Topology
December 7, 2004
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Basic Definitions
Definition 1.1 A topological space is a set X together with a family O of
subsets of X satisfying the following properties:
1. φ, X ∈ O;
2. O1 , O2 ∈ O ⇒ O1 ∩ O2 ∈ O;
3. S ⊂ O ⇒ ∪{O : O ∈ S} ∈ O.
Members of O are called open sets.
It follows immediately from the definition that intersection of finitely many open
sets is open.
Example 1.1 1. For any set X, O = {φ, X} defines a topology, called the
indiscrete topology on X.
2 For any set X, O = P(X) – the power set of X, defines a topology, called the
discrete topology on X.
3. Let X = {0, 1, 2} together with O = {φ, {1}, {0, 1}, {1, 2}, X} is a topological
space.
4. Rn with the usual open sets is a topological space.
Exercise 1.1 Prove that a set O in R is open if and only if it is a disjoint union
of countably many open intervals.[ For x, y ∈ O, define an equivalence relation
∼ on O by x ∼ y if and only if the interval [x, y] lies in O. Prove that each
equivalence class determined by ∼ is an open interval. Then show that there
are countably many equivalence classes.]
Definition 1.2 Let X be a topological space. A subset C of X is called closed
if its complement is open.
Proposition 1.1 Let C be the family of closed sets of a topological space X.
Then:
1. φ, X ∈ C;
2. C1 , . . . , Cn ∈ C ⇒ C1 ∪ . . . ∪ Cn ∈ C;
3. S ⊂ C ⇒ ∩{C : C ∈ S} ∈ C.
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Definition 1.3 Let X be a topological space. A family B of open sets is called
a base for the topology of X if for every open set O and x ∈ O, there exists
B ∈ B such that x ∈ B ⊂ O. A topological space is called second countable if it
has a countable base.
Example 1.2 1. In Rn with the usual topology, the family of open balls is a
base for the topology.
2. In example ??, the family {{0, 1}, {1, 2}} is not a base.
Proposition 1.2 A family B of open sets of a topological space X is a base if
and only if every open set of X is a union of some members of B.
Proposition 1.3 Let X be a set. A family B of subsets of X is a base for
some topology for X if and only if (i) union of sets in B is X, (ii) for every
two members B1 , B2 of B and each x ∈ B1 ∩ B2 there is B ∈ B such that
x ∈ B ⊂ B1 ∩ B2 .
Definition 1.4 Let X be a topological space and x ∈ X. A subset N of X is
called a neighborhood of x if there exists an open set O with x ∈ O ⊂ N .
Note that by definition every open set is a neighborhood of each of its members.
Exercise 1.2 Prove that a set U in a topological space is open if and only if
every point in U has a neighborhood contained in U .
Definition 1.5 Let X be a topological space and x ∈ X. A family B of neighborhoods of x is called a neighborhood base at x if for every neighborhood N of
x there exists B ∈ B such that B ⊂ N .
Proposition 1.4 Let X be a topological space and B a base for the topology.
Then for each x ∈ X, the family {B ∈ B : x ∈ B} is a neighborhood base at x.
Example 1.3 In Rn , the family of open balls centered at x is a neighborhood
base at x.
Definition 1.6 A topological space X is called first countable if every point in
X has a countable neighborhood base.
Proposition 1.5 Every second countable topological space is first countable.
Proof.
Let X be a second countable topological space and let T = {O1 , O2 , . . .} be a
countable base. Let x ∈ X. The (countable) family {O ∈ T : x ∈ O} forms a
neighborhood base at x. So X is first countable. Exercise 1.3 1. Prove that Rn with the usual topology is second countable.
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Definition 1.7 A subset A of a topological space is called dense if for every
nonempty open set O, O ∩ A 6= φ. A topological space is called separable if it
has a countable dense subset.
Exercise 1.4 1. Prove that every second countable topological space is separable.
2. Find a countable dense subset of Rn .
Definition 1.8 Let X be a topological space and A a subset of X. The interior
of A, denoted by int(A) is the largest open set contained in A, i.e.
int(A) = ∪{O : O open, O ⊂ A}
Dually, the closure of A, denoted by A, is the smallest closed set containing A,
i.e.
A = ∩{C : C closed, C ⊃ A}
Exercise 1.5 1. Let A be a subset of a topological space X. A point x is called
an interior point of A if there exists an open set O such that x ∈ O ⊂ A, i.e. A
is a neighborhood of x. Prove that int(A) is equal to the set of interior points
of A.
2. Prove that A = {x ∈ X : Nx ∩ A 6= φ for every neighborhood Nx of x}.
3. Prove Proposition 4.1.3 in textbook.
4. Prove Proposition 4.1.8 in textbook.
Let X is a topological space. Suppose to each x ∈ X there associates a neighborhood base Bx at x containing solely of open sets. Then:
(H1) each Bx is nonempty and x ∈ B for every B ∈ Bx ;
(H2) B1 , B2 ∈ Bx ⇒ ∃B ∈ Bx with B ⊂ B1 ∩ B2 ;
(H3) if B ∈ Bx and y ∈ B, there exists C ∈ By such that C ⊂ B.
Conversely if to each x ∈ X there associates a family Bx of sets satisfying H(1)(3), then there is a unique topology on X such that each Bx is a neighborhood
base at x. Moreover, each set in Bx is open.
Exercise 1.6 Prove the above statement.
Let X be a topological space with topology O, and Y ⊂ X. The family {O ∩ Y :
O ∈ O} defines a topology on Y , called the relative topology on Y induced by
(the topology of ) X.
Example 1.4 1. The relative topology on {0, 1, 2} induced by R with the usual
topology is discrete.
Proposition 1.6 Let X be a topological space and Y a subspace of X with
the relative topology. Let A be a subset of Y . The closure of A relative to the
relative topology of Y is equal A ∩ Y .
Definition 1.9 A topological space X is connected if it is not a union of two
disjoint nonempty open sets. A subset Y of X is connected if it is connected in
its relative topology.
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The following follows immediately from the definition.
Proposition 1.7 Let X be a topological space. The following are equivalent:
1. X is connected.
2. X is not a union of two disjoint nonempty closed sets.
3. The only subsets of X that are both open and closed are φ and X.
Proposition 1.8 A subset Y of X is connected if and only if there do not exist
disjoint open sets O1 , O2 such that Oi ∩ Y 6= φ, i = 1, 2 and Y ⊂ O1 ∪ O2 .
Proposition 1.9 Let X be a topological space and {Ci : i ∈ Λ} a family of
connected subsets of X. If ∩i∈Λ Ci 6= φ, then ∪i∈Λ Ci is connected.
Example 1.5 1. Any discrete space with more than one point is disconnected
(not connected).
2. The set Q of rational numbers with the relative topology induced by R is
disconnected.
3. A nonempty subset of R is connected if and only if it is a singleton or an
interval.
4. An open subset O of Rn is connected if and only if every two distinct points
in O can be joined by a polygon inside O.
Exercise 1.7 1. Prove Example 3 above.
2. Prove Example 4 above.
3. Prove that a subset of Rn is open if and only if it is a disjoint union of
countably many open connected sets.
Proposition 1.10 If A is a connected subset of a topological space and A ⊂
B ⊂ A, then B is also connected, in particular, A is connected.
Proof.
Let O1 , O2 be disjoint open subsets of X such that B ⊂ O1 ∪ O2 . We must show
that either B ⊂ O1 or B ⊂ O2 . Since A ⊂ B ⊂ O1 ∪ O2 and A is connected,
we have A ⊂ O1 or A ⊂ O2 . Without loss of generality, assume A ⊂ O1 . Then
A ⊂ O1 ∩ B. Since the closure of A with respective to B is A ∩ B = B and
O1 ∩ B is closed relative to B (because the complement of O1 ∩ B relative to B
is O2 ∩ B) , we have B ⊂ O1 ∩ B, which is the same as B ⊂ O1 . This proves
that B is connected. Example 1.6 We will see later that continuous image of connected set is connected and therefore A = {(x, sin 1/x) : 0 < x ≤ 1} is connected. Since
A = A ∪ {(0, y) : −1 ≤ y ≤ 1}, the above proposition implies that A ∪
any subset of {(0, y) : −1 ≤ y ≤ 1} is connected. This is somewhat counter
intuitive.
Let X, Y be topological spaces with topologies τX and τY , respectively. The
unique topology on X × Y determined by the base {O × Q : O ∈ τX , Q ∈ τY }
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is called the product topology on X × Y . The product topology on the product
of a finitely many topological spaces is defined similarly.
Let X be a topological space. A family F of open sets of a topological space X
is called an open cover if X = ∪{O : O ∈ F}. X is called compact if every open
cover of X has a finite subcover.
A family F of subsets of X is said to have finite intersection property if the
intersection of any finitely many members of F is nonempty.
A family F of subsets of X is called a filter if (i) φ ∈
/ F ; (ii) A, B ∈ F ⇒
A ∩ B ∈ F ; (iii) A ∈ F and B ⊃ A ⇒ B ∈ F.
A family F of subsets of X is called a filter base if (i) φ ∈
/ F ; (ii) A, B ∈ F ⇒
∃C ∈ F such that C ⊂ A ∩ B.
By definition every filter is a filter base. It is easy to prove that a filter base
generates a filter in the following sense:
Proposition 1.11 If F is a filter base, then {B : B ⊃ A for some A ∈ F} is a
filter.
An x ∈ X is called a limit point of a filter base F if
x ∈ ∩{F : F ∈ F}.
Theorem 1.1 Let X be a topological space. The following are equivalent:
1. X is compact;
2. Every family of closed sets of X with finite intersection property has a
nonempty intersection;
3. Every filter base in X has a limit point.
If X is second countable, each of these conditions is also equivalent to:
4. Every sequence in X has a convergent subsequence.
Proof.
1. ⇒ 2: If F is a family of closed sets with finite intersection property, then
the family G = {F 0 : F ∈ F} is a family of open sets any finite number of its
members do not cover X. Consequently G does not cover X since X is compact.
By DeMorie’s theorem, the family F has a nonempty intersection.
2. ⇒ 3: Let B be a filter base. Then the family {B : B ∈ B} is a family of closed
sets satisfying condition 2 (why?) and hence has a nonempty intersection. This
means B has a limit point.
3. ⇒ 2. (Exercise)
2 ⇒ 1. (Exercise)
Suppose X is second countable.
3. ⇒ 4.(we need only X be first countable for this part of the proof.) Let
xn , n = 1, 2, . . . be a sequence in X. Let Bn = {xk : k ≥ n}. The family
{Bn : n = 1, 2, . . .} is a filter base, so it has a limit point x. Let {O1 , O2 , . . .}
be a neighborhood base at x. We may assume that O1 ⊃ O2 ⊃ . . .. Since
x ∈ B1 , there exists n1 ≥ 1 such that xn1 ∈ O1 . Since x ∈ Bn1 +1 , there exists
n2 > n1 such that xn2 ∈ O2 . Continuing this process, we obtain a subsequence
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of {xn : n = 1, 2, . . .} that converges to x.
4. ⇒ 3. Suppose X is second countable. Then every open cover of X has a
countable subcover (why). Let this countable subcover be O1 , O2 , . . .. Now we
need to show that this countable subcover has a finite subcover. Suppose not.
Then for each n, there exists xn in the complement of O1 ∪ O2 ∪ . . . On . By
4, the sequence {xn , n = 1, 2, . . .} has a convergent subsequence with limit x.
Since O1 , O2 , . . . cover X, x ∈ Ok for some integer k ≥ 1. Since x is a subsequential limit of xn , n = 1, 2, , . . ., there exists l > k such that xl ∈ Ok . This is
a contradiction since xl is in the complement of O1 ∪ O2 . . . Ok ∪ . . . Ol . Exercise 1.8 1. Define compact subset of a compact topological space.
2. Prove that every closed subset of a compact topological space is compact.
3. Prove that product of two (and hence finitely many) compact spaces is
compact.
4. Prove that product of two (and hence finitely many) connected spaces is
connected.
5. (Heine-Borel) A subset of R is compact if and only if it is closed and bounded.
6. A subset of Rn is compact if and only if it is closed and bounded.
Definition 1.10 A topological space X is called Hausdorff if every two distinct
points x, y in X can be separated by two disjoint open sets, i. e. there exist
disjoint open sets O1 , O2 such that x ∈ O1 and y ∈ O2 .
A topological space X is called Normal if every two disjoint closed sets C1 , C2
in X can be separated by two disjoint open sets, i. e. there exist disjoint open
sets O1 , O2 such that C1 ⊂ O1 and C2 ⊂ O2 .
Proposition 1.12 1. Every singleton in a Hausdorff space is closed.
2. Every compact subset of a Hausdorff space is closed.
Definition 1.11 A mapping (or function) f : X → Y from a topological space
X to another topological space Y is continuous if f −1 (O) is open for every open
set O of Y .
Theorem 1.2 Let f : X → Y be a continuous function from a topological space
X to another topological space Y . Let A be a subset of X.
1. If A is compact, so is f (A).
2. If A is connected so is f (A).
Theorem 1.3 (Extreme Value Theorem) Let X be a compact topological space
and f : X → R be a continuous map. Then f attains its maximum and minimum values.
Theorem 1.4 (Intermediate Value Theorem) Let X be a connected topological
space and f : X → R be a continuous map. Then f (X) is either a singleton or
an interval.
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Definition 1.12 Let X be a set. A function d : X × X → R+ is called a metric
on X if it satisfies the following properties:
1. d(x, y) = 0 if and only if x = y;
2. d(x, y) = d(y, x)∀x, y ∈ X;
3. d(x, z) ≤ d(x, y) + d(y, z)∀x, y, z ∈ X.
(X, d) is called a metric space.
Example 1.7 1. Let X be a nonempty set. Define d(x, y) = 1 if x 6= y, = 0 if
x = y. d is a metric on X.
2. Let X = Rn . The Euclidean distance is a metric on X.
If (X, d) is a metric space, and x ∈ X. For each real number r > 0, the open
ball centered at x with radius r is defined to be the set
B(x, r) = {y ∈ X : d(x, y) < r}
A subset O of X is called open if for every x ∈ O, there exists r > 0 such that
B(x, r) ⊂ O. The collection of open sets of X defines a topology on X, called
the topology induced by the metric d.
Exercise 1.9 Let (X, d) be a metric space. Prove the following:
1. Every open ball is open.
2. The function d : X × X → R is continuous.
3. Let A be a nonempty subset of X. For each x ∈ X, define d(x, A) =
inf{d(x, y) : y ∈ A}. Then d(x, A) is a continuous function on X. [|d(x, A) −
d(y, A)| ≤ d(x, y)]. d(x, A) = 0 if and only if x ∈ A.
4. X is normal. [If A, B are disjoint closed sets, OA = {d : d(x, A) < d(x, B)}
and OB = {x : d(x, B) < d(x, A)} are disjoint open sets separating A and B.]
5. X is first countable. It is second countable if and only if it is separable. [For
an example of a separable first countable space that is not second countable,
see Kelley, chapter 5, problem M.]
Let (X, d) be a metric space. A sequence {xn : n = 1, 2, . . .} in X is called
Cauchy if for every > 0 there exists an integer N > 0 such that d(xm , xn ) < for all m, n ≥ N . X is called complete if every Cauchy sequence in X converges.
A subset A of a metric space (X, d) is called bounded if there exists M > 0 such
that d(x, y) ≤ M for all x, y ∈ A. If A is bounded, the diameter of A is defined
to be diam A = sup{d(x, y) : x, y ∈ A}.
Let > 0. A subset N of X is called an -net if the family {B(x, ) : x ∈ N }
covers X. X is called totally bounded if it contains a finite -net for every > 0.
Every totally bounded metric space is bounded.
Theorem 1.5 Let (X.d) be a metric space. The following are equivalent:
1. X is compact.
2. X is complete and totally bounded.
3. Every sequence in X has a convergent subsequence.
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Proof.
We shall prove that 1 ⇒ 3, 3 ⇒ 2, 2 ⇒ 3 and then 2 and 3 ⇒ 1.
1 ⇒ 3 is the proof of 3 ⇒ 4 in Theorem ??.
3 ⇒ 2: Since every Cauchy sequence with a convergent subsequence converges,
the proof for completeness is immediate. If X is not totally bounded, then
there exists > 0 such that no finite -nets exist. Let x1 ∈ X be an arbitrary
point. Since {x1 } is not an -net, there exists x2 ∈ X such that d(x1 , x2 ) ≥ .
Since {x1 , x2 } is not an -net, there exists x3 such that d(x3 , xi ) ≥ , i = 1, 2.
Continuing in this way, we find a sequence xn , n = 1, 2, . . . such that d(xi , xj ) ≥ for all distinct positive integers i, j. This sequence has clearly no convergent
subsequences.
2 ⇒ 3: Let xn , n = 1, 2, . . . be a sequence in X. Since X is covered with finitely
many balls of radius 1, some subsequence of xn is contained in a ball of radius 1.
Again, for the same reason, this subsequence has a subsequence that is contained
in a ball of radius 1/2. Continuing in this way, and using the standard diagonal
process, we obtain a subsequence whose tail starting from nth term is contained
in a ball of radius 1/n, making it a Cauchy subsequence. By completeness, the
subsequence converges.
2 and 3 ⇒ 1: For each positive integer n, let Bn be a finite 1/n-net of X. Then
the union of these Bn , n = 1, 2, . . . is clearly a dense subset of X. Thus, X is
separable and hence second countable. By 3 and Theorem ??, X is compact. A subset A of a topological space X is called nowhere dense if the closure of A
contains no nonempty open set, i.e. has empty interior. We say X is of first
category if it is a countable union of nowhere dense sets, otherwise, X is said to
be of second category.
Theorem 1.6 (Baire category theorem) Every complete metric space is of second category.
For proof, see p.356 in textbook.
Let (X, d) be a metric space. A mapping f : X → X is called a contraction if
there exists 0 ≤ λ < 1 such that
d(f (x), f (y)) ≤ λd(x, y)
for all x, y ∈ X. A point x ∈ X is called a fixed point of f if f (x) = x.
Theorem 1.7 (Banach contraction principle) Let (X, d) be a complete metric
space and f : X → X a contraction. Then f has a unique fixed point p.
Moreover, for every x ∈ X the sequence {f n (x) : n = 1, 2, . . .} converges to p.
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