On Continuously Urysohn Spaces
by
Asl? G?uld?urdek
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful?llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
August 9, 2010
Keywords: Urysohn, Continuously Urysohn
Copyright 2010 by Asl? G?uld?urdek
Approved by:
Gary F. Gruenhage, Chair, Professor of Mathematics and Statistics
Stewart L. Baldwin, Professor of Mathematics and Statistics
Overtoun M. Jenda, Professor of Mathematics and Statistics
Thomas H. Pate, Professor of Mathematics and Statistics
Abstract
We study the properties of weakly continuously Urysohn (denoted by wcU) and con-
tinuously Urysohn (denoted by cU) spaces. The class of continuously Urysohn spaces is
known to contain the class of metrizable, submetrizable, and nonarchimedean spaces. In
this work, by using the scattering process, we show that the class of proto-metrizable spaces
is also contained in the class of continuously Urysohn spaces. We show that being a (weakly)
continuously Urysohn space is not a multiplicative property, and that this property is not
preserved under perfect maps. However, being a weakly continuously Urysohn space is pre-
served under perfect open maps. We give a proof that the topological sum of (weakly)
continuously Urysohn spaces is also (weakly) continuously Urysohn and that any paracom-
pact locally continuously Urysohn ordered space is also continuously Urysohn. We prove that
a well-ordered space is continuously Urysohn if and only if it is hereditarily paracompact and
we obtain a result which characterizes when the linear extension of a separable GO-space is
continuously Urysohn.
ii
Acknowledgments
I would like to express the deepest appreciation to my advisor, Dr. Gary Gruenhage,
for his great patience, and excitement in regard to mentoring. Besides Mathematics, he
taught me ?not to give up?, and ?patience? are essential not only in research, but also in
life. Without his guidance and persistent help this dissertation would not have been possible.
I would like to thank my committee members, Dr. Stewart Baldwin, Dr. Overtoun
Jenda, and Dr. Thomas Pate, for their support. I also would like to thank Dr. Richard
Sesek for serving as the outside reader.
My sincere appreciation also is extended to Dr. Michel Smith, and our lovely secretaries:
Gwen, Carolyn, and Lori for creating an excellent work environment and for being supportive.
I am thankful to my friends: Suna, Y?ucel, Necibe, `Ilknur, Yasemin, G?uven, Sean, and
Erkan. During this journey, whenever I needed a shoulder, they were present. They showed
me, good friends (like them) make things much easier, and they can be your family far from
home.
I also would like to thank Dr. Ferhan At?c?, Dr. Oya ?Ozbak?r, and Dr. Tom Richmond.
They did not only start my United States journey, but also followed and supported me
throughout each step on my way.
Last but not least thanks for my family. My father: MehmetAli G?uld?urdek, my mother:
Ay?se G?uld?urdek, my sister: Ay?sen G?uld?urdek, and my brother (in law): Serdar Samsun.
During this work, my family was at the other side of the ocean. However, I never lived it
this way. They were always as close as my heart. I do not know what would I do without
their endless love and support. Because of my family, I always feel special, lucky, and loved.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Notation and Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Some Classes Which Are Continuously Urysohn . . . . . . . . . . . . . . . . . . 8
4 Some Properties of Continuously Urysohn Spaces . . . . . . . . . . . . . . . . . 19
5 Continuously Urysohn Ordered Spaces . . . . . . . . . . . . . . . . . . . . . . . 25
6 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
iv
Chapter 1
Introduction
In 1993, Stepanova [16] de?ned a property, and showed that this property is equivalent
to metrizability for paracompact p-spaces. Then, in 2002, Halbeisen and Hungerb?uhler [9]
named the spaces which carry this property as continuously Urysohn spaces. Bennett and
Lutzer [3] studied continuously Urysohn spaces in the class of ordered spaces. In 2007, Zenor
[19] de?ned weakly continuously Urysohn spaces and gave the following characterization: A
Hausdor? space is weakly continuously Urysohn if and only if continuous functions de?ned
on the compact subsets can be continuously extended to the continuous functions de?ned on
the space itself. Thus continuously Urysohn and weakly continuously Urysohn spaces have
been shown to be interesting and useful classes of spaces. In this dissertation we continue
their study.
A topological space X is said to be continuously Urysohn(cU) if there is a continuous
function ? : X2\ ?? C(X) such that ?(x,y)(x)?= ?(x,y)(y), where C(X) is the space of
all bounded real-valued functions endowed with the norm topology, and ? ={(x,x) : x?X}
is the diagonal. This notion ?rst appeared in [16], and found its name in [9]. After this, in
[19], the de?nition of a weakly continuously Urysohn space appeared. A space X is said to
be weakly continuously Urysohn(wcU) if there is a continuous function ? : (X2\?)?X?R
such that ?(x,y,x)?= ?(x,y,y).
In [18], the class of continuously perfectly normal spaces, continuously normal spaces,
and continuously completely regular spaces were de?ned, and studied by Phillip Zenor for
the ?rst time. When we look at these de?nitions, they carry the notion of continuity with the
corresponding separation axiom. For instance, one may think that a continuously normal
space is nothing di?erent than a normal space, which has a continuous operator, separating
1
the disjoint closed subsets. After this, in [7], some results concerning the local properties of
continuously normal spaces, and continuously completely regular spaces are obtained.
Alexander V. Arhangel?ski??, in [1], introduced a new type of space, namely a paracompact
p-space, and characterized this type as the class of preimages of metric spaces under perfect
surjections. In [16] and [17], E.N. Stepanova called a mapping ? : X2\?? C(X) separating
if it satis?es the conditions for continuity, and ?(x,y)(x)?= ?(x,y)(y) for any point (x,y) of
X2\?. She showed that the existence of a continuous separating mapping is a necessary
and su?cient condition for a paracompact p-space to be metrizable. In [9], L. Halbeisen and
N. Hungerb?uhler named the spaces carrying this property as continuously Urysohn spaces.
Following the hierarchy of separation axioms, naming a space with this property as
continuously Urysohn space is not surprising. Recall that a space X is a Urysohn space if
for any (x,y) ?X2\?, there exists a continuous function f : X ? R such that f(x) = 0
and f(y) = 1. Also, it is easy to see that every continuously Urysohn space is Urysohn.
When we look at the de?nition of continuously Urysohn space, it is clear that metric
spaces are examples of continuously Urysohn spaces. In [9], Halbeisen and Hungerb?uhler,
examine the continuously Urysohn spaces with the additional property that the functions
?(x,y) depend on just one parameter. They show that spaces with this property are spaces
with a weaker metric topology, namely submetrizable spaces. It is clear that any submetriz-
able space is also continuously Urysohn.
After that, Bennett and Lutzer studied continuously Urysohn linearly ordered and gen-
eralized ordered spaces in [3]. They show that a continuously Urysohn generalized ordered
space is hereditarily paracompact. Also, they proved that for a separable generalized ordered
space, being continuously Urysohn, submetrizable, and having a G?-diagonal are equivalent.
In [19], a weakly continuously Urysohn space is de?ned, and the following theorem is
proved. A space X is weakly continuously Urysohn if and only if there is a continuous
function e : {f : f ? C(H), H is a compact subset of X}? C(X), where e(f) is a
2
continuous extension of f and{f : f ?C(H), H is a compact subset of X}is endowed with
Vietoris topology.
In [8], G. Gruenhage and P. Zenor worked on the properties of weakly continuously
Urysohn spaces. They showed that every weakly continuously Urysohn w?-space has a base
of countable order, and separable weakly continuously Urysohn spaces are submetrizable,
hence continuously Urysohn. It is also shown that monotonically normal weakly continuously
Urysohn spaces are hereditarily paracompact, and no linear extension of any uncountable
subspace of the Sorgenfrey line is weakly continuously Urysohn. Finally, they showed that
any nonarchimedean space is continuously Urysohn.
In this dissertation, we examine weakly continuously Urysohn and continuously Urysohn
spaces in more detail.
We know that the class of nonarchimedean spaces are contained in the class of con-
tinuously Urysohn spaces. In chapter 3, by using the scattering process, we generalize this
by showing that the class of proto-metrizable spaces is a subclass of continuously Urysohn
spaces. We also prove that the topological sum of continuously Urysohn spaces and ultra-
paracompact, locally continuously Urysohn spaces are indeed continuously Urysohn.
In chapter 4, we prove that being a (weakly)continuously Urysohn space is not a mul-
tiplicative property. With a counterexample, we show that the image of a continuously
Urysohn space under a perfect map is not always continuously Urysohn. On the other hand
we prove that the perfect-open image of a weakly continuously Urysohn space is also weakly
continuously Urysohn.
In chapter 5, ordered spaces are studied. From [3] we know that for separable gener-
alized ordered spaces, being continuously Urysohn implies submetrizability. It is clear for a
separable linearly ordered space, being continuously Urysohn is equivalent to metrizability.
We answer the question: When is a linear extension of a separable generalized ordered space
X continuously Urysohn? We also give a characterization of when separable generalized
ordered spaces are weakly continuously Urysohn. For the well-ordered space case, we prove
3
that the space is continuously Urysohn if and only if it is hereditarily paracompact. Finally,
we give a proof which shows that a paracompact, locally continuously Urysohn linearly
ordered space is indeed continuously Urysohn.
In the ?nal chapter, we give a corollary to the free set lemma, and a theorem about the
relation between a special partition of X2\? and the function ? which witnesses continuously
Urysohn on X. At the end, an example is provided.
4
Chapter 2
Notation and Background
C(X) will denote the set of bounded real valued functions de?ned on the space X. ?norm
will denote the norm topology on C(X).
For any f ?C(X), the family{Ui(f)}?i=1, where
Ui(f) ={g?C(X) : there exists an a < 1/i such that|f(x)?g(x)|< a for x?X},
is a base at the point f for the space C(X) endowed with norm topology.
De nition 2.1. For any topological space X letA(X) denote a family of non-empty subsets
of X. If{U1,...,Un}is a ?nite collection of open subsets of X, then the set
< U1,...,Un >={H?A(X) : H?
n?
i=1
Ui, and if 1?i?n, then H?Ui?=?}is a basic
open set for the Vietoris topology onA(X).
For a topological space X, the diagonal is denoted by ? ={(x,x) : x?X}.
De nition 2.2. A subset A of the space X a regular G?-set if there is a sequence {Un} of
open sets in X such that A =
??
i=1
Ui =
??
i=1
?Ui. Here, ?U denotes the closure of the set U. We
will say that X has a regular G?-diagonal if ? is a regular G?-set in X2.
De nition 2.3. A topological space X is said to have a zero-set diagonal if there exists a
continuous function F : X2 ?[0,1] such that F?1(0) = ?.
De nition 2.4. A space X is submetrizable(contractible onto a metric space) provided that
there exists a one-to-one and continuous map from X onto a metric space.
De nition 2.5. A topological space X is said to be separable if there exists a countable
D?X such that ?D = X.
5
Theorem 2.6. ([12]) A separable space is submetrizable if and only if it has a zero-set
diagonal.
De nition 2.7. ([4]) A space X is a w?-space provided that there is a sequence {Wi} of
open covers of X such that if, for each i, x?ui ?Wi and yi ?ui, then {yi} has a cluster
point.
The notion of a base of countable order is due to Arhangel?ski?? ([1]). Chaber, ?Coban,
and Nagami in [5] show that a space (X,?) has a base of countable order if there is a function
? : X) : x?X}covers X.
(2) If x??(< x0,x1,...,xn >), then ?(< x0,x1,...,xn,x >)??(< x0,x1,...,xn >).
(3){?(< x0,x1,...,xn,x >) : x??(< x0,x1,...,xn >)}covers ?(< x0,x1,...,xn >).
(4) If x and the sequence x0,x1,... satisfy x,xn+1 ??(< x0,x1,...,xn >) for all n, then
{?(< x0,x1,...,xi >)}i?? is a base at x. Here, X: n < ?,xi?X}
De nition 2.8. A continuous mapping f : X ?Y is perfect if X is a Hausdor? space, f
is a closed mapping and all ?bers f?1(y) are compact subsets of X.
Searching for generalized metric spaces, Arhangel?ski?? introduced in [1] a certain type of
spaces, which could be characterized as the class of preimages of metric spaces under perfect
surjections, and which are paracompact p-spaces.
De nition 2.9. A space X is nonarchimedean if X has a base B which is a tree under
reverse inclusion.
De nition 2.10. A set T is transitive if every element of T is a subset of T. A set is an
ordinal number (an ordinal) if it is transitive and well-ordered by?.
If ? = ? + 1, then ? is a successor ordinal. If ? is not a successor ordinal, then it is
called a limit ordinal.
6
If the set of all ordinal numbers smaller than a limit ordinal ? contains a subset A of
type ? such that for every ? < ? there exists a ???A satisfying ? < ?? < ?, then we say
that the ordinal ? is co?nal with ?.
An in?nite ordinal ? is an initial ordinal if ? is the smallest among all ordinal numbers
? satisfying|? |=|?|. An initial ordinal ? is regular if there is no ? < ? which is co?nal
with ?.
De nition 2.11. An ordinal ? is called a cardinal number (a cardinal) if|?|?=|?|for all
? < ?.
For every cardinal number ? there exists an initial ordinal ? such that | ? |= ? and
this ? is unique. A cardinal number ? is regular if the initial ordinal ? mentioned above is
regular.
If X is a set of ordinals and ? > 0 is a limit ordinal then ? is a limit point of X if
sup(X??) = ?.
De nition 2.12. Let ? be a regular uncountable cardinal. A set C?? is a closed unbounded
subset of ? if C is unbounded in ? and if it contains all its limit points less than ?.
A set S?? is stationary if S?C?=?for every closed unbounded subset C of ?.
De nition 2.13. Let X be a topological space. The function ?A : X ? {0,1} de?ned
for any subset A of X by setting ?A(x) = 1 if x ? A, and ?A(x) = 0 if x /? A is called
characteristic function on A or characteristic of A.
Note that, ?A is a continuous function if and only if A is an open and closed (clopen)
subset of X.
7
Chapter 3
Some Classes Which Are Continuously Urysohn
The journey of this dissertation started with the search of a space which is weakly
continuously Urysohn but not continuously Urysohn. After Halbeisen and Hungerb?uhler?s
naming the spaces with a certain property as continuously Urysohn and Zenor?s de?nition for
weakly continuously Urysohn spaces, this search was natural. The de?nition of continuously
Urysohn contains a continuous function from the o?-diagonal of a space to the space of all
continuous, bounded real valued functions on that space. On the other hand, the de?nition
of weakly continuously Urysohn contains a continuous function from the o? diagonal times
the space itself to the real numbers.
Before providing any proofs, the following useful fact is presented. If X is continuously
Urysohn space and ? is the witnessing function, then for every (x,y) ? X2 \?, we can
assume that ?(x,y)(x) = 0, and ?(x,y)(y) = 1.
Indeed, the mapping ?? : X2\??C(X), where ??(x,y) is de?ned by:
??(x,y)(t) = (?(x,y)(t)??(x,y)(x))/(?(x,y)(y)??(x,y)(x))
also witnesses for X being a continuously Urysohn space. Note that ??(x,y)(x) = 0, and
??(x,y)(y) = 1. There is a similar result for weakly continuously Urysohn spaces.
We will start with a known theorem. The class of metric spaces is a subclass of cU-
spaces.
Theorem 3.1. Any metric space X is also a cU-space.
Proof. Suppose d is the metric function on space X. Then we can de?ne the function
? : X2\??C(X) as follows:
8
?(x,y)(z) = d(x,z) for all (x,y) in X2\?, and every z in X.
Claim 1: For each (x,y), ?(x,y) is a continuous function.
Proof of Claim 1. For ?xed (x,y) from X2\?, suppose ? > 0 and z?X are given.
Note that|?(x,y)(z)??(x,y)(t)|=|d(x,z)?d(x,t)|. Since d is a metric function,
it is also continuous. We can set the neighborhood G of z such that|d(x,z)?d(x,t)|< ?
for every t?G.
Claim 2: ? is a continuous function.
Proof of Claim 2. Suppose ? > 0 and (x,y)?X2\? are given. For any z?X,
|?(x,y)(z)??(u,v)(z)|=|d(x,z)?d(u,z)|<|d(x,u)|.
In [16], there is a stronger result.
Theorem 3.2. A submetrizable space is continuously Urysohn.
We also know that a space with a zero-set diagonal is weakly continuously Urysohn [17].
De nition 3.3. ([9]) If X is a continuously Urysohn space, then we call the corresponding
family{?(x,y) : (x,y)?X2\?}a continuous separating family for X.
The following proposition characterizes topological spaces with continuous separating
families that depend on only one of its parameters.
Proposition 3.4. ([9]) A space X admits a continuous separating family
{?(x,y) : (x,y)?X2\?}
that depends on just one parameter if and only if X is submetrizable.
Recall that a space X is nonarchimedean if X has a baseBwhich is a tree under reverse
inclusion. In [8], Gruenhage and Zenor present a result related to nonarchimedean spaces.
Theorem 3.5. ([8]) Any nonarchimedean space is cU.
9
This work continues with the investigation of other classes which are contained in the
class of cU-spaces.
IfKis a class of spaces, then de?neS(K) to be the class of spaces which are obtained by
the following scattering process: take any space inK, isolate all the points of some subset,
replace each such point by a space inK, and repeat trans?nitely, taking some subspace of the
inverse limit at limit ordinals. In [14], Nyikos has shown that the class of proto-metrizable
spaces is precisely the classS(M) whereMdenotes the class of metrizable spaces.
We will give a proof to a theorem stating that the class of proto-metrizable spaces is
a subclass of continuously Urysohn spaces. Before this proof, we present a theorem for an
easier case.
Suppose X is a cU-space, P is a subset of X, and the space X? is de?ned as follows:
Isolate the points of P, and replace each p in P by a cU-space Xp. The underlying set X?
can be seen as: X? =?{{p}?Xp : p?P}?(X\P).
Let us de?ne the function ? : X??X as follows: ?(x) =
?
??
??
x, if x?X\P,
p, if x?{p}?Xp, for some p?P.
Note that if x?{p}?Xp for some p?P, then x = (p,a) for some a?Xp. In X? we
declare an open neighborhood of a point x = (p,a)?X? to be{p}?U where U is an open
neighborhood of a in Xp, and an open neighborhood of a point x?X\P is U? = ??1(U)
where U is an open neighborhood of x in X.
Theorem 3.6. Every space X? obtained from a cU-space (respectively, from a wcU-space) X
by ?rst isolating the points of some subset P, and then replacing these with cU-spaces(respectively,
with wcU-spaces) is also a cU-space (respectively a wcU-space).
Proof. Since X is a cU-space, there exists a continuous function ? : X2\? ?C(X) such
that ?(x,y)(x)?= ?(x,y)(y). Also for every Xp there is a function ?p witnessing cU on Xp.
We de?ne the function ? : (X?)2\??C(X?) as follows:
10
?(x,y)(z) =
?
???
???
??
???
???
??
?(?(x),?(y))(?(z)), if ?(x)?= ?(y),
?p(a,b)(c), if p = ?(x) = ?(y) = ?(z)
where x = (p,a),y = (p,b),and z = (p,c),
0, if p = ?(x) = ?(y)?= ?(z).
It is clear that ?(x,y)(x)?= ?(x,y)(y).
First we are going to show that ? is a continuous function.
Claim 1: ? is a continuous function.
Proof of Claim 1.
Case 1: ?(x)?= ?(y).
Note that under Case 1 we have four subcases:
Subcase 1: x,y?X\P.
x,y?X. Since X is a cU-space, for every ? > 0 there exist U,V neighborhoods of x,y
such that
?(u,v)?U?V and?z?X,|?(x,y)(z)??(u,v)(z)|< ?.
If we take U? = ??1(U) and V? = ??1(V) neighborhoods of x and y in the space X?,
then for all (u?,v?)?U??V? and z?X?;
|?(x,y)(z)??(u?,v?)(z)|=|?(x,y)(?(z))??(u?,v?)(?(z))|< ?.
Subcase 2: x?{p}?Xp for some p?P, y?{q}?Xq for some q?P, and p?= q.
x = (p,a), and y = (q,b) where a?Xp, and b?Xq. Take any neighborhood U of a in
Xp, and de?ne U? ={p}?U neighborhood of x, and take any neighborhood V of b in Xq,
similarly de?ne V? ={q}?V neighborhood of y.
Let us look at ?(u?,v?) for all (u?,v?)?U??V?:
?(u?,v?)(z) = ?(p,q)(?(z)), and ?(x,y)(z) = ?(p,q)(?(z)). That is,
|?(x,y)(z)??(u?,v?)(z)|= 0.
11
Subcase 3: x?{p}?Xp for some p?P, and y?X\P.
x = (p,a) for some a ? Xp. Since X is a cU-space, then there exists a neighborhood
U?V of (p,y) such that:
?(u,v)?U?V and z?X,|?(p,y)(z)??(u,v)(z)|< ?. We set a neighborhood U? of
x as U? = ??1(U), and V? of y as V? = ??1(V). For all (u?,v?)?U??V?, and z?X?
|?(x,y)(z)??(u?,v?)(z)|=|?(p,y)(?(z))??(u,v)(?(z))|< ?.
Subcase 4: x?X\P, and y?{p}?Xp for some p?P.
This case can be proven similar to subcase 3.
Case 2: ?(x) = ?(y) = p.
x = (p,a), and y = (p,b) for some p?P, where a,b?Xp.
Since a?= b, and Xp is a cU-space, there exists a U?V neighborhood of (a,b) such that
for all (u,v)?U?V and c?Xp,|?p(a,b)(c)??p(u,v)(c)|< ?.
If we set U? ={p}?U, as a neighborhood of x, and V? ={p}?V as a neighborhood
of y, for all (u?,v?)?U??V?, and z?X?;
|?(x,y)(z)??(u?,v?)(z)|=
?
???
??
????
?
|?p(a,b)(c)??p(u,v)(c)| ?(z) = p
where z = (p,c), u? = (p,u),and v? = (p,v),
|0?0| ?(z)?= p.
Now we will show that for all (x,y), ?(x,y) is a continuous function.
Claim 2: For every (x,y)?(X?)2\?, ?(x,y) is a continuous function.
Proof of Claim 2.
Case 1: ?(x)?= ?(y).
Since ?(?(x),?(y)) is a continuous function on X, for every ? > 0, and every ?(z)?X
there exists a neighborhood G of ?(z) such that
|?(?(x),?(y))(?(z))??(?(x),?(y))(p)|< ?
12
for every p?G.
So, for given any ? > 0 and z?X? we take G? = ??1(G) open neighborhood of z, where
G is the open neighborhood of ?(z) as mentioned previously. Then, for every t in G? we
have;
|?(x,y)(z)??(x,y)(t)|=|?(?(x),?(y))(?(z))??(?(x),?(y))(?(t))|< ?.
Case 2: ?(x) = ?(y).
Subcase 1: ?(x) = ?(y) = ?(z) = p.
Note that ?(x,y)(z) = ?p(a,b)(c). Since ?p(a,b) is a continuous function on Xp for any
? > 0, and c?Xp there exists a neighborhood of G of c such that for every d?G,
|?p(a,b)(c)??p(a,b)(d)|< ?.
With the help of this setting, for given any ? > 0 and z = (p,c)?X? where;
?(x) = ?(y) = ?(z) = p, we set G? neighborhood of z as {p}?G, where G is the
neighborhood of c mentioned before.
For every t in G?, t = (p,d) and|?(x,y)(z)??(x,y)(t)|=|?p(a,b)(c)??(a,b)(d)|< ?.
Subcase 2: ?(x) = ?(y)?= ?(z).
That means there exists a p?P such that x,y?{p}?Xp, and ?(x,y)(z) = 0.
If z?X\P, take a neighborhood G of z such that p /?G, and de?ne the neighborhood
G? of z in X? as G? = ??1(G). If z ?{q}?Xq, z = (q,c) for a q ? P, then de?ne the
neighborhood of z as G? ={q}?G where c?Xq and G is a neighborhood of c.
Note that same proof works for a wcU-space X, if one replaces the points of P ?X by
wcU-spaces. For this case by using the function ? which witnesses X being a wcU-space, we
13
de?ne the function ? on ((X?)2\?)?X? as follows:
?(x,y,z) =
?
???
???
??
???
???
??
?(?(x),?(y),?(z)) if ?(x)?= ?(y),
?p(a,b,c) if p = ?(x) = ?(y) = ?(z)
where x = (p,a), y = (p,b),and z = (p,c),
0 if p = ?(x) = ?(y)?= ?(z).
Now that we have this theorem, we will use it to prove that the classes of continuously
Urysohn and weakly continuously Urysohn spaces are closed under the scattering process.
Theorem 3.7. If cU is the class of all continuously Urysohn spaces, then S(cU) is a subclass
of cU. That is, the class of continuously Urysohn spaces is closed under the scattering process.
Proof. Let ? be an ordinal and X0 be a cU-space. We start the scattering process by isolating
a subset of X0, and replacing these points with cU-spaces. Let us name the resulting space
X1, and de?ne the map ?1,0 : X1 ?X0 naturally. On the second step, we isolate a subset of
X1 and replace these points with cU-spaces. We name this space as X2, and de?ne the map
?2,1 : X2 ?X1 naturally. Similarly we can de?ne Xn and ?n,n?1 for all n < ?. Also for any
m < n we de?ne the map ?n,m : Xn?Xm as ?n,m = ?k,m??n,k where m < k < n. By using
the previous theorem we know that for every n < ?, Xn is a cU-space.
At the ?rst limit ordinal step, we de?ne the space X? = lim??Xn such that,
??x =< x0,x1,... >?lim
??Xn if and only if x0 = ?1,0(x1), x1 = ?2,1(x2), ...
We also de?ne the map ??,n as the nth projection map from X? into Xn. At the next
step we de?ne the space X?+1 by repeating the scattering process. For all n < ?, the maps
??+1,? and ??+1,n are de?ned similarly.
Finally, by replacing points with cU-spaces at successor ordinal steps, and by taking
the limit space at the limit ordinal steps, we form a space X? = lim??X?. Note that for
14
every ? < ?, X? is a cU-space with the witnessing continuous function ??. WLOG, we may
suppose ??(x,y)(x) = 0, and ??(x,y)(y) = 1.
By using ?? :?X??X?, ?th projection map, we de?ne the continuous map
??? = ??|X : X? ?X?, and {(???)?1(U?) : U? is open subset of X?} forms a base for
the limit space X?.
Let us de?ne the function ? : X2? \??C(X?) as follows:
?(??x,??y )(??z ) = ??(x?,y?)(z?), where ? is the least ordinal such that x? ?= y?. It is
clear that ?(??x,??y )(??x) = 0, ?(??x,??y )(??y ) = 1, and ?(??x,??y ) is continuous for every (??x,??y )
in X2?\?. We only need to show the continuity of ?. Suppose ? > 0 and (??x,??y ) are given.
By examining the ordinal ?, where??x di?ers from??y for the ?rst time, we have 3 cases.
Case 1: ? = 0.
Since X0 is a cU-space there exists a U0?V0 neighborhood of (x0,y0) such that;
|?0(x0,y0)(z0)??0(u0,v0)(z0)|< ? for every (u0,v0) in U0?V0, and z0 in X0.
We set U?V = (??0)?1(U0)?(??0)?1(V0). Then for every (??u,??v ) in U?V,
|?(??x,??y )(??z )??(??u,??v )(??z )|=|?0(x0,y0)(z0)??0(u0,v0)(z0)|< ?.
Case 2: ? > 0 is a successor ordinal.
Note that x? = (x??1,a)?X?, and y? = (y??1,b)?X? where a,b are in the cU-space
which has been replaced with x??1 = y??1 ? X??1. Since X? is a cU-space, there exists
U??V? neighborhood of (x?,y?) such that,
| ??(x?,y?)(z?)???(u?,v?)(z?) |< ? for every (u?,v?) in U? ?V?, and z? in X?.
WLOG, we can choose U?, and V? so that ??,??1(U?) = ??,??1(V?) = x??1 = y??1.
We set U?V = (???)?1(U?)?(???)?1(V?). Then for every (??u,??v ) in U?V,
|?(??x,??y )(??z )??(??u,??v )(??z )|=|??(x?,y?)(z?)???(u?,v?)(z?)|< ?.
Case 3: ? > 0 is a limit ordinal.
Since ? is a limit ordinal, for every x? ? X?, x? =< x? >? 0, and z?X are given.
Case 1: There exists a s?S such that x,y?Xs.
Subcase 1: z?Xs.
Then ?(x,y)(z) = ?s(x,y)(z). Since ?s(x,y) is a continuous function there exists a
neighborhood Gs of z in Xs such that;
|?s(x,y)(z)??s(x,y)(t?)|< ? for every t??Gs. We take G = Gs open neighborhood
of z in X, and for every t?G,
|?(x,y)(z)??(x,y)(t)|=|?s(x,y)(z)??s(x,y)(t)|< ?.
Subcase 2: z /?Xs.
Note that there exists a p?S such that z?Xp, and ?(x,y)(z) = 0. We set G = Xp as
an open neighborhood of z. Then for every t?G,
|?(x,y)(z)??(x,y)(t)|=|0?0|.
Case 2: x?Xs and y?Xs?, s?= s?.
Then ?(x,y)(z) = ?Xs?(z). If z?Xs? we take G = Xs?, and if z /?Xs? we take G = Xcs?
as a neighborhood of z. So, for every t?G we have;
|?(x,y)(z)??(x,y)(t)|= 0.
Claim 3: ? : X2\??C(X) is a continuous function.
Proof of Claim 3.
Suppose ? > 0, and (x,y)?X2\? are given.
Case 1: There exists a s?S such that x,y?Xs.
Since Xs is a cU-space with the function ?s there exists a neighborhood Us?Vs of (x,y)
in Xs so that for every (u?,v?)?Us?Vs, and z?Xs
17
|?s(x,y)(z)??s(u?,v?)(z)|< ?.
We set U = Us, and V = Vs. Then for every (u,v)?U?V;
|?(x,y)(z)??(u,v)(z)|=
?
??
??
|0?0|= 0, if z /?Xs,
|?s(x,y)(z)??s(u,v)(z)|< ?, if z?Xs.
Case 2: x?Xs, and y?Xs?, s?= s?.
We take U = Xs, and V = Xs?. Then for every (u,v)?U?V;
|?(x,y)(z)??(u,v)(t)|=|?Xs?(z)??Xs?(z)|= 0.
By using a function de?ned similar to the function used in the proof of previous theorem,
we have a result on ultraparacompact locally cU-spaces. Remember that a space X is
ultraparacompact if for every open coverU of X, there exists a locally ?nite pairwise disjoint
open re?nementV. Also, if every point in the space has an open neighborhood which is cU
we say that the space is locally cU-space.
Corollary 3.10. If X is a ultraparacompact and locally cU-space, then X is a cU-space.
The analogues of Theorem 3.9, and corollary 3.10 also hold for wcU-spaces.
18
Chapter 4
Some Properties of Continuously Urysohn Spaces
The function witnessing being a cU-space appears stronger than the function witness-
ing being a wcU-space. It is clear that any cU-space is a wcU-space. We know that any
submetrizable space is continuously Urysohn, and a zero-set diagonal implies being a weakly
continuously Urysohn space. Furthermore, if a separable space is also a wcU-space, then it
is also a cU-space.
While searching for an example which distinguishes wcU and cU, some new properties
of these spaces were examined.
Firstly, we looked at the property of being wcU in product spaces. We discovered that
being a wcU-space is not a multiplicative property. In order to have that result we proved
the following lemma and two theorems.
Lemma 4.1. If X is a space with a zero-set diagonal, then the product space Y = X?M,
where (M,d) is a metric space, also has a zero-set diagonal.
Proof. Since X has a zero-set diagonal, there exists a continuous function F : X2 ?[0,1]
such that, F?1(0) = ?. Let us now de?ne the function G : Y 2 ?R as follows:
G((x,m),(y,n)) = F(x,y) +d(m,n).
Claim 1: G is continuous.
Proof of Claim 1. Given any ? > 0 and ((x,m),(y,n)) ? Y 2, we choose the open
neighborhood A = (U1?V1)?(U2?V2). Here, U1?U2 is the neighborhood of (x,y) such
that|F(x,y)?F(z,t)|< ?2 for every (z,t)?U1?U2, and V1?V2 is the neighborhood of
(m,n) such that|d(m,n)?d(p,q)|< ?2 for every (p,q)?V1?V2.
19
Then
|G((x,m),(y,n))?G((z,p),(t,q))|?|F(x,y)?F(z,t)|+|d(m,n)?d(p,q)|< ?,
for every ((z,p),(t,q))?A.
Claim 2: G?1(0) = ?.
Proof of Claim 2. Suppose, G((x,m),(y,n)) = 0. This is true if and only if F(x,y) = 0
and d(m,n) = 0, and this is true if and only if x = y and m = n.
Theorem 4.2. Let X be a topological space, and ?+1 = [0,?] a convergent sequence. Then
Y = X?[0,?] is a wcU-space if and only if X has a zero-set diagonal.
Proof. First, suppose that X is a space with a zero-set diagonal. By Lemma 4.1, the product
space Y = X?(?+1) has a zero-set diagonal. Since zero-set diagonal implies being a wcU-
space, Y is a wcU-space. Now, suppose Y = X?(?+1) is a wcU-space. Then, there exists
a continuous function ? : (Y 2\?)?Y ?R such that;
?((x,n),(y,m),(x,n)) ?= ?((x,n),(y,m),(y,m)).
WLOG, we can assume that ?((x,n),(y,m),(x,n)) = 0, and ?((x,n),(y,m),(y,m)) = 1,
and 0??((x,n),(y,m),(z,p))?1 for all (x,n),(y,m),(z,p)?Y. For n < ?, let?s de?ne
the function fn : X2 ?R as follows:
fn(x,y) = ?((x,n),(y,?),(y,n)).
Note that for all n, fn is a continuous function. Finally we can de?ne the function
F : X2 ? R such that; F(x,y) = ??n=1 fn(x,y)2n . Since each fn is a continuous function,
F is a continuous function. And also, F?1(0) = ?. Indeed, if x = y, then for all n,
fn(x,y) = ?((x,n),(x,?),(x,n)) = 0. So, F(x,y) = 0. If x ?= y and F(x,y) = 0, then
fn(x,y) = 0 for all n. That means, ?((x,n),(y,?),(y,n)) = 0. If we look at the sequence
20
((x,n),(y,?),(y,n)), it converges to the point ((x,?),(y,?),(y,?)). Since ? is a continu-
ous function, ?((x,n),(y,?),(y,n)) converges to the point ?((x,?),(y,?),(y,?)). However,
?((x,n),(y,?),(y,n)) = 0 for all n, and ?((x,?),(y,?),(y,?)) = 1, so this is a contradic-
tion. That is, F?1(0) = ?, and the space X has a zero-set diagonal.
Corollary 4.3. If X?(? + 1) is wcU, then X has a regular G?-diagonal.
Corollary 4.4. If X has a non-G? point, then X?(? + 1) is not a wcU-space.
Example 4.5. There is cU-space X such that X?(? + 1) is not wcU.
Proof. Let X be L(?1) where L(?1) is the one-point Lindel?o?cation of ?1. That is, every
? < ?1 is isolated, and the neighborhoods of ?1 are the sets with countable complements.
Since L(?1) is a nonarchimedean space, by using Theorem 3.5 it is also a cU-space. On the
other hand, by using Corollary 4.4 and the non-G? point ?1 in L(?1), L(?1)?(?+1) is not
a cU-space.
Example 4.6. Suppose X? is the topological sum of L(?1) and ? + 1. Let us form the
quotient space X of X? obtained by identifying ?1 in L(?1) with ? in ? + 1. Then X is not
even a wcU-space.
Proof. Suppose X is a wcU-space. Then there exists a continuous function
?: (X2\?)?X ?R such that ?(x,y,x)?= ?(x,y,y).
We can assume that ?(x,y,x) = 0 and ?(x,y,y) = 1. For every n < ?, ?(p,n,p) = 0,
where p = ? = ?1. So, for every n < ?, there exists a neighborhood Gn of p such that
?(q,n,p) < 1/2 for all q?Gn. Note that ?Gn ?={p}. There exists a point q0 ??Gn and
q0 ?= p. Then ?(q0,n,p) < 1/2 for all n < ?. When we take the limit as n??; ?(q0,n,p)?
?(q0,p,p). Since ?(q0,n,p) < 1/2 and ?(q0,p,p) = 1 we have a contradiction.
Corollary 4.7. The cU-space and wcU-space properties are not always preserved under
perfect maps.
21
Proof. Let X and X? be as in Example 4.6. The quotient map q : X??X is perfect, and
X? is a cU-space. However, X is not even wcU.
We will show, however, that open-perfect maps preserve weakly continuously Urysohn
spaces. Since we have an open-perfect map, we will use a theorem of Michael?s from [13].
Before the statement of this theorem we will give some de?nitions that he mentions before
that theorem.
De nition 4.8. ([13]) Let f : X?Y be onto, andA(X) ={E?X : E?=?}. We de?ne
(1) f?1? : Y ?A(X) by f?1?(y) = f?1(y),
(2) f?1?? :A(Y)?A(X) by f?1??(E) = f?1(E).
Theorem 4.9. ([13]) Let X, Y be topological spaces, and f : X ?Y be onto. Then with
the Vietoris topology onA(X), andA(Y) we have:
(1) f?1? is continuous if and only if f is open and closed,
(2) f?1?? is continuous if and only if f?1? is continuous.
Now, we can prove our theorem related to weakly continuously Urysohn spaces.
Theorem 4.10. The perfect-open image of a weakly continuously Urysohn space is also
weakly continuously Urysohn.
Proof. Suppose f : X?Y is a perfect-open map from a wcU-space X onto a space Y.
If X admits a continuous separating function ? : (X2\?)?X ?R, by using Zenor?s
Lemma 2 from [19], there is a continuous function ?? :M(X)?X?[0,1] such that
(1) if x?H, then ??(H,K,x) = 0, and if x?K,then ??(H,K,x) = 1,
(2) if (H,K), (H?,K?)?M(X) with H?H? and K??K, then
??(H?,K?,x)? ??(H,K,x).
HereM(X) ={(H,K)?K(X)?K(X) : H?K =?}, andK(X) denotes the space of
compact subsets of X endowed with the Vietoris topology.
Suppose X is a wcU-space with the function ?, and let us de?ne the function
22
? : (Y 2\?)?Y ?R by using the functions f, and ??.
?(x,y,z) = max{??(f?1(x),f?1(y),c) : c?f?1(z)}.
Since f is a perfect map, f?1(z) is a compact subset of X for every z ?Y. Also note
that for every z from Y there exists a c1 in f?1(z) such that;
?(x,y,z) = max{??(f?1(x),f?1(y),c) : c?f?1(z)}= ??(f?1(x),f?1(y),c1).
Claim 1: ?(x,y,x)?= ?(x,y,y).
Proof of Claim 1. It is clear that ?(x,y,x) = 0, and ?(x,y,y) = 1 for every (x,y) from
Y 2\?.
Claim 2: ? : (Y 2\?)?Y ?R is a continuous function.
Proof of Claim 2. Suppose ? > 0, and (x,y,z) are given. Since ?? is a continuous
function, for every c ? f?1(z) there are open neighborhoods Hc, Kc, and Gc of f?1(x),
f?1(y), and c respectively, such that;
| ??(f?1(x),f?1(y),c)? ??(A,B,c?)|< ? for every A?Hc, B?Kc, and c??Gc.
By using the compactness of f?1(z) we can select a ?nite subset{c1,c2,...,cn}of f?1(z)
so that{Gci : i = 1,2,...,n}covers f?1(z).
Since neighborhoods of f?1(x), and f?1(y) are open with respect to Vietoris topology,
for every 1?i?n, we can assume that;
Hci =< U1,U2,...,Ur > is a neighborhood of f?1(x), and Kci =< V1,V2,...,Vs > is a
neighborhood of f?1(y).
So, we take H =
n?
i=1
Hci neighborhood of f?1(x), K =
n?
i=1
Kci neighborhood of f?1(y),
and < Gc1,Gc2,...,Gcn > neighborhood of f?1(z). Note that by using the de?nition of
Vietoris topology, we can restate H, and K respectively as follows:
H =< U1,U2,...,Um >, and K =< V1,V2,...,Vk >. Then there exist neighborhoods U
of x, V of y, and W of z such that f?1(U)?
m?
i=1
Ui, f?1(V)?
k?
i=1
Vi, and f?1(W)?
n?
i=1
Gci.
Under this setting of neighborhoods, let us look at | ?(x,y,z)??(u,v,w) | for every
u?U, v?V, and w?W.
Recall that ?(x,y,z) = max{??(f?1(x),f?1(y),c) : c?f?1(z)}= ??(f?1(x),f?1(y),c1).
23
Case 1: Suppose?(u,v,w) = max{??(f?1(u),f?1(v),t) : t?f?1(w)}= ??(f?1(u),f?1(v),t1)
where t1 ?Gc1. Then;
|?(x,y,z)??(u,v,w)|=| ??(f?1(x),f?1(y),c1)? ??(f?1(u),f?1(v),t1)|.
Since f?1(u)?< U1,U2,...,Um >=
n?
i=1
Hci, and f?1(v)?< V1,V2,...,Vk >=
n?
i=1
Kci we
have f?1(u)?Hc1, and f?1(v)?Kc1. By using the fact that ?? is a continuous function and
t1 ?Gc1 we conclude that:
|?(x,y,z)??(u,v,w)|< ?.
Case 2: Suppose?(u,v,w) = max{??(f?1(u),f?1(v),t) : t?f?1(w)}= ??(f?1(u),f?1(v),s1)
where s1 /?Gc1.
Then there exists an i?n, such that s1 ?Gci. By using the de?nition of ?, we know
that ??(f?1(x),f?1(y),c1)? ??(f?1(x),f?1(y),ci). Since ?? is a continuous function,
??(f?1(x),f?1(y),ci)?? < ??(f?1(u),f?1(v),s1) < ??(f?1(x),f?1(y),ci) +?. That is;
??(f?1(u),f?1(v),s1) < ??(f?1(x),f?1(y),c1) +?...(?).
We also know that if t?Gc1?f?1(W), then ??(f?1(u),f?1(v),t)? ??(f?1(u),f?1(v),s1).
By using the continuity of ?? one more time we have;
??(f?1(x),f?1(y),c1)?? < ??(f?1(u),f?1(v),t) < ??(f?1(x),f?1(y),c1) +?. That is;
??(f?1(x),f?1(y),c1)?? < ??(f?1(u),f?1(v),s1)...(??).
With the combination of (?), and (??) we ?nally have:
|?(x,y,z)??(u,v,w)|< ?.
24
Chapter 5
Continuously Urysohn Ordered Spaces
In this chapter we study weakly continuously Urysohn and continuously Urysohn prop-
erties on linearly ordered topological spaces(LOTS) and generalized ordered spaces(GO-
spaces). Recall that a LOTS is a linearly ordered set with the usual open interval topology.
A GO-space is a Hausdor? space with a linear order < such that the topology of the space
has a base consisting of convex sets. Equivalently, GO-spaces are subspaces of LOTS.
In [3], Bennett and Lutzer show that for a separable GO-space X, being continuously
Urysohn, submetrizable, and having a G?-diagonal are equivalent properties. The behavior
of the linear extensions of separable GO-spaces in that manner was not known. We have
a theorem which characterizes exactly when a linear extension of a separable GO-spaces is
continuously Urysohn. Let us start with the de?nition of linear extension of a GO-space.
De nition 5.1. Let X = (X,?,?) be a GO-space, where (X,?) is a linearly ordered set
and ? is a topology on X such that:
(a) ?(?)??, where ?(?) is the open-interval topology of?,
(b) every point of X has a local ?-base consisting of (possibly degenerate) intervals of
X.
De?ne a subset X? = (X,?,?)? of X?Z by
X? = (X ?{0}) ?{(x,n) : [x,? [? ? \? and n ? 0}?{(x,m) :] ?,x] ? ? \?
and m ? 0}. Then we order X? = (X,?,?)? lexicographically, and it carries the usual
open-interval topology of this lexicographic order.
X? is the smallest LOTS which contains X as a closed subspace.
25
Theorem 5.2. Let X be a separable GO-space, X? be any linear extension of X, and de?ne
A={x?X|{(y,x] : y < x} base at x}?{x?X|{[x,y) : x < y} base at x}. Then, A is
countable if and only if X? is continuously Urysohn.
Proof. Let us ?rst assume that X is a separable GO-space, X? is a linear extension of X,
and A={x|{(y,x] : y < x} base at x}?{x|{[x,y) : x < y} base at x} is countable. We
will show that X? is a cU-space.
Case 1: Ais countable, and X has no isolated points.
Claim: X is metrizable.
Proof of Claim. Since regular and second countable spaces are metrizable all we need
to show is the second countability of X. By the de?nition of a GO-space without isolated
points, the collection:
{]?,x] : x?L}?{[x,?[: x?R}??(?), where L,R are disjoint subsets of X, and
?(?) is the open interval topology on linearly ordered set X, is a subbase for X.
X is separable, so there exists a countable dense subset D of X. If we take the collection:
S ={(?,x] : x?L}?{[x,?) : x?R}???(?), where ??(?) is a subcollection of ?(?),
with intervals having their endpoints from the set D as the subbase, then the resulting base
is a countable base and gives the same topology for X. That ?nishes the proof of claim.
We then conclude that metrizability of X implies the metrizability of X?, and that
implies cU for X?.
Case 2: Ais countable and X has isolated points:
First we construct a LOTS extension of X by assuming no isolated points. Let us call
this space Y. In the previous case we showed that Y is a cU-space. Then we apply the
scattering process to the space Y. Since Y is a cU-space, and the class of all cU-spaces is
closed under scattering process X? the linear extension of X is also a cU-space.
Let us now assume that X is a separable GO-space, and the cU hence wcU-space X? is
a linear extension of X.
26
Note that since for GO-spaces ccc implies hereditarily Lindel?ofness, X is also a heredi-
tarily Lindel?of space.
Claim: Ais countable.
Proof of Claim. SupposeAis uncountable.
Case 1: R ={x?X : [x,y) is a base at x}is uncountable.
By using hereditarily Lindel?ofness of X, and removing countably many points if neces-
sary we may assume that every x?R is a limit point of (x,?)?R.
Given any r from R there must be a point r? ?X?\R such that r? < r, and s < r? for
any s from R with s < r.
For U ?R let U? ={r? : r?U}. Since being a wcU-space is hereditary R?R? should
also be a cU-space. Suppose ? witnesses that R?R? is wcU.
Let C?R be countable and dense in R. Note that each r?R has a local base of sets
of the form [r,c), where c?C and the interval computed in X?.
WLOG, we may assume ?(x,y,x) = 0 and ?(x,y,y) = 1 for all x?= y ?R?R?. Then
for each r?R there is a point cr ?C such that;
?(r?,[r,cr)2) > 3/4.
Let Ac ={r?R : cr = c}. We choose a point c?C such that|Ac |> ?. Since X is
hereditarily Lindel?of space, and Ac is an uncountable subset of R there exist x,rn,r ? Ac
such that rn?x from the right and r > rn for all n.
Also r?n ? x, and (r,rn) ? [rn,c)2. Thus ?(r?n,r,rn) > 3/4, but (r?n,r,rn) ? (x,r,x)
and ?(x,r,x) = 0.
So, we have a contradiction.
Case 2: L ={x : (y,x] is a base at x}is uncountable.
Proof of this case is identical to the proof of case 1.
After this theorem we have a similar characterization for separable GO-spaces being
weakly continuously Urysohn.
27
Theorem 5.3. Let X be a separable GO-space. Then X is weakly continuously Urysohn if
and only ifAis countable, whereA={x : x has an immediate successor or predecessor}.
Proof. First we suppose X is a separable, wcU, GO-space. Then there exists a continuous
function ? : (X2\?)?X?R such that:
?(x,y,x)?= ?(x,y,y), and we can assume ?(x,y,x) = 0, and ?(x,y,y) = 1.
Suppose A is uncountable. WLOG we will assume that the set R = {r ? X : r has
an immediate successor} is uncountable. So for every r from R there exists r? immediate
successor of r. Let us de?ne R? ={r? : r ?R}. Since being wcU-space is hereditary then
R?R? is also wcU.
We have ?(r?,r,r) = 1. Then for every r?R there exists a basic neighborhood Gr of r
such that, ?(r?,G2r) > 3/4.
As a result of the separability assumption on X there exists a countable dense subset
D. So for every r?R there exists a cr ?Gr. We can also assume that (cr,r]?Gr. Note
that there exists a c that repeats for uncountably many r.
Let us de?ne P ={r : c?Gr}where|P |> ?. Since X is hereditarily Lindel?of, P is
also a Lindel?of space.
Then there exist x,rn,r?P such that rn ?x and c < r < rn for every n < ?. Note
that r?n converges to x.
We have ?(r?n,r,rn) > 3/4. However (r?n,r,rn)?(x,r,x) and ?(x,r,x) = 0 is a contra-
diction.
Now we supposeA={x : x has an immediate successor or predecessor}is countable.
Let us form a baseB={{x}: x isolated in X?}?{K : K convex-open with endpoints
inA?D}; where X? is the topology on GO-space and D is the countable dense subset. It is
clear that this is a countable base for a weaker order topology on X. So, X is submetrizable,
hence cU.
28
In [3], Bennett and Lutzer prove that a stationary subset S of a regular uncountable
cardinal ? is not continuously Urysohn, and for any monotonically normal space being con-
tinuously Urysohn implies being hereditarily paracompact. Also Gruenhage and Zenor in [8]
show that any stationary subset of a regular uncountable cardinal cannot even be weakly con-
tinuously Urysohn. With the help of these results, we give a characterization for well-ordered
spaces in the following theorem.
Theorem 5.4. A well-ordered space is continuously Urysohn if and only if it does not contain
a subset homeomorphic to a stationary subset of a regular uncountable cardinal. Equivalently,
a well-ordered space is continuously Urysohn if and only if it is hereditarily paracompact.
Proof. With the result of Bennett and Lutzer?s mentioned above, we know that well-ordered
cU-spaces cannot contain a stationary subset of a regular uncountable cardinal. All we need
to show is that the cU property holds when a well-ordered set does not contain the subsets
mentioned.
Since a subset of ?1 contains a stationary subset if and only if it is nonmetrizable, claim
is true for subspaces of ?1.
Let ? be any ordinal. Suppose the claim is true for the subsets of ? for every ? < ?.
That is if B ? ? does not contain a stationary subset of a regular uncountable cardinal,
then B is a cU-space.
Claim : If A?? does not contain a stationary subset of a regular uncountable cardinal,
then A is cU-space.
Proof of Claim.
Case 1: ? is a limit ordinal.
Subcase 1: Co?nality of ? is greater than ?. That is cf? > ?.
Suppose cf? = ? > ?. Note that ? is a regular cardinal. Also there exists a subset K
of ?, which is a copy of ?. We know that K is closed, order isomorphic to ?, and co?nal in
?.
29
On the other hand; since A does not contain a stationary subset of any uncountable
regular cardinal there exists a club subset C of K such that C?A = ?. WLOG; we can
assume K = C and K?A =?. Since Kc is the union of disjoint convex subsets, Kc =
?
?
I?.
That is, Kc is the topological sum of cU-spaces. By using Theorem 3.9, we conclude that
Kc is cU. Hence A?Kc is cU.
Subcase 2: Co?nality of ? is equal to ?. That is cf? = ?.
There exists a K ={kn}n 0, and a point z are given.
Case 1: x,y,z?I?.
Since ??(x,y) is a continuous function, there is a neighborhood G? of z in such that;
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G?. We set G = G?. Then,
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
Case 2: x,y?I?, and z /?I?.
Subcase 1: z?= ?. There exists a I? such that z?I?.
We set G = I?. Then,
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
Subcase 2: z = ?.
We set G = (c?+1,?]?A?. Then,
|?(x,y)(?)??(x,y)(t)|< ?, for every t?G.
Case 3: Other cases:
(1) x < y.
Subcase 1: x,z?I?.
We set G = (c?,c?+1)?A?. Then,
31
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
Subcase 2: x?I?, and z /?I?.
We set G = [[c?,c?+1]?A?]c.
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
(2) y < x
Subcase 1: y,z?I?.
We set G = (c?,c?+1)?A?.
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
Subcase 2: y?I?, and z /?I?.
We set G = [[c?,c?+1]?A?]c.
|?(x,y)(z)??(x,y)(t)|< ?, for every t?G.
Claim 3: ? is a continuous function.
Proof of Claim 3.
Suppose an ? > 0, and (x,y) are given.
Case 1: x,y?I?.
We de?ne ??(x,y) for (x,y). By using the continuity of ?? we know that there is a
neighborhood U??V? of (x,y) in I? such that
|??(x,y)(z)???(u,v)(z)|< ? for every (u,v) from U??V?.
We take neighborhoods U = U? and V = V?.
If z?I?, then|?(x,y)(z)??(u,v)(z)|=|??(x,y)(z)???(u,v)(z)|< ?, and if z /?I?,
then ?(x,y)(z) = 0 and ?(u,v)(z) = 0.
Case 2: Other cases:
(1) x < y.
We take U = I?, and V = (c?,?]?A?. Note that x?I?. Then,
?(x,y)(z) = 1??(c ,c +1](z), and ?(u,v)(z) = 1??(c ,c +1](z).
(2) y < x.
We take U = (c?,?]?A? and V = I?. Note that y?I?. Then,
32
?(x,y)(z) = ?(c ,c +1](z), and ?(u,v)(z) = ?(c ,c +1](z).
Subcase 2: ? = ? + 1, for some ordinal ?. We de?ne the set A? = [0,?]?A?{?}.
Notice that [0,?] is cU, and ? is isolated. So, we have a topological sum of cU-spaces.
After these results we want to examine the locally continuously Urysohn property in
ordered spaces. The question was: locally continuously Urysohn and what other assumption
on the ordered spaces imply continuously Urysohn?.
It is known that for many cases, paracompactness and having a property locally imply
this property for the space itself. We do not know if this is true for cU in general, but
we prove that a locally continuously Urysohn paracompact LOTS is continuously Urysohn.
Before proving this theorem, we will prove a useful lemma.
Lemma 5.5. Assume X is a LOTS, a,b?X, and a < b. If (?,b) and (a,?) are continu-
ously Urysohn, then X is continuously Urysohn.
Proof. Suppose that the functions which witness cU on (?,b), and (a,?) are ?1, and ?2
respectively. By using Proposition 2.1. from [3], we can assume that if x < y then ?1(x,y)
sends (?,x] to 0, and [y,?) to 1. Similarly, ?2(x,y) sends (?,x] to 0, and [y,?) to 1.
In addition to these assumptions normality gives a function ? : X?[0,1] such that:
?(x) = 0 for x?a, and ?(x) = 1 for x?b.
In order to prove this lemma, we will de?ne a function ? : X2\??C(X) as follows:
33
If x < y, de?ne
?(x,y)(z) =
?
???
??
???
??
?(z), if z?x,
?(x).?2(x,y)(z) + (1??(y)).?1(x,y)(z) +?(z), if x < z < y,
?(x) + (1??(y)) +?(z), if y?z.
If y < x, de?ne
?(x,y)(z) =
?
???
??
???
??
(1??(x)) +?(y)??(z), if z?y,
(1??(x)).?1(x,y)(z) +?(y).?2(x,y)(z)??(z), if y < z < x,
??(z), if x?z.
Claim 1: ?(x,y)(x)?= ?(x,y)(y) for every (x,y)?X2\?.
Proof of Claim 1. Suppose x < y. Then ?(x,y)(x) = ?(x), and ?(x,y)(y) = ?(x) + 1.
Similarly, if we assume y < x, then ?(x,y)(x) =??(x), and ?(x,y)(y) = 1??(x).
Claim 2: ?(x,y) is continuous for every (x,y)?X2\?.
Proof of Claim 2. Suppose ? > 0, z?X are given, and WLOG we may assume x < y.
Case 1: z?x. Then ?(x,y)(z) = ?(z).
Subcase 1: z < x.
Since ? is a continuous function there exists a neighborhood G? of z such that
|?(z)??(t)|< ?, for every t?G?. We take G = G??(?,x) neighborhood of z. Then
for every t?G,
|?(x,y)(z)??(x,y)(t)|=|?(z)??(t)|< ?.
Subcase 2: z = x, and x = z < y?a.
Since ?, and ?1(x,y) are continuous functions there exist neighborhoods G?, and G1 of
z = x such that|?(t)|< ? for every t?G?, and|?1(x,y)(t)|< ? for every t?G1.
We set G = (?,y)?G??G1.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|0?0|= 0.
If t?G also satis?es t > x = z, then|?(x,y)(x)??(x,y)(t)|=|?1(x,y)(t)|< ?.
34
Subcase 3: z = x, and x = z?a < y < b.
Since ?, and ?1(x,y) are continuous functions there exist neighborhoods G?, and G1 of
z = x such that|?(t)|< ?\2 for every t?G?, and|?1(x,y)(t)|< ?\2 for every t?G1.
We set G = (?,y)?G??G1.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|0?0|= 0.
If t?G also satis?es t > x = z, then
|?(x,y)(x)??(x,y)(t)|=|(1??(y))?1(x,y)(t)??(t)|?
|(1??(y))|.|?1(x,y)(t)|+|?(t)|< ?.
Subcase 4: z = x, and x = z?a < b?y.
Since ? is a continuous function, and ?(x) = 0 there exists a neighborhood G? of z = x
such that|?(t)|< ? for every t?G?.
We set G = (?,y)?G?.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|0?0|= 0.
If t?G also satis?es t > x = z, then|?(x,y)(x)??(x,y)(t)|=|?(t)|< ?.
Subcase 5: z = x, and a?x = z < y < b.
Since ?, ?1(x,y), and ?2(x,y) are continuous functions there exist neighborhoods G?,
G1, and G2 of z = x such that|?(x)??(t)|< ?\3 for every t?G?,|?1(x,y)(t)|< ?\3 for
every t?G1, and|?2(x,y)(t)|< ?\3 for every t?G2.
We set G = (?,y)?G??G1?G2.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|?(x)??(t)|< ?.
If t?G also satis?es t > x = z, then
|?(x,y)(x)??(x,y)(t)|=|?(x)??(x)?2(x,y)(t)?(1??(y))?1(x,y)(t)??(t)|?
|?(x)??(t)|+|?(x)|.|?2(x,y)(t)|+|(1??(y))|.|?1(x,y)(t)|< ?.
Subcase 6: z = x, and a?x = z < b < y.
Since ?, and ?2(x,y) are continuous functions there exist neighborhoods G?, and G2 of
z = x such that | ?(x)??(t) |< ?\2 for every t ? G?, and | ?2(x,y)(t) |< ?\2 for every
t?G2.
35
We set G = (?,y)?G??G2.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|?(x)??(t)|< ?.
If t?G also satis?es t > x = z, then
|?(x,y)(x)??(x,y)(t)|=|?(x)??(x)?2(x,y)(t)??(t)|?
|?(x)??(t)|+|?(x)|.|?2(x,y)(t)|< ?.
Subcase 7: z = x, and b?x = z < y.
Since ?, and ?2(x,y) are continuous functions there exist neighborhoods G?, and G2 of
z = x such that
|?(x)??(t)|< ? for every t?G?, and|?2(x,y)(t)|< ? for every t?G2.
We set G = (?,y)?G??G2.
If t?G also satis?es t < x = z, then|?(x,y)(x)??(x,y)(t)|=|?(x)??(t)|< ?.
If t?G also satis?es t > x = z, then
|?(x,y)(x)??(x,y)(t)|=|?(x)|.|?2(x,y)(t)|< ?.
Case 2: x < z < y. Then ?(x,y)(z) = ?(x)?2(x,y)(z) + (1??(y))?1(x,y)(z) +?(z).
Subcase 1: x < y?a.
Since ?1(x,y) is a continuous function there exists a neighborhood G1 of z such that
|?1(x,y)(z)??1(x,y)(t)|< ?.
We set G = (x,y)?G1. Then,
|?(x,y)(z)??(x,y)(t)|=|?1(x,y)(z)??1(x,y)(t)|< ? for every t?G.
Subcase 2: x?a < y < b.
Since ?1(x,y) is a continuous function there exists a neighborhood G1 of z such that
|?1(x,y)(z)??1(x,y)(t)|< ?.
We set G = (x,y)?G1. Then,
|?(x,y)(z)??(x,y)(t)|=|(1??(y))|.|?1(x,y)(z)??1(x,y)(t)|< ? for every t?G.
Subcase 3: x?a, and b?y.
Since ? is a continuous function there exists a neighborhood G? of z such that
|?(z)??(t)|< ?.
36
We set G = (x,y)?G?. Then,
|?(x,y)(z)??(x,y)(t)|=|?(z)??(t)|< ? for every t?G.
Subcase 4: a < x < y < b.
Since ?, ?1(x,y), and ?2(x,y) are continuous functions there exist neighborhoods G?,
G1, and G2 of z such that
|?(z)??(t)|< ?\3 for every t?G?,|?1(x,y)(z)??1(x,y)(t)|< ?\3 for every t?G1,
and|?2(x,y)(z)??2(x,y)(t)|< ?\3 for every t?G2.
We set G = (x,y)?G??G1?G2.
Then
|?(x,y)(x)??(x,y)(t)|?|?(x)|.|?2(x,y)(z)??2(x,y)(t)|+
|1??(y)|.|?1(x,y)(z)??1(x,y)(t)|+|?(z)??(t)|< ?.
Subcase 5: a < x < b?y.
Since ?, and ?2(x,y) are continuous functions there exist neighborhoods G?, and G2 of
z such that
|?(z)??(t)|< ?\2 for every t?G?, and|?2(x,y)(t)|< ?\2 for every t?G2.
We set G = (x,y)?G??G2. Then,
|?(x,y)(z)??(x,y)(t)|?|?(x)|.|?2(x,y)(z)??2(x,y)(t)|+|?(z)??(t)|< ?.
Subcase 6: b?x < y.
Since ?2(x,y) is a continuous function there exists a neighborhood G2 of z such that
|?2(x,y)(z)??2(x,y)(t)|< ? for every t?G2.
We set G = (x,y)?G2. Then,
|?(x,y)(z)??(x,y)(t)|=|?2(x,y)(z)??2(x,y)(t)|< ? for every t?G.
Case 3: y?z. Then ?(x,y)(z) = ?(x) + (1??(y)) +?(z).
The proof for this case is identical to the proof of case 1. And also the for the case
y < x the proof works similarly.
Claim 3: ? : X2\??C(X) is a continuous function.
37
Proof of Claim. Suppose ? > 0, and (x,y)?X2\? are given. And WLOG we assume
x < y.
Since ? is a continuous function, there exists a neighborhoods U?, and V? of x and y so
that|?(x)??(u)|< ?\4 for every u?U?, and|?(y)??(v)|< ?\4 for every v?V?. Also
note that|?(x)|?1 for every x?X.
If the location of (x,y) with respect to a, and b is examined, it is one of the following:
only ?1(x,y) can be de?ned, only ?2(x,y) can be de?ned or they both can be de?ned.
Also we know that ?1, and ?2 are continuous functions.
Then there exists a neighborhood U1?V1 of (x,y) such that:
|?1(x,y)(z)??1(u,v)(z)|< ?\4 for every (u,v)?U1?V1, and z.
Similarly, there exists a neighborhood U2?V2 of (x,y) such that:
|?2(x,y)(z)??2(u,v)(z)|< ?\4 for every (u,v)?U2?V2, and z.
So, we set U = U1?U2?U??(?,y), and U = V1?V2?V??(x,?), depending on the
location of (x,y). Note that WLOG, we may assume that U1?V1 = U2?V2 =?, and u < v
for every u?U, and v?V.
In addition to these, we assume that |?i(x,y)(z)|?1 for every (x,y)?X2\?, and
z?X, where i = 1,2.
Case 1: z?x.
That is ?(x,y)(z) = ?(z). Under this case two subcases can occur.
Subcase 1: z?u. That is ?(u,v)(z) = ?(z), and we are done.
Subcase 2: u < z < v.
That is ?(u,v)(z) = ?(u).?2(u,v)(z) + (1??(v)).?1(u,v)(z) +?(z). Then,
|?(x,y)(z)??(u,v)(z)|=|??(u)?2(u,v)(z) + (1??(v))?1(u,v)(z)|?
|?(u)|.|?2(u,v)(z)??2(x,y)(z)|+|1??(v)|.|?1(u,v)(z)??1(x,y)(z)|< ?.
Case 2: x < z < y.
That is ?(x,y)(z) = ?(x).?2(x,y)(z) + (1??(y)).?1(x,y)(z) +?(z).
Subcase 1: z?u. That is ?(u,v)(z) = ?(z). Then,
38
|?(x,y)(z)??(u,v)(z)|=|??(x)?2(x,y)(z) + (1??(y))?1(x,y)(z)|?
|?(x)|.|?2(x,y)(z)??2(u,v)(z)|+|1??(y)|.|?1(x,y)(z)??1(u,v)(z)|?
|?2(x,y)(z)??2(u,v)(z)|+|?1(x,y)(z)??1(u,v)(z)|< ?.
Subcase 2: u < z < v.
That is ?(u,v)(z) = ?(u).?2(u,v)(z) + (1??(v)).?1(u,v)(z) +?(z). Then,
|?(x,y)(z)??(u,v)(z)|=|?(x)?2(x,y)(z) + (1??(y))?1(x,y)(z)??(u)?2(u,v)(z)?
(1??(v))?1(u,v)(z)|?
|?(x)?2(x,y)(z)??(u)?2(u,v)(z)??(x)?2(u,v)(z)|+
| (1??(y))?1(x,y)(z)?(1??(v))?1(u,v)(z)?(1??(y))?1(u,v)(z)|?
|?(x)|.|?2(x,y)(z)??2(u,v)(z)|+|?2(u,v)(z)|.|?(x)??(u)|+
|1??(y)|.|?1(x,y)(z)??1(u,v)(z)|+|?1(u,v)(z)|.|?(y)??(v)|?
|?2(x,y)(z)??2(u,v)(z)|+|?(x)??(u)|+|?1(x,y)(z)??1(u,v)(z)|+
|?(y)??(v)|< ?.
Subcase 3: v?z. That is ?(u,v)(z) = ?(u) + (1??(v)) +?(z). Then,
|?(x,y)(z)??(u,v)(z)|=|?(x)?2(x,y)(z)+(1??(y))?1(x,y)(z)??(u)?(1??(v))|?
|?(x)?2(x,y)(z)??(u)??(x)|+| (1??(y))?1(x,y)(z)?(1??(v))?(1??(y))|?
|?(x)|.|?2(x,y)(z)?1|+|?(x)??(u)|+|1??(y)|.|?1(x,y)(z)?1|+
|?(y)??(v)|?
|?2(x,y)(z)?1??2(u,v)(z)|+|?(x)??(u)|+|?1(x,y)(z)?1??1(u,v)(z)|+
|?(y)??(v)|?
|?2(x,y)(z)?1?2(u,v)(z)|+|?(x)??(u)|+|?1(x,y)(z)??1(u,v)(z)|+
|?(y)??(v)|< ?.
Case 3: y?z.
That is ?(x,y)(z) = ?(x) + (1??(y)) +?(z).
Subcase 1: v?z. That is ?(u,v)(z) = ?(u) + (1??(v)) +?(z). Then,
|?(x,y)(z)??(u,v)(z)|=|?(x)??(y)??(u) +?(v)|?
|?(x)??(u)|+|?(y)??(v)|< ?.
39
Subcase 2: u < z < v.
That is ?(u,v)(z) = ?(u).?2(u,v)(z) + (1??(v)).?1(u,v)(z) +?(z). Then,
|?(x,y)(z)??(u,v)(z)|=|?(x)+(1??(y))??(u)?2(u,v)(z)?(1??(v))?1(u,v)(z)|?
|?(x)??(u)?2(u,v)(z)??(u)|+|1??(y)?(1??(v))?1(u,v)(z)?(1??(v))|?
|?(x)??(u)|+|?(u)|.|1??2(u,v)(z)|+|1??(v)|.|1??1(u,v)(z)|+
|?(y)??(v)|?
|?(x)??(u)|+|1??2(u,v)(z)??2(x,y)(z)|+|1??1(u,v)(z)??1(x,y)(z)|+
|?(t)??(v)|?
|?(x)??(u)|+|?2(x,y)(z)??2(u,v)(z)|+|?1(x,y)(z)??1(u,v)(z)|+
|?(y)??(v)|< ?.
After this lemma, we are going to prove that a paracompact, locally continuously
Urysohn LOTS is continuously Urysohn. Here a space X is locally continuously Urysohn
provided that every x?X has an open neighborhood which is also a continuously Urysohn
space.
Theorem 5.6. If X is a paracompact, locally continuously Urysohn LOTS, then X is con-
tinuously Urysohn.
Proof. Suppose X is a paracompact, locally cU LOTS.
Since X is locally cU, then for every x ? X there exists a neighborhood Ux which is
also a cU-space. The collectionU ={Ux : x?X}is an open cover for the space.
By using paracompactness, we ?ndV ={I? : ??A}locally ?nite open re?nement of
U such that for every ??A:
(1) I? = (a?,b?),
(2) ?I? is cU,
(3)|{??A : I??I? ?=?}|?2.
Let us de?ne an equivalence relation?as follows:
40
I??I? if and only if there exists a ?nite chain link between I?, and I?. If we examine
the equivalence classes [I?], it is clear that|[I?]|??. Suppose ?I?=
?
I ?[I ]
I?. Then for every
?,??A:
(i) ?I? is clopen,
(ii) ?I? is convex, and
(iii) ?I?? ?I?=?for ??= ?.
Claim : ?I? is cU for every ??A.
Proof of Claim.
Case 1: [I?] is ?nite.
Then by using the previous lemma, we are done.
Case 2: [I?] is countably in?nite.
Let us denote this connected collection as [I?] ={I0,I1,...}, and pick x,y??I? such that
x < y. We will show that [x,?), and (?,y] are cU.
These proofs work identically, so we will only prove that [x,?) is cU. Note that ?I? can
be densely embedded into a compact LOTS ?X. Let us say X? = ?X\{lastpoint}.
Subcase 1: X?\ ?I? is co?nal in X?. That means for every a ? X? there exists a
b ? X?\ ?I? such that a < b. Then, if I0 = (a0,b0), there exists an x0 ? X?\ ?I? so that
b0 < x0, if I1 = (a1,b1), there exists an x1 ? X?\ ?I? so that b1 < x1, and by following
this idea we have a sequence x0,x1,.... Note that (?,x0]?[x,?) is either empty or can
be covered by ?nitely many of I??s which implies cU. Similarly, [xn,xn+1]?[x,?) is either
empty or cU for every n < ?. That tells us [x,?) is cU.
Subcase 2: X?\ ?I? is not co?nal in X?. That means there exists a p?X? such that
q < p for every q?X?\ ?I?. So WLOG, we assume x < p, and pick a?X? such that p < a.
Then, [x,?) = [x,?)?{[x,a)?(p,?).
So far we we showed that [x,?) is cU for the countable and connected collection ?I?, and
x??I?. With a similar proof one can show that (?,y] is also cU. Combining these together
41
we say ?I? is cU. Since ?I??s make a partition of X, we have a topological sum of cU spaces,
which is also cU.
Note also that if X has a last point m, in addition to this proof we only need to look
at the collection ?I? which contains m. Pick x??I?. By revisiting subcase 1 and subcase 2,
[x,m] can be covered by ?nitely many I??s. So it is cU.
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Chapter 6
Miscellaneous
A set S?X is said to be free for a set function F if x,y?S with x?= y imply x /?F(y).
Free Set Lemma ([11]) If F : X?[X] 0, and z?X are given. For ?xed (x,y);
If (x,z)?A?B we set G = B. Then,
|?(x,y)(z)??(x,y)(t)|=|?B(z)??B(t)|= 0 for every t?G.
If (x,z) /?A?B we set G = Bc. Then,
|?(x,y)(z)??(x,y)(t)|=|?B(z)??B(t)|= 0 for every t?G.
(3) ? is a continuous function. Indeed,
Suppose ? > 0, and (x,y) ? X2 \? are given. We will set U = ?1(P) ? A, and
V = ?2(P)?B, where ?1 is the ?rst projection map from X2 onto X, and ?2 is the second
projection map from X2 onto X. Then,
|?(x,y)(z)??(u,v)(z)|=|?B(z)??B(z)|= 0.
We now suppose that X is a cU-space with the function ?, and ?(x,y)(z) = 0 or
?(x,y)(z) = 1 for every (x,y), and z. WLOG we may assume ?(x,y)(x) = 0, and
?(x,y)(y) = 1.
Let us ?x ? = 1\2, and pick (x,y)?X2\?. By using the continuity of ?, there exists
a neighborhood U?xy?V?xy of (x,y) such that
|?(x,y)(z)??(u,v)(z)|< 1\2 for every (u,v)?U?xy?V?xy, and z?X.
44
Since ? is a function which carries every (x,y)?X2\? to a function which is always
0 or 1 then ?(x,y) = ?(u,v) for every (u,v)?U?xy?V?xy.
For a ?xed (x,y)?X2\? we de?ne the set P ={(u,v)?X2\? : ?(u,v) = ?(x,y)}.
Since ?(x,y) and ?(u,v) are continuous functions, P is closed. In addition to this, by using
the fact that ? is a locally constant function, P is also open. It is clear that for any two
sets P1 and P2 which are de?ned this way, either P1 = P2 or P1?P2 =?. So, we have our
partition P.
Finally, for a point (x,y), the sets (?(x,y))?1(0), and (?(x,y))?1(1) are clopen sets.
Also, P ={(u,v)?X2\? : ?(u,v) = ?(x,y)}?(?(x,y))?1(0)?(?(x,y))?1(1).
Although we do not have an example which is weakly continuously Urysohn, but not
continuously Urysohn there is a candidate. The next example, due to Reed [15]. In [2] it
is veri?ed that X is a continuously symmetrizable space. That implies a zero-set diagonal,
which implies wcU.
In [2] it is also proved that X is not submetrizable. Since submetrizability implies cU,
there is still hope about this space not being cU.
Example 6.3. Let X = X0?X1?U, where X0 = R?{0}, X1 = R?{?1}, and U =
R?(0,?). If x = (a,0)?X0, then x? denotes (a,?1)?X1. For n?N, and x = (a,0)?X0
we let Vn(x) ={x}?{(s,t)?U : (t = s?a)?(0 < t < 1n)}, and Vn(x?) ={x?}?{(s,t)?
U : (t = a?s)?(0 < t < 1n)}.
The topology ? on X is induced by isolating all elements of U, and using the collections
{Vn(x) : n ? ?,n ? 1} and {Vn(x?) : n ? ?,n ? 1} as bases of the topology at x and x?,
respectively.
45
Bibliography
[1] A. V. Arhangel?ski??, On a class of spaces containing all metric and all locally bicompact
spaces, Dokl. Akad. Nauk SSSR 151(1963)751-754.
[2] A. V. Arhangel?ski??, D. K. Burke, Spaces with a regular G?-diagonal, Topology and
Appl. 153(2006)1917-1929.
[3] Harold R. Bennett, David J. Lutzer, Continuously separating families in ordered spaces
and strong base conditions, Topology Appl. 119(2002)305-314.
[4] C. J. R. Borges, On metrizability of topological spaces, Canad. J. Math. 20(1968)795-803.
[5] J. Chaber, M. M. ?Coban, K. Nagami, On monotonic generalizations of Moore spaces,
?Cech complete spaces and p-spaces, Fund. Math. 84(2)(1974)107-119.
[6] R. Engelking, General Topology, Revised and completed ed., Sigma Series in Pure Math-
ematics, no.6, Heldermann Verlag, Berlin, 1989.
[7] G. Gruenhage, Continuously perfectly normal spaces and some generalizations, Trans.
Amer. Math. Soc. 224(2)(1976)323-338.
[8] G. Gruenhage, P. Zenor, Weakly continuously Urysohn spaces, Topology Appl.
156(2009)1957-1961.
[9] L. Halbeisen, N. Hungerb?uhler, On continuously Urysohn and strongly separating spaces,
Topology Appl. 118(3)(2002)329-335.
[10] T. Jech, Set Theory: The Third Millennium Edition, Revised and Expanded, Springer,
Berlin 2002.
[11] I. Juh?asz, Cardinal Functions In Topology-Ten Years Later, Mathematical Centre Tracts
123, Amsterdam 1980.
[12] H. Martin, Contractibility of topological spaces onto metric spaces, Paci?c J. Math.
61(1)(1975)209-217
[13] Ernest Michael, Topologies on spaces of subsets, Transactions of the American Mathe-
matical Society, 71(1951)152-182.
[14] P. J. Nyikos, Some surprising base properties in topology, Part 1, in Studies in topology,
Stavrakas & Allen, ed., Acad. Press, 1975, pp.427-450, Part 2, in Set-theoretic topology,
M.Reed, ed., Acad. Press, 1977, pp.277-305.
46
[15] G. M. Reed, On normality and countable paracompactness, Fun. Math. 110(1980), 145-
152.
[16] E. N. Stepanova, Continuation of continuous functions and the metrizability of
paracompact-p spaces, Math. Zametki 53(3)(1993)92-101(in Russian); translation
in:Math Notes 53(3-4)(1993)308-314.
[17] E. N. Stepanova, On the metrizability of paracompact-p spaces, Vestnik Moskov Univ.
Ser. I Mat. Mekh. 6(1994)69-71(in Russian);translation in: Moscow Univ. Math. Bull.
49(6)(1994)41-43,(1995).
[18] P. Zenor, Some continuous separation axioms, Fund. Math. 90(2)(1975/76)143-158.
[19] P. Zenor, Continuously extending partial functions, Proc. Amer. Math. Soc.
135(1)(2007)305-312
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