f ∗ relative to the basis {vi }.
If f is self-adjoint (that is, f ∗ = f ), and ⟨f (x) , x⟩ = 0 for all
x ∈ F n , then
0 = ⟨f (x + ya) , x + ya⟩
= ⟨f (x) , ya⟩ + ⟨f (ya) , x⟩
= ⟨f (x) , y⟩ a + ⟨f (x) , y⟩ a
for all x, y ∈ F n and a ∈ F . By taking suitable values of a, we find
that ⟨f (x) , y⟩ = 0 for all x, y ∈ F n , and therefore f = 0.
Now, we see that a linear map f : F n → F n is an isometry if and
only if
t
∥f (x)∥ = ∥x∥ ⇔ ⟨f (x) , f (y)⟩ = ⟨x, y⟩ ⇔ f ∗ ◦ f = 1 ⇔ A A = I.
It follows that the elements of O (n), U (n) and Sp (n) correspond precisely to the linear isometries of Rn , Cn and Hn , respectively. Thus
O (n) may be considered as the group of the linear isometries of Rn ,
U (n) as the group of the linear isometries of Cn and Sp (n) as the group
of the linear isometries of Hn . Accordingly, the linear isometries of F n ,
F = R, C or H, are referred to as the orthogonal, unitary or symplectic
transformations, respectively. We will see later that these three families of topological groups O (n), U (n) and Sp (n) are, in fact, related.
Observe that O (1) ≈ S0 , U (1) ≈ S1 and Sp (1) ≈ S3 as topological
spaces.
Turning to the general discussion, we prove
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Elements of Topology
Proposition 12.2.1 Let H be a subgroup of the topological group G.
Then
(a) the closure H of H in G is a subgroup of G, and
(b) if H is normal in G, then so is H.
Proof. (a): Let x, y ∈ H and let U be a nbd of xy. By the continuity
of the multiplication function µ : G × G → G, there exist nbds V of
x and W of y such that V W ⊂ U . Since V ∩ H ̸= ∅ ̸= W ∩ H, U
intersects H. We see from this that xy ∈ H. Similarly, x−1 ∈ H, and
H is a subgroup of G.
(b): Suppose that H is normal in G. Given an element g ∈ G, let
γ : G → G be the conjugation map determined by (g, )that is, γ (x) =
gxg −1 . By the continuity of γ, we have gHg −1 = γ H ⊆ γ (H) ⊆ H.
Therefore H is normal in G.
♢
Proposition 12.2.2 If H is a commutative subgroup of a Hausdorff
topological group G, then H is also commutative.
Proof.
Let) µ denote the multiplication map in G, and let µH =
(
µ| H × H . Also, define µ′ : H × H → G by µ′ (x, y) = yx = µ (y, x).
Then µH and µ′ are continuous, and agree on H ×H, by our hypothesis.
Since H × H is dense in H × H, the conclusion follows from Corollary
4.4.3.
♢
If H is an open subgroup of a topological group G, then each coset
of H in G is open, and the complement G − H is the union of cosets
of H. So H is also closed in G. Conversely, if H is a closed subgroup
of G having a finite index, then it is also open.
The following fact for compact groups will be required later.
Proposition 12.2.3 Let G be a compact Hausdorff group and H ⊂ G
a closed subgroup. Then, for any g ∈ G, gHg −1 = H if and only if
gHg −1 ⊆ H.
Proof. Suppose that gHg −1 ⊆ H. Consider the mapping ϕ : G × G →
G defined by ϕ (x, y) = xyx−1 . If A = {g n |n =
( 0, 1, .). .}, then
ϕ (A × H) ⊆ H and continuity of ϕ implies that ϕ A × H ⊆ H. If
we can show that g −1 ∈ A, then we have g −1 Hg ⊆ H, which implies
that H ⊆ gHg −1 . Thus it suffices to show that A is a subgroup of G.
Put B = {g n |n ∈ Z}. This is obviously a subgroup of G, and hence
TOPOLOGICAL GROUPS
327
B is a subgroup of G. If the identity element e is an isolated point of
B, then B is discrete. Being compact and discrete, B is finite; consequently, g n = e for some integer n > 0. If e is not an isolated point of
B, then for any symmetric nbd V (of e in G, )there is an integer n > 0
such that g n ∈ V. Then g n−1 ∈ g −1 V ∩ A . Since the sets g −1 (V )
form a nbd basis at g −1 , it follows that g −1 ∈ A. This implies that
A = B, and the proposition follows.
♢
From the proof of the preceding proposition, it follows that if g
is an element of a compact Hausdorff group G, then the closure of
the set {g n |n = 0, 1, . . .} is a subgroup of G. It should be noted that
this statement and the preceding proposition are not true without the
condition of compactness on G.
In the next two examples, we discuss the closed subgroups of R and
S1 .
Example 12.2.3 A proper nontrivial closed subgroup of R is infinite
cyclic. Let H ̸= {0} be a proper closed subgroup of R. Put k =
inf {h ∈ H|h > 0}. If k = 0, then we assert that every real number
is a limit point of H. It is obvious that 0 is a limit point of H, and
so is every element of H, since a translation by an element of H is a
homeomorphism of R onto itself and H is invariant under such maps.
Next, suppose that x > 0 and x ̸∈ H. Given a real number r > 0, we
choose an h ∈ H such that 0 < h < r. Let n > 0 be the least integer
such that x < nh. Then x − r < nh < x + r, and therefore x is a limit
point of H. Since H is invariant under the homeomorphism x 7→ −x of
R onto itself, it follows that every real number x < 0 is a limit point
of H. Hence our assertion. Since H is closed, we have H = R, contrary
to our hypothesis. Therefore k ̸= 0. Obviously, k ∈ H = H whence
nk ∈ H for all n ∈ Z. If there is an h ∈ H that is not a multiple of k,
then we find an integer n such that nk < h < (n + 1) k. Accordingly,
we have 0 < h − nk < k. Since H is a subgroup, h − nk ∈ H, which
contradicts the definition of k. So H = {nk|n ∈ Z}.
Example 12.2.4 A closed proper subgroup of S1 is finite and cyclic.
Let H be a closed proper subgroup of S1 . We first observe that H is
a discrete subset of S1 . If H is not discrete, then 1 ∈ S1 is a limit
point
for each integer n > 0, the open set Un =
{ ıθ of H. Consequently,
}
e |0 < θ < 1/n contains a point z ∈ H. Given x ∈ S1 , we find
a largest integer m such that arg(z m ) ≤ arg(x). Then arg(z m+1 ) >
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Elements of Topology
arg(x). If both z m and z m+1 do not belong to the open ball B(x; 1/n)
of S1 , then arg(z m+1 ) − arg(z m ) ≥ 2/n, a contradiction. So each open
ball in S1 with radius 1/n contains a power of z, and H is dense in S1 .
Since H is closed, we obtain H = S1 , contrary to our hypothesis. Being
a closed discrete subset of a compact space, H is finite. To see that H
is cyclic, suppose that H ̸= {1}. Then there is an element z = eıθ ∈ H
such that 0 < θ ≤ π and |z − 1| is minimum. We assert that H = ⟨z⟩. If
H had an element x that was not a power of z, then there would exist
an integer n > 0 such that |z −n x − 1| < |z − 1| and 0 < arg(z −n x) ≤ π.
This contradicts the definition of z, and hence our assertion.
To see the connectedness property of a topological group, we notice
that the components of a topological group G are the left (or right)
cosets of the component G0 of the identity element, since the left translations λg : x 7→ gx are homeomorphisms. A similar result holds for
path components of G.
Proposition 12.2.4 The component of the identity element of a topological group is a closed normal subgroup.
Proof. Let G be a topological group with the identity element e and
G0 be the component of e. Since components are closed, G0 is closed
in G. By the continuity of the inversion function x 7→ x−1 , G−1
is
0
−1
connected and contains e. Therefore G0 ⊆ G0 and this implies that
(
)−1
−1
G0 = G−1
⊆ G−1
0
0 . Thus we have G0 = G0 . For x ∈ G, xG0 is the
continuous image of G0 under the translation map λx , and therefore
connected. Now, if x, y ∈ G0 , then xG0 contains both x and xy, so
xy ∈ G0 , and hence G0 is a subgroup of G. To show that it is normal, let
x ∈ G be arbitrary. Then x−1 G0 x is the image of G0 under conjugation
by x, and therefore connected. As e ∈ x−1 G0 x, we have x−1 G0 x ⊆ G0 .
This implies that G0 is normal in G.
♢
Example 12.2.5 The subgroup of GL (n, R) consisting of all matrices
with determinant 1 is path-connected. It is called the special linear
group, and is denoted by SL (n, R). Let Ei (λ) denote the elementary
matrix obtained by multiplying the ith row of the identity matrix I
by λ (̸= 0), and let Eij (λ) denote the elementary matrix obtained by
adding λ times the jth row of I to the ith row. Clearly, every matrix
A ∈ SL (n, R) can be written as a product of elementary matrices
Eij (λ) or Ei (λ), and in such
(
)a factorisation of A if Ei (λ) is one of
the factors, then so is Ej λ−1 for some j. For each 0 ≤ t ≤ 1, put
TOPOLOGICAL GROUPS
329
ϕ (t) = (1 − t) λ + t if λ > 0, and ϕ (t) = (1 − t) λ − t if λ < 0.
Then t 7→ Ei (ϕ (t)) is a path joining Ei (λ) to Ei (±1), according as
λ > 0 or λ < 0. It follows that
)t 7→ Ei (ϕ (t)) Ej (1/ϕ (t)) is a path in
(
SL (n, R) joining Ei (λ) Ej λ−1 to I, since the matrix multiplication
is continuous. Also, it is obvious that t 7→ Eij ((1 − t) λ), 0 ≤ t ≤ 1,
is a path in SL (n, R) joining Eij (λ) to I. Therefore every matrix
A ∈ SL (n, R) can be joined to I by a path in SL (n, R) , and SL (n, R)
is path-connected.
The topological group GL (n, R) is disconnected, since R − {0} is
disconnected and the function det : GL (n, R) → R − {0} is a continuous surjection. We observe that the open subgroup GL+ (n, R) ⊂
GL (n, R) consisting of all matrices with positive determinants and
the set GL− (n, R) of all matrices in GL (n, R) having negative determinants are path components of GL (n, R). If A ∈ GL+ (n, R), then
δ −1 A ∈ SL (n, R), where δ = det(A). As seen above, there is a path f
in SL (n, R) joining δ −1 A and I. So t 7→ δf (t) is a path in GL+ (n, R)
joining A and δI. Clearly, the path t 7→ ((1 − t) δ + t) I joins δI and
I. Hence GL+ (n, R) is path-connected. The path-connectedness of
GL− (n, R) follows from the fact that it is the image of GL+ (n, R) under the left translation λE , where E is the elementary matrix E1 (−1) .
The subgroup of GL (n, C) consisting of all matrices with determinant 1 is denoted by SL (n, C). As above, one can prove that the topological groups GL (n, C) and SL (n, C) are also path-connected. The
connectedness property of O (n), U (n) and Sp (n) will be discussed in
Chapter 13.
Exercises
1. Prove:
(a) R has no proper open subgroup.
(b) A connected subgroup of R1 is either {0} or R.
(c) Every nontrivial discrete subgroup of R is infinite cyclic.
2. Prove that a subgroup of a topological group G is discrete ⇔ it has an
isolated point.
3. Let G be a topological group with the identity element e. Prove the
closure of {e} in G has the trivial topology.
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Elements of Topology
4. Let G be a finite connected topological group. Show that G has the
trivial topology.
5. Let G be a topological group. Prove that a subgroup of G with nonempty
interior is open.
6. Prove that a subgroup H of a topological group G is either closed or
H − H is dense in H.
7. Let G be a connected topological group with the identity element e. If
a subgroup H ⊆ G contains an open nbd of e, then G = H.
8. Let G be a connected topological group∪with the identity element e and
U be an open nbd of e. Show that G = n≥1 U n (that is, every element
of G is a finite product of elements of U ; thus G is generated by U ).
9. • Let H be a discrete subgroup of a Hausdorff group G. Show that H
is closed in G. Deduce that H ∩ K is finite for every compact subset
K ⊆ G.
10. Let G be a topological group and H a subgroup of G. If U is a nbd of
e in G such that H ∩ U is closed in G, show that H is closed in G.
11. Prove that every locally compact subgroup of a Hausdorff topological
group is closed.
12. Let G be a topological group, K a compact subset and U an open subset
of G such that K ⊆ U . Prove that there exists an open nbd W of e such
that KW ⊆ U .
13. Let G be a Hausdorff topological group and V be a compact and open
nbd of e. Prove that V contains a compact open subgroup H.
14. Prove that the subgroup of the topological group in Exercise 12.1.5,
which consists of all isometries of X onto itself, is compact.
15. Let G be a topological group and H ⊂ G a dense subgroup. If N is a
normal subgroup of H, show that N is a normal subgroup of G.
16. Prove that the intersection of all nontrivial closed (or open) subgroups
of a topological group G is a normal subgroup of G.
17. If G is a locally compact topological group, show that the identity component of G is the intersection of all open subgroups.
18. Prove that a locally compact group G is connected ⇔ G has no proper
open subgroup.
19. Let G be a locally compact Hausdorff and totally disconnected group.
Prove that the open subgroups of G form a local base at e.
20. If H is closed
in the topological
group G, prove that its normaliser
{
}
N (H) = g ∈ G|gHg −1 = H is closed in G.
TOPOLOGICAL GROUPS
331
21. Let G be a topological group with the identity element e. Show that
the path component of e is a normal subgroup of G, and the cosets of
this subgroup are precisely the path components of G.
22. Let G be a Hausdorff topological group with the identity e. If every
nbd U of e contains an open subgroup of G, show that G is totally
disconnected.
23. Prove that the group Z (resp. Q) with the p-adic topology is totally
disconnected.
24. Let G be a Hausdorff topological group with the identity element e.
Prove that Z (G) = {z ∈ G|zx = xz for all x ∈ G} is a closed normal
subgroup of G. (This is called the centre of G.)
25. Let G be a connected topological group and H be a totally disconnected
normal subgroup of G. Show that H is contained in the centre of G.
(Such a subgroup of G is called central.)
26. Prove:
(a) Z (U (n)) ∼
= S1 ,
(b) Z (SU (n)) is a cyclic group of order n, and
(c) Z (Sp (n)) = {1, −1}.
27. Prove:
(a) If a matrix M in O(2) commutes with every matrix in SO(2), then
M ∈ SO(2);
(b) Z (SO (2k)) = {1, −1}, k > 1;
(c) Z (SO (2k + 1)) = {1}, k ≥ 1.
12.3
Isomorphisms
QUOTIENT GROUPS
Let G be a topological group and H ⊆ G a subgroup. Let G/H be
the set of left cosets xH, x ∈ G, and π : G → G/H be the natural
map x 7→ xH. The set G/H with the quotient topology determined
by π is called the left coset (or factor) space of G by H. Similarly,
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Elements of Topology
the space of right cosets Hx is defined. In general, the space of left
cosets of H in G is different from the space of right cosets of H in
G, but the two spaces are homeomorphic (the mapping xH 7→ Hx−1
gives a homeomorphism). For x ∈ G, the translation map λx : G → G
induces a homeomorphism of G/H onto itself; consequently, G/H is a
homogeneous space.
Proposition 12.3.1 Let G be a topological group and H ⊂ G a subgroup. Then the natural projection π : G → G/H is continuous and
open in the quotient topology for G/H. If H is compact, then π is
closed.
Proof. By the definition of topology on G/H, π is continuous. If U
is any open subset of G, then π −1 (π (U )) = U H is open in G, and
therefore π (U ) is open.
If H is compact and F ⊆ G is closed, then π −1 (π (F )) = F H is
closed, by Proposition 12.1.4. So π (F ) is closed for every closed set
F ⊆ G, and the proof is complete.
♢
It is clear that the factor space G/H is discrete if and only if H is
open in G. By Proposition 12.1.6, the coset-space G/H is Hausdorff if
and only if H is closed in G.
If N is a normal subgroup of the topological group G, then we
know that the set G/N of left (or right) cosets of N in G is endowed
with a group structure. The group G/N together with the quotient
topology determined by the canonical projection π : G → G/N is a
topological group. By the definition of the group operations in G/N ,
we have commutative diagrams
G×G
µ
- G
π×π
G
π
?
µ′
G/N × G/N
(a)
?
- G/N
ι
- G
π
π
?
G/N
?
ι′ G/N
(b)
FIGURE 12.1: Continuity of the multiplication and the inversion functions
on a quotient group of a topological group.
TOPOLOGICAL GROUPS
333
where µ′ is induced by the multiplication µ and ı′ is induced by the
inversion ı in G. Since π is open, so is π × π. If O∗ ⊆ G/N is open, then
−1
(π × π) µ′−1 (O∗ ) = µ−1 π −1 (O∗ ) is open in G × G, for µ and π are
continuous. Since π × π is an identification, µ′−1 is open. It follows that
the multiplication µ′ in G/N is continuous. Similarly, ı′ is continuous,
and G/N is a topological group. We call it the quotient or the factor
group of G by the normal subgroup N .
As seen above, the quotient of a topological group G modulo a
closed normal subgroup N is a Hausdorff topological group under the
induced multiplication and topology. Moreover, if G is compact (resp.
locally compact or second countable), then so is G/N , by Proposition
12.3.1.
Proposition 12.3.2 Let G0 be the component of the identity element
e of a topological group G. Then the quotient group G/G0 is totally
disconnected.
Proof. Let C be a connected subset of G/G0 . We show that C is a
singleton set; equivalently, π −1 (C) is a coset of G0 in G, where π :
G → G/G0 is the canonical map. Since the components of G are cosets
of G0 in G, it suffices to prove that π −1 (C) is connected. Assume on
the contrary that π −1 (C) is disconnected, and let π −1 (C) = A ∪ B
be a disconnection. Then both A and B are nonempty open subsets of
π −1 (C), and A ∩ B = ∅. So there exist open sets U and V in G such
that A = U ∩π −1 (C) and B = V ∩π −1 (C) . Obviously, we have π (A) =
π (U )∩C and π (B) = π (V )∩C. Since π is open, π (A) and π (B), both,
are nonempty open subsets of C such that C = π (A)∪π (B). Since C is
connected, we must have π (A)∩π (B) ̸= ∅. So there exist points a ∈ A
and b ∈ B such that π (a) = π (b) . Then E = aG0 = bG0 ⊂ π −1 (C) .
Consequently, E = (aG0 ∩ A)∪(bG0 ∩ B) is a disconnection of E. This
is a contradiction, for E is obviously connected. Therefore π −1 (C) is
connected. This completes the proof.
♢
HOMOMORPHISMS
Definition 12.3.3 A homomorphism of a topological group G into another topological group G′ is a function f : G → G′ that is continuous
as well as an algebraic homomorphism.
It is obvious that an inclusion map of a subgroup of the topolog-
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Elements of Topology
ical group G and the natural projection of G onto a quotient group
are homomorphisms (in this sense). The next proposition often simplifies the problem of checking the continuity or openness of algebraic
homomorphism between topological groups.
Proposition 12.3.4 Let G and G′ be topological groups, and let f :
G → G′ be a homomorphism in the algebraic sense. Then
(a) f is continuous if it is continuous at the identity element e of G,
and
(b) f is open if it carries nbds of e to nbds of the identity element e′
of G′ .
Proof. (a): Let x ∈ G, and let V be an open nbd of f (x) = x′ in
G′ . Then ρx′−1 (V ) is an open nbd of the identity element e′ of G′ . As
f (e) = e′ , we find an open nbd U of e such that f (U ) ⊆ ρx′−1 (V ).
This implies that f (ρx (U )) = ρx′ f (U ) ⊆ V. Since ρx (U ) is an open
nbd of x, f is continuous at x.
(b): Let U be an open subset of G, and x ∈ U. Then ρx−1 (U ) is
an open nbd of e. By our hypothesis, f (ρx−1 (U )) is a nbd of e′ . If
f (x) = x′ , then ρx′−1 (f (U )) = f (ρx−1 (U )), and therefore f (U ) is a
nbd of x′ . Since x ∈ U is arbitrary, f (U ) is open.
♢
The following examples show that two topological groups may have
essentially the same algebraic structures but different topological structures, and vice versa.
Example 12.3.1 The group (R, +) can be given the usual topology
as well as the discrete topology to produce two different topological
groups.
[
]
a 0
Example 12.3.2 Let G be the set of all matrices
and G′ be the
0 b
]
[ a
e
b
, where a, b ∈ R. Then G is an abelian
set of all matrices
0 e−b
group under the addition of matrices, and G′ is a nonabelian group
under the multiplication of matrices. With the usual topologies, the
underlying spaces of the topological groups G and G′ are homeomorphic to the euclidean space R2 .
TOPOLOGICAL GROUPS
335
Definition 12.3.5 Let G and G′ be topological groups, and f : G →
G′ be a homomorphism. We say that f is
(a) a monomorphism if it is injective,
(b) an epimorphism if it is surjective, and
(c) an isomorphism if it is bijective and open.
Thus, an isomorphism between two topological groups G and G′ is
an algebraic homomorphism G → G′ which is also a homeomorphism.
We call G and G′ isomorphic (written G ∼
= G′ ) if there is an isomo′
morphism G → G . If N is a normal subgroup of a topological group
G, then the natural projection π : G → G/N is an epimorphism. As in
ordinary group theory, the kernel of f is defined as the inverse image
of the identity element of G′ , and is denoted by ker(f ). The homomorphism f is a monomorphism if ker(f ) = {e}, where e ∈ G is the identity
element, and an epimorphism if its image f (G) equals G′ . It is easily
verified that K = ker(f ) is a normal subgroup of G, and the induced
homomorphism f¯ : G/K → G′ , xK 7→ f (x), is a monomorphism of
topological groups. The continuity of f¯ follows from the factorisation
f¯π = f and the fact that the natural projection π : G → G/K is
an identification map. The remaining properties of f¯ are well-known
from group theory. Furthermore, if f is an open epimorphism, then f¯ is
clearly an isomorphism. Conversely, if f¯ is an isomorphism (of topological groups), then f must be an open epimorphism, for π : G → G/K
is open. Thus we have established
Proposition 12.3.6 Let f : G → G′ be a homomorphism between
topological groups. Then the induced homomorphism f¯ : G/ker(f ) →
G′ is an isomorphism if and only if f is an open epimorphism.
Corollary 12.3.7 A homomorphism of a compact topological group
onto a Hausdorff topological group is open.
Proof. Let G be a compact topological group, G′ a Hausdorff topological
group and f : G → G′ be an epimorphism. Then G/ker(f ) is compact,
and hence the induced map f¯ : G/ker(f ) → G′ is an isomorphism
between topological groups. By the preceding proposition, f is open.♢
Example 12.3.3 The exponential map p : R1 → S1 , t 7→ e2πıt , is an
open epimorphism of topological groups with the kernel Z, the group
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Elements of Topology
of integers (refer to Exercise 2.1.18c). Therefore the quotient group
R1 /Z is isomorphic to S1 .
Example 12.3.4 For F = R and C, the restriction of the determinant
function
det : M (n, F ) → F
to GL (n, F ) is an epimorphism GL (n, F ) → F0 , where F0 is the
multiplicative group of the field F, with the usual topology. We assert that it is open, too. We prove the assertion in the complex case;
the argument for the real case is similar. Let U be an open subset of
M (n, C) and 0 ̸= A ∈ U . Then there exists a real r > 0 such that
B (A; r) ⊆ U . If z ∈ C and |z − 1| < r/ ∥A∥ = ϵ, then zA ∈ U . We
have det(zA) = z n det(A). It is a fact in Complex Analysis that the
function z 7→ z n , n > 0 an integer, is an open mapping on C. It follows
that W = {z n : |z − 1| < ϵ} is an open set in C with 1 ∈ W . Thus,
there is an open ball about 1 (in C) contained in W . Consequently,
the set of numbers z n det(A), as z ranges over the set {z : |z − 1| < ϵ},
contains an open ball about det(A). This ball (with centre det(A))
is contained in the image of U under the function det, and therefore
√ n
det(U ) is open. If A = 0, then the open ball of radius ϵ = (r/ n)
about 0 ∈ C is contained in det(U ), for the scalar matrix
 √
n
z

 0

 .
 ..

0
0 ···
√
n
z ···
..
.
..
···
···
.
0


0 

.. 
. 

√
n
z
belongs to U when |z| < ϵ. Thus the determinant function from
M (n, F ) to F is open. Since GL (n, F ) is an open subset of M (n, F ),
it follows that the function det : GL (n, F ) → F0 is open. The kernel of
the homomorphism det is the special linear group SL (n, F ) . By Proposition 12.3.6, the quotient group GL (n, F ) /SL (n, F ) is isomorphic to
the topological group F0 .
For F = R or C, the group
SL (n, F ) = {A ∈ GL (n, F ) |det(A) = 1}
is closed, since 1 ∈ F is closed and the function det is continuous. We
TOPOLOGICAL GROUPS
337
write SO (n) = O (n) ∩ SL (n, R) and SU (n) = U (n) ∩ SL (n, C), and
call SO (n) the special orthogonal group and SU (n) the special unitary
group. It follows that the groups SO (n) and SU (n) are closed in O (n)
and U (n), respectively, and hence compact for every n. Obviously,
both the topological groups SO (1) and SU (1) are trivial groups. The
geometry of SO (2) and SU (2) is described by the following.
Proposition 12.3.8 SO (2) ∼
= S1 and SU (2) ∼
= S3 as topological
groups.
Proof. Clearly, the multiplication by each z ∈ S1 determines a linear
isometry µz : C → C. Also, the canonical vector space isomorphism ξ :
C → R2 preserves the norms, and therefore the composition ξ ◦ µz ◦ ξ −1
is a linear isometry λz : R2 → R2 for each z ∈ S1 . Hence there is a
monomorphism ψ : S1 → O (2) which takes z ∈ S1 into the matrix of
λz relative to the standard basis of R2 . If z = cos θ + ı sin θ, then the
matrix of λz is
[
]
cos θ − sin θ
sin θ
cos θ
which obviously belongs to SO (2). On the other hand, if A ∈ SO (2),
then it is easily seen that the matrix A is of this form. Consequently,
ψ is the desired isomorphism between S1 and SO (2).
To find an isomorphism between topological groups S3 and SU (2),
we write q ∈ H as q = zq + wq ȷ, where zq , wq are in C. The canonical
map ξ : zq +wq ȷ 7→ (zq , wq ) is an isomorphism of the (right) vector space
H over C onto C2 . Also, it is norm preserving, and hence an isometry.
Observe that the left multiplication µx by x ∈ S3 is a linear isometry of
H. Consequently, we obtain a C-linear isometry λx : C2 → C2 defined
by λx = ξ ◦ µx ◦ ξ −1 . Now we can define a map ψ : S3 → U (2) by
setting ψ (x) = [λx ], the matrix of λx in the standard basis of C2 .
It is easily verified that ψ is a monomorphism
of topological
groups.
]
[
zx −wx
, which belongs
For x = zx + wx ȷ, the matrix of λx is
wx zx
[
]
a b
3
to SU (2) for all x ∈ S . Conversely, if the matrix A =
c d
[
]
[
]
a c
d −b
belongs to SU (2), then
=
. So d = a, c = −b
−c a
b d
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Elements of Topology
and 1 = det(A) = |a|2 + |b|2 . It follows that x = a − bȷ ∈ S3 , and we
have ψ (x) = A. Thus ψ is a monomorphism of S3 onto SU (2), and
this completes the proof.
♢
In the same vein, we observe that there is a monomorphism U (n) →
O (2n). Consider the canonical bijection ξ : Cn → R2n given by
(a1 + ıb1 , . . . , an + ıbn ) 7→ (a1 , b1 , . . . , an , bn ).
Since R ⊂ C, Cn can be regarded as a linear space over R. Then ξ
is a linear isomorphism; consequently, any C-linear map f : Cn →
Cn defines an R-linear map λn (f ) : R2n → R2n , in the obvious way.
Clearly, λn (gf ) = λn (g) ◦ λn (f ). If the matrix of f in the standard
basis of Cn has zrs as the (rs)th entry, then the matrix of λn (f ) in the
standard basis of Rn has the matrix
]
[
ars −brs
brs ars
at the rs-block, where zrs = ars + ıbrs . So there is a monomorphism
λn : GL (n, C) → GL (2n, R).
The matrices in the image of λn are referred to as the complex-linear
real matrices. The isomorphism ξ preserves the norms, and therefore
λn (f ) is an isometry whenever f is so. Hence the restriction of λn to
U (n) gives a monomorphism U (n) → O (2n). Thus U (n) is isomorphic
to the group of complex-linear real orthogonal matrices.
Similarly, there is a monomorphism Sp (n) → U (2n). Note that
C ⊂ H and each quaternion can be written as z + wȷ, z, w ∈ C. The
mapping z + wȷ 7→ (z, w) defines a bijection ξ : Hn → C2n , which is
C-linear, when Hn is considered as a right linear space over C. So each
linear map Hn → Hn gives a linear map C2n → C2n ; consequently, an
n × n matrix A over H determines a 2n × 2n matrix over C obtained
by replacing the element z + wj of A by the 2 × 2 matrix
[
]
z −w
.
w z
This correspondence is a monomorphism λn : GL (n, H) → GL (2n, C).
The matrices in the image of λn are referred to as the quaternioniclinear complex matrices. It is easily checked that if A ∈ Sp (n), then
TOPOLOGICAL GROUPS
339
λn (A) ∈ U (2n). Thus we obtain the desired monomorphism Sp (n) →
U (2n).
In order to apply the first isomorphism theorem for topological
groups (Proposition 12.3.6), it is important to know some conditions
which ensure the openness of homomorphisms between topological
groups. We prove the following.
Theorem 12.3.9 Let G be a second countable, locally compact Hausdorff group, and G′ be a locally compact Hausdorff group. Then a
homomorphism f of G onto G′ is an open map.
Proof. By Proposition 12.3.4, it suffices to prove that f (N ) is a nbd of
the identity element e′ of G′ for every nbd N of the identity element e of
G. We first prove that the interior of f (U ) is nonempty for every open
set U ⊆ G containing e. Since G is locally compact Hausdorff, there
exists an open set V ⊆ G such that the closure V of V is compact and
e ∈ V ⊆ V ⊂ U. For each x ∈ G, V x = ρx (V ) is an open nbd of x, and
the family {V x|x ∈ G} covers G. Since G is second countable, this open
covering has a countable
subcovering.( So we
) find points x1 , x2 , . . . in G
∪∞
such that G = 1 Vxn . Put Kn = f V xn for every
∪∞ n. Then each Kn
′
′
is compact, and hence closed in G . Also, G = 1 Kn . We claim that
◦
(Kn ) ̸= ∅ for some n. Assume the contrary, and choose a nonempty
open set W0 of G′ such that W0 is compact. By our assumption, there
is a point y1 ∈ W0 − K1 . Then we find an open nbd W1 of y1 such that
W1 ⊂ W0 − K1 , and W1 is compact. Again, by our assumption, there
is a point y2 ∈ W1 − K2 . Consequently, there is an open nbd W2 of
y2 such that W2 is compact, and W2 ⊂ W1 − K2 . Continuing in this
way, we obtain open sets Wn , n = 1, 2, . . . , such that Wn is compact,
and Wn ∩ Kn = ∅ and Wn+1 ⊆ Wn . Since∩the intersection of every
∞
finite subfamily of Wn is nonempty, we have 1∪ Wn ̸= ∅ (see Exercise
′
6.1.21). But, this is impossible because G = Kn( and
not)
) Kn does
( −1
◦
meet Wn . Hence our claim. If (Kn ) ̸= ∅, then f V = Kn f xn
◦
also has a nonempty interior. Since V ⊂ U, we have f (U ) ̸= ∅. Now,
let N be an nbd of e. Then there exists an open nbd V of e such that
◦
V V −1 ⊆ N. As seen above, f (V ) ̸= ∅. Let y be an interior point of
−1
f (V ) . Then f (V ) y is a nbd of e′ . Suppose(that y) = f (x), where
x ∈ V. Then V x−1 ⊆ N so that f (V ) y −1 = f V x−1 ⊆ f (N ). Thus
f (N ) is a nbd of e′ .
♢
The condition of second countability on G in the preceding theo-
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Elements of Topology
rem cannot be dropped. For, if G is the group of real numbers with
the discrete topology and G′ = R, the group of reals with the usual
topology, then the identity map G → G′ is continuous and surjective
but not open. Notice that the group G is not second countable. Also,
the following example shows that the condition of local compactness
on G′ in Theorem 12.3.9 is essential.
Example 12.3.5 Let α > 0 be an irrational number, and consider the
subgroup X = Z+αZ ⊂ R. If X has a least positive number c, say, then
X is generated by c (see Ex. 12.2.3). Consequently, α = mc and 1 = nc
for some m, n ∈ Z, which implies that α = m/n, a contradiction. So X
has no least positive number; accordingly, there is a strictly decreasing
sequence of positive numbers x1 > x2 > · · · in X. Now, given an open
interval (a, b), we find an integer n > 0 such that 0 < xn − xn+1 = u <
b − a. Next, we choose an integer m > 0 such that m − 1 < a/u < m.
From this and the inequality 1 < (b − a) /u, we obtain a < mu < b.
We have mu ∈ X, for X is a subgroup. It follows that X is dense in R,
and therefore X/Z is dense in the quotient space R/Z ≈ S1 . Obviously,
the restriction of the quotient map R → R/Z to αZ is an algebraic
isomorphism between αZ and X/Z. But, it is not a homeomorphism,
since αZ is a discrete space while every nonempty open subset of R/Z
contains infinitely many points of X/Z. It is easily seen that X/Z is
not locally compact.
Recall that the second isomorphism theorem of group theory states
that if G is a group, and M, N ⊆ G are subgroups and N is normal in
G, then the factor group M N/N is isomorphic to M/ (M ∩ N ) . The
preceding examples also show that the analogue of this theorem for
topological groups is not true. However, if G is a second countable,
locally compact Hausdorff topological group, and the subgroups M, N
and M N are all closed in G, then we have M N/N ∼
= M/ (M ∩ N )
as topological groups, by Theorem 12.3.9. The analogue of the third
isomorphism theorem of group theory is valid for topological groups,
and we leave the proof to the reader.
Exercises
1. Show that there is no monomorphism of S1 into R1 .
2. Prove that the quotient group C0 /S1 is isomorphic to the group R+ of
positive reals.
TOPOLOGICAL GROUPS
341
3. Show that the group of automorphisms of the circle group S1 is isomorphic to Z2 .
4. Let G ⊂ GL (n, R) be a compact group. Prove that every element of G
has determinant +1 or -1. Is G contained in O (n)?
5. Prove every discrete subgroup of O (2) is cyclic or dihedral.
6. Let C (I) be the topological group in Exercise 12.1.3. Prove that the
mapping et : C (I) → R given by et (f ) = f (t) is a open epimorphism
for each t ∈ I, and hence C (I) /ker (et ) is isomorphic to R.
7. Prove that a Lindelöf (or separable) first countable topological group is
second countable.
8. Let H be a subgroup of a topological group G. Prove that G/H has the
indiscrete topology ⇔ H is dense in G.
9. Let H be a subgroup of a topological group G. Show that the coset space
G/H is T3 . (It follows that a topological group satisfies the T3 -axiom
necessarily.)
10. Let H be a compact subgroup of a topological group G. Show that the
natural projection G → G/H is closed.
11. Prove that S1 is a closed subgroup of S3 . Also, show that S3 /S1 is
homeomorphic to S2 . Is S1 normal in S3 ?
12. Let H be a subgroup of a Hausdorff topological group G. Show that G is
compact (resp. connected) if and only if both H and G/H are compact
(resp. connected).
13. Let G be a topological group, and M and N be subgroups of G with
M ⊂ N. Show that the quotient topology for N/M coincides with the
subspace topology induced from G/M .
14. Let G be a topological group. If M and N are normal subgroups of G
with M ⊂ N , prove that (G/M ) / (N/M ) ∼
= G/N as topological groups.
12.4
Direct Products
Given a family of topological
groups Gα , α ∈ A, it is known
∏
that their direct product
Gα is a group under the coordinatewise
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Elements of Topology
group operations:
If x = (xα ) and y = (yα ), then xy = (xα yα ),
(
)
x−1 = x−1
,
and
e = (eα ) (where
eα is the identity ∏
element of
α
∏
Gα ) is the identity element of
Gα . We observe that
Gα with
the product (Tychonoff)
topology
is
a
topological
group.
The
mul∏
can
be
considered
as
the
composition
of
the
functiplication
in
G
∏
∏α
∏
tion η : Gα × Gα → (Gα × Gα ), which takes ((xα ) , (yα )) into
the
is (xα , yα ), and the product function
∏ point
∏ whose αth coordinate
∏
µα : (Gα × Gα ) → ∏Gα , where µα : Gα × Gα → Gα is the multiplication. The function µα is continuous by Theorem 2.2.10, and
the
∏ function η is clearly a homeomorphism. So the multiplication
∏ in
Gα is continuous. The continuity of the inversion function on Gα
is also
∏an easy consequence of Theorem 2.2.10. Clearly, each projection
pβ : Gα → Gβ is an open epimorphism. If ∏
H is a topological group,
then it is easy to see that a function f : H → Gα is a homomorphism
if and only if each composition pα ◦ f : H → Gα is a∏homomorphism.
In particular, we have monomorphisms qβ : Gβ →
Gα defined by
(qβ (xβ ))α = eα if α ̸= β, and (qβ (xβ ))β = xβ .
If G1 and G2 are two topological groups, then their direct product is denoted by G1 × G2 . Obviously, ker (p1 ) = q2 (G2 ) and
ker (p2 ) = q1 (G1 ). By Proposition 12.3.6, there are isomorphisms
(G1 × G2 ) /q2 (G2 ) ∼
= G1 and (G1 × G2 ) /q1 (G1 ) ∼
= G2 induced by
the projections p1 and p2 , respectively. There is another fact which
is of use at times. Let G be an abelian topological group. Then
∆ = {(x, x) |x ∈ G} is a normal subgroup of the direct product G × G.
We observe that the topological group (G × G) /∆ is isomorphic to G.
Note that the map ϕ : G × G → G, (x, y) 7→ xy −1 , has a continuous
right inverse G → G × G, x 7→ (x, e). Therefore ϕ is an identification (see Exercise 7.2.2). By Theorem 7.2.5, there is a homeomorphism
ψ : (G × G) /∆ → G such that ψπ = ϕ, where π : G×G → (G × G) /∆
is the natural projection. It is easily checked that ψ is an algebraic isomorphism.
Recall that in group theory we say that a group G decomposes
into the direct product of its subgroups M and N if both M and
N are normal in G, G = M N and M ∩ N = {e}. These conditions
are equivalent to the requirement that the mapping M × N → G,
(x, y) 7→ xy, is an isomorphism. If G is a topological group, and M
and N are its subgroups satisfying the above conditions, then this
isomorphism (in the sense of algebra) is obviously continuous; however,
it need not be a homeomorphism, in general (see Ex. 12.4.2 below).
TOPOLOGICAL GROUPS
343
Definition 12.4.1 A topological group G is said to decompose into the
direct product of its subgroups M and N if the mapping M × N → G,
(x, y) 7→ xy, is an isomorphism of topological groups.
It is obvious that if a compact Hausdorff topological group G decomposes into the direct product of two closed subgroups in the algebraic sense, then it decomposes into the direct product of these subgroups in the sense of the preceding definition. More generally, this
proposition remains valid if G is a second countable, locally compact
Hausdorff topological group, by Theorem 12.3.9.
If a topological group G decomposes into the direct product of the
subgroups M and N , then the quotient group G/N is isomorphic to the
topological group M . For, the isomorphism (x, y) 7→ xy of M × N onto
G carries {e}×N onto N, and we have G/N ∼
= (M × N ) / ({e} × N ) ∼
=
M, as topological groups.
Example 12.4.1 Let C0 denote the punctured plane C − {0}. Then we
have a mapping η : C0 → R+ × S1 defined by η (z) = (|z|, z/|z|) , where
R+ is the set of positive reals. Note that each element z ∈ C0 can be
uniquely expressed as z = |z| (z/|z|). Since the absolute value function
z 7→ |z| is an open homomorphism, it follows that η is an isomorphism
of topological groups.
Similarly, the punctured quaternionic line H0 = H − {0} is isomorphic to S3 × R+ .
Example 12.4.2 Let α > 0 be a real number, and consider the subgroups N = Z × Z and H = {(x, αx) |x ∈ R} of R2 . Then G = H + N
is a subgroup of the topological group R2 . If α is an irrational number,
then G is the the direct sum of H and N in the algebraic sense. However, this is not a decomposition of the topological group G into the
subgroups H and N. To see this, assume otherwise. Then the quotient
group G/H is homeomorphic to the discrete group N . We observe
that G/H is also homeomorphic to Z + αZ. Consider the mapping
f : R2 → R1 , (x, y) 7→ x − α−1 y. This is clearly a continuous open
surjection, and hence an identification. It is easily checked that the
decomposition space of f is the coset space R2 /H. By Theorem 7.2.5,
f indices a homeomorphism R2 /H ≈ R1 , which carries the quotient
space G/H onto Z + αZ. Thus, being a discrete subgroup, Z + αZ is
closed in R1 (by Exercise 12.2.9). But, by Ex. 12.3.5, it is dense in R1 .
This contradiction proves our claim.
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Elements of Topology
On the other hand, a topological group homeomorphic to the direct
product of two of its subgroups, may fail to decompose into them. As an
illustration of this fact, we consider the group I (Rn ) of all isometries of
the euclidean space Rn , the composition of mappings being the group
multiplication. If f ∈ I (Rn ) and f (0) = b, then the composition of f
and the translation ρ−b : x 7→ x − b is an isometry of Rn , which fixes
the origin 0 of Rn . We show that g = ρ−b ◦ f is linear.
Lemma 12.4.2 An isometry of Rn , which fixes the origin, is a linear
map.
Proof. First, observe that if ⟨x, x⟩ = ⟨y, y⟩ = ⟨x, y⟩ , then x = y; this can
be seen by expanding ⟨x − y, x − y⟩. Now suppose that g : Rn → Rn is
an isometry with g (0) = 0, where 0 is the origin of Rn . Then ∥g(x) −
g(y)∥ = ∥x − y∥ for all x, y ∈ Rn . In particular, ∥g(x)∥ = ∥x∥ for every
x ∈ Rn . Consequently, we have ⟨g(x), g(y)⟩ = ⟨x, y⟩ for all x, y ∈ Rn .
If u = x + y, then ⟨u, u⟩ = ⟨u, x⟩ + ⟨u, y⟩ = ⟨x, x⟩ + 2 ⟨x, y⟩ + ⟨y, y⟩.
Therefore
⟨g (u) , g (u)⟩ = ⟨g (u) , g (x) + g (y)⟩ = ⟨g (x) + g (y) , g (x) + g (y)⟩
which implies that g (x + y) = g (u) = g (x) + g (y). Similarly, we find
that g (xa) = g (x) a, a ∈ R, and this completes the proof.
♢
Obviously, f is the composition of g = ρ−b ◦ f with the translation
ρb : x 7→ x + b. It is immediate from the preceding lemma that g is
an othogonal transformation (refer to §2). Thus f is an (invertible)
affine map on Rn , being the composition of a linear transformation
and a translation. It follows that the group I (Rn ) is a subgroup of
the topological group Af fn (R) and, therefore, can be assigned the
topology of the product space GL(n, R) × Rn (refer to Exercise 12.1.7).
Alternatively, I (Rn ) can be regarded as a subgroup of GL (n + 1, R).
In fact, each isometry f in I (Rn ) determines the matrix
[
]
A(f ) b
M (f ) =
0
1
in the group GL (n + 1, R), where b = f (0) and A(f ) is the matrix of
the orthogonal transformation x → f (x) − b. Moreover, the euclidean
space Rn can be identified with the subspace Rn × {1} ⊂ Rn+1 by the
isometry x 7→ (x, 1). With this identification, we have M (f )x = f (x).
TOPOLOGICAL GROUPS
345
The correspondence f 7→ M (f ) is a monomorphism of the group I (Rn )
into GL (n + 1, R). Therefore I (Rn ) can be identified with a subgroup
of GL (n + 1, R) and, thus, receives the relative topology induced from
GL (n + 1, R) . This topology for I (Rn ) is referred to as the metric
topology. Endowed with this topology, the group I (Rn ) is called a euclidean group.
We note that the set M of all orthogonal transformations of Rn
forms a subgroup of I (Rn ), and the set N of all translations of Rn is a
normal subgroup of I (Rn ), which is isomorphic to the topological group
Rn . The euclidean group I (Rn ) is homeomorphic to the direct product
O (n) × Rn . A homeomorphism is given by the mapping f 7→ ([g], b),
where b = f (0) and g is the orthogonal transformation ρ−b ◦ f and ρ−b
is the right translation of Rn by −b. Clearly, M ∼
= O (n) and N ∼
= Rn
n
under this homeomorphism. But, the topological group I (R ) does not
decompose into M and N , since the subgroup M ⊂ I (Rn ) is in general
not normal. We say that the group I (Rn ) is the semidirect product of
O (n) and Rn .
Exercises
1. Prove:
(a) The topological group R is isomorphic to the topological group
R+ of positive reals.
(b) The topological group R0 = R − {0} is isomorphic to S0 × R+ .
2. If g is a homomorphism of R into itself, prove that g (x) = xg (1) for
every x ∈ R. Deduce that the group of automorphisms of R is isomorphic
to R × Z2 .
3. Prove that the factor group Rn /Zn is isomorphic to the direct product
S1 × · · · × S1 (n factors).
4. Prove that the orthogonal group O (n) is isomorphic to the group of all
isometries of Sn−1 onto itself topologised by the metric
{
}
ρ (f, g) = sup ∥f (x) − g (x)∥ : x ∈ Sn−1 .
5.
(a) Prove that O (n) is homeomorphic to SO (n) × Z2 . Are they isomorphic as topological groups?
(b) Prove that U (n) is homeomorphic to SU (n) × S1 . Are they isomorphic as topological groups?
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Elements of Topology
6. Let F = R, C or H. Prove that Af fn (F ) is isomorphic to the subgroup
{[
]
}
A v
n
: A ∈ GL (n, F ) and v ∈ F
0 1
of GL (n + 1, F ). Deduce that Af fn (F ) considered as a subgroup of
GL (n + 1, F ) is closed. Is it compact?
7. Show that the topological
group in Exercise 12.1.2 is not isomorphic to
∏
the direct product Rr of copies of R, one copy for each real number
r in R, although the underlying groups of both topological groups are
the same.
8. Let G be a second countable, locally compact Hausdorff topological
group. Prove that G decomposes into a direct product of N and M
if and only if N and M are closed normal subgroups of G such that
N ∩ M = {e} and G = N M .
Chapter 13
TRANSFORMATION GROUPS
13.1
13.2
13.1
Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Orbit Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
347
365
Group Actions
The theory of transformation groups is a beautiful area of topology
which has connections with different parts of mathematics. Generally
speaking, this deals with symmetries (automorphisms) of mathematical objects such as vector spaces, topological spaces, manifolds, cellular
complexes, etc. Thus, one studies actions of groups on spaces in this
theory. In this section, the rudiments of this theory and some elementary results from its point-set topology are discussed. Connectedness
of some topological groups will also be ascertained. We also introduce
here a geometric definition of (proper) rigid motion and see the justification for describing the elements of the group SO(n) as rotations of
Rn .
Definition 13.1.1 Let G be a topological group and X a space. A left
action of G on X is a continuous map ϕ : G × X → X such that
(a) ϕ (g, ϕ (h, x)) = ϕ (gh, x) for all g, h ∈ G and x ∈ X,
(b) ϕ (e, x) = x for all x ∈ X, where e ∈ G is the identity element.
The space X, together with a left action ϕ of G on X, is called a left
G-space, and the triple (G, X, ϕ) is called a topological transformation
group.
Often, we denote the topological transformation group (G, X, ϕ)
just by the pair (G, X), regarding ϕ as understood. It is convenient to
denote the image of (g, x) under ϕ by gx. Then conditions (a) and (b)
become g (hx) = (gh) x and ex = x.
347
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Elements of Topology
There is an analogous notion of right action of G on X. This is a
continuous map X × G → X, (x, g) 7→ xg, such that (xg) h = x (gh)
and xe = x. The space X, together with a right action of G on X,
is called a right G-space. Any right action defines a left action in a
canonical way: gx = xg −1 , and vice versa. Thus, for most purposes the
choice of left or right action is a matter of taste. However, we will see
later some situations in which an action appears naturally as a right
action. Unless we specify otherwise, we shall mean (for brevity) by an
action of G on X a left action.
Let ϕ be an action of a topological group G on a space X. For each
g ∈ G, the map ϕg : x 7→ gx is a homeomorphism of X with ϕg−1 as its
inverse. Therefore the associated function g 7→ ϕg defines an algebraic
homomorphism G into the group Homeo(X) of all homeomorphisms
of X onto itself (with the composition of mappings as the group operation). This homomorphism is continuous relative to the compact-open
topology for Homeo (X) (see Theorem 11.2.10). With this topology on
it, the group Homeo (X) is a topological group for many nice spaces
X, as discussed below. Thus, an action of the topological group group
G on such spaces X is a homomorphism of G into the topological group
group Homeo(X).
Denote the group Homeo (X) with the compact-open topology by
Homeo (X)co . If X is locally compact Hausdorff, then the multiplication function
Homeo (X)co × Homeo (X)co → Homeo (X)co ,
(g, h) 7→ gh,
is continuous. To see this, let (K, U ) be a subbasic open set in
Homeo (X)co and suppose that gh ∈ (K, U ). Then K ⊆ X is compact, U ⊆ X is open and gh (K) ⊆ U . So h (K) ⊆ g −1 (U ). Since X is
locally compact Hausdorff, for each x ∈ K, there exists an open set Vx
in X such that h (x) ∈ Vx ⊂ Vx ⊂ g −1 (U ) and Vx is compact. Since
h (K) is compact,
we find finitely many points x1 , . . . , xn in K( such)that
∪n
h (K) ⊂ 1 Vxi = V, say. Consequently, h ∈ (K, V ) and g ∈ V , U , for
V
and contained in g −1 (U ) . (It is clear
( is compact
)
) that both (K, V ) and
V , U are open in Homeo (X)co , and V , U (K, V ) ⊆ (K, U ). Hence
the multiplication in Homeo (X)co is continuous. However, there exist
locally compact spaces X such that the inversion function
Homeo (X)co → Homeo (X)co ,
g 7→ g −1,
is not continuous. If X is compact Hausdorff, then the continuity of the
−1
inversion function ı on Homeo (X)co is immediate, since (K, U ) =
TRANSFORMATION GROUPS
349
(X − U, X − K) . Thus, in this case, Homeo (X)co is a Hausdorff topological group.
More generally, we have the following.
Theorem 13.1.2 Let X be a locally connected, locally compact Hausdorff space. Then the group Homeo (X) with the compact-open topology is a topological group.
Proof. In view of the above remarks, we need only to establish the
continuity of the inversion function
ı : Homeo (X)co → Homeo (X)co ,
h 7→ h−1.
Consider a subbasic open set (K, U ) of Homeo (X)co with f ∈
−1
(K, U ) . Then we have K ⊆ f (U ). Since X is locally connected and
locally compact and f (U ) is open, for each x ∈ K, there exists a connected open set Vx such that x ∈ Vx ⊆ Vx ⊆ f (U ) and Vx is compact.
By the compactness of K, we find finitely many points x1 , . . . , xn in K
)−1
∪n
∩n (
such that K ⊆ j=1 Vxj . Then f ∈ j=1 Vxj , U
. Thus it suffices to
)−1
(
show that V , U
is open in Homeo((X)co) for every connected open
( )
set V such that V (is )compact. Let g ∈ V , U be arbitrary. Since g V
is compact and (g )V ⊆ U , there exists an open set W such that W is
compact and g V ⊆ W ⊆ W( ⊆) U. For the same reason, we find another open set O such that g V ⊆
( O ⊆ O ⊆−1W. Now, )choose a point
x0 ∈ g (V ) and set N = (x0 , V ) ∩ W − O, g (U ) − V . Then g −1 ∈
(
)−1
N(. We observe
that N ⊆ V , U
. If h ∈ N , then h(x0 )(∈ V and
)
)
h W − O ⊆ g −1 (U ) − V , which implies that V ⊆ h(O) ∪ h X − W .
Since V is connected, we have either V ∩ h(O) = ∅ or V ⊆ h(O). As
x0 ∈ g(V ) ⊂ O, we have h(x0 ) ∈ h(O). Thus V ∩ h(O) ̸= ∅, and there( )
(
)−1
fore V ⊆ h(O). This forces V ⊆ h O ⊂ h(U ), and so h ∈ V , U
.
(
)−1
−1
Since h ∈ N is arbitrary, we have g ∈ N ⊆ V , U
. It follows that
(
)−1
V ,U
is open in the compact-open topology for Homeo (X), and
therefore ı is continuous.
♢
If X is just locally compact Hausdorff, then Homeo (X) is a topological group with the topology generated by the subbase which consists of all sets ⟨K, U ⟩, K ⊆ X compact and U ⊆ X open, and their
inverses. If ϕ is an action of a topological group G on the space X, then
the homomorphism G → Homeo (X) , g 7→ ϕg , is continuous relative
to this topology for Homeo (X) also. Conversely, a homomorphism
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Elements of Topology
ϕ̂ : G → Homeo (X) defines an action ϕ : G × X → X given by
ϕ(g, x) = ϕ̂(g)(x), by Theorem 11.2.11. Thus an action of a topological
group G on a locally compact Hausdorff space X may be defined as a
homomorphism of G into the topological group Homeo (X).
Recall that any group can be considered as a topological group
with the discrete topology, and as such it is called a discrete group.
Accordingly, we can also consider the actions of nontopologised groups
on topological spaces. Occasionally, the purely algebraic aspect of a
transformation group is needed. Given a group G and a set X, an
action of G on X means an action in the sense of Definition 13.1.1,
where both G and X are given the discrete topology. It follows that
an action of G on X can be interpreted as a homomorphism G into
the permutation group Symm (X) of X (referred to as a permutation
representation of G). In this case, we say that X is a G-set. Notice that
any G-space can be considered as a G-set by forgetting the topology
of G and X.
An action of G on the space X is called effective if the homomorphism G → Homeo (X) is injective, that is, for each g ̸= e in G, there
exists a point x ∈ X such that gx ̸= x. Clearly, an effective topological
transformation group on X is (upto isomorphism) an admissible group
of continuous functions of X into itself with composition as the group
operation.
An action of G on X is called free if gx = x for any x ∈ X implies
g = e, that is, each nontrivial element of G moves every point of X.
Notice that a free action is effective.
Definition 13.1.3 Let (G, X) be a topological transformation group.
For each x ∈ X, the set Gx = {g ∈ G|gx = x} is clearly a subgroup of
G, called the stabilizer of x. The subgroup Gx is also referred to as the
isotropy subgroup (or stability subgroup) of G at x.
The mapping G → X, g 7→ gx, is clearly continuous. It is immediate
from Corollary 4.4.3 that if X is Hausdorff, then Gx is closed in G. It
is also clear that an action of G on X is free if Gx is trivial for every
x ∈ X. It is trivial if Gx = G for all x ∈ X.
Definition 13.1.4 Given a G-space X, the set G (x) = {gx|g ∈ G} is
called the orbit of x (under G).
TRANSFORMATION GROUPS
351
Clearly, the orbit of x is the image of the set G × {x} under the
given action. The action of G on X is called transitive if there is only
one orbit, the entire space X itself; equivalently, for every pair of points
x, y ∈ X, there is a g ∈ G such that gx = y.
The following terminology will be occasionally used in this book,
although it is an important concept in the theory of transformation
groups: A point x of a G-space X is called a fixed point (or stationary
point) if Gx = G. Notice that a point x ∈ X is a fixed point of G ⇔
G(x) = {x}.
Example 13.1.1 Any topological group G acts on itself by left translation; more specifically, the mapping G×G → G, (g, x) 7→ gx, defines an
action of G on itself. This is obviously free and transitive. Similarly, the
right translations (x, g) 7→ xg determine a (right) free and transitive
action of G on itself.
Another action of G on itself is defined by conjugation: (g, x) 7→
gxg −1 . The kernel of this action is clearly the centre of G. The mapping
(g, x) 7→ g −1 xg of G × G into G defines a right action.
Example 13.1.2 Let H be a subgroup of a topological group G, and
G/H be the space of left cosets xH, x ∈ G. The multiplication µ :
(g, x) 7→ gx in G induces a map θ : G × G/H → G/H, (g, xH) 7→ gxH,
which makes the following diagram commutative.
G×G
1×π
?
G × G/H
µ
- G
π
?
θ G/H
FIGURE 13.1: Continuity of the function θ in Example 13.1.2.
−1
Accordingly, we have θ−1 (O) = (1 × π) (πµ) (O) for any O ⊆ G/H.
By Proposition 12.3.1, the natural projection π : G → G/H is continuous and open in the quotient topology for G/H. So the mapping
1×π : G×G → G×G/H is also continuous and open, and the continuity of θ is now clear. It is easily verified that θ is an action of G on G/H.
It is also apparent that xH is carried to yH by the map θyx−1 so that
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Elements of Topology
∩
G is transitive on G/H. The kernel of this action is g∈G gHg −1 . The
G-space G/H is called a homogeneous space. Notice that the action
described in Ex. 13.1.1 is a particular instance of the present one.
Example 13.1.3 The natural action of the general linear group
GL (n, R) on Rn is defined by (A, x) 7→ Ax, this being multiplication of
the n × n matrix A and the n × 1 matrix x ∈ Rn (obtained by writing
x as a column matrix). This action is effective, and the origin 0 ∈ Rn
is a fixed point. We observe that there are only two orbits: Rn − {0}
and {0}. If x ̸= 0 in Rn , then there is a basis {b1 , . . . , bn } of Rn with
x = b1 . Let {e1 , . . . , en } be the standard
basis of Rn . Then there exist
∑n
real numbers aij such that bj = i=1 aij ei , j = 1, . . . , n. We see that
the matrix A = (aij ) ∈ GL (n, R), and x = Ae1 . This shows that every
x ̸= 0 is in the orbit of e1 so that Rn − {0} is the another orbit.
We have similar actions of GL (n, C) on Cn , and GL (n, H) on Hn .
Let X be a G-space and Gx be the isotropy group at x ∈ X. The
mapping G → X, g 7→ gx, is clearly continuous, and constant on cosets
gGx . Therefore it induces the continuous mapping αx : G/Gx → X
given by αx (gGx ) = gx. We have the factorisation
θ
α
x
x
G −→
G/Gx −→
X
of g 7→ gx, where θx is the canonical projection of G onto the homogeneous space G/Gx . Obviously, αx is injective and has the image
G (x). Accordingly, if X is a transitive G-space, then the mapping
αx : G/Gx → X is a continuous bijection between G/Gx and X. But
a transitive G-space X need not be homeomorphic to the coset space
G/Gx , as shown by Ex. 13.1.4 below. A necessary and sufficient condition for αx to be a homeomorphism is that the mapping g 7→ gx
of G into X be open. Another hypothesis which guarantees this is
that G be compact and X be Hausdorff. In this case, the mapping
αx : G/Gx → X is closed, and thus a homeomorphism. A transitive
G-space X such that αx is a homeomorphism for every x ∈ X is called
a topological homogeneous space (of the topological group G).
Example 13.1.4 Let α > 0 be an irrational number. Then G = Z, the
group of integers, acts on the circle S1 by rotating through 2nαπ. By
Ex. 12.3.5, the orbit of 1 ∈ S1 under G is dense and therefore the orbit
of any x ∈ S1 is dense. It follows that G(x) cannot be a coset space of
G.
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353
Here is another example showing that the continuous bijection αx :
G/Gx → G (x) may not be a homeomorphism.
Example 13.1.5 (Irrational Flow) Let p : R1 → S1 denote the exponential map r 7→ e2rπı . Given a fixed number α ∈ R, we define an action
of G = R on the torus T = S1 × S1 by
(r, (x, y)) 7→ (p(r)x, p(αr)y).
When α is an irrational, the stabilizer of every point in T is the trivial
group. For, (x, y) = (p(r)x, p(αr)y) implies that both r and αr are
integers. Since α is an irrational number, this is possible only if r =
0. Thus the action of G on T in this case is free; it is referred to
as an “irrational flow” on the torus. From the multiplication in T ,
it follows immediately that the orbits through different points of T
are translations of the orbit through the point v = (1, 1) , and this
orbit is the image of the line L = {(r, αr) |r ∈ R} under the canonical
map π : R2 → T, (r, s) 7→ (p (r) , p (s)). Note that L is a subgroup
of R2 . So π (L) is a subgroup of T. We show that it is dense in T.
Observe that π −1 π (L) = L + N = L + H, where N = Z × Z and
H = {(0, n − mα)|n, m ∈ Z}. Since L + N is a subgroup of R2 and
contains both L and H = {0} × R, it contains L + H = R2 . It follows
that L + N is dense in R2 , and so π (L) is dense in T. The orbits of
this transformation group are referred to as the “flow lines.” As the
isotropy subgroup of G at the point v ∈ T is the trivial group, G/Gv
is locally compact. But the orbit G (v) = π (L) is not locally compact.
For, if K is a compact subset of π (L), then K is closed in T, and if K ◦
is nonempty, then there would be limit points of K in its complement,
a contradiction.
Observe that if α is a rational number, say α = m/n, where
(m, n) = 1, then π (L) is closed in T, being the image of I under
the continuous map t 7→ (p(nt), p(mt)).
Let X and Y be G-spaces with the actions denoted by ϕ and ψ,
respectively. A continuous function f : X → Y is called an equivariant
map (or a G-map) if f (ϕg (x)) = ψg (f (x)) for all g ∈ G and x ∈ X.
An equivariant map f : X → Y which is also a homeomorphism is
called an equivalence of G-spaces. If f : X → Y is an equivalence of
G-spaces, then f −1 : Y → X is also equivariant:
(
)
ϕg f −1 (f (x)) = ϕg (x) = f −1 (ψg (f (x))).
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Elements of Topology
If X is a G-space, then the mapping αx : G/Gx → X, gGx 7→ gx,
is equivariant with respect to the left translation of G on G/Gx for
every x ∈ X. Moreover, if X is Hausdorff and a compact group G acts
on X transitively, then each αx is an equivalence of G-spaces. Thus it
is desirable to determine the equivariant maps on the coset spaces of
G. Clearly, the stabilizer of gx is gGx g −1 , and the isotropy subgroups
at points in the same orbit constitute a complete conjugacy class of
subgroups of G.
We now discuss two elementary methods of constructing new transformation groups out of old ones. First, given a G-space X, consider
a subgroup H ⊂ G. Then X is also an H-space in the obvious way.
The action of H on X is just the restriction of the action of G on
X to H × X. We say that it is obtained by restricting the group G
to H. Obviously, the restriction of an effective action is effective. Thus
two actions which are equivalent can not be topologically distinguished
from one another, and they are regarded as essentially the same.
Example 13.1.6 By restricting the natural action of GL (n, R) on Rn to
the orthogonal group O (n) ⊂ GL (n, R), we obtain an effective action
of O (n) on Rn . Unlike Ex. 13.1.3, there are many orbits in this transformation group. Because orthogonal transformations preserve lengths
of vectors, an element of O (n) sends a vector x ∈ Rn to a vector of
the same length. Conversely, if the nonzero vectors x, y ∈ Rn have the
same length, then there are orthonormal bases {ui } and {vi } of Rn
with u1 = x/ ∥x∥ and v1 = y/ ∥y∥. The linear map f : ui 7→ vi is
an orthogonal transformation on Rn . So its matrix A in the standard
basis is an element of O (n) and carries x into y. Thus the orbits are
concentric spheres (the origin being a sphere of radius zero).
Next, consider the unit sphere Sn−1 ⊂ Rn . The discussion in the
preceding example shows that Ax ∈ Sn−1 for all A ∈ O (n) and x ∈
Sn−1 . So there is an action of O (n) on Sn−1 obtained by restricting
the action of O (n) to Sn−1 . This suggests another method of obtaining
transformation groups from a given one.
Definition 13.1.5 Let X be a G-space. A subspace Y ⊂ X is called
invariant under G (or a G-subspace) if gy ∈ Y for all g ∈ G and y ∈ Y.
We write G (Y ) = {gy|g ∈ G and y ∈ Y }. With this notation, Y is
invariant ⇔ G (Y ) = Y. If this is the case, then we have an induced
TRANSFORMATION GROUPS
355
action of G on Y ; thus Y becomes a G-space. Note that the orbits in
a G-space X are the smallest invariant subsets of X.
Example 13.1.7 The action of the special orthogonal group SO (n) on
Sn−1 obtained by restricting the standard action of O (n) (by matrix
multiplication) to SO (n) is transitive, except for the case n = 1. For
n = 2, we have already seen that SO (2) ≈ S1 . Notice that if x =
cos θ + ı sin θ, then the matrix
[
]
cos θ − sin θ
sin θ cos θ
belongs to SO (2) and carries e1 = (1, 0) to x. If n > 2 and x ∈
Sn−1 , then we choose an orthonormal basis {x = f1 , . . . , fn } of Rn ,
and consider the matrix A of this basis relative to the standard basis
{e1 , . . . , en }. It is clear that either A or the matrix B obtained from
A by interchanging the last two columns belongs to SO (n) , and x =
Ae1 = Be1 . This shows that the orbit of e1 is all of Sn−1 .
It is immediate from the preceding example that the orbits of the
standard action of SO (n) on Rn are concentric spheres for n > 1.
We now turn to discuss connectivity of the topological groups
SO (n) , SU (n), U (n) and Sp (n). All these groups are connected.
First, consider the group SO (n). Obviously, this is true for n = 1,
since SO (1) = {1}. The proof for n > 1 is by induction on n. Once
we know the stabilizer of e1 = (1, 0, . . . , 0) ∈ Sn−1 , we obtain a relation between SO (n) and Sn−1 , since SO (n) acts transitively on Sn−1 .
Suppose that A = (aij ) ∈ SO (n) , and Ae1 = e1 . It is readily checked
that a11 = 1, ai1 = 0 = a1i for 2 ≤ i ≤ n. Thus A has the form
(
)
1 0
A=
,
0 B
where B ∈ SO (n − 1) . Conversely, it is clear that any matrix of this
form fixes e1 . This gives an isomorphism between SO (n − 1) and the
isotropy subgroup of SO (n) at e1 . Accordingly, we can identify this
subgroup of SO (n) with SO (n − 1), and regard SO (n − 1) ⊂ SO (n).
Hence the coset space SO (n) /SO (n − 1) ≈ Sn−1 . Since the spheres of
all positive dimensions are connected, the following lemma shows that
SO(n) is connected for every n.
Lemma 13.1.6 Let G be a topological group and H ⊂ G a subgroup.
If H and the coset space G/H are both connected, then so is G.
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Elements of Topology
Proof. Suppose that G = U ∪ V is a disconnection of G. By Proposition
12.3.1, the quotient map π : G → G/H, π (g) = gH, is open. So both
π (U ) and π (V ) are open in G/H. Since G/H is connected, we find an
element g ∈ G such that π (g) ∈ π (U ) ∩ π (V ). Then U ∩ gH ̸= ∅ ̸=
V ∩ gH. As gH ⊂ U ∪ V , we have a disconnection of gH. But gH,
being a continuous image of the connected set H, is connected. This
contradiction proves the lemma.
♢
To establish the connectivity of the groups SU (n), U (n) and
Sp (n), we proceed as above. Notice that SU (1) = {1}. U (1) = S1 ,
and Sp (1) = S3 . As in the case of SO (n), we see that the standard action of SU (n) (n > 1) on the sphere S2n−1 is transitive,
and the isotropy subgroup at (1, 0, . . . , 0) ∈ S2n−1 is isomorphic to
SU (n − 1). Similarly, U (n) acts on S2n−1 transitively with the isotropy
subgroup isomorphic to U (n − 1) at (1, 0, . . . , 0). Therefore we have
U (n) /U (n − 1) ≈ S2n−1 ≈ SU (n) /SU (n − 1). It is also easily seen
that the action of Sp (n) on S4n−1 by matrix multiplication is transitive, and the isotropy subgroup at (1, 0, . . . , 0) ∈ S4n−1 is Sp (n − 1).
So Sp (n) /Sp (n − 1) ≈ S4n−1 . It follows that the induction applies in
each case.
In the case of O(n), the coset space O (n) /O (n − 1) ≈ Sn−1 , since
the action of O (n) on Sn−1 is transitive and the isotropy subgroup
of O (n) at e1 is isomorphic to O (n − 1). However, the components of
O(n) are SO(n) and its complement {A ∈ O(n)|det(A) = −1}.
In the literature, the group SO(n) is usually referred to as the
“rotation group” of Rn . In fact, a rotation of the euclidean space Rn is,
by definition, an element of SO(n). We give here a geometric meaning
to this terminology based on direct observation of the motion of a rigid
body, and provide a justification for the above definition of rotation.
The position of a rigid body is completely determined by describing the
location of one of its particles and the orientation of the body. To find
the change in the orientation of a moving rigid body, one of the several
ways is to study the change in a cartesian set of coordinate axes fixed
in the body relative to the coordinate axes parallel to the reference
frame fixed in the space and with the same origin as that of the body
set of axes. The obvious facts about the physical motion of a rigid body
are that the distance between any pair of particles of the body remains
constant throughout the motion, by definition, and each particle of the
body traverses a continuous curve (path). Also, it occurs over a period
TRANSFORMATION GROUPS
357
of time. A ‘rigid motion’ of a euclidean space (of dimension ≥ 3) is an
abstraction of the motion of a rigid body. Accordingly, we introduce
the following.
Definition 13.1.7 A proper rigid motion of the euclidean space Rn is
a function ϕ from an interval [0, b] ⊆ R to the group I (Rn ) of isometries
of Rn such that ϕ(0) is the translation by ϕ(0)(0) and, for each x ∈ Rn ,
the mapping [0, b] → Rn , t 7→ ϕ(t)(x), is continuous.
Since the elements of I (Rn ) are continuous functions Rn → Rn , it
can also be assigned the relative topology induced from the pointwise
topology for C(Rn , Rn ). It is immediate that ϕ is continuous relative
to this topology for I (Rn ). We show that it coincides with the metric
topology on I (Rn ).
As we have seen, for each f ∈ I (Rn ) , there is an (n × n) orthogonal
matrix A(f ) such that f (x) = A(f )x + f (0) for every x ∈ Rn . Then
the metric on I (Rn ) is given by
√
∑ g
∑ g
2
2
ci − cfi ,
ai,j − afi,j +
d(g, f ) =
i
i,j
(
where A(f ) =
)
( )
afij , f (0) = cfi , etc. Let ex : I (Rn ) → Rn ,
f → f (x), be the evaluation map at x ∈ Rn . Then the sets e−1
x (U ),
where U ⊆ Rn is open and x ∈ Rn , form a subbase for the pointwise topology for I (Rn ). If f ∈ e−1
x (U ), then there is a real ϵ > 0
such that B(f (x); ϵ) ⊆ U. Clearly, d(g, f ) < ϵ/(1 + ∥x∥) implies that
∥g(x) − f (x)∥ ≤ ∥A(g) − A(f )∥ ∥x∥ + cg − cf < ϵ. So g(x) ∈ U, and
it follows that e−1
x (U ) contains an open ball about f. Thus the metric topology for I (Rn ) is finer than the pointwise topology. To see the
converse, consider an open ball B(f ; r) about f in I (Rn ). For each
j
1 ≤ j ≤ n, denote the unit vectors (0, . . . 0, 1, 0, . . . , 0) in Rn by uj and
the zero vector by u0 . Then for any real ϵ > 0, the set
V = {g ∈ I (Rn ) : ∥g (uj ) − f (uj )∥ < ϵ for all 0 ≤ j ≤ n}
358
Elements of Topology
is a basic nbd of f in the pointwise topology for I (Rn ). We have
v
u n
2
u∑ g
t
aij − afij
= ∥(A(g) − A(f )) uj ∥
i=1
≤
(A(g) − A(f )) uj + cg − cf + cg − cf
= ∥g(uj ) − f (uj )∥ + ∥g(0) − f (0)∥
for each
√ j = 1, . . . , n. Consequently, V is contained in B(f ; r) when
ϵ = r/ 4n + 1. It follows that the pointwise topology for I (Rn ) is finer
than the metric topology, and we conclude that the two topologies for
I (Rn ) are identical.
Thus, we have proved the following.
Proposition 13.1.8 A proper rigid motion of the euclidean space Rn
is a continuous function ϕ from an interval [0, b] to the euclidean group
I (Rn ) such that ϕ(0) is the translation by ϕ(0)(0).
We remark that the group I (Rn ) can also be given the compactopen topology and, by Theorem 13.1.2, it is a topological group (since
its elements are the homeomorphisms f : Rn → Rn such that ∥f (x) −
f (y) ∥ = ∥x − y∥). But this, too, agrees with the two topologies for
I (Rn ) discussed above. In fact, we already know that the compact-open
topology is finer than the pointwise topology. Moreover, the metric
topology for I (Rn ) is finer than the compact-open topology, since the
evaluation mapping e : I (Rn ) × Rn → Rn is continuous in the the
metric topology on I (Rn ).
Before we give a technical definition of “rotation” of Rn , let us think
of the literal meaning of the term. A rotation of an object is a motion
in which all its particles follow circles about a common fixed line or
point. Accordingly, we define
“a rotation of Rn to be a proper rigid motion ϕ such that a point
x0 ∈ Rn remains fixed under every transformation ϕ(t), 0 ≤ t ≤ b.”
If the fixed point x0 of the motion is chosen as the origin, then each
isometry ϕ(t) is an orthogonal transformation, by Lemma 12.4.2. Also,
ϕ(0) is the identity map. Since the determinant function is continuous
and det[ϕ(0)] = 1, each ϕ(t) is a special orthogonal transformation.
Thus the matrix of ϕ(t) relative to an orthonormal basis is an element of
SO (n). It follows that a rotation of Rn about the origin is a continuous
map ϕ from an interval [0, b] to the topological group SO(n), where
TRANSFORMATION GROUPS
359
SO(n) acts on Rn by the matrix multiplication and ϕ(0) is the identity
matrix.
In the plane R2 , the above definition of rotation carries the usual
sense of the term. To see this, suppose that ϕ : [0, b] → SO(2) is a continuous function such that ϕ(0) is the identity matrix. Since each transformation ϕ(t) is distance-preserving, the motion of R2 determined by
ϕ can be described by considering the motion of any one of the points
of R2 . Let P be an arbitrary point of R2 other than the origin. If the
transformation ϕ(t) takes the point P into a point P (t), then t 7→ P (t)
is a continuous function from [0, b] to R2 and ∥P (t)∥ = ∥P ∥. Also,
P (0) = P , for ϕ(0) is the identity matrix. It follows that the point
P (t) moves continuously in a circle starting from its initial position. So
ϕ determines a rotation of R2 .
On the other way round, a rotation of the plane R2 about the
origin O (in the usual sense) gives a continuous function of an interval
into SO(2). To see this, consider such a rotation through an angle
β > 0. Then, for each 0 < α ≤ β, there is an isometry fα which
fixes the origin O and sends a point P (̸= O) to the point P ′ such
that the measure of the directed angle from the ray through P to the
ray through P ′ is α. If the point P has the coordinates (x, y) with
respect to a cartesian coordinate system, then its image P ′ under fα
has coordinates (x cos α − y sin α, x sin α + y cos α). It follows that the
transformation fα is just the left multiplication by the matrix
[
]
cos α − sin α
sin α
cos α
which has determinant 1. Thus we have the function α 7→ [fα ] of [0, β]
into SO(2), where the transformation f0 corresponding to the angle
zero is the identity mapping of R2 . Clearly, this function is continuous.
We remark that the resultant of two successive rotations is defined
as follows: Suppose that ϕ : [0, b] → SO(n) and ψ : [0, c] → SO(n)
are two successive rotations of Rn . Then their resultant is the map
ρ : [0, b + c] → SO(n) defined by
{
ϕ(t)
for 0 ≤ t ≤ b, and
ρ(t) =
ψ(t − b) · ϕ(b) for b ≤ t ≤ b + c.
The following theorem justifies the description of the elements of
SO(n) as “rotations.”
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Elements of Topology
Theorem 13.1.9 For every n, SO(n) is path-connected.
Proof. The proposition is proved by induction on n. It is known that
SO(1) = {1} and SO(2) ∼
= S1 , which are obviously path-connected.
Assume that n > 1 and SO(n − 1) is path-connected. Let A ∈ SO(n)
be arbitrary. It suffices to show that there is a path in SO(n) joining
the identity matrix I and [A. Let e1], . . . , en be the standard basis of
1
0
Rn . If Ae1 = e1 , then A =
, where A1 ∈ SO(n − 1). By our
0 A1
induction assumption, there is a path in SO(n − 1) joining I and A1 .
The composition
[
] of this path with the embedding SO(n−1) → SO(n),
1 0
M 7→
, gives us a desired path in SO(n). So assume that
0 M
Ae1 ̸= e1 . Then the vectors e1 and Ae1 determine a plane P. Since
∥Ae1 ∥ = ∥e1 ∥ = 1, and SO(2) acts transitively on S1 , there exist
matrices B ∈ SO(n) and C ∈ O(n) such that Be1 = Ae1 and

cos θ − sin θ 0 · · ·

 sin θ


−1
0
C BC = 

 .
 ..

0
cos θ
0 ···
0
1 ···
..
.
..
.
0
0 ···
..
.
0


0 


0 .

.. 
. 

1
Notice that the vectors Be2 , . . . , Ben are all orthogonal to Be1 = Ae1
and Ae2 , . . . , Aen are all orthogonal to Ae1 . Accordingly, they all lie
in the (n − 1)-dimensional euclidean subspace of Rn othogonal to Ae1 .
The transformation Bei 7→ Aei is obviously orthogonal and if D denotes the matrix of this transformation relative
to] the ordered basis
[
1
0
Be1 , . . . , Ben , then D has the form D =
and BD = A. So
0 D1
det(D) = +1; consequently, D1 ∈ SO(n − 1). By our induction assumption, there is a path in SO(n − 1) joining I and D1 , and hence a
path in SO(n) joining I and D. Since the multiplication in SO(n) is
continuous, we have a path in it connecting B to BD = A. Moreover,
TRANSFORMATION GROUPS
there is a path in SO(n) connecting I to B. To

cos tθ − sin tθ 0 · · ·

 sin tθ cos tθ 0 · · ·


0
0
1 ···
Bt = C 

 .
..
.. . .
 ..
.
.
.

0
0
0 ···
361
see this, put

0

0 


0  C −1

.. 
. 

1
for each 0 ≤ t ≤ 1. Then t 7→ Bt is a path in SO(n) connecting I to
B. The product of this path and the path connecting B to A is a path
in SO(n) connecting I to A. This completes the proof.
♢
In the same vein, we understand the geometry of SO (3).
Proposition 13.1.10 The topological group SO (3) is homeomorphic
to the real projective 3-space RP3 .
Proof. Since the space RP3 is just the quotient group S3 / {−1, 1}, we
show that SO (3) ∼
= S3 /{1, −1} as topological groups. Let S3 be regarded as the group of unit quaternions. We define a continuous homomorphism of S3 onto SO (3). Given x ∈ S3 , consider the mapping
fx : H → H defined by fx (q) = xqx̄. This is an R-isomorphism of H
onto itself, for q 7→ x̄qx is its inverse. Since ∥x∥ = 1, and the norm of the
product is the product of the norms, ∥fx (q)∥ = ∥q∥. Hence the transformation fx is orthogonal; in fact, it is a special orthogonal transformation of R4 ∼
= H (verify). We can identify R3 with the subspace of H consisting of quaternions q whose real part is zero. Then for every q ∈ R3 ,
we have 2Re (xqx̄) = xqx̄+xq̄x̄ = x (q + q̄) x̄ = 0. So xqx̄ ∈ R3 , and the
restriction of fx to R3 defines an isomorphism gx : R3 → R3 , q 7→ xqx̄.
Since fx (1) = 1, and R3 is orthogonal to the vector 1, it follows that
detgx = detfx = 1. Thus gx is a special orthogonal transformation of
R3 . Let [gx ] denote the matrix of gx in the basis {ı, ȷ, k} of R3 . Then we
have a mapping ϕ : S3 → SO (3) defined by ϕ (x) = [gx ]. If x, y ∈ S3 ,
then gx (q) − gy (q) = (x − y) qx̄ + yq (x̄ − ȳ), which implies that
∥gx (q) − gy (q)∥ ≤ ∥x√− y∥ ∥q∥ for every q ∈ R3 . From this, we deduce
that ∥[gx ] − [gy ]∥ ≤ 2 3 ∥x − y∥, and the continuity of ϕ follows. This
preserves multiplications, for gxy (q) = xyqxy = xyq ȳx̄ = (gx ◦ gy ) (q).
Thus ϕ is a (continuous) homomorphism of S3 into SO (3). Since S3 is
compact and SO (3) Hausdorff, we have a continuous closed injection
S3 /ker(ϕ) → SO(3) induced by ϕ.
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Elements of Topology
Next, we determine the kernel of ϕ. Observe that a quaternion
commutes with the pure quaternions ı, ȷ and k if and only if it is real.
Accordingly, gx = 1 ⇔ x = ±1, so the kernel of ϕ is {−1, 1}.
Finally, it remains to show that ϕ is surjective. Let Hı , Hȷ and Hk
be the subgroups of SO (3) leaving fixed the quaternions ı, ȷ and k,
respectively. We observe that any element A ∈ SO (3) is a product of
elements from these subgroups. The vector Ak can be rotated into the
(ı, k)-plane by a rotation B in Hk . Then we can send (BA) k to k by
a rotation C in Hȷ . Thus (CBA)k = k so that CBA = B ′ belongs to
Hk . This implies that A = B −1 C −1 B ′ . Since ϕ is a homomorphism,
it suffices to show that each of these subgroups Hı , Hȷ and Hk is
contained in the image of ϕ. If A in Hı , then


1
0
0



0
cos
θ
−
sin
θ
A=


0 sin θ
cos θ
for some 0 ≤ θ < 2π. Now, for x = cos θ/2 + ı sin θ/2 ∈ S3 , we have
gx (ı) = ı, gx (ȷ) = ȷ cos θ + k sin θ and gx (k) = −ȷ sin θ + k cos θ so that
[gx ] = A. Similarly, it can be shown that every element in Hȷ and Hk
has a preimage in S3 under ϕ. It follows that SO (3) ∼
= S3 /ker(ϕ), and
this completes the proof.
♢
It is noteworthy that rotations and translations in R3 are physically
performable motions. On the other hand, there are rigid motions of R3
which are not physically performable. To see such an example, consider
a hyperplane H = {x ∈ Rn |⟨x, u⟩ = p} (i.e., a translation of an (n − 1)dimensional subspace) in Rn , where ∥u∥ = 1. If x ∈ Rn , then the
projection of x into H is a vector v such that v + pu lies in H and
x = v + (p + q) u for some q ∈ R. We define ϕ (x) by ϕ (x) = x − 2qu.
Since ⟨v, u⟩ = 0 and ⟨x, u⟩ = p + q, we have ϕ (x) = x − 2 (⟨x, u⟩ − p) u.
It is easily verified that ϕ is an isometry of Rn onto itself, called the
reflection in the plane H. The reflection through the line Ru is given by
ψ (x) = x−2v = x−2 (x − (p + q) u) = −x+2 (p + q) u = −ϕ (x)+2pu.
(See Figure 13.2 below.)
If the hyperplane H passes through the origin (i.e., p = 0),
then Rn = H ⊕ Ru, where u is the unit vector orthogonal to H.
In this case, reflection in the hyperplane H is the transformation
ϕ : (v, w) 7→ (v, −w). Clearly, ϕ is an orthogonal transformation of
Rn . If {u1 , u2 , . . . , un−1 } is an orthonormal basis of H, then the matrix
TRANSFORMATION GROUPS
363
Á(x)
v+pu
x
pu
v
H
Ã(x)
0
FIGURE 13.2: Reflection in a hyperplane.
of the reflection ϕ through H in the basis {u1 , u2 , . . . , un−1 , un = u} is





A=




1 0 ···
0
0 1 ···
0
..
.
.
..
.
0 0 ···
1
0 0 ···
0 −1
..
.
..
0


0 

.. 

. 

0 

and the matrix of the reflection ψ through Ru is −A. Let {e1 , . . . , en }
be the standard basis of Rn and B be the matrix of the basis {ui } with
respect to {ei }. Then B ∈ O (n) and the matrix of the reflection ϕ in
the basis {ei } is the matrix BAB −1 . Conversely, any such matrix fixes
the orthogonal complement of the vector Ben , and can be visualised
as a reflection through this hyperplane. It is obvious that a reflection
defines a Z2 -action on Rn with every point of the hyperplane being
fixed.
The composition of a reflection through a line and a translation
along that line is called a glide reflection. Notice that reversing the
order of combining gives the same result. We may consider a reflection
a special case of glide reflection, where the translation vector is the zero
vector. In the euclidean plane R2 , the isometry (x, y) 7→ (x + 1, −y)
consists of the reflection on the x-axis, followed by translation of one
unit parallel to it. The isometry group generated by a glide reflection in
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Elements of Topology
R2 is an infinite cyclic group. It fixes a system of parallel lines. Observe
that combining two equal glide reflections gives a pure translation with
a translation vector that is twice that of the glide reflection, so the even
powers of the glide reflection form a translation group.
Exercises
1. Prove that the group of all translations of the topological group S1
furnished with the compact-open topology is isomorphic to S1 .
2. Let G be a topological group and X be a G-space. For H ⊂ G and
A ⊂ X, set H (A) = {ha|h ∈ H, a ∈ A}. Prove:
(a) If H ⊂ G and A ⊂ X are open, then H (A) is open in X.
(b) If H ⊂ G is compact and A ⊂ X is closed, then H (A) is closed in
X.
(c) If H ⊂ G and A ⊂ X are compact, then H (A) is compact.
3. If G is a compact Hausdorff group, prove that every nbd U of e in G
contains a nbd V of e which is invariant under conjugation.
4. Let X be a G-space. Prove that Ggx = gGx g −1 for every g ∈ G.
5. Let X be a transitive G-set and x, y ∈ X. Prove that
(a) Gx = Gy ⇔ there exists an equivariant bijection f : X → X such
that f (x) = y, and
(b) the quotient group N (Gx )/Gx is isomorphic to the group of all
equivariant bijections X → X.
6. Let G be a topological group and H, K be subgroups of G. Prove that
there exists an equivariant map G/H → G/K if and only if H is conjugate to a subgroup of K.
7. Let G be a compact Hausdorff group and H, K ⊂ G be closed subgroups.
If there exist equivariant maps G/H → G/K and G/K → G/H, prove
that each is an equivalence and H is conjugate to K.
8. If G is a compact Hausdorff group and H ⊂ G is a closed subgroup,
prove that each equivariant map G/H → G/H is a right translation by
an element of N (H), the normalizer of H in G, and is an equivalence
of G-spaces.
9. Prove that the action (gH, aH) 7→ gaH of N (H) /H on G/H is free.
10. If G is a compact Hausdorff group and H ⊂ G is a closed subgroup, show
that the topological groups HomeoG (G/H)co of all self-equivalences of
the G-space G/H and N (H) /H are isomorphic.
TRANSFORMATION GROUPS
13.2
365
Orbit Spaces
With each transformation group, there is associated a new space,
called the “orbit space.” Let X be a G-space. For any two points x, y ∈
X, it is easily checked that the orbits G (x) and G (y) are either equal or
disjoint. Consequently, there is an equivalence relation ∼ on X whose
equivalence classes are precisely the orbits of the given action. Thus the
relation ∼ is given by x ∼ y if there exists a g ∈ G such that y = gx.
Definition 13.2.1 The orbit space of a G-space X is the quotient
space X/ ∼ whose elements are the orbits of the action, and is denoted
by X/G.
This section concerns the study of topological relationships between
the total space X and the orbit space X/G.
Let G be a topological group and H ⊆ G a subgroup. Then the
right action G × H → G, (g, h) 7→ gh, of H on G is free and its orbits
are precisely the left cosets of H in G. So the orbit space G/H is the
space of left cosets gH. Notice that the notation G/H stands for the
same space whether interpreted as an orbit space or as the space of left
cosets of H in G. Similarly, the orbit space of the left action (h, g) 7→ hg
of H on G is the space of the right cosets Hg.
Example 13.2.1 The group Z of integers acts on R1 by translation:
(n, x) 7→ n + x. The orbit space R1 /Z is the circle S1 . The mapping
R1 → S1 , x 7→ e2πıx , induces a desired homeomorphism.
Example 13.2.2 The group G = Z ⊕ Z acts on R2 by ϕ(m,n) : (x, y) 7→
(m + x, n + y). The orbit space R2 /G is homeomorphic to the torus.
The group G may be considered as the group of isometries of R2 generated by two translations λ(1,0) and λ(0,1) .
It is obvious that if the action is transitive, then the orbit space
is a one-point space. The orbit space of each of the transformation
groups in Example 13.1.3 is the Sierpinski space. The orbit space of the
transformation group in Example 13.1.4 and that of Example 13.1.5,
both, have the trivial topology.
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Elements of Topology
Example 13.2.3 In Example 13.1.6, the mapping Rn /O (n) → [0, ∞)
which assigns to each sphere its radius is a one-to-one correspondence.
This mapping is induced by the continuous open map Rn → [0, ∞),
x 7→ ∥x∥, and hence is a homeomorphism between the orbit space
Rn /O (n) and the ray [0, ∞). We obtain similar results for the actions
of SO (n) on Rn (n > 1), U (n) on Cn , and Sp (n) on Hn .
Example 13.2.4 The group G = Z2 acts freely on Sn by the antipodal
map and the orbit space Sn /G is the real projective space RPn .
Example 13.2.5 Let p and q be relatively prime integers, and regard
S3 as the subspace {(z0 , z1 ) ||z0 |2 + |z(1 |2 = 1} ⊂ C2 . Then
the map
)
h : S3 → S3 defined by h (z0 , z1 ) = e2πı/p z0 , e2qπı/p z1 is a homeomorphism with hp = 1. Thus h generates a cyclic group of order p and
we have a free action of G = Zp on S3 given by g (z0 , z1 ) = hn (z0 , z1 ),
where g is the residue class of
( the3 )integer n modulo p. The orbit space
of the transformation group G, S is called a lens space and is denoted
by L (p, q).
We now return to the general discussion. Let X be a G-space. Then
the natural map π : X → X/G, x 7→ G (x), is usually referred to as
the orbit map. This is continuous, by the definition of the topology for
X/G. We also have
Proposition 13.2.2 Let X be a G-space. The orbit map π : X →
X/G is open.
Proof. Let U ⊆ X be open. If ϕ denotes the action of G ∪
on X, then
−1
gU = ϕg (U ) is open in X for every g ∈ G. So π π (U ) = g∈G gU is
open in X. By the definition of the topology on X/G, π (U ) is open,
and the proposition follows.
♢
It follows from the preceding proposition that some topological
properties, for example, compactness, connectedness, local compactness, local connectedness, being a first or second countable space, etc.
are transmitted from X to X/G. For actions of compact groups, the
following theorem shows that some more properties such as Hausdorffness, regularity, complete regularity, normality, paracompactness, etc.
are also transmitted from X to X/G.
Theorem 13.2.3 Let G be a compact group, and X be a G-space.
Then the orbit map π : X → X/G is proper.
TRANSFORMATION GROUPS
367
Proof. Since G is compact, so is each orbit G (x), x ∈ X. By Theorem
6.5.2, it suffices to prove that π is a closed mapping. If A ⊆ X, then we
have π −1 π (A) = G (A). So we need to establish that G (A) is closed
for every closed A ⊆ X. Assume that A is closed and let x ∈ X − G (A)
be arbitrary. Then, for each g ∈ G, g (A) = {ga|a ∈ A} is closed in
X; consequently, X − g (A) is open. By the continuity of the action
G × X → X, we find open nbds Ng of e ∈ G and Ug of x such that
Ng (Ug ) ⊂ X − g (A), where Ng (Ug ) denotes the image of Ng × Ug
under the action. This implies that Ug does not meet Ng−1 g (A). Since
G is compact, the open covering {Ng−1 g|g ∈ G} has finite subcovering.
Accordingly,
many elements g1 , . . . , gn in G such
∪n there exist finitely ∩
n
that G = i=1 Ng−1
g
.
Put
V
=
i
i=1 Ugi . Then V is obviously a nbd
i
−1
of x, and V ∩ Ngi gi (A) = ∅ for every i = 1, . . . , n. So V ∩ G (A) = ∅,
which implies that X − G (A) is a nbd of x. It follows that G (A) is
closed, and this completes the proof.
♢
In view of the results of §7.6, we see that some properties like compactness, localcompactness, paracompactness, etc., are also acquired
by X from X/G.
Next, we see that it is sufficient to consider only effective actions for
most purposes. If ϕ is an action of G on X and K is the kernel of the
associated homomorphism ϕ̂ : G → Homeo (X), then G/K acts on X
in a canonical way. In fact, we define ψ : G/K × X → X by setting
ψ (gK, x) = ϕ (g, x) . It is readily verified that ϕ is well-defined, and
satisfies the conditions (a) and (b) of Definition 13.1.1. To check the
continuity of ψ, we note that ϕ = ψ ◦ (π × 1), where π : G → G/K is
the natural projection, and 1 is the identity map on X. Consequently,
ψ −1 (U ) = (π × 1) ϕ−1 (U ) for every U ⊆ X. Since π is open, it follows
that ψ −1 (U ) is open whenever U ⊆ X is open. So ψ is continuous, and
defines an action of G/K on X. This canonical action of G/K on X
is clearly effective. The subgroup K ⊆ G is referred to as the “kernel
of the action ϕ.” We note that if X is Hausdorff, then K is a closed
normal subgroup of G. For, if ⟨gα ⟩ is a net in K converging to g in G
and x ∈ X, then gα x → gx, by the continuity of the action. So gx = x
for all x ∈ X ∩whence g ∈ K. Also, observe that the kernel K of the
action is just x∈X Gx .
We close this chapter by establishing the fact that the orbit space
of a G-space X and that of the action of G/K on it, where K is the
kernel of the action, are the same upto homeomorphism. To this end,
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Elements of Topology
let N be a normal subgroup of G. Then, by restricting the action of
G to N , we can consider X as an N -space. Also, G/N , equipped with
the quotient structure, is a topological group. We observe that there
is an induced action of G/N on the orbit space X/N . To see this, we
define a function θ : G/N × X/N → X/N by θ (gN, N (x)) = N (gx) .
Clearly, θ is single-valued, and we have the commutative diagram
- X
G×X
π
?
θ
G/N × X/N
?
- X/N
FIGURE 13.3: The canonical action of G/N on the orbit space of the Gspace X with respect to N .
Since the mappings G × X → X and X → X/N are continuous, and
the map G × X → G/N × X/N is open and surjective, we see that θ
is continuous. It is routine to verify that θ satisfies the conditions of
Definition 13.1.1, and thus defines an action. This is referred to as an
induced action of G/N on the orbit space X/N . We have the following.
Proposition 13.2.4 Let X be a G-space and N a normal subgroup
of G. Then the orbit space X/G is homeomorphic to the orbit space of
X/N under the induced action of G/N .
Proof. Denote the orbit of N (x) under G/N by [N (x)], and define a
map ξ : X/N
G/N → X/G by ξ ([N (x)]) = G (x). It makes the diagram
X/N πG/N
πN
X
πG
?
(X/N )/(G/N )
?
ξ X/G
FIGURE 13.4: Proof of Proposition 13.2.4.
TRANSFORMATION GROUPS
369
commutative, where πN , πG and πG/N are all natural projections. Since
these maps are continuous, open surjections, it follows that ξ is continuous and open. It is obvious that ξ is a bijection and, therefore, a
homeomorphism.
♢
Exercises
1. Find actions of Z2 on the torus such that the orbit space is (a) sphere,
(b) cylinder, (c) Klein bottle and (d) torus.
2. Let G be connected group and X be a G-space. If X/G is connected,
show that X is also connected.
3. Show that CPn (resp. HPn ) is the orbit space of a free action of S1
(resp. S3 ) on S2n+1 (resp. S4n+3 ).
4. Let G be a topological group, and let f : X → Y be a continuous
G-map. Prove:
(a) f induces a continuous map fˆ : X/G → Y /G.
(b) If f is an equivalence, then so is fˆ.
5. Let G be a compact group, and X be a locally compact Hausdorff Gspace. If f : X → X is an equivariant map such that πf = π, where
π : X → X/G is the orbit map, show that f is an equivalence.
6. Let X be a G-space and A be an invariant subset of X. Prove that the
canonical mapping A/G → X/G is an embedding.
7. Let G be a compact group and X be a G-space. If A is a closed invariant
subset of X, prove that each nbd of A contains a G-invariant nbd. (Note
the particular case A = G(x).)
8. Suppose that X and Y are G-spaces and f : X → Y is an equivariant
map. Show that, for every x ∈ X, Gx ⊆ Gf (x) and the equality holds if
and only if f is injective on the orbit G(x).
9. Let X be a G-space. A cross section for the orbit map π : X → X/G is
a continuous right inverse of π. A subset F ⊂ X is called a fundamental
domain of this G-space if the restriction of π to F is a bijection between
F and X/G.
Let G be a compact group and X be a G-space. Prove that a cross
section for the orbit map π : X → X/G exists if and only if there exists
a closed fundamental domain of X.
10. Let G be a compact group, and X and Y be G-spaces. Suppose that F is
a closed fundamental domain of X, and f ′ : F → Y is a continuous map
such that Gx ⊆ Gf ′ (x) for all x ∈ F . Then there is a unique equivariant
map f : X → Y such that f |F = f ′ .
Chapter 14
THE FUNDAMENTAL GROUP
14.1
14.2
14.3
14.4
14.5
14.1
Homotopic Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Fundamental Groups of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Some Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Seifert–van Kampen Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
371
383
397
408
424
Homotopic Maps
A standard problem in topology is the classification of spaces and
continuous functions up to homeomorphisms. For example, one would
like to know whether or not euclidean spaces of different dimensions
are homeomorphic; alternatively, one may be interested in finding a
topological property which can be used to distinguish between a circular disc and an annulus (i.e., a disc with a hole). An answer to such a
question is generally based on the intuitive idea of a continuous deformation. This concept is technically known as “homotopy.” Although
the relation of homotopy equivalence is cruder than the topological
equivalence, it is of central importance in topology because the understanding of homotopy types is a base from which to attack more subtle
questions involving topological type. In this section, we shall deal with
some elementary properties of this concept.
Definition 14.1.1 Let X and Y be spaces. Two continuous maps f, g :
X → Y are called homotopic, denoted by f ≃ g, if there exists a
continuous function
H : X × I → Y,
I = [0, 1],
such that H (x, 0) = f (x), H (x, 1) = g (x) for every x ∈ X. The map
H is referred to as a homotopy connecting f and g, and we often write
H : f ≃ g.
371
372
Elements of Topology
As we have seen in Chapter 11, given a homotopy H : X × I → Y,
for each t ∈ I, there is a continuous map ht : X → Y defined by
ht (x) = H (x, t) for every x ∈ X. Accordingly, a homotopy connecting
f and g gives a one-parameter family of continuous maps ht : X → Y
indexed by the numbers t ∈ I with h0 = f and h1 = g. If we consider
the parameter t as representing time, then, as time varies (over I),
the point ht (x), for each x ∈ X, moves continuously from f (x) to
g (x). Hence a homotopy from f to g describes the intuitive idea of
continuously deforming f into g, and ht may be thought of as describing
the deformation at time t. If the space C (X, Y ) of all continuous maps
X → Y is assigned the compact-open topology, then the function I →
C (X, Y ), t 7→ ht , becomes continuous, and the maps f and g lie in a
path-component of C (X, Y )co . Conversely, if X is a locally compact
Hausdorff space, then any continuous function h : I → C (X, Y )co
defines a homotopy between h (0) and h (1), by Theorem 11.2.11. Thus,
in this case, two continuous maps f, g : X → Y are homotopic if and
only if they belong to the same path-component of the space C (X, Y )co .
We remark that this is true even for k-spaces X (ref. Exercise 11.2.24).
Example 14.1.1 Let Y ⊆ Rn be a convex space and f, g : X → Y be
two continuous maps. Then H : X × I → Y , defined by H (x, t) =
(1 − t) f (x) + tg (x), is a homotopy connecting f to g.
Given two continuous maps f, g : X → Y, it is not always true that
f and g are homotopic.
Example 14.1.2 Let X = Y = Sn (the n-sphere), f be the identity
map, and g a constant map. Then f and g are not homotopic. This
claim is difficult to prove and needs more knowledge of algebraic topology than we intend to introduce in this book. However, we will justify
this for n = 1 later on.
The problem whether or not two given continuous maps f, g : X →
Y are homotopic is actually a special case of the extension problem.
For, if we define a map
h : Z = (X × {0}) ∪ (X × {1}) → Y
by setting h (x, 0) = f (x), h (x, 1) = g (x) for every x ∈ X, then h is
continuous, and an extension H of h over X × I is a homotopy between
f and g, and vice versa. So f ≃ g if and only if h has an extension over
X × I.
THE FUNDAMENTAL GROUP
373
Proposition 14.1.2 Let X and Y be spaces, and C (X, Y ) the set of
all continuous maps X → Y. The relation ≃ on C (X, Y ) is an equivalence relation.
Proof. To show that the relation ≃ is reflexive, given a continuous map
f : X → Y define H : X × I → Y by H (x, t) = f (x) for every x ∈ X
and t ∈ I. Then H : f ≃ f , obviously.
To prove the symmetry, suppose that F : f ≃ g, where f, g : X →
Y are continuous. Define H : X × I → Y by H (x, t) = F (x, 1 − t) for
every x ∈ X and t ∈ I. Then H : g ≃ f.
Finally, to prove transitivity, suppose that f, g, h ∈ C (X, Y ) and
F : f ≃ g and G : g ≃ h. Define H : X × I → Y by
{
F (x, 2t)
for 0 ≤ t ≤ 1/2;
H (x, t) =
G (x, 2t − 1)
for 1/2 ≤ t ≤ 1.
By the Gluing lemma, H is continuous, and therefore H : f ≃ h.
♢
Definition 14.1.3 If f : X → Y is continuous, then its equivalence
class [f ] = {g : X → Y |g is continuous and g ≃ f } is referred to as the
homotopy class of f .
We will denote by [X; Y ] the totality of [f ], where f ∈ C (X, Y ). If
X is a k-space and C (X, Y ) is given the compact-open topology, then
[f ] is just the path-component of f .
Homotopy behaves well with respect to composition of maps in the
sense that if f, g : X → Y are homotopic, then hf ≃ hg for any
continuous map h : Y → Z, and f k ≃ gk for any continuous map
k : Z → X. In fact, if H : f ≃ g is a homotopy, then h ◦ H : hf ≃ hg
and H ◦ (k × 1) : f k ≃ gk, where 1 is the identity map on I. An easy
consequence of this is
Proposition 14.1.4 Composites of homotopic maps are homotopic.
Proof. Suppose that f0 , f1 : X → Y , and g0 , g1 : Y → Z are homotopic
maps. Then g0 f0 ≃ g0 f1 and g0 f1 ≃ g1 f1 , as observed above. By the
transitivity of the relation ≃, we have g0 f0 ≃ g1 f1 .
♢
Definition 14.1.5 A continuous map f : X → Y is called a homotopy
equivalence if there is a continuous map g : Y → X such that gf ≃ 1X
and f g ≃ 1Y . Such a map g is referred to as the homotopy inverse of f .
We say that X and Y have the same homotopy type, written X ≃ Y , if
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Elements of Topology
there exists a homotopy equivalence X → Y . In this case, we also say
that X and Y are homotopically equivalent.
Obviously, a homeomorphism is a homotopy equivalence, so homeomorphic spaces have the same homotopy type. It is easily observed
that the relation X ≃ Y is an equivalence relation between spaces.
Much of algebraic topology is concerned with the study of homotopy
invariants, that is, those properties which are preserved by homotopy
equivalences.
Definition 14.1.6 A space X is called contractible if it is homotopically equivalent to the one-point space.
Proposition 14.1.7 A space X is contractible if and only if the identity map on X is homotopic to a constant map of X into itself.
Proof. Assume that X is contractible and let Y = {y0 } be a one-point
space such that X ≃ Y . Let f : X → Y be a homotopy equivalence
with homotopy inverse g : Y → X. Then 1X ≃ gf , where gf is a
constant map. Conversely, suppose that 1X ≃ c, where c : X → X
is a constant map. Let Y = {x0 } = c (X). Then the inclusion map
i : Y ,→ X and the map r : X → Y satisfy ir = c and ri = 1Y so that
X ≃Y.
♢
If X is contractible, then a homotopy connecting 1X and the constant map of X at x0 ∈ X is called a contraction of X to x0 .
Example 14.1.3 Any convex subspace of Rn is contractible; in particular, Rn itself, the unit cube I n and the unit disc Dn are contractible.
Example 14.1.4 The punctured sphere Sn − {pt}, being homeomorphic
to Rn , is contractible. For the same reason, a hemisphere is contractible.
These examples show that a contractible space need not be convex.
Example 14.1.5 The “comb space” X = ({1/n|n = 1, 2, . . .} × I) ∪
{0} × I ∪ I × {0} is contractible (ref. Ex. 3.3.3). Let c : X → X be the
constant map at the point (0, 0). Then a homotopy H : X × I → X
between 1X and c is given by
{
(x, (1 − 2t) y)
for 0 ≤ t ≤ 1/2,
H ((x, y) , t) =
(2 (1 − t) x, 0)
for 1/2 ≤ t ≤ 1.
By the Gluing lemma, H is continuous and hence a contraction of X
to (0, 0) .
THE FUNDAMENTAL GROUP
375
These examples suggest that the notion of homotopy equivalence is
weaker than that of topological equivalence.
Definition 14.1.8 A continuous map from a space X to a space Y is
called null homotopic or inessential if it is homotopic to some constant
map X → Y.
With this terminology a space X is contractible if 1X is null homotopic.
Proposition 14.1.9 Two contractible spaces have the same homotopy type.
Proof. Let X and Y be contractible spaces, and P be a one-point space.
Then X ≃ P and Y ≃ P . Suppose that f : X → P and g : Y → P
are homotopy equivalences with homotopy inverses f ′ : P → X and
g ′ : P → Y . Then (f ′ g) (g ′ f ) = f ′ (gg ′ ) f ≃ f ′ 1P f = f ′ f ≃ 1X , and
similarly (g ′ f ) (f ′ g) ≃ 1Y . So g ′ f : X → Y is a homotopy equivalence
with f ′ g : Y → X as its homotopy inverse. Accordingly, X ≃ Y .
♢
Note that a contractible space is necessarily path-connected, because if H is a contraction of X to x0 , then , for each x ∈ X,
t 7→ H (x, t) is a path in X joining x to x0 . Accordingly, a discrete
space with more than one point is not contractible. An annulus and
the n-sphere Sn are non-trivial examples of spaces which are not contractible. We will establish the first fact later, but the proof of the
second fact needs results from algebraic topology which are beyond
the scope of this book.
Next, we observe that two null homotopic maps need not be homotopic, for if X is connected, Y is not, and points y0 , y1 belong to
distinct path components of Y , then the constant maps x 7→ y0 and
x 7→ y1 of X into Y are obviously not homotopic. On the other hand,
if f is a path joining y0 to y1 , then H (x, t) = f (t) for all x ∈ X is
a homotopy between the constant maps at y0 and y1 . Thus it is only
when Y is path-connected that all null homotopic maps are homotopic.
Lemma 14.1.10 Any two continuous maps of a space X into a contractible space Y are homotopic, and if X is also contractible, then any
continuous map X → Y is a homotopy equivalence.
Proof. By Proposition 14.1.7, there exists a constant map c : Y → Y
such that 1Y ≃ c. If f, g : X → Y are continuous maps, then f =
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Elements of Topology
1Y f ≃ cf , and similarly, g ≃ cg. Since cf = cg, and ≃ is an equivalence
relation, we have f ≃ g.
To prove the last statement, let ϕ : X → Y be a continuous map. By
Proposition 14.1.9, there exists a homotopy equivalence ψ : X → Y . If
ψ ′ : Y → X is a homotopy inverse of ψ, then we have ψ ′ ϕ ≃ ψ ′ ψ ≃ 1X
and ϕψ ′ ≃ ψψ ′ ≃ 1Y . So ϕ is a homotopy equivalence with ψ ′ as a
homotopy inverse.
♢
It follows that [X; Y ] is a singleton set when Y is contractible.
Also, the identity map on a contractible space Y is homotopic to any
constant map of Y into itself, and any two constant maps of Y into
itself are homotopic.
The next result describes a convenient property of the null homotopic maps of spheres.
Theorem 14.1.11 A continuous map f : Sn → Y is null homotopic
if and only if f can be continuously extended to the disc Dn+1 .
Proof. Suppose that H : f ≃ c, where c : Sn → Y is the constant
map at y0 ∈ Y. Consider the mapping ϕ : Sn × I → Dn+1 defined
by ϕ (x, t) = (1 − t/2) x. It is clearly an embedding with the image
C = Dn+1 − B, where B = B (0; 1/2). The inverse ψ : C → Sn × I of ϕ
is given by ψ (z) = (z/ ∥z∥ , 2 (1 − ∥z∥)). Notice that ψ maps ∂B onto
Sn × 1, and Sn onto Sn × 0; accordingly, the composition Hψ : C → Y
agrees with f on Sn , and is constant on ∂B at y0 . So, we can define a
continuous map F : Dn+1 → Y by setting F (B) = y0 and F |C = Hψ.
Thus f extends to a continuous map over Dn+1 .
Conversely, suppose that f has a continuous extension g : Dn+1 →
Y . Define H : Sn × I → Y by H (x, t) = g ((1 − t) x + tx0 ), where x0 ∈
Sn is a fixed point. Then H (x, 0) = g (x) = f (x) and H (x, 1) = g (x0 )
for all x ∈ Sn . If c : Sn → Y is a constant map at g (x0 ), then H : f ≃ c,
and the proof is complete.
♢
As an immediate consequence of the preceding theorem, we see that
any continuous map of Sn into a contractible space can be continuously
extended over Dn+1 .
In many cases, we are required to deal with maps f : X → Y
having special properties, and homotopies deforming one such map
to the other in a certain way. For instance, consider the question of
distinguishing between an annulus and a circular disc. An intuitive
answer to this question is obtained by considering the possibility, or
THE FUNDAMENTAL GROUP
377
otherwise, of shrinking to a point a loop drawn on the two spaces.
A loop such as f on the annulus cannot be shrunk to a point of the
annulus without going outside of it, while this is possible for any loop
g on the circular disc (see Figure 14.1).
f
g
•
•
x
x
Disk
Annulus
FIGURE 14.1: Differentiation between a annulus and a disc.
Technically speaking, by a loop in a space X, we mean a path f :
I → X with f (0) = f (1). We also say that f is a closed path, and
the point f (0) = f (1) is referred to as the base point of f . Suppose
that the annulus A is bounded by two concentric circles C1 and C2
centered at the origin, and having radii 1 and 2, respectively. Consider
the loop f : I → A defined by f (s) = re2πıs , where 1 < r < 2. Then
H : I × I → A defined by H (s, t) = re2πıs(1−t) is a homotopy between
the loop f and the constant loop base at (r, 0). This example shows
that simply asking whether a loop is homotopic to a constant loop is
not sufficient to detect holes, and we need to modify the definition of
homotopy.
Observe that the terminal points of intermediate paths ht in the
preceding example don’t stay fixed. This situation is avoided by the
following.
Definition 14.1.12 Let A be a subspace of X, and f, g : X → Y be
continuous maps such that f |A = g|A. We say that f is homotopic to
g relative to A if there exists a continuous map
H :X ×I →Y
such that H (x, 0) = f (x), H (x, 1) = g (x) for every x ∈ X, and
H (a, t) = f (a) = g (a) for all a ∈ A and t ∈ I. Such a map H is called
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Elements of Topology
a homotopy from f to g relative to A, and we write H : f ≃ g rel A to
indicate this.
Example 14.1.6 In Ex. 14.1.1, if f and g agree on A ⊆ X, then the
homotopy H is relative to A.
Example 14.1.7 Let X = Y = D2 , f : X → Y be the identity map and
g : X → Y be the map z 7→ −z. Then H : X × I → Y defined by
H (z, t) = zeπıt is a homotopy from f to g relative to the origin.
Example 14.1.8 Consider the map r : X = Rn − {0} → Sn−1 defined
by r (x) = x/ ∥x∥. If i : Sn−1 ,→ X is the inclusion map, then 1X ≃ ir
rel Sn−1 . Such a homotopy H : X × I → X is given by H (x, t) =
(1 − t) x + tx/ ∥x∥. Also, X ≃ Sn−1 , since ri = 1Sn−1 .
Let X and Y be spaces and suppose that A ⊂ X. As in Proposition 14.1.2, we see that the homotopy relative to A is an equivalence
relation in the set of all continuous functions X → Y which agree on
A. Moreover, in this term, the extended version of 14.1.4 is
Proposition 14.1.13 If f, g : X → Y are homotopic relative to A,
h, k : Y → Z are homotopic relative to B and f (A) ⊆ B, then hf ≃
kg rel A.
The proof is similar to that of Proposition 14.1.4.
The homotopy given in Ex. 14.1.8 can be seen as a gradual collapsing of the space X onto the subspace Sn−1 . This sort of homotopy
equivalence can thus be visualised geometrically, although homotopy
equivalences are hard to visualise, in general. These special maps are
introduced in
Definition 14.1.14 Let A be a subspace of a space X and let i : A ,→
X be the inclusion map. We say that
(a) A is a retract of X if there is a continuous map r : X → A with
ri = 1A . Such a map r is called a retraction of X onto A.
(b) A is a deformation retract of X if there is a retraction r of X
onto A such that 1X ≃ ir. A homotopy H : 1X ≃ ir is called a
deformation retraction of X onto A.
THE FUNDAMENTAL GROUP
379
(c) A is a strong deformation retract of X if there is a retraction r of
X onto A such that 1X ≃ ir rel A. A homotopy H : 1X ≃ ir rel A
is called a strong deformation retraction of X onto A.
We can rephrase the condition of a deformation retraction as follows: A continuous map H : X × I → X is a deformation retraction if
and only if H (x, 0) = x, H (x, 1) ∈ A for all x ∈ X, and H (a, 1) = a
for every a ∈ A. It is a strong deformation retraction if H (a, t) = a for
all t ∈ I and a ∈ A.
Obviously, a one-point subspace of any space is a retract of the
space, while a discrete subspace (with more than one point) of a connected space is not. The sphere Sn−1 , n ≥ 1, is a strong deformation
retract of Rn − {0}, as is shown by Ex. 14.1.8. Similarly, it is a strong
deformation retraction of the punctured disc Dn − {0}.
It is clear that if A is a strong deformation retract of X, then A is
also a deformation retract of X, and if A is a deformation retract of
a space X, then A ≃ X. The following examples show that neither of
these implications is reversible.
Example 14.1.9 Let X denote the “comb space.” We observe that X
is not a retract of the unit square I 2 ; however, these are homotopically
equivalent, since both the spaces are contractible. Consider the point
q = (0, 1/2). Let V = X ∩ B (q; ϵ), where ϵ = 1/4. If there is a continuous map r : I 2 → X with r (q) = q, then we must have an open ball
B (q; δ) such that U = I 2 ∩ B (q; δ) is mapped into V by r. Since U
is connected, r (U ) is contained in the set {(0, t) |1/4 < t < 3/4}, the
component of V containing q. However, for n sufficiently large, we have
the point p = (1/n, 1/2) in U ∩ X, which is clearly moved by r. So r
cannot be a retraction of I 2 onto X.
Example 14.1.10 Consider the subspace Y = {0} × I of the “comb
space” X. Then Y is a deformation retract of X, but not a strong
deformation retract. The function H : X × I → X defined by

(x, (1 − 3t)y)
for 0 ≤ t ≤ 1/3,



((2 − 3t)x, 0)
for 1/3 ≤ t ≤ 2/3,
H ((x, y), t) =



(0, (3t − 2)y)
for 2/3 ≤ t ≤ 1
is a homotopy between 1X and ir, where i : Y ,→ X is the inclusion
map and r : (x, y) 7→ (0, y) is the retraction of X onto Y . So Y is a
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Elements of Topology
deformation retract of X. To establish the second claim, we assume,
on the contrary, that there is a homotopy F : X × I → X such that
F (p, 0) = p, F (p, 1) ∈ Y for all p ∈ X, and F (q, t) = q for every q ∈ Y
and all t ∈ I. Let q be a point of Y other than the point (0, 0) and let
B (q; ϵ) be an open ball in R2 , which does not meet the set I ×{0} ⊂ X.
Then W = X ∩ B (q; ϵ) is a nbd of q in X, and {q} × I ⊂ F −1 (W ). By
the Tube Lemma (6.1.14), there is an open nbd U of q in X such that
U × I ⊂ F −1 (W ). So, for each p ∈ U , F ({p} × I) is contained in the
component of W containing p, and this component is the intersection
of B (q; ϵ) with the “tooth” containing p. This contradicts the fact that
F (p, 1) ∈ Y for every p ∈ X, and hence our claim.
By the proof of the preceding example, it is clear that there is no
contraction of X to q = (0, 1) leaving the point q fixed, though X is
contractible to q.
Later, we will see that S1 is not contractible, so a one-point subspace
of S1 is not its deformation retract. The simple proof of the following
proposition is left to the reader.
Proposition 14.1.15 A subspace A of a space X is a retract of X if
and only if, for any space Y , every continuous map f : A → Y has a
continuous extension over X.
Notice that A is a retract of X if and only if the identity map on A
has an extension over X. The following result shows that any extension
question can always be formulated in this particular form.
Proposition 14.1.16 Let A ⊂ X be a closed subset and f : A → Y
be a continuous map. Then f extends over X continuously if and only
if Y is a retract of the adjunction space X ∪f Y .
Proof. Let π : X + Y → X ∪f Y = Z be the quotient map. If r : Z → Y
is a retraction, then the composition F = r ◦ (π|X) is a continuous
extension of f , for F (x) = r (π (x)) = r (f (x)) = f (x) when x ∈ A.
Conversely, let F : X → Y be a continuous map such that F |A = f .
Since Z is the disjoint union of X − A and Y , we can define a function
r : Z → Y by setting
{
F (z)
if z ∈ X − A, and
r (z) =
z
if z ∈ Y.
THE FUNDAMENTAL GROUP
381
Then the composition rπ = g is continuous, since g|X = F and g|Y =
1Y . Hence r is continuous, and it is clear that r is a retraction of Z
onto Y.
♢
Exercises
1. Let X be a k-space. Prove that the path components of C (X, Y )co are
precisely the homotopy classes.
2.
(a) If X ≃ Y and X is path-connected, show that Y is also pathconnected.
(b) If X ≃ Y , show that there is a one-to-one correspondence between
their path-components.
3.
(a) If a continuous map f : X → Y has a left homotopy inverse g and a
right homotopy inverse h, show that f is a homotopy equivalence.
(b) If one of the continuous maps f : X → Y and g : Y → Z, and
their composition gf : X → Z are homotopy equivalences, show
that the other map is also a homotopy equivalence.
4. Prove: A space X is path-connected ⇔ every two constant maps of X
into itself are homotopic.
5. If X is contractible, show that any continuous map X → Y is null
homotopic. If Y is path-connected, then any two continuous maps X →
Y are homotopic (and each is null homotopic).
6.
(a) If X and Y are contractible, show that X × Y is also contractible.
7.
(b) Let X be a contractible space and Y any space. Show that X ×Y ≃
Y.
(
)
(a) Is the function H : S1 × I → S1 defined by H e2πιs , t = e2πιst ,
(s ̸= 1), a homotopy between the identity map on S1 and a constant map?
(b) Suppose that a continuous map f : S1 → S1 is not homotopic to
the identity map. Prove that there is a point x ∈ S1 such that
f (x) = −x.
8. Prove that the identity map on S1 is homotopic to the antipodal map
x 7→ −x. Generalise this result to the sphere Sn , n is odd.
9. If f, g : X → Sn are two continuous maps such that f (x) ̸= −g (x) for
every x ∈ X, show that f ≃ g. Deduce that a continuous map X → Sn
which is not surjective is null homotopic.
10. Let f : Dn → Dn be a homeomorphism such that f (x) = x for every
x ∈ Sn−1 . Show that there is a homotopy H : f ≃ 1 rel Sn−1 such that
each ht : t 7→ H (x, t) is a homeomorphism.
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Elements of Topology
11. If A is a retract of a Hausdorff space X, show that A is closed in X.
12.
(a) Show that A is a retract of X ⇔ there exists a continuous surjection r : X → A with r (x) = r (r (x)) for every x ∈ X.
(b) Show that a retraction r : X → A is an identification map.
13. Assume that X ⊂ Y ⊂ Z. If X is a retract (resp. deformation retract or
strong deformation retract) of Y , and Y is a retract (resp. deformation
retract or strong deformation retract) of Z, show that X is a retract
(resp. deformation retract or strong deformation retract) of Z.
14. If A is a retract (resp. deformation retract or strong deformation retract)
of X, and B is a retract (resp. deformation retract or strong deformation
retract) of Y , prove that A × B is a retract (resp. deformation retract
or strong deformation retract) of X × Y .
15. Let A ⊂ X and f : X → A be a continuous map such that f |A ≃ 1A . If
X is contractible, show that A is contractible. Deduce that the retract
of a contractible space is contractible.
16.
(a) Prove that X is contractible ⇔ it is a retract of CX, the cone on
X.
(b) Prove that a continuous map f : X → Y is nullhomotopic ⇔ f
can be continuously extended over CX.
17. If f, g : S1 → X are homotopic maps, show that D2 ∪f X and D2 ∪g X
have the same homotopy type.
18. Let X be a triangle (with the interior) having vertices v0 , v1 , v2 . Consider the orientation of the edges in the direction from v0 to v1 , from
v1 to v2 , and from v0 to v2 . The quotient space of X obtained by identifying the edges with one another in the given orientation is called the
‘dunce cap’ space.
Prove that the ‘dunce cap’ has the homotopy type of a disc (hence
this is a contractible space).
19.
(a) Prove that the ‘center’ circle in the Möbius band is its strong
deformation retract.
(b) Show that the circle S1 × {0} is a strong deformation retract of
the cylinder S1 × I.
20.
(a) Prove that the disc Dn is a strong deformation retract of Rn .
(
)
(b) Prove that (Dn × {0})∪ Sn−1 × I is a strong deformation retract
of Dn × I.
21. Let X be the ‘punctured’ torus (that is, the space obtained from a
torus by deleting a point). Show that X contains a subspace A which
is homeomorphic to S1 ∨ S1 , and is a strong deformation retract of X.
THE FUNDAMENTAL GROUP
383
22. Prove that a subspace A ⊂ X is a deformation retract of X if and only if,
for every space Y , each continuous map f : A → Y can be continuously
extended over X, and any two continuous maps f, g : X → Y are
homotopic whenever f |A ≃ g|A.
23. Suppose that A is a strong deformation retract of a compact space X,
B is a closed subspace of a Hausdorff space Y , and f : X → Y is
a continuous map such that f | (X − A) is a homeomorphism between
X − A and Y − B. Show that B is a strong deformation retract of Y .
24.
(a) Let A be a retract of X, and Y a space. Show that, for each
continuous map f : A → Y , the subspace Y is a retract of X ∪f Y .
(b) If A is a strong deformation retract of X, show that Y is also a
strong deformation retract of X ∪f Y .
25.
(a) Let f : X → Y be a continuous map, and Mf be the mapping
cylinder of f . Show that Y , regarded as a subspace of Mf , is a
strong deformation retract of Mf .
(b) If f is a homotopy equivalence, show that X is a deformation
retract of Mf .
26. Let X and Y be spaces, and η be the the natural injection of Y into
C (X, Y )co (i.e., η (y) : X → Y is the constant map at y). Show that
η (Y ) is a retract of C (X, Y )co .
27. Let X be a contractible locally compact Hausdorff space.
(a) Show that η (Y ) is a strong deformation retract of C (X, Y )co .
(b) If {x0 } is a strong deformation retract of X, prove that that e−1
x0 (y)
is contractible to η (y) for every y ∈ Y .
14.2
The Fundamental Group
We study here a basic method of associating a group to each topological space so that the process always attaches isomorphic groups
to homeomorphic spaces, that is, the attached group is a topological
invariant. This algebraic invariant was introduced by the great French
mathematician Henri Poincaré in 1895, and is called the Fundamental
Group or Poincaré Group of the space.
The construction of the fundamental group of a space X uses loops
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Elements of Topology
in the space. Recall that a loop in X is a continuous function f : I → X
with f (0) = f (1). Since I is contractible, any path in X is null homotopic. Therefore, to make better use of paths, we generally consider
the homotopies relative to {0, 1}. A necessary condition for this is that
homotopic paths must have the same end points. Intuitively speaking,
a homotopy rel {0, 1} between two paths having the same end points
deforms one into the other, keeping the end points fixed throughout
the deformation (see Figure 14.2). As remarked in the previous section,
the relation of homotopy rel {0, 1} is an equivalence relation on the set
of all paths in X having common end points. The resulting equivalence
classes are referred to as homotopy classes or path classes. If f is a
path in X, the homotopy class containing it is denoted by [f ]. Notice
that two paths in the same homotopy class have the same initial point
and the same terminal point. So one can define the initial and terminal
points of [f ] to be those of f .
g
^
^
x1
x0
^
f
I£I
FIGURE 14.2: Relative homotopy.
For x0 ∈ X, the set of all homotopy classes of loops in X based at x0
is denoted by π (X, x0 ). This is the underlying set of the fundamental
group of X based at x0 . To turn this set into a group, we must define
an operation of “multiplication.” This is done via the multiplication of
loops, given in
Definition 14.2.1 If f and g are paths in X such that f (1) = g (0),
then there is a path f ∗ g in X obtained by juxtaposing f and g. This
THE FUNDAMENTAL GROUP
is defined by the formula
{
(f ∗ g) (s) =
f (2s)
for 0 ≤ s ≤ 1/2,
g (2s − 1)
for 1/2 ≤ s ≤ 1.
385
Observe that f ∗ g is a path from f (0) to g (1); in particular, if f
and g are loops, then f ∗ g is a loop. So we can define a multiplication
in π (X, x0 ) by setting [f ] · [g] = [f ∗ g]. Of course, we must check that
this multiplication is well-defined. This follows from
Lemma 14.2.2 Let f, f ′ , g and g ′ be paths in X such that f ≃ f ′
rel {0, 1}, g ≃ g ′ rel {0, 1}, and f (1) = g (0). Then f ∗ g ≃ f ′ ∗ g ′ rel
{0, 1}.
Proof. Since f ′ (1) = f (1) = g (0) = g ′ (0), the product f ′ ∗g ′ is defined,
and has the same end points as f ∗ g. Let F : f ≃ f ′ and G : g ≃ g ′
be homotopies relative to {0, 1}. Then define H : I × I → X by the
formula
{
F (2s, t)
for 0 ≤ s ≤ 1/2,
H (s, t) =
G (2s − 1, t)
for 1/2 ≤ s ≤ 1.
(see Figure 14.3).
g’
f’
t
F
G
f
g
I£ I
FIGURE 14.3: A homotopy between the products of two pairs of homotopic
paths.
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Elements of Topology
Since F (1, t) = f (1) = g (0) = G (0, t) for all t ∈ I, the Gluing
lemma shows that H is continuous. We leave the verification of the
statement H : f ∗ g ≃ f ′ ∗ g ′ rel {0, 1} to the reader.
♢
Now, we establish
Theorem 14.2.3 The set π (X, x0 ) forms a group under the multiplication [f ] · [g] = [f ∗ g].
The proof of this theorem is fairly lengthy and will be divided into
three lemmas.
Lemma 14.2.4 Suppose that f , g and h are paths in X such that
f (1) = g (0) and g (1) = h (0). Then (f ∗ g) ∗ h ≃ f ∗ (g ∗ h) rel {0, 1}.
Proof. By definition, we have

f (4s)



g (4s − 1)
((f ∗ g) ∗ h) (s) =



h (2s − 1)

f (2s)



g (4s − 2)
(f ∗ (g ∗ h)) (s) =



h (4s − 3)
for 0 ≤ s ≤ 1/4,
for 1/4 ≤ s ≤ 1/2,
for 1/2 ≤ s ≤ 1;
for 0 ≤ s ≤ 1/2,
for 1/2 ≤ s ≤ 3/4,
for 3/4 ≤ s ≤ 1.
An intuitive idea for constructing a homotopy H connecting
(f ∗ g) ∗ h and f ∗ (g ∗ h) is obtained by drawing the slanted lines
in I × I, according to the domains of g in the above equations. Thus
I × I is divided into three closed sets. We construct suitable continuous
functions on each of these pieces and glue them together. On horizontal
lines AB, BC and CD at a height t from the base (Figure 14.4 below),
these functions are reparametrisations of maps f, g and h. Specifically,
we define H : I × I → X by

f (4s/ (1 + t))
for 0 ≤ s ≤ (1 + t) /4,



g (4s − t − 1)
for (1 + t) /4 ≤ s ≤ (2 + t) /4,
H (s, t) =



h (4s − t − 2/ (2 − t)) for (2 + t) /4 ≤ s ≤ 1.
THE FUNDAMENTAL GROUP
g
f
A
h
g
B
f
h
g
ff
387
h
C
g
D
h
I£I
FIGURE 14.4: A homotopy between the products of three paths.
Since f and g agree on the line t = 4s − 1, and g and h agree on the
line t = 4s − 2, the Gluing lemma shows that H is continuous. It is
now routine to check that H : (f ∗ g) ∗ h ≃ f ∗ (g ∗ h) rel {0, 1}.
♢
In particular, if f , g and h are loops in X based at x0 , then we
have [(f ∗ g) ∗ h] = [f ∗ (g ∗ h)], which proves the associativity of the
multiplication in π (X, x0 ).
Next, we show the existence of the identity element in π (X, x0 ). In
fact, we prove the following.
Lemma 14.2.5 Let f be a path in X with f (0) = x0 , and f (1) = x1 .
If cx0 and cx1 are the constant paths at x0 and x1 , respectively, then
cx0 ∗ f ≃ f rel {0, 1} and f ∗ cx1 ≃ f rel {0, 1}.
Proof. By definition, (cx0 ∗ f ) (s) is x0 for
for 1/2 ≤ s ≤ 1. Accordingly, we define H
{
x0
H (s, t) =
f ((2s + t − 1) / (1 + t))
0 ≤ s ≤ 1/2, and f (2s − 1)
: I × I → X by
for 0 ≤ s ≤ (1 − t) /2,
for (1 − t) /2 ≤ s ≤ 1.
(See Figure 14.5(a)). It is easily checked that H : cx0 ∗ f ≃ f rel {0, 1}.
Similarly, the function H : I × I → X defined by
{
f (2s/ (1 + t))
for 0 ≤ s ≤ (1 + t) /2,
H (s, t) =
x1
for (1 + t) /2 ≤ s ≤ 1
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Elements of Topology
is a homotopy relative to {0, 1} from f ∗ cx1 to f .
f
f
x0
x1
f
f
x0
f
f
x1
I£I
I£I
(a) H : cx0 ∗ f ≃ f
(b) H:f ∗ cx1 ≃ f
FIGURE 14.5: Homotopies between a path and its products with constant
paths.
♢
In particular, if f is a loop in X based at x0 and cx0 is the constant
loop at x0 , then we have [f ∗ cx0 ] = [f ] = [cx0 ∗ f ], which shows that the
homotopy class [cx0 ] acts as the identity element for the multiplication
in π (X, x0 ). To complete the proof of Theorem 14.2.3, it remains to
show the existence of the inverse of each element. We define the inverse
(or reverse) of a path f : I → X to be the path f −1 given by f −1 (s) =
f (1 − s). (Note that f −1 is not the inverse of the mapping f ; this just
retraces f from its terminal point to its initial point.) Obviously, if f
is a loop based at x0 , then so is f −1 . We show that [f −1 ] is the inverse
of [f ] in π (X, x0 ); equivalently, f ∗ f −1 ≃ cx0 ≃ f −1 ∗ f rel {0, 1}. This
is a particular case of the following.
Lemma 14.2.6 Let f be a path in X with the initial point x0 and
the terminal point x1 . Then f ∗ f −1 ≃ cx0 rel {0, 1} and f −1 ∗ f ≃ cx1
rel {0, 1}, where cx denotes the constant path at x.
THE FUNDAMENTAL GROUP
Proof. A homotopy H : f ∗ f −1 ≃ cx0


 x0



 f (2s − t)
H (s, t) =

f (2 − 2s − t)





x0
389
rel {0, 1} is defined by
for 0 ≤ s ≤ t/2,
for t/2 ≤ s ≤ 1/2,
for 1/2 ≤ s ≤ 1 − t/2,
for 1 − t/2 ≤ s ≤ 1.
The intuitive idea behind this construction is to pull the terminal point
of the path f along it to the initial point. Accordingly, on the horizontal
line AE at height t above the base, a length t/2 at each end should
be mapped into x0 while the segment BC should be mapped into X
in the same way as f ∗ f −1 maps the interval from 0 to 1/2 − t/2 on
the base, and the segment CD in the same way as f ∗ f −1 maps the
interval from 1/2 + t/2 to 1 on the base. Thus we remain at x0 during
the time AB and DE, and traverse the path f in opposite directions
during BD.
x0
x0
x0
A
C
B
f
1/2
D
E
f -1
I£I
FIGURE 14.6: A homotopy between the product of a path with its inverse
and a constant path.
The second homotopy relation f −1 ∗ f ≃ cx1 follows from the relation f ∗ f −1 ≃ cx0 by interchanging the roles of f and f −1 , since
( −1 )−1
f
= f.
♢
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Elements of Topology
This completes the proof of Theorem 14.2.3. The group π (X, x0 )
constructed therein is called the fundamental group of X at the base
point x0 . The class [cx0 ] is the identity element (denoted by 1), and the
class [f −1 ] is the inverse of [f]. In general, the fundamental group of a
space X depends on the choice of the base point (see Ex. 14.3.1 below).
However, it is independent of the base point when X is path-connected.
This is proved in the following.
Theorem 14.2.7 If X is a path-connected space, then π (X, x0 ) is
isomorphic to π (X, x1 ) for two points x0 , x1 ∈ X.
Proof. Let h : I → X be a path in X form x0 to x1 , and let k be its
inverse path. Then k (t) = h (1 − t) for t ∈ I. If f is a loop in X based
at x0 , then (k ∗ f ) ∗ h is a loop based at x1 (see Figure 14.7).
^
f
•
x1
^
x0
h
^
•
FIGURE 14.7: The loop h−1 ∗ f ∗ h in the proof of Theorem 14.2.7.
And, if f ′ is another loop in X based at x0 such that f ≃ f ′ rel {0, 1},
then (k ∗ f ) ∗ h ≃ (k ∗ f ′ ) ∗ h rel {0, 1}, by Lemma 14.2.2. Hence there
is a mapping ĥ : π (X, x0 ) → π (X, x1 ) defined by ĥ ([f ]) = [k ∗ f ∗ h].
By Lemma 14.2.6, h ∗ k ≃ cx0 rel {0, 1}, where cx0 is the constant loop
at x0 ; consequently, [h ∗ k] is the identity element of π (X, x0 ). For [f ]
and [g] in π (X, x0 ), we have [f ] · [g] = [f ] · [h ∗ k] · [g] = [f ∗ h ∗ k ∗ g]. So
ĥ ([f ] · [g]) = [k ∗ f ∗ h ∗ k ∗ g ∗ h] = [k ∗ f ∗ h] · [k ∗ g ∗ h] = ĥ ([f ]) · ĥ ([g]).
Thus ĥ is a homomorphism. Similarly, the inverse path k of h induces
a homomorphism k̂ : π (X, x1 ) → π (X, x0 ) given by k̂ ([f ]) = [h ∗ f ∗ k],
[f ] ∈ π (X, x1 ). We have k̂ ĥ ([f ]) = k̂ ([k ∗ f ∗ h]) = [h ∗ k ∗ f ∗ h ∗ k] =
[h ∗ k] · [f ] · [h ∗ k] = [f ]. So k̂ ĥ is the identity map on π (X, x0 ). By
symmetry, ĥk̂ is the identity map on π (X, x1 ), and therefore ĥ is an
isomorphism with k̂ as its inverse.
♢
THE FUNDAMENTAL GROUP
391
By the preceding theorem, the fundamental groups π (X, x0 ) of a
path-connected space X at various base points x0 ∈ X may be considered abstractly the same group, and denoted by π (X). On the other
hand, we must note that the isomorphism ĥ : π (X, x0 ) ∼
= π (X, x1 )
constructed in Theorem 14.2.7 is not canonical, so these groups cannot
be identified. To see this, let h1 and h2 be two paths in X from x0 to x1 .
Then there are isomorphisms ĥ1 , ĥ2 : π (X, x0 ) → π (X, x1 ) induced by
h1 and h2 . The product h1 ∗ h−1
2 is a loop in X based at x0 and hence
−1
determines an element
α
=
[h
1
(
) ∗h2 ] of π (X, x0 ). For any element [f ] in
−1
−1
−1
π (X, x0 ), we have ĥ−1
.
1 ◦ ĥ2 ([f ]) = [h1 ∗h2 ∗f ∗h2 ∗h1 ] = α[f ]α
This implies that ĥ2 = ĥ1 ◦ iα , where iα is the inner automorphism of
π (X, x0 ) determined by α. By Lemma 14.2.2, ĥ1 = ĥ2 when h1 ≃ h2 rel
{0, 1}. So the isomorphism induced by a path joining two base points
depends on its homotopy class, and if the fundamental group at some
point (and hence all points) is abelian, then the isomorphism connecting different base points is natural. However, the fundamental group
need not be abelian, in general.
The definition of the fundamental group is by no means constructive and the problem of computing group effectively is quite difficult,
in general. We will give here some simple examples and discuss the
interesting cases in the following section and the next chapter.
Example 14.2.1 A discrete space X has a trivial fundamental group,
that is, π (X, x0 ) = {1}.
Example 14.2.2 The fundamental group of the euclidean space Rn is
the trivial group. For, if f is a loop in Rn based at the origin 0, and
c0 is the constant loop at 0, then a homotopy H : f ≃ c0 rel {0, 1} is
given by H (x, t) = (1 − t) f (x).
More generally, any convex subset X of Rn has a trivial fundamental group, for each loop in X can be shrunk to the constant loop at
the base point by a straight-line homotopy. It follows that the closed
disk Dn , the n-cube I n , and an open ball in Rn each has a trivial
fundamental group. All such spaces are given a name.
Definition 14.2.8 A path-connected space X is called simply connected if π (X) = {1}.
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Elements of Topology
We now come to see the effect of continuous mappings on the fundamental group. Let ϕ : X → Y be a continuous mapping, and let
x0 ∈ X. If f is a loop in X based at x0 , then ϕ◦f is a loop in Y based at
y0 = ϕ (x0 ). Moreover, if F : f ≃ f ′ rel {0, 1}, then ϕ ◦ F : ϕ ◦ f ≃ ϕ ◦ f ′
rel {0, 1}. So there is a well-defined function ϕ# : π (X, x0 ) → π (Y, y0 )
given by ϕ# [f ] = [ϕ ◦ f ]. For any two loops f and g in X based at x0 ,
we have ϕ (f ∗ g) = (ϕf ) ∗ (ϕg); this is immediate from the definition
of path product. So
ϕ# ([f ] · [g]) = [ϕ (f ∗ g)] = [ϕf ] · [ϕg] = ϕ# [f ] · ϕ# [g].
So ϕ# is a homomorphism, referred to as the homomorphism induced by
ϕ (on the fundamental groups). The following “functorial properties”
of the induced homomorphism are crucial in applications.
Proposition 14.2.9 (a) The identity map on X induces the identity automorphism of π (X, x0 ).
(b) If ϕ : X → Y and ψ : Y → Z are continuous maps, then the
composition
ϕ#
ψ#
π (X, x0 ) −→ π (Y, ϕ (x0 )) −→ π (Z, ψϕ (x0 ))
agrees with the homomorphism (ψϕ)# .
The simple proofs are left to the reader.
As an immediate consequence of the preceding proposition, we see
that two homeomorphic spaces have isomorphic fundamental groups.
Thus the fundamental group is a topological invariant of the space;
accordingly, we can use it to distinguish between two spaces by proving that their fundamental groups are not isomorphic. We must add,
however, that the fundamental group cannot characterise a topological space. It will be seen in the following section that two nonhomeomorphic spaces may well have isomorphic fundamental groups.
At the end of this section, we shall see that the fundamental group is,
in fact, a homotopy invariant of the space, that is, it is preserved by
homotopy equivalences.
Next, we examine the homomorphism induced by the inclusion map
of a path component.
THE FUNDAMENTAL GROUP
393
Proposition 14.2.10 Let A be a path component of X, and x0 ∈ A.
Then the homomorphism i# : π (A, x0 ) → π (X, x0 ) induced by the
inclusion i : A ,→ X is an isomorphism.
Proof. For a loop g in X based at x0 , we have g (I) ⊂ A. So we can
define a loop f : I → A by setting f (t) = g (t) for all t ∈ I. Obviously,
g = if and i# is surjective. To prove that i# is injective, suppose
that f1 and f2 are loops in A based at x0 such that F : if1 ≃ if2 rel
{0, 1}. Since F (I × I) is path-connected and contains x0 = F (0, 0),
we have F (I × I) ⊂ A. Therefore the mapping F defines a homotopy
G : I × I → A between f1 and f2 . It follows that i# is injective, and
this completes the proof.
♢
It follows that the fundamental group of X based at x0 can give
information only about the path component of X containing x0 , and
we may restrict ourselves to path-connected spaces. There is another
important property of the induced homomorphisms given in
Proposition 14.2.11 Let X and Y be spaces, and x0 ∈ X. If continuous maps ϕ, ψ : X → Y are homotopic relative to {x0 }, then
ϕ# = ψ# : π (X, x0 ) → π (X, ϕ (x0 )).
Proof. This is immediate from the fact that if ϕ ≃ ψ rel {x0 }, then
ϕ ◦ f ≃ ψ ◦ f rel {0, 1} for every loop f in X based at x0 .
♢
Corollary 14.2.12 If A is a strong deformation retract of X, and x0 ∈
A, then the inclusion i : A ,→ X induces an isomorphism π (A, x0 ) ∼
=
π (X, x0 ).
Proof. Let r : X → A be a retraction such that 1X ≃ i ◦ r rel A.
Then, by the preceding proposition, i# ◦ r# = (1X )# is the identity automorphism of π (X, x0 ). And, since r ◦ i = 1A , we have
r# ◦ i# = (1A )# , the identity automorphism of π (A, x0 ). Therefore,
i# : π (A, x0 ) → π (X, x0 ) is an isomorphism with r# as its inverse. ♢
In particular, the preceding corollary shows that the fundamental
group of a space X, which is contractible to a point that is a strong
deformation retract of X, is trivial. Here, the stronger condition on the
contraction that it does not move the base point is actually unnecessary.
This is shown by
Theorem 14.2.13 A contractible space is simply connected.
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Elements of Topology
Proof. Let X be a contractible space, and let H : X × I → X be
a contraction of X to x0 . Then, for each x ∈ X, t 7→ H (x, t) is a
path from x to x0 , and therefore X is path-connected. We show that
π (X, x0 ) = {1}. The function g : I → X defined by g (t) = H (x0 , t) is
a loop based at x0 , so determines an element of π (X, x0 ). Let cx0 be
the constant loop at x0 . For any loop f in X based at x0 , we find that
f ≃ (g ∗ cx0 ) ∗ g −1 rel {0, 1}. A desired homotopy connecting the two
loops is given by

g (4s)
for 0 ≤ s ≤ t/4,



H (f ((4s − t) / (4 − 3t)) , t)
for t/4 ≤ s ≤ 1 − t/2,
F (s, t) =


 −1
g (2s − 1)
for 1 − t/2 ≤ s ≤ 1.
(See Figure 14.8.) It is now immediate that [f ] = [g ∗ cx0 ∗ g −1 ] =
[g] · [cx0 ] · [g −1 ] = [g] · [g]−1 = 1, the identity element of π (X, x0 ).
g
g
g -1
c x0
H(f;¢)
g -1
f
I£I
FIGURE 14.8: The homotopy F in the proof of Theorem 14.2.13.
♢
By Theorem 14.2.13, we see that the fundamental groups of the
“comb space” and the “dunce cap space” are trivial.
Finally in this section, we establish the homotopy invariance property of the fundamental group: Two homotopically equivalent spaces
have isomorphic fundamental groups. With this end in view, we prove
THE FUNDAMENTAL GROUP
395
Theorem 14.2.14 Let ϕ, ψ : X → Y be homotopic maps and suppose
that x0 ∈ X. Then there is a path g in Y from ϕ (x0 ) to ψ (x0 ) such
that ψ# = ĝ ◦ ϕ# , that is, the following diagram
ϕ#
π (X, x0 )
Q
- π (Y, ϕ(x0 ))
Q
ψ# QQ
Q
s
Q
ĝ
π (Y, ψ(x0 ))
commutes, where ĝ is the isomorphism [h] → [g −1 ∗ h ∗ g] induced by
the path g.
Proof. Let H : ϕ ≃ ψ be a homotopy connecting ϕ to ψ. Define g :
I → Y by g (t) = H (x0 , t) , t ∈ I. Then g is a path in Y joining ϕ (x0 )
to ψ (x0 ). By the proof of Theorem 14.2.7, g induces the isomorphism
ĝ : [h] → [g −1 ∗ h ∗ g] from π (Y, ϕ(x0 )) to π (Y, ψ(x0 )). It remains
to show that ĝ[ϕ ◦ f ] = [ψ ◦ f ] for any loop f in X based at x0 .
This follows immediately from the relation g −1 ∗ ϕ ◦ f ∗ g ≃ ψ ◦ f rel
{0, 1} ⇐⇒ ϕ ◦ f ≃ g ∗ ψ ◦ f ∗ g −1 rel {0, 1}. A homotopy F : I × I → Y
between ϕ ◦ f and (g ∗ ψ ◦ f ) ∗ g −1 is given by

g (4s)
for 0 ≤ s ≤ t/4,



H (f ((4s − t) / (4 − 3t)) , t)
for t/4 ≤ s ≤ 1 − t/2,
F (s, t) =


 −1
g (2s − 1)
for 1 − t/2 ≤ s ≤ 1
(cf. Figure 14.8). The continuity of F is seen by invoking the Gluing
lemma.
♢
Corollary 14.2.15 If ϕ : X → Y is a homotopy equivalence, then for
any x0 ∈ X, ϕ# : π (X, x0 ) → π (Y, ϕ (x0 )) is an isomorphism.
Proof. Suppose that ϕ is a homotopy equivalence, and let ψ : Y → X be
a homotopy inverse of ϕ. Put y0 = ϕ (x0 ) , ψ (y0 ) = x1 and y1 = ϕ (x1 ).
Since ψ ◦ ϕ ≃ 1X , the preceding theorem guarantees the existence of
a path g in X from x0 to x1 such that (ψϕ)# = ĝ, where ĝ is the isomorphism [f ] 7→ [g −1 ∗ f ∗ g]. From the equation ψ# ◦ ϕ# = (ψϕ)# = ĝ
(Figure 14.9(a) below), it follows that ϕ# is a monomorphism. Similarly, the relation ϕ◦ψ ≃ 1Y gives a path h in Y from y0 to y1 such that
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Elements of Topology
ϕ# ◦ ψ# = ĥ (see Figure 14.9(b) below). Since ĥ is an isomorphism,
ϕ# is an epimorphism, and the corollary follows.
π (X, x0 )
1X#
- π (X, x0 )
@
(ψϕ)#@
@
R
@
π (X, x1 )
(a)
ĝ
π (Y, y0 )
ψ#
- π (X, x1 )
@
ĥ@
@
R
@
ϕ#
π (Y, y1 )
(b)
FIGURE 14.9: Proof of Corollary 14.2.15.
♢
The preceding corollary establishes the homotopy invariance property of the fundamental group: Two path-connected spaces of the same
homotopy type have isomorphic fundamental groups. This will be used
as an aid in the determination of fundamental groups of certain spaces.
In particular, Corollary 14.2.12 remains valid even if A is just a deformation retract of X.
Exercises
1. Let Ω (X, x0 ) be the subspace of C (I, X)co consisting of all loops based
at x0 . Prove that π (X, x0 ) is just the set of all path components in
Ω (X, x0 ).
2. Let f, g : I → X be two paths with initial point x0 and terminal point
x1 . Show:
(a) f ≃ g rel {0, 1} if and only if f ∗ g −1 ≃ cx0 rel {0, 1}.
(b) If f ≃ g rel {0, 1}, then f −1 ≃ g −1 rel {0, 1}.
3. Let x and y be distinct points of a simply connected space X. Prove
that there is a unique path class in X with initial point x and terminal
point y.
4. Find the fundamental group of an indiscrete space.
5. Let r : X → A be a retraction and i : A ,→ X be the inclusion map.
If i# π (A) is a normal subgroup of π (X), show that π (X) = im (i# ) ⊕
ker (r# ).
THE FUNDAMENTAL GROUP
397
6. If ϕ, ψ : X → Y are homotopic maps and ϕ (x0 ) = ψ (x0 ) for some x0 ∈
X, prove that the homomorphisms ϕ# , ψ# : π (X, x0 ) → π (Y, ϕ (x0 ))
differ by an inner automorphism of π (Y, ϕ (x0 )).
7. If ϕ : (X, x0 ) → (Y, y0 ) is (free) null homotopic, show that ϕ# :
π (X, x0 ) → π (Y, y0 ) is the trivial homomorphism.
14.3
Fundamental Groups of Spheres
The section is devoted to computation of the fundamental group of
a few simple spaces such as spheres, the punctured plane, the Möbius
band, torus etc. In the next chapter, we shall determine the structure
of the fundamental group of some more spaces (e.g., projective plane,
Klein bottle and Lens spaces).
The 0-sphere S0 , being a discrete space, has a trivial fundamental
group. The computation of the fundamental group of S1 is an involved
one, so we determine first the fundamental group of the n-sphere Sn ,
n ≥ 2, which is rather simple. With this end in view, we prove a special
case of the Seifert–van Kampen theorem.
Proposition 14.3.1 Let X be a space and {Uα } be an open cover of
X such that (a) each Uα is simply connected,∩(b) Uα ∩ Uβ is pathconnected for any pair of indices α, β, and (c) Uα ̸= ∅. Then X is
simply connected.
∩
Proof. Choose a point x0 ∈ Uα . Each point x ∈ X belongs to some
Uα . Since Uα is path-connected, there is a path in Uα ⊆ X joining x to
x0 . Hence X is path-connected. We show that π (X, x0 ) = {1}. Let f be
a loop in X based at x0 . Then the family {f −1 (Uα )} is an open cover
of the compact metric space I, and therefore has a Lebesgue number
δ, say. Consider a partition
0 = t0 < t1 < · · · < tn = 1
of I so that tj −tj−1 < δ for every j = 1, . . . , n. Then f maps the subinterval [tj−1 , tj ] into some Uαj , 1 ≤ j ≤ n. For notational convenience,
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Elements of Topology
we will write Uj for Uαj . Now, define a path gj : I → X by setting
gj (s) = f ((1 − s) tj−1 + stj ), 0 ≤ s ≤ 1, for every j = 1, . . . , n. It is
clear that the path gj lies entirely in Uj (Figure 14.10). We observe
that f ≃ (g1 ∗ · · · ∗ gn−1 ) ∗ gn rel {0, 1}. Divide the interval I into n
subintervals by the points 21−n , 22−n , . . . , 2−1 , and consider the map
h : I → I defined by
{ n−1
2
st1
if 0 ≤ s ≤ 21−n , and
h (s) =
(
)
(
)
2 − 2n−j s tj + 2n−j s − 1 tj+1 if 2j−n ≤ s ≤ 21+j−n .
Then f h = (g1 ∗ · · · ∗ gn−1 ) ∗ gn , and H : (s, t) 7→ (th (s) + (1 − t) s),
s, t ∈ I, is a homotopy between the identity map on I and h relative to
{0, 1}. So the composition f H is a homotopy between f and f h holding
the base point fixed. Notice that, for every 1 ≤ j < n, f (tj ) ∈ Uj ∩Uj+1 ,
accordingly, there is a path γj from x0 to f (tj ) = xj lying entirely in
f(t2)
U2
g2
^
f
f(t1)
°2
^
^
°n−1
^
^
f(tn)
gn
g1
U1
^
Un
°1
x0
FIGURE 14.10: Proof of Proposition 14.3.1.
Uj ∩ Uj+1 , by our hypothesis. By Lemma 14.2.6, we have γj−1 ∗ γj ≃ cxj
rel {0, 1}. So gj ∗gj+1 ≃ gj ∗γj−1 ∗γj ∗gj+1 rel {0, 1} for every 1 ≤ j < n,
−1
and hence f ≃ g1 ∗ γ1−1 ∗ γ1 ∗ g2 ∗ · · · ∗ γn−1
∗ γn−1 ∗ gn rel {0, 1}. This
−1
−1
implies that [f ] = [g1 ∗ γ1 ] · [γ1 ∗ g2 ∗ γ2 ] · · · [γn−1 ∗ gn ]. Notice that
each term in the right hand side of this equation is a homotopy class
of a loop lying in some Uj , and therefore coincides with the homotopy
class [cx0 ], for each Uj is simply connected. So [f ] = [cx0 ] ,the identity
element of π (X, x0 ). This proves the proposition.
♢
THE FUNDAMENTAL GROUP
399
Corollary 14.3.2 For n ≥ 2, the n-sphere Sn has the trivial fundamental group.
Proof. Let p = (0, . . . , 0, 1) ∈ Sn and q = −p. Then both U = Sn − {p}
and V = Sn − {q} are open subsets of Sn such that Sn = U ∪ V. Since
U ≈ Rn , U is simply connected. Clearly, V ≈ U under the reflection
map (x0 , . . . , xn ) 7→ (x0 , . . . , −xn ); accordingly, V is also simply con1
(x0 , . . . , xn−1 )
nected. The homeomorphism f : (x0 , . . . , xn ) 7→ 1−x
n
n
between U and R takes q into the origin 0. Hence f induces a homeomorphism between U ∩ V = Sn − {p, q} and Rn − {0}. Since the
latter space is path-connected for n > 1, so is U ∩ V . By the preceding
proposition, we see that Sn is simply connected.
♢
Now, we turn to compute the fundamental group of the circle S1 .
To this end, we recall some elementary properties of the exponential
map p : R1 → S1 , t 7→ e2πıt . We know that p (t + t′ ) = p (t) p (t′ ) for
any t, t′ ∈ R1 and p−1 (1) = Z, the group of integers. If 0 < t < 1,
then p (t) ̸= 1 and for each z ∈ S1 , there is unique t ∈ [0, 1) such
that p (t) = z. The continuity of trigonometric functions sin t and cos t
shows that p is continuous. Also, we have the following.
Lemma 14.3.3 The exponential map p : R1 → S1 is open.
Proof. Since the open intervals of length less than 1 form a base for R1 ,
it suffices to prove that the images of all such open intervals under p are
open in S1 . Let q denote the restriction of p to the interval [−1/2, 1/2].
Then q is a continuous closed surjection. As q −1 (−1) = {−1/2, 1/2},
the mapping r : (−1/2, 1/2) → S1 −{−1} defined by p is also a continuous closed surjection. Since p(t) ̸= 1 for 0 < t < 1, r is injective and thus
a homeomorphism. It follows that the restriction of p to (−1/2, 1/2) is
an open mapping, for S1 − {−1} is open in S1 . We next observe that p
maps every open interval of length less than 1 onto an open subset of
S1 . It is obvious that each open interval (a, b) in R1 with b − a < 1 is
sent to some open interval contained in (−1/2, 1/2) by the translation
map τ : t 7→ t + t0 , where t0 = −(a + 1/2). For every a < t < b, we have
p(t) = p(−t0 )p(τ t). Since the multiplication by p(−t0 ) is a homeomorphism of S1 onto itself, and τ (a, b) is an open subset of (−1/2, 1/2), we
see that p(a, b) is open in S1 .
♢
Proposition 14.3.4 Each point z ∈ S1 has an open neighbourhood
U in S1 such that p−1 (U ) is the disjoint union of open intervals each
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Elements of Topology
of which is mapped homeomorphically onto U by the exponential map
p.
Proof. Define the sets
{
}
H1 = (x, y) ∈ R2 |x > 0 ,
{
}
H3 = (x, y) ∈ R2 |x < 0 ,
{
}
H2 = (x, y) ∈ R2 |y > 0 ,
{
}
H4 = (x, y) ∈ R2 |y < 0 ,
and put Ui = S1 ∩ Hi , i = 1, 2, 3, 4. Then each Ui is open in S1 , and
(
(
(
(
U2
(
U1
(
(
(
U3
U4
FIGURE 14.11: Open subsets Ui ⊂ S1 in the proof of Proposition 14.3.3.
their union equals S1 (see Figure 14.11). We observe that the set Ui
satisfy the requirements of the proposition. Consider, for example,
the subset U1 . Since cos θ ≤ 0 for π/2 ≤ θ ≤ 3π/2, U1 can be expressed as U1 = {eıθ
( < 1θ < π/2}
) = {p (t) | − 1/4 < t < 1/4}.
∪| − π/2
1
−1
p (U
(Clearly,
) 1 ) = n∈Z n − 4 , n + 4 and p maps each open interval
n − 41 , n + 14 bijectively onto U1 . (Since p is continuous
and open, the
)
1
1
restriction
of
p
to
the
open
interval
n
−
,
n
+
is
a
homeomorphism
4
4
)
(
between n − 14 , n + 14 and U1 .
Similar arguments apply to other cases.
♢
To describe another required property of the map p, we introduce
the following terminology: A subset X of Rn is starlike with respect to
a point x0 ∈ X if all the line segments joining x0 to any other point of
X lie entirely in X. It is obvious that a convex subset of Rn is starlike
with respect to any of its points; however, the converse is not true.
Theorem 14.3.5 Let X be a compact subspace of a euclidean space
Rn , and suppose that it is starlike with respect to the origin 0. If
f : X → S1 is a continuous map such that f (0) = 1, then, for each
integer m, there exists a unique continuous map f˜ : X → R1 such that
pf˜ = f and f˜ (0) = m, where p is the exponential map.
THE FUNDAMENTAL GROUP
401
Proof. Since X is compact, f is uniformly continuous. So there is a real
δ > 0 such that |f (x) − f (x′ )| < 2 whenever ∥x − x′ ∥ < δ. Since X is
bounded, we find a positive integer n such that ∥x∥ < nδ for all x ∈ X.
Then, for every x ∈ X and each integer 0 ≤ i < n, we have
i+1
i
x − x < δ.
n
n
)
(i )
) (i )
(
(
So f i+1
< 2 whence f i+1
n x − f nx
n x /f n x ̸= −1 for all x ∈
X and every 0 ≤ i < n. For each
( i = )0, 1,(. .i . ,)n−1, define a mapping gi :
X → S1 − {−1} by gi (x) = f i+1
n x /f n x . It is clear that each gi is
continuous, (gi (0) =) 1 and f = g0 g1 · · · gn−1 . By Lemma 14.3.3, p maps
the interval − 21 , 12 homeomorphically onto S1 −{−1}. Let q denote the
restriction of p to (−1/2, 1/2). Then q −1 : S1 − {−1} → (−1/2, 1/2) is
continuous and, therefore, the mapping f˜ : X → R1 given by f˜ (x) =
m + q −1 g0 (
(x) + · · · + q −1 gn−1
is continuous. Obviously, we have
) (x)
∑n−1 −1
∏n−1
˜
q gi (x) = 0 gi (x) = f (x) for every x ∈ X
pf (x) = p
0
and f˜ (0) = m, for q −1 (1) = 0.
To see the uniqueness of f˜, suppose that there is a continuous map
g̃ : X → R1 satisfying pg̃ = f and g̃ (0) = m. Then the mapping
h̃ (: X → R1 defined
by h̃ = g̃ − f˜ is continuous and satisfies ph̃ (x) =
)
p g̃ (x) − f˜ (x) = 1 for all x ∈ X, and h̃ (0) = 0. As p−1 (1) = Z,
we have h̃ (x) ∈ Z for every x ∈ X. Since X is connected, and h̃ is
continuous, we conclude that h̃ (x) = 0 for all x ∈ X. This completes
the proof of the theorem.
♢
The map f˜ in the above theorem is called a lifting of f relative to
the exponential map p. The two most important cases of this theorem
arise for X = I and X = I × I. It is obvious that the unit interval I is
a starlike subset of R1 with respect to 0, and the unit square I × I is
a starlike subset of R2 with respect to the point (0, 0). We emphasize
these instances by stating them below as
The Path Lifting Property: If f : I → S1 is a path with f (0) =
1, then there exists a unique path f˜ : I → R1 such that pf˜ = f and
f˜ (0) = 0.
The Homotopy Lifting Property: If F : I × I → S1 is a homotopy with F (0, 0) = 1, then there is a unique homotopy Fe : I × I → R1
such that pFe = F and Fe (0, 0) = 0.
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Elements of Topology
These properties of the exponential map p generalise to a very important class of spaces, namely, the “covering spaces,” which will be
studied in the next chapter. We shall see there constructive proofs of
the above statements.
Definition 14.3.6 Let f be a loop in S1 based at 1, and f˜ be the
lifting of f with the initial point 0 ∈ R1 . The degree of f , denoted by
deg f , is the number f˜ (1) (the terminal point of f˜).
Note that f˜ (1) is always an integer, since pf˜ (1) = f (1) = 1.
Let f be a loop in S1 based at 1. We observe that deg f depends only
on the homotopy class [f ] of f . To see this, suppose that F : f ≃ g
rel {0, 1}. By the Homotopy Lifting Property, there is a homotopy
Fe : I × I → R1 such that Fe (0, 0) = 0 and pFe = F , where p is the
exponential map. We have pFe (0, t) = 1 for all t ∈ I; accordingly,
Fe (0, t) is an integer for each t. Since {0} × I ≈ I is connected, and
Fe (0, 0) = 0, we must have Fe (0, t) = 0 for all t ∈ I. Similarly, Fe (1, t)
is also a fixed integer. We define f˜ (s) = Fe (s, 0) and g̃ (s) = Fe (s, 1),
s ∈ I. Then pf˜ (s) = f (s), pg̃ (s) = g (s) and f˜ (0) = 0 = g̃ (0). Thus f˜
and g̃ are the liftings of f and g, respectively, having the same initial
e
point 0. So deg f = f˜ (1) =( Fe (1,
) 0) = F (1, 1) = g̃ (1) = deg g. Hence
1
we have a function deg : π S , 1 → Z defined by deg[f ] = deg f .
Theorem
(
) 14.3.7 The function [f ] 7→ deg f is an isomorphism of
π S1 , 1 onto the group Z of integers.
Proof. To prove that deg is onto, let n be any integer and consider the
path f˜n : I → R1 defined by f˜n (s) = ns, s ∈ I. Then fn = pf˜n is a
loop in S1 based at 1 with deg fn = f˜n (1) = n, and therefore deg is
onto.
To prove that deg is one-to-one, suppose that deg f = deg g, where
f and g are loops in S1 based at 1. We need to show f ≃ g rel {0, 1}.
Let f˜ and g̃ be the liftings of f and g, respectively, with f˜ (0) = 0 =
g̃ (0). Our assumption implies that f˜ (1) = g̃ (1). Consider the mapping
H : I ×I → R1 defined by H (s, t) = (1 − t) f˜ (s)+tg̃ (s) for all s, t ∈ I.
It is clear that H : f˜ ≃ g̃ rel {0, 1}, and so pH : f ≃ g rel {0, 1}.
Finally, we prove that deg is a homomorphism. Given loops f and
g in S1 based at 1, let f˜ and g̃, respectively, be their liftings with
f˜ (0) = g̃ (0). We show that deg (f ∗ g) = deg f + deg g = f˜ (1) + g̃ (1).
THE FUNDAMENTAL GROUP
403
To establish this equation, we define a path h̃ : I → R1 by
{
f˜ (2s)
for 0 ≤ s ≤ 1/2,
h̃ (s) =
f˜ (1) + g̃ (2s − 1) for 1/2 ≤ s ≤ 1.
It is easily checked that h̃ is a lifting of f ∗ g and h̃ (0) = 0. So
deg (f ∗ g) = h̃ (1) = f˜ (1) + g̃ (1), and this completes the proof.
♢
By the proof of the preceding theorem, it is clear that each loop in
S1 based at 1 is homotopically equivalent to fn for unique integer n
and, intuitively, the map fn : t → e2πınt wraps the interval I around
S1 |n| times (anticlockwise if n > 0 and clockwise if n < 0). Thus the
degree of a loop in S1 based at 1, roughly speaking, tells how many
times the path is wrapped around the circle, and is referred to as the
“winding number” of the loop.
The following example justifies our remark made earlier that
π (X, x0 ) generally depends on the choice of the base point x0 .
Example 14.3.1 Let A = {(x, sin 1/x) |0 < x ≤ 1}, B = {0} × I, and C
be a circle lying in the second quadrant and tangent to the y-axis at
the point (0, 1). Consider the subspace X = A∪B ∪C of R2 (see Figure
14.12 below). It is easily verified that π (X, x0 ) = {1} for x0 ∈ A, and
π (X, x0 ) = Z for x0 = (0, 1). Notice that the space X is connected.
+1
–
C
A
B
·
0 ¦
¦
¦
−1 -
FIGURE 14.12: The space X in Example 14.3.1.
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Elements of Topology
Example 14.3.2 The fundamental group of the punctured plane R2 −
{pt} is Z, the group of( integers. For,
S1( is )a strong deformation retract
)
of R2 − {0} so that π R2 − {pt} ∼
= π S1 = Z, by Corollary 14.2.12.
The next theorem determines the fundamental group of the product
of finitely many spaces in terms of the fundamental groups of its factor
spaces.
Theorem 14.3.8 Let X and Y be spaces, and x0 ∈ X, y0 ∈ Y . Then
π ((X × Y ), (x0 , y0 )) ∼
= π (X, x0 ) ⊕ π (Y, y0 ).
Proof. Let pX : X × Y → X, and pY : X × Y → Y denote the
projection maps. Then we have induced homomorphisms (pX )# :
π ((X × Y ), (x0 , y0 )) → π (X, x0 ), and (pY )# : π ((X × Y ), (x0 , y0 )) →
π (Y, y0 ). Hence there is a homomorphism
ϕ : π ((X × Y ), (x0 , y0 )) → π (X, x0 ) ⊕ π (Y, y0 )
given by ϕ ([h]) = ([pX h], [pY h]), where h is a loop in X × Y based at
(x0 , y0 ). We show that ϕ is an isomorphism. If f is a loop in X based
at x0 , and g is a loop in Y based at y0 , then h : I → X × Y defined by
h (t) = (f (t) , g (t)) is a loop based at (x0 , y0 ), since pX h = f , pY h = g.
Obviously, ϕ ([h]) = ([f ], [g]), and so ϕ is onto. To see that ϕ is oneto-one, suppose that ϕ ([h]) = ϕ ([k]), h, k are loops in X × Y based at
(x0 , y0 ). Then there exist homotopies F : pX h ≃ pX k rel {0, 1}, and
G : pY h ≃ pY k rel {0, 1}. Accordingly, H : I × I → X × Y , defined
by H (s, t) = (F (s, t) , G (s, t)), is a homotopy between h = (pX h, pY h)
and k = (pX k, pY k) rel {0, 1}. So [h] = [k], and ϕ is one-to-one. Thus
ϕ is an isomorphism.
♢
As an immediate consequence of the preceding theorem, we see that
the fundamental group of the torus S1 × S1 is Z ⊕ Z, and that of solid
torus D2 × S1 is Z.
We conclude this section by exemplifying two uses of the fundamental group.
Theorem 14.3.9 (Brouwer’s Fixed-Point Theorem) A continuous map f of Dn into itself has at least one fixed point, that is, f (x) = x
for some x ∈ Dn .
THE FUNDAMENTAL GROUP
405
Proof. By our knowledge thus far, we can prove the theorem for n ≤
2. Let f : Dn → Dn be a continuous map. Assume that f has no
fixed points, that is, f (x) ̸= x for each x ∈ Dn . Let r (x) denote the
point of intersection of Sn−1 = ∂Dn with the ray emanating from f (x)
and going through the point x. Then r(x) is uniquely determined, by
Lemma 6.3.6. Thus we have a mapping r : Dn → Sn−1 . Obviously,
r (x) = x for every x ∈ Sn−1 , and it is not difficult to see that r is
continuous because f is continuous. So r is a retraction of Dn onto
Sn−1 . This is a contradiction for n = 1, since D1 is connected, and S0
is not. For n = 2, let i : S1 ,→ D2 denote the inclusion map. Then we
have ri = 1, the identity on S1 . Then, for x0 ∈ S1 , the composition
)
(
) i# (
) r# (
π (S1 , x0 ) −→ π D2 , x0 −→ π S1 , x0 is the
) automorphism of
( identity
π S1 , x0 . (This is)clearly impossible, for π D2 , x0 is the trivial group,
whereas π S1 , x0 is nontrivial. Therefore, there is a point x ∈ D2 such
that f (x) = x, and the theorem follows.
♢
A proof of this theorem in full generality requires higher dimensional analogs of the fundamental group. Observe that while establishing the preceding theorem, we have proved that Sn−1 is not a retract of
Dn (n ≤ 2). This result is known as Brouwer’s No-retraction Theorem.
Remarks 14.3.10 Before giving another application of the fundamental group, we extend the notion of the degree to any continuous
map g : S1 → S1 . Note that if g is a continuous mapping of S1 into itself, then there is a unique continuous mapping g ′ of S1 into itself such
that g ′ (1) = 1 and g = g(1)g ′ . Obviously, the composition g ′ q is a loop
in S1 based at 1, where q : I → S1 is the identification map t 7→ e2πıt .
This loop in S1 is uniquely determined by g. Therefore, one can define
the degree of g to be the integer deg g ′ q, where g ′ (z) = g(z)/g(1).
We observe that the “degree” of a continuous map g : S1 → S1 is
a homotopy invariant. First, notice that if g (1) = eıθ , then (z, t) 7→
e−ıtθ g (z), z ∈ S1 , t ∈ I, is obviously a homotopy between g and g ′ .
So two continuous maps g, h : S1 → S1 are homotopic if and only
if g ′ ≃ h′ . Moreover, if G : g ≃ h, then H : S1 × I → S1 defined
by H (z, t) = G (z, t) /G (1, t) is a homotopy relative to {1} connecting g ′ and h′ . Thus
( there
) ( is a )one-to-one correspondence between the
set [S1 ; S1 ] and [ S1 , 1 ; S1 , 1 ], the set of homotopy classes relative
to {1} of continuous maps S1 → S1 . Next, we note that each loop f
in S1 based at 1 clearly induces a continuous map g : S1 → S1 with
g(1) = 1 and f = gq. So the correspondence g 7→ gq between the set
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Elements of Topology
((
) (
))
C S1 , 1 ; S1 (, 1 of
g : S1 → S1 with g(1) = 1
) all continuous maps
1
1
and the set L S , 1 of all loops in S based at 1 is a bijection. Also,
if F : gq ≃ hq rel {0, 1}, then the function G : S1 × I → S1 , defined
by G(q(s), t) = F (s, t), is a homotopy relative to {1} between g and
h, since q × 1 : I × I → S1 × I is an identification.
( 1 ) Conversely,
( 1 ) if G is
a homotopy between continuous maps g, h : S , 1 → S , 1 relative
to {1}, then G ◦ (q × 1) : I × I → S1 is a homotopy between gq and
hq relative
It follows(that )the bijection g ↔ gq between the
(( to) {0,
( 1}. ))
sets C S1 , 1 ; S1 , 1 and L S1 , 1 is homotopy preserving.
( 1 ) (There)
1
fore [g](7→ [gq]
is
a
one-to-one
correspondence
between
[
S
,
1
;
S
,
1
]
)
1
−1
and π S , 1 . Accordingly,
(
) the correspondence [g] 7→ [g(1) gq]) between [S1 ; S1 ] and π S1 , 1 is a bijection. It is now immediate that two
continuous maps S1 → S1 are homotopic if and only if they have the
same degree.
Note, in particular, that the mapping z 7→ z n of S1 into itself has
degree n for each integer n.
Theorem 14.3.11 (The Fundamental Theorem of Algebra)
Each nonconstant polynomial with complex coefficients has at least
one complex root.
Proof. Let f (z) = a0 + a1 z + · · · + an z n be a polynomial of degree
n > 0, where ai ’s are complex numbers. We show that f (z) = 0 for
some z ∈ C. Obviously, it suffices to consider the case an = 1. Assume,
on the contrary, that f (z) ̸= 0 for all z ∈ C. Then we can consider
f as a continuous map of C into C − {0}. For each real r > 0, let
Cr denote the circle of radius r centred at the origin 0. The mapping
F : Cr × I → C − {0} defined by F (z, t) = f ((1 − t) z), z ∈ Cr , t ∈ I,
is a homotopy between f |Cr and the constant map at a0 . Next, we
show that, for large r, f |Cr is homotopic to the restriction of the map
g : z 7→ z n to Cr in
− {0}. Define a mapping G : Cr × I → C − {0}
∑C
n−1
by G (z, t) = z n + 0 (1 − t) ai z i . We need to check that G (z, t) ̸= 0
for every z ∈ Cr and t ∈ I. If G (z, t) = 0 for some z with |z| = r, then
rn = −
∑n−1
0
(1 − t) ai z i ≤ (1 − t)
∑n−1
0
|ai | ri ≤
∑n−1
0
|ai | ri .
∑n−1
This implies that if r > max{1, 0 |ai |}, then G (z, t) ̸= 0 for every
z ∈ Cr and t ∈ I. So, for such values of r, G determines a continuous
function. It is easy to check that G : f |Cr ≃ g|Cr . Since the homotopy
relation is an equivalence relation, we see that g|Cr is null homotopic.
THE FUNDAMENTAL GROUP
407
We observe that the composition
g
S1 → Cr → C − {0} → S1 ,
where the first mapping is z 7→ rz, and the last mapping is z 7→
z/ ∥z∥, coincides with the mapping h : z 7→ z n . It follows that h is
freely homotopic to a constant map, and hence n = deg h = 0 (by
the preceding remarks 14.3.10). This contradicts the hypothesis that
deg f > 0.
♢
Exercises
1. Determine the fundamental groups of the following spaces: (a) Möbius
band, (b) cylinder (closed), (c) an annulus (i.e., a region in the plane R2
bounded by two concentric circles of different radii) and (d) a punctured
disc.
2. Let f : D2 → D2 be a homeomorphism. Show that f maps S1 onto itself,
and B (0; 1) onto itself.
(
)
3. let ϕ : S1 , 1 → (X,
( x0 ) )be a continuous map. Show that ϕ is null
homotopic ⇔ ϕ# : π S1 , 1 → π (X, x0 ) is the trivial homomorphism.
4. If ϕ : S1 → S1 is null homotopic, show that ϕ has a fixed point and
ϕ (x) = −x for some x ∈ S1 .
(
)
5. If (f : S)1 → S1 is a continuous function, show that f# : π S1 , 1 →
π S1 , 1 is the multiplication by deg f.
6. Let f be a loop in S1 based at 1. If f is not surjective, show that
deg f = 0. Give an example of a surjective loop f with deg f = 0.
7. Prove that a continuous map f : S1 → S1 of degree 1 is homotopic to
the identity map.
8. Prove that S1 is not a retract of D2 .
9. Prove that S1 × {x} is a retract of the torus S1 × S1 , but it is not a
strong deformation retract for any x ∈ S1 .
10. Determine the fundamental groups of (a) a punctured n-space Rn −{pt}
(n > 2), (b) Sm × Sn , m, n ≥ 2, and (c) R3 with a line removed and (d)
a punctured n-disc, n > 2.
11. Prove that R2 is not homeomorphic to Rn for n ̸= 2.
408
12.
Elements of Topology
(a) Find a starlike set that is not convex.
(b) If X is starlike, show that X is simply connected.
14.4
Some Group Theory
In the previous section, we have seen that the fundamental group of
the product of two spaces X and Y is isomorphic to the direct product
of their fundamental groups. Naturally, one would contemplate computing the fundamental group of the union, the wedge, etc. of X and
Y in terms of the fundamental groups of these spaces. The structure
of the fundamental group of such spaces and the technique involved
in the computation of these groups entail certain concepts of group
theory which are not generally discussed at the undergraduate level.
We postpone the discussion of fundamental group to the next section
and collect here the necessary background material in this regard.
FREE PRODUCTS
∏
As we know, the direct product
Gα of an indexed family
{Gα |α ∈ A} of groups is characterised by the universal property: Given
any group H and and a family of homomorphisms fα : H → Gα∏
, one
for each index α, there exists a unique homomorphism
h
:
H
→
Gα
∏
such that pα ◦ h = fα for all α, where pβ : Gα → Gβ denotes the
projection map. We use the dual of this property to introduce the following.
Definition 14.4.1 The free product of an indexed family of groups
Gα , α ∈ A, is a group F together with a family of homomorphisms
ıα : Gα → F , one for each α ∈ A, such that for any group G and
a family of homomorphisms fα : Gα → G, α ∈ A, there is a unique
homomorphism h : F → G such that h ◦ ıα = fα , that is, the following
diagram commutes for every α.
THE FUNDAMENTAL GROUP
Gα
409
ıα F
Q
Q
fαQ
h
Q
Q
s ?
Q
G.
The free product of the groups Gα , α ∈ A, is denoted by
⋆ {Gα |α ∈ A} or in the finite case by G1 ⋆ · · · ⋆ Gn . It is worth noticing that each homomorphism ıβ : Gβ → ⋆ {Gα |α ∈ A} is, in fact, a
monomorphism. For, taking G = Gβ , fβ = 1Gβ and fα : Gα → G to be
the trivial homomorphism for all α ̸= β, we obtain a homomorphism
h : ⋆ {Gα |α ∈ A} → G such that h ◦ ıβ = 1Gβ and h ◦ ıα is the trivial
homomorphism for every α ̸= β. Then the equality h ◦ ıβ = 1Gβ implies
that ıβ is injective. Moreover, for α ̸= β, we have ıα (Gα ) ∩ ıβ (Gβ ) =
{ε}, the identity element of ⋆ {Gα |α ∈ A}. For, if xα ∈ Gα , xβ ∈ Gβ
and ıα (xα ) = ıβ (xβ ), then xβ = (h ◦ ıβ ) (xβ ) = (h ◦ ıα ) (xα ) = eβ . So
ıβ (xβ ) = ε, and it follows that ıα (Gα )∪∩ ıβ (Gβ ) = {ε}. Further, we
ıα (Gα ). Let H be the subsee that ⋆ {Gα |α ∈ A} is generated by α ∪
group of ⋆ {Gα |α ∈ A} = F generated by α ıα (Gα ). For each index
α, let ȷα : Gα → H be the homomorphism defined by ıα . Then there
exists a unique homomorphism k : F → H such that k ◦ ıα = ȷα for
all α ∈ A. Denote the inclusion homomorphism H ,→ F by ℓ. Then
obviously ℓ ◦ k ◦ ıα = ℓ ◦ ȷα = ıα for all α. Since 1F ◦ ıα = ıα also holds
for every α, we have ℓ ◦ k = 1F , by the definition of F . This implies
that ℓ is onto, and so H = F .
Since each homomorphism ıα : Gα → ⋆ {Gα |α ∈ A} is injective,
we usually identify Gα with its image under ıα , and regard it as a
subgroup of ⋆ {Gα |α ∈ A}. Then ⋆ {Gα |α ∈ A} is generated by the
subgroups Gα . So each element ξ in ⋆ {Gα |α ∈ A} can be written as
ξ = x1 · · · xn , xi ∈ Gαi . For many purposes, it is important to have a
unique factorisation of each element ξ of ⋆ {Gα |α ∈ A} as a product
of elements of the subgroups Gα . We will resume this discussion later,
and turn to see the uniqueness and the existence of ⋆ {Gα |α ∈ A}.
The free product of the groups Gα , if it exists, is unique up to
isomorphism. For, suppose that F together with the homomorphisms
ıα : Gα → F , and F ′ together with the homomorphisms ı′α : Gα → F ′
are free products of the groups Gα . Then there are unique homomorphisms h : F → F ′ and h′ : F ′ → F such that h◦ıα = ı′α and h′ ◦ı′α = ıα
for all α. Thus h′ ◦ h ◦ ıα = ıα and h ◦ h′ ◦ ı′α = ı′α , that is, the diagram
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Gα
@
ıα@
ıα
- F
h′ h
@
R
@
F
commutes for all α. It is obvious that the identity map on F also makes
the above diagrams commutative. So, by the uniqueness property of
the homomorphism F → F , we have h′ ◦ h = 1F . Similarly, we have
h ◦ h′ = 1F ′ , and it follows that h : F → F ′ is an isomorphism with h′
as its inverse.
The following theorem establishes the existence of ⋆ {Gα |α ∈ A}.
Theorem 14.4.2 Given a collection {Gα |α ∈ A} of groups, their free
product exists.
Proof. For each integer n ≥ 0, we define a word of length n in the
groups Gα to be a finite sequence∑(or an ordered n-tuple) (x1 , . . . , xn )
of elements of the disjoint union
Gα . If n = 0, the word (x1 , . . . , xn )
is considered to be the empty sequence and denoted by (). By a reduced
word in the Gα , we mean a word such that any two consecutive terms in
the word belong to different groups and no term is the identity element
of any Gα . The empty word is (vacuously) a reduced word. Let W be
the set of all reduced words in the Gα . For each x ∈ Gα , we define a
permutation σx of W as follows.
For x ̸= eα (the identity element of Gα ) and w = (x1 , . . . , xn ) ∈ W ,
we write

if x1 ̸∈ Gα ,

 (x, x1 , . . . , xn )

(xx1 , . . . , xn )
if x1 ∈ Gα and xx1 ̸= eα ,
σx (w) =



(x2 , . . . , xn )
if x1 ∈ Gα and xx1 = eα
and σx () = (x). For x = eα , we set σx (w) = w for all w ∈ W. A simple
checking of various cases shows that σx σy = σxy for all x, y ∈ Gα . In
particular, we have σx σx−1 = σeα = ι, the identity permutation of W ,
for every x ∈ Gα . Putting x−1 for x in this equation, we get σx−1 σx = ι.
So each σx is a permutation of W and there is a homomorphism iα :
Gα → Symm(W ) given by iα (x) = σx . We note that iα is injective.
For, if x ̸= eα in Gα , then σx () = (x) and σeα () = (). Denote the image
of iα in Symm(W ) by G′α and let S be the subgroup of Symm(W )
THE FUNDAMENTAL GROUP
411
∪
generated by α G′α . Then we have monomorphism iα : Gα → S for
each α ∈ A. We claim that the group S is the free product of the Gα
with respect to the homomorphisms iα .
To this end, we first observe that G′α ∩ G′β = {ι} for α ̸= β. Assume
that σx = σy , where x ∈ Gα , y ∈ Gβ and α ̸= β. Then x ̸= y. If x ̸= eα ,
then σx () = (x). As σy () = () or (y), according to y = eβ or y ̸= eβ ,
we see that σx ̸= σy , in either case. Hence x = eα and σy = σx = ι.
Next, it is obvious from the definition of S that every element θ ∈ S
can be expressed as a finite product of elements of the groups G′α . In
an expression θ = σx1 · · · σxn , if two consecutive factors σxi and σxi+1
belong to the same group, we can replace them by a single letter, viz.,
their product σxi σxi+1 , to obtain a similar expression for θ with fewer
factors. Moreover, if θ ̸= ι and a factor in the above expression for θ is
ι, then we can delete it from the given expression, obtaining a shorter
one. Applying these reduction operations repeatedly, an element θ ̸= ι
can be written so that no two consecutive factors belong to the same
group G′α and no factor is the identity permutation. Such an expression
of θ is said to be in the reduced form. We observe that the expression
θ = σx1 · · · σxn in reduced form is unique for any element θ ̸= ι of S.
In fact, if xj ∈ Gαj , then no xj = eαj and αj ̸= αj+1 in this case. So
the word (x1 , . . . , xn ) belongs to W , and we have θ() = (x1 , . . . , xn ).
Similarly, if θ = σy1 · · · σym is another expression in reduced form, then
θ() = (y1 , . . . , ym ). Thus (x1 , . . . , xn ) = (y1 , . . . , ym ), which implies
that n = m and xj = yj for every j.
Now, to prove our claim, let G be any group and fα : Gα → G
be a homomorphism for every α ∈ A. We need to construct a unique
homomorphism h : S → G such that h ◦ iα = fα for every α. Notice
that if there is a homomorphism h : S → G satisfying h ◦ iα = fα for
all α, and θ = σx1 · · · σxn , where xj ∈ Gαj , is an element of S, then we
have
h(θ) = h (σx1 ) · · · h (σxn ) = fα1 (x1 ) · · · fαn (xn ).
Accordingly, if σx1 · · · σxn is the reduced form of θ ̸= ι, then we define h(θ) by this equation and put h (ι) = e, the identity element of
G. Since the expression θ = σx1 · · · σxn of θ in the reduced form is
unique, h is a mapping. Also, the equality h ◦ iα = fα is obvious for
every α. It remains to see that h is a homomorphism. Suppose that
θ, ζ ∈ S. If θ or ζ is ι, then we obviously have h(θζ) = h(θ)h(ζ). So
assume that θ ̸= ι ̸= ζ, and let θ = σx1 · · · σxn and ζ = σy1 · · · σym
be the expressions in reduced form, where xj ∈ Gαj and yk ∈ Gβk .
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Elements of Topology
Then the product θζ = σx1 · · · σxn σy1 · · · σym is not in the reduced
form when αn = β1 . Therefore, we cannot evaluate h at θζ directly. To remedy this problem, we consider an auxiliary function ψ
from the set W of all words in the groups Gα to G. Notice that if
w = (z1 , . . . , zq ) is a word in W with q > 0, then there is a unique
index νl ∈ A such that zl ∈ Gνl . So w determines a unique element
ϑ = iν1 (z1 ) · · · iνq (zq ) = σz1 · · · σzq of S. Also, if w = (z1 , . . . , zq ) is
a reduced word, then the expression ϑ = iν1 (z1 ) · · · iνq (zq ) is in the
reduced form. We set ψ(w) = fν1 (z1 ) · · · fνq (zq ), and ψ() = e. Then
ψ defines a mapping W → G such that ψ(w) = h(ϑ) if w is a reduced
word determining ϑ, and ψ() = h(ι). Note that if w = (z1 , . . . , zq ) ∈ W,
where zl ∈ Gνl , and a word w′ is obtained from w by deleting some
zl = eνl , then ψ(w) = ψ(w′ ), for fνl (zl ) = e. If νl = νl+1 and a
word w′ is obtained from w by replacing its consecutive terms zl and
zl+1 with their product zl zl+1 , then also we have ψ(w) = ψ(w′ ), for
fνl (zl ) fνl+1 (zl+1 ) = fνl (zl zl+1 ). Thus if the reduced word w′ is obtained from a word w ∈ W by applying finitely many times the preceding reduction operations, then ψ(w) = ψ(w′ ). Moreover, both w and
w′ determine the same element of S, since each iν is a homomorphism.
It follows that if an element ϑ ∈ S is determined by a word w, then
h(ϑ) = ψ(w). Since θζ = iα1 (x1 ) · · · iαn (xn ) iβ1 (y1 ) · · · iβm (ym ), we
see that
h(θζ) = ψ (x1 , . . . , xn , y1 , . . . , ym )
= fα1 (x1 ) · · · fαn (xn ) fβ1 (y1 ) · · · fβm (ym )
= h(θ)h(ζ),
and h is a homomorphism.
Finally, the homomorphism h is unique, for S is generated by the
images of iα ’s, and this establishes our claim.
♢
We return to the question of uniqueness of the representation of an
element ξ of ⋆ {Gα |α ∈ A} as a finite product of elements of the Gα .
By the proof of the preceding theorem, such an expression for ξ ̸= ε
(the identity element of ⋆ {Gα |α ∈ A}) can be put in reduced form
(that is, so that no two consecutive factors belong to the same Gα and
no factor is the identity element). We observe that this expression of ξ
is unique. Assume that x1 · · · xn = ξ = y1 · · · ym are two factorisations
in reduced form of ξ, where xj ∈ Gαj and yk ∈ Gβk . We need to
show that n = m and xj = yj for 1 ≤ j ≤ n. With the notations
THE FUNDAMENTAL GROUP
413
of the preceding theorem, there exists a unique homomorphism h′ :
⋆ {Gα |α ∈ A} → S such that h′ |Gα = iα for every α. So we have
h′ (ξ) = iα1 (x1 ) · · · iαn (xn ) = iβ1 (y1 ) · · · iβm (ym ). Since iαj (xj ) = σxj
and iβk (yk ) = σyk , it follows that h′ (ξ) = σx1 · · · σxn = σy1 · · · σym . As
no xj = eαj and αj ̸= αj+1 , etc., these expressions of h′ (ξ) are in the
reduced forms. Therefore n = m and xj = yj for 1 ≤ j ≤ n. We note
that any expression for the identity element ε of ⋆ {Gα |α ∈ A} as a
product of elements of the Gα is not in the reduced form.
There is another way to construct the free product of the groups
Gα . In fact, one can turn the set W of all reduced words in the Gα itself
into a group so that it satisfies the “universal property” stipulated in
Definition 14.4.1. The multiplication is defined by concatenation
(x1 , . . . , xn ) (y1 , . . . , yn ) = (x1 , . . . , xn , y1 , . . . , yn )
and then reduction. The empty word acts as the identity
element and
( −1
)
the inverse of a word (x1 , . . . , xn ) in W is the word xn , . . . , x−1
.
1
For each group Gα , there is a canonical monomorphism ıα : Gα →
W defined by ıα (x) = (x). Apparently this approach is simpler; but,
the verification of the axioms for a group in W , particularly that of
associativity, is tedious. To avoid this tedium, the former proof for the
existence of free products, due to B. L. van der Waerden, has been
preferred. Clearly, there is an isomorphism between S and W given
by θ 7→ θ() and ι 7→ (), the empty word. Under this isomorphism, a
word (x1 , . . . , xn ) in W corresponds to the permutation σx1 · · · σxn in
S. Also, the homomorphism
h′ : ⋆ {Gα |α ∈ A} → S,
x1 · · · xn 7→ σx1 · · · σxn ,
is actually an isomorphism. Thus an element x1 · · · xn in ⋆ {Gα |α ∈ A}
corresponds to the unique word (x1 , . . . , xn ) in W . Accordingly, by
abuse of the terminology, we will also call such a product a word. In
particular, the empty word is considered the reduced form of ε.
Example 14.4.1 Let G1 = {e1 , g1 } and G2 = {e2 , g2 } be two disjoint
copies of the cyclic group Z2 . Then an element of Z2 ⋆Z2 other than the
identity element has a unique representation as a product of alternating
g1 ’s and g2 ’s. For example,
g1 , g1 g2 , g1 g2 g1 , g1 g2 g1 g2
g2 , g2 g1 , g2 g1 g2 , g2 g1 g2 g1
and
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Elements of Topology
are some of the elements of Z2 ⋆ Z2 . Note that g1 g2 ̸= g2 g1 and both
are of infinite order.
FREE PRODUCTS WITH AMALGAMATED SUBGROUPS
The following concept will be useful in expressing the fundamental
group of a space in terms of the fundamental groups of two open subsets which cover the space, and whose fundamental groups are known.
Let G1 , G2 and A be groups, and f1 : A → G1 , f2 : A → G2 be
homomorphisms. The free product of G1 and G2 amalgamated via f1
and f2 is defined to be the quotient group (G1 ⋆ G2 ) /N , where N is
the normal subgroup of G1 ⋆ G2 generated by the words f1 (a)f2 (a)−1 ,
a ∈ A. This is denoted by G1 ⋆A G2 . It is clear that there is a homomorphism A → G1 ⋆A G2 which sends a into f1 (a)N = f2 (a)N .
So G1 ⋆A G2 can be thought of as the group of all words in G1 and
G2 with the relations f1 (a) = f2 (a), a ∈ A. The free product of G1
and G2 amalgamated via f1 and f2 is also universal for homomorphisms from G1 and G2 . For, given homomorphisms hj : Gj → H,
j = 1, 2, such that h1 f1 = h2 f2 , there exists a unique homomorphism
ϕ : G1 ⋆ G2 → H such that ϕ|Gj = hj , by the universal property of
free products. Clearly, ϕ maps each word f1 (a)f2 (a)−1 , a ∈ A, to the
identity element of H. Therefore N ⊆ ker(ϕ), and there is an induced
homomorphism ϕ̄ : G1 ⋆A G2 → H such that ϕ̄ ◦ı̄j = hj , where j = 1, 2
and each ı̄j is the inclusion ıj : Gj → G1 ⋆ G2 followed by the natural projection of G1 ⋆ G2 onto G1 ⋆A G2 . Thus we have the following
commutative diagram:
G1
3
Q
Q
f1
Q h1
ı̄1
Q
Q
Q
?
s
Q
ϕ̄
- G 1 ⋆A G 2
- H
A
Q
3
Q
6
Q
ı̄2
Q
h2
f2 Q
Q
s
Q
G2
FIGURE 14.13: The universal property of free product with amagamation.
THE FUNDAMENTAL GROUP
415
FREE ABELIAN GROUPS
Recall that a group G is said to be generated by a set X ⊆ G if every
element of G can be written as a product of elements of X and their
inverses. The empty set generates the trivial group. In case the group G
is abelian, the expression of an element g ∈ G in terms of a generating
set X ̸= ∅ simplifies as g = xn1 1 · · · xnk k , where xi ∈ X and ni ∈ Z. In
general, one can not assert that the elements x1 , . . . , xk in X and the
exponents n1 , . . . , nk are uniquely determined by g. An abelian group
G is called free on a subset X ⊂ G if every element g ∈ G different
from the identity element e can be uniquely written as g = xn1 1 · · · xnk k .
If G is free on X, then X generates G and is linearly independent in
the sense that a relation xn1 1 · · · xnk k = e with distinct xi ∈ X holds if
and only if all the exponents ni are zero. Such a subset of G is called
a basis for G. If X is a basis for G, then the cyclic subgroup Fx = ⟨x⟩
generated by each x ∈ X is infinite, and G is the direct sum of the
subgroups Fx , x ∈ X. It follows that a free abelian group is the direct
sum of isomorphic copies of Z, the group of integers.
Given a nonempty set X, there exists an abelian group F (X) which
is free on X. To construct F (X), we form an infinite cyclic group
⟨x⟩ = {xn |n ∈ Z} for every x ∈ X. The multiplication in ⟨x⟩ is defined
by xn xm = xn+m , 1 = x0 is taken as the identity element and x−n as
the inverse of xn . It is easily checked that ⟨x⟩ is an an infinite cyclic
group generated by x1 . The element x is identified with x1 , and we say
that ⟨x⟩ is the infinite
cyclic group generated by x. Denote ⟨x⟩ by Fx ,
⊕
and let F (X) = x∈X Fx (the external direct sum). Then
∪ F (X) is an
abelian group and consists of the functions f : X → x∈X Fx such
that f (x) ∈ Fx and f (x) = 1 for all but
∪ finitely many x’s. For each
x ∈ X, consider the function ux : X → x∈X Fx defined by
{
x
if y = x, and
ux (y) =
1 (in Fy ) for all y ̸= x.
Then ux ∈ F (X). We observe that B = {ux |x
∪ ∈ X} is a basis of F (X).
Let f ∈ F (X) be arbitrary. Then f : X → x∈X Fx such that f (x) ̸=
x0 for at most a finite number of x ∈ X. Suppose that f has the values
xn1 1 , . . . , xnk k with nonzero exponents on x1 , . . . , xk and is 1 elsewhere.
Then we have f = unx11 . . . unxkk . This shows that F (X) is spanned by B.
Next, if unx11 · · · unxkk = e, the identity
( element)of F (X), and x1 , . . . , xk
are all distinct, then 1 = e(xi ) = unx11 · · · unxkk (xi ) = xni i . This implies
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Elements of Topology
that ni = 0 for every i = 1, . . . , k and B is linearly independent. Thus
we see that B is a basis of F (X). Obviously, the canonical mapping
λ : X → F (X), x 7→ ux , is an injection. So we can identify X with
B = λ(X) and, thus, regard X a subset of F (X). Then X is a base of
F (X).
The free abelian groups are characterized by the following universal
mapping property:
Theorem 14.4.3 If G is a free abelian group with basis X, then any
function from X to an abelian group H can be extended to a unique
homomorphism G → H, and conversely.
Proof. Suppose that an abelian group G is free with basis X. Let H
be an abelian group and ϕ : X → H be a set function. Then each
g in G distinct from the identity element eG can be uniquely written
as g = xn1 1 · · · xnk k , where the xi are distinct elements of X and the
integers ni ̸= 0. We put ϕ̃(g) = ϕ(x1 )n1 · · · ϕ(xk )nk and ϕ̃ (eG ) = eH .
The uniqueness of the expression for g shows that ϕ̃ is a well defined
mapping G → H. It is easily verified that ϕ̃ is a homomorphism and
ϕ̃(x) = ϕ(x) for all x ∈ X. Next, if ψ̃ : G → H is another homomorphism such that ψ̃|X = ϕ, then ψ̃ = ϕ̃, since X spans G.
Conversely, suppose that X is a subset of an abelian group G such
that, for each abelian group H and each function ϕ : X → H, there
exists a unique homomorphism ϕ̃ : G → H such that ϕ̃(x) = ϕ(x) for
all x ∈ X. Let F (X) be the free abelian group generated by X with
the canonical injection λ : X → F (X). Taking H = F (X), we get a
homomorphism λ̃ : G → F (X) such that λ̃(x) = λ(x) for all x ∈ X. If
µ : X ,→ G is the inclusion map, then there exists a homomorphism
µ̃ : F (X) → G such that µ̃λ(x) = µ(x) for all x ∈ X. So µ̃λ̃(x) = x
for all x ∈ X. Since the identity automorphism 1G of G also satisfies
these equations, we see that µ̃λ̃ = 1G . This implies that λ̃ is injective. Since {λ(x)|x ∈ X} generates F (X), λ̃ is surjective. Thus λ̃ is an
isomorphism with inverse µ̃. Since {λ(x)|x ∈ X} is a basis of F (X),
X = {µ̃λ(x)|x ∈ X} is a basis of G. So G is free on X.
♢
From the preceding theorem, it is clear that every abelian group is
a homomorphic image of a free abelian group. For, given an abelian
group G, we can always find a set X of generators of G. Then the
homomorphism F (X) → G, which extends the inclusion map X ,→ G,
is surjective.
It is easily seen that two free abelian groups whose bases have the
THE FUNDAMENTAL GROUP
417
same cardinality are isomorphic. Furthermore, we see that the cardinal
number |X| of a basis X of a free abelian group F is uniquely determined by it. It is easily checked that N = {u2 |u ∈ F } is a normal subgroup of F , and the quotient F/N is a vector space over Z2 with the obvious scalar multiplication. Clearly, the set B = {xN |x ∈ X} generates
F/N over Z2 . If B were linearly dependent over Z2 , then a finite product x1 · · · xn of distinct elements xi ∈ X would be square of some ele2km
ment of F . Accordingly, we have an equation x1 · · · xn = y12k1 · · · ym
,
where yj ∈ X and the integers kj ̸= 0. This contradicts the linear independence of X over Z. So B is linearly independent over Z2 , and forms
a basis of F/N . If B is finite, then |F/N | = 2|B| . And if B is infinite,
then the family of all finite subsets of B has the same cardinality as B
(for countably infinite B, see Theorem A.5.12, and for uncountable B,
refer to Theorem A.8.4 (in the appendices)) so that |F/N | = |B|. We
also note that |B| = |X|. In fact, the mapping x ↔ xN is a bijection
between X and B. This is immediate from the linear independence of
X over Z. Moreover, if |X| is finite, then every basis for F is finite. To
see this, suppose that Y is another basis of F . Then each element of
X is a product of integral powers of finitely many elements in Y . Thus
we find a finite subset S ⊂ Y which also generates F . As Y is a basis
of F , we have Y = S which is finite. It follows that |X| is independent
of the choice of the basis X. We call |X| the rank of F.
FREE GROUPS
We use the property appearing in Theorem 14.4.3 to describe free
groups.
Definition 14.4.4 A group F is said to be free on a subset X ⊂ F
(or freely generated by X) if, for every group G and every function
f : X → G, there exists a unique homomorphism h : F → G such that
h|X = f. The subset X is referred to as a basis for F .
We remark that the requirement of uniqueness on the extensions
of maps X → G is equivalent to condition that X generates F . By the
universal mapping property of F (that is, any function from X to a
group G can be extended in just one way to a homomorphism F → G),
no relation amongst the elements of X holds except the trivial relations
that are valid for any set of elements in any group. This justifies the
use of the adjective ‘free’ to F .
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Elements of Topology
Suppose that F is a free group with basis X ̸= ∅. Then each
mapping f from X to a group G extends to a unique homomorphism
h : F → G. Let x ∈ X be a fixed element. Consider the function
f : X → Z which sends x to 1 and maps the other elements of X
into 0. By our hypothesis, f extends to a homomorphism f˜ : F →
Z. Consequently, the order of x must be infinite; so the cyclic group
⟨x⟩ ⊆ F is infinite for every x ∈ X. We observe that F is the free
product of the groups ⟨x⟩, x ∈ X. Given a family of homomorphisms
hx : ⟨x⟩ → G, x ∈ X, we have a function f : X → G obtained by setting
f (x) = hx (x). Then there is a unique homomorphism ϕ : F → G such
that ϕ(x) = f (x) = hx (x) for every x ∈ X. Since the group ⟨x⟩ is
generated by x, we have ϕ| ⟨x⟩ = hx for every x ∈ X. It follows that F
is the free product of cyclic groups ⟨x⟩, x ∈ X. In particular, the free
group with basis a singleton {x} is the infinite cyclic group ⟨x⟩; this is
the only case where a free group is abelian.
The above property of free groups provides us a clue for the construction of a free group with a given set X as basis. If X is nonempty,
then we consider the free product P of the infinite cyclic groups
Fx = ⟨x⟩, x ∈ X. There is a natural injection X → P which sends
an element x ∈ X to the word x ∈ P. Thus, X can be regarded as a
subset of P , and then P is actually generated by X. We show that the
group P is free on X. Let G be a group and f : X → G a mapping.
Then we have a family of homomorphisms fx : Fx → G, x ∈ X, defined
by fx (xn ) = f (x)n . By the universal mapping property for free products, there exists a unique homomorphism h : P → G which extends
each fx . Then f (x) = fx (x) = h(x) for all x ∈ X, and P is free on X.
We denote P by FX . The free group generated by the empty set ∅ is,
by convention, the trivial group {1}. Thus we have established
Theorem 14.4.5 Given a set X, there exists a free group FX with
basis X.
{
}
We note that X −1 = x−1 |x ∈ X ⊂ FX is disjoint from X, and
x ↔ x−1 is a bijection between X and X −1 . Some authors construct the
free group FX on the given set X by taking another set X −1 which is
disjoint from and equipotent to X. A one-to-one correspondence X →
X −1 is indicated by x 7→ x−1 . A word on X ∪ X −1 is a formal product
w = y1 · · · yn , where each yi ∈ X ∪ X −1 . An elementary reduction of
a word w consists of deletion of parts xx−1 and x−1 x, x ∈ X. By
applying elementary reductions on w, one can assume that for any
THE FUNDAMENTAL GROUP
419
i = 1, . . . , n − 1, the part yi yi+1 of w is not of the form xx−1 or x−1 x.
Then w is called a reduced word. Under the above reduction process,
some words, such as xx−1 , lead to the empty word which has no factors.
This is regarded as a reduced word. The set of all reduced words on
X ∪X −1 is taken as the underlying set for FX and the binary operation
is juxtaposition and then elementary reductions. The empty word acts
as the identity element and the inverse of the word y1 · · · yn is defined
to be yn−1 · · · y1−1 .
Proposition 14.4.6 Let X be a subset of a group G such that X ∩
X −1 = ∅. Then the subgroup ⟨X⟩ generated by X is free with X as a
basis if and only if no product w = x1 · · · xn is the identity element e
of G, where n ≥ 1, xi ∈ X ∪ X −1 and all xi xi+1 ̸= e.
Proof. Assume first that ⟨X⟩ is free with basis X. Then for each x ∈ X,
the cyclic subgroup ⟨x⟩ is infinite and ⟨X⟩ is the free product of these
subgroups. If w = x1 · · · xn , where n ≥ 1, xi ∈ X ∪ X −1 and all
m
xi xi+1 ̸= e, then w can also be written as w = xkj11 · · · xkjm
, where
xji ∈ X, xji ̸= xji+1 and ki ∈ Z are all nonzero. This expression in
⋆ {⟨x⟩ |x ∈ X} is obviously in reduced form, and therefore w ̸= e.
Conversely, suppose that no such product is e. Consider a set Y
that is in one-to-one correspondence with X, and let f : Y → X be
a bijection. Let FY be the free group on Y , which exists by Theorem
14.4.5. Then there exists a homomorphism h : FY → G such that
h|Y = f . Also, if u ∈ FY is different from the identity element, then
we have u = y1n1 · · · yknk , where yi ∈ Y , yi ̸= yi+1 and ni ∈ Z are
all nonzero. So h(u) = h(y1 )n1 · · · h(yk )nk which can be expanded as
x1 · · · xm , xj ∈ X ∪X −1 . Since h|Y is injective and yi ̸= yi+1 , xj xj+1 ̸=
e for all j = 1, . . . , m − 1. By our hypothesis, h(u) ̸= e and h is a
monomorphism. Since h(Y ) = f (Y ) = X, h maps FY onto ⟨X⟩. Thus
h : FY → ⟨X⟩ is an isomorphism carrying Y onto X, and therefore
⟨X⟩ is free on X.
♢
If X is a basis of a free group F , then each element w ∈ F can
be written as w = xn1 1 · · · xnk k , xi ∈ X and ni ∈ Z. It can be further
expanded as a product of elements of X ∪ X −1 . If in this expression
of w, certain part is of the form xx−1 or x−1 x, then we can omit it
without affecting w. By carrying out all such simplifications for w ̸= 1
(the identity element of F ), we may assume that w = x1 · · · xm , where
xi ∈ X ∪ X −1 and all xi xi+1 ̸= 1. Such an expression for w is said
to be in reduced form. If the expressions for two elements v and w of
420
Elements of Topology
F as products in reduced form of elements of X ∪ X −1 are distinct,
then vw−1 has clearly a nonempty expression in reduced form. By the
preceding proposition, vw−1 ̸= 1 which implies that v ̸= w. It follows
that each element of F has a unique expression as a product in x’s and
their inverses in reduced form.
Next, we consider an important relation between the free group and
the free abelian group generated by the same set. Recall that if x, y are
two elements of a group G, then the element xyx−1 y −1 in G is denoted
by [x, y] and called the commutator of x and y. The subgroup of G
generated by the set of all commutators in G is called the commutator
subgroup and denoted by [G, G] = G′ . It is easily seen that G′ is a
normal subgroup of G and the factor group G/G′ is abelian.
Proposition 14.4.7 Let F be a free group on a set X. Then the
quotient F/F ′ is a free abelian group with basis {xF ′ : x ∈ X} .
Proof. Let H be an abelian group and f : {xF ′ : x ∈ X} → H be a
function. If π : F → F/F ′ is the natural projection, then the composition f ◦ (π|X) is a function from X to H. Since F is free on X, there
exists a unique homomorphism ϕ : F → H such that ϕ(x) = f π(x) for
every x ∈ X. Clearly, ϕ maps F ′ into the identity element of H. Hence
there is an induced homomorphism ϕ̄ : F/F ′ → H such that ϕ̄ ◦ π = ϕ.
So ϕ̄ (xF ′ ) = f (xF ′ ) for every x ∈ X and, by Theorem 14.4.3, F/F ′ is
a free abelian group with basis {xF ′ : x ∈ X}.
♢
Now, suppose that F1 and F2 are free groups with bases X1 and
X2 , respectively, and F1 ∼
= F2 . Then the free abelian groups F1 /F1′
′
and F2 /F2 are also isomorphic, and therefore have the same rank.
By the preceding proposition, the rank of Fi /Fi′ is the cardinality of
{xFi′ : x ∈ Xi } for i = 1, 2. So | {xF1′ : x ∈ X1 } | = | {xF2′ : x ∈ X2 } |.
We observe that | {xFi′ : x ∈ Xi } | = |Xi |. An element of Fi′ is a finite product of commutators in Fi , and a commutator in Fi other
than the identity element involves at least four elements when expressed as a product in reduced form of elements of Xi ∪ Xi−1 . Consequently, xy −1 ∈
/ Fi′ for any two distinct elements x, y ∈ Fi . It follows
that x 7→ xFi is a bijection between Xi and {xFi′ : x ∈ Xi }, and thus
|X1 | = |X2 |. In particular, we see that all bases of a given free group
have the same cardinality, called its rank.
On the other hand, a free group is determined (upto isomorphism)
by its rank. To see this, let F1 and F2 be free groups with bases X1
and X2 , respectively, and suppose that |X1 | = |X2 |. Then there is a
THE FUNDAMENTAL GROUP
421
bijection f : X1 → X2 . Let f1 : X1 → F2 and f2 : X2 → F1 be functions
defined by f and f −1 , respectively. There are homomorphisms ϕ1 :
F1 → F2 and ϕ2 : F2 → F1 such that ϕi |Xi = fi , i = 1, 2. Obviously,
the restriction of the composition ϕ2 ◦ ϕ1 to X1 is the inclusion map
i1 : X1 ,→ F1 . Since the identity map 1F1 also extends i1 , we have
ϕ2 ◦ ϕ1 = 1F1 , by uniqueness of extension. Similarly, ϕ1 ◦ ϕ2 = 1F2 ,
and so ϕ1 is an isomorphism with ϕ2 as its inverse. Thus, we have
established the following
Theorem 14.4.8 Two free groups are isomorphic if and only if they
have the same rank.
As another application of the universal mapping property of free
groups, we have
Proposition 14.4.9 Every group G is the homomorphic image of a
free group.
Proof. The proof is similar to the abelian case.
♢
We also observe that if the groups G and H are freely generated by
the sets {xα |α ∈ A} and {yβ |β ∈ B}, respectively, then G ⋆ H is freely
generated by the disjoint union of {xα |α ∈ A} and {yβ |β ∈ B}. This
follows immediately from the following
Proposition 14.4.10 Suppose that G is the free products of the
groups Gα , α ∈ A, and H is the free products of the groups Hβ ,
β ∈ B. Then G ⋆ H is the free product of all the Gα and the Hβ .
Proof. Let iα : Gα → G and iβ : Hβ → H be the canonical monomorphisms. Then, given a group K and homomorphisms fα : Gα → K,
α ∈ A, and fβ : Hα → K, β ∈ B, we have unique homomorphisms
ϕ : G → K and ψ : H → K such that ϕ ◦ iα = fα for every α ∈ A, and
ψ◦iβ = fβ for every β ∈ B. Let ν : G → G⋆H and µ : H → G⋆H be the
canonical monomorphisms. Then there exists a unique homomorphism
η : G ⋆ H → K such that ην = ϕ and ηµ = ψ. So η ◦ ν ◦ iα = ϕ ◦ iα = fα
for every α ∈ A and η ◦ µ ◦ iβ = ψ ◦ iβ = fβ for every β ∈ B. It follows
that G ⋆ H is the free product of the groups {Gα |α ∈ A} + {Hβ |β ∈ B}
relative to the monomorphisms ν ◦ iα , µ ◦ iβ .
♢
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Elements of Topology
GENERATORS AND RELATIONS
We now discuss an economical way to describe any abstract group.
In order to describe a group, we need to list all elements of the group
and write down its multiplication table. This description can be shortened by giving a generating set of the group in place of a complete list
of its elements, and writing down a set of equations between products
of elements of the given generating set, which enables us to construct
the multiplication table. For example, consider the group V of plane
symmetries of a rectangle (not a square). The nontrivial elements of V
are two reflections ρ and σ in the axes of symmetry parallel to the sides
and a rotation τ about its centre through 180◦ . It is called Klein’s four
group. Clearly, τ = ρ◦σ so that ρ and σ generate V . Also, ρ2 = σ 2 = ε,
the identity element of V , and ρσ = σρ; all other relations in the group,
such as τ 2 = ε, can be derived from these three relations. So V can be
described as the group with two generators ρ, σ satisfying the relations
ρ2 = σ 2 = ε, ρσ = σρ.
Given a group G, we find a generating set X for G and construct
the free group F on X. Let ϕ : F → G be the epimorphism which
extends the inclusion map X ,→ G and N be its kernel. Then G ∼
=
F/N . We recall that the elements of F are products in reduced form of
elements of X or X −1 . Since ϕ is an extension of the inclusion X ,→ G,
every element of F is sent by ϕ to the corresponding product in G. In
particular, every element of N reduces in G to the identity element.
The elements of N other than the identity element are called relators
for G. If two reduced words s and t in F define the same element in G,
then the equation s = t is called a relation among the generators X.
Observe that N is infinite, unless G is free on X. So it is not possible
to list all elements of N. However, we can specify N by a subset R ⊂ N
such that N is the smallest normal subgroup of F containing R (i.e.,
N is generated by the elements of R and their conjugates). We call R
a set of defining relators or a complete set of relators for the group G.
Thus, G can be described by specifying a set X of generators and a
set R of defining relators. This description of G is called a presentation
and we write G = ⟨X|R⟩ . As noted above, each r ∈ R reduces in G to
the identity element 1. The equations r = 1, r ∈ R, are referred to as
the defining relations for G.
We say that a presentation G = ⟨X|R⟩ is finitely generated (resp.
finitely related) if X (resp. R) is finite. If G has a presentation which
is both finitely generated and finitely related, then we say that it is
THE FUNDAMENTAL GROUP
423
finitely presented. In this case, we write G = ⟨x1 , . . . , xn |r1 , . . . , rm ⟩
or G = ⟨x1 , . . . , xn |r1 = · · · = rm = 1⟩. Sometimes, a group is given
by using a hybrid of relators and relations. For example, each of the
following
⟨
⟩
a, b|a2 , b2 , aba−1 b−1 ,
⟨
⟩
a, b|a2 , b2 , ab = ba ,
⟨
⟩
a, b|a2 = b2 = 1, ab = ba
is interpreted as a presentation of the Klein’s four group V. The free
group of rank two might be described as the group generated by two
elements a, b with the empty set of defining relations. Thus it has the
presentation ⟨a, b⟩.
In order to find the group G described by a presentation, we need
to construct an isomorphism of G into a known group. The following
simple result is quite useful for this purpose.
Theorem 14.4.11 (Dyck) Let G be group with the presentation
G = ⟨X|R⟩ . If H is a group and there is a map θ : X → H such
that the relations r(θ(x)) = 1 holds in H for all r ∈ R, then there
exists a unique homomorphism ϕ : G → H such that ϕ(x) = θ(x) for
every x ∈ X.
Proof. Let Y be a set equipotent to X and F be a free group on Y. If
N is the normal subgroup of F generated by the words {r(y)|r ∈ R},
then F/N ∼
= G under the isomorphism which sends yN to α(y), where
α is a bijection between Y and X. By the universal mapping property
of F, there exists a unique homomorphism ψ : F → H given by ψ(y) =
θ(α(y)). Clearly, ψ (r(y)) = r(θ(α(y))) = 1. So r(y) belongs to ker(ψ)
for every r ∈ R. Since N is generated by the words r(y), we have N ⊆
ker(ψ). Hence there is an induced homomorphism ψ̄ : F/N → H. The
composition of this homomorphism with the isomorphism G ∼
= F/N ,
α(y) ↔ yN , is the desired homomorphism.
♢
From the preceding theorem, it is clear that two groups having
the same presentation are isomorphic; but two completely different
presentations
can describe the same
both presen⟨
⟩ group.
⟨ 6For example,
⟩
3
2
tations a, b|a = b = 1, ab = ba and a|a = 1 describe the cyclic
group of order 6. Furthermore, there is no effective procedure to determine whether or not the groups given by two finite presentations are
isomorphic. Nevertheless, a presentation allows efficient computation
in many cases and gives a concrete way of describing abstract groups.
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Elements of Topology
As an illustration, we determine the group described by the presentation G = ⟨x, y|xy = yx⟩. We observe that it is the free abelian group
of rank two. By the preceding theorem, there exists a homomorphism
ϕ : G → Z × Z such that ϕ(x) = (1, 0) and ϕ(y) = (0, 1). Since G
is generated by the elements x, y, every element of G is a product of
powers of x and y. Using the relation xy = yx, we can express each
element of G in the form xm y n and see that G is abelian. Clearly,
ϕ(xm y n ) = (m, n) which shows that ϕ is bijective. Thus G ∼
= Z × Z.
We conclude this section with the remark that if G and H
are finitely generated (resp. presented) groups, then so is their free
product G ⋆ H. Indeed, if G = ⟨x1 , . . . , xn |rj , j ∈ J⟩ and H =
⟨y1 , . . . , ym |sk , k ∈ K⟩, then
G ⋆ H = ⟨x1 , . . . , xn , y1 , . . . , ym |rj , sk , j ∈ J, k ∈ K⟩.
14.5
The Seifert–van Kampen Theorem
In this final section, we return to the discussion of fundamental
group. We will study a powerful theorem, proved independently by
H. Seifert (1931) and E. R. van Kampen (1933), which enables us to
determine the structure of the fundamental group of a path-connected
space that can be decomposed as the union of two path-connected,
open subsets whose intersection is also path-connected. Applying the
result proved here, we will compute the fundamental groups of some
more spaces.
Suppose that a topological space X is the union of two pathconnected open subsets U and V such that U ∩ V is also (nonempty)
path-connected. Choose a point x0 ∈ U ∩ V to be the base point for
all fundamental groups under consideration. Denote the composition
of the monomorphism ıU : π(U ) → π(U ) ⋆ π(V ) and the canonical projection π(U ) ⋆ π(V ) → π(U ) ⋆π(U ∩V ) π(V ) = G by ı̄U , and define ı̄V
similarly. For S = U, V or U ∩ V , let k S : S ,→ X and lS : U ∩ V ,→ S
U
U ∩V
V
denote the inclusion maps. Then k#
◦ lU # = k#
= k#
◦ lV # . So,
THE FUNDAMENTAL GROUP
425
by the universal property of free products with amalgamation, there is
a homomorphism Φ : G → π(X) which makes the following diagram
U
of groups and homomorphisms commutative that is, Φ ◦ ı̄U = k#
and
V
Φ ◦ ı̄V = k# .
π(U )
Q
3
Q kU
lU # Q #
ı̄
U
Q
Q
?
s
Q
Φ
π(U ∩ V )
G − − − − − → π(X)
Q
3
6
Q
Q
ı̄V
V
lV #QQ
k#
s
Q
π(V )
FIGURE 14.14: The homomorphism Φ in the proof of the Seifert–van Kampen theorem.
We show that Φ is an isomorphism. For surjectivity of Φ, it suffices
V
U
. Let f be
and k#
to show that π(X) is generated by the images of k#
a loop in X based at x0 . By the Lebesgue covering lemma, there exists
a partition
0 = t0 < t1 < · · · < tn = 1
of I such that f maps each subinterval [tj−1 , tj ] into either U or V .
If two consecutive subintervals [tj−1 , tj ] and [tj , tj+1 ], say, are mapped
by f into the same set U or V, then we omit the common end point tj
and amalgamate these two subintervals to have the partition
0 = t0 < t1 < · · · < tj−1 < tj+1 < · · · < tn = 1
of I. By repeating this process finitely many times, we obtain a partition
0 = s0 < s1 < · · · < sm = 1
of I such that f takes each subinterval [sj−1 , sj ] into either U or V
and f (sj ) ∈ U ∩ V for every j = 1, . . . , m − 1. Denote the composite
of the homeomorphism I → [sj−1 , sj ], r 7→ (1 − r)sj−1 + rsj , and
the restriction of f to [sj−1 , sj ] by fj for every j = 1, . . . , m. Then
f ≃ f1 ∗ · · · ∗ fm rel {0, 1}, by the proof of Proposition 14.3.1. Since
U ∩ V is path-connected, we find a path γj from x0 to f (sj ) in U ∩ V.
426
Elements of Topology
Clearly, we have
f
−1
≃ f1 ∗ (γ1−1 ∗ γ1 ) ∗ f2 ∗ · · · ∗ (γm−1
∗ γm−1 ) ∗ fm
−1
−1
≃ (f1 ∗ γ1 ) ∗ (γ1 ∗ f2 ∗ γ2 ) ∗ · · · ∗ ∗(γm−1 ∗ fm )
holding the base point fixed. Thus
[f ] = [f1 ∗ γ1−1 ][γ1 ∗ f2 ∗ γ2−1 ] · · · [γm−1 ∗ fm ].
Notice that each of f1 ∗ γ1−1 , γ1 ∗ f2 ∗ γ2−1 , . . . , γm−1 ∗ fm is a loop based
at x0 lying completely in either U or V . Therefore
( )the homotopy
( ) class
U
V
[f ] belongs to the subgroup generated by im k# and im k#
, and
Φ is onto.
The injectivity of Φ is rather involved and, with this end in view,
we prove the following
Theorem 14.5.1 With the notations having the above meanings, suppose that ΘU : π(U ) → H and ΘV : π(V ) → H are homomorphisms
such that ΘU ◦ lU # = ΘV ◦ lV # . Then there exists a unique homomorphism Ψ : π(X) → H which makes the diagram
π(U )
π(U ∩ V )
3
Q
U
Q k#
lU # Q
ΘU
Q
QQ
?
s
Ψ
Q
Q
H ← − − − − − π(X)
Q
ΘV
lV # Q
Q
s
Q
6
π(V )
3
kV
#
V
U
= ΘU and Ψ ◦ k#
= ΘV .
commutative, that is, Ψ ◦ k#
Proof. The uniqueness of Ψ is immediate
from
( )
( the
) fact that π(X) is
U
V
generated by the subgroups im k# and im k# of π(X), and any
two homomorphisms π(X) → H satisfying the desired conditions agree
on these subgroups.
To prove the existence of Ψ, we first define a mapping Ψ′ on the
family L of all loops in X based at x0 as follows:
If a loop f ∈ L lies completely in U (or V ), then we denote the
THE FUNDAMENTAL GROUP
427
homotopy class of a loop f in U (resp. V ) by [f ]U (resp. [f ]V ). Put
Ψ′ (f ) = ΘU ([f ]U ) or Ψ′ (f ) = ΘV ([f ]V ), according to f lies completely
in U (or V ). If f lies in U ∩ V , then we have
ΘU ([f ]U ) = ΘU lU # ([f ]U ∩V ) = ΘV lV # ([f ]U ∩V ) = ΘV ([f ]V ) .
So, for a loop f in U or in V , Ψ′ (f ) is a uniquely determined element
of H. For any two such loops f and g, it is clear that
(a) Ψ′ (f ∗ g) = Ψ′ (f )Ψ′ (g), and
(b) if f ≃ g rel {0, 1}, then Ψ′ (f ) = Ψ′ (g).
Now, let f ∈ L be arbitrary. For each x ̸= x0 in X, choose a path γx
from x0 to x in U or V , preferably in U ∩ V when x ∈ U ∩ V , and
denote the constant path at x0 by γx0 . Then choose a subdivision
0 = s0 < s1 < · · · < sm = 1
(1)
of I such that f maps each subinterval [sj−1 , sj ] into U or V . Let
fj , 1 ≤ j ≤ m, be the composite of the linear homeomorphism I ≈
[sj−1 , sj ], t 7→ (1 − t)sj−1 + tsj , and the restriction of f to [sj−1 , sj ].
Then each fj is a path lying in U or V , and we have
f ≃ f1 ∗ · · · ∗ fm
rel {0, 1}.
For each j = 1, . . . , m, put xj = f (sj ). Then
f
∗ γxm−1 ) ∗ fm ∗ γx−1
∗ γx1 ) ∗ f2 ∗ · · · ∗ (γx−1
≃ γx0 ∗ f1 ∗ (γx−1
m
m−1
1
)
)
(
(
.
(2)
∗ · · · ∗ γxm−1 ∗ fm ∗ γx−1
≃ γx0 ∗ f1 ∗ γx−1
m
1
Observe that each bracketed term in the r.h.s. of (2) is a loop in U or
V based at x0 ; accordingly, it determines an element of π(U ) or π(V ).
Thus
(
)
Ψ′ (f ) = Ψ′ (γx0 ∗ f1 ∗ γx1 ) · · · Ψ′ γxm−1 ∗ fm ∗ γx−1
m
is an element of H. We assert that this element is independent of the
choice of a particular subdivision of I. For, if
0 = t0 < t1 < · · · < tn = 1
is another subdivision of I such that f maps each subinterval [ti−1 , ti ]
into U or V , then there is a subdivision of I which has all the points
428
Elements of Topology
sj and ti . So it suffices to show that Ψ′ (f ) remains unchanged if the
subdivision in (1) is refined by dividing a subinterval [sj−1 , sj ] into two
subintervals [sj−1 , t] and [t, sj ]. In this case, if g and h are the paths
≈
f
I −→ [sj−1 , t] −→ X
and
≈
f
I −→ [t, sj ] −→ X,
respectively, and x = f (t), then fj ≃ g ∗ h rel {0, 1} and the term
γxj−1 ∗ fj ∗ γx−1
in (2) would be replaced by the product of two loops
j
−1
. Notice that γx lies in U (or V ), according
γxj−1 ∗g ∗γx and γx ∗h∗γx−1
j
to fj lies in U (or V ). So we have
(
)
(
)
′
−1
Ψ′ γxj−1 ∗ fj ∗ γx−1
=
Ψ
γ
∗
g
∗
h
∗
γ
xj−1
xj
j
)
(
= Ψ′ γxj−1 ∗ g ∗ γx−1 ∗ γx ∗ h ∗ γx−1
j
(
)
)
(
,
= Ψ′ γxj−1 ∗ g ∗ γx−1 Ψ′ γx ∗ h ∗ γx−1
j
and hence our assertion.
Moreover, we show that if two loops f and g in L are homotopic
relative to {0, 1}, then Ψ′ (f ) = Ψ′ (g). This is the most important part
of the proof. Suppose that F : f ≃ g rel {0, 1}. By the Lebesgue
covering lemma, there are two subdivisions
0 = s0 < s1 < · · · < sm = 1
and
0 = t0 < t1 < · · · < t n = 1
of I such that F maps each rectangle [si−1 , si ] × [tj−1 , tj ] into U
or V . Consider the loops hj (all based at x0 ) defined by hj (s) =
F (s, tj ), j = 0, 1, . . . , n. Then h0 = f , hn = g and hj−1 ≃ hj
rel {0, 1}. A homotopy between hj−1 and hj is given by the mapping (s, t) 7→ F (s, (1 − t)tj−1 + ttj ). So, it suffices to establish that
Ψ′ (hj−1 ) = Ψ′ (hj ) for every j = 1, . . . , n. By induction, we may
assume n = 1. Then, for every i = 1, . . . , m, F maps the rectangle
[si−1 , si ] × I into U or V (Figure 14.15). We define the paths fi and
gi by restricting F to [si−1 , si ] × {0} and [si−1 , si ] × {1}, respectively,
and put xi = F (si , 0) and yi = F (si , 1). Then there is a homotopy
E : fi ≃ gi defined by F , and we have
f = f1 ∗ · · · ∗ fm
and
g = g1 ∗ · · · ∗ gm .
THE FUNDAMENTAL GROUP
gi
U
429
^
gi
δi ^
I£I
^
^
δ i¡1
δi
δ i−1
^
^
fi
>
F
X
•
V
x0
fi
FIGURE 14.15: Proof of Theorem 14.5.1
Therefore,
)
(
) }
(
· · · Ψ′ γxm−1 ∗ fm ∗ γx−1
Ψ′ (f ) = Ψ′ γx0 ∗ f1 ∗ γx−1
m
1
,
)
(
)
(
′
−1
·
·
·
Ψ
γ
∗
g
∗
γ
Ψ′ (g) = Ψ′ γy0 ∗ g1 ∗ γy−1
y
m
y
m−1
m
1
(3)
where x0 = y0 . Now, let δi be the path defined by δi (t) = F (si , t) for
every i. Then fi ≃ δi−1 ∗ gi ∗ δi−1 rel {0, 1}; a desired homotopy D
connecting fi to δi−1 ∗ gi ∗ δi−1 is given by


δi−1 (4r)


 (
)
4r−t
D(r, t) =
E 4−3t
,t



 δ −1 (2r − 1)
i
for
0 ≤ r ≤ t/4,
for t/4 ≤ r ≤ 1 − t/2 and
for
1 − t/2 ≤ r ≤ 1.
Consequently, we have
(
)
( −1
) −1
∗
γ
fi ≃ δi−1 ∗ γy−1
yi−1 ∗ gi ∗ γyi ∗ γyi ∗ δi
i−1
(
) (
) (
)
≃
δi−1 ∗ γy−1
∗ γyi−1 ∗ gi ∗ γy−1
∗ γyi ∗ δi−1
i−1
i
rel {0, 1}.
By the conditions (a) and (b), we see that
(
)
Ψ′ γxi−1 ∗ fi ∗ γx−1
=
i
) (
(
) ′(
)
Ψ′ γyi−1 ∗ gi ∗ γy−1
Ψ γyi ∗ δi−1 ∗ γx−1
.
Ψ′ γxi−1 ∗ δi−1 ∗ γy−1
i
i
i−1
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Elements of Topology
Since the path product of the loops γyi ∗ δi−1 ∗ γx−1
and γxi ∗ δi ∗ γy−1
is
i
i
homotopic to the constant loop cx0 ,
(
) ′(
)
Ψ′ γyi ∗ δi−1 ∗ γx−1
Ψ γxi ∗ δi ∗ γy−1
= Ψ′ (cx0 ) = e,
i
i
the identity element of H. It follows from equations (3) that Ψ′ (f ) =
Ψ′ (g), and therefore there is a function Ψ : π(X) → H given by Ψ[f ] =
Ψ′ (f ).
U
V
Clearly, Ψ ◦ k#
= ΘU and Ψ ◦ k#
= ΘV , and it remains to show
that Ψ is a homomorphism. Let f and g be any two loops in X based
at x0 . Then we can clearly choose a subdivision
0 = s0 < s1 < · · · < sj = 1/2 < sj+1 < · · · < sm = 1
of I such that f maps each subinterval [2si−1 , 2si ] into U or V for
1 ≤ i ≤ j, and g maps each subinterval [2si−1 − 1, 2si − 1] into U or V
for j < i ≤ m. Then
0 = s0 < s1 < · · · < sm = 1
is a subdivision of I such that each subinterval [si−1 , si ] is mapped by
≈
f
f ∗ g into U or V . Denote the composition I −→ [2si−1 , 2si ] −→ X by
g
≈
fi for 1 ≤ i ≤ j, and the composition I −→ [2si − 1, 2si+1 − 1] −→ X
by gi−j for j < i ≤ m. Then
f ∗ g ≃ f1 ∗ · · · ∗ fk ∗ g1 ∗ · · · ∗ gm−k
rel {0, 1}.
Put xi = f (2si ), 0 ≤ i ≤ j and xi = g(2si − 1) for j < i ≤ m. Then
Ψ′ (f ∗ g)
(
) (
)
(
)
′
−1
′
−1
= Ψ′ γx0 ∗ f1 ∗ γx−1
·
·
·
Ψ
γ
∗
f
∗
γ
Ψ
γ
∗
g
∗
γ
x
j
x
1
xj
xj+1
j−1
j
1
)
(
′
−1
· · · Ψ γxm−1 ∗ gm−j ∗ γxm
= Ψ′ (f )Ψ′ (g).
It is now immediate that Ψ is a homomorphism, and the proof is complete.
♢
Theorem 14.5.2 (Seifert-van Kampen Theorem) Suppose that
a topological space X is the union of two path-connected open subsets U and V such that U ∩ V is also (nonempty) path-connected.
Choose a point x0 ∈ U ∩ V to be the base point for all fundamental groups under consideration. Then there is an isomorphism
Φ : π(U ) ⋆π(U ∩V ) π(V ) ∼
= π(X).
THE FUNDAMENTAL GROUP
431
Proof. With the notations having above meanings, there exists a surjective homomorphism Φ : π(U ) ⋆π(U ∩V ) π(V ) → π(X) such that
U
V
Φ ◦ ı̄U = k#
and Φ ◦ ı̄V = k#
. By the definition of the amalgated
product, ı̄U ◦ lU # = ı̄V ◦ lV # . Accordingly, the preceding theorem applies, and we have a homomorphism Ψ : π(X) → π(U ) ⋆π(U ∩V ) π(V )
U
V
such that Ψ ◦ k#
= ı̄U and Ψ ◦ k#
= ı̄V . It follows that (Ψ ◦ Φ ◦ ı̄U )
agrees with ı̄U on π(U ), and (Ψ ◦ Φ ◦ ı̄V ) agrees with ı̄V on π(V ). Since
π(U ) ⋆π(U ∩V ) π(V ) is generated by the images of ı̄U and ı̄V , we have
ΨΦ = id. Therefore Φ is one-to-one, and this completes the proof. ♢
The Seifert-Van Kampen theorem is a powerful tool for computing
the fundamental groups of many spaces. Most applications of this theorem are in special cases in which one of the sets U , V or U ∩ V is
simply connected.
Corollary 14.5.3 Let U and V be open path-connected subsets of a
topological space X such that X = U ∪V . If U ∩V is simply connected,
then π(X) ∼
= π(U ) ⋆ π(V ).
Corollary 14.5.4 Suppose that a topological space X = U ∪ V with
path-connected open subsets U and V such that U ∩ V ̸= ∅ pathconnected. If V is simply connected, then π(X) ∼
= π(U )/N , where N
is the normal subgroup of π(U ) generated by the image of π(U ∩ V ).
We illustrate the usefulness of the preceding results by computing
the fundamental group of some spaces.
Example 14.5.1 The fundamental group of S1 ∨ · · · ∨ S1 . First, consider
the wedge of two circles. Suppose that (X1 , x1 ) and (X2 , x2 ) are two
pointed spaces, where X1 ≈ S1 ≈ X2 . Put Wj = Xj − {−xj } for
j = 1, 2. Then each Wj is open in Xj . So U = p (X1 + W2 ) and V =
p (W1 + X2 ) are open in X, where p : X1 + X2 → X1 ∨ X2 is the
quotient map. Clearly, xj is a strong deformation retract of Wj . Let
Fj : Wj × I → Wj be a strong deformation retraction of Wj onto xj .
Then the mapping G1 : (X1 + W2 ) × I → X1 + W2 defined by
{
x
for x ∈ X1 , t ∈ I, and
G1 (x, t) =
F2 (x, t)
for x ∈ W2 , t ∈ I
induces a strong deformation retraction of U onto X1 . Similarly, X2 is
a strong deformation retract of V . Also, H : (W1 + W2 )×I → W1 +W2
432
Elements of Topology
defined by H(x, t) = Gj (x, t) for x ∈ Wj , t ∈ I induces a strong deformation retraction of U ∩ V onto x0 = [xj ] ∈ X. Since U ∩ V is
contractible, Corollary 14.5.3 implies that π(X) ∼
= π(U ) ⋆ π(V ). The
injections X1 → U and X2 → V , being homotopy equivalences, induce
isomorphisms π (X1 , x1 ) ∼
= π (U, x0 ) and π (X2 , x2 ) ∼
= π (V, x0 ). There∼
fore π(X) = π (X1 ) ⋆ π (X2 ) which is a free group on two generators.
Now, suppose that n > 2 and X = X1 ∨ · · · ∨ Xn , where each
Xj ≈ S1 . By induction, we assume that X1 ∨ · · · ∨ Xn−1 is a free
group on n − 1 generators. Let xj be the base point of Xj for every
j and denote the base point [xj ] of X by x0 . Put Wj = Xj − {−xj },
j = 1, . . . , n. Then each Wj is open. Set U = p (X1 + · · · + Xn−1 + Wn )
and V = p (W1 + · · · + Wn−1 + Xn ). Then both U and V are open
path-connected subsets
∨ of X such that X = U ∪ V and U ∩ V =
p (W1 + · · · + Wn ) = Wj . (See Figure 14.16 below for n=4.)
-x1
-x1
-x2
-x2
-x4
-x4
-x3
-x3
U
V
U∩V
FIGURE 14.16: Proof of the second step in Example 14.5.1.
We observe that x0 is a strong deformation retract of U ∩ V . Suppose
that Fj : Wj × I∑→ Wj is a strong
∑ deformation retraction of Wj onto
xj . Then H ∑
: ( Wj ) × I →
Wj is clearly a strong deformation
retraction of
Wj onto {x1 , . . . , xn }. It induces a strong deformation
retraction of U ∩ V onto {x0 }. Thus U ∩ V is simply connected, and
hence∨π(X) is the free product of π(U ) and π(V ). As above, one sees
n−1
that 1 Xj is a strong deformation retract of U and Xn is a strong
deformation retract of V . Consequently, π(U ) is a free group on n − 1
generators and π(V ) is an infinite cyclic group. It follows
π(X) is a
∨ that
∼
free group on n generators. Since the isomorphism
π
(
X
)
Z⋆· · ·⋆Z
=
j
∨
is induced by inclusion of each Xj into Xj , we can write explicit
THE FUNDAMENTAL GROUP
433
generators of∨this free group. In fact, if fj denote the standard loop in
Xj , then π ( Xj ) is just the free group ⟨[f1 ], . . . , [fn ]⟩.
Example 14.5.2 The fundamental group of the Klein bottle K 2 . Recall that K 2 is the quotient space I 2 / ∼, where (s, 0) ∼ (s, 1) and
(0, t) ∼ (1, 1 − t) for all s, t ∈ I.[ Choose two] reals 0 < q < r < 1/2
1
1
2
2
and let
2 )− q, 2 + q . Put U = I − J and
( 1J be 1the closed
) ( 1interval
1
2
V = 2 − r, 2 + r × 2 − r, 2 + r (Figure 14.17). If p : I → K 2
is the quotient map, then p(U ), p(V ) and p(U ) ∩ p(V ) = p(V ) − J 2
are all open path-connected subsets of K 2 such that K 2 = p(U )∪p(V ).
^
•
^
V
•
•
^
(1⁄2,1⁄2)
z0
^
^
^
J×J
^
S
^
I×I
FIGURE 14.17: Proof of Example 14.5.2.
Clearly,( p(V ))is homeomorphic to V , and V is simply connected. Therefore π K 2 , ∗ = π (p(U ), ∗) /N , where N is the normal subgroup of
π (p(U ), ∗) generated by the image of π (p(U ) ∩ p(V ), ∗) under the homomorphism induced by the inclusion λ : p(U ) ∩ p(V ) ,→ p(U ). We
first find π(p(U )) at the base point x1 = p((1, 1)). Observe that the
quotient space of the boundary ∂I 2 of I 2 modulo the restriction of the
2
relation ∼ to
as well as the
( ∂I2 ) is homeomorphic to the ‘figure 8’ space
subspace p ∂I ⊂ p(U ). So we can identify p(∂I 2 ) with the ‘figure
8’ space. Now, we show that p(∂I 2 ) is a strong deformation retract of
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Elements of Topology
p(U ). The mapping ρ : U → ∂I 2 defined by
)
 (
(1−2s)t

s
+
,
0
for t ≤ s ≤ 1 − t and t < 1/2 − q,

2t−1


(
)



 1, t + (s+2t−1−2st)
for 1 − s ≤ t ≤ s and s > 1/2 + q,
2s−1
(
)
ρ(s, t) =


s + (2s+t−1−2st)
,1
for 1 − t ≤ s ≤ t and t > 1/2 + q,

2t−1


(
)


 0, t + (1−2t)s
for s ≤ t ≤ 1 − s and s < 1/2 − q
2s−1
is clearly a retraction of U onto ∂I 2 . Now, define a mapping F : U ×I →
U by F (u, t′ ) = (1 − t′ )u + t′ ρ(u). It is easily verified(that F) is a strong
deformation retraction of U to ∂I 2 .( The) composite p|∂I 2 (◦ ρ induces
)
2
a continuous map ρ̄ : p(U ) → p ∂I 2 such that
ρ̄p
=
p|∂I
◦ρ
( 2)
on U . Obviously, ρ̄ is a retraction of p(U ) onto p ∂I . Also, observe
that if u1 ̸= u2 are in U and p(u1 ) = p(u2 ), then F (u1 , t′ ) = u1
and F (u2 , t′ ) = u2 for all t′ ∈ I. Consequently, the composition p ◦
F factors through p(U ) × I. Since the mapping p|U × 1 : U × I →
p(U ) × I is an identification, there is an induced continuous map G :
′
p(U ) × I → p(U ) given by G (p(s, t), t′ ) = pF
( (s,2t,
) t ). Clearly,
( 2G) is
a strong deformation retraction of p(U ) to p ∂I . Since
p
( ( 2 )∂I ) is
∼
homeomorphic to the ‘figure 8’ space, π (p(U ), x1 ) = π p ∂I , x1 is
the free group generated by two elements α, β say.
Next, we determine generators for N . We choose a point z0 ∈ U ∩V
on the line segment from (1/2, 1/2) to (1, 1) and take the point x0 =
p(z0 ) as the base point ∗ of p(U ) ∩ p(V ) and p(U ). Let S denote the
square with centre (1/2, 1/2), passing through the point z0 and having
sides parallel to that of J 2 . Then ∂S ≈ S1 is a strong deformation
retract of U ∩ V . Since p(U ) ∩ p(V ) ≈ U ∩ V , π (p(U ) ∩ p(V ), x0 ) is Z
generated by the homotopy class of a loop g which goes once around
p(∂S). Clearly, the retraction ρ maps ∂S homeomorphically
onto ∂I 2 ;
( 2)
accordingly, p(∂S) is wrapped twice around p ∂I by the retraction ρ̄.
Thus the loop ρ̄λg goes twice around one of the circles of ‘figure 8’ in the
same direction and traverses the other circle in the opposite directions.
It follows that the homotopy
class [µ̄ρ̄λg] ∈ π (p(U ), x1 ) is the word
(
)
αβα−1 β, where µ̄ : p ∂I 2 ,→ p(U ) is the inclusion map. If γ is the path
t′ → G (x0 , 1 − t′ ) in p(U ), then γ̂ : π (p(U ), x1 ) → π (p(U ), x0 ), [ω] 7→
[γ −1 ∗ ω ∗ γ], is an isomorphism, and the composition γ̂ ◦ µ̄# ◦ ρ̄# is the
identity isomorphism on π (pU, x0 ), since the inverse of G is a homotopy
between µ̄ρ̄ and 1p(U ) . Writing γ̂α = a and γ̂β = b, we see that [λg] =
γ̂[µ̄ρ̄λg] is the word aba−1 b. It follows that N is generated by the word
THE FUNDAMENTAL GROUP
435
(
) ⟨
⟩
aba−1 b (as a normal subgroup) and π K 2 , x0 = a, b|aba−1 b = 1 . (In
the next chapter, we shall see another presentation of the fundamental
group of the Klein bottle (refer to Ex. 15.4.4).)
The technique involved in the above calculation can be used to
prove the following more general result.
Theorem 14.5.5 Let X be the space obtained by attaching D2 to
a path-connected Hausdorff space Y by a continuous map f : S1 →
Y . Let j : Y → X be the canonical embedding. Suppose that z0 ∈
S1 and y0 = f (z0 ). Then π (X, j(y0 )) is isomorphic to the quotient
group π (Y, y0 ) /ker(j# ), and ker(j
( 1 # ) )is the smallest normal subgroup
containing the image of f# : π S , z0 → π (Y, y0 ).
(
)
Proof. Let p : D2 + Y → X be the quotient map and 0 denote the
center of D2 . By Theorem 7.5.4, X is Hausdorff so that U = X −{p(0)}
is open in X. It is also path-connected, since both D2 − {0} and Y are
path-connected and p(y)(̸= )p(0) for all y ∈ Y . Since j(Y ) is closed
◦
in X, V = X − j(Y ) ≈ D2 is open and path-connected. We have
( 2 )◦
( )◦
X = U ∪ V and U ∩ V ≈ D
− {0}. Since D2 is contractible, V is
!
●
°
<
x1
Á
U
<
V
>
●
^
x0
q
<
°
®
<
^
z0 ● ¾
● z1
!
<
°
●
°
^
0
x1
r
D2–{0}
U∩V
FIGURE 14.18: Proof of Theorem 14.5.5.
436
Elements of Topology
simply connected. It follows that the homomorphism k# : π (U, x1 ) →
π (X, x1 ) induced by the inclusion k : U ,→ X is surjective, and ker(k# )
is the smallest normal subgroup of π (U, x1 ) containing the image of
π (U ∩ V, x1 ) under the homomorphism induced by the inclusion U ∩
V ,→ U . Of course, we need to change the base point from x1 to
x0 = j(y0 ) = p(z0 ). Assume that z1 is on the radius through z0 and
consider the straight-line path σ in D2 from z1 to z0 . Let ϕ and ψ be
paths in U and X, respectively, defined by σ. (See Figure 14.18). Then
ψ = pσ and ϕ = qσ, where q : D2 − {0} → U is the map defined by p.
So we have the following commutative diagram
π(U, x1 )
k# -
ϕ̂
π(X, x1 )
ψ̂
?
π(U, x0 )
k#
?
- π(X, x0 )
FIGURE 14.19: Surjectivity of k# in the proof of Theorem 14.5.5.
where ϕ̂ and ψ̂ denote the isomorphisms defined by the path ϕ and ψ,
respectively. Since ψ̂ ◦ k# = k# ◦ ϕ̂ and the top arrow in Figure 14.19
is an epimorphism, so is the bottom arrow.
Now, we determine π (U, x0 ) and the kernel of the homomorphism
k# : π (U, x0 ) → π (X, x0 ). We first( observe )that j(Y ) is a strong
deformation retract of U . Let F : D2 − {0} × I → D2 − {0} be
a strong deformation retraction of D2 − {0} to S1 . Define a function
H : U × I → U by
{
pF (z, t)
if x = p(z) for some z ∈ D2 and
H(x, t) =
x
if x = p(y) for some y ∈ Y .
(
)
Notice that p(Y ) and p D2 − {0} are closed in U and pF (z, t) =
p(z) = p(y) for z ∈ S1 and f (z) = y. So H is a continuous function.
It is now easily checked that H is a strong deformation retraction
of U to j(Y ). Hence i# : π (Y, y0 ) → π (U, x0 ) is an isomorphism,
where i : Y ,→ U is defined by( j. )Next, to see the kernel of ker (k# ),
◦
we choose a point z1 ̸= 0 in D2 and take x1 = p (z1 ) ∈ U ∩ V .
Let C denote the concentric circle in D2 passing through z1 , and γ
denote the loop which goes once around C. Observe that the mapping
THE FUNDAMENTAL GROUP
437
( )◦
r : D2 − {0} → U ∩ V defined by p is a homeomorphism. Therefore
U ∩V ≃ S1 and hence π (U ∩ V, x1 ) is an infinite cyclic group generated
by the homotopy class of ω = rγ. Thus the kernel of the homomorphism
k# : π (U, x1 ) → π (X, x1 ) is the smallest normal subgroup of π (U, x1 )
generated by the homotopy class of ω in U . Now, it is immediate that
the kernel of the homomorphism k# : π (U, x0 ) → π (X, x0 ) is the
smallest normal
subgroup
generated by the class ϕ̂[ω]. If the class [α]
(
)
generates π S1 , z0 , then σ −1 ∗ γ ∗ σ ≃ α±1 rel {0, 1} in D2 − {0} ⊂
p−1 (U ). So, in U , we have ϕ−1 ∗ ω ∗ ϕ = ϕ−1 ∗ rγ ∗ ϕ ≃ qα±1 rel {0, 1}.
It follows that ϕ̂[ω] = [qα]±1 in π (U, x0 ). Obviously, q|S1 = if , so that
q# [α] = i# f# ([α]). Also, j# = k# ◦ i# and thus we have the following
commutative diagram.
f#
- π(Y, y0 ))
i# q#
j#
?
?
k#
- π(X, x0 )
π(U, x0 )
π(S1 , z0 )
FIGURE 14.20: The last step in the proof of Theorem 14.5.5.
Since i# is an isomorphism, it is immediate that the inverse image
of ker (k# ) under the isomorphism i# : π (Y, y0 ) → π (U, x0 ) is the
smallest normal subgroup containing im(f# ). This completes the proof.
♢
Exercises
1. Let Gα , α ∈ A, be a collection of abelian groups, and F be their free
product with respect to monomorphisms ıα : Gα → F . If F ′ is the
commutator subgroup of F , show that F/F ′ is the direct sum of the
groups Gα .
2. Suppose that a group G is generated by a family of subgroups Gα ,
α ∈ A. If, for every α ̸= β, Gα ∩ Gβ = {e} (the identity element of G)
and the empty word is the only reduced form of e as a finite product of
elements of the Gα , show that G is the free product of the groups Gα .
3. Prove that the smallest normal subgroup of a group
{ G containing a sub}
set X ⊂ G is generated as a subgroup by the set gxg −1 |g ∈ G, x ∈ X .
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Elements of Topology
4. Let F be the free group on two elements a and b. Prove that (i) the
commutator subgroup F ′ is not finitely generated, and (ii) the normal
subgroup generated by the single commutator aba−1 b−1 in F is F ′ .
5. Suppose that G1 and G2 are groups and N is the smallest normal subgroup of the free product G1 ⋆ G2 containing G1 . Prove that
(G1 ⋆ G2 ) /N ∼
= G2 .
6. Let F be a free group with basis X and ϕ be a homomorphism of a
group G onto F . Suppose that there is a subset S ⊂ G which is mapped
bijectively into X by ϕ. Show that the subgroup ⟨S⟩ of G is free on S.
7. Show that
of a regular n-gon has presen⟨ the group Dn of symmetries
⟩
tation a, b|an = b2 = (ab)2 = 1 .
8. Compute the fundamental group of (i) two circles joined by a line segment, (ii) the figure θ space, (iii) the doubly punctured plane R2 −{p, q},
(iv) the punctured torus, (v) the complement of the three coordinate
axes in R3 , (vi) the unit 2-sphere S2 with the unit disc in xy-plane, and
(vii) the unit 2-sphere S2 with a diameter.
9. Let τ : S1 → S1 be rotation through the angle 2π/n, n > 1, and
X be the quotient space of D2 obtained by identifying the points
x, τ (x), . . . , τ n−1 (x) for every x ∈ S1 . Show that π(X) is the cyclic
group of order n. (Some authors call X the n-fold dunce cap).
10. Find a presentation of the fundamental group of the torus.
11. Determine the fundamental group of the adjunction space of Dn to a
path-connected Hausdorff space Y by a continuous function f : Sn−1 →
Y, n > 2.
12. Let X1 , . . . , Xn be pointed spaces with base points x1 , . . . , xn , respectively. Suppose that each {xi } is a strong deformation retract of an open
subset Wi ⊆ Xi . Prove that
π (X1 ∨ · · · ∨ Xn , x0 ) ∼
= π (X1 , x1 ) ∗ · · · ∗ π (Xn , xn ) ,
where x0 denote the element determined by the xi .
Chapter 15
COVERING SPACES
15.1
15.2
15.3
15.4
15.5
15.1
Covering Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Lifting Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Universal Covering Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Deck Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Existence of Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
439
448
459
468
480
Covering Maps
The concept of “Covering Space” provides a powerful tool for the
computation of the fundamental group of various spaces and is, thus,
intimately related to its study. In this section, we discuss some basic
properties and examples of covering maps. A familiar example of such
maps is the exponential map p : R1 → S1 , t 7→ e2πıt . In §14.3, we have
already studied some properties of this map, and the most essential
one is captured in the following.
e be spaces. A continuous map p :
Definition 15.1.1 Let X and X
e
X → X is called a covering map (or projection) if each point x ∈ X
has an open nbd U such that p−1 (U ) is a disjoint union of open sets
eα in X,
e and p|U
eα : U
eα → U is a homeomorphism for every index α.
U
Such an open set U is called admissible or evenly covered.
e → X is a covering map, then X is called the base space of p,
If p : X
e
and X is called a(covering) space of X. If x ∈ X and U is an admissible
nbd of x, then p p−1 (U ) = U . So x ∈ im(p) and p is surjective.
e are homeomorphic spaces, then any homeExample 15.1.1 If X and X
e → X is a covering map, so X
e is a covering space of X.
omorphism X
We use the term “trivial” to refer to such covering maps.
Example 15.1.2 By Proposition 14.3.4, the exponential map p : R1 →
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Elements of Topology
S1 , t 7→ e2πıt , is a covering map. The covering space R1 may be looked
at as an infinite spiral over S1 (see Figure 15.1).
R1
…
…
^
p
S1
FIGURE 15.1: An infinite spiral as a covering space of S1 .
Example 15.1.3 Let G be a Hausdorff topological group, and H be
a discrete subgroup of G. Let G/H denote the space of left cosets (or
right cosets) with the quotient topology determined by the natural map
π : G → G/H. Then π is a covering map. To see this, we first observe
that π is an open mapping. If O ⊆ G is open, then Oh is open in G for
every h ∈ H, since the right translation
x 7→ xh
∪
∪ is a homeomorphism
of G onto itself. Thus π −1 π (O) = x∈O xH = h∈H Oh is open in G.
Since π is an identification map, π (O) is open in G/H.
Next, we note that it is enough to prove the existence of an admissible nbd of ē ∈ G/H, where e is the identity element of G. Recall that,
for each g ∈ G, the left translation λg : G → G, x 7→ gx, is a homeomorphism. This clearly induces a homeomorphism θg : G/H → G/H
given by θg (x̄) = gx, x̄ ∈ G/H. It is easily checked that θg (U ) is an
admissible nbd of ḡ, if U is that of ē.
Finally, we show that there is an admissible nbd U of ē in G/H.
Since H is discrete, there exists an open nbd N of e in G such that
N ∩ H = {e}. By the continuity of the map (x, y) 7→ x−1 y on G × G,
we find an open nbd V of e such that V −1 V ⊆ N. Set U = π (V ). Then
U is obviously an open nbd of ē in G/H. We
∪ assert that U is evenly
covered by π. It is clear that π −1 (U ) = h∈H V h. As above, each
COVERING SPACES
441
V h is open in G. To see that these sets are mutually disjoint, suppose
that V h ∩ V h′ ̸= ∅, where h, h′ ∈ H. Then there exists v, v ′ ∈ V
such that vh = v ′ h′ ⇒ v ′−1 v = h′ h−1 ∈ V −1 V ∩ H ⊆ N ∩ H = {e}.
So h = h′ , and the sets V h are pairwise disjoint. It remains to prove
that π|V h maps V h homeomorphically onto U. Being the restriction
of a continuous and open map to an open set, π|V h : V h → U is also
continuous and open. If v, v ′ ∈ V and π (vh) = π (v ′ h), then vh = v ′ hh′
for some h′ ∈ H. So v ′−1 v = hh′ h−1 ∈ V −1 V ∩ H = {e}, which implies
that v = v ′ . Thus π|V h is injective. Also, since xhH = xH for every
x ∈ G and h ∈ H, we have π (V h) = U, and π|V h maps V h onto U .
This completes the proof.
Example 15.1.4 Consider the topological group S1 , the unit circle. If
n is a positive integer, then the map pn : S1 → S1 which maps z ∈ S1
into z n is a covering map and S1 itself is a covering
S1 . To see
{ space
}
1
1
1 n
this, given z0 ∈ S , let U = S − {−z0 } and F = z ∈ S |z = −z0 .
Then U is a nbd of z0 and p−1
) = S1 − F . If z0 = eıθ0 , 0 ≤ θ0 < 2π,
n (U{
}
ı(θ0 +π)
ı(θ0 +(2j+1)π)/n
then −z0 = e
so
|j = }
0, 1, . . . , n − 1 .
{ that F = e
<θ<
For each j, put Vj = eıθ | θ0 +(2j+1)π
n
θ0 +(2j+3)π
n
.
V2
Ä
•
z0
Ä
•
Ä
Ä
• V1
p4
U
V3 •
Ä
Ä
•
Ä
Ä
Ä
Ä
V4
FIGURE 15.2: An admissible nbd in Example 15.1.4 (for n = 4).
Then each Vj is the intersection of S1 with a suitable open disc in R2 ; in
fact,Vj is an open arc on S1 subtending an angle of 2π/n at the centre.
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Elements of Topology
∪n−1
Obviously, p−1
Vj . Notice that each of the mappings
n (U ) =
0
(
)
θ0 +(2j+3)π
Vj → θ0 +(2j+1)π
,
, eıθ 7→ θ,
n
n
(θ0 − π, θ0 + π) → S1 − {−z0 }, θ 7→ eıθ , and
(
)
θ0 +(2j+1)π θ0 +(2j+3)π
,
→ (θ0 − π, θ0 + π), t 7→ nt − 2(j + 1)π,
n
n
is a homeomorphism, and their composition is the map z 7→ z n . Thus
pn |Vj is a homeomorphism of Vj onto U .
Example 15.1.5 The plane R2 is a covering
space) of the torus T =
(
S1 × S1 with the covering map (s, t) 7→ e2πıs , e2πıt .
This is immediate from the following.
e → X and q : Ye → Y are covering maps,
Proposition 15.1.2 If p : X
e × Ye → X × Y , (x̃, ỹ) →
then so is p × q : X
7 (p (x̃) , q (ỹ)).
Proof. Given (x, y) in X × Y , let U and V be admissible nbds of x and
y, respectively. We observe that U × V is an admissible nbd of (x, y).
−1
Since (p × q) (U × V ) = p−1 (U ) × q −1 (V ), the product of a sheet
over U and a sheet over V is clearly a sheet over U × V . This completes
the proof.
♢
Note that the infinite cylinder R1 ×S1 or the torus T itself can serve
as a covering space of T. The following example shows that Proposition
15.1.2 cannot be generalised to arbitrary products.
en = R1 , Xn = S1 and
Example 15.1.6 For each integer n ≥ 1, let X
en → Xn be the exponential map t 7→ e2πıt . We have already seen
pn : X
∏
∏ e
∏
that pn is a covering map.∏
But pn : X
Xn is not a covering
n → ∏
en is not injective.
map, for the restriction of pn to an open set in X
Definition 15.1.3 A continuous map f : Y → X is called a local
homeomorphism if each point y ∈ Y has an open nbd V such that
f (V ) is open in X and f |V is a homeomorphism of V onto f (V ).
e → X is a local homeWith this terminology, a covering map p : X
omorphism. Consequently, any local property of X, such as local come It follows
pactness, local path-connectedness, etc., is inherited by X.
from the following proposition that p is an identification. Thus the base
space of a covering map is a quotient space of its covering space.
COVERING SPACES
443
Proposition 15.1.4 A local homeomorphism is an open mapping.
Proof. Let p : Y → X be a local homeomorphism, and V be an open
subset of Y . If x ∈ p (V ), then we find a y ∈ V such that p (y) = x. By
our hypothesis, there exists an open neighbourhood U of x in X, and
an open nbd W of y in Y such that p maps W homeomorphically onto
U . Since V ∩W is open in W , and U is open in X, p (V ∩ W ) is open in
X. Obviously, x ∈ p (V ∩ W ) ⊂ p (V ). Thus p (V ) is a neighbourhood
of x, and the proposition follows.
♢
However, a local homeomorphism need not be a covering map, as
shown by the following.
Example 15.1.7 Let n > 1 be an integer and X be the open interval
(0, n) with the usual topology. Consider the map p : X → S1 given
by p (t) = e2πıt . Being the restriction of a local homeomorphism to an
open subset, p is a local homeomorphism. Clearly, p is a surjection;
but it is not a covering map because 1 ∈ S1 has no admissible nbd (see
Figure 15.3 for n = 4). The space X may be regarded as an open finite
spiral over S1 .
0
X
( )
1
( ı )
S1
2
( ı )
3
( ı )
4
( )
•
•
•
( •1
(
FIGURE 15.3: Proof of Example 15.1.7.
e → X is a covering map, and the open set U ⊆ X is evenly
If p : X
eα ⊂ p−1 (U ) which are mapped homeomorphicovered, then the sets U
cally onto U by p are referred to as the sheets over U . The inverse
image of a point x ∈ X under p is called the fibre over x. Clearly, the
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Elements of Topology
e The number of
fibre p−1 (x) over any point x is a discrete subset of X.
points in a fibre is locally constant in the sense that each point x ∈ X
has an open nbd U such that the cardinality of p−1 (u) is the same for
eα is a sheet
every u ∈ U . For, if U is an admissible open set in X, and U
−1
eα contains exactly one point of each fibre p (x), x ∈ U,
over U , then U
−1
and therefore
{ there
} is a one-to-one correspondence between p (x) and
eα of all sheets over U. Since a locally constant function
the family U
on a connected space is constant, we see that the cardinality of p−1 (x)
is the same for all x ∈ X if X is connected. In this case, the number
of sheets (or the multiplicity) of p is defined to be the cardinal number of a fibre p−1 (x). Covering maps with n (finite) sheets are often
called “n-fold” coverings; in particular, a two-fold covering is termed a
“double covering.”
Example 15.1.8 The quotient map q : S2 → RP2 which identifies each
pair of antipodal points is a double covering. We first observe that π
is an open map. Let O be any open set in S2 . Since the antipodal map
f : S2 → S2 , f (x) = −x, is a homeomorphism, f (O) is open in S2 .
Obviously, q −1 (q (O)) = O ∪ f (O) is open, and therefore q (O) is open
in RP2 . To show the existence of admissible nbds, let ξ ∈ RP2 . Suppose
that ξ = q (x) = q (−x), and consider an open set O in S2 containing
x which does not contain any pair of antipodal points (O may be
taken to be a small disc centred at x). Then U = q (O) is an open set
containing ξ, and q|O : O → U is a bijection. Being the restriction of a
continuous open map to an open set, q|O is also continuous and open.
Thus q|O is a homeomorphism between O and U. Similarly, q maps
f (O) homeomorphically onto U . Clearly, q −1 (U ) is the disjoint union
of the sets O and f (O) whence U is an admissible nbd of ξ. So S2 is a
covering space of RP2 with π as the covering map.
Similarly, the quotient map Sn → RPn which identifies the antipodal points is a double covering.
For locally path-connected spaces, the study of covering maps reduces to the study of covering maps with (path) connected base space.
This follows from the following.
Theorem 15.1.5 Let X be a locally path-connected space. Then a
e → X is a covering map if and only if, for each
continuous map p : X
path component C of X, p|p−1 (C) : p−1 (C) → C is a covering map.
COVERING SPACES
445
e → X is a covering map and x ∈ C. Let U
Proof. Suppose that p : X
be an admissible open nbd of x ∈ X, and V be the path component of
U containing x. Then V ⊆ C, for C is a path component of X. Since
X is locally path-connected, V is open in X, and therefore open in C.
Clearly, V is evenly covered by q = p|p−1 (C), and q is a covering map.
Conversely, assume that the map q : p−1 (C) → C, x 7→ p (x), is a
covering map for each path component C of X. Given x ∈ X, let C be
the path component of X which contains x. By our hypothesis, there
is an open nbd U of x in C, which is evenly covered by q. Since X is
locally path-connected, the path component C is open in X, and hence
e It is
U is open in X. Also, every open subset of p−1 (C) is open in X.
now clear that U is p-admissible, and p is a covering map.
♢
Recall that the path components of a locally path-connected space
coincide with the components. The next result increases the supply of
examples of covering spaces.
e → X be a covering map. If X is locally
Proposition 15.1.6 Let p : X
e
e then p(C)
e is a path
path-connected and C a path component of X,
e:C
e → p(C)
e is a covering map.
component of X and p|C
e a path
Proof. Suppose that X is locally path-connected, and let C
e
e
component of X. To prove that p(C) is a path component of X, it
suffices to establish that it is a component, since the components and
e is conpath components of X are identical. It is obvious that p(C)
e is clopen in X. Let x ∈ cl(p(C)).
e Since
nected. We observe that p(C)
X is locally path-connected, there is an admissible path-connected
e over U is path-connected.
open nbd U of x. Accordingly, each sheet U
−1
e ∩ p (U ) ̸= ∅, for U ∩ p(C)
e ̸= ∅. So U
e ∩C
e ̸= ∅ for
We have C
e
e
e
some sheet U over U . As C is a path component of X, we have
e ⊆ C.
e So U = p(U
e ) ⊆ p(C)
e and x ∈ int(p(C)).
e
U
This implies that
e
e
e
cl(p(C)) ⊆ int(p(C)), and therefore p(C) is both closed and open. It
e is a path component of X.
follows that p(C)
e : C
e → p(C)
e is a covering map. Let
Now, we show that q = p|C
e
x ∈ p(C) and U be an admissible path-connected open nbd of x in X.
e If U
e is a sheet over U and U
e ∩C
e ̸= ∅, then U
e ⊆ C.
e
Then U ⊆ p(C).
e over U
It follows that q −1 (U ) is the disjoint union of those sheets U
e
which intersect C. Thus U is evenly covered by q, and q is a covering
map.
♢
The converse of the preceding proposition is false, as shown by the
following.
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Elements of Topology
Example 15.1.9 Let X be a countable product of unit circles S1 and,
en = Rn × X. Then the mapping pn : X
en → X
for each integer n ≥ 1, X
defined by
(
)
pn (t1 , . . . , tn , z1 , z2 , . . .) = e2πıt1 , . . . , e2πıtn , z1 , z2 , . . .
∑ e
en = pn .
is clearly a covering map. Let p : ∑
Xn → X be defined by p|X
en are the sets X
en . If an open
Note that the path components of
X
set O ⊆ X is evenly covered by p, then there is a basic connected
open set G ⊆ O which is also evenly covered by p, since X is locally
connected (ref. Exercise 7). Such a set G must be evenly covered by
∩k
each pn . On the other hand, suppose that G = 1 πi−1 (Ui ), where
πi : X → S1 is the projection onto the ith factor of X and the Ui ⊆ S1
are connected open sets. Let π
ei be the projection onto the ith factor
en and ex : R1 → S1 denote the map t → e2πıt . Then, for n > k,
of X
∩k
−1
p−1
ei ) (Ui ).
n (G) =
1 (ex ◦ π
1
In this case, the (k + 1)st factor of each component of p−1
n (G) is R .
−1
Consequently, pn is not injective on any component of pn (G). Thus,
G is not evenly covered by pn . This contradiction shows that no open
subset of X is admissible under p, and p is not a covering map.
Exercises
e be the product of a space X with a discrete space. Show that
1. Let X
e → X is a covering map.
the projection X
2. Let X be the wedge of two circles.
{
}
e = (x, y) ∈ R2 : x or y is an integer
(a) Prove that the subspace X
is a covering space of the subspace X ⊂ R2 .
(b) Is the quotient map π : S1 → X which identifies the points 1 and
−1 a covering map?
3. • Let A, B and C denote the circles of radius 1 centred at 0, 2 and 4,
e = A∪B ∪C
respectively, in the complex plane C. Let X = A∪B and X
e
be subspaces of C. Define a mapping p : X → X by

for z ∈ A,
 z
2
p (z) =
2 − (z − 2) for z ∈ B and

4−z
for z ∈ C.
Show that p is a covering map.
COVERING SPACES
447
4. Prove that the mapping z 7→ z n of C − {0} onto itself, where n ̸= 0, is
a covering map.
5. Consider
∑∞ n the exponential function exp : C → C given by exp (z) =
0 z /n!. Show that C is a covering space of C − {0} relative to exp.
6. Show that there is a two-sheeted covering of the Klein bottle by the
torus.
e → X be a covering map. If X is locally path-connected,
7. • Let p : X
show that each point of X has a path-connected open nbd U such that
each path component of p−1 (V ) is mapped homeomorphically onto V
by p.
8. Let X be a connected space. Show that a locally constant function on
X is constant.
eν → Xν , ν ∈ N, be a collection of covering maps. Show that
9. Let pν : X
∑ e
∑
e
the mapping p : ν X
ν →
ν Xν , defined by p|Xν = pν , is a covering
map.
e
10. If N is finite and, for each
∑ eν ∈ N, pν : Xν → X is a covering map, prove
that the function q : ν Xν → X, defined by q|Xν = pν , is a covering
map.
e → X be a covering map, and suppose that X is connected.
11. Let p : X
e is compact if and only if X is compact and the multiplicity
Prove that X
of p is finite.
12. Let f : X → Y be a local homeomorphism, and suppose that X is
compact. Prove:
(a) f −1 (y) is finite for every y ∈ Y.
(b) If Y is connected and both X and Y are Hausdorff, show that f
is a covering map.
13. Suppose that p : Y → X and q : Z → Y are covering maps such
that p−1 (x) is finite for every x ∈ X. Show that the composite pq is a
covering map.
e → X be a covering map. If X is Hausdorff, show that X
e is
14. Let p : X
also Hausdorff.
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15.2
Elements of Topology
The Lifting Problem
e → X and f : Y → X are
Suppose that continuous maps p : X
given. By a lifting (or lift) of f relative to p, we mean a continuous
e such that p ◦ f˜ = f , that is, the following diagram
map f˜ : Y → X
f˜ Y
f
3
e
X
p
?
- X
is commutative. If there is such a map f˜, then we say that f can be
e An important problem in topology is to determine
lifted to f˜ in X.
e This is referred to as the lifting
whether or not f has a lifting in X.
problem for the given map f .
We begin by solving the lifting problem for maps from the unit
interval I to X, and use this fact to show that every homotopy from a
locally connected space into X can also be lifted. These are straightforward generalisations of the path lifting property and the homotopy
lifting property of the exponential map from R1 to S1 for a covere → X. Finally, we will establish a simple criterion for
ing map p : X
the existence of liftings of continuous maps from a connected, locally
path-connected space into the base space of p purely in terms of the
fundamental groups of the spaces involved. We also study here a technique for computing the fundamental group of (some
) spaces which is
analogous to one used in the determination of π S1 .
e → X
Theorem 15.2.1 (The Path Lifting Property) Let p : X
be a covering map and let f : I → X be a path with the origin x0 .
e with
If x̃0 ∈ p−1 (x0 ), then there exists a unique path f˜ : I → X
f˜ (0) = x̃0 and pf˜ = f .
Proof. Suppose that [a, b] ⊆ I and U is an admissible nbd of x = f (a)
such that f ([a, b]) ⊆ U . If x̃ ∈ p−1 (x), then x̃ lies in a unique sheet, say
e . It is easy to see that g̃ : [a, b] → X
e defined by g̃ = (p|U
e )−1 ◦(f |[a, b])
U
is continuous and satisfies pg̃ = f |[a, b], and g̃ (a) = x̃.
Now, for each t ∈ I, let Ut be an admissible nbd of f (t). Then
COVERING SPACES
449
{ −1
}
f (Ut ) : t ∈ I is an open covering of I. Since I is a compact metric
space, this open cover of I has a Lebesgue number δ, say. Consider a
partition of I with points
0 = t0 < t1 < · · · < tn = 1
so that ti+1 − ti < δ for 0 ≤ i < n. Then each subinterval [ti , ti+1 ] is
mapped by f into some Ut . By our initial remarks, there is a continuous
e with pg̃1 = f |[0, t1 ] and g̃1 (0) = x̃0 . Similarly,
map g̃1 : [0, t1 ] → X
e with pg̃2 = f |[t1 , t2 ] and
there is a continuous map g̃2 : [t1 , t2 ] → X
g̃2 (t1 ) = g̃1 (t1 ); indeed for each 1 ≤ i < n, there is a continuous map
e with pg̃i+1 = f |[ti , ti+1 ] and g̃i+1 (ti ) = g̃i (ti ). By
g̃i+1 : [ti , ti+1 ] → X
the Gluing lemma, we may combine the functions g̃i into a continuous
e where f˜ (t) = g̃i (t) for ti−1 ≤ t ≤ ti . The path f˜
map f˜ : I → X,
clearly satisfies the requirement of the theorem.
The uniqueness of the lifting f˜ follows from the following more
general result.
e →X
Theorem 15.2.2 (The Unique Lifting Property) Let p : X
be a covering map, Y a connected space, and f : Y → X be continuous.
e and y0 ∈ Y such that f (y0 ) = p (x̃0 ), there is at most
Given x̃0 ∈ X
e with pf˜ = f and f˜ (y0 ) = x̃0 .
one continuous map f˜ : Y → X
e
Proof. Assume that there are two continuous maps f˜1 , f˜2 : Y → X
such
that pf˜1 = f }= pf˜2 and {f˜1 (y0 ) = x̃0 = f˜2 }
(y0 ). Let A =
{
y ∈ Y |f˜1 (y) = f˜2 (y) and B = y ∈ Y |f˜1 (y) ̸= f˜2 (y) . It is obvious
that Y = A ∪ B, A ∩ B = ∅. We show that A and B, both, are open.
This forces that B = ∅, since Y is connected, and A ̸= ∅ (for y0 ∈ A).
So f˜1 = f˜2 .
e be
Given a ∈ A, let U be an admissible open nbd of f (a), and let U
−1 e
−1 e
˜
˜
˜
˜
the sheet over U containing f1 (a) = f2 (a). Then O = f1 (U )∩ f2 (U )
e . So
is an open nbd of a in Y . If y ∈ O, then f˜1 (y) and f˜2 (y) belong to U
˜
˜
˜
˜
e is
pf1 (y) = f (y) = pf2 (y), which implies that f1 (y) = f2 (y), since p|U
a homeomorphism. It follows that y ∈ A, and O ⊆ A. Thus A is a nbd
of each of its points, and therefore open. To show that B is open, let
b ∈ B be arbitrary. Consider an admissible open nbd U of f (b). Since
e1 and U
e2 over U containing
f˜1 (b) ̸= f˜2 (b), there are distinct sheets U
−1 e
˜
˜
˜
e2 ) is an
f1 (b) and f2 (b), respectively. Obviously, N = f1 (U1 ) ∩ f˜2−1 (U
e1 and U
e2 are disjoint, f˜1 (y) ̸= f˜2 (y) for every
open nbd of b. Since U
y ∈ N , so we have N ⊆ B. Therefore B is also open, and the proof is
complete.
♢
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Elements of Topology
Theorem 15.2.3 (The Covering Homotopy Theorem) Let p :
e → X be a covering map. If f˜ : Y → X
e is a continuous map of
X
e
a locally connected space Y into X, and F : Y × I → X is a homotopy
such that F (y, 0) = pf˜ (y) for every y ∈ Y , then there exists a unique
e such that Fe (y, 0) = f˜ (y) for every y ∈ Y
homotopy Fe : Y × I → X
e
and pF = F , that is, the following diagram
f˜
e
- X
3
j
p
Fe
? ?
F
Y ×I
X
Y
commutes, where j maps y into (y, 0). Moreover, if F is a homotopy
rel A for some A ⊂ Y , then so is Fe.
Proof. If a homotopy Fe having the desired properties exists, then, for
each y ∈ Y , f˜y : t 7→ Fe (y, t) is a lifting of the path t 7→ F (y, t) starting
at f˜ (y) and, by Theorem 15.2.1, such a lift exists and is unique. So we
define Fe (y, t) = f˜y (t). Then we have Fe (y, 0) = f˜ (y) and pFe (y, t) =
pf˜y (t) = F (y, t) for every y ∈ Y and t ∈ I. The uniqueness of Fe
is immediate once we prove that it is continuous.
Let U be an
{
} open
covering of X by admissible open sets. Then F −1 (U ) |U ∈ U is an
open covering of Y × I. Choose a covering V of Y × I consisting of
basic open sets each of which is contained in some F −1 (U ), U ∈ U.
For each y ∈ Y , {y} × I is compact, and therefore the open covering
{V ∩ ({y} × I) |V ∈ V} of {y} × I has a Lebesgue number ϵ, say. We
choose a partition 0 = t0 < t1 < · · · < tn = 1 of I such that ti −ti−1 < ϵ
for every i = 1, . . . , n. Then there is a member Vi of V such that {y} ×
[ti−1 , ti ] ⊂ Vi . Let Ny be the intersection of the first factors of the Vi .
Then Ny ×[ti−1 , ti ] is also contained in Vi so that F maps Ny ×[ti−1 , ti ]
into some U ∈ U. Since Y is locally connected, we may assume that
Ny is connected. We show, by induction on i, that Fe is continuous on
each Ny × [ti−1 , ti ]. Note that Fe is continuous on Ny × {0}, since it
agrees there with the continuous map f˜. Assume that Fe | (Ny × {ti−1 })
is continuous. Then Fe (Ny × {ti−1 }) is connected, for Ny is so. As seen
above, there is a U ∈ U such that F (Ny × [ti−1 , ti ]) ⊂ U . Being a
e (say) over
connected set, Fe (Ny × {ti−1 }) is contained in a sheet U
COVERING SPACES
451
(
)
U . Now, if z ∈ Ny and ti−1 ≤ t ≤ ti , then we have p f˜z (t) =
F (z, t) ∈ U . Since the subinterval [ti−1 , ti ] is connected and f˜z (ti−1 ) =
e , we have f˜z ([ti−1 , ti ]) ⊂ U
e . So Fe (Ny × [ti−1 , ti ]) ⊂ U
e.
Fe (z, ti−1 ) ∈ U
e
e
e
Since p|U is a homeomorphism of U with U , and pF (z, t) = F (z, t), we
e )−1 (F (z, t)) for every z ∈ Ny and ti−1 ≤ t ≤ ti .
see that Fe (z, t) = (p|U
This implies that Fe is continuous on Ny × [ti−1 , ti ]. Observe that this
induction argument also shows that Fe is continuous on Ny × [t0 , t1 ]
because Fe is continuous on Ny × {t0 }. Since Ny × I is a finite union of
closed sets Ny × [ti−1 , ti ], 1 ≤ i ≤ n, the Gluing lemma shows that Fe is
continuous on Ny × I. The family {Ny × I : y ∈ Y } is an open covering
of Y × I, and hence Fe is continuous (refer to Exercise 2.1.9).
To see the last statement, suppose that a ∈ A and F (a, t) = pf˜ (a)
for every t ∈ I. Then Fe (a, t) ∈ p−1 (pf˜ (a)) for every t ∈ I. Since
Fe ({a} × I) is connected, and p−1 (pf˜ (a)) is discrete, we have Fe (a, t) =
f˜ (a) for every t ∈ I and each a ∈ A. This completes the proof.
♢
We remark that the preceding theorem holds without the condition
of local connectedness on Y , but the proof is slightly complex (ref.
Spanier [13] Theorem 3 on p. 67). Observe that if f, g : Y → X are
e → X,
homotopic and f has a lifting relative to the covering map p : X
e Thus, whether or not a continuous map
then g also has a lifting in X.
e
Y → X can be lifted to X is a property of the homotopy class of the
map. Also, notice that the “Path Lifting Property” of a covering map
can be derived from the “Covering Homotopy Theorem” by taking Y
to be the one-point space.
e → X
Theorem 15.2.4 (The Monodromy Theorem) Let p : X
be a covering map. Suppose that f and g are paths in X and f ≃ g
rel {0, 1}. If f˜ and g̃ are the liftings of f and g, respectively, with
f˜ (0) = g̃ (0), then f˜ (1) = g̃ (1) and f˜ ≃ g̃ rel {0, 1}.
Proof. Let F : f ≃ g rel {0, 1}. By Theorem 15.2.3, there exists a unique
e with pFe = F and Fe (s, 0) = f˜ (s),
continuous map Fe : I × I → X
Fe (0, t) = f˜(0) and Fe (1, t) = f˜ (1) for every s, t ∈ I. The function
e defined by ge1 (s) = Fe (s, 1), s ∈ I, is obviously a continuous
ge1 : I → X,
lifting of g with the origin ge1 (0) = g̃ (0). By Theorem 15.2.2, ge1 = g̃,
and we have g̃ (1) = Fe (1, 1) = f˜ (1). Thus Fe : f˜ ≃ g̃ rel {0, 1}, as
desired.
♢
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Elements of Topology
e → X be a covering map and x̃ belong to
Corollary 15.2.5 Let p : X
the fibre over x0 ∈ X. Then
e x̃) → π (X, x0 ) is a
(a) the induced homomorphism p# : π(X,
monomorphism, and
(b) the path class [f ] of a loop f in X based at x0 belongs to im(p# )
e based at x̃.
if and only if f lifts to a loop in X
e be a closed path in X
e at x̃, and p# ([f˜]) =
Proof. (a): Let f˜ : I → X
[pf˜] = 1 in π (X, x0 ). Then there is a homotopy F : pf˜ ≃ cx0 rel {0, 1},
where cx0 is the constant path at x0 . Since the constant path cx̃ at x̃ is
a lifting of cx0 and f˜ (0) = x̃, we have f˜ ≃ cx̃ rel {0, 1}, by the preceding
e x̃), and p# is a monomorphism.
theorem. Therefore [f˜] = 1 in π(X,
e based at x̃ and pf˜ = f , then clearly [f ] ∈
(b): If f˜ is a loop in X
im (p# ). Conversely, suppose that [f ] ∈ im (p# ). Then there exists a
e based at x̃ such that [f ] = [pg̃]. Consequently, we have
loop g̃ in X
a homotopy F : pg̃ ≃ f rel {0, 1}. By Theorem 15.2.3, there exists a
e rel {0, 1} such that pFe = F and Fe (s, 0) =
homotopy Fe : I × I → X
e defined by f˜ (s) = Fe (s, 1), is
g̃ (s), s ∈ I. Then the path f˜ : I → X,
e
a lift of f and has the same end points as g̃. Thus f˜ is a loop in X
based at x̃, and there is no other lifting of f starting at x̃, by Theorem
15.2.2.
♢
We now apply the preceding results to give a general solution to
the lifting problem for covering maps.
e → X be a
Theorem 15.2.6 (The Lifting Theorem) Let p : X
covering map, and let Y be a connected and locally path-connected
space. If f : (Y, y0 ) → (X, x0 ) is continuous, and x̃0 ∈ p−1 (x0 ),
e x̃0 ) of f if and only if
then there exists a lifting f˜ : (Y, y0 ) → (X,
e x̃0 ). Moreover, f˜ is unique.
f# π (Y, y0 ) ⊆ p# π(X,
Proof. If there is a lifting f˜ of f with f˜ (y0 ) = x̃0 , then p# ◦ f˜# = f# ,
e x̃0 ), and the
for pf˜ = f . So f# π (Y, y0 ) = (p# ◦ f˜# )π (Y, y0 ) ⊆ p# π(X,
necessity of the condition follows.
e x̃0 ). We construct
Conversely, suppose that f# π (Y, y0 ) ⊆ p# π(X,
e as follows. Given y ∈ Y , there is a path h in Y with the
f˜ : Y → X
origin y0 and the end y, since Y is path-connected. Accordingly, f h is
a path in X with the origin f (y0 ) = x0 and the end f (y). By Theorem
e which lifts f h and has the origin x̃0
15.2.1, there exists a path ϕ̃ in X
e is
(see Figure 15.4 below). We observe that the point ϕ̃ (1) of X
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453
~
Á
^
0
~
Á(³)
~
x0
1
~
X
h
~
p
f
y
f
^
y0
^
Y
X
x0
FIGURE 15.4: Proof of Theorem 15.2.6.
independent of the choice of path h. Choose another path k in Y from
e that lifts f k and starts at x̃0 .
y0 to y, and let ψ̃ be the path in X
(
)
−1
−1
Then (h ∗ k ) is a loop in Y based at y0 ; consequently,
f
◦
h
∗
k
[
( −1 )] =
−1
(f h)∗ f k
is a loop in X at x0 . By our hypothesis, (f h) ∗ f k
=
(
)
−1
e
f# [h ∗ k ] ∈ f# π (Y, y0 ) ⊆ p# π(X, x̃0 ), and therefore there exists
(
)
e based at x̃0 such that (f h) ∗ f k −1 = pg̃, by Corollary
a loop g̃ in X
15.2.5. It is easily checked that ϕ̃ (t) = g̃ (t/2), and ψ̃ (t) = g̃ (1 − t/2)
for 0 ≤ t ≤ 1. So ϕ̃ (1) = g̃ (1/2) = ψ̃ (1), and we may define a mapping
e by setting f˜ (y) = ϕ̃ (1). By the definition of the function
f˜ : Y → X
f˜, it is obvious that pf˜ (y) = pϕ̃ (1) = f h (1) = f (y) for every y ∈ Y ,
and f˜ (y0 ) = x̃0 .
f be an open
We now show that f˜ is continuous. Let y ∈ Y and let W
˜
nbd of f (y) = x̃. Consider an admissible open nbd U of f (y) = x ∈ X.
e over U containing x̃. Replacing W
f
As x̃ ∈ p−1 (U ), we find a sheet U
f ∩U
e , if necessary, we may assume that W
f⊆U
e . Since p is an open
by W
f
map, W = p(W ) is an open nbd of x with W ⊆ U . By the continuity of
f , f −1 (W ) is an open nbd of y in Y . Since Y is locally path-connected,
there is a path-connected open nbd V of y with V ⊆ f −1 (W ). We claim
f . Let h : I → Y be a path in Y from y0 to y, and let ϕ̃
that f˜ (V ) ⊆ W
be the lifting of f h with the origin x̃0 . Then ϕ̃ (1) = x̃, by the definition
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Elements of Topology
of f˜. If z ∈ V , then there is a path g : I → V from y to z. Clearly,
(f g) (I) ⊆ W ⊆ U so that the composition of f g with the inverse of the
e :U
e → U is defined. Write ψ̃ = (p|U
e )−1 ◦ (f g).
homeomorphism p|U
f . Since
Then ψ̃ is the lifting of f g with ψ̃ (0) = x̃ and ψ̃ (1) ∈ W
ϕ̃(1) = x̃, ϕ̃ ∗ ψ̃ is defined. Obviously, h ∗ g is a path in Y from y0 to
z, and p ◦ (ϕ̃ ∗ ψ̃) = (pϕ̃) ∗ (pψ̃) = (f h) ∗ (f g) = f ◦ (h ∗ g). As ϕ̃ ∗ ψ̃
f , and hence our
starts at x̃0 , we have f˜ (z) = (ϕ̃ ∗ ψ̃) (1) = ψ̃ (1) ∈ W
claim. It follows that f˜ is a lift of f with f˜ (y0 ) = x̃0 .
Finally, f˜ is unique, by Theorem 15.2.2.
♢
Notice that the construction of the lift f˜ in the preceding theorem
is a forced one, because if f˜ exists, then for any path h in Y from y0
to y, f˜h is a lift of f h starting at x̃0 . We also note that the condition
of local path-connectedness on Y in the preceding theorem is essential.
This is shown by the following.
Example 15.2.1 Consider the exponential map p : R1 → S1 , t 7→ e2πıt .
Let S be the closed topologist’s sine curve (ref. Ex. 3.1.7) and C be
a path in R2 with the end points (0, 1) and (1, sin 1) that intersects S
only at those two points (see Figure 15.5).
1
~
f
-1
Y
f
0
R1
•
---
p
^
---
• S1
FIGURE 15.5: Illustration of Example 15.2.1.
It is easily seen that Y = S∪C is simply connected. Also, it is clear that
the quotient
space Y /S can be identified with S1 , and we may assume
[ ]
that S = 1 ∈ S1 . Let f be the quotient map Y → S1 . We take 0 ∈ R1 ,
1 ∈ S1 , and (0, −1) ∈ Y as base points. If f˜ is constructed as in the
proof of the preceding theorem, then the points (0, y), −1 ≤ y ≤ 1,
COVERING SPACES
455
are mapped to 0 under f˜ and, by the continuity of path liftings, every
point (x, sin 1/x), 0 < x ≤ 1, maps to −1. But, then f˜ is discontinuous,
for the point (0, −1) is a limit point of S.
As an immediate consequence of the Lifting Theorem, we see that
if Y is a locally path-connected and simply connected space, then
)
( each
e x̃0
continuous map f : Y → X has a unique lifting f˜ : (Y, y0 ) → X,
e → X, where f (y0 ) = p (x̃0 ). It is
relative to a covering map p : X
noteworthy that the preceding theorem reduces a topological problem
to a purely algebraic question; this is the general strategy of algebraic
topology.
( )
The technique used in the computation of π S1 can be adapted
to determine the fundamental group of many more spaces. With this
end in view, we now discuss an action of the fundamental group of the
e → X on the fibre F = p−1 (x0 )
base space of a covering map p : X
over the base point x0 in X. This will also play an important role in
the classification of all covering spaces of a given space.
Let α = [f ] ∈ π (X, x0 ), and x̃ ∈ F . Then there is a unique lift f˜ of
f with f˜ (0) = x̃. Observe that f˜ (1) ∈ F , for pf˜ (1) = f (1) = x0 . So
we can define x̃ · α = f˜ (1). By the Monodromy theorem, this definition
does not depend on the choice of representative of α. Suppose now
that β = [g] is another element of π (X, x0 ), and let g̃ be the lifting of
g with g̃ (0) = f˜ (1). Clearly, f˜ ∗ g̃ is the lifting of f ∗ g which starts
at x̃ and ends at g̃ (1). Since αβ is the path class of f ∗ g, we have
x̃ · (αβ) = (f˜ ∗ g̃) (1) = g̃ (1) = (x̃ · α) · β. If ε ∈ π (X, x0 ) is the
identity element, then x̃ · ε = x̃, for the lift of the constant loop at
x0 is the constant loop at x̃. Therefore π (X, x0 ) acts on the right on
the set F (as a group of permutations of F ). This action is called the
“monodromy” action.
e is path-connected, then the above action of π (X, x0 ) on F is
If X
e
transitive. For, given any two points x̃, ỹ of F, there is a path f˜ in X
˜
from x̃ to ỹ. Now the composition pf is a loop f in X based at x0 so
that α = [f ] ∈ π (X, x0 ). Obviously, f˜ is the lift of f with the initial
point x̃, so x̃ · α = f˜ (1) = ỹ.
e → X be a covering map, where X
e is
Proposition 15.2.7 Let p : X
path-connected. If p (x̃0 ) = x0 , then the multiplicity of p is the index
e x̃0 ) in π (X, x0 ).
of p# π(X,
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Elements of Topology
Proof. We show that there is a bijection between F = p−1 (x0 ) and
e x̃0 ) in π (X, x0 ) = Γ. Let Γx̃ denote
the set of right cosets of p# π(X,
0
the isotropy subgroup of Γ at x̃0 . Since Γ acts transitively on F , the
function Γ/Γx̃0 → F , Γx̃0 α 7→ x̃0 · α, is a bijection. Thus we need
e x̃0 ). Let α ∈ Γx̃ and f be a loop in
to prove that Γx̃0 = p# π(X,
0
˜
X representing α. If f is the lifting of f with f˜ (0) = x̃0 , then x̃0 =
e based at x̃0 . Thus we have [f˜] ∈
x̃0 · α = f˜ (1), so f˜ is a loop in X
˜
e
e x̃0 ), and the inclusion Γx̃ ⊆
π(X, x̃0 ) whence α = [f ] = [pf ] ∈ p# π(X,
0
e x̃0 ) follows. To see the reverse inclusion, assume that α = [pg̃]
p# π(X,
e x̃0 ). Since g̃ is the lift of pg̃ starting at x̃0 , we
for some [g̃] ∈ π(X,
have x̃0 · α = g̃ (1) = x̃0 . This implies that α ∈ Γx̃0 , and the equality
Γx̃0 = im (p# ) holds.
♢
It is immediate from the preceding proposition that the multiplicity
e → X with X
e simply connected is the order of
of a covering map p : X
the group π (X, x0 ).
Example 15.2.2 If n ≥ 2, then π (RPn ) ∼
= Z/2Z.
Proof. We know that Sn is a covering space of RPn relative to the
identification map which identifies every pair of antipodal points in
Sn . The multiplicity of this covering map is obviously 2. Since Sn is
simply connected for n ≥ 2, we have |π (RPn )| = 2, and therefore
π (RPn ) ∼
♢
= Z/2Z.
Now, we establish a theorem par excellence.
Theorem 15.2.8 (Borsuk–Ulam Theorem) If f : Sn → Rn , n ≥
1, is continuous, then there exists a point x ∈ Sn such that f (x) =
f (−x).
If f (x) ̸= f (−x) for every x ∈ Sn , then we have a continuous map
g : Sn → Sn−1 defined by g (x) = [f (x) − f (−x)]/ ∥f (x) − f (−x)∥,
which satisfies g (−x) = −g (x). Therefore, it suffices to establish the
following.
Theorem 15.2.9 There is no continuous map Sn → Sn−1 , n ≥ 1,
which sends antipodal points to antipodal points.
Proof. We will prove the theorem only for n ≤ 2 (the proof for n > 2
requires higher dimensional analogues of the fundamental group). The
case n = 1 is obvious, for S1 is connected but S0 is not. To prove the case
COVERING SPACES
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n = 2, assume that there is a continuous map f : S2 → S1 such that
f (−x) = −f (x) for all x ∈ S2 . For n = 1, 2, let qn : Sn → RPn , x 7→
[x], denote the quotient map which identifies the antipodes. Since f is a
relation preserving map, it induces a continuous map f¯ : RP2 → RP1 ,
[x] 7→ [f (x)]. Thus we have q1 f = f¯q2 . Choose a base point x0 ∈ S2
and write f (x0 ) = y0 . Then q1# ◦ f# = f¯# ◦ q2# , that is, the following
diagram commutes:
)
(
π S2 , x 0
f#
q2#
(
- π S1 , y 0
)
q1#
?
(
)
π RP2 , [x0 ]
f¯#
?
(
)
- π RP1 , [y ]
0
Let g be a path in S2 from x0 to −x0 . Then f g is a path in S1 from y0 to
−y0 , since f is antipode preserving. Also, q2 g and q(1 f g are closed
paths
)
in RP2 and RP1 , respectively. So α = [q2 g] ∈ π RP2 , [x0 ] and β =
(
)
[q1 f g] ∈ π RP1 , [y0 ] . By the definition of action of the fundamental
group on fibers, we have x0 ·α = −x0 and y0 ·β = −y0 . This (shows that)
both α and β are different from the identity elements of π RP2 , [x0 ]
(
)
and π RP1 , [y0 ] , respectively. The equality q1 f g = f¯q2 g implies that
(
)
f¯# (α) = β. But, this contradicts the fact that f¯# : π RP2 , [x0 ] =
(
)
Z2 → π RP1 , [y0 ] = Z is the trivial homomorphism, and the theorem
follows.
♢
Since a continuous function into Rn is determined by n real-valued
continuous functions, the Borsuk–Ulam theorem can be rephrased as
follows: If fi : Sn → R1 , 1 ≤ i ≤ n, are continuous maps, then there
exists a point x ∈ Sn such that fi (−x) = fi (x) for i = 1, . . . , n. Here
is a beautiful consequence of this form of the Borsuk–Ulam theorem.
Let f1 (x) denote the temperature, and f2 (x) denote the atmospheric
pressure at a particular instant at a point x on the earth’s surface.
Assuming that both the temperature and the pressure are continuous
functions of the point x, we find a pair of diametrically opposite points
on the earth’s surface having temperature and pressure simultaneously
the same.
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Elements of Topology
Exercises
e → X be a covering map and f : I → X be a loop based
1. Let p : X
at x0 . Prove:
e is a
(a) If f ≃ cx0 rel {0, 1}, then any lift of f to a path in X
loop and is homotopic to a constant loop rel {0, 1}.
e at x̃0 , then any loop homotopic to
(b) If f lifts to a loop in X
e at x̃0 .
f rel {0, 1} also lifts to a loop in X
2. If a space X has a nontrivial path-connected covering space, show
that π (X) is nontrivial.
e → X be a covering map, where X
e is path3. Let p : X
(
)
e x̃0 =
connected. Prove that p is a homeomorphism ⇔ p# π X,
π (X, p (x̃0 )).
e → X be an n-sheeted covering map. If X
e is simply
4. Let p : X
connected, and n is a prime, prove that π (X, x0 ) ∼
= Z/nZ.
5. Given a connected and locally path-connected space X, prove
that a continuous map f : X → S1 can be lifted to a continuous
map f˜ : X → R1 relative to the exponential map t 7→ e2πıt ⇔ f
is null homotopic.
6. Prove that any two continuous maps from a simply connected
and locally path-connected space to S1 are homotopic.
7. Give an example of a simply connected space that is not locally
path-connected.
8. Prove that any continuous map of the projective space RPn , n >
1, into the circle S1 is homotopic to a constant map.
9. Show that any continuous map RP2 → RP2 which is nontrivial
on the fundamental group can be lifted to a continuous map
f˜ : S2 → S2 such that f˜ (−x) = −f˜ (x) for all x ∈ S2 .
10. Prove that the following statements are equivalent:
(a) The Borsuk–Ulam Theorem (15.2.8).
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459
(b) If f : Sn → Rn is a continuous map such that f (−x) =
−f (x) for every x ∈ Sn , then there is a point x ∈ Sn such
that f (x) = 0.
(c) Theorem 15.2.9.
11. Prove that no subset of Rn is homeomorphic to Sn .
12. Prove that there does not exist a continuous injection from Rn+1
to Rn for any n ≥ 1.
15.3
The Universal Covering Space
This and the next two sections concern mainly a classification of
covering spaces of the spaces with some nice properties. The various
possible covering spaces of a given space are classified by means of fibre
preserving homeomorphisms. In this section, the problem of “equivalence” for connected covering spaces of a connected and locally pathconnected space X is resolved. We see that two simply connected covering spaces of X are equivalent. We also see that a simply connected
covering space of a space X covers every other covering space of X. For
this reason, such a covering space of X is called a “universal covering
space.” A natural question arises about the existence of such a covering
space of a given space X. It will be seen that a space X satisfying some
mild conditions has a universal covering space.
The Lifting Theorem provides a simple criterion for the existence
of fibre preserving continuous maps between covering spaces of a given
space in terms of the fundamental groups of the spaces involved. We
shall assume hereafter that all spaces are path-connected and locally
path-connected. To save words, we will not repeat this assumption.
ei → X, i =
Theorem 15.3.1 (The Covering Theorem) Let pi : X
e
1, 2, be covering maps, and suppose that x̃i ∈ Xi and p1 (x̃1 ) = x =
e1 , x̃1 ) ⊆ p2# π(X
e2 , x̃2 ), then there exists a unique
p2 (x̃2 ). If p1# π(X
e
e
continuous map q : X1 → X2 such that p2 ◦ q = p1 and q (x̃1 ) = x̃2 ,
that is, the triangle
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Elements of Topology
q
e1 , x
e2 , x
(X
e1 ) − − −− → (X
e2 )
Z
Z
p1 Z
p2
Z
~ Z
=
(X, x)
commutes. Moreover, q is a covering map.
e1 , x̃1 ) ⊆ p2# π(X
e2 , x̃2 ), the Lifting Theorem shows
Proof. Since p1# π(X
e1 , x̃1 ) → (X
e2 , x̃2 ) of p1 relative to
that there is a unique lifting q : (X
p2 . So it is the last statement which needs to be established. We first
e2 be arbitrary and f be a path
observe that q is surjective. Let ỹ2 ∈ X
e
in X2 from x̃2 to ỹ2 . Then p2 ◦ f is a path in X with the origin x. By
e1 beginning at
the path lifting property of p1 , there is a path g in X
x̃1 such that p2 ◦ f = p1 ◦ g = p2 ◦ q ◦ g. Since f (0) = q (x̃1 ) = qg (0),
we have f = qg, by Theorem 15.2.2. So ye2 = f (1) = q (g (1)), and q
e2 ,
is surjective. Now, we show that q is a covering map. Given ye2 ∈ X
put y = p2 (e
y2 ). Let U1 and U2 be open nbds of y, which are evenly
covered by p1 and p2 , respectively. Since X is locally path-connected,
we can find a path-connected open set U such that x ∈ U ⊆ U1 ∩ U2 .
Then U is evenly covered by both p1 and p2 . So p−1
2 (U ) is a disjoint
union of open sets Vλ , each of which is mapped homeomorphically onto
U by p2 . Since U is path-connected, each Vλ is path-connected, and
−1
hence a path component of p−1
2 (U ). Denote the component of p2 (U )
that contains ye2 by V . We observe that V is evenly
∪ covered by q.
−1
Clearly, q −1 (V ) ⊂ p−1
(U
).
Suppose
that
p
(U
)
=
Wµ , where each
1
1
Wµ is open and mapped homeomorphically onto U by p1 . Note that
each Wµ is path-connected so that q (Wµ ) is also path-connected. Also,
q (Wµ ) ⊆ p−1
2 (U ), for p1 = p2 q. Since the sets Vλ are open and disjoint,
we have q (Wµ ) ⊆ V whenever q (Wµ ) ∩ V ̸= ∅. It follows that q −1 (V )
is the union of those Wµ for which q (Wµ ) intersects V nontrivially. It
remains to show that q|Wµ : Wµ → V is a homeomorphism. But this
is clear from the fact that p1 |Wµ and p2 |V are homeomorphisms, and
p1 |Wµ = (p2 |V ) ◦ (q|Wµ ).
♢
As an immediate consequence of the preceding theorem, we obtain
e is a simply connected covering space of a space
Corollary 15.3.2 If X
X relative to the map p, then for any covering map r : Ye → X, there
e → Ye such that the diagram
exists a covering map q : X
COVERING SPACES
461
q
e − − − − −− → Ye
X
Z
pZ
r
Z
Z
~
=
X
is commutative, that is, p = r ◦ q.
It follows that a simply connected covering space of the space X
is a covering space of any other covering space of X. This justifies the
adjective “universal” in the following.
Definition 15.3.3 A simply connected covering space of the space X
is called a universal covering space of X.
The real line R1 is a universal covering space of S1 ; the plane R2
is that of the torus S1 × S1 ; and the n-sphere Sn is that of RPn , n ≥
2. However, not every connected, locally path-connected space has a
universal covering space, as shown by the following
Example 15.3.1 Let X be the “Hawaiian earring” (ref. Ex. 7.4.2).
Clearly, X is connected and locally path-connected. We observe that
e
X has no universal covering space. Assume on the contrary that X
is a universal covering space of X relative to the map p. Let x0 ∈ X
denote the origin of R2 and U be a nbd of x0 that is evenly covered
e be the sheet over U
by the covering map p. For x̃0 ∈ p−1 (x0 ), let U
e is a homeomorphism between U
e and U , each
containing x̃0 . Since p|U
˜
e
e is
loop f in U based at x0 lifts to a loop f in U based at x̃0 . Since X
simply connected, there is a base point preserving homotopy between
f˜ and the constant path at x̃0 . Hence f is homotopic in X to the constant path at x0 . It follows that the inclusion i : U ,→ X induces the
trivial homomorphism i# : π (U, x0 ) → π (X, x0 ). For large n, we have
Cn ⊂ U , where Cn is the circle with radius 1/n and centre (1/n, 0).
If j : Cn ,→ U and k : Cn ,→ X are the inclusion maps, then we
have k = ij. It is easily checked that Cn is a retract of X, and therej#
i#
fore the composition π (Cn , x0 ) → π (U, x0 ) → π (X, x0 ) is injective.
This implies that i# is not trivial, a contradiction. Therefore X has no
universal covering space.
Thus, a natural question arises: When does a space X have a unie As seen in the above example, a space X
versal covering space X?
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Elements of Topology
which has a universal covering space satisfies the following condition:
Each point x ∈ X has an open nbd U such that the homomorphism
π(U, x) → π(X, x) induced by the inclusion map U ,→ X is trivial
(that is, all loops in U are null homotopic in X).
A space X which satisfies the above condition is called locally relatively simply connected or semilocally 1-connected.
The next theorem shows that a space with the above property does
have a universal covering space.
Theorem 15.3.4 A connected, locally path-connected and semilocally 1-connected space X has a universal covering space.
Proof. We choose a point x0 ∈ X and consider the set P of all paths in
X with origin x0 . For each f ∈ P, put [f ] = {g ∈ P|g ≃ f rel {0, 1}}
e = {[f ]|f ∈ P}. We construct a base for a topology on X.
e
and let X
Notice that the family B = {U ⊆ X|U is open and path-connected} is
a base of the space X. For U ∈ B and f ∈ P with f (1) ∈ U , let ⟨f, U ⟩
e such that g(1) ∈ U and
denote the set of all homotopy classes [g] ∈ X
g ≃ f ∗ γ rel {0, 1} for some path γ in U (see Figure 15.6 below).
f
U
>
>
x0
°
>
g
FIGURE 15.6: The set ⟨f, U ⟩.
e = {⟨f, U ⟩ |U ∈ B and f ∈ P} is a basis for
We claim that the family B
e
a topology on X. We first observe the following properties of the sets
⟨f, U ⟩:
(a) [f ] ∈ ⟨f, U ⟩.
(b) For U, V in B and f ∈ P satisfying f (1) ∈ U ⊂ V ,
⟨f, U ⟩ ⊆ ⟨f, V ⟩.
(c) [g] ∈ ⟨f, U ⟩ ⇒ ⟨g, U ⟩ = ⟨f, U ⟩.
The proofs of (a) and (b) are obvious. To see (c), suppose that
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463
[h] ∈ ⟨g, U ⟩. Then there is a path γ in U such that g(1) = γ(0) and
h ≃ g ∗ γ rel {0, 1}. By our hypothesis, there is a path δ in U such that
f (1) = δ(0) and g ≃ f ∗ δ rel {0, 1}. So h ≃ (f ∗ δ) ∗ γ ≃ f ∗ (δ ∗ γ)
rel {0, 1}. Clearly, the path δ ∗ γ lies in U and we have [h] ∈ ⟨f, U ⟩.
Thus ⟨g, U ⟩ ⊆ ⟨f, U ⟩. To see the reverse inclusion, we note that f ≃
g ∗ δ −1 rel {0, 1}. So the roles of f and g in the above argument can be
interchanged, and we obtain ⟨g, U ⟩ ⊇ ⟨f, U ⟩. Now, we can prove our
e covers X.
e To check
claim. From (a), it is obvious that the family B
the other requirement of a basis, suppose that h ∈ ⟨f, U ⟩ ∩ ⟨g, V ⟩. By
(c), ⟨f, U ⟩ = ⟨h, U ⟩ and ⟨g, V ⟩ = ⟨h, V ⟩. So [h] ∈ ⟨h, U ⟩ ∩ ⟨h, V ⟩. Since
B is a basis of X and h(1) ∈ U ∩ V , we find a W ∈ B such that
h(1) ∈ W ⊆ U ∩ V . By (b), we have [h] ∈ ⟨h, W ⟩ ⊆ ⟨h, U ⟩ ∩ ⟨h, W ⟩,
and hence our claim.
e → X defined by p ([f ]) = f (1).
Next, we consider the mapping p : X
We show that p is a covering projection. Since X is path-connected, p
is surjective. If U ∈ B and [f ] ∈ p−1 (U ), then p (⟨f, U ⟩) ⊆ U , by the
e
definition. Since B is a base of X, p is continuous. Next, if ⟨f, U ⟩ ∈ B,
then for any x ∈ U , there is a path γ in U with origin f (1) and end
x, since U is path-connected. So, for g = f ∗ γ, we have [g] ∈ ⟨f, U ⟩
e is a basis of X,
e we see
and p([g]) = x. Thus p (⟨f, U ⟩) = U . Since B
that p is an open mapping. Next, we observe that every point of X
has an admissible nbd relative to p. Since X is locally path-connected
and semilocally 1-connected, for each x ∈ X, there exists a U ∈ B such
that x ∈ U ∪and all loops in U are homotopically trivial in X. Clearly,
p−1 (U ) = {⟨f, U ⟩ |f (1) ∈ U }, and any two distinct sets ⟨f, U ⟩ and
⟨g, U ⟩ are disjoint, by (c). As seen above, the restriction of p to ⟨f, U ⟩
is a continuous open mapping of ⟨f, U ⟩ onto U . To see that it is also
injective, suppose that [g], [h] ∈ ⟨f, U ⟩ and g(1) = h(1). Then, by (c),
⟨g, U ⟩ = ⟨f, U ⟩ = ⟨h, U ⟩. Consequently, g ∈ ⟨h, U ⟩, so there exists a
path γ in U such that h(1) = γ(0) and g ≃ h∗γ rel {0, 1}. In particular,
we have γ(1) = g(1) = h(1) so that γ is a loop in U . By the choice of
U, γ ≃ ch(1) rel {0, 1}. Therefore g ≃ h ∗ γ ≃ h ∗ ch(1) ≃ h rel {0, 1}
whence [g] = [h]. It follows that p| ⟨f, U ⟩ is injective, and U is evenly
covered by p. Thus p is a covering projection.
e is simply connected. To see that X
e is pathFinally, we show that X
e and put x̃0 = [cx ]. It is
connected, let [f ] be an arbitrary point of X
0
e For each t ∈ I, let ft be
enough to construct a path from x̃0 to [f ] in X.
e
the path in X defined by ft (s) = f (st). Then ft ∈ P so that [ft ] ∈ X.
e by f˜(t) = [ft ]. Then we obviously have
We define a function f˜ : I → X
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Elements of Topology
f˜(0) = x̃0 and f˜(1) = [f ]. To see the continuity of f˜, suppose that
t0 ∈ I and let ⟨g, U ⟩ be a basic nbd of f˜ [t0 ]. Then ⟨ft0 , U ⟩ = ⟨g, U ⟩. By
the continuity of f , there exists an open (or half-open) interval V ⊂ I
such that t0 ∈ V and f (V ) ⊆ U. So f ((1 − s)t0 + st) ∈ U for all s ∈ I
and all t ∈ V . For each t ∈ V , the function γt : I → U defined by
γ(s) = f ((1 − s)t0 + st) is continuous. We assert that ft ≃ ft0 ∗ γt rel
{0, 1}. A desired homotopy is given by
{
f (2s((1 − u)t + ut0 )/(2 − u)) for 0 ≤ s ≤ (2 − u)/2,
F (s, u) =
f ((2 − 2s)t0 + (2s − 1)t)
for (2 − u)/2 ≤ s ≤ 1.
So f˜(t) = [ft ] ∈ ⟨g, U ⟩ and f˜ maps V into ⟨g, U ⟩. It follows that f˜
e from c to [f ]. Now,
is continuous at t0 , and thus f˜ is a path in X
e x̃0 ). To this end, consider the right action of Γ =
we determine π(X,
π (X, x0 ) on the fibre p−1 (x0 ). Notice that p (x̃0 ) = x0 and Γx̃0 =
e x̃0 ), by the proof of Proposition 15.2.7. Suppose that α = [f ] ∈
p# π(X,
Γ fixes x̃0 . Then x̃0 = x̃0 · α = f˜(1), where f˜ is the lifting of f starting
at x̃0 . Observe that f˜ is given by f˜(t) = [ft ], where ft : s 7→ f (st). So
e x̃0 ) = {1}. This implies that π(X,
e x̃0 ) =
we have x̃0 = [f ], and p# π(X,
{1}, since p# is a monomorphism. This completes the proof.
♢
Now, we see that a universal covering space of a space, when it
exists, is unique up to isomorphism in the following sense.
e1 and X
e2 of a given space
Definition 15.3.5 Two covering spaces X
ei → X, respectively, are said to be
X relative to projections pi : X
e1 → X
e2
equivalent or isomorphic if there exists a homeomorphism h : X
such that p1 = p2 ◦ h. The homeomorphism h is called an equivalence
or isomorphism of covering spaces.
e1 and X
e2 are two universal covering spaces of a
Suppose that X
space X relative to the covering projections p1 and p2 , respectively.
e1 and x̃2 ∈ X
e2 such that p1 (x̃1 ) = p2 (x̃2 ). Then
Choose points x̃1 ∈ X
e
e2 , x̃2 ). So, by the Lifting
it is obvious that p1# π(X1 , x̃1 ) = p2# π(X
e1 → X
e2 and k :
Theorem, there exist unique continuous maps h : X
e
e
X2 → X1 such that h (x̃1 ) = x̃2 , p1 = p2 ◦ h and k (x̃2 ) = x̃1 , p2 =
e1 → X
e1 satisfies kh (x̃1 ) = x̃1 and
p1 ◦ k. Then the composite kh : X
p1 = p1 ◦ (kh). By the unique-lifting property, we have kh = 1Xe1 .
Similarly, hk = 1Xe2 , and h is a homeomorphism with k as its inverse.
e1 and X
e2 are equivalent.
Thus X
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465
The proof of the uniqueness of the universal covering space of a
given space establishes “the sufficient part” of the following.
ei → X, i = 1, 2, be covering maps,
Proposition 15.3.6 Let pi : X
e
e2 satisfy p1 (x̃1 ) = p2 (x̃2 ). Then
and suppose that x̃1 ∈ X1 and x̃2 ∈ X
e1 → X
e2 such that h (x̃1 ) = x̃2 and
there exists a homeomorphism h : X
e
e2 , x̃2 ).
p2 ◦ h = p1 if and only if p1# π(X1 , x̃1 ) = p2# π(X
Proof. We need to prove only the necessity of the condition. Suppose
e1 → X
e2 is a homeomorphism with p1 = p2 ◦ h. Then we have
that h : X
e
e1 , x̃1 ) = p2# π(X
e2 , h (x̃1 )), for h# is an
p1# π(X1 , x̃1 ) = (p2# ◦ h# ) π(X
isomorphism. This completes the proof.
♢
However, the preceding proposition does not solve the equivalence
question in general. In fact, this is a stronger form of equivalence,
that one for which one specifies base points, which, of course, must
be preserved under the mappings. It may be possible that a space X
e1 and X
e2 , but there is no
does have two equivalent covering spaces X
e1 → X
e2 which takes the base point x̃1 ∈ X
e1 to the base
equivalence X
e
e1 → X
e2
point x̃2 ∈ X2 , even if p1 (x̃1 ) = p2 (x̃2 ). Notice that if h : X
is an equivalence, then h (x̃1 ) belongs to the fibre over p2 (x̃2 ). And,
e → X and a point x̃ ∈ X,
e it is obvious
given a covering map p : X
−1
e
that p# π(X, x̃) depends on the choice of x̃ ∈ p (x0 ), where x0 =
p (x̃). Thus, we need to know how the subgroups p# π(X̃, x̃) ⊆ π(X, x0 )
are related for various choices of x̃ ∈ p−1 (x0 ). There are no relations
e x̃) in general, unless the points x̃ can
among the subgroups p# π(X,
e If X
e is path-connected, then we know that
be joined by paths in X.
∼
e
e
π(X, x̃) = π(X, ỹ) so that their images under the monomorphism p#
are isomorphic. Our next lemma shows that any two such subgroups
can differ by a conjugation only.
e → X be a covering map. For x0 ∈ X, the
Lemma 15.3.7 Let p : X
e x̃) ⊆ π (X, x0 ), x̃ ∈ p−1 (x0 ), constitute a conjugacy
subgroups p# π(X,
class in π (X, x0 ).
e from x̃ to
Proof. Suppose that x̃, ỹ ∈ p−1 (x0 ). Let f˜ be a path in X
˜
ỹ. Then f = pf is a loop in X based at x0 so that [f ] is an element of
π (X, x0 ). There are isomorphisms
e x̃) → π(X,
e ỹ),
η : π(X,
[g̃] 7→ [f˜−1 ∗ g̃ ∗ f˜], and
θ : π (X, x0 ) → π (X, x0 ) ,
[g] 7→ [f −1 ∗ g ∗ f ].
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Elements of Topology
For any [g̃] ∈ π(X̃, x̃), we have
(p# ◦ η) [g̃] = p# [f˜−1 ∗ g̃ ∗ f˜] = [p ◦ (f˜−1 ∗ g̃ ∗ f˜)]
[
]
= f −1 ∗ pg̃ ∗ f = (θ ◦ p# ) [g̃].
So p# ◦ η = θ ◦ p# , that is, the diagram
e x̃0 )
π(X,
η
p#
e x̃1 )
- π(X,
p#
?
π (X, x0 )
θ
?
- π (X, x1 )
FIGURE 15.7: Proof of Lemma 15.3.7.
e x̃) and p# π(X,
e ỹ)
commutes. Since η is surjective, we see that p# π(X,
are conjugate subgroups of π (X, x0 ). On the other hand, we observe
that for each subgroup S in the conjugacy class of the subgroup
e x̃0 ), where p (x̃0 ) = x0 , there is a point x̃1 ∈ p−1 (x0 ) such
p# π(X,
e x̃1 ). Suppose that S = [f ]−1 p# π(X,
e x̃0 )[f ], where
that S = p# π(X,
˜
[f ] ∈ π (X, x0 ). Let f be the lift of f starting at x̃0 . If f˜ (1) = x̃1 ,
then p (x̃1 ) = f (1) = x0 ; thus x̃1 ∈ p−1 (x0 ). For any element [g̃] in
e x̃0 ), we have [f ]−1 · p# ([g̃]) · [f ] = (p# ◦ η) [g̃], which implies that
π(X,
e x̃0 ) = p# π(X,
e x̃1 ), as desired.
S = (p# ◦ η) π(X,
♢
The preceding result asserts that a path-connected covering space of
X determines a conjugacy class of subgroups of the fundamental group
π (X) of X. We shall later prove that, under the conditions on X for
Theorem 15.3.4, each conjugacy class of subgroups of π (X) completely
determines a covering space of X up to isomorphism. The following
theorem completely solves the uniqueness question for covering spaces
up to isomorphism.
e1 and X
e2 of a space X relaTheorem 15.3.8 Two covering spaces X
e
tive to covering projections pi : Xi → X, respectively, are isomorphic
e1 , x̃1 ) and p2# π(X
e2 , x̃2 ), where
if and only if the subgroups p1# π(X
p1 (x̃1 ) = p2 (x̃2 ) = x, are conjugate in π (X, x).
e1 and X
e2 are equivalent, and let h : X
e1 → X
e2
Proof. Assume first that X
e1 , x̃1 ) =
be a homeomorphism with p1 = p2 ◦ h. Then we have p1# π(X
COVERING SPACES
467
e1 , x̃1 ) = p2# π(X
e2 , h (x̃1 )). As p2 (x̃2 ) = x = p2 (h (x̃1 )),
(p2# ◦ h# ) π(X
e2 , h (x̃1 )) and p2# π(X
e2 , x̃2 ) are conjugate in
the subgroups p2# π(X
π (X, x), by Lemma 15.3.7.
e1 , x̃1 ) is conjugate in
Conversely, assume that the subgroup p1# π(X
e
π (X, x) to the subgroup p2# π(X2 , x̃2 ). Again, by Lemma 15.3.7, there
e
e
is a point ỹ2 ∈ p−1
2 (x) such that p1# π(X1 , x̃1 ) = p2# π(X2 , ỹ2 ). By
e1 , x̃1 ) →
the preceding proposition, there is a homeomorphism h : (X
e
(X2 , ỹ2 ) such that p1 = p2 ◦ h, and the proof is complete.
♢
Theorem 15.3.8 can be used to determine all inequivalent connected
covering spaces of a connected and locally path-connected space. We
illustrate this by the following examples.
( )
Example 15.3.2 The connected covering
spaces of S1 . Since π S1 ∼
= Z,
( 1)
the non-trivial subgroups of π S are isomorphic to infinite cyclic
subgroups nZ ⊆ Z, n = 1, 2, . . .. For each integer n ≥ 1, we have
the covering map pn : S1 → S1 , z 7→ z n . If g denotes the loop
t (7→ e2πıt
) , t ∈ I, then the homotopy class [g] is a generator of
1
π S , 1 . We observe that pn ◦ g ≃ g ∗ · · · ∗ g (n-terms) rel {0, 1}.
To see this, [let h : ]I [→ I be the] homeomorphism
which maps the
[ −1 ]
1−n
1−n 2−n
subintervals 0, 2
, 2
,2
, . . . , 2 , 1 linearly onto [0, 1/n],
[1/n, 2/n], . . . , [(n − 1) /n, 1], respectively. It is easily checked that the
composite pn ◦ g ◦ h is the path product g ∗ · · · ∗ g. Now, a homotopy between pn ◦ g and pn ◦ g ◦ h is given by the mapping I × I → S1 ,
(s, t) 7→ (pn ◦ g) ((1 (− t) s)+ th (s)). So [g]n = [g∗· · ·∗g] is a generator
of
( ( 1 ))
1
the subgroup pn# π S , 1 , and it follows that the group pn# π S , 1
is isomorphic to nZ. Thus the covering space corresponding to nZ is
S1 itself relative to the map pn . The real line R1 is the universal covering space S1 relative to the exponential map t 7→ e2πıt . These are
essentially all connected covering spaces of S1 .
Example 15.3.3 The connected covering spaces of RPn . Recall that
e
π (RPn ) ∼
= Z2 for n ≥ 2. Therefore every connected covering space X
n
n
of RP is equivalent to the universal covering space S relative to the
natural projection Sn → RPn or the trivial covering space RPn relative
to the identity map.
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Elements of Topology
Exercises
1. Let p : Y → X and q : Z → Y be continuous maps and assume that
the spaces X, Y and Z are all connected and locally path-connected.
(a) If the composition pq : Z → X is a covering map, prove that q is
a covering map ⇔ p is so.
(b) For each integer n ≥ 1, let pn : S1 → S1 denote the covering map
z 7→ z n . Given two positive integers m and n, show that there is
a covering map q : S1 → S1 such that pn = pm ◦ q if and only if
m|n. When this is the case, prove that q = pk , where n = mk.
(c) If p : Y → X and q : Z → Y are covering maps and X has
a universal covering space, show that pq : Z → X is a covering
map.
2. Prove that an infinite product of circles S1 has no universal covering
space.
3. Let X be a connected, locally path-connected space and x0 ∈ X. Prove
ei , pi ) (i = 1, 2) of X are equivalent if and
that two covering spaces (X
only if the fibers p−1
(x
)
and
p−1
0
1
2 (x0 ) are isomorphic π(X, x0 )-set.
15.4
Deck Transformations
In this section, we study the self-isomorphisms of a covering space
of a given space. We will explore an important relation between the
group of automorphisms of a covering map and the fundamental groups
of the base space and the covering space. This gives a description of
the fundamental group of certain spaces which requires no choice of
base point. We will also discuss a kind of group action which plays an
important role in the existence of covering spaces of a given space. All
the spaces considered in this section are connected and locally pathconnected.
e → X be a covering map. A homeomorDefinition 15.4.1 Let p : X
e
e
phism h : X → X with ph = p is called a deck transformation of the
covering.
COVERING SPACES
469
In the literature, deck transformations are also called automorphisms or covering transformations of the covering. Note that a deck
e → X permutes the “decks” of X;
e
transformation h of a covering p : X
that is, if U is an admissible open set in X, then h permutes the copies
of U in p−1 (U ). If h is a deck transformation, then ph−1 = phh−1 = p
so that h−1 is also a deck transformation. Also, it is clear that the
composition of two deck transformations of p is a deck transformation.
It follows that the set ∆ (p) of all deck transformations of the covering
p is a group under composition of functions. The group ∆ (p) is also
referred to as the automorphism group or the covering group of the
covering.
e → X be a covering map. If h, h′ ∈
Proposition 15.4.2 Let p : X
e with h (x̃) = h′ (x̃), then h = h′ . In
∆ (p) and there exists x̃ ∈ X
e then h = 1 e .
particular, if h ∈ ∆ (p) has a fixed point in X,
X
e is connected, the proposition follows immediately
Proof. Since X
from Theorem 15.2.2.
♢
Example 15.4.1 Consider the covering map p : R1 → S1 , t 7→ e2πıt . The
group ∆ (p) is Z, the group of integers. It is clear that the translations
R1 → R1 , x 7→ x + n, where n ∈ Z, are deck transformations. And,
if h ∈ ∆ (p), then n = h (0) ∈ Z, and therefore h coincides with the
translation x 7→ x + n.
Example 15.4.2 The group ∆ (p) of deck transformations of the covering map p : Sn → RPn , n ≥ 2, consists of just two elements, namely,
the identity map of Sn and the antipodal map.
It is clear from the preceding proposition that the group ∆ (p)
e (on the left). Under this action of ∆ (p), each fibre
acts freely on X
−1
p (x) , x ∈ X, is obviously invariant. In Section 15.2, we have discussed a right transitive action of the group π (X, x) on p−1 (x). The
two actions on p−1 (x) commute in the following sense.
e → X be a covering map, and x ∈ X
Proposition 15.4.3 Let p : X
be fixed. If h ∈ ∆ (p), and α ∈ π (X, x), then (hx̃) · α = h (x̃ · α) for all
x̃ ∈ p−1 (x).
Proof. Let g be a loop in X (based at x), which represents α. If g̃ is
the lift of g with g̃ (0) = x̃, then x̃ · α = g̃ (1), by the definition of
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Elements of Topology
action of π (X, x). Also, hg̃ is a lift of g with the initial point hx̃. So
hx̃ · α = hg̃ (1) = h (x̃ · α).
♢
With the notations of the preceding proposition, we have
(
)
(x̃ · α) · β = x̃ · α ⇔ ((x̃ · α) · β) · α−1 = x̃ ⇔ x̃ · α · β · α−1 = x̃,
for every β ∈ π (X, x). Thus, if Γ denotes the group π (X, x), then we
have Γx̃·α = α−1 Γx̃ α, where Γx̃ is the isotropy subgroup of Γ at x̃, etc.
The next result involves the following algebraic notion: If H is a
subgroup of a group Γ, the normaliser of H in Γ is the set N (H) of all
elements γ ∈ Γ such that γ −1 Hγ = H. This is the largest subgroup of
Γ in which H is normal.
e → X be a covering map, and x̃1 , x̃2 ∈
Proposition 15.4.4 Let p : X
e be two points in the same fibre p−1 (x). Then the following stateX
ments are equivalent:
(a) There exists an h ∈ ∆ (p) such that h (x̃1 ) = x̃2 .
e x̃1 ) = p# π(X,
e x̃2 ).
(b) p# π(X,
e x̃1 )
(c) There exists an α in the normaliser of the subgroup p# π(X,
in π (X, x) such that x̃1 · α = x̃2 .
Proof. (a) ⇐⇒ (b): This is a special case of Proposition 15.3.6.
e x̃i ) =
(b) ⇒ (c): By the proof of Proposition 15.2.7, we have p# π(X,
−1
Γx̃i , where Γ = π (X, x). Since the action of Γ on p (x) is transitive,
there exists α ∈ Γ such that x̃1 · α = x̃2 . We need to show that α ∈
N (Γx̃1 ). By (b), Γx̃1 = Γx̃2 = Γx̃1 ·α = α−1 Γx̃1 α, which implies that
α ∈ N (Γx̃1 ).
(c) ⇒ (b): By our hypothesis, Γx̃1 = α−1 Γx̃1 α = Γx̃1 ·α = Γx̃2 , and
(b) holds.
♢
e → X be a covering map. Then ∆ (p) acts
Corollary 15.4.5 Let p : X
−1
e x̃)
transitively on the fibre p (x) if and only if the subgroup p# π(X,
is normal in π (X, x), where p(x̃) = x.
Proof. Suppose that ∆ (p) is transitive on p−1 (x). Let α ∈ π (X, x) = Γ,
and put x̃ · α = ỹ. Then ỹ ∈ p−1 (x) so that there exists h ∈ ∆ (p) such
that h(x̃) = ỹ. Since h is a homeomorphism and p = ph, we have
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471
e x̃) = (ph)# π(X,
e x̃) = p# π(X,
e ỹ) = Γỹ = α−1 Γx̃ α, which
Γx̃ = p# π(X,
shows that Γx̃ is normal in Γ.
Conversely, suppose that Γx̃ is normal in Γ. Let ỹ ∈ p−1 (x) be
arbitrary. Then there exists α ∈ Γ such that x̃·α = ỹ, for Γ is transitive
on p−1 (x). Again, by Proposition 15.4.4, there exists h ∈ ∆ (p) such
that h(x̃) = ỹ, and therefore ∆ (p) acts transitively on p−1 (x).
♢
e x̃) ⊆ π (X, p (x̃)) deterIt is to be noted that the subgroup p# π(X,
e → X depends on the choice of base
mined by a covering map p : X
e
point x̃ ∈ X, in general. But, any two such subgroups can differ by
a conjugation only (ref. Lemma 15.3.7). There is an important special
case in which the subgroup does not depend on the choice of base point.
e → X is called regular (or
Definition 15.4.6 A covering map p : X
e x̃) is a normal subgroup of π (X, p (x̃)) for every
normal) if p# π(X,
e
x̃ ∈ X.
e x̃) is independent of the choice
It is clear that the subgroup p# π(X,
of base point x̃ in the fibre over p (x̃) when p is a regular covering,
because the only subgroup conjugate to a normal subgroup H is H
e → X with X
e
itself. It is also immediate that a covering map p : X
simply connected is regular. We remark that, unlike the action of the
e → X, the
fundamental group π (X) on a fibre of a covering map p : X
action of the covering group ∆ (p) on fibers is not transitive, in general.
By Corollary 15.4.5, the group ∆ (p) acts transitively on each fibre of
p if and only if p is regular.
The following proposition shows that it suffices to check the condie in order to detertion of regularity for some convenient point x̃0 ∈ X
e
mine whether a covering map p : X → X is regular.
e → X is regular if, for some
Proposition 15.4.7 A covering map p : X
e p# π(X,
e x̃0 ) is normal in π (X, p (x̃0 )).
x̃0 ∈ X,
e x̃0 ) is a normal subgroup of π (X, x0 ),
Proof. Assume that p# π(X,
e be arbitrary and put p (x̃1 ) = x1 .
where x0 = p (x̃0 ). Let x̃1 ∈ X
e is path-connected, there is a path f˜ in X
e from x̃0 to x̃1 . Then
Since X
˜
f = pf is a path in X from x0 to x1 , and there are isomorphisms
e x̃0 ) → π(X,
e x̃1 ),
η : π(X,
θ : π (X, x0 ) → π (X, x1 ) ,
[g̃] 7→ [f˜−1 ∗ g̃ ∗ f˜],
[g] 7→ [f −1 ∗ g ∗ f ].
and
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Elements of Topology
It is easily checked that p# ◦ η = θ ◦ p# (see Figure 15.7). From this,
e x̃0 ) onto p# π(X,
e x̃1 ). Since θ
we see that θ maps the subgroup p# π(X,
e x̃0 ) is normal in π (X, x0 ), p# π(X,
e x̃1 )
is an isomorphism and p# π(X,
is also normal in π (X, x1 ), and the proposition follows.
♢
The next result gives a topological criterion for regular coverings.
e → X is regular if and only if
Theorem 15.4.8 A covering map p : X
for any closed path g in X, either every lifting of g is closed or none is
closed.
Proof. Assume first that p is regular, and let g be a closed path in
e based at x̃. If h̃ is a
X based at x having a closed lifting g̃ in X
lifting of g with h̃ (0) = ỹ, then p (x̃) = x = p (ỹ) and the subgroups
e x̃) and p# π(X,
e ỹ) are conjugate in π (X, x), by Lemma 15.3.7.
p# π(X,
e
e x̃) = p# π(X,
e ỹ).
Since p# π(X, x̃) is normal in π(X, x), we have p# π(X,
e ỹ) and so there exists a loop f˜ in
Consequently, [g] = p# [g̃] ∈ p# π(X,
e based at ỹ such that p# [f˜] = [g]. It follows that ph̃ = g ≃ pf˜ rel
X
{0, 1}. By the Monodromy Theorem, we see that h̃ (1) = f˜ (1) = ỹ,
and thus h̃ is a closed path.
Conversely, suppose that either every lifting relative to p of a closed
path in X is closed or none is closed. To see that p is regular, consider
e x̃) ⊆ π(X, x). Then, by Lemma 15.3.7 again, a
a subgroup p# π(X,
e x̃) in π(X, x) is the subgroup p# π(X,
e ỹ) for some
conjugate of p# π(X,
e x̃) = p# π(X,
e ỹ) whenỹ ∈ p−1 (x). So it suffices to prove that p# π(X,
e x̃). By the Path Lifting Property of p,
ever p (x̃) = p (ỹ). Let [g̃] ∈ π(X,
e
there is a path h̃ in X with ph̃ = pg̃ and h̃ (0) = ỹ. Since g̃ is a closed
e ỹ)
lifting of pg̃, h̃ must be closed (by our assumption). Thus [h̃] ∈ π(X,
e
e
and p# [h̃] = p# [g̃]. This implies that p# π(X, x̃) ⊆ p# π(X, ỹ). Simie ỹ) ⊆ p# π(X,
e x̃), and the equality holds.
larly, p# π(X,
♢
e → X, a loop in X may have
We remark that, given a covering X
e that is a loop and another which is a path with distinct
one lift in X
end points.
Example 15.4.3 Let X = A ∨ B be the wedge of two circles, and let
e be the subspace of R2 , as pictured in Figure 15.8 (below). Let p be
X
e once around
the mapping which winds the first and fifth circles of X
circle A, the third twice around circle A, and the second and fourth
twice around circle B. Then p is a covering map. Consider the loop in
COVERING SPACES
473
~
X
=
~
x
0
•
•
^
^^
^^
^^
^
X based at x0 represented by the circle A taken counterclockwise. Its
lift starting at x̃1 is not a loop, while it lifts to a loop based at x̃0 .
~
x
1
^^
^^
^^
^
p
X
=
A
•
x0
B
FIGURE 15.8: The covering map in Example 15.4.3.
We now prove a theorem which completely determines the group of
deck transformations in terms of the fundamental groups of the covering space and the base, and can be used to compute the fundamental
groups of certain spaces from properties of their coverings.
e → X be a covering map and x̃ ∈ X.
e The
Theorem 15.4.9 Let p : X
group ∆ (p) of deck transformations of the covering p is isomorphic to
the quotient group
(
)
e x̃) /p# π(X,
e x̃),
N p# π(X,
(
)
e x̃) is the normaliser of p# π(X,
e x̃) in π (X, p (x̃)).
where N p# π(X,
e x̃) is the isotropy subProof. Write Γ = π (X, p (x̃)). Then p# π(X,
group Γx̃ . Let α ∈ N (Γx̃ ) be arbitrary and put x̃ · α = ỹ. By Proposition 15.4.4, there exists hα ∈ ∆ (p) such that hα (x̃) = ỹ and, by
e which
Proposition 15.4.2, this is a unique deck transformation of X
maps x̃ into x̃ · α. So there is a mapping Ψ : N (Γx̃ ) → ∆ (p) defined by Ψ (α) = hα . We first show that Ψ is a homomorphism. Let
α, β ∈ N (Γx̃ ). Then hαβ (x̃) = x̃ · (αβ) = (x̃ · α) · β = hα (x̃) · β =
hα (x̃ · β) (by Proposition 15.4.3) = hα (hβ (x̃)) = (hα ◦ hβ ) (x̃). By
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Elements of Topology
Proposition 15.4.2, again, we have hαβ = hα ◦ hβ . Thus Ψ is a homomorphism. Next, we observe that Ψ is surjective. Let h ∈ ∆ (p) be arbitrary, and suppose that h (x̃) = ỹ. By Proposition 15.4.4, again, there
exists α ∈ N (Γx̃ ) such that ỹ = x̃·α = Ψ (α) (x̃). Hence h = Ψ (α), and
Ψ is surjective. Finally, we have α ∈ ker(Ψ) ⇔ hα = 1Xe ⇔ hα (x̃) =
x̃ ⇔ x̃ · α = x̃ ⇔ α ∈ Γx̃ . So ker(Ψ) = Γx̃ , and the theorem follows. ♢
(
)
e → X is a regular covering, then N p# π(X,
e x̃) =
If p : X
π (X, p (x̃)), and therefore we have
e → X is a regular covering and x̃ ∈ X,
e
Corollary 15.4.10 If p : X
∼
e
then ∆ (p) = π (X, p (x̃)) /p# π(X, x̃).
e is a simply connected covering space of X, then
In particular, if X
e →X
the group ∆ of the deck transformations of the covering map X
is isomorphic to π (X), and the order of the group π (X) is equal to the
multiplicity of the covering (ref. Proposition 15.2.7). This result gives
a description of the fundamental group of X, which requires no choice
of base point.
( )
As an application of the preceding result, we determine π S1 ,
alternatively. Since R1 is simply connected, the covering p : R1(→ )S1 ,
t 7→ e2πıt , is universal. By Ex. 15.4.1, ∆ (p) ∼
= Z, and therefore π S1 ∼
=
Z.
As another instance, consider the covering map p : Sn → RPn , n ≥
2. Since Sn is simply connected for n ≥ 2, and the group ∆ (p) of deck
transformations of p is Z2 (ref. Ex. 15.4.2), we see that π (RPn ) ∼
= Z2
for n ≥ 2.
PROPER ACTIONS OF DISCRETE GROUPS
By Corollaries 15.3.2 and 15.4.10, we see that if a space X has
a universal covering space Ye , then the fundamental group of a pathe of X is isomorphic to the group ∆(q) of
connected covering space X
e Moreover,
deck transformations of a regular covering map q : Ye → X.
e is homeomorphic to the orbit space Ye /∆ (q), because the orbits of
X
e and both mappings Ye →
∆ (q) are precisely the fibers q −1 (x̃), x̃ ∈ X,
e and Ye → Ye /∆ (q) are identifications. This suggests a possibility of
X
e of X, with a given group G ⊆ π(X)
constructing a covering space X
as its fundamental group, via an action of G on the universal covering
COVERING SPACES
475
space of X. On the other hand, it is easily seen that the orbit map
associated with an arbitrary action of a group on a space may not be
a covering map.
To see the conditions under which the orbit map of a transformation
group could be a covering map, let us examine more closely the action
e →X
of the group of deck transformations of a given covering map p : X
e First, notice that ∆(p) acts freely on X,
e by
on its covering space X.
e
Proposition 15.4.2. Next, suppose that x̃, ỹ ∈ X and p (x̃) = p (ỹ). Then
we choose a connected admissible nbd U of p (x̃) , and consider the
sheets Ux̃ and Uỹ over U containing x̃ and ỹ, respectively. It is obvious
that both Ux̃ and Uỹ are components of p−1 (U ). Suppose that gUx̃ ∩
Uỹ ̸= ∅ ̸= hUx̃ ∩ Uỹ for g, h ∈ ∆(p). Since gUx̃ is a connected subset of
p−1 (U ), it must be contained in one of its components. It follows that
gUx̃ ⊆ Uỹ . Since p|Uỹ is injective, we have gx̃ = ỹ. Similarly, hx̃ = ỹ
and we obtain gx̃ = hx̃. This implies that g = h, since the action of
e is free. It follows that the set {h ∈ ∆(p)|h (Ux̃ ) ∩ Uỹ ̸= ∅}
∆(p) on X
is a singleton. In the case x = p (x̃) ̸= p (ỹ) = y, this set is empty if X
is assumed Hausdorff. For, then there are disjoint open nbds Vx and
Vy of x and y in X, respectively. So Ux̃ = p−1 (Vx ) and Uỹ = p−1 (Vy )
e respectively. Since h (Ux̃ ) ⊆ Ux̃
are disjoint open nbds of x̃ and ỹ in X,
for every h ∈ ∆(p), we have h (Ux̃ ) ∩ Uỹ = ∅.
This property of the group of deck transformations of a covering
map with the Hausdorff base space is abstracted in the following.
Definition 15.4.11 A discrete group G is said to act properly on a
space X if for each pair of points x, y ∈ X, there exist open nbds Ux
and Uy of x and y, respectively, in X such that {g ∈ G|gUx ∩ Uy ̸= ∅}
is finite.
Interestingly, any discrete group G acting freely and properly on a
connected Hausdorff space X is the group of deck transformations of
a covering projection with the covering space X. To see this, we first
note an important property of proper actions of discrete groups.
Lemma 15.4.12 If a discrete group G acts properly on a Hausdorff
space X, then the isotropy group Gx at x ∈ X is finite and there exists
an open nbd U of x such that gU ∩ U = ∅ for g ∈
/ Gx .
Proof. Suppose that G is a discrete group and acts properly on a Hausdorff space X. Let x ∈ X be arbitrary. Then there exists an open
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Elements of Topology
nbd U0 of x such that H = {g ∈ G|gU0 ∩ U0 ̸= ∅} is finite. Clearly,
Gx ⊆ H and is, therefore, finite. Write H − Gx = {h1 , . . . , hn } and
put xi = hi x for every i = 1, . . . , n. Then x ̸= xi for every i. Since
X is Hausdorff, we find nbds Vi of x and Wi of xi in X such that
Vi ∩ Wi = ∅. Clearly, each Ui = Vi ∩ h−1
i (Wi ) is a nbd
∩n of x and satisfies the condition Ui ∩ hi Ui = ∅. Then the set U = i=0 Ui is an open
nbd of x and satisfies the requirement of the lemma.
♢
In particular, if a discrete group G acts freely and properly on a
Hausdorff space X, then there exists an open nbd U of x such that
gU ∩ U = ∅ for all g ̸= e, the identity element of G. In the literature,
a continuous action of a discrete group having this property is often
called properly discontinuous. (This is a confusing terminology and we
will avoid it.) Now, we come to see a theorem which plays an important
role in the next section in establishing the existence of a covering space
of a space X with the given subgroup of π(X) as its fundamental group.
Theorem 15.4.13 If a discrete group G acts freely and properly on
a connected Hausdorff space X, then the orbit map q : X → X/G
is a regular covering map. Moreover, G ∼
= ∆(q), the group of deck
transformations of q.
Proof. We have already seen in §13.2 that q is continuous, open and
surjective. By the preceding lemma, for each x ∈ X, there exists an
open nbd U of x such that gU ∩U = ∅ for all g ̸= e, the identity element
of G. Set V = q(U ). Then V is an open nbd of q(x), and q −1 (V ) is
the union of the disjoint open sets gU , g ∈ G. We observe that q|gU is
a homeomorphism between gU and V. Since g|U is a homeomorphism
between U and gU for every g ∈ G and (q|gU )◦(g|U ) = q|U, it is enough
to show that q|U is a homeomorphism between U and V. Obviously, it
is continuous, open and surjective. To see the injectivity, suppose that
q(x) = q(x′ ) for x, x′ ∈ U. Then x′ = gx for some g ∈ G. Accordingly,
gU ∩ U ̸= ∅, and therefore g = e. Thus x = x′ and q|U is injective. It
follows that V is an admissible nbd of q(x), and q is a covering map.
For every g ∈ G, let θg denote the homeomorphism x 7→ gx of
X. Since the action of G on X is effective, G is isomorphic to the
group Γ = {θg |g ∈ G}. Since q(gx) = q(x), Γ ⊆ ∆(q). Conversely, if
h ∈ ∆(q), then q(h(x)) = q(x) for every x ∈ X. Choose an x ∈ X.
Then there exists a g ∈ G such that h(x) = gx = θg (x). Thus the
deck transformations h and θg agree at a point. Since X is connected,
we have h = θg . So ∆(q) ⊆ Γ, and the equality holds. It follows that
COVERING SPACES
477
the group ∆(q) is transitive on each fibre, so q is a regular covering
projection, by Corollary 15.4.5. This completes the proof.
♢
We remark that the proof of the preceding theorem does not require
the condition of local path-connectedness on the space X. Also, notice
that this result and its following corollary hold good even for properly
discontinuous actions.
Corollary 15.4.14 If a discrete group G acts freely and properly on
a simply connected Hausdorff space X, then π(X/G) ∼
= G.
Proof. Let q : X → X/G denote the orbit map. By the preceding
proposition, q is a regular covering projection and G is canonically isomorphic to the group ∆(q) of deck transformations of q. So we can
consider the elements of G as the deck transformations of q. Choose a
point x0 ∈ X. Then, given α ∈ π (X/G, q(x0 )), there exists a unique
gα ∈ G such that gα (x0 ) = x0 · α, since X is connected and G is
transitive on q −1 (x0 ). So there is a function Ψ : π (X/G, q(x0 )) → G
defined by Ψ(α) = gα . As seen in Theorem 15.4.9, Ψ is a homomorphism with ker(Ψ) = q# π (X, x0 ) = {0}. Since π (X/G, q(x0 )) is transitive on q −1 (x0 ), given g ∈ G, we find an α ∈ π (X/G, q(x0 )) such that
x0 ·α = gx0 . This shows that Ψ is surjective, and the proof is complete.
♢
The preceding corollary can be used to determine the fundamental
groups of many spaces. By way of illustration, we compute the fundamental groups of the Klein bottle and the lens space L (p, q).
Example 15.4.4 Consider the following isometries of R2 : The reflection
ρ : (x, y) 7→ (x, −y) , the translations τX : (x, y) 7→ (x + 1, y) and
τY : (x, y) 7→ (x, y + 1) . Then σ = τX ◦ τY ◦ ρ is a glide reflection of R2 ,
for τY ◦ ρ is a reflection in the line y = 1/2. Let G be the subgroup of
the euclidean group E (2) generated by σ and τY . It is easily verified
that σ −1 ◦ τY ◦ σ = τY−1 , so G is a nonabelian group. Obviously, the
subgroup generated by ⟨τY ⟩ is infinite cyclic and normal in G. We also
note that G is spanned by the two parallel glide reflections σ and τY σ
2
which satisfy σ 2 = (τY σ) (a translation parallel to the x-axis). Since
the topology of E (2) coincides with the product topology of O (2)×R2 ,
we see that the subgroup of E (2) generated by ρ, τX and τY has the
discrete topology, and therefore G is a discrete group.
Next, we show that the action of G on R2 is properly discontinuous. Using the relation τY ◦ σ = σ ◦ τY−1 , each element g of G can be
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Elements of Topology
written as g = τYm ◦ σ n . Obviously, one can write σ n (x, y) = (x + n, z).
Then g(x, y) = (x + n, z + m). Thus, with U = B ((x, y); 1/3), we have
gU ∩ U = ∅ for every g ̸= e. So the action of G on R2 is properly discontinuous and hence the fundamental group of the orbit space R2 /G
is G.
Finally, we see that R2 /G is homeomorphic to the Klein bottle.
Notice that the image of a unit square X having vertices at the lattice
points (points in the plane R2 whose coordinates are integers) under
each element of G is such a square, and the images of X under all
elements of G fill out R2 . Moreover, every nontrivial element of G maps
a point in the interior of X into the interior of another such square. It is
also clear that τY maps the bottom side of X = I ×I homeomorphically
onto its top side while σ maps the left side of X homeomorphically onto
its right side, as shown in Figure 15.9. It follows that the orbit space
R2 /G is homeomorphic to the Klein bottle (ref. Ex. 7.1.5).
y
^
^
^
X
^
^
^
x
FIGURE 15.9: Proof of Example 15.4.4.
Recall that we have already computed the fundamental group of
the Klein bottle in the previous chapter; we have obtained here another
presentation of it.
Example 15.4.5 Let p and q0 , . . . , qn be positive integers and suppose
that each qi is relatively prime
+ 1)-sphere
S2n+1
{ to p. The (2nn+1
} can be
∑
2
considered as the subspace (z0 , . . . , zn ) ∈ C
| |zi | = 1 of Cn+1 .
COVERING SPACES
479
There is a free action of the cyclic group Zp on S2n+1 . Let h : S2n+1 →
S2n+1 be the homeomorphism defined by
h (z0 , . . . , zn ) = (ϵq0 z0 , . . . , ϵqn zn ),
where ϵ = e2πι/p is a primitive pth root of unity. Then h generates a
cyclic group of order p, and therefore determines an action of the group
Zp on S2n+1 . This is clearly free and proper. The orbit space S2n+1 /Zp
is called a (generalised) lens space and denoted by L (p; q0 , . . . , qn ). The
particular case n = 1 and q0 = 1 is usually denoted by L(p; q1 ). Since
S2n+1 is simply connected, the fundamental group of L (p; q0 , . . . , qn )
is Zp , by Corollary 15.4.14.
Exercises
e → X be the covering map of Exercise 15.1.3. Show
1. Let p : X
that p is a regular covering and determine the group ∆ (p).
e → X be the covering map and Γ = π(X, x0 ). Show that
2. Let p : X
∆(p) is isomorphic to the group of all Γ-equivalences of p−1 (x0 ).
3. If H is a discrete subgroup of a connected Hausdorff group G,
prove that the quotient map G → G/H is a regular covering
projection with the covering group H.
4. Let G be a simply connected Hausdorff topological group and
H ⊂ G a discrete subgroup. Prove that π (G/H) ∼
= H.
e → X be a regular covering map. Show that X is
5. Let p : X
e
homeomorphic to the orbit space X/∆(p).
6. Show that a free action of a finite group on a Hausdorff space is
properly discontinuous.
7. Find an example of a free action of an infinite group on some
space that is not properly discontinuous.
8. Prove that the set U in Lemma 15.4.12 can be so chosen that it
is invariant under Gx and the canonical map U/Gx → X/G is a
homeomorphism onto an open subset of X/G.
480
15.5
Elements of Topology
The Existence of Covering Spaces
In Section 15.3, we have seen that each path-connected covering
space of a space X picks out a conjugacy class of subgroups of the
fundamental group of X, and obtained a criterion for equivalence of
such covering spaces of X in terms of conjugacy of subgroups of the
fundamental group of X induced by the covering maps. In view of these
facts, one can expect a classification of the covering spaces of a space
X by means of conjugacy classes of subgroups of π (X). To this end, we
need to settle only the existence problem. This entails the construction
of a covering space of X corresponding to each subgroup of π (X).
We first prove the following.
Lemma 15.5.1 Let p : Y → X and q : Z → Y be continuous maps
and suppose that all the three spaces X, Y and Z are connected and
locally path-connected. If q and the composition pq : Z → X are
covering maps, then p is also a covering map.
Proof. First, notice that p is a surjection, for pq is so. Next, given x ∈ X,
find a path-connected open nbd U of x which is evenly covered by pq.
Assume that Vα , α ∈ A, are the sheets over U relative ∪
to pq. Then
the Vα are disjoint open subsets of Z and (pq)−1 (U ) = α Vα . Also,
(pq)|Vα is a homeomorphism between Vα and U, so∪each Vα is pathconnected. Since q is surjective, we have p−1 (U ) = α q (Vα ) . Since q
is continuous and open, each q (Vα ) is path-connected and open in Y .
We assert that these sets are path components of p−1 (U ). Clearly, it
suffices to show that each q (Vα ) is also closed in p−1 (U ). To this end,
fix an arbitrary index α0 and suppose that y belongs to the closure
of q (Vα0 ) in p−1 (U ). Since Y is locally path-connected, there exists a
path-connected open nbd W of y which is contained in p−1 (U ) and
is evenly covered
∪ by q. Then each sheet over W is path-connected,
∪
and q −1 (W ) ⊂ α Vα . Since Vα are path components of α Vα , each
sheet over W is contained in some Vα . Also, q −1 (W ) ∩ Vα0 ̸= ∅, for
W ∩ q (Vα0 ) ̸= ∅. So Vα0 contains some sheet over W and we have
y ∈ W ⊂ q (Vα0 ). Thus q (Vα0 ) is closed in p−1 (U ) and our assertion
follows. Now, it is clear that either q (Vα ) = q (Vβ ) or q (Vα )∩q (Vβ ) = ∅
for every α, β ∈ A. Finally, we observe that each q (Vα ) is a sheet over
U . Since (pq)|Vα is a homeomorphism between Vα and U , it is clear that
q|Vα is injective and thus a homeomorphism between Vα and q (Vα ).
COVERING SPACES
481
From the equality (pq)|Vα = (p|q(Vα )) ◦ (q|Vα ), we see p|q (Vα ) is a
homeomorphism between q (Vα ) and U. It follows that U is evenly
covered by p, and p is a covering map.
♢
Now, we come to see the main theorem.
Theorem 15.5.2 Let X be a Hausdorff, connected, locally pathconnected and semilocally 1-connected space and x0 ∈ X. Then,
for each subgroup H ⊆ π(X, x0 ), there exists a covering projection
e → X such that p# π(X,
e x̃0 ) = H.
p:X
Proof. By Theorem 15.3.4, X has a universal covering space Ye . Let
r : Ye → X be the covering projection and ∆(r) be the group of deck
transformations of r. As seen above, ∆(r) acts freely and properly on Ye
and X ≈ Ye /∆(r). Choose a point ỹ0 ∈ r−1 (x0 ). By Corollary 15.4.10,
there is an isomorphism Ψ : π(X, x0 ) → ∆(r) given by Ψ(α) (ỹ0 ) =
ỹ0 · α. Put D = Ψ(H). Then D ∼
= H and acts freely and properly
on Ye , since this is true of ∆(r). By Theorem 15.4.13, the orbit map
e = Ye /D and x̃0 = q (ỹ0 ).
q : Ye → Ye /D is a covering map. We write X
e
Obviously, there is a continuous map p : X → Ye /∆(r) ≈ X such that
r = pq, that is, the following diagram
Ye
q
Z
Z
rZ
Z
=
~ e
- X
p
X
commutes. Since r and q are covering projections, p is also a covering
projection, by the preceding lemma. We have x̃0 ∈ p−1 (x0 ) and show
e x̃0 ). Obviously, α ∈ H ⇐⇒ Ψ(α) ∈ D. Suppose
that H = p# π(X,
that α = [f ], where f is a loop in X based at x0 . Let Fe be the lifting
of f relative to the covering map r with the origin ỹ0 . Then Fe (1) =
ỹ0 · α = Ψ(α) (ỹ0 ). So, for Ψ(α) ∈ D, Fe(0) and Fe(1) are in the same
e based at x̃0 . Since
orbit of D; consequently, f˜ = q ◦ Fe is a loop in X
e x̃0 ) for every α ∈ H. Conversely,
pf˜ = rFe = f , we have α ∈ p# π(X,
e based at x̃0 , f = pf˜ is a loop in X based at x0 . Let
given a loop f˜ in X
e
F be the lifting of f˜ relative to the covering map q with the origin ỹ0 .
Then rFe = pf˜ = f ; accordingly, Fe is the lifting of f relative to r with
the origin ỹ0 . So, for α = [f ] ∈ π(X, x0 ), Ψ(α) (ỹ0 ) = ỹ0 · α = Fe(1).
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Elements of Topology
e ỹ0 and Ψ(α) (ỹ0 ) must be in the same orbit
Since f˜ = q Fe is a loop in X,
of D. So there exists an element ϕ ∈ D such that ϕ (ỹ0 ) = Ψ(α) (ỹ0 ).
By Proposition 15.4.2, Ψ(α) = ϕ ∈ D. Hence we have the equality
e x̃0 ) = H.
p# π(X,
♢
Thus, we see that if X is a Hausdorff, connected, locally pathconnected and semilocally 1-connected space, then there is a one-toone correspondence between the equivalence classes of covering spaces
of X
the
classes
of π(X). The correspondence is given
[ and
] conjugacy
[
]
p
e
e
by X → X ←→ p# π(X) .
Exercise
1. Determine all covering spaces of the lens space L(p, q), up to
equivalence.
Appendix A
Set Theory
A.1
A.2
A.3
A.4
A.5
A.6
A.7
A.8
A.1
Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Finite and Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Ordinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
483
485
488
491
492
500
507
512
Sets
The purpose of this appendix is to introduce some of the basic ideas
and terminologies from set theory which are essential to our present
work. In this naive treatment, we don’t intend to give a complete and
precise analysis of set theory, which belongs to the foundations of mathematics and to mathematical logic. Rather, we shall deal with sets on
an intuitive basis. We remark that this usage can be formally justified.
Intuitively, a set is a collection of objects called members of the set.
The terms “collection’, and “family” are used as synonyms for “set.”
If an object x is a member of a set A, we write x ∈ A to express
this fact. Implicit in the idea of a set is the notion that a given object
either belongs or does not belong to the set. The statement “x is not
a member of A” is indicated by x ∈
/ A. The objects which make up a
set are called the elements or points of the set.
If X and Y are sets, we say that X is a subset of Y, written X ⊆ Y,
if every element of X is also an element of Y. The sets X and Y are
said to be equal, written X = Y if X ⊆ Y and Y ⊆ X. We call X
a proper subset of Y (written X ⊂ Y ) if X ⊆ Y but X ̸= Y . The
empty or null set which has no element is denoted by ∅. We have the
inclusion ∅ ⊆ X for every set X. The set whose only element is x is
called a singleton, denoted by {x}. Note that ∅ ̸= {∅}. If P (x) is a
483
484
Elements of Topology
statement about elements in X, that is either true or false for a given
element of X, then the subset of all the x ∈ X for which P (x) is true
is denoted by {x ∈ X|P (x)} or {x ∈ X : P (x)}.
In a particular mathematical discussion, there is usually a set which
consists of all primary elements under consideration. This set is referred
to as the “universe.” To avoid any logical difficulties, all the sets we
consider in this section are assumed to be subsets of the universe X.
The difference of two sets A and B is the set A − B = {x ∈
A|x ∈
/ B}. If B ⊆ A, the complement of B in A is A − B. Notice that the complement operation is defined only when one set is
contained in the other, whereas the difference operation does not
have such a restriction. The union of two sets A and B is the set
A ∪ B = {x|x belongs to at least one of A, B}. The intersection of two
sets A and B is the set A∩B = {x|x belongs to both A and B}. When
A ∩ B = ∅, the sets A and B are called disjoint; otherwise we say that
they intersect.
Proposition A.1.1
(a) X − (X − A) = A.
(b) B ⊆ A ⇔ X − A ⊆ X − B, and A = B ⇔ X − A = X − B.
Proposition A.1.2 A ∪ A = A ∩ A.
Proposition A.1.3 A ∪ B = B ∪ A, A ∩ B = B ∩ A.
Proposition A.1.4 A ⊆ B ⇔ A = A ∩ B ⇔ B = A ∪ B.
Proposition A.1.5 A ∪ (B ∪ C) = (A ∪ B) ∪ C, A ∩ (B ∩ C) =
(A ∩ B) ∩ C.
Proposition A.1.6 A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), A ∪ (B ∩ C) =
(A ∪ B) ∩ (A ∪ C).
(a) A − B = A ∩ (X − B).
}
X − (A ∪ B) = (X − A) ∩ (X − B)
(De Morgan’s laws)
X − (A ∩ B) = (X − A) ∪ (X − B).
Proposition A.1.7
(b)
(c) (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).
(d) If X = A ∪ B and A ∩ B = ∅, then B = X − A.
Set Theory
485
Let A be a nonempty set, and suppose that a set Yα is given each
α ∈ A. Then the collection of sets {Yα |α ∈ A} is called an indexed
family of sets, and A is called an indexing set for the family. The union
of this family is the set
∪
Yα = {x|x ∈ Yα for some α in A},
α∈A
and the intersection is the set
∩
Yα = {x|x ∈ Yα for every α in A}.
α∈A
∪
∩
We also write {Yα |α ∈ A} for the union of the Yα , and {Yα |α ∈ A}
for their intersection.
If there is no ambiguity
about the indexing set,
∪
∩
we simply use Yα for the union and Yα for
∪n the intersection. If
A = {1, . . . , n}, n >∩0 an integer, then we write α=1 Yα for the union
n
of Y1 , . . . , Yn , and α=1 Yα for their intersection. Observe that any
nonempty collection C of sets can be considered an indexed family of
sets by “self-indexing”: The indexing set is C itself and one assigns to
each S ∈ C the set S. Accordingly, the foregoing definitions become:
∪
{S : S ∈ C} = {x|x ∈ S for some S in C} and
∩
{S : S ∈ C} = {x|x ∈ S for every S in C}.
∪ If we allow the collection
∩ C to be the empty set, then, by convention,
{S : S ∈ C} = ∅ and {S : S ∈ C} = X, the specified universe of
the discourse.
Proposition A.1.8 Let Yα , α ∈ A, be a family of subsets of a set X.
∪
∩
∩
∪
(a) X − α Yα = α (X − Yα ), X − α Yα = α (X − Yα ).
∪
∪
∩
∩
(b) If B ⊂ A, then β∈B Yβ ⊆ α∈A Yα and β∈B Yβ ⊇ α∈A Yα .
A.2
Functions
With each two objects x, y, there corresponds a new object (x, y),
called their ordered pair. This is another primitive notion that we
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Elements of Topology
will use without a formal definition. Ordered pairs are subject to
the condition: (x, y) = (x′ , y ′ ) ⇔ x = x′ and y = y ′ . Accordingly,
(x, y) = (y, x) ⇔ x = y. The first (resp. second) element of an ordered pair is called the first (resp. second) coordinate. Given two sets
X1 and X2 , their cartesian product X1 × X2 is defined to be the
set of all ordered pairs (x1 , x2 ), where xi ∈ Xi for i = 1, 2. Thus
X1 ×X2 = {(x1 , x2 ) |xi ∈ Xi for every i = 1, 2}. Note that X1 ×X2 = ∅
if and only if X1 = ∅ or X2 = ∅. When both X1 and X2 are nonempty,
X1 × X2 = X2 × X1 ⇔ X1 = X2 .
Let X and Y be two sets. A function f from the set X to Y (written
f : X → Y ) is a subset of X × Y with the following property: for each
x ∈ X, there is one and only one y ∈ Y such that (x, y) ∈ f. A function
is also referred to as a mapping (or briefly, a map). We write y = f (x)
to denote (x, y) ∈ f, and say that y is the image of x under f or the
value of f at x. We also say that f maps (or carries) x into y or f or
sends (or takes) x to y. A function f from X to Y is usually defined by
specifying its value at each x ∈ X, and if the value at a typical point
x ∈ X is f (x), we write x 7→ f (x) to give f. We refer to X as the
domain and Y as the codomain of f. The set f (X) = {f (x) |x ∈ X},
also denoted by im(f ), is referred to as the range of f.
The identity function on X, which sends every element of X to itself,
is denoted by 1 or 1X . A map c : X → Y which sends every element
of X to a single element of Y is called a constant function. Notice that
the range of a constant function consists of just one element. If A ⊂ X,
the function i : A ,→ X, a 7→ a, is called the inclusion map of A into
X. If f : A → X and A ⊂ X, then the restriction of f to A is the
function f |A : A → Y defined by f |A (a) = f (a) ∀ a ∈ A. In the other
direction, if A ⊂ X and g : A → Y is a function, then an extension of
g over X is a function G : X → Y such that G|A = g. The inclusion
map i : A ,→ X is the restriction of the identity map 1 on X.
Proposition A.2.1
∪ Let Y be a set and {Xα |α ∈ A} a family of subsets of Y with Y = Xα . If, for each α ∈ A, fα : Xα → Z, is a function
such that fα | (Xα ∩ Xβ ) = fβ | (Xα ∩ Xβ ) for all α, β ∈ A, then there
exists a unique function F : Y → Z which extends each fα .
Proof. Given y ∈ Y, there exists an index α ∈ A such that y ∈ Xα . We
put F (y) = fα (y) if y ∈ Xα . If y ∈ Xα ∩ Xβ , then fα (y) = fβ (y), by
our hypothesis. Thus F (y) is uniquely determined by y, and we have
a single-valued function F : Y → Z. It is clearly an extension of each
Set Theory
487
fα . The uniqueness of F follows from the fact that each y ∈ Y belongs
to some Xα so that any function Y → Z satisfying the requirement of
the proposition will have to assume the value fα (y) at y.
♢
We say that a function f : X → Y is surjective (or a surjection
or onto) if Y = f (X). If f (x) ̸= f (x′ ) for every x ̸= x′ , then we say
that f is injective (or an injection or one-to-one). A function is bijective
(or a bijection or a one-one correspondence) if it is both injective and
surjective.
If f : X → Y and g : Y → Z are functions, then the function
X → Z which maps x into g (f (x)) is called their composition and
denoted by g ◦ f or simply gf.
Proposition A.2.2 Suppose that f : X → Y and g : Y → X satisfy
gf = 1X . Then f is injective and g is surjective.
Given a function f : X → Y, a function g : Y → X such that
gf = 1X and f g = 1Y is called an inverse of f . If such a function g
exists, then it is unique and we denote it by f −1 . It is clear from the
preceding result that f has an inverse if and only if it is a bijection.
(
)−1
Also, it is evident that f −1
= f.
Let f : X → Y be a function. For each A ⊆ X, the subset f (A) =
{f (x) |x ∈ X} of Y is called the image of A under f, and for each
B ⊆ Y , the subset f −1 (B) = {x ∈ X|f (x) ∈ B} is called the inverse
image of B in X under f.
The power set of a set X is the family P (X) of all subsets of X.
A function f : X → Y induces a function P (X) → P (X), A 7→ f (A).
It also induces a function P (Y ) → P (Y ), B 7→ f −1 (B). The main
properties of these functions are described in the following.
Proposition A.2.3 Let f : X → Y be a function.
(a) A1 ⊆ A2 ⇒ f (A1 ) ⊆ f (A2 ),
(∪
) ∪
(b f
A
= j f (Aj ),
j
j
(c) f
(∩
) ∩
A
⊆ j f (Aj ), and
j
j
(d) Y − f (A) ⊆ f (X − A) ⇔ f is surjective.
Proposition A.2.4 Let f : X → Y be a function.
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Elements of Topology
(a) B1 ⊆ B2 ⇒ f −1 (B1 ) ⊆ f −1 (B2 ),
(∪
) ∪
−1
(b) f −1
B
(Bj ),
j j =
jf
(c) f −1
) ∩
⊆ j f −1 (Bj ), and
B
j
j
(∩
(d) f −1 (Y − B) ⊆ X − f −1 (B).
Proposition A.2.5 Let f : X → Y be a function. Then:
(a) For each A ⊆ X, f −1 (f (A)) ⊇ A; in particular, f −1 (f (A)) = A
if f is injective.
( −1
)
(b) For
each B
( −1
) ⊆ Y, f f (B) = B ∩ f (X); in particular,
f f (B) = B if f is surjective.
A.3
Cartesian Products
In the previous section, we have defined the cartesian product of
two sets. For this operation, the following statements are easily proved.
(a) X1 × (X2 ∪ X3 ) = (X1 × X2 ) ∪ (X1 × X3 ).
Proposition A.3.1
(b) X1 × (X2 ∩ X3 ) = (X1 × X2 ) ∩ (X1 × X3 ).
(c) X1 × (X2 − X3 ) = (X1 × X2 ) − (X1 × X3 ).
(d) For Yi ⊆ Xi , i = 1, 2,
(X1 × X2 ) − (Y1 × Y2 ) = X1 × (X2 − Y2 ) ∪ (X1 − Y1 ) × Y2 =
[(X1 − Y1 ) × (X2 − Y2 )] ∪ [Y1 × (X2 − Y2 ) ∪ [(X1 − Y1 ) × Y2 ].
(∪
) ∪
∪
(a) ( α Xα ) ×
Y
β β =
α,β (Xα × Yβ ).
Proposition A.3.2
(b) (
∩
α Xα ) ×
(∩
) ∩
Y
= α,β (Xα × Yβ ).
β
β
Set Theory
489
The cartesian product of n (an integer > 2) sets X1 , . . . , Xn is
defined by induction: X1 ×· · ·×Xn = (X1 × . . . × Xn−1 )×Xn . A typical
element of X1 ×· · ·×Xn is written as (x1 , . . . , xn ), and xi is called its ith
coordinate. It should be noted that (X1 × X2 )× X3 ̸= X1 × (X2 × X3 ).
We now extend the definition of the cartesian product to an arbitrary indexed family of sets {Xα |α ∈ A}. Of course, the new definition of the product of the sets Xα for α ∈ {1, 2} must reduce to
the earlier notion of the cartesian product of two sets. Observe that
each ordered pair (x1 , x2 ) in X1 × X2 may be considered as defining
a function x : {1, 2} → X1 ∪ X2 with x (1) = x1 and x (2) = x2 .
Accordingly,
the cartesian product
∏
∏ of the Xα is defined to be the
∪ set
X
(or
simply
written
as
X
)
of
all
functions
x
:
A
→
α
α α
α Xα
such that
Xα for each ∏
α ∈ A. Occasionally, we denote the
∏x (α) ∈ ∏
product α Xα by α∈A Xα or {Xα |α ∈ A} to avoid any ambiguity
about
∏ the indexing
∏ set. We call Xα the αth factor (or coordinate set)
of Xα . If x ∈ Xα , then x (α) is called the αth coordinate of x, and
is often denoted by xα .
If the family {Xα } has n sets, n a positive integer, then it may be
indexed by the set {1, . . . , n}. In this case, we have two definitions of
its
∏2present sense will be denoted by
∏ncartesian product; the one in the
7→ (x1 , x2 ),
i=1 Xi → X1 × X2 , x ∏
i=1 Xi . Notice that the function
is a bijection. This shows that the notion of the product α Xα is a
reasonable generalisation of the notion of the cartesian product of two
sets.
∏n Similarly, the mapping x 7→ (x1 , . . . , xn ) gives a bijection between
i=1 Xi → X1 × · · · × Xn , and allows us to identify
∏n the two products
of X1 , . . . , Xn . We usually call an∏element x ∈ i=1 Xi an ordered ntuple. In general, an element x ∈ α∈A Xα is referred to as an
∏A-tuple
and written as (xα ). It is obvious that two elements x, y in α∈A Xα
are equal if and only∏if xα = yα for all α ∈ A. If one of the sets Xα
is empty, then so is Xα . On the other hand, if {Xα }∏is a nonempty
family of nonempty sets, it is not quite obvious that α Xα ̸= ∅. In
fact,
∏ a positive answer to the question of the existence of an element
in α Xα is one of the set-theoretic axioms, known as
Theorem A.3.3 (The Axiom of Choice) If {Xα |α ∈ A} is a
nonempty family of nonempty sets, then there exists a function
∪
c:A→
Xα
α
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Elements of Topology
such that c (α) ∈ Xα for each α ∈ A (c is called a choice function for
the family {Xα }.)
This axiom is logically equivalent to a number of interesting propositions; one such proposition is
Theorem A.3.4 (Zermelo’s Postulate) Let {Xα |α ∈ A} be a
nonempty family of nonempty pairwise disjoint sets. Then there exists a set C consisting of exactly one element from each Xα .
A few more equivalent reformulations of the axiom of choice will be
discussed later.
It should
∏ be noted that the sets Xα in the definition of the cartesian
product α∈A Xα need not be different from one another; indeed, it
may happen that they are all the same set X. In this case, the product
∏
α∈A Xα may be called the cartesian product of A copies of X or the
cartesian Ath power of X, and denoted by X A . Notice that X A is just
the set of all functions A → X.
∏
Proposition
A.3.5 If Yα ⊆ Xα for every ∏
α ∈ A, then
α Yα ⊆
∏
∏
X
.
Conversely,
if
each
X
=
̸
∅
and
Y
⊆
X
α
α α
α α
α α , then
Yα ⊆ Xα for every α.
Proposition A.3.6 Let {Xα |α ∈ A} be a family of nonempty sets,
and let Uα , Vα ⊆ Xα for every α. Then we have
∏
∏
∏
(a)
Uα ∪ Vα ⊆ (Uα ∪ Vα ).
∏
∏
∏
(b)
Uα ∩ Vα = (Uα ∩ Vα ).
∏
For each β ∈ A, we have the mapping pβ : α∈A Xα → Xβ given
by x 7→ xβ . It is easy to see that each of these maps is surjective.
The map pβ is referred to as
∏ the projection onto the βth factor. For
−1
Uβ ⊆ Xβ , p∏
(U
)
=
{x
∈
Xα such that xβ ∈ Uβ } is referred to as
β
β
the slab in Xα .
∩
∏
Proposition A.3.7 If B ⊂ A, then β∈B p−1
Yα , where
β (Uβ ) =
Yα = Xα if α ∈ A − B while Yα = Uα for α ∈ B.
Set Theory
A.4
491
Equivalence Relations
A relation on a set X is a subset R ⊆ X × X. If (x, y) ∈ R, we write
xRy. The relation ∆ = {(x, x) |x ∈ X} is called the identity relation
on X. This is also referred to as the diagonal. If R is a relation on X,
and Y ⊂ X, then R ∩ (Y × Y ) is called the relation induced by R on
Y.
Given a set X, a relation R on X is
(a) reflexive if xRx ∀ x ∈ X, (equivalently, ∆ ⊆ R),
(b) symmetric if xRy ⇒ yRx ∀ x, y ∈ X, and
(c) transitive if xRy and yRz ⇒ xRz, ∀ x, y, z ∈ X.
An equivalence relation on the set X is a relation which is reflexive,
symmetric and transitive. An equivalence relation is usually denoted by
the symbol ∼, read “tilde.” Suppose that ∼ is an equivalence relation
on X. Given an element x ∈ X, the set [x] = {y ∈ X|y ∼ x} is
called the equivalence class of x. It is clear that X equals the union
of all the equivalence classes, and two equivalence classes are either
disjoint or identical. A partition of a set X is a collection of nonempty,
disjoint subsets of X whose union is X. With this terminology, the
family of equivalence classes of X determined by ∼ is a partition of X.
Conversely, given a partition E of X, there is an equivalence relation
∼ on X such that the equivalence classes of ∼ are precisely the sets of
E. This relation is obtained by declaring x ∼ y if both x and y belong
to the same partition set.
Given an equivalence relation ∼ on X, the set of all equivalence
classes [x], x ∈ X, is called the quotient set of X by ∼, and is denoted
by X/ ∼. Thus we have another important method of forming new sets
from old ones. The map π : X → X/ ∼ defined by π (x) = [x] is called
the projection of X onto X/ ∼.
If R is a binary relation on a set X, then there is an equivalence
relation ∼ on X defined by x ∼ y if and only if one of the following is
true: x = y, xRy, yRx, or there exist finitely many points z1 , . . . , zn+1
in X such that z1 = x, zn+1 = y and either zi Rzi+1 or zi+1 Rzi for all
1 ≤ i ≤ n. It is called the equivalence relation generated by R.
492
A.5
Elements of Topology
Finite and Countable Sets
For counting objects in a set, we use the natural numbers (or positive integers) 1, 2, 3, . . .. When it is feasible, the process of counting a set
X requires putting it in one-one correspondence with a set {1, 2, . . . , n}
consisting of the natural numbers from 1 to n. Also, counting of sets in
essence serves to determine if one of two given sets has more elements
than the other. For this purpose, it may be easier to pair off each member of one set with a member of the other and see if any elements are
left over in one of the sets rather than count each. The following definition makes precise the notion of “sets having the same size.” We say
that two sets X and Y are equipotent (or have the same cardinality) if
there exists a bijection between them. Clearly, the relation of equipotence between sets is an equivalence relation on any given collection of
sets.
We shall assume familiarity with the set of natural numbers N =
{1, 2, 3, . . .} (also denoted by Z+ ), and the usual arithmetic operations
of addition and multiplication in N. We shall also assume the notion of
the order relation “less than” < on N and the “well-ordering property”:
Every nonempty subset of N has a smallest element. For any n ∈ N the
set {1, 2, . . . , n} consisting of the natural numbers from 1 to n will be
denoted by Nn . The set Nn is referred to as an initial segment of N. A
set X is finite if it is either empty or equipotent to a set Nn for some
n ∈ N. X is called infinite if it is not finite. When X is equipotent to a
set Nn we say that X contains n elements or the cardinality of X is n.
The cardinality of the empty set is 0. Of course, we must justify that
the cardinality of a finite set X is uniquely determined by it. To see
this, we first prove the following.
Proposition A.5.1 For any natural numbers m < n, there is no injection from Nn to Nm .
Proof. We use induction on n to establish the proposition. If n = 2,
then m = 1, and the proposition is obviously true. Now, let n > 2
and assume that the proposition is true for n − 1. We show that it is
true for n. If possible, suppose that there is an injection f : Nn → Nm ,
where m < n. If m is not in the image of Nn−1 , then f |Nn−1 is an
injection Nn−1 → Nm−1 , contrary to our inductive assumption. So we
may further assume that f (j) = m, j ̸= n. Then f (n) ̸= m. We define
Set Theory
a mapping g : Nn−1 → Nm−1 by
{
f (i)
g (i) =
f (n)
493
for i ̸= j
for i = j.
It is clear that g is an injection which, again, contradicts our inductive hypothesis. Therefore there is no injection Nn → Nm , and this
completes the proof.
♢
The preceding proposition is sometimes called the “Pigeonhole
Principle.” The contrapositive of the above proposition states that if
there is an injection Nm → Nn , then m ≤ n. As an immediate consequence of this, we see that the cardinality of a finite set X is uniquely
determined. For, if f : X → Nn and g : X → Nm are bijections, then
the composite f g −1 : Nm → Nn is a bijection. Hence m ≤ n and n ≤ m
which implies that m = n. It is evident that two finite sets X and Y
are equipotent if and only if they have the same number of elements.
Notice that for infinite sets, the idea of having the same “number of elements” becomes quite vague, whereas the notion of equipotence retains
its clarity.
Next, we observe the following useful property of finite sets.
Proposition A.5.2 An injective mapping from a finite set to itself is
also surjective.
Proof. Let X be nonempty finite set, and f : X → X be injective. Let
x ∈ X be an arbitrary point. Write x = x0 , f (x0 ) = x1 , f (x1 ) = x2 ,
and so on. Since X is finite, we must have xm = xn for some positive
integers m < n. Because f is injective, we obtain x0 = xn−m ⇒ x =
f (xn−m−1 ). Thus f is surjective.
♢
Proposition A.5.3 A proper subset of a finite set X is finite and has
cardinality less than that of X.
Proof. If X has n elements, then there is a bijection X → Nn . Consequently, each subset of X is equipotent to a subset of Nn , and thus it
suffices to prove that every proper subset of Nn is finite and has at most
n elements. We use induction on n to show that any proper subset of
Nn is finite and has at most n−1 elements. If n = 1, then N1 = {1} and
its only proper subset is the empty set ∅. Obviously, the proposition is
true in this case. Suppose now that every proper subset of Nk (k > 1)
has less than k elements. We show that each proper subset Y of Nk+1
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Elements of Topology
has at most k elements. If k +1 does not belong to Y, then we are done,
by the induction assumption. If k + 1 is in Y, then Y − {k + 1} ⊆ Nk .
By the induction hypothesis again, either Y = {k + 1}, or there exists
an integer m < k and a bijection f : Y − {k + 1} → Nm . In the latter
case, we define g : Y → Nm+1 by setting g (y) = f (y) for all y ̸= k + 1,
and g (k + 1) = m + 1. It is clear that g is a bijection. Accordingly, Y
is finite and has m + 1 ≤ k elements. This completes the proof.
♢
The contrapositive of this proposition states that if a subset Y ⊆ X
is infinite, then so is X.
Theorem A.5.4 For any nonempty set X, the following statements
are equivalent:
(a) X is finite.
(b) There is a surjection Nn → X for some n ∈ N.
(c) There is an injection X → Nn for some n ∈ N.
Proof. (a)⇒(b): Obvious.
(b)⇒(c): Let f : Nn → X be a surjection. Then f −1 (x) ̸= ∅ for
every x ∈ X. So, for each x ∈ X, we can choose a number c (x) ∈
f −1 (x) . Since {f −1 (x) |x ∈ X} is a partition of Nn , the function
x 7→ c (x) is an injection from X to Nn .
(c)⇒(a): If there is an injection f : X → Nn , then X is equipotent
to f (X) ⊆ Nn . By the preceding proposition, there exists a positive
integer m ≤ n and a bijection f (X) → Nm . It follows that X is
equipotent to Nm , and therefore finite.
♢
Proposition A.5.5 If X is a finite set, then there is no one-to-one
mapping of X onto any of its proper subsets.
Proof. Let X be a finite set and Y a proper subset of X. Suppose
that X has n elements. Then there is a bijection f : X → Nn for
some n ∈ N. Also, there is a positive integer m < n and a bijection
g : Y → Nm , by Proposition A.5.3. If h : X → Y is a bijection, then the
composite ghf −1 : Nn → Nm would be an injection, which contradicts
Proposition A.5.1.
♢
By the preceding proposition, a finite set cannot be equipotent to
Set Theory
495
one of its proper subsets. As n 7→ n + 1 is a bijection between N and
N−{1}, the set N is not finite. In fact, this is the characteristic property
of infinite sets. To establish this, we need the following.
Theorem A.5.6 (Principle of Recursive Definition) Let X be a
set and f : X → X be a function. Given a point x0 ∈ X, there is
a unique function g : N → X such that g (1) = x0 and g (n + 1) =
f (g (n)) for all n ∈ N.
This is one of the most useful ways of defining a function on N,
which we will take for granted.
Theorem A.5.7 Let X be a set. The following statements are equivalent:
(a) X is infinite.
(b) There exists an injection N → X.
(c) X is equipotent to one of its proper subsets.
Proof. (a)⇒(b): Let F be the family of all finite subsets of X. By the
axiom of choice, there exists a function c from the collection of all
nonempty subsets Y ⊆ X to itself such that c (Y ) ∈ Y. Since X is not
finite, X −F ̸= ∅ for all F ∈ F. So c (X − F ) ∈ X −F. It is obvious that
F ∪ {c (X − F )} is a member of F for every F ∈ F. Accordingly, F 7→
F ∪ {c (X − F )} is a function of F into itself. Set G (1) = {c (X)} and
G (n + 1) = G (n)∪{c (X − G (n))} for every n ∈ N. By the principle of
recursive definition, G is a function N → F. We show that the function
ϕ : N → X defined by ϕ (n) = c (X − G (n)) is an injection. It is
easily seen, by induction, that G (m) ⊆ G (n) for m ≤ n. Therefore,
for m < n, we have ϕ (m) = c (X − G (m)) ∈ G (m + 1) ⊆ G (n), while
ϕ (n) ∈ X − G (n).
(b)⇒(c): Let ϕ : N → X be an injection, and put Y = ϕ (N).
Consider the mapping ψ : X → X defined by
)
{ (
ϕ 1 + ϕ−1 (x)
if x ∈ Y , and
ψ (x) =
if x ̸∈ Y.
x
We assert that ψ is a bijection between X and X − {ϕ (1)}. The
surjectivity of ψ is clear. To show that it is injective, assume that
ψ (x) = ψ (x′ ) . Then x and x′ , both, are either in Y or in X − Y. If
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Elements of Topology
(
)
(
)
x, x′ ∈ Y, then we have ϕ 1 + ϕ−1 (x) = ϕ 1 + ϕ−1 (x′ ) , and it follows from the injectivity of ϕ that x = x′ . And, if x, x′ ∈ X − Y, then
x = x′ , by the definition of ψ.
(c)⇒(a): This is contrapositive of Proposition A.5.5.
♢
A set A is called countably infinite (or denumerable, or enumerable)
if A is equipotent to the set N of natural numbers. A is called countable
if it is either finite or denumerable; otherwise, it is called uncountable.
For example, the set Z of all integers is countably infinite, since
f : N → Z, defined by
{
n/2
if n is even, and
f (n) =
− (n − 1) /2
if n is odd,
is a bijection.
We conclude from Theorem A.5.7 that every infinite set contains
a countably infinite subset. The following proposition shows that no
uncountable set can be a subset of a countable set.
Proposition A.5.8 Every subset of N is countable.
Proof. Let M ⊆ N. If M is finite, then it is countable, by definition. If
M = N, then M is a countably infinite set, since the identity function on N is a bijection of N onto itself. So assume that M is an
infinite proper subset of N. We define a function from N to M by
recursion as follows: Let f (1) be the smallest integer in M, which exists by the well-ordering principle for N. Suppose that we have chosen
integers f (1) , f (2) , . . . , f (k). Since M is infinite, M − f {1, 2, . . . , k}
is nonempty for every k ∈ N. We define f (k + 1) to be the smallest integer of this set. By the principle of recursive definition, f (k)
is defined for all k ∈ N. Clearly, f (1) < f (2) < · · · so that f is injective. Now, we see that f is surjective, too. For each m ∈ M, the
set {1, 2, . . . , m} is finite. Since the set {f (k) |k ∈ N} is infinite, there
is a k ∈ N such that f (k) > m. So we can find the smallest integer
l ∈ N such that f (l) ≥ m. Then f (j) < m for each j < l; accordingly,
m ̸∈ f {1, . . . , l − 1}. By the definition of f (l), we have f (l) ≤ m, and
the equality m = f (l) follows. Thus f : N → M is a bijection, and this
completes the proof.
♢
It follows that a subset of N is either finite or countably infinite.
Note that any set which is equipotent to a countable set is countable.
Accordingly, any subset of a countable set is countable.
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497
Theorem A.5.9 For any nonempty set X, the following statements
are equivalent:
(a) X is countable.
(b) There is a surjection N → X.
(c) There is an injection X → N.
Proof. (a)⇒(b): If X is countably infinite, then there is a bijection
between N and X, and we are done. If X is finite, then there exists an
n ∈ N and a bijection f : {1, . . . , n} → X. We define g : N → X by
setting
{
f (m)
for 1 ≤ m ≤ n, and
g (m) =
f (n)
for m > n.
Clearly, g is a surjection.
(b)⇒(c): Let g : N → X be a surjection. Then g −1 (x) is nonempty
for every x ∈ X. So it contains a unique smallest integer h (x), say.
Then x 7→ h (x) is a mapping h : X → N, which is injective, since
g −1 (x) ∩ g −1 (y) ̸= ∅ whenever x ̸= y.
(c)⇒(a): Suppose that h : X → N is an injection. Then h : X →
h (X) is a bijection. By Proposition A.5.8, h (X) is countable, and
therefore X is countable.
♢
By definition, a countable set is the range of a (finite or infinite)
sequence, and the converse follows from the preceding theorem. Thus
the elements of a countable set X can be listed as x1 , x2 , . . ., and such
a listing is called an enumeration of X. Observe that the range of a
function of a countable set is countable, and the domain of an injective
function into a countable set is countable.
Lemma A.5.10 For any finite number factors, N×· · ·×N is countably
infinite.
Proof. Let 2, 3, . . . , pk be the first k prime numbers, where k is the
number of factors in N × · · · × N. Define a mapping f : N × · · · × N → N
by f (n1 , n2 , . . . , nk ) = 2n1 3n2 · · · pnk k . By the fundamental theorem of
arithmetic, f is injective, and hence N × · · · × N is countable. It is
obvious that n 7→ (n, 1, . . . , 1) is an injection N → N × · · · × N so that
N × · · · × N is infinite.
♢
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Elements of Topology
It is immediate from the preceding lemma that a finite product of
countable sets Xi is countable, for if fi : N → Xi are surjections for
1 ≤ i ≤ k, then so is the mapping (n1 , . . . , nk ) 7→ (f1 (n1 ) , . . . , fk (nk ))
of N×· · ·×N into X1 ×· · ·×Xk . Notice∏that if each Xi is nonempty and
k
infinite, then 1 X)i is also countably infinite,
some Xj is countably
( 0
since xj 7→ x1 , . . . , x0j−1 , xj , x0j+1 , . . . , x0k , where x0i ∈ Xi are fixed
∏k
elements for i ̸= j, is an injection Xj → 1 Xi . We also observe that if
∏k
∏k
each Xi is finite, then so is 1 Xi . For, if any Xj = ∅, then 1 Xi =
∏k
∅, and, if Xi has mi > 0 elements, then 1 Xi contains m1 · · · mk
elements. By induction on k, it suffices to establish this proposition
in the case k = 2. Let Nmi = {1, 2, . . . , mi } and fi : Nmi → Xi be
bijections, i = 1, 2. Then f1 × f2 : Nm1 × Nm2 → X1 × X2 , (a, b) 7→
(f1 (a) , f2 (b)) is a bijection. Define a mapping g : Nm1 ×Nm2 → Nm1 m2
by g (a, b) = (a − 1) m2 + b. It is easily verified that g is a bijection, so
X1 × X2 has m1 m2 elements.
But, an infinite product of even finite sets is not countable. For
example,
let X = {0, 1}, and consider the countable product X N =
∏∞
N
1 Xn , where each Xn = X. Note that an element in X is a sequence
N
s : N → X. Let f : N → X be any function. Write tn = 1 − f (n)n .
Then t = ⟨tn ⟩ ∈ X N and t ̸= f (n) for all n ∈ N. Thus f is not
surjective. Since f is an arbitrary function N → X N , X N cannot be
countable, by Theorem A.5.9.
This method of proof was first used by G. Cantor, and is known as
Cantor’s diagonal process. We can use this technique to prove:
The set R of all real numbers is uncountable.
In fact, we show that the unit interval I ⊂ R is uncountable. By
Theorem A.5.9, it suffices to show that there is no surjective mapping
f : N → I. Let f : N → I be any function. We use the decimal representation of real numbers to write f (n) = 0.an1 an2 · · · . Here each
ani is a digit between 0 and 9. This representation of a number is not
necessarily unique, but if a number has two different decimal representations, then one of these representations repeats 9’s from some place
onwards and the other repeats 0’s from some place onwards. We define
a new real number r whose decimal representation is 0.b1 b2 · · · , where
bn = 3 if ann ̸= 3, and bn = 5 otherwise. It is clear that r has a unique
decimal representation and differs from f (n) in the nth decimal place
for every n ∈ N. So r ̸= f (n) for all n ∈ N. As r belongs to I, f is not
surjective.
Set Theory
499
As another consequence of the preceding lemma, we obtain the
following.
Proposition A.5.11 The union of a countable family of countable
sets is countable.
Proof. Let A be a countable set,∪and suppose that for every α ∈ A,
Eα is a countable set. Put X = Eα . We show that X is countable.
If X = ∅, there is nothing to prove. So we assume that X ̸= ∅. Then
A ̸= ∅, and we may further assume that Eα ̸= ∅ for every α ∈ A,
since the empty set contributes nothing to the union of the Eα . Since
A is countable, there is a surjection N → A, n 7→ αn . Since Eαn is
a nonempty countable set, there is a surjection fn : N → Eαn . Write
fn (m) = xnm for every m ∈ N. Then we have a mapping ϕ : N×N
∪ →X
defined by ϕ (n, m) = xnm . Clearly, ϕ is surjective, for X = n Eαn .
By Lemma A.5.10, N × N is countable, and hence X is countable. ♢
If the indexing set A and the sets Eα in the preceding
proposition
∪
are all finite, then it is easily verified that the union α Eα is finite.
Example A.5.1 The set Q of all rational numbers is countably infinite.
Let Q+ denote the set of all positive rationals, and Q− denote the set of
all negative rationals. Then Q = Q+ ∪Q− ∪{0}. Obviously, Q+ and Q−
are equipotent, and therefore it suffices to prove that Q+ is countably
infinite. As N ⊂ Q+ , Q+ is infinite. There is a surjective mapping
g : N × N → Q+ defined by g (m, n) = m/n. By Lemma A.5.10, there
is a bijection f : N → N × N, so the composition gf : N → Q+ is
surjective. By Theorem A.5.9, Q+ is countable.
Since R is uncountable, we see that the set R − Q of all irrational
numbers is uncountable.
Theorem A.5.12 The family of all finite subsets of a countable set is
countable.
Proof. Let X be a countable set, and F (X) be the family of all finite
subsets of X. If X is a finite set having n elements, then every subset
of X is finite, and F (X) has 2n members. If X is countably infinite,
then there is a bijection between F (X) and the family F (N) of all finite
subsets of N. So it suffices to prove that F (N) is countably infinite. It is
obvious that n 7→ {n} is an injection N → F (N), so F (N) is infinite. To
see that it is countable, consider the sequence⟨2, 3, . . . , pk , . . .⟩ of prime
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Elements of Topology
numbers. If F = {n1 , n2 , . . . , nk } ⊂ N, then the ni can be indexed so
that n1 < n2 < · · · < nk . Put ν (F ) = 2n1 3n2 · · · pnk k . Then F 7→ ν (F )
defines an injection F (N) − {∅} → N, by the fundamental theorem of
arithmetic. By Theorem A.5.9, F (N) − {∅} is countable, so F (N) is
countable.
♢
It must be noted that the family of all subsets of a countably infinite
set is not countable. This follows from
Theorem A.5.13 (Cantor (1883)) For any set X, there is no surjection X → P (X).
Proof. Assume that there is a surjection f : X → P (X) and consider
the set S = {x ∈ X|x ∈
/ f (x)}. Then S ⊆ X so that there exists an
x ∈ X such that S = f (x). Now, if x ∈ S, then x ∈
/ f (x), and if x ∈
/ S,
then x ∈ f (x). Thus, in either case, we obtain a contradiction, and the
theorem follows.
♢
We state the following theorem without proof.
Theorem A.5.14 (Bernstein–Schröeder) Let X and Y be sets. If
there exists injections X → Y and Y → X, then there exists a bijection
X → Y.
A.6
Orderings
Definition A.6.1 Let X be a set. An order (or a simple order) on X
is a binary relation, denoted by ≺, such that
(a) if x, y ∈ X, then one and only one of the statements x = y, x ≺ y,
y ≺ x is true, and
(b) x ≺ y and y ≺ z ⇒ x ≺ z.
X together with a definite order relation defined in it is called an
ordered set.
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501
The statement “x ≺ y ′′ is read as “x precedes y” or “y follows
x.” We also say that “x is less than y” or “y is greater than x.” It
is sometimes convenient to write y ≻ x to mean x ≺ y. The notation
x ≼ y is used to indicate x ≺ y or x = y, that is, the negation of y ≺ x.
It should be noticed that an order relation is not necessarily reflexive.
However, if ≺ is an order relation on X, then the relation ≼ satisfies
the following conditions.
(a) x ≼ x ∀ x ∈ X (reflexivity);
(ib x ≼ y and y ≼ x ⇒ x = y (antisymmetry);
(c) x ≼ y and y ≼ z ⇒ x ≼ z (transitivity); and
(d) x ≼ y or y ≼ x ∀ x, y ∈ X (comparability).
A relation having the properties (a)–(c) is called a partial ordering. A set together with a definite partial ordering is called a partially
ordered set.
Suppose that (X, ≼) is a partially ordered set. Two elements x, y ∈
X are called comparable if x ≼ y or y ≼ x. A subset Y ⊆ X such
that any two elements in Y are comparable is called a chain in X. A
partially ordered set that is also a chain is called a linearly or totally
ordered set. Thus we have associated with each simple order a total
(or linear) order relation. Conversely, there is a simple order relation
associated with each total ordering. In fact, to any partial order relation
≼ on X, there is associated a unique relation ≺:
x ≺ y ⇔ x ≼ y and x ̸= y.
It is transitive and has the property that (a) for no x ∈ X, the relation
x ≺ x holds, and (b) if x ≺ y, then y ̸≺ x for every x, y ∈ X. A
transitive relation with this property is called a strict partial order.
If ≼ totally orders the set X, then ≺ is clearly a simple order (or a
strict total ordering) on X. It follows that the notions of a total (resp.
partial) order relation and a strict total (resp. partial) order relation
are interchangeable. If ≼ is a partial order, we use the notation ≺ to
denote the associated strict partial order, and conversely. The relation
≤ is a total ordering on the set R of real numbers, while the inclusion
relation ⊆ on P (X) is not. Note that the ordering by inclusion in P (X)
is always a partial order.
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It is obvious that the induced ordering on a subset of a partially ordered set is a partial ordering on that subset. If (X1 , ≼1 ) and (X2 , ≼2 )
are partially (totally) ordered sets, then the dictionary (or lexicographic) order relation on X1 × X2 defined by (x1 , x2 ) ≼ (y1 , y2 ) if
x1 ≺1 y1 or if x1 = y1 and x2 ≼2 y2 is a partial (total) ordering.
Let (X, ≼) be a partially ordered set. If Y ⊆ X, then an element
b ∈ X is an upper bound of Y if for each y ∈ Y, y ≼ b. If there exists
an upper bound of Y, then we say that Y is bounded above. A least
upper bound or supremum of Y is an element b0 ∈ X such that b0 is
an upper bound of Y and if c ≺ b0 , then c is not an upper bound of
Y. Observe that there is at most one such b0 , by antisymmetry, and
it may or may not exist. We write b0 = sup Y when it exists. Notice
that sup Y may or may not belong to Y. If sup Y ∈ Y , it is referred to
as the largest (or last) element of Y. Analogously, an element a ∈ X
is a lower bound of Y if a ≼ y for all y ∈ Y. If there exists a lower
bound of Y, we say that Y is bounded below. A greatest lower bound
or infimum of Y is an element a0 ∈ X which is a lower bound of Y
and if a0 ≺ c, then c is not a lower bound of Y. It is clear that there
is at most one such a0 , and it may or may not exist. When it exists,
we write a0 = inf Y . If inf Y ∈ Y , we call it the smallest (or least or
first) element of Y. It is obvious that sup Y is the smallest element of
the set {x ∈ X|y ≼ x ∀ y ∈ Y }, and inf Y is the largest element of the
set {x ∈ X|x ≼ y ∀ y ∈ Y }.
A partially ordered set X is said to have the least upper bound
property if each nonempty subset Y ⊆ X with an upper bound has a
least upper bound. Analogously, X is said to have the greatest lower
bound property if each nonempty subset Y ⊆ X which is bounded
below has a greatest lower bound. If X is a set, then P (X) ordered by
inclusion has the least upper bound property.
Proposition A.6.2 If an ordered set X has the least upper bound
property, then it also has the greatest lower bound property.
An ordered set X which has the least upper bound (equivalently,
the greatest lower bound) property is called order complete.
Let (X, ≼) be a partially ordered set. An element a ∈ X is called
a minimal element of X if for every x ∈ X, x ≼ a ⇒ x = a, that is, no
x ∈ X which is distinct from a precedes a. Similarly, an element b ∈ X
is called a maximal element of X if for every x ∈ X, b ≼ x ⇒ x = b,
that is, no x ∈ X which is distinct from b follows b. It should be
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noted that if X has a largest (smallest) element, then that element is
the unique maximal (minimal) element of X. If ≼ is a total order, a
maximal element is a largest element, but there are partially ordered
sets with unique maximal elements which are not largest elements.
A partially ordered set X is called well-ordered (or an ordinal) if
each nonempty subset A ⊆ X has a least element (also called first or
smallest element). The set N of the positive integers is well-ordered by
its natural ordering ≤. On the other hand, the set R of real numbers is
not well-ordered by the usual ordering ≤. And, if X has more than one
element, then the partial ordering ⊆ in P (X) is not a well-ordering.
The empty set ∅ is a well-ordered set. Any subset of a well-ordered
set is well-ordered in the induced ordering, and if (X1 , ≼1 ) and (X2 , ≼2 )
are well-ordered sets, then the dictionary order relation on X1 × X2
is a well-ordering. Also, notice that every well-ordered set X is in fact
totally ordered, since each subset {x, y} ⊆ X has a first element whence
either x ≼ y or y ≼ x.
Theorem A.6.3 The following statements are equivalent:
(a) The axiom of choice.
(b) Well-ordering principle: Every set can be well-ordered.
(c) Zorn’s lemma: Let X be a partially ordered set. If each chain
in X has an upper bound, then X has a maximal element.
(d) Hausdorff Maximal Principle: If X is a partially ordered set,
then each chain in X is contained in a maximal chain, that is,
for each chain C in X, there exists a chain M in X such that
C ⊆ M and M is not properly contained in any other chain
which contains C.
We do not wish to prove this theorem and refer the reader to
Dugundji [3] and Kelley [6]. We remark that there are several other
statements each equivalent to the axiom of choice given in these references.
ORDINALS
Now, we turn to discuss a few elementary properties of ordinals
which are necessary for our purposes. Let (W, ≼) be a well-ordered set,
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and x, y ∈ W. If x ≺ y, we say that x is a predecessor of y (or y is a
successor of x). If x ≺ y and there is no z ∈ W such that x ≺ z ≺ y,
then we say that x is an immediate predecessor of y (or y an immediate
successor of x). Each element x of a well-ordered set (W, ≼) which is
not the largest element of W has an immediate successor. This is first
element of the set {y ∈ W |x ≺ y}, usually denoted by x + 1. However,
x need not have an immediate predecessor. For example, consider the
set N of positive integers with the usual ordering ≤. Choose an element
∞ ̸∈ N, and put W = N ∪ {∞}. Define a relation ≼ on W by x ≼ y if
x, y ∈ N and x ≤ y, x ≼ ∞ for all x ∈ N, and ∞ ≼ ∞. Then (W, ≼)
is a well-ordered set, and ∞ has no immediate predecessor in W. It is
obvious that ∞ is the last element of W . We say that N ∪ {∞} has
been obtained from N by adjoining ∞ as the last element.
Let (W, ≼) be a well-ordered set. For each x ∈ W, the set S (x) =
{w ∈ W |w ≺ x} is called the section (or initial segment) determined
by x. The first element of W is usually denoted by 0; it is obvious that
S (0) = ∅. Each section S (x) in W obviously satisfies the condition:
y ∈ S (x) and z ≼ y ⇒ z ∈ S (x). Conversely, if S is a proper subset of
W satisfying this condition, then S is a section in W. For, if x is the
least element of W − S, then y ≺ x ⇔ y ∈ S. So S = S (x). It is clear
that a union of sections in W is either a section in W or equals W.
Also, it is immediate that an intersection of sections in W is a section
in W.
A.6.4 (Principle of Transfinite Induction) Let (W, ≼) be a wellordered set. If X is a subset of W such that S (y) ⊆ X implies y ∈ X
for each y ∈ W, then X = W.
Proof. The first element 0 of W is in X, since ∅ = S (0) ⊂ X. If
W − X ̸= ∅, then it has a first element y0 , say. So S (y0 ) ⊆ X, which
forces y0 ∈ X, by our hypothesis. This contradiction establishes the
theorem.
♢
The preceding theorem is generally used in the following form: For
each x ∈ W, let P (x) be a proposition. Suppose that (a) P (0) is true,
and (b) for each x ∈ W, P (x) is also true whenever P (y) is true for
all y ∈ S (x). Then P (x) is true for every x ∈ W.
Since each element of N other than 1 has an immediate predecessor,
the induction principle on N is equivalent to the following statement.
Let P (n) be a proposition defined for each n ∈ N. If P (1) is true,
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and for each n > 1, the hypothesis “P (n − 1) is true” implies that
P (n) is true, then P (n) is true for every n ∈ N.
Let (W, ≼) and (W ′ , ≼′ ) be well-ordered sets. A mapping f : W →
W is said to be order-preserving if f (x) ≼′ f (y) whenever x ≼ y in
W. We call W and W ′ isomorphic (or of the same order type) if there
is a bijective order-preserving map f : W → W ′ ; such a map f is called
an isomorphism. An order-preserving injection f : W → W ′ is called a
monomorphism. It should be noted that an isomorphism of a partially
ordered set X onto a partially ordered set X ′ is a bijection f : X → X ′
such that x ≼ y ⇔ f (x) ≼′ f (y). However, if X and X ′ are totally
ordered, one needs only the implication x ≼ y ⇒ f (x) ≼′ f (y). The
reverse implication f (x) ≼′ f (y) ⇒ x ≼ y, for y ≺ x ⇒ f (y) ≺′ f (x),
contradicting the antisymmetry of ≼′ .
The ordinal N is isomorphic to the well-ordered set {0, 1, 2, . . .} of
nonnegative integers in its natural order. We note that a well-ordered
set may be isomorphic to one of its proper subsets. For example, the
well-ordered set N is isomorphic to the set E of even integers; n 7→
2n is a desired isomorphism. However, no well-ordered set W can be
isomorphic to a section in it; this is immediate from the following fact.
′
Proposition A.6.5 If f is a monomorphism of a well-ordered set W
into itself, then f (w) ≽ w for every w ∈ W.
Proof. Let X = {x ∈ W |f (x) ≺ x}. If X ̸= ∅, then it has a first
element, say x0 . As x0 ∈ X, we have f (x0 ) ≺ x0 whence f (f (x0 )) ≺
f (x0 ) . This implies that f (x0 ) ∈ X, which contradicts the definition
of x0 . Therefore X = ∅.
♢
Corollary A.6.6 Two sections in a well-ordered set W are isomorphic
if and only if they are identical.
Proof. Suppose that S (x) and S (x′ ) are isomorphic sections in W and
let f : S (x) → S (x′ ) be an isomorphism. If S (x) ̸= S (x′ ), then we
have either x ≺ x′ or x′ ≺ x. By interchanging S (x) and S (x′ ), if
necessary, we may assume that x′ ≺ x. Then x′ ∈ S (x) which implies
that f (x′ ) ≺ x′ . This contradicts Proposition A.6.5.
♢
Corollary A.6.7 The only one-to-one order-preserving mapping of a
well-ordered set onto itself is the identity mapping.
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Proof. Suppose that W is a well-ordered set and f : W → W is an
isomorphism. If the set S = {x ∈ W |f (x) ̸= x} is nonempty, then S
has a least element a, say. By A.6.5, we have a ≺ f (a). Since f is
surjective, a = f (x) for some x ∈ W . So f (x) ≺ f (a) ⇒ x ≺ a.
Then we have f (x) = x, by the definition of a. Thus a = x ≺ a, a
contradiction. Therefore S = ∅, and f is the identity mapping.
Theorem A.6.8 If X and Y are well-ordered sets, then exactly one
of the following statements holds.
(a) X is isomorphic to Y.
(b) X is isomorphic to a section in Y.
(c) Y is isomorphic to a section in X.
Proof. We first show that the three possibilities are mutually exclusive.
If (a) and (b) occur, then Y is isomorphic to a section in it, a contradiction. For the same reason, (a) and (c) cannot occur. If f : X → S (y)
and g : Y → S (x) are isomorphisms, then gf : X → S (x) is a
monomorphism with gf (x) ≺ x. This contradicts A.6.5.
Now, we prove that one of these possibilities does occur. Suppose
that neither of the possibilities (a) and (b) holds. Let Σ be the family
of all sections S (xα ) in X for which there are isomorphisms fα of
S (xα ) onto sections in Y or Y itself. For every pair of indices α and β,
S (xα )∩S (xβ ) is a section in X, and fα (x)∪= fβ (x) for all
∪ x ∈ S (xα )∩
S (xβ ) . Therefore there is a function f : α S (xα ) → α fα (S (xα ))
defined by f (x) = fα (x) ∪
if x ∈ S (xα ). It is easily checked that f is
an
isomorphism.
Clearly,
α S (xα ) is either X
∪
∪ or a section in it, and
f
(S
(x
))
is
either
Y
or
a
section
in
Y
.
If
α
α α
α S (xα ) =∪X, then we
have alternative (a) or (b), contrary to our assumption. So α S (xα ) =
S (x0 ) for some x0 ∈ X, and S (x0 ) ∈ Σ. We assert that f : S (x0 ) → Y
is surjective. Assume otherwise. Then we have f (S (x0 )) = S (y0 ) for
some y0 ∈ Y. Obviously, we can extend f to an isomorphism S (x0 ) ∪
{x0 } → S (y0 ) ∪ {y0 } by defining f (x0 ) = y0 . Note that S (y0 ) ∪ {y0 }
is either a section in Y or coincides with Y (if y0 is the last element of
Y ). And, by our assumption again, x0 is not the last element of X so
that S (x0 ) ∪ {x0 } is a member of the family Σ. From the definition of
S(x0 ), it follows that x0 ∈ S(x0 ), a contradiction. Hence our assertion,
and we have the alternative (c).
♢
Corollary A.6.9 Any subset of a well-ordered set W is isomorphic to
either a section in W or W itself.
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Proof. Let X be a subset of W. Then X is a well-ordered set under
the induced order. If there is an isomorphism f of W onto a section
S (x0 ) in X, then there exists a monomorphism g : X → X defined
by f, which satisfies g (x0 ) ≺ x0 . This contradicts Proposition A.6.5.
Therefore one of the two possibilities (a) and (c) in Theorem A.6.8
must hold.
♢
A.7
Ordinal Numbers
It is evident that the relation of isomorphism between ordinals is
an equivalence relation. Ordinal numbers are objects uniquely associated with the isomorphic classes of well-ordered sets. A natural way to
define them is to consider these isomorphic classes themselves as the
ordinal numbers. The main drawback with this definition is that isomorphic classes of well-ordered sets are unfortunately not sets; accordingly, logical contradictions arise when such large classes are collected
into sets. To avoid such difficulties, ordinal numbers are defined to be
well-ordered sets such that each isomorphic class of well-ordered sets
contains exactly one ordinal number.
Definition A.7.1 An ordinal number is a well-ordered set (Σ, ≼) such
that
(a) if X ∈ Σ and x ∈ X, then x ∈ Σ,
(b) for every X, Y ∈ Σ, one of the possibilities: X = Y , X ∈ Y or
Y ∈ X holds, and
(c) X ≼ Y ⇔ X ∈ Y or X = Y .
The fact that order relation ≼ defined by conditions (b) and (c)
in the above definition is actually a well-ordering follows from one of
the axioms in set theory: every nonempty set X contains an element y
such that x ∈
/ y for all x ∈ X. As a consequence of this, we see that no
nonempty set can be a member of itself, and both the relations X ∈ Y
and Y ∈ X cannot hold simultaneously for any two sets X, Y . So only
one of the three alternatives in condition (b) can occur. Now, it is
obvious that ≼ is reflexive and antisymmetric. To see the transitivity,
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suppose that X ≼ Y and Y ≼ Z in Σ. Then we must have one of the
relations X = Z, X ∈ Z or Z ∈ X. Assume that Z ∈ X. Then we
can not have X = Y or Y = Z, for Y ∈
/ X and Z ∈
/ Y. So X ∈ Y
and Y ∈ Z. But, this leads to the violation of the above axiom by the
set {X, Y, Z}. Hence X ≼ Z. Clearly, every subset of Σ has a least
member, and ≼ is actually a well-ordering for Σ.
The empty set ∅ is obviously an ordinal number. To construct a
few more ordinal numbers, we observe that if Σ is an ordinal number,
then so is Σ ∪ {Σ}. Thus {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}, etc. are
ordinal numbers. Henceforth, we will generally denote ordinal numbers
by small Greek letters.
Proposition A.7.2 Every element of an ordinal number is an ordinal
number.
Proof. Let β be an ordinal number and α ∈ β. We show that α is also
an ordinal number. Suppose that x ∈ y and y ∈ α. Then we have either
x = α or x ∈ α or α ∈ x. If x = α, then we would have x ∈ y and
y ∈ x, a contradiction. Now, if α ∈ x, then the subset {x, y, α} ⊂ β
fails to have a least element. So x ∈ α, and condition A.7.1 (a) holds.
Next, suppose that x, y ∈ α. Then both x, y are members of β. So one
of the possibilities x = y, x ∈ y or y ∈ x must hold. Thus condition (b)
in A.7.1 is satisfied by α.
♢
Lemma A.7.3 If α and β are ordinal numbers, then α ∈ β if and only
if α ⊂ β.
Proof. If α ∈ β, then x ∈ α ⇒ x ∈ β, by definition. Moreover, α ̸= β;
otherwise, we have α ∈ α. So α ⊂ β. Conversely, assume that α ⊂ β.
Then β − α is nonempty and has a least element γ, say. Then γ ⊆ α,
by the definition of γ. Also, for any x ∈ α, x ̸= γ and γ ∈
/ x, since
γ∈
/ α. So x ∈ γ, and we have α = γ ∈ β.
♢
Proposition A.7.4 (a) If α and β are any two ordinal numbers,
then α ⊆ β or β ⊆ α.
(b) Every nonempty set of ordinal numbers has a least element with
respect to inclusion.
Proof. (a): Suppose that α and β are two ordinal numbers. Then γ =
α ∩ β = {x|x ∈ α and x ∈ β} is an ordinal number. If γ ̸= α, β, then
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γ ⊂ α and γ ⊂ β. By the preceding lemma, we have γ ∈ α and γ ∈ β.
Thus γ ∈ α ∩ β = γ, a contradiction. So γ = α or γ = β; accordingly,
we have α ⊆ β or β ⊆ α.
(b): Let X be a nonempty set of ordinal numbers. Choose α ∈ X.
If Y = α ∩ X is empty, then x ∈ X implies that x ̸⊂ α, by Proposition
A.7.3. Hence α ⊆ x, and α is the least element of X. If Y ̸= ∅, then Y
contains an element z such that z ∈ y or z = y for every y ∈ Y . Thus,
for every y ∈ Y, we have z ⊆ y. Also, if x ∈ X − Y, then z ⊂ α ⊆ x.
Obviously, z ∈ X, and we see that z is the least element of X.
♢
It follows that the class O of all ordinal numbers is well-ordered by
≤, where α ≤ β if and only if α ⊆ β, and an ordinal number α consists
of all those ordinal numbers that precede it. Also, we see that any two
distinct ordinal numbers cannot be isomorphic. For, if ordinal numbers
α, β are distinct and α < β, then α is the section {γ ∈ β|γ < α} in β.
So α cannot be isomorphic to β, by A.6.5.
Theorem A.7.5 (a) The union of a set of ordinal numbers is an
ordinal number.
(b) The class O of all ordinal numbers is not a set.
∪
Proof. (a): Let Γ be a set of ordinal numbers. Then Σ = α∈Γ α is a set.
It is obvious that condition (a) of A.7.1 holds in Σ. Next, if x, y ∈ Σ,
then there exist ordinal numbers α, β ∈ Γ with x ∈ α and y ∈ β. By
the preceding proposition, we have α ⊆ β or β ⊆ α. To be specific,
suppose that α ⊆ β. Then x, y ∈ β, and therefore condition (b) of
A.7.1 is satisfied by x and y. Finally, we show that Σ is well ordered by
the relation x ≼ y ⇔ x ∈ y or x = y. Clearly, this is a linear ordering
on Σ. If T is a nonempty subset of Σ, then T ∩ α ̸= ∅ for some α ∈ Γ.
Let z be the least element of T ∩ α. Then z is the least element of T,
as in the proof of the preceding proposition.
∪ (b): Assume, on the contrary, that O is a set. Then, by (a), Σ =
α∈O α is an ordinal number. Hence T = Σ∪{Σ} is an ordinal number.
Clearly, we have Σ < T ≤ Σ, a contradiction.
♢
Theorem A.7.6 Every well-ordered set W is isomorphic to a unique
ordinal number.
Proof. If W is isomorphic to two ordinal numbers λ and µ, then they
themselves are isomorphic, and therefore λ = µ.
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Now, let X be the collection of all elements x ∈ W for which there
are ordinal numbers αx (depending upon x) such that the section S (x)
in W is isomorphic to αx . Put Ψ(x) = αx . We observe that Ψ is singlevalued on X. Suppose that Ψ(x) = αx , Ψ(y) = αy and x = y. Then
the ordinal numbers αx and αy are isomorphic, and therefore αx = αy .
By an axiom in set theory, it follows that X is a set, and so is im(Ψ). If
Ψ (x) = Ψ (y), then S (x) and S (y) are isomorphic. By Corollary A.6.6,
S (x) = S (y), and therefore x = y. Thus Ψ is a bijection between X
and im(Ψ).
Next, suppose that x ∈ X, y ∈ W and y ≺ x, where ≼ denote the ordering in W. Let f : S (x) → Ψ (x) be an isomorphism.
Then f (y) < Ψ (x), and f induces an isomorphism of S (y) onto
{α ∈ Ψ (x) |α < f (y)} = f (y). So y ∈ X and Ψ (y) = f (y) < Ψ (x).
It follows that either X is a section in W or coincides with W. Also,
Ψ is order preserving. On the other hand, if β = Ψ (y) < Ψ (x), and
g : Ψ (x) → S (x) is an isomorphism, then g (β) ≺ x and g|β is an isomorphism of β onto S (g (β)). So Ψ (g (β)) = Ψ (y). Since Ψ is injective,
we have y = g (β) ≺ x. Thus Ψ is an isomorphism of X onto im(Ψ).
Since the class O of all ordinal numbers is not a set, we have O ̸= im(Ψ).
By A.7.4, there is a least ordinal number λ such that λ ∈
/ im(Ψ). We
observe that im(Ψ) = {α ∈ O|α < λ}. Clearly, α < λ ⇒ α ∈ im(Ψ). To
see the reverse inclusion, suppose that α ∈ im(Ψ), and Ψ (x) = α. Then
there is an isomorphism g : α → S (x). If β < α, then y = g (β) ≺ x,
and β = Ψ (y) ∈ im(Ψ), as above. This implies that α < λ, and the
equality holds. Thus im(Ψ) = λ so that Ψ is an isomorphism of X onto
λ.
Finally, we see that X = W. For, otherwise, we have X = S (w) for
some w ∈ W, and Ψ (w) = λ. Consequently, w ∈ X, a contradiction.
Hence our assertion.
♢
If an ordinal number α is nonempty, then the least element a of
α must be ∅, for x ∈ a would imply that x ∈ α and x ≺ a. So ∅
is the least ordinal number, denoted by 0. For each positive integer
n, we denote the ordinal number determined by the well-ordered set
{0, 1, . . . , n − 1} by n (note the abuse of the notation n). Thus 1 =
{∅}, 2 = {∅, {∅}}, 3 = {∅, {∅}, {∅, {∅}}}, and so on. Clearly, 1
is the immediate successor of 0, 2 is the immediate successor of 1,
etc. As we have noted earlier, if α is an ordinal number, then so is
α ∪ {α}. In fact, α ∪ {α} is the least ordinal number > α, usually
denoted by α + 1. Thus α + 1 is the immediate successor of α. However,
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there are ordinal numbers
∪ which are not of this form, for example, the
ordinal number ω = n<ω n; this is isomorphic to the well-ordered
set of all nonnegative integers in its natural order. Obviously, ω has
no immediate predecessor in O, and such ordinal numbers are called
limit ordinal numbers. Note that each ordinal number n < ω contains
a finite number of elements, in fact n elements. These are referred
to as finite ordinal numbers. The ordinal number ω is the smallest
ordinal number greater than every finite ordinal number; accordingly,
it is called the first infinite ordinal number. Using the construction of
successors, we obtain the sequence
of ordinal numbers ω + 1, ω + 2, . . ..
∪
By A.7.5(i), 2ω = ω + ω = n>0 (ω + n), 3ω = 2ω + ω, etc. are ordinal
numbers. Thus, we obtain a sequence of ordinal numbers 0, 1, . . . , ω, ω+
1, ω + 2, . . . , 2ω, 2ω + 1, . . .. Notice that these are all countable ordinal
numbers and ω + 1 ̸= 1 + ω.
We end this section by establishing the existence of an uncountable ordinal number. By Theorem A.7.6, it suffices to construct an
uncountable well-ordered set.
Proposition A.7.7 There is an uncountable ordinal (W, ≼) which has
a last element Ω such that for each w ∈ W other than Ω, the section
S (w) = {x ∈ W |x ≺ w} is countable.
Proof. Let X be any uncountable set. By the well-ordering principle,
there is a well-ordering ≼ for X. If X does not have a last element, we
choose an element ∞ ̸∈ X and construct a well-ordered set X ∪ {∞}
by adjoining ∞ to X as a last element. This is done by extending the
relation on X to X ∪ {∞} by defining x ≼ ∞ for all x ∈ X. So we
can assume that the well-ordered set (X, ≼) has the last element. Let
Y be the set of elements y ∈ X such that the set {x ∈ X|x ≼ y} is
uncountable. Then Y is nonempty, and therefore has a least element.
If Ω denotes the least element of Y, then W = {x ∈ X|x ≼ Ω} is the
desired set.
♢
The ordinal W in the preceding proposition is unique in the sense
that, if W ′ is any ordinal with the same properties, then there is an
isomorphism of W onto W ′ . For, if f : W → W ′ is a monomorphism,
then f (Ω) is the last element of f (W ). If f (Ω) is distinct from the
last element Ω′ of W ′ , then f (W ) , being a section S (w′ ) for some
w′ < Ω′ , would be countable. This contradicts the fact that f (W )
is uncountable. It follows from Theorem A.6.8 that W and W ′ are
isomorphic. Also, it is clear that the section S (Ω) is uncountable, and
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any countable ordinal is isomorphic to a section S (x) in W for some
x ≺ Ω. Another notable fact about W is that if A ⊂ W is countable
and Ω ∈
/ A, then sup A ≺ Ω. To see this, note that the set Ea = {x ∈
W |x ∪
≼ a} is countable for each a ∈ A. Since A is countable, so is
B = a Ea . Let y0 be the least element of the set {y ∈ W |y ̸∈ B}.
Then x ∈ B ⇔ x ≺ y0 . Thus y0 has only a countable number of
predecessors, so y0 ≺ Ω. As y0 is an upper bound for A, sup A ≺ Ω.
The ordinal number determined by W is called the first (or least)
uncountable ordinal number, and is denoted by Ω. The section [0, Ω) is
the set of all countable ordinal numbers, and has no largest element.
Clearly, for any α ∈ [0, Ω), its immediate successor α + 1 ∈ [0, Ω]. But
α + 1 ̸= Ω for any α < Ω, since [0, Ω) is uncountable. Thus Ω is also
a limit ordinal; in fact, there are uncountably many limit ordinals in
[0, Ω].
A.8
Cardinal Numbers
As seen in §A5, two finite sets X and Y are equipotent if and only
if they have the same number of elements. We associate with each set
X an object |X| such that two sets X and Y are quipotent (that is, X
and Y have the same cardinality) if and only if |X| = |Y |.
Isomorphic ordinals are obviously equipotent, and we observe that
the converse is also true for finite ordinals. We first show that an ordinal
X consisting of n elements, n any nonnegative integer, is isomorphic
to a section S (n) in ω. For, if X = ∅, then it is S (0). So assume
that X ̸= ∅, and define a mapping f : X → ω as follows. Map the
first element x1 of X into the integer 0. If X − {x1 } ̸= ∅, denote the
first element of X − {x1 } by x2 and put f (x2 ) = 1. If X ̸= {x1 , x2 },
denote the first element of X − {x1 , x2 } by x3 and put f (x3 ) = 2.
Since X consists of finitely many elements only, this process terminates
after finitely many steps. Thus, we obtain a positive integer n such
that X = {x1 , . . . , xn }, and there is a mapping f : X → ω given
by f (xi ) = i − 1, i = 1, 2, . . . , n. Clearly, x1 < · · · < xn so that f
is an isomorphism of X onto the section S (n) in ω. It follows that
any two finite ordinals both having the same number of elements are
isomorphic. However, this may not be true for ordinals having infinitely
Set Theory
513
many elements. For example, if ω ∪ {q} is the ordinal obtained from
the ordinal ω by adjoining q to it as the last element, then ω ∪ {q} is
not isomorphic to ω, by Proposition A.6.5. But ϕ : ω ∪ {q} → ω defined
by ϕ (n) = n + 1, ϕ (q) = 0 is a bijection.
By Theorem A.6.3, every set X can be well-ordered and, by Theorem A.7.6, has the cardinality of an ordinal number. We define |X| to
be the least ordinal number λ such that X is equipotent to λ. Clearly,
the object |X| is uniquely determined by X, and is called the cardinal
number of X.
If X is finite and has n elements, then |X| is the ordinal number n.
The cardinal number |N| is denoted by ℵ0 , and the cardinal number |R|
is denoted by ℵ1 and also by c, called the cardinality of the continuum.
The mapping x 7→ x/(1 − |x|) is a bijection between the open interval
(−1, 1) and R, and the mapping x 7→ (2x − a − b)/(b − a) is a bijection
between an open interval (a, b) and (−1, 1). Therefore the cardinality
of any open interval (a, b) is c. By Theorem A.5.14, we see that the
cardinality of a closed interval or a half open interval is also c.
Given two sets X and Y , we have |X| < |Y | if and only if there
exists an injection X → Y but there is no bijection between X and
Y . Accordingly, it is logical to say that X has fewer elements than Y
when |X| < |Y |. As seen in §A5, there is no surjection from N to I
while there is an injection N → I, since R and I are equipotent. So we
have ℵ0 < c.
The sum of two cardinal numbers α and β, denoted by α + β, is the
cardinality of the disjoint union of two sets A and B, where |A| = α
and |B| = β.
Obviously, N = {1, 2, . . . , n} ∪ {n + 1, n + 2, . . .} and so n + ℵ0 = ℵ0 .
Since N is the union of disjoint sets {1, 3, 5, . . .} and {2, 4, 6, . . .}, we
have ℵ0 + ℵ0 = ℵ0 . Similarly, we obtain ℵ0 + · · · + ℵ0 = ℵ0 .
Considering the interval [0, 2) as the union of the intervals [0, 1)
and [1, 2), we see that c + c = c. Therefore, for any integer n ≥ 0,
c ≤ n + c ≤ ℵ0 + c ≤ c + c = c
which implies that n + c = ℵ0 + c = c + c = c. If ∑
M is a set and for
each m ∈ M , we have a cardinal number αm , then m∈M αm denotes
the cardinal number of the union of pairwise disjoint sets Am , where
|Am | = αm .
For example, the set N can be written as the disjoint union of
countably many countable sets:
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Elements of Topology
1 3 5 7 ···
2 6 10 24 · · ·
4 12 20 28 · · ·
8 24 40 56 · · ·
.. ..
..
..
. .
.
. ···
So ℵ0 + ℵ0 + · · · = ℵ0 . Thus, for all cardinal numbers 1 ≤ αn ≤ ℵ0 ,
n = 1, 2, 3, . . ., we have α1 + α2 + · · · = ℵ0 . This is also immediate from
Proposition A.5.11.
The decomposition of the interval [0, ∞) into the intervals [n−1, n),
where n = 1, 2, . . ., shows that c + c + · · · = c.
If α is an infinite cardinal number, then α + ℵ0 = α. To see this,
we first observe that every infinite set X contains a countably infinite
subset. Choose an element x1 ∈ X. Since X ̸= {x1 }, we can choose an
element x2 ∈ X such that x2 ̸= x1 . We still have X − {x1 , x2 } =
̸ ∅.
So we can choose an element x3 ∈ X such that x3 ̸= x1 , x2 . Assume
that we have chosen n distinct elements x1 , x2 , . . . , xn of X. Then X ̸=
{x1 , x2 , . . . , xn }, since X is infinite. Therefore we can find a xn+1 ∈
X − {x1 , x2 , . . . , xn }. Thus we have a sequence ⟨x1 , x2 , . . .⟩ of distinct
points of X. The set {x1 , x2 , . . .} constructed in this way is obviously
countably infinite. Now, given an infinite cardinal number α, find a set
X with α = |X|. Let Y be a countably infinite subset X. Then the
equality X = Y ∪ (X − Y ) implies α = ℵ0 + β, where β = |X − Y |. So
α + ℵ0 = β + ℵ0 + ℵ0 = β + ℵ0 = α.
The product of two cardinal numbers α and β, denoted by αβ, is
the cardinality of the cartesian product of two sets A and B, where
|A| = α and |B| = β. If µ is a cardinal number, then the µth power of
α, denoted by αµ , is the cardinal number of the set AM , where M is a
set with |M | = µ.
Suppose that
∑for every m ∈ M , there
∏ is the sameµ cardinal number
αm = α. Then m∈M αm = αµ and m∈M αm = α∪.
For the first equation, we note that A × M = m∈M Am , where
Am = A × {m}. Obviously, Am is equivalent to∑A, and the family
{A
|Am | and we have
∑ m |m ∈ M } is pairwise disjoint. So |A × M | =
α
=
αµ.
The
second
formula
is
immediate
from the equality
∏ m∈M m
M
A
=
A
,
where
A
=
A
for
all
m
∈
M
.
m
m∈M m
Set Theory
515
Thus, we have
2ℵ0
= ℵ0 + ℵ0
= ℵ0 ,
nℵ0
= ℵ0 + · · · + ℵ0
= ℵ0 ,
ℵ0 ℵ0 = ℵ0 + ℵ0 + · · ·
= ℵ0 ,
nc
= c + ··· + c
= c,
ℵ0 c
= c + c + ···
= c,
where n is any positive integer. We also have
Proposition A.8.1 c = 2ℵ0 = |2N |.
Proof. As seen in §A5, every real number in the interval [0, 1] has one
or at most two decimal representations as 0.a1 a2 · · · an · · · , where each
an is one of the digits 0,1,. . . ,9. Similarly, we can express each x in
[0, 1] as x = 0.a1 a2 · · · an · · · , where each an is either 0 or 1. This is
referred to as the binary or dyadic expansion of x. For any cardinal
number α, αℵ0 is the cardinality of all sequences ⟨a1 , a2 , . . .⟩ in a set A
with cardinality α. In particular, 2ℵ0 is the cardinality of the set of all
sequences which have terms the digits 0 and 1 only (that is, the dyadic
sequences). As each real number in [0, 1] has at least one and at most
two binary expansions, we have c ≤ |2N | ≤ 2c = c.
♢
Note that ℵ20 = ℵ0 ℵ0 = ℵ0 and ℵn0 = ℵ0 · · · ℵ0 = ℵ0 , by induction
on n. Similar results can be proved for any transfinite cardinal number
α. To this end, we first establish the following.
Proposition A.8.2 For any infinite cardinal number α,
2α = α + α = α.
Proof. Let X be an infinite set with |X| = α. Denote the two-point set
{0, 1} by 2. Then, for any set A, 2×A is the union of disjoint sets {0}×A
and {1} × A and so we have 2|A| = |A| + |A| = |2 × A|. Now, consider
the family F of all pairs (A, f ) such that A ⊆ X and f : A → 2 × A is
a bijection. By Theorem A.5.7, X contains a countably infinite set A
and, by Propoposition A.5.11, the set 2 × A is also countable. So there
is a bijection f between A and 2×A. Obviously, the pair (A, f ) belongs
to F, and thus F is nonempty. Partially order F by (A, f ) ≤ (B, g) if
A ⊆ B and f = g|B. It is easily verified that every chain in F has
an upper bound in F so that the Zorn’s lemma applies. Let (M, h)
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Elements of Topology
be a maximal member in F. Then |M | + |M | = |M |. We show that
|M | = |X|. If X − M is infinite, then it contains a countably infinite
set B. Suppose that g is a bijection between B and 2×B. Then we have
a bijection k : B ∪ M → 2 × (B ∪ M ) defined by k|B = g and k|M = h.
This contradicts the maximality of M . Therefore X −M must be finite;
accordingly, M is infinite. If Y ⊆ M is a countably infinite set (which
does exist), then |Y ∪ (X − M )| = |Y | + |X − M | = |Y | and we obtain
|X| = |Y ∪ (X − M ) ∪ (M − Y )|
= |Y ∪ (X − M )| + |M − Y |
= |Y | + |M − Y | = |M |.
♢
As an immediate consequence of this result, we see that if α is
an infinite cardinal number, then, by induction, nα = α for every
integer n > 0. And, for a cardinal number β ≤ α, α + β = α, since
α ≤ α + β ≤ 2α = α.
Proposition A.8.3 For any infinite cardinal number α, α2 = αα = α.
Proof. Let X be an infinite set with |X| = α and let F bethe family
of all pairs (A, f ) such that A ⊆ X and f : A → A × A is a bijection.
Then X contains a countably infinite set A and, by Lemma A.5.10, the
set A × A is also countable. So there is a bijection f between A and
A × A. Thus the pair (A, f ) belongs to F and F is nonempty. Partially
order F by (A, f ) ≤ (B, g)∪if A ⊆ B and f = g|B. Given a chain
C = {Ai , fi } in F, put B = Ai and define f : B → B × B by setting
f (x) = fi (x) if x ∈ Ai . Then f is clearly a bijection so that the pair
(B, f ) belongs to F. It is obvious that (B, f ) is an upper bound for the
chain C. By the Zorn’s lemma, we have a maximal member (M, h) in
F. Since h : M → M × M is a bijection, |M | = |M ||M |. We observe
that |X| = |M |. As M ⊆ X, |M | ≤ |X|. If |M | < |X|, then we find that
|M | < |X −M |. For, |X −M | ≤ |M | implies that |X| = |M |+|X −M | ≤
|M | + |M | = |M |, by the preceding proposition. This contradicts our
assumption that |M | < |X|. Let j : M → X − M be an injection and
put Y = j(M ). Then |Y | = |M |. By the preceding proposition, we have
3|Y | = |Y | and, therefore, we can find three disjoint subsets A, B, C of
Y such that Y = A ∪ B ∪ C and |Y | = |A| = |B| = |C|. It follows that
|A| = |M × Y |, |B| = |Y × M | and |C| = |Y × Y |; accordingly, there is
a bijection k : Y → (M × Y ) ∪ (Y × M ) ∪ (Y × Y ). Since M ∩ Y ̸= ∅,
Set Theory
517
we have a bijection M ∪ Y → (M ∪ Y ) × (M ∪ Y ) which extends h (and
k). This contradicts the maximality of M , and hence |M | = |X|.
♢
It is now clear by induction on n that if α is an infinite cardinal
number, then αn = α for every integer n > 0. And, for a cardinal
number β ≤ α, αβ = α, since α ≤ αβ ≤ αα = α.
Theorem A.8.4 Let X be an infinite set and F be the family of all
finite subsets of X. Then |F| = |X|.
Proof. Since the mapping X → F, x 7→ {x}, is an injection, we have
|X| ≤ |F|. To see the opposite inequality, for each integer n > 0, let
Fn denote the family of all those subsets of X, which contain exactly
n elements. Now, for each set F ∈ Fn , choose an element ϕ(F ) in
X × · · · × X = X n (n copies), which has all its coordinates in F . Of
course, there are several choices for ϕ(F ), we pick any one of these.
Then ϕ : Fn → X n is an∪injection, and so |Fn | ≤ |X n | = |X|n = |X|.
Obviously, we have F = n Fn and therefore
∑
∑
|F| ≤ n |Fn | ≤ n |X| = ℵ0 |X| = |X|.
♢
This completes the proof.
Let X be a set and A ⊆ X. The function fA : X → {0, 1} defined
by
{
fA (x) =
0
1
for x ̸∈ A, and
for x ∈ A
is called the characteristic function of A. The function which is zero
everywhere is the characteristic function of the empty set, and the
function which is identically 1 on X is the characteristic function of
X. The set of all functions : X → {0, 1} is denoted by 2X . Obviously,
every element of 2X is a characteristic function on X.
Proposition A.8.5 For any set X, there is a bijection between the
set P (X) of all its subsets and 2X .
Proof. For any subset A ⊆ X, we have the function fA : X → {0, 1}
defined by
{
0
for x ̸∈ A, and
fA (x) =
1
for x ∈ A.
So the mapping ϕ : 2X → P (X), f 7→ f −1 (1), is surjective. This is
injective, too. For, if f ̸= g are in 2X , then we have f (x) ̸= g (x) for
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Elements of Topology
some x ∈ X. This implies that x belongs to one of the sets f −1 (1) and
g −1 (1) , but not the other. So ϕ (f ) ̸= ϕ (g), and ϕ is a bijection.
♢
It follows that |P (X) | = 2|X| . The function x 7→ {x} is obviously an
injection from X into the set P (X), but there is no bijection between
these, by Theorem A.5.13. So |X| < |P (X) |, and we have established
Proposition A.8.6 For any set X, |X| < 2|X| .
Appendix B
Fields R, C and H
B.1
B.2
B.3
The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
519
521
523
In this appendix, we will discuss the fundamental properties of the
fields of real numbers, complex numbers and quaternions.
B.1
The Real Numbers
We shall not concern ourselves here with the construction of the
real number system on the basis of a more primitive concept such
as the positive integers or the rational numbers. Instead, we assume
familiarity with the system R of real numbers as an ordered field which
is complete (that is, it has the least upper bound property). We review
and list the essential properties of R.
With the usual addition and multiplication, the set R has the following properties.
Theorem B.1.1 (a) R is an abelian group under addition, the number 0 acts as the neutral element.
(b) R − {0} is an abelian group under multiplication, the number 1
acts as the multiplicative identity (unit element).
(c) For all a, b, c ∈ R, a (b + c) = ab + ac.
A field is a set F containing at least two elements 1 ̸= 0 together
with two binary operations called addition and multiplication, denoted
by + and · (or juxtaposition), respectively, which satisfy B.1.1. The
element 0 is the identity element for addition, and 1 acts as the mul519
520
Elements of Topology
tiplicative identity. A skew-field satisfies all field axioms except the
commutativity of multiplication; this is also called a division ring.
With this terminology, the set R is a field under the usual addition
and multiplication. This field has an order relation < which satisfies
(a) a + b < a + c if b < c, and (b) ab > 0 if a > 0 and b > 0. A
field which also has an order relation satisfying these two conditions
is called an ordered field. The set Q of all rational numbers is another
example of an ordered field.
We call an element a in an ordered field positive if a > 0, and
negative if a < 0.
Theorem B.1.2 The following statements are true in every ordered
field.
(a) a > 0 ⇒ −a < 0, and vice versa.
(b) a > 0 and b < c ⇒ ab < ac,
a < 0 and b < c ⇒ ab > ac.
(c) a ̸= 0 ⇒ a2 > 0. In particular 1 > 0.
(d) a > 0 ⇒ a−1 > 0,
a < 0 ⇒ a−1 < 0.
(e ab > 0 ⇒ either both a > 0 and b > 0 or both a < 0 and b < 0.
(f) ab < 0 ⇒ either both a < 0 and b > 0 or both a > 0 and b < 0.
The ordered field R has the least upper bound property: If S ⊆ R
is nonempty and bounded above, then sup S exists in R. This is also
called the completeness property of R.
Thus R is a complete ordered field. Using this property, it can be
shown that every nonempty set of real numbers with a lower bound
has a infimum. Another important consequence of the completeness
property of R is the following.
Theorem B.1.3 If a real number x > 0, then given any real number
y, there exists a positive integer n such that nx > y.
This is called the archimedean property of R. Using this property
of R, one can prove that, for any two real numbers x < y, there is a
rational number r ∈ Q such that x < r < y. This fact is usually stated
by saying that Q is dense in R.
Fields R, C and H
521
An ordered set (X, ≺) containing more than one point is called a
linear continuum if it is order complete, and has no gaps (i.e., for any
two points x ≺ y in X, there exists z ∈ X such that x ≺ z ≺ y). It
follows that R is a linear continuum.
The absolute value of a real number x is defined by
{
x if x ≥ 0,
|x| =
-x if x < 0.
Notice that |x| ≥ 0 for all x ∈ R.
Proposition B.1.4 For any x, y ∈ R, we have
(a) |x| = 0 ⇔ x = 0.
(b) | − x| = |x|.
(c) |xy| = |x||y|.
(d) for y ≥ 0, |x| ≤ y ⇔ −y ≤ x ≤ y.
(e) − |x| ≤ x ≤ |x|.
(f) |x + y| ≤ |x| + |y|.
(g) ||x| − |y|| ≤ |x − y|.
(h) |x − y| ≤ |x| + |y|.
By (c), |x| =
B.2
√
x2 .
The Complex Numbers
The set R2 of all ordered pairs (x, y) of real numbers turns into a
field under the following addition and multiplication:
(x, y) + (x′ , y ′ ) = (x + x′ , y + y ′ ),
(x, y) (x′ , y ′ )
= (xx′ − yy ′ , xy ′ + yx′ ).
The element (0, 0) acts as the neutral element for addition, and the
element (1, 0) plays the role of multiplicative identity. It is routine to
check that R2 is a field under these definitions. It is usually denoted by
C, and its elements are referred to as the complex numbers. It is readily
verified that (x, 0) + (y, 0) = (x + y, 0), and (x, 0) (y, 0) = (xy, 0). This
shows that the complex numbers of the form (x, 0) form a subfield of
C, which is isomorphic to R under the correspondence x 7→ (x, 0). We
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Elements of Topology
can therefore identify this subfield of C with the real field and regard
R ⊂ C.
Writing ı = (0, 1), we have ı2 = −1 and (x, y) = (x, 0) +
(y, 0) (0, 1) = x + yı, using the identification x ↔ (x, 0). Thus C =
{x + yı|x, y ∈ R}, where ı2 = −1. If z = x + yı is a complex number, then we call x the real part of z (denoted by Re(z)), and y the
imaginary part of z (denoted by Im(z)).
If z = x + yı ∈ C, its conjugate is defined to be the complex number
z̄ = x − yı. For any complex numbers z and w, we have
z + w = z̄ + w̄,
zw = z̄ w̄.
Observe that z z̄ is a positive real number unless z = 0.
√
The absolute value |z| of a complex number z is defined to be z z̄
(the nonnegative square root). It is immediate from the definition that
|z| > 0 except when z = 0, and |0| = 0. Also, it is obvious that
|Re (z) | ≤ |z| and |z| = |z̄|. If z and w are any two complex numbers,
then it is easily checked that |zw| = |z||w| and |z + w| ≤ |z| + |w|.
B.2.1 (Schwarz Inequality) If z1 , . . . , zn and w1 , . . . , wn are complex numbers, then we have
(∑n
) ( ∑n
)
∑n
2
2
.
| 1 zj w̄j |2 ≤
1 |zj |
1 |wj |
∑n
∑n
∑n
Proof. Put α = 1 |zj |2 , β = 1 |wj |2 and γ = 1 zj w̄j . We need
to show that αβ − |γ|2 ≥ 0. If α = 0 or β = 0, then the conclusion is
trivial. We therefore assume that α ̸= 0 ̸= β. We have
∑n
∑n
2
=
1 |βzj − γwj |
1 (βzj − γwj ) (β z¯j − γwj )
∑n
2 ∑n
2
= β
1 zj w̄j −
1 |zj | − βγ̄
∑
∑n
n
2
2
βγ 1 z¯j wj + |γ|
1 |wj |
= β 2 α − β|γ|2
(
)
= β αβ − |γ|2 .
Obviously, the left-hand side is nonnegative and β > 0. So αβ−|γ|2 ≥ 0,
and the desired inequality holds.
♢
Fields R, C and H
B.3
523
The Quaternions
By using the usual scalar and (non-associative) vector product in
R3 , Hamilton defined (in 1843) a multiplication in R4 , which together
with componentwise addition makes it into a skew-field (that is, a division ring). This field has proven to be fundamental in several areas
of mathematics and physics. We discuss this field here.
The mapping R4 → R × R3 , (x0 , x1 , x2 , x3 ) 7→ (x0 , (x1 , x2 , x3 )), is
a bijection. If we define the vector space structure on R × R3 over R
componentwise:
(a, x) + (b, y) = (a + b, x + y) and c (a, x) = (ca, cx),
then this mapping becomes an isomorphism. So we can identify R ×
R3 = H with R4 , and call its elements quaternions. If q = (a, x),
we refer to a as the real part of q and x as the vector part of q.
There are canonical monomorphisms R → H, a 7→ (a, 0), and R3 → H,
x 7→ (0, x), of vector spaces. Accordingly, we identify a with (a, 0),
and x with (0, x), and write (a, x) = a + x. Then, for any quaternions
q = a+x, r = b+y and a real c, we have q+r = a+b+x+y, cq = ca+cx.
If x, y ∈ R3 ⊂ H, we first define xy = −x · y + x × y, where · is the
usual scalar product, and × is the usual vector product in R3 . Notice
that xy is in general an element of H. It is easily checked that this
multiplication of vectors in H is associative. As the multiplication in H
ought to be distributive, for q = a + x and r = b + y, we set
qr = ab + ay + bx + xy.
We leave it to the reader to verify the following conditions:
q (cr) = c (qr) = (cq) r,
q (r + s) = qr + qs,
q (rs) = (qr) s,
(r + s) q = rq + sq
for all quaternions q, r, s and real c. The element 1 ∈ H acts as the
identity: 1q = q = q1 for any q ∈ H. To prove that H is a skew-field, it
remains to verify that every nonzero quaternion q has a multiplicative
inverse. For this, we define the conjugate of q = a + x by q̄ = a − x.
Observe that q + r = q̄ + r̄, cq = cq̄, qr = r̄q̄ for any quaternions q, r, s
and real c. Also, it is straightforward to see that q q̄ = q̄q = a2 + x · x
√
(a real). We define the modulus of q to be |q| = q q̄. Thus |q| is
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the euclidean norm of q when it is considered as an element of R4 . If
q, r ∈ H, then we have
|qr|2 = (qr) (qr) = qrr̄q̄ = q|r|2 q̄ = |r|2 q q̄ = |r|2 |q|2 .
So |qr| (= |q||r|.
that
) Also, note
(
) q = 0 ⇔ |q| = 0. Obviously, if q ̸= 0,
then q q̄/|q|2 = 1 = q̄/|q|2 q. Thus q̄/|q|2 is the inverse of q in H,
usually denoted by q −1 .
Observe that x ∈ R3 is a unit vector ⇔ x · x = 1 ⇔ x2 = −1,
and two vectors x, y ∈ R3 are orthogonal ⇔ x · y = 0 ⇔ xy = −yx.
A right-handed orthonormal system in R3 is an ordered triple ı, ȷ, k of
vectors in R3 such that ı, ȷ, k are of unit length, mutually orthogonal
and ı × ȷ = k. So, if ı, ȷ, k form a right-handed orthonormal system,
then ı2 = ȷ2 = k 2 = −1 and ıȷk = −1. Conversely, these conditions
imply that ı, ȷ, k are of unit length, and ıȷ = k whence ı · ȷ = 0 and
ı × ȷ = k. Thus ı, ȷ, k form a right-handed orthonormal system.
Suppose now that ı, ȷ, k is a right-handed orthonormal system in R3 .
Then any vector x ∈ R3 can be written uniquely as x = x1 ı+x2 ȷ+x3 k,
xi ∈ R; accordingly, any quaternion q can be expressed uniquely as
q = q0 + q1 ı + q2 ȷ + q3 k, qi ∈ R.
Clearly, we have ıȷ = k = −ȷı, ȷk = ı = −kȷ, kı = ȷ = −ık. Using these
rules, we obtain the following formula for the product of two elements
q = q0 + q1 ı + q2 ȷ + q3 k, q ′ = q0′ + q1′ ı + q2′ ȷ + q3′ k in H:
qq ′ = (q0 q0′ − q1 q1′ − q2 q2′ − q3 q3′ ) + (q0 q1′ + q1 q0′ + q2 q3′ − q3 q2′ ) ı +
(q0 q2′ + q2 q0′ + q3 q1′ − q1 q3′ ) ȷ + (q0 q3′ + q3 q0′ + q1 q2′ − q2 q1′ ) k.
√
We also note that q̄ = q0 −q1 ı−q2 ȷ−q3 k and |q| = (q02 + q12 + q22 + q32 ).
For any unit vector x ∈ R3 , the set of quaternions a + bx, a, b ∈ R,
is a subfield of H isomorphic to C under the mapping a + bx 7→ a + bı.
In particular, the subfield of quaternions with no ȷ and k components is
identified with C, and we regard C as a subfield of H. Thus, we have field
inclusions R ⊂ C ⊂ H. It is obvious that any real number commutes
with every element of H. Conversely, if a quaternion q commutes with
every element of H, then q ∈ R. We emphasize, however, that the
elements of C do not commute with the elements of H.
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