AN ABSTRACT OF THE THESIS OF
Lynn Taylor Winter
for the
Master of Arts
(name)
Mathematics
in
(degree)
presented on
July 29, 1969
(major)
Title:
(date)
FOUR FUNCTION SPACE TOPOLOGIES
Abstract approved:
Redacted for Privacy
(pp-ftd"sof B.H. Arnbld)
This paper defines four function space topologies,
characterizes two of them in terms of more familiar
concepts, and compares the four topologies.
Then in the
cases of the two less familiar topologies we have considered several common properties of topological spaces
and attempted to answer the following question:
If the
range space has a given property, does the function space
necessarily have the same property?
FOUR FUNCTION SPACE TOPOLOGIES
by
Lynn Taylor Winter
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the
requirements
for
degree of
Master of Arts
June 1970
the
APPROVED:
Redacted for Privacy
Professor of 4ethematics
in charge of major
Redacted for Privacy
Acting Chairman of the Departpent of Mathematics
Redacted for Privacy
Dean"rot Graduate School
Date thesis is presented:
July 29, 1969
Typed by Charlene Laski for
Lynn Taylor Winter
ACKNOWLEDGEMENT
I would like to express my gratitude to Professor
B.H. Arnold for his proposing of this topic and for his
guidance and encouragement during the writing of this
paper.
TABLE OF CONTENTS
Chapter
Page
I
INTRODUCTION
II
FOUR TOPOLOGIES FOR FUNCTION SPACES
1
1.
General Procedure
2
2.
Four Specific Topologies
2
3.
Characterizations and Comparisons
3
of these topologies
4.
III
The Composition Map
19
RELATIONSHIPS BETWEEN A TOPOLOGY
FOR
F(X,Y)
AND THE TOPOLOGY OF
X
OR
Y
1.
29
2.
Separation of Points
Separation Involving Sets
3.
Compactness
47
4.
Countability Properties
A9
BIBLIOGRAPHY
37
FOUR FUNCTION SPACE TOPOLOGIES
CHAPTER I
Introduction
In Chapter II we shall introduce a method for topologizing a space of functions and consider four topologies
given by this method.
We then characterize two of these
topologies in terms of more familiar concepts and compare
the four topologies.
The composition map is also consid-
ered in relation to all four topologies.
In Chapter III we consider the two less familiar
function space topologies in regard to the following
natural question.
What properties of the range space carry
over to the function space?
Generally some of the separa-
tion properties are inherited by the function space and
in one case some subspaces of the function space are shown
to inherit a certain property when the whole space does
not.
For the other properties considered, examples are
given to show that the range space can have the property
without the function space having it.
A knowledge of
basic topological concepts is assumed; these topics are
discussed in the references listed in the Bibliography.
2
CHAPTER II
FOUR TOPOLOGIES FOR FUNCTION SPACES
1.
General Procedure
If
X
Y
and
topologizing
into
X
all functions mapping
Y.
Given a topological space
has for a subbase the col-
F(X,Y)
{f E F(X,Y)If[A] c G}
lection of all sets of the form
A E G
and
G E 3.
by
{f E F(X,Y)If[E] c:H}
to be a subset of
X
and
then the cor-
X
and a collection G of subsets of
responding topology on
where
There are many ways of
however in this paper we will con-
F(X,Y),
sider only the following method.
(Y,8)
denote the set of
F(X,Y)
are sets let
It will be convenient to denote
where
(E,H)
H
is understood
E
a subset of
Y.
Since any
collection of subsets is a subbase for some topology, this
F(X,Y).
method does define a topology on
2.
Four Specific Topologies
In this paper we are going to be concerned with four
different collections
6
of subsets of
X.
Letting
Gi = ({x} Ix E X ) we have that an arbitrary subbasic open
set in the topology corresponding to Gi
all those functions in
x
0
E X
F(X,Y)
into a given open set
the topology on
F(X,Y)
will consist of
which map a given point
G
0
c Y.
For this reason
corresponding to Gi
the point-open topology. G2
is called
will be the collection of all
3
infinite subsets of
the corresponding topology is then
X;
called the infinite-open topology.
Now suppose that
a set with a topology, and let
be the set of all com-
pact subsets and let
of
4
l3
is
be the set of all closed subsets
The topologies corresponding to
X.
X
G3
and
G4
are
known as the compact-open and the closed-open, respectively.
For convenience we may write
P.O., I.O., C.O.
and
C1.0.
as abreviations for point-open, infinite-open, compact-open
and closed-open, respectively.
3.
Characterizations and Comparisons of these Topologies
The compact-open and the point-open topologies have
been studied extensively and most of the results concerning
them can be found in the texts listed in the Bibliography.
The point-open topology has two equivalent characterizations
one of which is given in the following definition and
theorem.
Definition 2.1.
on
F(X,Y)
The topology of pointwise convergence
is that topology obtained by defining conver-
gence in the following manner:
converges to
{f A
(x)}
AEA
f E F(X,Y)
converges to
Theorem 2.1.
A net
{fx}AEA in
iff for every
f(x)
in
x E X
F(X,Y)
the net
Y.
The point-open topology is equivalent to
the topology of pointwise convergence on
F(X,Y).
4
in
Proof:
Suppose that
F(X,Y)
which converges to
A AEA
So
P.O.-topology and
({x0}, G)
X
0
for which
E A
X
>
A
Therefore
fx(xo) E G.
hence there exists a
implies
{f
}
X
f(x0)
is an open set in the
({x0}, G)
0
containing
Y
in
G
be arbitrarily given.
f E
in the P.O.-topology.
f
and any open set
xo E X
Let
is any net of functions
{f,}
f
0
converges to
XEA
or
({x }, G),
E
x
in the
f
topology of pointwise convergence.
Now suppose
{f
A
}
converges to
AEA
of pointwise convergence, and let
the P.O.-topology which contains
basic P.O.-open set
({x1}, B1) n .
form
sets in
A
hence
{fa
> Ai
of
Y
for
}AEA
The set
has the
B
where the
are open
Bi
for
1
A0 E A
< i < p.
1 < i < p
f
E A
1
Now since
A
such that
A
Therefore
which implies
converges to
F(X,Y)
there is a
fx(xi) E Bi.
implies
> Ai
fx(xi) E Hi
Then there exists a
Since the convergence is pointwise, corres-
Y.
A
be any open set in
f E B c H.
n ({xp}, Bp)
directed set there exists a
implies
f.
for which
B
porldingtheachx1 ,,1 < i < p,
that
H
in the topology
f
X
such
is a
> A0
> A0
implies
fx E B c G,
in the P.O.-topology.
is the same as the Cartesian product
with itself with each element of
one factor in the product.
X
Associated with a product space
is a natural map called the projection map.
map corresponding to an
giving rise to
x E X
is denoted
The projection
Trx
and is
5
defined by
for all
Trx(f) = f(x)
f E F(X,Y).
As we know
the product topology is the smallest topology on
F(X,Y)
with respect to which the projection maps are all continuous.
Hence the class of all inverse images of open sets
is a subbase for the product topology.
open set in
({x}, G)
then
Y
for all
7
But if
G
is any
[G) = {f E F(X,Y)If(x) E G}
x
Hence we have proved
x E X.
The point-open topology i5 equivalent to
Theorem 2.2.
F(X,Y).
the product topology on
We shall now compare the point-open topology with the
Note that if
other three topologies we have described.
has the indiscrete topology then given any collection
of subsets of
and any
X
sets corresponding to
(A,Y) = F(X,Y),
to
C
(A,4) =
are
so the topology on
is the indiscrete topology.
and
(I)
F(X,Y)
corresponding
Hence for comparison
purposes we shall avoid the case where
discrete topology.
G
the only subbasic open
A E
A
Y
Y
has the in-
For a similar reason we shall only
consider the case where
X
infinite-open topology.
If
is infinite in dealing with the
X
were finite the subbasis
would be empty and using the convention that the empty
union is the empty set and the empty intersection is the
whole space we obtain the indiscrete topology on
Theorem 2.3.
F(X,Y).
The compact-open topology always contains
6
the point-open topology and coincides with it iff the only
are finite.
X
compact subsets of
Since singletons are compact every subbasic
Proof:
open set in the P.O.-topology is also a subbasic open set
Now
in the C.O.-topology which proves the first assertion.
suppose that the only compact subsets of
are finite and
X
(K,G)
is an arbitrary subbasic open set of the C.O.-topo-
logy.
Since
is finite we can write
K
K =
q
q
so(K,G).({,x1,...dy,G)=.(M({xi }, G)
which is an
i=1
open set in the point-open topology.
Then let
Y.
which is not finite, call it
X
a compact subset of
be an open subset of
G
We claim the set
x
'
E X
p
open in
Y
K0
that
xi
y0 E Y
is in
G
G
11
...,G
which is not all of
there would exist
f E (K0,G)
p
Y
such that the
are
Gi
n ({xp}, GP) c (K0,G).
f E ({x1}, Gl)n
and
But since
x0
and
Y
K0.
is not a point-open open set.
(K0,G)
If it were then for any
x
Now suppose there is
is infinite there is some
x
0
E K0
such
and there is some point
x = x0
y0,
so that the function f0 (x) =
for
1 < i < p
( {x1 }, G) n.-. fl (ix 1, G
P
which is a contradiction.
)
P
but not in
(K0, G)
Hence we have proved the second
assertion.
Theorem 2.4.
Except in the degenerate cases of
having the indiscrete topology or
X
Y
being finite, the
7
P.0.-topology and the I.0.-topology are not comparable.
First we shall exhibit a set which is open in
Proof:
Let
the P.O.-topology but not open in the I.0.-topology.
G
be any open set in
x0
be any point of
then define
suppose
is a basic open set in the I.0.-topology such
B
g).
By definition of a basic open
'
and open sets
set there exist infinite sets F. c X
such that
Y
G.
1
B = (E
)
1
fl ...n (E
and
G
y1
ylf Gp.
But
f E B
a contradiction since
E
by
f
mapped inside
Gp
x0.
At least
).
q
B
will not he
possibly by
Since
f (x0) E G1 fl ...n Gs C G.
we must have
there exists a
G
El,...,Es,
renumbering, be the one; containing
f E B C ( {x0 }, G)
,
or else
x0
so let
({xo),G),
contained in
G
must contain
Ei
one of the
Since
G,
,
f E B c (ix 0 1
that
and let
Y
is the set we
({x0}, G)
Then
X.
# G
cl)
y0, yl E Y such that y0 E G and y1 E Y
x0
0' x
Cy7i
Now
so f E ({x0}, G).
f(x) =
x0
x
Pick
seek.
such that
Y
p
implies
such that
f E
1 < p < s
(Ep, Gp)
which is
is infinite but the only point
is
x0.
Therefore
({x0}, G)
is
not open in the I.0.-topology.
G
On the other hand with
as above let
(E,G)
subbasic open set in the I.0.-topology, and let
point of
(E,G).
Suppose
H
point-open topology such that
f
be any
be any
is a basic open set in the
f E H c (E,G).
Then
H
has
8
the form
({x1}, H1) fl... fl ({x }, G
infinite there exists an
xo
fx1,...,x(11.
g(x)
=
then
,
(E,G)
hence
Therefore
(E,G)
such that
g E H
since
f E H
yet
x0
H
(E,G)
which is a contradiction.
is not open in the P.O.-topology.
A topological space
Definition 2.2.
space iff for any
x
E E
is
Now consider the function
x
of
0
E
x # xo
x),
g
x
and since
)
x, y E X,
not containing
x / y
(X,a)
Tl
there exists a nbhd.
and a nbhd. of
y
is a
y
not containing
x.
It is easy to show that a space is
Tl
iff all single-
tons are closed sets and that there are spaces which are
not
Tl.
So since singletons are not necessarily closed
and closed sets are not necessarily finite the point-open
topology and the closed-open topology are not in general
comparable.
However the following result does hold:
Theorem 2.5.
If
X
contains the point-open
is a finite
Proof:
T
T1
space, the C1.0-topology
topology and they coincide iff
X
space.
1
If
is a
X
is a
T
1
space then singletons are
closed and any subbasic open set in the P.O.-topology is
also a subbasic open set in the C1.0.-topology which proves
the first part.
Now suppose
X
is a finite
Tl
space.
9
Let
be any arbitrary subbasic open set in the
(C,G)
C1.0.-topology.
Then since
finite so suppose
X
is finite
C = {x1,...,xr},
({x1}, G)n ..-n ({xr}, G)
must be
C
hence
(C,G) =
which is open in the P.O.-topo-
logy so the topologies coincide.
In the discussion pre-
ceding this theorem we ruled out the case where
not
T
so to prove the "only if" part suppose that
1
not finite.
G
(1)
was
X
Then if
then
Y,
G
is an open set in
Y
X
is
such that
is open in the C1.0.-topology but
(X,G)
not open in the P.O.-topology as was shown in the proof
of Theorem 2.4.
We shall now characterize the compact-open topology in
terms of a useful concept of analysis, namely that of uniform convergence on compact subsets.
Definition 2.3.
Let
(Y,d)
be a metric space and let
00
X
be an arbitrary topological space.
in
is said to converge to
F(X,Y)
A sequence
f E F(X,Y)
on every compact subset if for every compact
for every
e
> 0
d(f(c), f
that
n
there is an integer
(c))
<
E
for every
}
n n=1
uniformly
K c X
N = N(K,E)
n > N
{f
and
such
and every
c E K.
We shall let
C(X,Y)
denote the subset of
consisting of the continuous functions.
Now
F(X,Y)
C(X,Y)
is to
be given the relative topology and a subbase for the relative topology on
C(X,Y)
consists of all intersections of
10
with elements of a subbase for the topology on
C(X,Y)
Therefore for convenience we shall write
F(X,Y).
for
Restricting attention to this sub-
fl C(X,Y).
(K,G)
(K,G)'
space the following theorem describes convergence in the
C.O.-topology.
Let
Theorem 2.6.
X be
be a metric space and let
(Y,d)
c0
A sequence
any topological space.
converges to an
f E C(X,Y)
subset iff
f
f
n
{f
C(X,Y)
in
}
n n=1
uniformly on each compact
C(X,Y).
in the compact-open topology on
For the proof we shall need
If
Lemma 2.l.
A
such that
(X,d)
is compact,
A,B C X
is a metric space and
is closed and
B
A fl B =
d(A,B) > 0.
then
Suppose
Proof:
xn E A
exist points
then given
d(A,B) = 0
and
yn E B
such that
1
=
E
d(x
there
y
n
<
)
n
2
,
n
Now since a compact subset of a metric space is countably
00
compact we have that the sequence
fx
has a conver-
J.
n n=1
x
is also a limit point of
can choose
that
Hence
we have
x
is a limit point of
x E B.
diction, so
Therefore
d(A,B)
> 0.
1
e
E
d(x,yn) < d(x,xn) 1.- d(xn,yn) <
A fl
+
1
so, since
B
B
(I)
<
e
< 7
17-
E
E
4.
B
we
> 0,
e
d(x,xn) < f and
so large that
n
since given
B,
Claim
x E A.
gent subsequence which converges to a point
so
. E.
is closed,
which is a contra-
11
Proof of Theorem 2.6:
each compact subset of
Suppose
Let
X.
fn
uniformly on
f
be any subbasic
(K,G)1
open set in the C.O.-topology which contains
is continuous
is compact and
f[K]
f[K] c G
Lemma 2.1 tells us that there is an
d(f[K], Y
d(f(c), fn (c))
n
E
Now pick
G) = e.
<
E
for all
for all
(K,G)1
so that
N
So
n > N
implies
then clearly
c E K;
Therefore
n > N.
and
such that
> 0
E
f
f[K]
so
are a compact set and a disjoint closed set.
Y - G
f
Since
f.
f
in the
f
n
compact-open topology.
Now suppose
be any compact set and
that
f
uous
f[K]
f
n
points
in the C.O.-topology and let
f
fn -0-
e > 0
uniformly on
1 < i < n
1
in
Gi = {y E Yld(f(xi), y)
1
compact.
Again since
K
cover
f(x))
<
4 1.
<
E
n
g E
Now
Let
f[K].
and let
Now since
f
is continu-
is a closed subset of a compact set and hence is
n
We want to show that
f E
(K.
l
i=1
that
is contin-
f
such that
1 < i < n,
K. =Knf -1 rL{37EY1d(y,
K.
We want to show
be given.
is compact so there exists a finite number of
{y E Yld(y, f(xi)) < P,
ous
K.
K
B
implies
d(f(c), g(c)) <
f[Ki] c {.y E Yld(f(xi), y)
c {y E Yld(f(xi), y)
<
341
for
e
1,
G.) = B,
for all
and
c E K.
<
1 < i < n
hence
f E B.
12
Let
c
be any point of
g
is any element of
c E K.
such that
Since the
B.
cover
Ki
there exists an index
K
3
d(f(c), f(xp)) <
Thus
g(c) E Gp = {y E Yld(f(xp), y)
1 < p < n
p,
and
So
}.
<
if
K,
2e
d(f(xp), g(c)) < T and the triangle inequality yields
e
d(f(xp), g(c)) < 7 +
d(f(c), g(c)) < d(f(c), f(xp))
Therefore, for any
n > N
implies
for all
c E K
e > 0
which implies
fn E B
which implies
fn
cp
and it is defined by
topology on
F(X,Y),
uniformly on
F(X,Y) X X
There is a cannonical map from
called the evaluation map.
d(f(c), fn (c))
f
<
e
K.
Y
to
We shall denote this map by
cp(f,x) = f(x).
then
= E.
such that
N
there exists an
2e
3'
If
3'
is a given
is said to be admissible in
case the evaluation map is continuous where
F(X,Y) x X
Preceding the next theorem
is given the product topology.
we need some definitions.
Definition 2.4.
A topological space
be locally compact if every point in
X
(X,U)
is said to
has some compact
nbhd.
Definition 2.5.
A subset
S
of a topological space
is said to be relatively compact if its closure,
S,
is
compact.
In a locally compact Hausdorff space
X
the following
two conditions are equivalent to local compactness:
13
and each nbhd.
x E X
For each
(1)
exists a relatively compact open set
H
of
there
x
for which
G
xEGcffcH, and
ing
and open set
H
contain-
there is a relatively compact open set
G
with
For each compact set
(2)
C
C
C c G c G c H.
Theorem 2.7.
If
is a locally compact Hausdorff
X
space, then the compact-open topology on
C(X,Y)
is ad-
missible and it is the smallest of all admissible topologies
on
C(X,Y).
Suppose we are arbitrarily given
Proof:
a point
x E X,
and an open set
By continuity,
f(x).
taining
f
-1
[H]
in
H
is an open set in
exists a relatively compact open set
ClfT].
Then B= (ff,H)
set in the C.O.-topology containing
con-
X
(f,x)
and
(2)
such that
G
is a subbaic open
To show that
f.
is continuous it will suffice to show
nbhd. of
containing
Y
By the remark preceding this theorem there
x.
x E G c_ "ffc
f E C(X,Y),
B x G
(1)
cp[B x G] c H.
(1)
cp
is a
is clear by
definition of the product topology and (2) holds since if
(g,y)
E B x G
then
g E B
g.[] c H = g[G] c H = cp(g,y)
which implies
E H.
Hence
cp
is continuous
and the compact-open topology is admissible.
To prove the second assertion let
3
be any admissible
14
topology on
C(X,Y).
If
is any subbasic open set
(K,H)
in the compact-open topology we want to show that
So take
any point in
f
for any
cp,
x E X
x E K
x
x
1"
B =
there exists an open nbhd.
Since
cP(B xxGx )c H.
xl
in
n
then by continuity of
(K,H),
and a a-open nbhd.
B
of
x
then
Bxn'
g E B
suppose that
that
f
and
E B
then
CP(g, x) E
x
,
g[K] = cl)[{g}
than
U.
Let
.
n
is a 3 -open set.
For this,
x K] c H
since if
1 < p < n,
g E Bx
such
which implies
p
p
(K,H)
p,
implies
p
cp[Bx x
fore
g E B
now
B
B c (K,H).
then there exists an index
x E G
of
such that
E C(X,Y)
1
n
Gx
is compact there exist points
K
So it only remains to show that
x E K
f
such that KCG
U U Gx
x
K
(K,H) E U.
Gx
c H
hence
cP[{g}x K] c H.
There-
p
is g-open and the C.O.-topology i8 smaller
We now search for some conditions under which we can
compare the compact-open topology with the infinite-open
and closed-open topologies.
Theorem 2.8.
The compact-open topology is never
smaller than the infinite-open topology and they are comparable iff every subset of
X
is compact in which case
the C.O.-topology is strictly larger than the 1.0.topology.
15
Suppose that there existed spaces
Proof:
and
X
Y
such that the C.O.-topology was smaller than the I.0.then by Theorem 2.3 we would have
F(X,Y),
topology on
the P.O.-topology smaller than the C.O.-topology smaller
than the I.0.-topology which would contradict Theorem 2.4.
So we have the first part of the theorem.
Now suppose every subset of
be any subbasic I.0.-open set, then
(A,G)
A
is compact and let
X
is compact hence
A c X
implies
is a C.O.-open set which shows
(A,G)
that the C.O.-topology is greater than or equal to the
I.0.-topology.
Note that since singletons are compact the
C.O.-topology is strictly larger than the I.0.-topology.
To prove the "only if" part we shall assume that there
A c X
exists an
A
such that
is not compact and show
that the assumption leads to a contradiction if the C.O.
topology is larger than the I.0.-topology.
open set in
Y
such that
Pick any
G / Y.
q)
Let
be any
G
yi E G
x E A,
and
y2 E Y - G
f(x) =
and define
First we treat the case where
I.0.-topology and
f E
a basic C.O.-open set
Hi
1
open in
Y.
Clearly
(A,G)
)
with
K
i
LjKi = A
i=1
A.
A
is not
is open in the
By assumption there exists
such that
B
n( n ,H n
B = (K ,H )n
1
(A,G).
Since
A = X.
A must be infinite so
compact,
y2, x E X
f E B C (A,G).
compact in
or else
B
X
Now
and
(A,G)
16
n
UK.
but also
A
A
since
is not compact and we have
i=1 1
Now we treat the general case where
a contradiction.
is properly contained in
As before
X.
the I.0.-topology, and UKi D A
is open in
(A,G)
or else
A
and
(A,G)
B
i=1
n
also UK.
A
because
A
is not compact.
So let
i=1 1
K
1
"
K
s
plies that
be the ones which intersect
This im-
X - A.
n
U K.
is properly contained in
A
so there
i=s+1
existsanx0EA-UK..Hence the function
f0
i =s +l
defined by
fo(x).
(f(x),
y2,
(A,G),
x
xo
x
x0
is in
B
but not in
a contradiction, and the theorem is proved.
The following result is easily established and gives
some conditions under which we can compare the C.O.
topology and the C1.0.-topology.
Theorem 2.9.
If
X
is a Hausdorff space then the
C.O.-topology is smaller than the C1.0.-topology.
The
C1.0.-topology is smaller than the C.O.-topology iff
is a compact space.
And if
X
X
is a compact Hausdorff
space then the C.O.-topology and the C1.0.-topology coincide.
Proof:
The first statement holds since any compact
subset of a Hausdorff space is closed so that the subbasis
17
of the C.O.-topology is contained in the subbasis of the
C1.0.- topology.
For the second statement, if
X
is compact then every
closed subset is compact hence the subbase for the C1.0.topology is contained in the subbase for the C.O.-topology.
On the other hand suppose that the C1.0.-topology is
Since
smaller than the C.O.-topology.
the whole space
we choose
(P
is open in the C1.0.-topology where
(X,G)
G
Y.
is closed being
X
By assumption then
(X,G)
is a
C.O.-open set so there exists a basic C.O.-open set
which is contained in
(K1,111) n... fl (K
of
X
and the
n
,H
H.
n
)
(X,G).
B
(I)
has the form
B
where the
Ki
are compact subsets
are open subsets of
Y.
Since
n
B c (X,G)
LjBi = X
we must have
which implies that
X
i=1
is compact, being the finite union of compact sets.
The third statement is clear by the first two since
if the C.O.-topology is both larger and smaller than the
I.0.-topology, then it must coincide with it.
The last remaining comparison is between the two
topologies on which we are going to center our attention
in Chapter III.
Theorem 2.10.
If
X
is a
T
1
space then the C1.0.-
topology is larger than the I.0.-topology iff
the discrete topology.
X
has
Certainly if
Proof:
has the discrete topology
X
then every subset is closed so every subbasic open set in
the I.0.-topology is also a subbasic open set in the C1.0.topology which proves the "if" part.
For the "only if" part suppose
is a
X
space and
T1
that the I.0.-topology is contained in the C1.0.-topology.
Let
be an open subset of
G
y1 E G
Pick any
y2 E Y
and
such that
Y
G
cl)
Y.
and define
G
x E E
f(x) =
x E X
y2,
If we suppose that
where
E
X
does not have the discrete topology
is not closed.
E
C1.0.-open set
X
with
open subsets of
Hi
and
f E B c (E,G).
n(Kn,Hn)
(K1,H1)n
form
X
E
4
By assumption there exists a basic
such that
B
such that
E c X
then there exists a subset
and
is chosen as follows.
E
'
Y.
has the
B
closed subsets of
Ki
B c (E,G)
Since
we have
n
Ec()
N/ K.aricildecarinothaveequalitysincetheIC1
1
1=1
So let
closed.
K
1
"
K
be the ones which intersect
s
n
U K.
This implies that
X - E.
are
is properly contained
i =s +l 1
in
E
so there exists an
x
0
E E -
U K.
Hence the
.
i=s+1
x
function
fo
defined by
is in
f0(x) =
Y2'
but not in
(E,G).
x
x0
This is a contradiction so
have the discrete topology.
X
must
B
19
4.
The Composition Map
There is another canonical map associated with function
spaces called the composition map.
are function spaces then the composition map,
into
F(X,Y) x F(Y,Z)
w(f,g) = g°f
where
F(X,Z)
f
w,
g E P(Y,Z).
and
and
C(Y,Z)
maps
We shall
in relation to the topo-
w
logies which we are considering in this paper on
C(X,Z)
F(Y,Z)
and is defined by
E P(x,Y)
investigate the continuity of
and
F(X,Y)
If
where the product
C(X,Y)
C(X,Y),
x C(Y,Z)
is given the product topology.
Definition 2.5.
Let
X, Y,
and
be topological
Z
be a function mapping
X x Y
spaces and let
f
We say that
is continuous in the first variable if for
any fixed
f
y0
the map
E Y
fl(x) = f(x,y0)
fl: X -*
Z
into
Z.
defined by
Continuity in the second
is continuous.
variable is defined similarly and
f
is said to be con-
tinuous in both variables if it is continuous with respect
to the product topology on
Lemma 2.2.
X x Y.
If f:XxY-+Z is continuous in both
variables then it is continuous in each variable separately.
Proof:
Suppose
f
is continuous in both variables we
shall show that it is continuous in the first variable.
Let
y0
be any fixed point in
Y
and define
20
Li: X
by
Z
tion 1.5 we must show that
is continuous.
f1
be any convergent net in
{xx}AEA
has limit
Now
x0.
that the n e t
{ ( x
X x Y.
Now according to Defini-
f1 (x) = f(x,y0).
X
{x,A }
'
170) }
converges to
XEA
By the continuity of
that
f(x0, y0)
definition of
f
Therefore
is continuous.
f
1
1
{f
1
(x
A
xo
implies
in
(x0 ,y0)
we know that the net
f
converges to
{f(xx, y0)}xEA
and suppose that it
X
converging to
AEA
Let
)}
which means by
converges to
AEA
f
1
(x
0
).
Continuity in the second
variable is proved in a similar manner.
Theorem 2.11.
If
C(X,Y), C(Y,Z)
given the point-open topology then
w
and
are
C(X,Z)
is continuous in
the first and second variables separately but not necessarily continuous in both variables.
Proof:
we let
let
(F,G)' = (F,G) n c(x,Y).
( {x0 }, G)'
defined by
Fix
and
g1 E C(Y,Z)
be an arbitrary subbasic open set in
As in Definition 2.5,
C(X,Z).
{f
Following the convention adopted in Theorem 2.6
w1(f) = w(f,g1).
wi: C(X,Y)
Then
C(X,Z)
is
w11({x0}, G)'] =
E C(X,Y)Igio f E ({x0}, G)'}
= ff E C(X,Y)If(x0) E gl[G]} =
Since
g1
is continuous,
-1
g 1[G]
({x0},
gl[G]P.
is open so we have that
the inverse image of any subbasic open set under
wi
is an
Therefore because set operations are preserved
open set.
by inverse functions we have that
open set
H.
Hence
w
1
w
is open for any
[H]
is continuous or
1
is continuous
w
in the first variable.
Now fix
and let
f2 E C(X,Y)
bitrary subbasic open set in
w2: C(Y,Z)
Then
and let
C(X,Z),
be defined by
C(X,Z)
be an ar-
( {x0 }, G)'
w2(g) = w(f2, g).
w 2[({x0}, G)'] = {g E C(Y,Z)1g0f2 E ({x0}, GPI
= {g E C(Y,Z)1g(f2(x0))
E G}
= ({f2(x0)}, GP
which is an open set in
and so
w
hence
C(Y,Z)
w
is continuous
2
is continuous in the second variable.
The following example will show that with respect to
the point-open topology
be the real line with
X = Y = Z
Let
in both variables.
is not necessarily continuous
w
the usual topology, and consider the set
which is open in
C(X,Z).
Now if
was continuous there
w
would exist a nonempty basic open set
such that
w[B] c D.
product topology
sets in
where
1
C G1
and
B
2
EG 2
and
G1
and
w[B
Bl
1
x B
and
]
2
for which
B2
C w[G
are open
G2
Hence there
respectively.
C(Y,Z)
exist nonempty basic open sets
B
C(X,Y) x C(Y,Z)
B
is a basic open set in the
B
B = G1 x G2
and
C(X,Y)
Since
D = (ill, (1,2)P
1
x G
]
2
E D.
22
Now suppose
Bi
l
Bi
'' "' n
B
= (B
1
are
{1}
which is infinite.
then
B
1
E
1
) n
1
B
n (B
1
,
1
El°
..,B
1
are all equal to
p
In either case
is an infinite subset of
Y.
f E B2
for which
C(Y,Z)
H = {f(1)If E B1}
Now if we are to have
(1,2))'
(1,2))',
({l},
{l}
which is a nonempty open
0 El
set so is also infinite.
w[B 1 x B 2 ]
If none of
En).
B [{1}] = {f(1)If CB1 } = Y
then
If
is
[{1}]
1
B
must be a subset of
2
f[H] C (1,2).
implies
But
this is a contradiction since as was shown in the proof of
Theorem 2.4 sets of the form
empty basic open sets when
H
(H,
(1,2))'
contain no non-
is infinite.
Hence the
composition map with respect to the point-open topology is
continuous in each variable separately but not continuous
in both variables.
With respect to the compact-open topology
Theorem 2.12.
is always continuous in each variable separately.
X, Y,
case
compact then
Proof:
and
w
Let
Z
are Hausdorff spaces and
1
1
is locally
is continuous in both variables.
w
1
be defined as before and let
be any subbasic open set in
w
Y
In
C(X,Z).
Then
P = {f E C(X,Y)Ig 1 of E (K,G)P
[(K,G)
= {f E C(X,Y)If[K] c gl[G]l
(K,G)t
23
= (K, g i[G])'
which is an open set in
since the continuity of
C(X,Y)
Therefore
open set.
1
implies
gl
is an
g 1[G]
is continuous in the first variable.
w
For proving continuity in the second variable let
be defined as before and let
set in
C(X,Z).
Then
be any subbasic open
(K,G)'
w32-[(K,G) ' ]
w2
=
{g E C(Y,Z)Igof, E (K,G)]} = {g E C(Y,Z) Ig E (fl[K], G)}
which is an open set in
= (fl [K]
since
C(Y,Z)
fl[K]
is compact being the continuous image of a compact set.
Hence
w
is continuous in the second variable.
For proving continuity in both variables fix
and
fl E C(X,Y)
gl E C(Y,Z)
subbasic open set containing
uous
f
f
1
1
[K]
[G] c:Y
g
1
[G].
giofl.
g
Since
1
Y
f
1
be any
(K,G)
Since
is open and since
is compact and
[K] cg
and let
fl
E (K,G)'
gl
is contin-
is continuous
implies
is a locally compact Hausdorff
space we know by the remarks preceding Theorem 2.7 that
there exists a relatively compact open set
f
1
[K] CH
H
cgl[G].
1
Therefore
are subbasic open sets containing
and it is clear that
suppose that
E
w[(K,H)' X
(K,H) '
fl
and
B
and
gl
(T, G)'
respectively,
(H, G)'] c (K,G)'.
is any open set containing
there is a basic open set
such that
H
such that
gio fl
Finally
then
glo fl EBcE
24
of
K.
and open subsets
X
wi(K
such that
1
E (K., H
i
H
and
)'
is an open set in
P
(Th (H., Gi)'
of
G.
As above for each
Z.
Hi
there exists a relatively compact open
Ki, Gi
pair
f
for some compact subsets
B = (K1, Gi) ' n ---n(Kp, Gp) '
and
)
g
(g
x
'
i
1
') c (K.
Gi)
E (Hi, Gi)
containing
C(X,Y)
f
and
1
C(Y,Z)
H.)' x n(17.
1
1
i=1
w[r) (K.
1
i=1
= w[r1{(K., H.)' X
A = n(K
Hence
H )'
i=1
is an open set in
and furthermore
.
where
G1)
B =
containing
gl
G.).]
1
GiP1
G.P}] c rTh cot (Ki,Hi)
1=1
1=1
(K., G1)' = B c E.
Therefore given any open set
1=1
containing
E c C(X,Z)
set, namely
(fl, g
)
1
n
there exists a basic open
f1
P
(K., Hi)' x
a.
fl
containing
(H., G1)',
1=1
which is mapped inside
E
by
ci),
hence
w
is
continuous in both variables.
With respect to the infinite-open
Theorem 2.13.
topology
w
is continuous in the first variable but not
necessarily continuous in the second variable.
Proof:
Let
w
1
be defined as before and let
be any subbasic open set in
C(X,Z).
Then
(A, G)'
25
1
1
w
[(A,G)'] = {f E C(X,Y)Ig 1 0
E (A,G)'}
f
gi[G] l
= {f E C(X,Y)If[A]
= (A, g
C(X,Y)
w
g 1[G]
since
1
which is an open set in
[G])1
is open by continuity of
Hence
g.
is continuous in the first variable.
We now give an example to show that
Let
tinuous in the second variable.
real line with the discrete topology.
the function
f
2
(x) = 0
for all
any nonempty proper subset of
basic open set in
C(X,Z),
Z.
but
may not be con-
w
X = Y = Z
Let
x E X,
all be the
f2 E C(X,Y)
and
let
Then (X,G)'
wl[(X, G) '
]
G
be
be
is a sub=
E C(Y,Z)1g0f2 E (X, G)'} = {g E C(Y,Z) Ig E ({0}, G)}
{g
which we showed in the proof of Theorem 2.4
= ({0}, G)'
is not an infinite-open open set.
tinuous or
w
Hence
w
2
is not con-
is not continuous in the second variable.
We note that in view of Lemma 2.2
w
will not, in
general, be continuous in both variables.
Theorem 2.14.
w
With respect to the closed-open topology
is always continuous in the first variable but not nec-
essarily continuous in the second variable.
26
Proof:
Let
w
I
be defined as before and let
be any subbasic open set in
{f E C(X,Y)1g10 f
C(X,Y)
w-1[(F,G)'] =
=
E
= if E C(X,Y)If[F]
is open in
Then
C(X,Z).
(F,G)'
=
hence
w
(F,
gl[G]s'
which
)
is continuous in the first
variable.
The following example shows that
tinuous in the second variable.
w
may not be con-
X = Y = Z
Let
real line with the usual topology and let
all be the
x
f2(x) =
x
Then if we take
C(X,Z)
{g
w2
is
(-1,1))}
E C(Y,Z)1g0f2 E ([0,c0),
= {g E c(Y,Z)Ig E (f2[[0,00)],
= {g E C(Y,Z) 1g E ([0,1),
+ 1
for our open set in
(0,00), (-1,1))'
the inverse image under
2
(-1,1))1
(-1,1))1.
We now claim that this set is not open in the closed-open
topology, thus
w2
E = ([0,1), (-1,1))'
is not continuous.
Suppose
was open in the closed-open topology.
Then there would exist a basic open set
(id)E
B CE where
(id)
B
such that
denotes the identity function on
the reals which is a point of
E.
B
has the form
(CI,G1)'11-11(Ck,G0'wherethe.C.are closed subsets
ofY=R1andtheGiare
open subsets of
Z = R
1
.
27
k
k
LJC.
Clearly
contains
and since
[0,1)
is
1
i=1
i=1
aclosedsetatleastoneoftheC.must contain the
pointl.IfalloftheC.'s contain
1
if not then choose
e = 7 ,
containing
c
0
>
Then for
1.
C
1 <
,
there exists an
< p
j
min
Now
} }.
C
l<3<p {e.
'
henceGp+1 , .
1
be the ones not
Ci,...,Cp
(1 -e, 1 + e.) n Ci =
such that
takec=min{-12
in the following manner:
e
possibly by renumbering let
then let
1
.
. ,Gk
p+1"
in which case
cl)
..,C
all contain
k
are all open sets containing
n
1
k
since
Therefore
E B.
(id)
is also an open set
G.
i=p+1 1
containing
and as such there exists a
1
n
for which
> 0
cS
k
6, 1 + 6) C
(1
i=p+1
define
G.
Now let
11
= min{e,6}
and
1
as the following function:
h
x, x< 1 n
h(x) =
2x +
-,_
1 < x
or
(1)-1), 1 - n < x < 1
1
We claim
.
2 < x < 1
is a continuous function which is contained in
B
Continuity and the fact that
h
contained in
E.
easily checked, so let us show that
h E B.
but not
are
E
The only values
of
x
for which there is any question are those between
1
n
and
1
since for all other values of
with the identity function which is in
x
0
E (1
n, 1).
Since
n
h
n
1
,
li
<
e
B.
x,
h
agrees
So suppose
we know that the only
C.
28
x
that
but since
h(x0)
E
p + 1 <
can be in are those fOr which
j
< k,
0
n <
S
we have that
(1 - n, 1 + n) c (1 - 6,
P+ 1< i< k
hence
h E B.
1 + 6) c G.
Therefore
B
for
E
and this
contradiction proves the theorem.
Again Lemma 2.2 rules out continuity in both variables.
29
CHAPTER III
F(X,Y)
RELATIONSHIPS BETWEEN A TOPOLOGY FOR
AND THE TOPOLOGIES OF
OR
X
Y
We now ask the question as to how the properties of
when
F(X,Y)
influence the space
Y
and
X
the topological spaces
is given the infinite-open or the
F(X,Y)
In this connection we specifically
closed-open topology.
attempt to answer the question:
perty then does the function space
has a given pro-
Y
If
also have this
F(X,Y)
property?
1.
Separation of Points
is a
T
o
space iff
is a
Y
f, g E F(X,Y)
with
point
for which
and
fore
E X
{f(xo )} c
{f(x0)}c
denotes
Case 1:
set.
Y
Y
{f(x0)}.
x0
E S,
Now
is
g
{f(x0)}
T1
,
Y
there-
where
We now have three cases:
(S,{f(x0)}
open open set which contains
Y
{f(x0)}c}
S = {x E Xlg(x) E
Then since
since
there exists a
are open sets in
{g(x0)}c
and
g
f
space and
T1
f(x0) / g(x0).
are closed in
{g(x0)}
is a
Y
Since
g.
f
space.
T1
First suppose that
Proof:
xo
with the infinite-open topology
F(X,Y)
Theorem 3.1.
c
)
but not
is an infinite
is an infinitef.
30
In this case
but not
is an open set containing
(T, {g(x0)}c)
f
g.
Both
Case 3:
and
S
consider the set
X
is infinite.
T = {x E Xlf(x) E {g(x0)}c}
Case 2:
It contains
g(x) = f(x 0
not contain
since
f
)
x E Sc
0
is equal to
finite number of points and hence
finite set into
{g(x0)}c.
then
However this set does
E {g(x )}c
f
is finite and
S
is infinite so this set
Sc
since if
g
f(x 0
and
)
Since
(Sc, {g(x0)}c).
is infinite we have that
is open.
are finite in which case
T
except at a
g(x0)
cannot map any in-
f
Therefore
F(X,Y)
is a
TO
space.
with the infinite-open topology and that
any two distinct points of
set containing
yl,
f (x) =
x0
x
x
and
but not
y2
0
is any fixed point of
containing
F(X,Y)
(S
infinite subsets of
S.
1
,H
X
1
but not
f
X
contains
x
0
H,
Suppose
being a basic open
H
) nn (S n ,H n
and the
g.
then there is a basic
G,
such that fEHcG.
set has the form
Now if no
are
We now show that there is
there was such an open set, say
Y.
y2
Consider the two functions
yl.
for all x E X.
no open set in
H
and
x00
g(x) = y2
open set
yl
space
To
Claim we can find an open
Y.
where
is a
F(X,Y)
On the other hand suppose that
)
where the
are
S,
are open subsets of
then clearly,
g
E H
since
31
except
g = f
but this implies
x0,
at
So possibly by renumbering let
is a contradiction.
be the ones containing
S1,...,Sp
which
g E G
but each
x0,
is
Si
infinite so they must all contain points different from
Since
x0.
E (S., H.)
f
J
1 <
for 1< j
g E (sj, Hi)
F(X,Y)
that
taining
g
T
is
but not
Pi,...,Pr
B.
such that
y2,
hence
which implies
g E H
and d
A
con-
such that gEBcA. Let
)
r > 1
xo, and
since
must be open sets containing
y2
but not
y2
is an open set containing
B4
or
So there exists a basic open set
f.
Bi,...,Br
n
H.,
Therefore since we have assumed
be the ones containing
Then
the sets
< p
j
there must exist an open set
0
B = (P1,B1) n..-n(p m ,B m
f
< n
a contradiction.
g E G,
1 <
must contain both
< p,
j
for
J
i=1
yl
or else
f
E B.
Hence
Y
T1
is
as desired.
The above proof actually shows more.
there is no open set containing
properties of
and
g
Y
f
In showing that
but not
g
no special
were used and since the two functions
can be constructed for any spaces
X
and
Y,
f
we
have also proved
Theorem 3.2.
is never a
T
1
F(X,Y)
with the infinite-open topology
space.
We now turn to the same question when
the closed-open topology.
F(X,Y)
is given
Remembering that in order to
define the closed-open topology it was necessary that
X
32
be a topological space we might suspect that the properties
of the topology on
X
might affect the results and we see
that they do.
Theorem 3.3.
a
T
space if
0
is a
X
The condition that
and if
Y
Y
space and
Tl
is a
T
but not
To
is
with the closed-open topology is
F(X,Y)
is a
Y
space.
To
space is also necessary
o
then
Tl
being
X
is both
Tl
necessary and sufficient.
First assume that
Proof:
is a
T
space, and that
0
f(x
)
Since
X
F(X,Y)
F(X,Y)
is
T
functions
f(x) = 171
Suppose that
is a
F(X,Y)
and
contain
is a
T
g
y2
0
B
,
G
g(x0).
but not
f
f,
T
0
is
so
g
space and let
Y.
Since
F(X,Y)
and not the other one.
then there is a basic open
at least one of
such that
131,...,Bp
f E B c G.
must not
but all are open sets containing
space.
such that
which contains one of the
g(x) = y2
set B= (C1, B1) n...n (cp, Bp)
Now since
E X
({x0}, H)
so
X
which contains
contains
G
0
but not
f(x0)
there exists an open set
0
x
there exists an open
0
be any two distinct points of
171, y2
Y
T0.
Second assume that
is
T
is
space and
1
are any two distinct
g
T1, {x0} is closed in
is
an open set in
that
Y
which contains say
H c Y
set
and since
g(x0),
)
0
and
T
Then there exists an
F(X,Y).
points of
f
is a
X
171
so
Y
33
Now we want to show that if
being
F(X,Y)
then
not
T
yi
y2
1
Y
implies
To
T
is
X
0
but not
T
Since
Y
Tl.
is
there exist two points
y1, y2 E Y
and any set containing
171
also contains
where
and
is any fixed point in
x0
there is no open set containing
cannot be separated from
F(X,Y)
is
T
and c G.
f E B c G
Bk.
there is an open set
0
y2
and not
Then since
f E
which means that
was arbitrary,
{x
Clearly
since
f
1
then
F(X,Y)
Example 3.2.
topology
not
T
1
a = fx,
B1,...,Br
we must have that
(Ck, Bk)
is a closed set in
T
must
Ck = {x0}
Since
X.
since
closed sets.
of
F(X,Y)
To
Let
he the set
{c}
If
f
X
without
{b,c}, {a,c },
= {a,c},
and
x0
0
space as we wanted to show.
1
can be
(1),
Since
Y.
The following example shows, however, that if
T
yl
or else gEBcG, say it is
y1,
is a
X
X.
such that
At least one of
}
x0
containing a basic
G
B = (Cl, yn .-.n (Cr, Br)
open set
contain
and not
g
x
by open subset3of
y2
y2.
xo
f(x) =
Y2
g(x) = 171
is
such that
x
Now consider the two functions
1
but
being
X
{a,b,c}
{c}, {b}}.
and
{a}
Y
is
T1.
with the
Then
{b}
X
is
are
are any two distinct points
g
then there is some point at which they differ.
First suppose that either
in which case since
Y
is
f(a)
T
1
g(a)
either
or
f(b)
g(h)
({a},{f(a)}c)
or
34
is an open set containing
({b}, {f(b)}c)
but not
g
so that
f(b) = g(b)
Next suppose
f(a) = g(a)
and
g(c).
Then either
f(b) = g(b) E {f(c)}c
f(c)
or both so either
f(b) = g(b) E {g(c)}c
which is an open set not containing
g
which is an open set not containing
f.
is
T
The function space
Theorem 3.4.
the closed-open topology iff
and that
Since
and
f
Y
T
is
g
but not
g,
y2
in
Tl.
is
Y
and
Y
({x
F(X,Y).
is a closed set in
is an open set in
{g(x )}c)
1
0
0
'
then there exist
171
suppose that
Y
is
is such that
T1
contain-
and
we ob-
g
is a
T
1
Tl
space.
and
y2
in
Y
space.
If
Y
such that
also contains
so we have a contradiction.
but
but
f
y2.
cannot be separated from
f(x) = 171
g(x) = y2
Y so
which contains
yl
X
F(X,Y)
is a
and any open set containing
the function
f
F(X,Y)
F(X,Y)
Tl
is
for which
By interchanging
f.
Hence the function
E X
Tl
is
is an open set in
so we have shown that
Tl
and
T1
E X
0
{x0}
0
Second suppose that
is not
x
{f(x )}c
1,
({x0}, {f(x0)}c)
tain the open set
x0
F(X,Y)
are any two distinct points of
g
Furthermore,
g(x0).
and since
Y1
Therefore
Tl
is
X
there exists an
g
f
f(x0)
not
( {b,c }, {g(c)}c)
g E ({b,c},{f(c)}c)
or
F(X,Y)
is
X
Suppose first that
Proof:
ing
f E
or
and it is not possible to improve on Theorem 2.3.
o
that
f.
X
is not
{x0} is not closed.
T1,
Now
so some point
Pick any
171, y2
35
distinct points in
and define the two functions
Y
xo
x
Since
ing
x0
x
Y2'
F(X,Y)
T
is
but not
f
there is an open set,
1
f
must contain
but not
y2
This implies that
{x }.
Let
then at least one
x0;
this implies that at least one of
E B
be
f E B C H.
with
be the ones containing
B1,...,Bp
of
contain-
H,
which contains a basic open set
g
B = (C1, B1) n...n (Cs, Bs)
C1,...,Cp
x E X.
for all
g(x) = y2
and
f(x) =
But since
y1.
must
C1,...,Cp
is closed which is a
{x0}
contradiction.
Theorem 3.5.
The space
is a
X
and
I
space.
Suppose
Proof:
f
T
is a Hausdorff space
Y
topology is a Hausdorff space iff
and
with the closed-open
F(X,Y)
X
Tl
is
and
is Hausdorff with
Y
F(X,Y).
any two distinct points of
g
a point in
for which
X
f(x0)
g(x0)
and since
a Hausdorff space there are disjoint open sets
with
f(x0) E F
and
g(x0)
sets
({x }, F)
and
({x }, G)
F(X,Y).
Because
which together with
implies that
G =
F fl
f E
X
F
is
be
xo
Y
is
and
G
Tl
the
are subbasic open sets in
we have
({x0}, F)
( {x0 },F) fl ({x0},G)
and
g E
({x0}, G)
is a Hausdorff space.
F(X,Y)
Assume now that
(I)
Now since
E G.
Let
Y
is not a Hausdorff space.
So there
36
are points
y2
y1
in
in
f(x) = yl
We claim that the functions
disjoint open sets.
cannot be separated by disjoint open sets
g(x) = y2
and
which cannot be separated by
Y
If they can be separated by disjoint open sets
F(X,Y).
then they can be separated by disjoint basic open sets so
suppose
and
B
are such that
E
E = (T1,E1) 11 ..-n (T
and
,E
f E (u Si, nBi) c B
Now
E = (I).
1=1
q
q
j=1
j=1 3
y0 E
Hence the function
is not
is not a Hausdorff space.
F(X,Y)
so
E
T
is a Hausdorff space but that
Y
and show that assuming
1
0
for which
E X
{x
x
xo
X
X
0
Y2'
Now define
is not a closed set.
}
and
f(x) =
to be Haus-
F(X,Y)
There is some point
dorff leads to a contradiction.
x
h (x) = yo
j=1 3
We now assume that
X
respectively, so there exists
y2
[(--) Bi] n [ nE.] .
B fl
are open
E.
j=1 3
q
1=1
is in
and
y1
p
a
q
and
1=1
sets containing
with
and
1=1
p
but n Bi
g E (UT., (1 Eo) c E
g E E
f E B,
q
q
B fl
and
)
n(Sp,Bp)
B = (S1,B1) fl
where
g(x) = y1
y1
and
y2
0
are any distinct points of
Since
Y.
is a
F(X,Y)
Hausdorff space there is a basic open set
B = (C1,131) n... fl (Cs,Bs)
C
of
...,C
p
f E B
are the ones containing
131,...,Bp
f E B
with
must contain
at least one of
C
x0,
and
C
p
B.
Suppose
then at least one
but not
y2
g
must be
Since
y1.
{x
0
}
which
37
implies that
2.
is closed which is a contradiction.
{x0}
Separation Involving Sets
We now consider two of the separation properties involving sets and ask if they carry over from
for any point
and any closed set
x
there exist disjoint open sets
x E G
F
and
G
F(X,Y).
to
is regular iff
X
A topological space
Definition 3.1.
Y
not containing
x
such that
H
F c H.
and
An alternate formulation which we shall find useful is
that a space
is regular iff for every point
X
containing
G
every open set
x
and
there is an open set
x
H
such that xEHcirc G.
In the next example we shoal that with the infinite-open
topology
may not be regular even if
F(X,Y)
Y
is a
regular space.
Let
Example 3.3.
topology and let
Y
be the set of real numbers.
X
the subbasic open set
f(x) = 0.
Then
closed set in
is in
TET
f
be the reals with the usual
and
F(X,Y).
1
and the constant
are a point and a disjoint
1,
x = h
g(x)
=
The function
0,
15
x
Sc
(
since as we showed in the proof of Theorem 2.1
there is no open set containing
TIT
1
(0,1), (-y, -2.))
Consider
g
but not
is the smallest closed set containing
f.
f
Since
and
38
g E
we see that
tfJ n Sc
is not regular.
F(X,Y)
Now we turn to the same question when
given the closed-open topology.
shows that
regular.
F(X,Y)
The following example
being regular does not imply
Y
is
F(X,Y)
is
We will show more than is needed here as it will
be useful in this form later.
A topological space
Definition 3.2.
for any two disjoint closed sets
disjoint open sets
and
G1
and
F1
is normal iff
X
such that
G2
there exist
F2
F1 s G1
and
F2 c G2.
Another formulation which is easily seen to he eauivalent is that a space
is normal iff for any closed
X
set
F
and open set
set
H
for which FcHcric G.
G
containing
there is an open
F
We give an example in which
Example 3.4.
is regular
Y
but
F(X,Y)
Let
X = {1,2,3,} with the discrete topology and
with the closed-open topology is not regular.
1 1 1
= {0,1,7,7,T,...},
where
topology from the reals.
Suppose
space.
The point
sets.
suppose that
and
F1
First we show that
F2
Fl.
1
fore
{0}
and
F1
This implies that
set since each singleton except
F
is a normal
Y
are any two disjoint closed
cannot be in both
0
0
is given the relative
Y
F1
F2
is an open
is an open set and
is the union of the singletons contained in it.
F
1
and
F
I
so
are disjoint open sets that
F
1
There-
c F1
and
39
F
2
c Fc
hence
,
1
is a normal space.
Y
Furthermore
Y
is a Hausdorff space being a subspace of a Hausdorff space
which implies that
f(n) =
be defined by
S = (X,Y
that
closed-open topology and
a set
and let
n E X,
is not regular since
F(X,Y)
f E F(X Y)
f EGccS. This will prove
such that
G
for all
-in=
Now let
We claim that there does not exist any
{0}).
open set
is also regular.
Y
Suppose there was such
E S.
f
is open in the
-S
then it would contain a basic open set
G
would also have the property fEEcfcS. And
the form
sets in
(K1, G
X
n... n (K
)
1
and the
G
s
,G
where the
)
s
are open sets in
i
we know that U Ki = X.
K
Y.
which
E
has
E
are closed
i
Since
F c S
Now pick a subsequence of the
i=1
sequence,
of positive integers which is
(1,2,3,...),
eitherinoroutofeachKoas follows:
step either
K1 n (1,2,3,...)
it is infinite take
N
= K
1
N
sequence of
(1,2,3,...)
1
N2 = K2 n N1,
take
N
so
is either in or out of
of
N
1
2
K1,
N1 n K2
s
and if it is
1
= N
N2
K2.
1
K
2
.
is a subK1.
is finite or infinite.
and if it is infinite
is a subsequence of
And since
it is also either in or out of
this process for
N
so
If
which is either in or out of
As the second step either
If it is finite take
is finite or infinite.
n (1,2,3,...)
1
= (1,2,3,...)
finite take
For the first
1
which
is a subsequence
N2
K
N1
.
Continuing
steps we obtain a subsequence
N = Ns
40
Now since the
which has the desired property.
cover
1 < i < s,
the subsequence
X
Ki
,
must be in at
N
leastoneoftheKRelabeltheK.'s if necessary so
where
f(x
)
Kl,...,Kp
is in
N
that
1
E G
xm
xj E Ki,
x E K.
,
Define a new function
x
N
We claim that
then
N
x
.
x E N,
so let
gk
B =
fl
1=1
containing
secting
N.
h.
set containing
Li n N
cl)
1
H)
i
Suppose
Then
0.
for
since
must be between
i
E G..
f (x)
,
Let
h
x 4 N
x E N
be
We
in the closed-open topology on
h
(L.
1
as follows:
g [K.] c G.
q
then
h(x) =
another function defined by
We now
In the first place if
1
landparuatherl
xq EG.so g (x)
claim that
gk
gq (x) = f(x) E G.
Secondly if
(Ki, Gi).
since
f E E.
and
S.
q = 1,2,3,...
and
and
= 1,2,3,...,
xEN
1
1 < i < S
m = 1,2,3,...,
for
<
RI
xk
then
10,
rf(x),
gk (x)
Kp+1,...,Ks
is out of
N
N = (xl,x2,x3,...)
n...n G
1
1 <
prove that
f E
Suppose
1 < P < S.
=
and
F(X,Y);
be an arbitrary basic open set
Li,...,Lq
are the ones inter-
fl Hg = U
0 E H1 (1
If there is no
then take U=
1).
j
which is an open
such that
44
The sequence
j j=1
decreases monotonically to
such that
m > M
implies
0
x
so there exists an
E U,
hence if
M
m > M
then
41
Therefore
(Li, Hi) .
gm E
which implies
gk
i=1
E fl
Sc
h E E
since
(I)
and since
h E Sc,
and
was
G
arbitrary we know that there is no open set containing
whose closure is contained in
S.
In particular we have shown that
not imply
under certain conditions
Y
being regular does
We can show, however, that
is regular.
F(X,Y)
with the closed-open
F(X,Y)
Let
topology does have some regular subspaces.
denote the subspace of
Cl(X,Y)
consisting of the closed
F(X,Y)
As before since a subbasis for the relative
mappings.
topology on
B = (C,G)
open in
fl
Y,
where
Cl(X,Y)
If
Let
is closed in
C
we shall denote
normal space then
Proof:
consists of the class of all
Cl(X,Y)
Theorem 3.6.
is a
X
by
B
be a point of
is a
Y
contained in
Cl(X,Y)
Now since
is a
(F,G)*.
closed mapping
is a closed set contained in
f*fFJ
By the normality of
g
f*[F] c E c E c G,
(F,E)*
g(x0)
f,
hence
then there exists an
hence
f*
there exists an open set
Y
is
is a regular space.
Cl(X,Y)
f*
G
(C,G)*.
space and
T1
and
X
the subbasic open set
that
f
({x6}, Y
f)*
f* E
x
0
(F,E)*.
E F
E
G.
such
If
such that
is a closed-open open
and is disjoint from
set which contains
g
we have shown that
(F,E)* = (F,E)* c (F,f)*c (F,G)*.
(F,E)*.
Thus
Now
42
let
G
be any open set containing
hence
f*,
B = (F1, G1) *n-"n (Fn, Gn) *
a basic open set
tainsf*.ToeachG,there corresponds an
and we set
We know that
for
< n
1 <
F
17 =
E ) * n ..-n (Fn, E
1'
1
,
f1) * II
Y
n
(F
E
1,
=
) *
(Fi, Ei)*
f* E
is always true.
c
)*n.-- n (F n' E n )*
(F
E
1
n (F
) * n
n'
c
E )*
n
n (Fn, Gn) * = B c G.
is regular.
Cl(X,Y)
be a topological space with a distinguished
point which we shall call
0.
of all continuous functions,
there exists a compact set
x E Kc
1
as above
E,
since
" 11 (Fn, fn)* c (F, G1) * n
Thus the subspace
Let
H c
as above and
Moreover, we find
(F
f* E H
which con-
and claim that
H = (F1,E1)*n... n (Fn, En) *
f* E H c E. c G.
contains
G
implies
We let
f,
from
K c X
be the class
Cl(X,Y)
X
to
Y
for which
with the property that
f(x) = 0.
Theorem 3.7.
If
X
is a Hausdorff space and
regular Hausdorff space then the subset
regular subspace of
F(X,Y)
Cl(X,Y)
is a
Y
is a
in the relative closed-open
topology.
Proof:
(F,G)0 = (F,G)nco(x,Y)
For convenience we let
denote an arbitrary subbasic open set in the relative
topology, where
Let
f*
F
is closed in
be an arbitrary point in
X
and
Cl(X,Y)
G
is open in
and let
K
Y.
be
43
Suppose
f*(x) = 0.
implies
x E KC
a compact set for which
is any subbasic open set containing
(F,G)0
we want to show that there is an open set
f*,
with
E
f* E E C frc (F,G)0.
Case 1:
set
K.
F fl
where
cp,
it is also
off of
0
K°.
is in the boundary of
c
.
Since
x.
x E K
Suppose
Now since
{0}
set containing
an open set
K
(F,G)0 we know
y E f*[F]
f*[F].
G
is an open
is a closed subset
and hence
F
is compact.
f*[F] c G
and the
f*
E = E
Yl
Since
so by regularity of
there exists an open set
Since
So we have
F
E 's
U UE Yr
E
Y,
such
form an open cover-
is continuous
is compact
f*CE1
covers
E
Y1
Letting
and
f*(x) = 0.
c (F,G)0.
and so some finite subcollection
f*[F].
which
Here
that yEE Y cfY cG,
ing of
Y
0 E E c E c G.
for which
(Kc) = (p.
of the compact set
for each
c
we know by regularity that there exists
0
E c Y
F n
intersects
x
we know that
is a closed set in
(F,E)0 c (F, E)0 c (F,E)
Case 2:
x
then
K°
K
then
K
is continuous it must preserve
f*
f*[Kc] = {0}
this limit and since
f* E
off of
0
Hence we can choose a net of points in
converges to
f* E
is
f*
so every nbhd. of
K,
and for
f *(F] = {0},
First we want to show that
K.
this it will do to show that if
K
is the interior of the
K°
Yn
we note that
44
f*[F] c E
yn
Y1
({x
0
and which
g
'
So if
(F,E)0.
is disjoint from
g
therefore
is an open set containing
E)0
Y
}
then there
(F,E)0,
g
g(x0) f E,
for which
x0 E F
exists an
yn
Y1
and
g E C0(X,Y)
Now suppose that
UUf c G.
UE
E = E
and
c (F,G)
(F,E)0 c (F,E)
(F,E)0,
hence
Case 3:
F n K°
and Fn (Kc)
cl)
then
(F,E)0
g 4
Since
ci).
is P
K
compact subset of a Hausdorff space it must he closed and
is open
since
K°
F n K
and
more
Now since
(K°)c) fl
of open sets
f* E
(F fl
K°
(1)
E'
and
= ((Fe K)
(K°) c, G)
K, G)0 n (F fl
= (F,G)0.
(F fl
are closed subsets of
F n (K°)c
(F fl
is also closed so we know that
(K°)c
(F n K) fl (Kc) =
X.
Further-
U (F
fl
(K°)c,G)0
and
(1)
cases 1 and 2 guarantee the existence
E"
such that
Y
in
(F fl K, E')0 c (F fl K, E')0 c (F fl K, f')0
K,
and
G)
c,
(F fl
,
f* E
E")0 c (F n
(Fn (K0 )c,
(K°)
c
,
arbitrary open set containing
open set
B = (F1, G1)
E") c (F n (K°)c, E") 0
G)0.
f*
Now suppose
such that
ne..n (Fn, Gn)
(Fi, Gi)0 's if necessary,
Now rearrange the
so that for
1 < k < p - 1
we have
Fk n K° =
and for
p < k < q - 1
we have
Fk fl
and for
q < k < n
we have
Fkfl K°
k
such that
is an
so there exists a basic
f* E B c G.
As above for
G
q < k < n
Kc
(I)
(1)
(P and Fk n
we can find
K
Ell(
C
FY
and
45
open sets in
E"
for which
Y
(Fk n K, E00 n (Fk n (K°)c, E00 c (Fk n K, Ek
n (Ko)c, E)(3 c (Fk n K, G)0 n (Fk n (K°)c, G)0 = (Fk, G)0.
Now we claim that the open set
11-1
H= n (F.,
c(Fi n K,Ei)0 n (Fin (Ko)c,EI)0)
n
E4)-0.]
i=q
has the desired property which is
That
H c B
=
f* E H
and
H c IT
f* E H c H c B c G.
are easily checked so we show
by the following computation:
n
c
(F.,E.)inp
0
._
1-q
(F.
1'
n
E.1
)]
n[
0
(F. E.)o n [
k(F.
,E!) 0
n (F. n (K°)c, B7)
1
nK,E!) 0
0
)1
n(F. n(K°)c, En0)1
n t(F, nK,E!) o n (Fin
(Ko)c, Eno ]
i=q
-1
n
(F.
1
,
ffi)
0
n
-q-1
n (F4 ,Gi)c]
n
[an
[n
i=q
=1
Therefore
n
i=q
KIT
n
)cq -1
[n
((F1 n
C
0
(X Y)
'
n
KC
(F. Gi)
,1)
0
n
(Fin
(K°) c,
Th)0)]
= B c G.
is a regular subspace of
F(X,Y).
46
We now turn to the question of normality and see how
Example 3.4 shows that
imply that
being a normal space does not
Y
is normal either with the infinite-open
F(X,Y)
or the closed-open topology.
With
Lemma 3.1.
and
X
1
f(n) = 171-
the function
an
n0
for which
F(X,Y).
and
g E F(X,Y)
Suppose
g(n
Example 3.4
is closed as a singleton in the
infinite-open topology on
Proof:
defined as in
Y
g
1
n
0
so there exists
We now have two
f(n )= --.
)
0
f
0
cases:
Case 1:
g-1[ {-1-}
this case the set
set containing
at
n
but
0
f.
is finite which implies
is infinite. Since
and
{i}c
no
0
f(n
n0 E g
since
G
f
=
)
0
1
n
0
is open the set
is an open set in
{-1}c)
n
no
g E G
but not
g
0
g-l[{ 1
}c]
no
G =
no
A.
}c.
{
n
-1
c
Therefore
set.
S
f
is
{f}
F(X,Y).
S = (X,Y
is open in the infinite-open topology since- X
and
g
0
Now in terms of Example 3.4 the set
{f}
Clearly
F(X,Y).
because
no
closed in the infinite-open topology on
set so
In
X.
is an infinite-open open
), {-1})
n0
(g-1 [{ -1-}
g-lr{ 1 ),
/j
n
Case 2:
is an infinite subset of
]
no
{0})
is an infinite
are a closed set contained in an open
When we showed that there was no open set
G
such
47
that
{f} cGcdcS _we were working with the closed-
open topology on
F(X,Y)
and the discrete topology on
Hence the closed subsets of
X
X.
were all the subsets so
the closed-open topology contained the infinite-open
Thus it is impossible to find such an open set
topology.
in the infinite-open topology, therefore
G
may
F(X,Y)
not be normal with the infinite-open topology even if
Y
is normal.
Example 3.4 also shows this result for the closed-open
topology on
is a
T
F(X,Y).
With the discrete topology on
space and recalling that
1
Theorem 3.4 says that
Therefore
topology.
shows that
{f}
and
F(X,Y)
{f}
Sc
is
is also a
Y
T
it
X
1
space,
in the closed-open
1
is a closed set so the example
are disjoint closed sets which
cannot be separated by disjoint open sets hence
F(X,Y)
is not normal in the closed-open topology.
Compactness
3.
We now give the following example to show that if
is a compact topological space then
F(X,Y)
Y
with the
infinite-open topology need not be compact.
Example 3.5.
and
f
f
Y = {a,b}
E F(X,Y)
-1
[{b}i
Let
X
be the set of positive integers
with the discrete topology.
either one or both of the sets
must be an infinite subset of
X.
Now for every
f-1[ {a}]
and
In case both
48
and
f-1[ {a }]
are infinite sets define
f-1[ {b}]
W(f) = (f
infinite then
if just
(f-1(ibl], {b}),
(f-1[ {a}], {a}) n
-1
f-1[ {b}]
is infinite then
f E W(f)
for every
W(f)
is
f-1[ {a }]
and if just
{a}),
Clearly
W(f) = (f-1({b}], {b}).
so U=
f E F(X,Y)
as
W(f)
is
fEF(X,Y)
an open cover of
It will suffice to show that
F(X,Y).
U
has no finite subcover so suppose that it does, say
Possibly by rearrang-
F(X,Y) E W(fl) U W(f2)U ...UW(fn) .
are
be the set of those
S1 = {f1,...,fp}
ing let
at only a finite number of points of
b
S2 = {fp+1,...,fq}
contain those
f's
S3 =
and let
X
and
X
f's which ate a at only
contain those
a finite number of points of
which
f's
which are not in
or
S1
52
At least
.
one of these sets must be nonempty so suppose it is
S1.
Let
subset of
defined by
xo EX
We claim that the function
B.
f0(x) =
a,
W(f1)U UW(fn)
none of
for
Also
f
..,f
1 < i < p
f
0
one point.
is a finite
B
being the finite union of finite sets so
X
there exists an
f0
Now
B = f-1 1 [ {b}] U ...Uf-1[{10}].
nO
x
no
Certainly
p
and
are
f0
W(f p+1 )U
If
x
51 = 4
b
maps
U W(f
but
W(f1)U
f0
at
thus
n0,
into
n0
)
is not in
since
S2
contradiction in the same way with
W(fp)
n
0
{b},
f
0
is
E f
1
i
not
b
since
[ {a}]
{a}.
at only
then we arrive at a
c
a
and
b
switched.
49
If
S1 =
S2 =
and
(I)
then none of the
(I)
contain
W(fi)
either constant function which again shows that
F(X,Y)
Therefore
W(fl) U...0 W(fn) .
F(X,Y) ,is not com-
pact in the infinite-open topology.
The following similar example shows that compactness
is not inherited from
in the case of the
F(X,Y)
Y
by
X
be the positive integers with the
closed-open topology.
Let
Example 3.6.
Y = {a,b}
discrete topology and
Then what we shall show is that the closed-open
topology.
is the discrete topology, so
F(X,Y)
topology on
Let
will not be compact.
f
-1
[{a}]
and
f
-1[
{b }]
all subsets are closed.
(f
1
also with the discrete
[{a}], {a}) n
(f
-1
are closed subsets of
X
since
Thus the set
[{b}], {b})
open topology and is exactly
the discrete topology.
then
F(X,Y)
be any point of
f
F(X,Y)
ff).
is open in the closedTherefore
has
F(X,Y)
We note that it does no good to
consider the subclass of continuous functions since all
functions are continuous when
4.
X
is discrete.
Countability Properties
We now seek to find out if either first or second countability is carried over to
F(X,Y)
from
Y
in case
F(X,Y)
has either the infinite-open or closed-open topology.
shall use
R to denote the cardinal number of the set
We
X.
50
Let
be the cardinal number of the set of natural num-
N.
bers and let
Recall that
c
be the cardinal number of the real numbers.
and if
= c
2
Y = {b,d}
has cardinal number
that when
Y
C
is
F(X,Y)
II'
let
X
Let
A = {f E F(X,Y)If-l[{b}]
sets of
X},
The next example shows
2a.
with the infinite-open
be the set of real numbers and
and
f-1[ {a }]
A > c.
then we claim that
B = {f E F(X,Y)If-1[{13}]
is finite} and
D = {f E F(X,Y)If-11{a}]
is finite}
A,B
the disjoint union of
=
=
F(X,Y) = A + B + D.
and
so that
D
=
D,
are infinite subLet
so that
F(X,Y)
is
and hence
There is a one-to-one correspondence
between the finite subsets of
or
If
with the discrete topology.
Y = {a,b}
and
Cll.
topology may fail to be
Example 3.7.
with R = a
F(X,Y)
Note that the set
2a > a.
is any cardinal number then
a
=
B = D
X
and the functions in
B
is equal to the cardinality of the
set of finite subsets of
It is known that the cardinal
X.
number of the set of finite subsets of a set of cardinality
c
is again
c.
Therefore
shows that A > c.
{f}
=
2c =c+c+A=c+A which
For every
f E A we have that
(f-1({a}], {a }) n (f-1[ {b}], {b})
is an open set
in the infinite-open topology and so the cardinal number
of the infinite-open topology on
c.
F(X,Y)
is greater than
However, there is a theorem (3, p. 82) which
states if
51
is a
(X,U)
C
space then
II
so we knew that
Y < c
F(X,Y)
Cll.
with the infinite-open topology is not
This same question can be answered more easily in the
case of the closed-open topology.
be the positive integers with the
X
Let
Example 3.8.
again with the discrete
Y = {a,b}
discrete topology and let
F(X,Y)
We showed in Example 3.6 that
topology.
This, together
closed-open topology is a discrete space.
F(X,Y)
with the fact that
is an uncountable set, since
implies that
F(X,Y) = 2 O = C,
with the
is not
F(X,Y)
C11.
The final property that we are going to consider is
Recall that
first countability.
is the first uncountable
is used to denote the set of all ordinals
ordinal and [0, 0)
The next example proves that the
O.
less than
Q
F(X,Y)
perty is not inherited by
from
Y
C1
pro-
in the case of
the infinite-open topology.
Let
Example 3.9.
the discrete topology.
X = [0,Q)
and
Y =
[0,52)
both with
Suppose for purposes of obtaining a
B = {G n In = 1,2,3,...} is a countable
open base of the identity function which we shall again
contradiction that
(id).
denote by
base of
B
n
=
a countable open
which is made up of basic open sets, and
(id)
B* = {BnIn = 1,2,3,...}
suppose that
Then
B
We can generate from
11
(Gn
,
Hi)
n
n
is such a base of (id).
Hk(n) )
(Gk (n) '
for some
Gr.1
52
infinitesubsetsofXandeopen subsets of
r){0}
intersections
consider the class of all
n = 1,2,3,...,
Now for each
IEP
where
Y.
is any subset of
P
1
With each intersection for a given
{1,2,3,...,k(n)}.
associate its least element and if the intersection is
empty associate the element
For each
0.
there are
n
Mn
only a finite number of such intersections so let
the maximum of the associated numbers.
M,
E =
B
contains (id) but contains no
be any member of
Now
B
= (G
0
1
,
n
and let
B*,
0
[M + l,Q).
1170
+ 1,52),
taining
x0.
If none of the
clearly
B
E.
) n ...n
G.
(id)
element, say
E B
m
n
)
no
the ones con-
0
x
contains
then
0
o
j=1 3
n0 cannot be empty
H.
hence this intersection has a least
0
M
which is less than or equal to
x
Therefore the function
E,
k(n 0
(Gk (n )'
G, ,...,Gq
The intersection
,
Let
.
0
q
no
[M + 1,Q))
be any point of
x0
'
rearrange if necessary to make
not in
which is
n = 1,2,3,...
no
since
's
Q.
We claim that the open set
B
n
Mn
The set of all
is then countable so it has a supremum, say
less than
he
hence no
(x) =
Bn
m
n
< M.
0
x0
x = x
iss in
0
is contained in
no countable base of the identity function.
E
B
n
but
0
so there is
This means
53
that
F(X,Y)
is not necessarily
open topology if
Y
CI
with the infinite-
CI.
is
The same example shows that first countability in
does not imply first countability in
F(X,Y)
Y
with the
closed-open topology.
Example 3.10.
crete topology.
Let
X = Y = (0,Q)
both with the dis-
Now since the discrete topology on
X
implies that all subsets are closed, and since we used no
propertiesofthesetsewhich appeared in Example 3.9
the same proof shows that
topology is not
CI.
F(X,Y)
with the closed-open
54
BIBLIOGRAPHY
1.
Dugundji, J.
1966.
2.
Topology. Boston, Allyn and Bacon,
447 p.
Hu, Sze-Tsen.
Elements of general topology.
San Francisco, Holden Day, 1964.
3.
Pervin, W.J.
204 p.
Foundations of general topology.
New York, Academic, 1964.
209 p.