AN ABSTRACT OF THE THESIS OF
William Albert Tryon for the
(Name)
Date thesis is presented
Title
M.S.
in
(Degree)
Mathematics
(Major)
December 6, 1968
PROPERTIES OF REAL VALUED CONTINUOUS FUNCTIONS
IN RELATION TO VARIOUS SEPRATION AXI9I'IS
Abstract approved
Redacted for Privacy
pt.Of es sokl
This paper defines and discusses some of the separation axioms of topological spaces.
In the cases con-
sidered, a search is made for sets of conditions which
would be equivalent in a space satisfying a given separation axiom to the existence of a family of real valued,
continuous functions which separates by functional values
the points and/or sets corresponding to the given separation axiom.
PROPERTIES OF REAL VALUED CONTINUOUS
FUNCTIONS IN RELATION TO VARIOUS
SEPARATION AXIOMS
by
William Albert Tryon
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1969
APPROVED:
Redacted for Privacy
Professor of kgthematics
in charge of major
Redacted for Privacy
Chairman of the Department of Mkthematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented December 6, 1968
Typed by Eileen Ash for
William Albert Tryon
ACKNOWLEDGEMENT
I would like to express my gratitude to Dr. B. H.
Arnold for his patience and many helpful suggestions, as
well as originally proposing this topic.
TABLE OF CONTENTS
Chapter
I
II
ITI
Page
INTRODUCTION
1
RELATIONSHIPS BETWEEN SEPARATION
AXIOMS
2
PROPERTIES OF REAL VALUED CONTINUOUS
FUNCTIONS IN RELATION TO VARIOUS
SEPARATION AXIOMS
25
BIBLIOGRAPHY
41
PROPERTIES OF REAL VALUED CONTINUOUS
FUNCTIONS IN RELATION TO VARIOUS
SEPARATION AXIOMS
CHAPTER I
INTRODUCTION
A topological space
Hausdorff
a
and
that
b
or
,
T
(X, T-)
is said to satisfy the
axiom iff for every two distinct points
2
U
there are disjoint open sets
a E U
and
b E V.
Thus, the points
can be "separated" by the open sets
and
a
V
and
and V .
U
question arises whether two such points
a
and
such
b
The
b
can
be separated by a function; that is, does there exist a
continuous, real valued function
f(a) = 0
and
f:
X
[0,1]
such that
f(b) = 1 ?
This paper poses the same type of question for topological spaces satisfying other separation axioms, as
well as Hausdorff spaces.
All of the implications between
these separation axioms, holding in arbitrary topological
spaces, are also proved in the paper.
2
RELATIONSHIP BETWEEN SEPARATION AXIOMS
CHAPTER II
Since different separation axioms are sometimes referred to by the same terminology, a precise definition
of terms is in order.
A topological space
Definition 2.1
to be a
T
(X,/)
is said
space iff for every pair of distinct points,
0
there is an open set which contains one but not the other.
A topological space
Dafin_Ltion 2.2
to be a
T
(X,()
is said
space iff for every ordered pair of distinct
1
points, there is an open set which contains the first but
not the second.
Obviously, a
space is a
T1
space; however, the
To
following space is frequently cited to demonstrate that
not all
T
spaces are
0
T
0
spaces.
1
X = {a,b}
Example 2.3
is a
T
= {0
space, but fails to be
T
must also contain
a
containing
b
Lemma 2.4
A space is
T1
,
{a}
(X,Z)
X}
,
since any open set
1
.
iff all singletons are
closed.
Assume
Proof:
member of
X
.
is an open set
(X,')
For all
Ub
is
b E X
such that
T1
and
such that
b E Ub
and
let
a
b
a
be a
,
there
a E X - Ub
.
3
Therefore
{a} =
X
ub
U
is an open set, so
{a}
is
bra
closed.
Now assume all singletons are closed, and let
X
be an ordered pair of distinct points of
X - {b}
is closed,
not
b
to be a
T
is an open set which contains
a
but
A topological space
(X,95)
is said
or Hausdorff space iff for every pair of
V
and
and
a
b
,
a E U
such that
there are disjoint open sets
b E V.
and
there are
T
1
spaces which fail to be
T1
cites the following space as a
X = [0,1]
Example 2.6
of countable sets }.
any pair
(a,b)
containing
a
,
(X,21)
where
but not
such that
a
b
separate any two points
V
space is
T2
Again it is obvious that a
and
{b}
2
distinct points
U
Since
.
Definition 2.5
U
.
(a,b)
b
.
a
a E U
,
X
but
Pervin [21
space.
all complements
{b} is an open set
X
,
,
,
space, since for
T1
is a
.
non-T2
,
al= {0
T2
T1
However, it is impossible to
and
by disjoint open sets
b
b E V.
and
If
then
U
is an
X - U
open set which contains
a
but not
Since
X
is itself an uncountable set,
a countable set.
the only open set contained in
X - U
which obviously cannot contain
b
Definition 2.7
to be
T22
or
b
,
is
is the empty set,
.
A topological space
(X,)
is said
Urysohn space iff for any pair of distinct
4
points
and
a
such that
b
G
there are disjoint closed sets
,
is a neighborhood of
F
neighborhood of
b
and
a
G
and
F
is a
.
By considering the definition of a neighborhood, we
see immediately that a
are
T
2
space is
T2
T2
however, there
;
spaces which fail to satisfy the
axiom.
T*,
Urysohn [3] gives the following example of such a space.
and
x
{(x, -y)
U {(0, -1)}
,
U {(x,0)
y E
where
and yE N} U
x
X= { (x,y)
Example 2.8
x E
I
U {(0,1)}
denotes the set of all positive
le
is defined in the
A basis for the topology
integers.
1\11-}
following manner.
{ (x,y) } E
if
0 # y
x
V (x,0) = {(x,0)} U ( U(x,y)) U ( U(x,-y)) E °X,
y>n
y>n
x E le
n E le
for
co
00
0
(0,1)
= {(0,1)} U
(
U (x
U
y=1 x=n
))
for
(x,-y))) E
x U=n
ll
(
(0,-1)
=
(0,-1)
U
(
n E N
co
co
Un
if
y=U 1
(
n E
for
We can easily verify that this does define a basis for a
topology, as claimed, since
X = 01
(0,1)
U
U1
(0,-1)
0(0,1) n U(07-1)=
,
U
(
U
xeN
+
for every
Vi
(x,0)
m
)
and
n E N
+
5
On
(0,1)
n vm
u
y>n+m {
(x,0)
n v (x,0)
un
(0,-1)
y>n+m f(x,-y)}
,
x >
n
{(x,y)} =
fl
(
0
1)
(I)
,
y <
0
,
,
y <
if n> x
Example 2.8 is a
Urysohn's axiom.
where
(A)
or
x> 0
and
0
if
n < x
and
x >
or x> 0
and
0
,
0
if n> x
y >
E
and
{ (x,y)}
(I)
(m + n, 0)
n < x
if
y > 0
{ (x,y) }
Un ,-1) n
n < x
if
,
n > x
if
{ (x,y) }
0
if n < x
,
n > x
if
,
(I)
(x,1) }
T2
0
.
space, but it fails to satisfy
Consider the following chain of relations.
(o7t)?1))
fl
(u771))
s (°110,1))
denotes the closure of the set
A .
(uTo,-1)"
Since the
intersection of the closures of any two basic open sets of
6
the points
and
(0,1)
respectively is nonempty,
(0,-1)
there can be no disjoint closed neighborhoods of the points
and
(0,1)
(0,-1)
.
A topological space
Definition 2.9
to be a
T
space iff it is
3
that given any closed set
disjoint open sets
U
T
is said
and has the property
and xEX-F, there are
F
V
and
1
(X,e)
such that
x E U
and
F c V .
Lemma 2.10
In a
T
space
3
(X,e), closed neigh-
borhoods form a basis for the topology
We need only show that given any open set
Proof:
x E 0, there is a closed neighborhood
and
such that
X - 0
G c 0 .
Since our space is
UcX-Vc0 .
T
3
and
U
If
TA.
X
,
is a
(X,/,()
are distinct points of
Using our
and
and
is a
(17)
,
0 .
but the converse impliT3
space and
we may consider
V
T3
a
and
which contain
b
and
b
as a point
a
a
.
axiom , we may choose disjoint open sets
and
a
{p}
respectively.
Lemma 2.10, we may find closed neighborhoods
a
x
This means that
{b} as a singleton closed set not containing
and
of
{x}
,
which is contained in
x
space is also
cation fails.
V .
Since X-V is closed,
closed neighborhood of
A
T3
of
G
0
are disjoint closed sets, which are contained in
the disjoint open sets
U
q:/
F
and
By
G
respectively such that aEFcU and
bEGcV. Since
U
and
V
are disjoint,
F
and
G
7
will be disjoint, closed neighborhoods of
and
a
b
respectively, as we desired.
Example 2.11
A = {(x,y)
B = f(x,0)
and
y > 0}
Let
X = A U B.
x
1
are real and
x,y
1
Our space shall be
is real }.
We shall define a basis for the topology in
the following manner.
n
O (x,y) = B((x,y),
vn
(
X, if
n A) U {(x,0)},
03((x,0),
x,0)
(x,0) E B, (and
B((x0,170),
and
T11-)
E A (and
(x,y)
n E N +)
is the usual Euclidean distance in
d
e)
if
where
d((x0,y0)
= f(x,Y) E R2
n E
(x,y)) <
,
R
2
12-1-}
Check-
.
ing the conditions needed for the existance of a basis, we
see immediately that we do have a topology.
Now let
and
(x1,171)
distinct points
1
d((x1,y1)
in our space; therefore
be
(x2,y2)
,
(x2,y2)) > 7
for
0
m0
some
4m
0
>
0
.
4m
0
))
1
4m
fl
(
(x°
=
2
(i)
if
Y1
if
y1
if
Y1 = y2
0
Y2
2
4m
o
n
1,
(vx
(o
0)
n
(o
(x
2'
(vx
4m°
(4m0
2'
y
,
)
(1)
2
0)
(1)
,
y2
8
Therefore, Example 2.11 is
space.
T2
This space, which is cited by Pervin, is not a
The point
space, however.
F = {(x,0)
(0,0)
x
there is an
,
n E N
is not in the closed set
(0,0)
is any open set containing
U
If
0}.
+
such that
Vn
(
4,0)
point
(
'2n'-'
Vn
taining
'n"
,
yet no
2n
U
.
Since any open set
(277,0)
V
V conof
Vm 1
there can be no open set containing
,
The
.
basic open set
must contain a basic open set
F
(21
F
c U
0,0)
has a nonempty intersection with
---,0)
and therefore with
,
0,0)
is a member of
1
of
Vni( 1
T3
which
,"(1)
fails to meet
U
.
Definition 2.12
A topological space
space iff it is a
T
T
1
F
tinuous function
X - [0,1]
f(F) = {l}
f:
x E X - F
T3
are open, disjoint sets which contain
]*,00[
There are
.
and
of the open sets
f
respectively, it is clear that a
F
f(x) = 0
.
and
and
a
there is a con-
,
such that
Since the inverse images under
]-00,2[
is a
space and has the property that
given any closed set
and
(X,(6
T3
spaces which fail to be
T32
space is
T32
how-
,
Before we give an example of such a space, we shall
ever.
first prove the following lemma.
Lemma 2.13
T
3
The product of two
T3
spaces is also a
space (with the product topology.)
Proof:
(x1,171)
and
Let
X
(x2,y2)
and
Y
be two
T
3
spaces, and let
be distinct points in
X x Y
.
9
Therefore, either
xl
or
x2
U C X
then there is an open set
x2 E X- U.
quite similar if
y1
distinct points in
y2
.
X x Y
,
If
.
such that
but not
(x1,171)
y2
x
(x2,y2)
x
1
xl E U
is an open set in
U x Y
Therefore
which contains
y1
2
'
and
X x Y
The argument is
.
Since for any ordered pair of
we can find an open set
0cXxY which contains the first ooint but not the
X x Y
second,
is a
Now, let
space,
T
be a closed set in
F
X x Y
and
(x,y) EXxY-F. Since F is closed, there is a basic
open set UxVcXxy-F such that (x,y) EUxV.
Since
set
X
is
X - U
,
T
and
3
there are disjoint open sets
x E A
such that
is not a member of the closed
x
and
(X - U) c B
(A)X
denotes the closure of
A
(-(5)1. c
in the space
F c X x Y that
X x Y
Y
V, where
.
Therefore,
X x Y0
(A
is a
Example 2.14
(17)Y
T
)
3
Let
.1- X
(A)
cX-Bc U,
Y
X
denotes the closure of
(x,y)
.
such that
E A x0
0
and
= X x Y - (A x0) X x Y , which means
space.
X
denote the set of all ordinals
less than or equal to the first uncountable ordinal
and let
B
in the space
Similarly, we can find an open set 0 c Y
y E 0 c
and
This means that
.
AcX-BCU. Since X-B is closed,
where
A
SZ
denote the set of all ordinals less than or
equal to the first infinite ordinal
w
.
We define the
10
bases for the topologies of
manner.
and
X
in the following
Y
X , we shall let each set consisting of a
In
single countable ordinal be open.
Neighborhoods of
will consist of all sets which contain
able complements.
Q
and have count-
0
Similarly, we shall let each set con-
sisting of a single finite ordinal be open in
well as all sets containing
Y
as
,
that have finite comple-
w
ments.
It is obvious that both spaces are
singletons are closed.
X
,
Now let
respectively.
X
and
F
is open, so
F
F
will be disjoint open sets containing F and x
X - F
and
since all
,
1
be a closed set in
F
and xEX-F. If C.2, then
and
T
{x}
If
SZ
E F, then
is open, so
{x}
{x}
will be disjoint open sets containing
respectively.
Therefore,
is a
X
A similar exercise will show that
T3
x
space.
is also a
Y
T
3
space.
By Lemma 2.13, we know that
X x Y
is also a
A quick check of the definition of the relative
space.
topology will verify that the subspace
X x Y
{.(0 w)}
Lemma 2.15
must also be
If
f:
Z
T3
R1
Z =
.
is a continuous function
with the property that there is a sequence
of distinct ordinals < w
n E N
T3
+
,
then there is a
such that
SO < Q
f(Q,i
{in}
n
)
such that
> r
n E N+
for all
f(d,w)
> r
11
for all
The lemma also holds if both the in-
> 60.
6
equalities are reversed.
Since
Proof:
n
N
f(6,i
+
there is an ordinal
1
n
> r - -
)
for all
sup {6 (m,n) I m,n E N +}l
6
>
6
Since
.
> 60
f(6,i
,
n
)
=
o
I m,n E e}
(m,n)
X - {Q}
6
60 <
,
S2
.
is
If
> sup {r 1- -ImEN + } =r. Since
is continuous,
f
Let
(m,n)
{6
and
such that
< Q
(m,n)
6
a countable set contained in
6
m
is continuous, for every
f
lim
f(6,i
)
= f(6,w) > r
for all
n->-co
6
>
The argument is quite similar if the inequalities
6
0
are reversed.
Now let
I
denote the space of all integers -
positive, negative and zero - with the discrete topology.
Forming the product
I x Z
,
we have a
T3
space in which
points may conveniently be renresented in the form
(n,6,i)
where nEI,
of the form
(n,Q,w)
and iEY. All points
6 E X
are omitted.
We now form a new space
K
by taking the quotient
space formed by the following equivalence relation
r = A
(IxZ)
x
(IxZ) U {(n,6,10),(n+1,6,w))
U {((n+1,6,w),(n,6,w))
I
n
I
n
r
.
is even}
is even}
U {((n,Q,i),(n+1,0,i)) In
is odd}
U {((n+1,0,i),(n,Q i)) In
is odd} , where
A
(Ixz)x(IxZ)
12
is the diagonal in the space
(I x Z)/
r
= K
I x
Let
Z
in which we defined
r
one or two points of
be a set in
[x] e K - F
I x Z
of
6
K - {[x]}
and
Z
,
[(n0,60,w)]
Since
that
[x]
I x Z
[x]
T
-1
E U'
(U')
is a
K
,
.
,
and
consists of only one point
where
(n,S,i)
Since all such points are open sets
.
If
Therefore
{[x]}
f
-1
and
[x]
consists of two points of
[x]
[x] by
[(n0,60,w)]
where
n0
where
n0
If
[x] =
is odd.
([x]) = {(n0,60,w)}U{(n0+1,60,0}
there is an open set
Therefore
f
-1
f
-1
([x])
U = {(m,60,i) Im = n0
.
I x Z
space, since singleton
1
is of the form
x
which contains
such that
f
the
F
[x]
will
([z])
be a closed set in
F
[(n0,Q,i0)]
,
-1
is closed, since
is a
K
we may represent
is even or by
f
are disjoint open sets containing
respectively.
I x
space.
3
By the manner
.
Therefore
.
is open as well as closed.
{[x]}
F
i < w
and
< 0
Z
([z])
If
.
then
,
x
Now let
sets are closed.
let
-1
Therefore
space.
3
f
I
K
T
will consist of either
[z]
,
G C K
consisting of either one or two points.
x Z
I
In either case,
T
is a
K
be an equivalence class in
[z]
Since
.
is open (or closed)
f-1(G)
We want to show that
.
(I x Z)
x
does form a quotient space, a set
will be open (or closed) iff
in
x Z)
(I
(U')
.
K - F
U'
is an open set in
There is an i
or
such
n0 +1
,
0
< w
i > i0}
Examining the equivalence relation
r
, we
.
13
see that
f(U) C U' c K - F
F
and
I x z
,
are disjoint open sets which contain
K - f(U)
and
and
[x]
Since
.
is the union of basic open sets in
I x z - U
f(U)
(f(U)) = U
f
respectively. The argument is in the other case.
Now let us adjoin two new points
Neighborhoods
of
Un(a+)
such that
(j,(5,i)E K
j
triples
(-j,d,i)
itself.
The families
along with
> n > 0
of
Un(a_)
such that
E K
{U n (a+
a_
to
)
I
a_
consist of all
> n > 0
j
along with
n E N+} and
{ U
n
(a
-
)11.1E1\1+}
K U {a +} U {a } = A , with
and the open neighborhoods for
and
a+
a_
.
space is the one to which Example 2.14 refers.
that (A,T)is a
This
It is clear
space, since every singleton is closed.
T1
be a closed set in A, and xEA-F. If x=
F
[(n,6,i)]
where
(5
<
SZ
and
will be disjoint open sets.
a So <
meet
a+
generated by the union of the topology for
the topology
Let
a
respectively.
a_
Let us denote the space
K
.
itself.
a+
are a bases for the neighborhood systems of the points
and
K
consist of all triples
a+
Similarly, neighborhoods
and
a+
such that
Q
F
.
A - 0
{x}
x = [(n,O.,i)]
d
1
and
,
A -{x}
there is
> (50]} fails to
may be written as the union of basic
open sets containing
x
If
0 = {[(n,6,i)]
A - 0
open sets, however, so
Similarly, if
w, then
i <
=
F
[(n
an
,
and
x
will be disjoint
0
respectively.
S ,w)]
,
a
+
or
a_
14
we can find an open set
A - 0
and
such that
0
x E 0
0 fl
,
F =
can be written as the union of basic open sets.
Since the various cases considered are the only possiblil-
(A,C)
ities,
is a
T
space.
3
-
We will now show that no continuous real valued function can separate
and
a+
a_
.
Since
{a
is a
}
closed set, this is sufficient to demonstrate that
is not a
f
:
T31
A + R1
then
f(a
It will suffice to show that if
space.
is a continuous function such that
)
= 0
Since
integer
f(a+) = 0
such that
n
,
and let
6
>
f(a+) = 0
f([(n,(5,w)]) = f([(n-1,8,w)]) < E/2
26
f([(n-1, 0 ,i)]) < -T
6
>
6
2
-
61
there is an odd
for all
for all
6
f([(n-1,6,w)]) = f([(n,6,w)]) >
where
6
<
2
< 61
26
-T.
>
f([(n, 61 + 62 + 1,w)])
< E/2
our conclusion must hold.
number of ordinals < w
,
26
i < w.
for
and
,
an obvious contradiction,
We can use a similar argument
f([(n-4,0,i)])
since
.
.
Since this would mean that
.
f([(n,61 + 62 + 1 ,w)])
< w
i
such that
< Q
for all but a finite number of
'
to show that
,
f([(n-2,0,i)]) =
From this we can conclude that
all
,
be arbitrary.
0
f([(n,Q,i)]) < 6/2
By Lemma 2.15, we know that there is a
If not, we have
= 0
)
is a continuous
A + R1
f :
is continuous, and
f
f(a
.
Let us assume that
function and
(A,r/6
46
< 77
for all but a finite
26
f([(n-2,Q,i)]) < 7r.
for
15
all but a finite number of ordinals < w
Continuing
.
by induction, we can show that for every even integer,
p >
0
,
there is a finite ordinal
f([(n- p,Q,i)])
<
is continuous,
f
> -e
f (a _ )
for all ordinals
p+1
f(a _
f(a ) = 0
,
T
4
e
i > i
.
Since
Since we can also show that
.
.
(X,()
A topological space
Definition 2.16
to be a
<
)
such that
< w
i
(X,2)
space iff
T
is
is said
and has the pro-
1
perty that given any pair of disjoint closed sets
G
,
there are disjoint open sets
and
F c U
V such that
and
U
G c V .
space, the
T1
Since singletons are closed in a
following lemma and theorem will demonstrate that a
T3i
space is also a
space
T4
(X,
closed set contained in the open set
Proof:
Lemma.
fil
p)
U
,
,
if
F
is a
then there is
such that FcVcVcU.
Let
and
F
be sets as described in the
U
Since FcU, F
and
X - U
are disjoint closed
This means that there are disjoint open sets
sets.
and
V
V
T4
space.
In a
Lemma 2.17
an open set
and
F
such that
and
c (X - 0) c U
and
F c V
(X - 0)
(X
U) c 0
.
0
Since
is closed,
FcVcVc (X - 0) cU.
Theorem 2.18
T
4
space, and
F
(Urysohn's Lemma)
and
G
If
r'
(X,L
)
is
a
i
are disjoint closed sets, then
16
there is a continuous function
f(F) = {0}
T
U
4
0
space
Let
(X, e)
F
and
.
By Lemma 2.17, there is an open set
such that cFU 0
G
c U0
Ordering the rationals in
11213
U1
smallest
be disjoint closed sets in the
c- X-G. Let
.
X - G = U
and invoking Lemma 2.17
as an open set such that
Ui
1
as follows:
]0,1[
={ri,r2,r3,...}
again, we define
c U2
such that
[0,1]
f(G) = {1}
and
Proof:
X
f:
U0 cHUi
We continue our induction by taking the
and largest
r,
such that
rk
and
j
k
are
lesstharimaricirk<rm<riand defining the open
such that U
cUUr c17Ur E U rj
Again,
rk
m
Lemma 2.17 guarantees us that such a set exists. We now
set
U
.
r
define a relation
inf {r
g(x) = n E N +
1
The relation
and
,
F s U0
I
n
if
x E Ur }
X - U
x
if
,
x E U
1
.
1
is obviously a well-defined function,
g
g(X) E [0,1]
since
in the following manner.
g(x)
and
Also
.
g(F) = {0}
G c X - U1
.
and
g(G) = {1}
Continuity is a con-
sequence of the following equations.
-1
g
([0,3[)
U
u
Ur
for
,
3 > 0
r <3
n
g
-1
(]a,1]) =
rn >a
(X -
r
,
n
for
a < 1
.
,
17
Since the inverse images of sub-basic open sets are open,
g
is continuous.
However, there are
the
T
1ff
The following space is often used as an
axiom.
4
spaces which fail to satisfy
T
example of such a space.
Our space
Example 2.19
same points as in Example 2.11.
A
sets
and
shall consist of the
X
We shall also denote the
We shall define a
in the same manner.
B
basis for our topology as follows.
0',,y) = B((x,y),
V(x,0) = B((x,
radius
a >
Now let
Since
F
center
and
0
F
111),.24.17
n X
)
,
if
E A
(x,y)
if
U {(x,0)}
(x,0)
E B
is the normal open Euclidean ball of
B((x,y), a)
where
131-:)
m
(x,y), with
be a closed set and
E X - F
(x0,y0)
is closed, there must he an
n
n E N+
and
E N
0
+
.
.
such that
n
0 0
(or
,y
0
)
V
(x
0
0'0
)
is contained
in
X
F
.
Let us
0
assume
( x0,170) E A
,
so our basic neighborhood will be
nil
0,`'
,y ).
0
In order to construct our function, we must extend
our notation slightly.
18
1
,
Oa
(x,y)
= B((x,y), -) i
X
fl
1
1
V (a
x,0)
= B((x,y407)-27dn {(x,0)}
(x,0) E B
a > 0
for
,
a
for
,
,
if
a > 0
(x,y)
E A
if
,
.
We shall now define the function which gives us the
desired separation.
11, if
E X - 0no
(x,y)
0'17 0 '
no
f(x,y) =
inf
(x,Y)
1
E d(x o,y0)
if
,
no
E 0 (x ,y
(x,y)
0
It is clear that
f(x
0'
y
)
and
= 0
)
0
is a well defined function and that
f
f(F) = {1}
Continuity follows from
.
0
the following equations.
-1
f
(]0,(3[) =
0
Sx n
0
f
-1
(la,1]) = X (0
for
0
,y
(
a.no
(x
0<
a <
1
0 <
a <
.
)
0
for
)
0,y0
)
1
.
°
Since the inverse images of sub-basic open sets are open,
f
is continuous.
The proof is quite similar for the case when
(x0,170)
E B
.
19
It remains to show that
Let
spce.
F = { (x,o)
is irrational}
x
.
is not a
f)()
is rational}
x
I
(X,
T4
and
G = { (x,0)
It is easy to show that
F
and
G
G
We
closed sets.
are disjoint
Now let
be any open set which contains
V
shalldefinea sequence of sets
{Gn }n
.
in the follow-
E N+
ing manner.
G
= { (x,0)
n
E G
1
Vn
(
Clearly
U
n E N+
Gn = G
c V}
x,0)
n E N
for
,
+
.
.
A famous theorem known as the Baire Category Theorem
states that the complement of a set of first category
Since
is dense.
R
1
R
is of the second category
with the usual relative topology from the metric
x {C}
space
G
2
there must be an
,
terior of
)
(
E N
0
+
such that the in-
is nonempty in the usual metric top-
0
ology.
(xo,°) E
This means that there is a rational
(G
n
,
)
where we are again talking about the
0
usual metric topology for
now shows that for any
(ym,0) E G
n
such that
0
any open set
U
such that
xo
R
1
x{0}
m E N+
V
,
m
(x
0'
0)
A short argument
there is a point
n
v
n
0
(
which contains the closed set
contain a basic open set of the form
But
m ,0)
V
(x
m0)
F
must
for some
20
m E N+
U
;
Therefore
.
V n U # (1)
for any such open set
,
ie, the space of Example 2.19 fails to be a
A topological space
Definition 2.20
to be a
T
space iff it is a
5
T
T4
is said
(X,Q))
space and has the
1
A
property that for every pair of separated subsets
B
,
there are disjoint open sets
A c U
iff
B c V.
and
(A n
A
and
such that
V
and
U
and
are separated
B
(T n B) = (1)).
U
"ff)
(Two sets
space.
Since any two disjoint closed sets clearly satisfy
the above relation, a
give an example of a
T
space is also
5
T4
Before we
.
space which fails to be
T4
T5
,
we
will first prove the following two lemmas.
A compact Hausdorff space
Lemma 2.21
Let
Proof:
the compact
and
F
space
T2
,
that
x E U
(x,/,r)
y E Vy
and
of a compact space,
.
Since
.
x E F
Fix
is
T4.
For every
.
and
U
V
such
is a closed subset
G
is also compact.
G
)
be disjoint closed sets in
G
there are disjoint open sets
y E G
(X,c(
Since GCU V
yEG
there is a finite subset
such that
n
n
G c U
{yi,y2,...,yri} C: G
V
i-1
is therefore an open set
U
.
i=1
Yi
Yi
n
containing
which fails to meet
x
U
i=1
(X,7=l)
is a
T
3
V
.
Therefore,
Yi
space.
Now, for every
x E F
,
there are disjoint open sets
,
21
U
V
and
x
such that
x
is compact and
F c
x E Ux
and
G c V
Since
.
x
F
Ux , there is a finite subset
U
xEF
{xl,x2,...,xm} c G
G C
such that
Ux
u
Therefore
.
i=1
m
m
U
U
i=1
contain
n
i=1
and
x.
1
and
F
is therefore a
Lemma 2.22
are disjoint open sets which
1
respectively.
G
T
V x.
Since
X
is
Tl
,
it
space.
4
(XX)
A topological space
space iff every subspace
T5
which the
T4
is
(X*4-!,*)
is a
relative topology.
Suppose
Proof:
F
and
(F
(X *, /,(*)
A c X*
set
fl
is a
)
space and let
T5
be disjoint (relatively) closed subsets of the
G
subspace
(X,
x*) n
by
G
Denoting the relative closure of a
.
IT* = A fl
x*
we have
,
F fl G=
= F fl
F n G =
Similarly,
.
Therefore, there are disjoint open sets(in
V
such that
F c U
G c V.
and
are disjoint open sets (in
F c U*
and
G c V*
perty,
(X *, Z *)
Since
.
is a
T4
T
B
X
.
and
and
such that
X*)
space.
be separated subsets in
an open subset of
U
is a hereditary pro-
1
Now assume that every subspace is
and
)
U* = U P X*
Then
V* = V n X*
X
F fl G= cp.
X
.
T4
and let
X* = X - (T n f)
Let us consider
space with the relative topology.
,
X* n
X*
A"
as a suband
x* n
A
is
22
will be disjoint closed subsets of
a
T
V
in
such that
X
respectively.
X*
is
T1
is
(x,i)r)
is open in
X*
X
is a
(x,9')
,
Example 2.23
(Tichonov)
,
X
U
V* = V fl X*
that contain
X*
are disjoint open sets in
V*
and
Since
and
X*
U* = U fl
are disjoint (open) sets in
U*
Since
.
space by hypothesis, there must be open sets
4
and
X*
A
and
B
however,
.
Since
space
T5
Let
X* = X U {a}
be the
one point compactification of an uncountable discrete space
Similarly, let
X .
be the one point com-
Y* = Y U
pactification of an infinite discrete space
both
and
X*
T
a
T
is not
and
F
X* x Y*
space, we only need exhibit a subspace
5
Z* = X* x Y* - { (a,13)}
Let
T4
.
G
in the following manner.
,
Since
x y*
x* x y*
By Lemma 2.21,
In order to show that
space.
4
.
are compact Hausdorff spaces,
Y*
is a compact Hausdorff space.
is a
Y
Z*
is not
which
and define
F = { (a,y) E z *}
G = { (x,(3) E z *}
Clearly
F
,
then
F
and
that
G
F
If
is an open set containing
{x0} x y*
that fails to meet
are disjoint sets.
G
(x0,y0)
(x0,y0)
A similar exercise will prove
.
is closed as well.
Now suppose that
that contain
F
and
U
G
and
V
are disjoint open sets
respectively.
Pick any infinite
23
sequence
each
fy ln
n
n E N+
the set
,
contained in
of distinct points in
N+
E
Vx = { (x,y)
1
V
tained in
)
.
For
must be
1 x E X}
for all but a finite number of terms.
U
Now, the sets
= {(x,y n
0
Y
can be entirely con-
y E Y*}
for only finitely many
x E X .
If not,
01
would have more than a finite number of terms in the complement of
U
Next, we see that the sets
.
all points but 1
belonging to
number.
U
If not,
01 U 02
we see that for any
n,
x
.
If not,
must also be finite in
Continuing by induction,
the number of
contains all but exactly
finite.
n
x E X
points in
must be
V
would not have
U On
01 U 02 U
such that
all but a finite number of terms contained in
X
have the property that
If this is true, then
finite.
containing
G
,
U , which
Therefore, only a countable number
is a contradiction.
of points in
that have
would not contain all but a finite
number of points of
V
V
Vx
vx n
since
Vx n(z - V)
is
cannot be an open set
V
(Z - V)
must be finite for
all of the points in the uncountable set
X
This
.
contradiction leads us to conclude that there cannot be
disjoint open sets that contain
is not
so
Z*
T
space
5
T4
.
Therefore
F
and
X* x Y*
respectively,
G
fails to be a
We have proven the followng implications
T5
T4
T31
T3
Ti
T2
TI
To
.
24
We also showed that the above implications are, in general,
not reversible.
Therefore, the above chain of implications
gives the only valid interrlation of the separation
axioms in arbitrary topological snaces.
It should be
noted, however, that a separation axiom along with an arl
ditional property will sometimes 7uarantee a stronger
axiom.
The topological spaces satisfying the various senaration axioms above are the ones with which we shall concern ourselves in this thesis.
It should be noted there
are other separation axioms, but we shall not examine them
in any detail.
25
CHAPTER III
PROPERTIES OF REAL-VALUED, CONTINUOUS FUNCTIONS
IN RELATION TO VARIOUS SEPARATION AXIOMS
(X,Z)
We know that if a topological space
then for any pair of distinct points
are disjoint open sets
such that
V
and
U
and
a
b
T2
is
,
,
there
a C U
and
Since topologists are often concerned with con-
b E V .
tinuous maps from arbitrary topological spaces into the
1
real line
R
,
it is natural to consider the possibility
of a continuous function
f(b)
X -> R
f:
1
such that
f(a)
.
If there is such a function
tinuous function
g(b) = 1
.
g:
such that
[0,1]
X -0-
then there is a con-
f
g(a) = 0
and
This is a consequence of the following com-
positions.
f(a)
g(x) = min{ max{ f(x)
f(b) - f(a)
If
and
is continuous, then
f
g
;
0}
11
;
must also be continuous,
g
has the desired bounds and values at
Similarly, if
ction such that
g:
g(a) = 0
pair of real numbers
[0,1]
X -0-
a
is a continuous function
and
h:
13
X ->
b
.
is a continuous fun-
g(b) = 1
and
and
a
,
such that
[a,.]
then for any
a <
S
,
there
which has the
26
property that
h(a) = a
h(b) = 6
and
.
This also fol-
lows from a composition of several continuous functions.
h(x) = minfmaxi((a)(1
g
is continuous,
h
g(x)) + 6-g(x)); al; 6}.
must also be continuous, and
tainly has the values at
and
a
If
h cer-
and the bounds that
b
we desired.
Therefore, it suffices to just consider the existence
of a continuous function
g(a) = 0
and
g(b) = 1
g:
.
X
such that
[0,1]
Throughout this chapter, we
shall restrict ourselves to functions corresponding to the
function
g
above, as they will exist iff the more arbi-
trary continuous, real-valued functions of the type discussed above exist.
We have examined several separation axioms, and now
wish to discuss them further in the manner mentioned at
the beginning of this chapter.
The next theorem will show
that we need only consider spaces which satisfy the
T21
separation axiom.
Theorem 3.1
Let
(X,2,7)
be a topological space.
If, for every pair of distinct points
there is a continuous function
f(a) = 0
and
Proof:
f(b) = 1
Let
topological space
a
,
and
(X,()()
f:
X
the space is
b
a
and
[0,1]
b
in
X
,
such that
Tr .
be distinct points in a
which possesses a family of
continuous, real-valued functions which separates points,
27
Let
f
be a member of this family such that
f(b) = 0
Therefore
.
f
-1
([0,4 ])
and
f
-1
f(a) = 0
3
(4,11)
and
are
disjoint closed sets which are neighborhoods of the points
and
a
b
respectively.
This leads one to wonder whether there are any nec-
essary and sufficient conditions in order to guarantee that
an arbitrary
space
T2,_
any pair of distinct points
and
f(b)
and
a
tinuous, real-valued function
f(a) = 0
has the property that for
(X, ()
f:
b
X
[0,1]
T
a
T
22
T
4
such that
.
We know by Lemma 2.21 that if a
pact, it is a
there is a con-
T2I
space is com-
Since singletons are closed in
space.
space, Theorem 2.18 assures us that every compact,
space has a family of continuous, real-valued fun -
ctions which separate points.
However, the condition of
compactness is by no means a necessary one, since
a non-compact
T4
R
1
is
space.
The following example by Hewitt demonstrates that
several other powerful properties are not enough to guarantee the existence of our family of functions.
Example 3.2
Our space shall be defined by the
following sets.
X = { (x,y)
x,y
x,y E ]0,10 U
are rational,
{ (0,0)} U { (1,0)
}
U {
(
,y)
!
y = r7
,
r
is
28
rational and
r,72- E 10,1[1 Uf (,17) I y = r1/7
,
is rational and r/ E ]0,1[} U
r
{4,y)
1
y = 1./5
is rational and r7 E
r
,
[0,1] }
A basis for the topology is defined in the following manner.
On
y<
(x,y) EXIO<
=
,
for
(0,0)
n E N
+
,
E X 13 < x <
= { (x,y)
V11
1
,
0 <
y <
,
for
(1,0)
n E N
+
n
E X
= { (x,y)
+
1
,
t
< x <
n E N+
for
,
3,-.
r1/7 -
n<
y< r1/7
.
All remaining points will have the usual Euclidean balls of
1
radii
,
for
n E N + , with center at the particular
point in question as a neighborhood basis.
A quick check shows that this does define a topology,
and that the resulting topological space is
T
,
2
how-
ever, no continuous, real-valued function separates the
points
(0,0)
and
(1,0)
a continuous function
0
and
f((1,0)) = 1
f-1([0,4]) = A
and
f:
.
Let us assume that there is
.
X
Since
[0,1]
f
such that
f((0,0)) =
is continuous,
f-1([3,1]) = B
are disjoint closed
29
sets which are neighborhoods of
m
rational
(O(0,0)) c A
r0
f((,r01/7))
and
such that
by
a
.
0 <
k E N
+
such that
in such a manner that
N
+
that
a >
3
0
.
0
V7)
Uk(1/2,r01/7)
and
C
B
/7 <
=
.
c C
.
meets
mo
N
+
Picking a
we denote
m0
-1
f
([a -
1
a + 7])
= C
Therefore there is
.
Since
r0
MO
0
pick-
is
for any
(0,0)
must also meet.
We can similarly show that
T .
a
5
8
A
the sets
,
r
(I,r01/7)
U (
-
0
(V (1,0))
Therefore,
is a closed neighborhood of
a
(1,0)
This implies that there is an
respectively.
such that
and
(0,0)
This means
B n C
,
so
Since this is an obvious contradiction, there can
be no continuous, real-valued function which separates the
points
and
(0,0)
Example
(1,0)
.
3.2 is a space which is countable, and there-
fore separable.
It is a
C
I
space, (i.e., has a countable
neighborhood base at each point) and is therefore a
CI,
space (i.e., has a countable base for the topology).
Since Example 3.2 was a
T22
space which failed to possess
the family of functions that we desired, none of the above
properties will be sufficient to guarantee the existence
of such a family.
Many other elementary properties fail to be
necessary for the existence of such a family.
Discreteness
30
is certainly sufficient, since any function from a discrete
space into any other space is continuous, but it is by no
means necessary.
Connectedness and the Lindeloff property
are neither necessary nor sufficient, since Example 3.2 is
a connected, Lindeloff space, and an uncountable discrete
space is neither connected nor Lindeloff.
is sufficient, since every metric soace is
Example 2.19 is a non-metrizable
T3
Metrizability
T4
, but
space.
We have not shown that there exists no set of conditions which is equivalent in a
T22
space to the exist-
ence of a family of real-valued, continuous functions
which separates points.
In fact, it would be very im-
practical to attempt such a feat, since we would have to
demonstrate that every conceivable set of properties fails
to be equivalent in a
T21
space to the existence of a
family of functions of the type we have discussed.
We
have, however, demonstrated that all the conditions considered, which constitute many of the basic properties of
topology, fail to have the equivalence that we desired.
Turning to the case of
T3
spaces, we ask ourselves
if there is any set of conditions which is necessary and
sufficient in a
T
3
space to the existence of a family
of continuous, real-valued functions which separates points
from disjoint closed sets.
(We shall use the terminology
above to mean that for any closed set
there is a function
f:
X
[0,1]
F
and
such that
x E X - F
f(x) = 0
,
31
and
f(F) = {1} .)
If we examine our proposal more close:-
ly we see that we are seeking a set of conditions that is
satisfied in every
T
3
space, as well as making every
T31
space that satisfies that set of conditions also
satisfy the
T
axiom.
3i
Theorem 3.3
A topological space
is a
(X, q)
3
space iff it is homeomorphic to a subset of a compact
Hausdorff space
(Y,
)
.
Let us assume that
Proof:
a subset of the compact
2.21, we know that
(X,Z)
space
T2
is a
(Y,9?)
is homeomorphic to a
(X, ,r)
that space by
(YI,
bijection from
X
Yl
.
x E X - F
,
to
By Lemma
.
space, and therefore
space is
T31
T3i,
Let us denote
space.
T
and let
1)
(Y,1?)
T4
Since every subspace of a
.
is homeomorphic to
be the bicontinuous
f
Therefore, if
we know that
is a closed
F
set in
X
and
f(x) = x
1
are a closed set and disjoint point in
Since
g:
Y
Y
1
is
[0,1]
1
T
such that
uous function taking
and
F
into
{1}
g(x
X
into
{f X}X
E A
Yl
.
1
)
= 0
and
gof
[0,1]
g(F
1
= {1}
)
will be a contin-
and maps
x
into
.
Now, let us assume that
shall let
and
there is a continuous function
,
3
Therefore the composition function
0
f(F) = F1
(X,
)
is a
T3i
space.
We
be the collection of all real-valued,
bounded, continuous functions defined on
X
,
and let
32
{IA
be acollection of closed and bounded intervals
A
such that fx(X) E Ix
Since the product of com-
}A
in
R1
.
pact spaces is compact, and the product of
also
T
,
AEA
define a mapping
A
X
g:
I
II
AEA
f
x E X
for all
(X)
x
,
by setting
A
and denote
g(x) =
Y = g(X)
I
II
AEA
To show that
is a homeomorphism from
g
only need show that
g
spaces is
2
is a compact, Hausdorff space. Let us
I
II
2
T
g
Y , we
is a continuous, open injection.
is continuous, since the
is continuous for all
to
X
A
A E A
Ath projection
.
If
and
x
HA° g = fx
are distinct
y
pointsinX,thereisanA0EAsuchthatf(x) = 0
AO
and
f
A
(y) = 1
g(x)
since the two
g(y)
0
points of
differ in the
Y
an open set in
is
Obviously,
.
T3yz
,
X
,
and let
there must be an
Now let
A0th place.
x
be
a C A
in
G
Since
.
such that
be
G
(X,,9)
fa(x) = 0
-1
and
fa (X - G) = {1} .
open set
z E g(G)
open in
in
.
If
Therefore
.
Y
Y
Clearly
z E U
,
Y n Ea (]-00,1[) = U
then
g
g(x) F g(U) Cg(G)
(z)
,
E G
and
,
g(G)
and
is
.
Clearly, if a topological space
(X, c)
is homeo-
morphic to a subspace of a compact Hausdorff space
then
is an
(X, i(1)
can be embedded in its closure in
which is a compact Hausdorff space.
(Y,
(Y, x2)
,
In light of Theorem 3.3
and the above remark, the following theorem is immediate.
33
Theorem 3.4
A
T
(X,5)
space
3
is
T3
iff
is homeomorphic to a subspace of a compact
(X,15)
Hausdorff space.
In the case of
T
spaces, we find that there is even
4
less to prove.
We know by Urysohn's Lemma (Theorem 2.18)
that every
space has a family of real-valued contin-
T4
uous functions that separates disjoint closed sets.
versely, if a
and
f
G
Con-
space possesses such a family, and
T1
F
are disjoint closed sets, then there is a function
in the family such that
Obviously
f
-1
and
(]-00,i[)
sets which contain
f(F) = {0}
f
and
F
-1
and
(h-, D
.
.
are disjoint open
respectively.
G
f(G) = {1}
The following
theorem is an immediate consequence of the preceding
remarks.
Theorem 3.5
space iff it is
A topological space
(X, Z)
is a
T4
and has a family of continuous, real-
T1
valued functions which separates disjoint closed sets,
In the case of
T
spaces, we are seeking a set of
5
conditions which is equivalent in a
T
5
space to the
existence of a family of continuous, real-valued functions
which separates separated subsets;
sets
A
and
B
such that
is a member of the family
(A fl
f
B)
(i.e., for any nonerpty
U
(A n f)
such that
(1)
,
f(A) = {0}
there
and
f(B) = {1}.)
Theorem 3.6
A
T
5
space
(X, Z')
has a family of
34
continuous real-valued functions which separates separated
subsets iff every pair of separated subsets have disjoint
closures.
Assume every pair of separated subsets have
Proof:
disjoint closures, and let
Therefore
sets.
is
T
T n B =
A
(1)
.
and
be separates sub-
B
Since
(X,95)
so there is a continuous function
,
4
such that
f(T) = {0}
and
f(-f) = {1}
T5
is
f:
it
,
X
[0,1]
.
Now assume there is a family of continuous, realvalued functions which separates separated subsets, and
let
A
and
B
be nonempty separated subsets of
hypothesis, there is a continuous function
such that
and
f
and
B
-1(
f(A) = {0}
and
f(B) = {1}
f:
Since
.
X
By
.
X
[0,1]
f-1({0})
are disjoint closed sets which contain
{l })
respectively,
A
and
B
A
will have disjoint
closures.
Even though we have found a set of conditions which
gives the desired equivalence in a
T
5
space, we do not
really have a good intuitive picture of what these conditions mean in a
T
5
space.
Before we dwell more on
this, let us first prove the following lemma and theorem.
Lemma 3.7
point of
points
X
{x
Let
be a
T
2
space.
If
is a
x
such that there is a sequence of distinct
n}n E N+
then the sets
(X, t)
ix
C x - {x}
2n
n E N+}
such that
and
Ix
2n-1
{ x }
n n
N+ -4-
N
n E N +}
x
are
35
separated subsets of
Let
Proof:
fx
nln E N+
show that
be a
(X, t)
{ x
2n
n E N
1
= {x
+
= { x
}
T
2n
1
m
N
+
and
n E N+} U {x}
and
since the two
,
belongs to neither set.
x
be a point in X- ({x2n 1 n E le} U {x})
y
and
x
It will suffice to
1nEN+sU{x}
Then there are disjoint open sets
y E U
space and let
2
,
2n-1
sets are disjoint and
Let
.
satisfy the hypothesiP.
HIEN + }
{x 2n-1
X
Since
x E V .
such that
x
{x-ril
E V
2k
n E
U
x
there is an
,
k < m
For each
Vk
and
k
such that
V
N+
k > m .
if
there are disjoint open sets
and
U
such that
m
y E Uk
xk E Vk
and
Therefore
.
U n
n
(
is an
Uk)
k=1
open set containing
U
{ x}
.
Since
x E fx
n E N+} U {x}
x2n
{x 2n-11 n E N
+}
=
Theorem 3.8
satisfying the
.
2n
1
n E N +}
{ x
1
2n
n E N +} =
Similarly, we can show that
2n-11 n E N
Let
CI
which fails to meet fx2n 1 n E N +}
y
(X, q1)
+1
{x}
.
be a topological space
and Hausdorff axioms.
Then every pair
of separated subsets have disjoint closures iff
q:
is the
discrete topology.
Proof:
(17
n B)
A = X
If
1,
fl (A n
and
B = F .
is the discrete topology, then
implies that Tn B =
(I)
,
since
36
Conversely, let us assume that
topology.
Therefore, there is an
x E (X - {x})
{xn}nE N+ E X
{x
such that
{x}
such that
By Lemma 3.7, we have that
n}n E N + + x .
and
x E X
This means that there is a sequence of
.
distinct points
{x
is not the discrete
9::
{x2n 1
+
2n-1
n E N } are separated subsets of
1
X
.
n E N
+
}
How-
ever, these two separated sets do not have disjoint
closures, since
x
is a member of the closures of both
Invoking the contrapositive, we have the result we
sets.
desired.
The preceding theorem shows us that a
CI
T5
,
space
has a family of continuous, real-valued functions that
separates separated subsets iff the space is a discrete
space.
We are tempted to try to "extend" this result to
arbitrary
T5
used to prove
spaces by using a method similar to the one
Theorem 3.8.
In a non - CI
space, however,
we must use a type of convergence more general than
sequences.
Instead, we must use nets or filters, which are
not necessarily indexed by either a countable or linearly
ordered set.
The method of proof used in proving Lemma
3.7 relied quite heavily on both these properties, which
often escape us when using nets or filters.
In fact, there are non- CI
,
non-discrete
T
5
spaces
which have the property that any pair of separated subsets
have disjoint closures.
37
Example 3.9
L*
Let
X
be the real line, and let
be the filter generated by the filter basis
n E N+1
One of the properties of filters (which
.
we will not discuss in any detail here) is that every
filter can be extended to an ultrafilter.
Accordingly,
let us extend
We shall define
to an ultrafilter
.
our topology Z in the following manner.
7j= {A cX10
A}U{FU{O} 1PE3 }
To show that this is a topology as claimed, we must demon-
strate that arbitrary unions of members of x and finite
intersections of members of g are also members of g'
{0 a
Let
If
0
U
0
aEA
0
0
a E A}
1
a E A, then
for all
a
E '.
be a collection of members of
If there is an
0
ao E A
U
0
,
so
a
aEA
such that 0 E O
,
a
0
= F U {0}
for some
F E 3
SinceFc0 a
.
0
U
aEA Oa
E 3
Now let
members of
and
0 4
Therefore
.
{o
1:
,
.
1
k
{0} U
U
0
k=1
0 E 0
k
for all
k
such that
k
A
0
a°
e 3
aEA Oa
be a finite collection of
1 < k < n}
0
u
(ayA Oa)
If there is an
then
aEU
m
such that
i° e.
n
'
1 < k <
k=1
n
,
0 < m < n
0
E t.
k
If
then
'This space was suggested to me by Darrell Kent, Professor
of Mathematics, Washington State University.
38
Ok = Fk U {0}
for all
a filter,
F1,
n
E 3
k , where
,
F
{0} U ( n
so
We now wish to show that
a
then
,
]0,-1 [
is a
(X, Z)
then there is an
,
Therefore,
.
but not
0
Ok
Fic)-=
15.
T
space.
1
If
is an open set which fails to contain
{a}
a = 0
If
.
is
be an ordered pair of distinct points.
(a,13)
0
Since
.
k=1
k=1
Let
E 3
k
N
n
+
such that
i
is
an open set containg
[0,-1,1
.
13
In order to show that our space is
T5
,
it will
suffice by Lemma 2.22 to show that every subspace is
Therefore,
and
A
let
(X*,P)
let
and
and
be a subspace of
B, then
A
and
0 X A
in
X
,
so they naturally will be open in
E A
,
then
B
0
will also be open
B
and
A
and
B
in
Again
X*
Now, let
T n E =
A
and
B
U (T n B) =
(A n TS)
(I)
Since
.
X*
If
.
will be open in
B
(X *, Z *)
X
,
is a
T4
X*
T1
which
property
space.
be separated subsets of
(1)
.)
0
E A.
If
0
TnE=Anf=
X
We want to show that
-
Case I:
E A, then A= A, so
.
so
is also closed,
Since the
respectively.
is a hereditary property,
(i.e.,
B
B
will be disjoint open sets in
X* - B
contain
.
X*
will be open
If
0
,
be disjoint, closed subsets in
B
0
(X, Z)
T4
39
Case II: OET-A. This means that
for all
F E 3.
Since
filter,
A E 3.
Clearly,
0 E B , then
B E 3
obvious contradiction.
Therefore
is open.
0
A.
B.
0
A
B
Example 3.9 is a
T
5
,
T c X- B.
,
then
If
However, the
is open, then
{0}
3
must
{0}
This means that {0}
.
must have a non-empty intersection, since
]0,i[
]0,2[ E 3
A .
space in which every pair of
be a member of the ultrafilter
and
{x}
and T n B= A n
separated subsets have disjoint closures.
space is not discrete.
,
is contained in the
is an open set which fails to meet
Therefore A= T
an
,
0
x E X- (A U {0})
If
If
A n B
Since
closed set X- B, and
Case III:
is an ultra-
3
so
,
(i)
By Theorem 3.8, this space cannot be a
CI
9
{o}
Clearly, this is a contradiction, so
.
space.
The attempt to generalize Theorem 3.8 to the case of
arbitrary
T5
spaces must fail, since
counter-example to such a theorem.
strict the number of
T
5
Example 3.9 is a
However, it does re-
spaces which will have the
equivalence that we desired in the case of
T5
Of the cases considered, the results of the
T
4
cases were already known.
were found in the case of
T
Zi
spaces.
T
3
and
Although no positive results
spaces, the properties
40
that we considered, which constitute many of the basic
properties of topological spaces, all either failed to be
necessary or sufficient.
In the case of
T
5
equivalence of the type we sought was found.
that this equivalence would exist in a
the space was discrete.
CI
,
spaces, an
We showed
T5
space iff
However, there are non CI
,
T
5
spaces which have the equivalence that we desired, but are
not discrete.
41
BIBLIOGRAPHY
1.
Hewitt, E.
On two problems of Urysohn.
Mathematics, Vol. 47: 503-509.
Annals of
1946.
2.
Pervin, W. J. Foundations of general topology.
New York, Academic Press, 1964.
209 p.
3.
Urysohn, R.
Uber die Machtigkeit der zusammanhangenden
Mangern.
Mathematische Annalen 94:
262-295.
1925.