I ntern. J. Math. & Math. Sci.
No. 4 (1980) 701-711
Vol.
701
K-SPACE FUNCTION SPACES
R. A. McCOY
Department of Mathematics
Virginia Polytechnic Institute
and State University
24061 U.S.A.
Blacksburg, Virginia
(Received January 28, 1980)
ABSTRACT.
A study is made of the properties on X which characterize when C (X)
is a k-space, where C
(X) is the space of real-valued continuous functions on X
having the topology of pointwise convergence.
Other properties related to the
k-space property are also considered.
KEY WORDS AND PHRASES. Function spaces, k-spaces, Sequential spaces, Fr@chet
spaces, Countable tightness, k-countable, w-countable.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
Primary, 54C35; secondary,
54D50, 54D55, 54D20.
1
INTRODUCTION
If X is a topological space, the notation C(X) is used for the space of all
real-valued continuous functions on X.
One of the natural topologies on C(X) is
the topology of pointwise convergence, where subbasic open sets are those of the
form
x,v
{fc(x)If(x)
v]
R.A. McCOY
702
for x
X and V open in the space of real numbers,
,
with the usual topology.
The space C(X) with the topology of pointwise convergence will be denoted by
c (x).
For a completely regular space X, C (X) is first countable, in fact metrlz-
able, if and only if X is countable [2].
to what extent this
The purpose of this paper is to show
result can be extended to properties more general than first
countability, such as that of being a k-space.
will be assumed to be completely regular
Throughout this paper
all spaces
Tl-.Spaces.
We first recall the definitions of certain generalizations of first countability.
The space X is a Frchet space if whenever x
sequence in A which converges to x.
A
= X,
there exists a
The space X is a s.equentia.l space if the
open subsets of X are precisely those subsets U such that whenever a sequence
converges to an element of U, the sequence is eventually in U.
Also X is a k-
space if the closed subsets of X are precisely those subsets A such that for
every compact’subspace K
tightness if whenever x
x
B.
X, A N K is closed in K.
A
= X,
Finally X has countable
there exists a countable subset B
=A
such that
The following diagram shows the implications between these properties.
first countable
Frchet
sequential
k-space
countable tightness
We will show that the Frchet space, sequential space, and k-space properties are equivalentfor C (X).
In order to characterize these properties for
C (X) in terms of internal properties of X, we will need to make some additional
definitions.
Let (X) be the set of all nonempty finite subsets of X.
A collec-
tlon I of open subsets of X is an open cover for finite subsets of X if for every
A
(X), there exists a U
such that A
tions of subsets of X, a strin from
[Un}
= U.
If
[n
is a sequence
is a sequence of collec-
[Un}
such that
Un
Un
K-SPACE FUNCTION SPACES
703
for every n 6 IN (I is the set of natural numbers).
that
[Un]
is residually
that for all n
THEOREM I.
In addition, we will say
if for every x 6 X, there exists an N 6 1 such
coverin$
N, x 6 U
n
The following are equivalent.
(a)
C (X) is a Frchet space.
(b)
C (X) is a sequential space.
(c)
C (X) is a k-space.
(d)
Every sequence of open covers for finite subsets of X has a residually
covering string.
(d)
PROOF.
=
(a).
Suppose that every sequence of open covers for finite
subsets of X has a residually covering string.
Let F be a subset of C (X), and
let f be an accumulation point of F in C (X).
Ix I
Xk 6 (X),
f
n,A
we may choose an
F x I’ (f(x)1 _In
Also define U(n,A)
Then for each n
,
[x
finite subsets of X.
define
n
[Un}
We wish to establish that
let
> 0.
(Xl +i)
n
Ifn,A(X)
6 X
Now
f
There is an N
E IN
Therefore
in C
(X).
(c)
{fn(X)}
(d).
(X’)
If n
> I, let
Ir
n
f
I
f
(Xk +i)
n
which is an open cover for
[U(n,An
so
n,A n
So let x
such that for every n
f(x)
<
N, x 6
X, and
U(n,An).
i < i
n
IN
so that
{f n
converges to f
Fr6chet space.
Suppose X has a sequence
{n 3
i
which is an open subset of X.
converges to f in C (X).
with N
must be a
such that no string from
n
},
converges to f(x) for every x 6 X
Hence C
=
1
(f(xk)
has a residually covering string
Ifn,A n (x)
f(x)
n
<
.xk
[U(n’A) IA6(X)}’
But then if n >_ N,
f (x)
.N
f(x)
IN, we may define f n
that for every n
q and A
Then for every n
[n
of open covers for finite subsets
is residually covering.
Let F
I
41,
and for each
be an open cover for finite subsets of X which refines both V
n-i
704
R.A. McCOY
n,A
n,A
(X)
and A
lq
C(X) be such that f
let f
f
E
For every n
n"
and
[ n].
E (X), let U(n,A)
n,A
}
(A)
f
=
such that A
fn
[n
n,A (X\U(n,A))
U(n,A), and
and
Then define
If n,A In
F
F\[Co
and also define F*
(X)}
1 and A
in C (X), where c
is the constant zero function.
o
First we establish that F* is not closed in C (X) by showing that c o is an
I
accumulation point of F in C (X).
W= Xl,Vl N... O Xk,Vk
To do this, let
x I,
be an arbitrary basic neighborhood of c in C (X). If A
o
i
WNF.
then
f
NVk,
n E IN such that
VI N
n,A
Xk}
and
We will then obtain that C (X) is not a k-space, as desired, if we can show
that the intersection of F* with each compact subspace of C (X) is closed in that
.
compact subspace.
To this end, let K be an arbitrary compact subspace of C (X).
Then for every x
X, the orbit
define
Ix
M(x)
XIM(x)
<
K,
[f(x)If
sup
m.
K
[f(x) If
is bounded in
q
and also for every m
q,
[Xmlm
Note that X
define X
m
n and V
[n].
First there exists a k
I,
i
Fk such that Xm
kl,
choose U
have been defined.
Xm+ I
c_
Vm+ I.
For each i
[U n ],
E X be arbitrary.
a j
that
m such that
[Un]
i and V
I
E
i
We define
i
km
+ I,
+ I
there exists a
such that X
= Ui
V
_c
Now suppose k
and
+
Fk
m
i and V
m+l
choose
Ui i
< n <
kj.
Then x
,Uk
m
such that
k
m
Xm c Xj c Vj c Un.
from
UI,
m+l
so that
Let n
such that x 6 X
m
[Un}
For each
I.
m
> k
km+ I
km+ I,
I
which we know to not be residually covering.
There is an m 6
kj_ I
I
E q,
by induction, a string
Fkl
I
so that V
Then there exists a
This defines string
x
c_ V.
Xm c Xm+I.
and that for every m,
Suppose, by way of contradiction, that for every m, n
k
E X,
For every x
Vm+ I
c U
Let
There is
But this says
is residually covering, which is a contradiction.
We have just established that there exist
k >_ n and for every V
Fk, Xm
V.
m, n
Then define M
I such that for every
max
[m n,
i.
let x o
E X
be
K-SPACE FUNCTION SPACES
arbitrary, and define W
C (X).
fk,A"
f
E W NF.
Suppose f
i
o)
Xm\U(k,A).
Since
exists an x
< f(x
I
Therefore WOFNK
,
i
i
Xo,(-, )
which is a neighborhood of c
E
Then there exists a k
<
,
I
M
then k
But then f(x
so that c
705
Thus
n.
I)
> M
k
lq
(X) such that
and A
Xm
> m
in
o
U(k,A), so that there
M(Xl)
K.
so that f
is not an accumulation point of F*NK in K.
o
Hence F*NK must be closed in K.
Since K was arbitrary, we obtain that
C(X)
is not a k-space.
THEOREM 2.
C (X) has countable tightness if and only if every open cover
for finite subsets of X has a countable subcover for finite subsets of X.
PROOF.
Suppose that every open cover for finite subsets of X has a count-
able subcover for finite subsets of X.
Let F be a subset of C (X), and let f
C(X).
be an accumulation point of F in
Then for each n
and A
Ix I
Xk
(X), choose
fn,A E F
x I, (f
Ix
Also let U(n,A)
Then for each n
X.
(Xl)
6 X
l’f(Xn +) N
Xk’ (f (Xk)
i
If n,A (x)
E IN, [U(n,A)
IA
f(x)
(X)
<
i
}
[U(n,A(n,i))li E
f (x
k) +)]]
which is an open subset of X
is an open cover for finite subsets of
So for each n 6 IN, there exists a sequence
that
’i
[A(n,i) li
] from (X) such
is a cover for finite subsets of X.
Then define G
{f n,A(n, i) In, i E I
[
To see that f 6 G, let W
f in C (X).
f(x.) +
3
A
f
c_
I)
n
Let A
c_
Ix I,
V. for each j
3
U(n,A(n,i)).
Xl,Vl N... N[ Xk,Vk be
Xk
So for each x
and choose n 6 IN so that
k.
i,
A,
a neighborhood of
Then there is an i 6
i
fn,A (n,i)(x)- f(x) <n--
(f(xj)
]q
i
n
such that
and hence
n,A(n, i) 6W
Conversely, suppose that C (X) has countable tightness, and let N be an
open cover for finite subsets of X.
that A c_ U(A).
For each A 6 (X), let U(A) E N be such
Also for each n 6 IN and A 6 (X), let f
C(X) be such that
n,A 6
R.A. McCOY
706
f
n,A
[
(A)
If n,A In
In}
fn ,A (X\U(A))
and f
n,A
(X)
c_
[l,n].
n
(X).
and A
is an accumulation point of F, then
Since the constant zero function, c
o
there is a countable subset G of F such that c
o
[ni
=
[Ai} = (X)
that [U(Ai) li
q and
To see
Ix i,
Xk
(-i,i)
[U(Ai) li
(X).
N
0
IN}
Then define F
so that G
}
If ni,A i li
is a cover for finite subsets of X, let A
Then there exists an i
xk,(-l,l)
There are sequences
G.
IN such that
But this means that A
= U(Ai),
fn i,Ai xi’
so that
is indeed a cover for finite subsets of X.
Let us now give names to the two properties of X which are expressed in
We will call X k-countable whenever C (X) is a k-space, and
Theorems 1 and 2.
we will call X
We state
T-countable whenever C (X) has countable tightness.
some immediate facts about these properties.
PROPOSITION 3.
Every countable space is k-countable.
PROPOSITION 4.
Every k-countable space is -countable.
PROPOSITION 5.
Every q-countable space is Lindelf.
PROOF.
be an open cover of X.
Let X be T-countable, and let
family of all finite unions of members of I.
Let If be the
Then F is an open cover for finite
subsets of X, so that it has a countable subcover [D for finite subsets of X.
Each member of
is a finite union of members of B, so that since
,
covers X,
then I has a countable subcover.
This means that if C (X) has countable tightness, X must be
particular,
C(o)
does not have countable tightness, where
o
is the space of
we see from the next proposition has countable tightness, where
PROPOSITION 6.
PROOF.
Let X
n
finite subsets of X
If X
is
be Lindelf for every n
IN, let
n
IN, and let
Un
oI
[i
.
q, then X is -countable.
Lindelf for every n
For each n
In
This is in contrast to C (), which
countable ordinals with the order topology
n
Lindelf.
c
xn IU
.
be an open cover for
Since I is an
_
open cover for finite subsets of X, then each
each n
707
K-SPACE FUNCTION SPACES
IN,
But then J
n
is an open cover of X
has a countable subcollection If n such that
[Fnln
E Fn ]
So for
covers X
n
is a countable subcollection of I which is a cover for
E IN]
finite subsets of X.
{unlu
n
[]
Every compact space is q-countable, and every separable metric
COROLLARY 7.
space is q-countable.
We now examine some properties of k-countable spaces.
Every closed subspace of a k-countable space is k-countable.
PROPOSITION 8.
PROOF.
Let
{Fn }
let U
n
Let X be a k-countable space, and let Y be a closed subspace of X.
[V U (X\Y) IV E
fn }’
has a residually covering string
{Vn]
is a residually covering string from
[V nj (X\Y)],
where each
Vn Fn"
Now
But then
{fn ]"
Every continuous image of a k-countable space is k-countable.
PROPOSITION 9.
PROOF.
lq,
which is an open cover for finite subsets of X.
[Un}
Y be a continuous surjection.
Let X be k-countable, and let f’X
Let
{Irn]
be a sequence of open covers for finite subsets of Y.
let
n
-I
{f (V) IV
[n ]
{Vn]
For each n
be a sequence of open covers for finite subsets of Y.
fn ]’
For each n E IN,
which is an open cover for finite subsets of X.
has a residually covering string
{f
-i
is a residually covering string from
(Vn)]’
Vn fn"
where each
Now
But then
{Vn].
In the next proposition, we use the term covering string, by which we mean
a string which is itself a cover of the space.
If X is k-countable, then every sequence of open covers of
PROPOSITION i0.
X has a covering string.
PROOF.
Let
[n}
f
n
[U
n.
!fUn+k+ II kE IN
which is an open cover for finite subsets of X.
covering string
[Vn}.
For each n
be a sequence of open covers of X.
Now V
I
U tl...IU
I
k
and each U
Thus
[Vn]
for some k
i
I
i
lq, let
Ui]
has a residually
IN.
Also
Vkl+l
R.A. McCOY
708
for some k
Ukl+l U. (JUk2
E
2
define an increasing sequence
This defines
let x
ki
X.
Un
for each n
E
with k
lq
.
[ki}
2
> k I.
UUk
i
i+l
i
is a covering string from
such that each V
k
To see that
Then there exists an N
[Un}
+l=Uk +IU"
such that for all n m N, x
is increasing, there is some i such that
N.
k.l
so that x is indeed in some
(JUki+l,
Uk.+IU"
We
Continuing by induction, we can
Then x
V
Since
n
Vk.+l
Un.
next give an important example of a space which is not k-countable.
EXAMPLE II. The closed unit interval, I, is not k-countable.
PROOF.
For each n E q, let
i
diameter less than
since
That is, 0
Unl,
I
be the set of all open intervals in I having
[Un
Suppose
2I is connected,
n
were to have a covering string
there would be a simple chain
[Unl ,..., U nk }
U
I, there
n
k
and for each i < i < k
is a t.
"
Un_
Then
from 0 to I.
U
n O Un i+l
But then
Ii
i <
<
1
tk_ll + Itk_l- tk_21 +.
2-
+
i
i
1
2nk_l
This is a contradiction, so that
+
+
1
2n-- +
+
It2 tll +Itll
1
i
[n}
cannot have a covering string.
by Proposition I0, I is not k-countable.
Therefore,
3
The next three results are consequences of Example II.
EXAMPLE 12.
PROOF.
The Cantor set,
,
is not k-countable.
Since there exists a continuous function from ( onto I, then
cannot be k-countable because of Proposition 9 and Example
Our next proposition then follows from Example 12
II. J
and Proposition 8.
PROPOSITION 13.
No k-countable space contains a Cantor set.
PROPOSITION 14.
Every k-countable space is o-dimensional.
K-SPACE FUNCTION SPACES
Let X be k-countable, let x
PROOF.
E X, and let U
be an open neighborhood
Since X is completely regular, there exists an f
of x in X.
f(x)
709
[I,
0, f(X\U)
and f(X)
= I.
E C(X)
such that
-Snce I is not k-countable by Example ii,
lf(X).
and since f(X) is k-countable by Proposition 9, then there exists a t
[O,t) N f(X)
Thus
is both open and closed in f(X), so that f
open and closed neighborhood of x contained in U.
-i
([0,t))
is an
[]
With all these necessary conditions which k-countable spaces must satisfy,
one might wonder whether there exists an uncountable k-countable space.
This is
answered by the next two examples.
We will call a space X virtually,
countable
if there exists a finite subset
F of X such that for every open subset U of X with F
= U,
X\U
it is true that
Notice that a first countable virtually countable space is
is countable.
c ount able.
PROPOSITION 15.
PROOF.
F
= U has
Every virtually countable space is k-countable.
Let F be a finite subset of X such that every open U in X with
countable complement, and let
finite subsets of X.
First let U
countable; say X\U I
[Xll,
F U
= U2-
[Xll]
Let U
3
Now
41
be a sequence of open covers for
be such that F
x12, x13,... ].
Let U
2
is also countable; say X\U
2
= U I.
X\U I
Then
is
42 be such that
Ix21, x22, x23,
[Xll, x12, x21] = U 3- Continuing by induction,
[Un] from [ln} such that for each n, Un X\[Xnl,Xn2,
.
U 3 be such that F U
we may define string
Xn3,
X\U 2
I
[4n}
]
and
Fll[Xll,
To see that every
elment
then x is residually in
such that x
U..
i
,Xln, x21,
,X2,n_ I,
of X is residually in
[Un].
If x
nO__l Un,
[Un],
,Xnl=Un+ I.
let x E X.
then let i be the first integer
x.. for some j, so that for every n
Then x
Therefore x is residually in
13
[Un}.
U
n=l n
If x
i
+
j,
xUn
R.A. McCOY
710
EXAMPLE 16.
The space of ordinals,
,
which are less than or equal to the
first uncountable ordinal is k-countable.
PROOF.
It is easy to see that
EXAMPLE 17.
is virtually countable.
[]
The Fortissimo space, IF, is k-countable, where IF is I with
the following topology:
each
It]
is open for t
# O, and the open
0 are the sets containing 0 which have countable complements.
sets containing
Also
]2
is not
Lindelf, which shows that the converse of Proposition 6 is not true.
PROOF.
Obviously IF is virtually countable.
However, an alternate proof
can be obtained from known properties of this space.
from
[3]
[I]
that C ()
In particular, it follows
is homeomorphic to a Z-product of copies of
that a E-product of first countable spaces is a
Frchet
,
and from
space.
The spaces in the previous two examples are not first countable.
This
raises the following question.
QUESTION 18.
Is every first countable k-countable space countable?
One well studied example of an uncountable first countable space which is
also a o-dimensional Lindelf space and which does not contain a Cantor set is
the Sorgenfrey line.
However, in our last example we show that this space is
not k-countable, and in fact is not even
EXAMPLE 19.
T-countable.
The Sorgenfrey line, S, is not q-countable.
This shows that
the converse of Proposition 5 is not true.
PROOF.
and let U(A)
A
Il[-ala A}.
{U21UI’
[(x,y)
cover of
For each A E (S), let 6(A)
U[a,a+6(A))la A}.
[la-a
Then define U
[a,a E
A, with a #a
[U()IA(S),
Clearly I is an open cover for finite subsets of S.
is an open cover of S
S2!x+y O on a
S 2. Therefore no
subsets of S.
"
min
2.
But each U
2,
finite set, so that
for U
[U21U
E
where
Then
l, intersects the set
U}
has no countable sub-
countable subcollection of l can cover all doubleton
K-SPACE FICTION SPACES
711
REFERENCES
i.
H. H. Corson, Normality in subsets of product spaces, Amer. J. Math. 81
(1959), 785-796.
2.
R. A. McCoy, Countabil, ity properties, of function spaces, to appear in
Rocky Mountain J. Math.
3.
N. Noble, The continuity of functions on Cartesian products, Trans. Amer.
Math. Soc. 149 (1970), 187-198.