Chapter VII Separation Properties

1. Introduction

“Separation” refers here to whether objects such as points or disjoint closed sets can be enclosed in disjoint open sets. In spite of the similarity of terminology, “separation properties” have no direct connection to the idea of “separated sets” that appeared in Chapter 5 in the context of connected spaces.

We have already met some simple separation properties of spaces: the XßX! " and X # (Hausdorff) properties. In this chapter, we look at these and others in more depth. As hypotheses for “more separation” are added, spaces generally become nicer and nicer especially when “separation” is combined with other properties. For example, we will see that “enough separation” and “a nice base” guarantees that a space is metrizable.

“Separation axioms” translates the German term Trennungsaxiome used in the original literature. Therefore the standard separation axioms were historically named XXXX!"#$ , , , ,and X % , each one stronger than its predecessor in the list. Once these had become common terminology, another separation axiom was discovered to be useful and “interpolated” into the list: X" Þ It turns out that the $ # X " spaces (also called Tychonoff spaces) are an extremely well-behaved class of spaces with some $ # very nice properties.

2. The Basics

Definition 2.1 A topological space \ is called a

1) X! -space if, whenever BÁC−\ , there either exists an open set YB−YCÂY with , or there exists an open set Z with C−ZBÂZ ,

2) X" -space if, whenever BÁC−\ , there exists an open set Y with B−YßCÂZ and there exists an open set Z with BÂYßC−Z

3) X# -space (or, Hausdorff space) if, whenever BÁC−\ , there exist disjoint open sets Y and Z in \ such that B−Y and C−Z .

It is immediately clear from the definitions that X# ÊX " ÊXÞ !

Example 2.2

1) \X is a ! -space if and only if: whenever BÁC , then aB Á a C that is, different points in \ have different neighborhood systems.

283 2) If \ has the trivial topology and l\l"\ , then is not a X ! -space.

3) A pseudometric space Ð\ß.Ñ is a metric space in and only if Ð\ß.Ñ is a X ! -space.

Clearly, a metric space is X! . On the other hand, suppose Ð\ß.ÑX is ! and that BÁCÞ Then for some % ! either BÂFÐCÑ% or CÂFÐBÑ % . Either way, .ÐBßCÑ % , so . is a metric.

4) In any topological space \ we candefine an equivalence relation BµC iffaB œ a C . Let 1À\Ä\εœ] by 1ÐBÑœÒBÓÞ Give ] the quotient topology. Then 1 is continuous, onto, open (not automatic for a quotient map!) and the quotient is a X! space:

If S is open in \ , we want to show that 1ÒSÓ is open in ] , and because ] has the quotient topology this is true iff 1" Ò1ÒSÓÓ is open in \ . But 1 " Ò1ÒSÓÓ œÖB−\À1ÐBÑ−1ÒSÓלÖB−\Àfor some C−S , 1ÐBÑœ1ÐCÑ× œÖB−\ÀB is equivalent to some point CSלS in .

If ÒBÓÁÒCÓ−] , then B is not equivalent to C , so there is an open set S©\ with (say) B−S and CÂS . Since 1 is open, 1ÒSÓ is open in ] and ÒBÓ−1ÒSÓ . Moreover, ÒCÓÂ1ÒSÓ or else C would be equivalent to some point of S implying C−SÞ

] is called the X! -identification of \ . This identification turns any space into a X ! -spaceby identifying points that have identical neighborhoods. If \ is a X! -space to begin with, then 1 is one- to-one and 1 is a homeomorphism. Applied to a X! space, the X ! -identification accomplishes nothing. If Ð\ß.Ñ is a pseudometric space, the X ! -identification is the same as the metric identification discussed in Example VI.5.6 because, in that case,aBœ a C if and only if .ÐBßCÑœ!Þ

w w 5) For 3œ!ß"ß#ÀÐ\ßÑ if g is a X3 space and gg ª is a new topology on \ , then Ð\ßÑ g is also a X3 space.

Example 2.3

1) (Exercise) It is easy to check that a space \ is a X " space

iff for each B − \ , ÖB× is closed iff for each B−\ÖBל , + ÖSÀS is open and B−S×

2) A finite X" space is discrete.

3) Sierpinski space \œÖ!ß"× with topology g œÖgßÖ"×ßÖ!ß"××Ñ is X! but not X " : Ö"× is an open set that contains " and not ! ; but there is no open set containing ! and not 1.

4) ‘ , with the right-ray topology, is X! but not X " : if B+C−‘ , then SœÐBß∞Ñ is an open set that contains CB and not ; but there is no open set that contains BC and not .

5) With the cofinite topology,  is X" but not X # because, in an infinite cofinite space, any two nonempty open sets have nonempty intersection.

284 These separation properties are very well-behaved with respect to subspaces and products.

Theorem 2.4 For 3 œ !ß"ß# :

a) A subspace of a X3 -space is a X 3 -space b) If \œ#α−E \Ágα , then \X is a 3 -space iff each \α is a X 3 -space.

Proof All of the proofs are easy. We consider here only the case 3 œ " , leaving the other cases as an exercise.

w a) Suppose +Á,−E©\ , where \X is a " space. If Y is an open set in \ containing B but not C , then YœY∩Ew is an open set in E containing BC but not . Similarly we can find an open set ZE in containing C but not BÞ Therefore EX is a " -space. b) Suppose \œ\#α−E α is a nonempty X" -space. Each \α is homeomorphic to a subspace of \ , so, by part a), each \XÞα is " Conversely, suppose each \Xα is " and that BÁC−\ . Then BÁCα α for some α . Pick an open set Y\αα in containing B αα but not C . Then Yœ+Y α is an open set in \BC containing but not . Similarly, we find an open set Z\CB in containing but not . Therefore \ is a X" -space. ñ

Exercise 2.5 Is a continuous image of a X3 -space necessarily a X 3 -space? How about a quotient? A continuous open image?

We now consider a slightly different kind of separation axiom for a space \ À formally, the definition is “just like” the definition of X# , but with a closed set replacing one of the points.

Definition 2.6 A topological space \ is called regular if whenever J is a closed set and BÂJß there exist disjoint open sets Y and Z such that B−Y and J©Z .

285 There are some easy equivalents of the definition of “regular” that are useful to recognize.

Theorem 2.7 The following are equivalent for any space \ :

i) \ is regular

ii) if S is an open set containing B , then there exists an open set Y©\ such that B−Y© cl Y©S

iii) at each point B − \ there exists a neighborhood base consisting of closed neighborhoods.

Proof i)Ê ii) Suppose \ is regular and S is an open set with B−SÞ Letting Jœ\S , we use regularity to get disjoint open sets YßZ with B−Y and J©Z as illustrated below:

Then B−Y© cl Y©S (since cl Y©\Z ).

ii)ÊR− iii) If aB , then B−SœR int . By ii), we can find an open set Y so that B−Y©Y©S©RÞcl Since cl Y is a neighborhood of B , the closed neighborhoods of B form a neighborhood base at B .

iii)Ê i) Suppose J is closed and BÂJ . By ii), there is a closed neighborhood OB of such that B−O©\J . We can choose YœO int and Zœ\O to complete the proof that \ is regular. ñ

Example 2.8 Every pseudometric space Ð\ß.Ñ is regular. Suppose +ÂJ and J is closed. We have a continuous function 0ÐBÑœ.ÐBßJÑ for which 0Ð+Ñœ-! and 0lJœ!Þ This gives us disjoint "- " - open sets with +−Yœ0 ÒÐß∞ÑÓ# and J©Zœ0 ÒÐ∞ßÑÓ# . Therefore \ is regular.

At first glance, one might think that regularity is a stronger condition than X# . But this is false: if Ð\ß .Ñ is a pseudometric space but not a metric space, then \ is regular but not even X ! .

286 To bring things into line, we make the following definition.

Definition A topological space \ is called a X\$ -space if is regular and X " .

It is easy to show that XÊX$#"! ( ÊXÊX ): suppose \ is X $ and BÁC−\ . Then JœÖC× is closed so, by regularity, there are disjoint open sets YZ , with B−Y and C−ÖCשZ .

Caution Terminology varies from book to book. For some authors, the definition of “regular” includes XÀ" for them, “regular” means what we have called “X $ .” Check the definitions when reading other books.

Exercise 2.10 Show that a regular X! space must be X $ ( so it would have been equivalent to use “X! ” instead of “ X " ” in the definition of “ X $ ” ).

# Example 2.11 XÊX#Î $ . We will put a new topology on the set \œ‘ . At each point :−\ , let a neighborhood base U: consist of all sets R of the form

RœFÐ:ÑÐ% a finite number of straight lines through :Ñ∪Ö:× for some % !Þ

(Check that the conditions in the Neighborhood Base Theorem III.5.2 are satisfied. ) With the resulting topology, \ is called the slotted plane . Note that F% Ð:Ñ − U: ( because “! ” is a finite # number ), so each F% Ð:Ñ is among the basic neighborhoods in U:  so the slotted plane topology on ‘ contains the usual Euclidean topology. It follows that \ is X # .

The set JœÖÐBß!ÑÀBÁ!ל “the B -axis with the origin deleted” is a closed set in \ (why? ).

If Y is any open set containing the origin Ð!ß!Ñ , then there is a basic neighborhood R with Ð!ß!Ñ−R©YÞUsing the % in the definition of R , we can choose a point :œÐBß!Ñ−J with !+B+Þ% Every basic neighborhood set of : must intersect R ( why?) and therefore must intersect Y . It follows that Ð!ß!Ñ and J cannot be separated by disjoint open sets, so the slotted plane is not regular (and therefore not X$ ).

287 Note: The usual topology in ‘# is regular. T his example shows that an “enlargement” of a regular (or X$ ) topology may not be regular (or X $ ). Although the enlarged topology has more open sets to work with, there are also more “point/closed set pairs Bß J ” that need to be separated. By contrast, it is easy to see that an “enlargement” of a X3 topology Ð3œ!ß"ß#Ñ is still XÞ3

Example 2.12 The Moore plane > (Example III.5.6) is clearly X# Þ In fact, at each point, there is a neighborhood base of closed neighborhoods. The figure illustrates this for a point T on the B -axis and a point U above the B -axis. Therefore > is X $ .

Theorem 2.13 a) A subspace of a regular (X$ Ñ space is regular ( X $ ). b) Suppose \œ#α−E \Ág\α . is regular ( X$ ) iff each \α is regular ( X $ ).

Proof a) Let E©\ where \ is regular. Suppose +−E and that J is a closed set in E that does not contain + . There exists a closed set Jw in \ such that J∩EœJÞ w Choose disjoint open sets Yww and Z in \ with +−Y www and J©ZÞ Then YœY∩E w and ZœZ∩E w are open in E , disjoint, +−Y , and J©Z . Therefore E is regular.

b) If \ œ#α−E \α Á g is regular, then part a) implies that each \α is regular ß because each \α is homeomorphic to a subspace of \ . Conversely, suppose each \ α is regular and that

Yœ+YßÞÞÞßYα" α 8 is a basic open set containing B . For each α3 , we can pick an open set Z α 3 in

\α3such that B−Z© αααα 3 3 cl Z©YÞ 3 3 Then B−Zœ+ZßÞÞÞßZ αα" 8 © cl Z

©+cl Zα" ßÞÞÞß cl Z α 8 ©YÞ (Why is the last inclusion true? ) Therefore \ is regular.

Since the X" property is hereditary and productive, a) and b) also hold for X$ -spaces ñ

The obvious “next step up” in separation is the following:

Definition 2.14 A topological space \ is called normal if, whenever EßF are disjoint closed sets in \, there exist disjoint open sets YZ\E©YF©ZÞ\ , in with and is called a X% -space if \ is normal and X1 Þ

288 Example 2.15 a) Every pseudometric space Ð\ß.Ñ is normal (so every metric space is XÑ% . .ÐBßEÑ In fact, if EF and are disjoint closed sets, we can define 0ÐBÑœ .ÐBßEÑ < .ÐBßFÑ . Since the " denominator cannot be !0 , is continuous and 0lEœ!0lFœ"Þ , The open sets YœÖBÀ0ÐBÑ+×# " and ZœÖBÀ0ÐBÑ×# are disjoint that contain EF and respectively. Therefore \ is normal. Note: the argument given is slick and clean. Can you show Ð\ß .Ñ is normal by directly constructing a pair of disjoint open sets that contain E and F ?

b) Let ‘ have the right ray topology gœÖÐBß∞ÑàB− ‘‘‘g ×∪Ögß × . Ð ß Ñ is normal because the only possible pair of disjoint closed sets is g and \ and we can separate these using the disjoint open sets YœgZœ\Þ and Also, ÐßÑ‘ g is not regular : for example " is not in the closed set JœÐ∞ß!Ó , butevery open set that contains J also contains " . So normal ÊÎ regular . But ÐßÑX‘ g is not " and therefore not XÞ %

When we combine “normal< X" ” into X % , we have a property that fits perfectly into the separation hierarchy.

Theorem 2.16 X% ÊX $ ÐÊX # ÊX " ÊXÑ !

Proof Suppose \XJ is % . If is a closed set not containing BÖB×J , then and are disjoint closed sets. By normality, we can find disjoint open sets separating ÖB×J and . It follows that \ is regular and therefore X$ . ñ

.

289 Exercises

E1. \ is called a door space if every subset is either open or closed. Prove that if a X# -space \ contains two points that are not isolated, then \ is not a door space, and that otherwise \ is a door space.

E2. A base for the closed sets in a space \ is a collection of Y of closed subsets such that every closed set J is an intersection of sets from Y . Clearly, Y is a base for the closed sets in \ iff UœÖSÀSœ\JßJ−× Y is a base for the open sets in \ .

For a polynomialT in 8 real variables, define the zero set of T as

8 ^ÐTÑœÖÐB"#8 ßB ßÞÞÞßB Ñ−‘ À T ÐB "#8 ßB ßÞÞÞßB Ñ œ!×

a) Prove that Ö^ÐTÑÀT a polynomial in 8 real variables × is the base for the closed sets of a topology (called the Zariski topology) on ‘8Þ

8 b) Prove that the Zariski topology on ‘ is X" but not X # .

c) Prove that the Zariski topology on ‘ is the cofinite topology, but that if 8  " , the Zariski topology on ‘8 is not the cofinite topology Þ

Note: The Zariski topology arises in studying algebraic geometry. After all, the sets ^ÐT Ñ are rather special geometric objects—those “surfaces” in ‘8 which can be described by polynomial equations TÐB" ßB # ßÞÞÞßB 8 Ñ œ! .

E3. A space \X is a &Î# space if, whenever BÁC−\ , there exist open sets YZ and such that B− Y, C − Z and cl YZœg ∩ cl .Clearly, ( T$ Ê XÊÞ&Î# T # )

a) Prove that a subspace of a X&Î# space is a X &Î# space.

b) Suppose \œ\Ág# α . Prove that \X is &Î# iff each \Xα is &Î# .

c) Let WœÖÐBßCÑ−‘# ÀC !× and PœÖÐBßCÑ−WÀCœ!×Þ Define a topology on W with the following neighborhood bases:

if :−WPÀ U: œÖFÐ:Ñ∩WÀ!×% % if :−P : U: œÖFÐ:Ñ∩ÐWPÑ∪Ö:×À% % !×

You may assume that these U: 's satisfy the axioms for a neighborhood base.

Prove that S is T&Î# but not T $ .

290 E4. Suppose E©\ . Define a topology on \ by

g œÖS©\ À SªE×∪Ög×

Decide whether or not Ð\ßg Ñ is normal.

E5. A function 0 À \ Ä ] is called perfect if 0 is continuous, closed, onto, and, for each C − ] ß " 0 ÐCÑ is compact. Prove that if \ is regular and 0 is perfect, then ] is regular à\X and that if is $ , the ] is also X$ Þ

E6. a) Suppose \ is finite . Prove that Ð\ßÑg is regular iff there is a partition U of \ that is a base for the topology.

b) Give an example to show that a compact subset O of a regular space \ need not be closed. However, show that if \ is regular then cl O is compact.

c) Suppose J is closed in a X$ -space \Þ Prove that

i) Prove that Jœ+ ÖSÀS is open and J©S×Þ

ii) Define BµC iff BœCBßC−JÞ or Prove that the quotient space \ε is Hausdorff.

d) Suppose F is an infinite subset of a X$ -space \ . Prove that there exists a sequence of open sets Y8 such that each Y∩FÁg 8 and that cl Y∩Yœg 87 cl whenever 8Á7 .

e) Suppose each point C] in a space has a neighborhood Z such that cl Z is regular. Prove that ] is regular.

291 3. Completely Regular Spaces and Tychonoff Spaces

The X$ property is well-behaved. For example, we saw in Theorem 2.13 that the X $ property is hereditary and productive. However, the X$ property is not sufficiently strong to give us really nice theorems.

For example, it's very useful if a space has many (nonconstant) continuous real-valued functions available to use. Remember how many times we have used the fact that continuous real-valued functions 0 can be defined on a metric space Ð\ß.Ñ using formulas like 0ÐBÑœ.ÐBß+Ñ or 0ÐBÑœ.ÐBßJÑ; when l\l" , we get many nonconstant real functions defined on Ð\ß .ÑÞ But a X$-space can sometimes be very deficient in continuous real-valued functions in 1946, Hewett gave an example of a infinite X$ -space L on which the only continuous real-valued functions are the constant functions.

In contrast, we will see that the X% property is strong enough to guarantee the existence of lots of continuous real-valued functions and, therefore, to prove some really nice theorems (for example, see Theorems 5.2 and 5.6 later in this chapter). The downside is that X% -spaces turn out also to exhibit some very bad behavior: the X% property is not hereditary ( explain why a proof analogous to the one given for Theorem 2.13b) doesn't work ) and it is not even finitely productive. Examples of such bad behavior are a little hard to find right now, but later they will appear rather naturally.

These observations lead us to look first at a class of spaces with separation somewhere “between X$ and X% .” We want a group of spaces that is well-behaved, but also with enough separation to give us some very nice theorems. We begin with some notation and a lemma.

Recall that GÐ\ÑœÖ0−‘\ À0 is continuous ל the collection of continuous real-valued functions on \

GÐ\ÑœÖ0−GÐ\ÑÀ0‡ is bounded ל the collection of continuous bounded real-valued functions on \

Lemma 3.1 Suppose 0ß1−GÐ\Ñ . Define real-valued functions 0”1 and 0•1 by

(0 ”1ÑÐBÑœ max Ö0ÐBÑß1ÐBÑ× Ð0 •1ÑÐBÑ œ min Ö0ÐBÑß1ÐBÑ×

Then 0”1 and 0•1 are in GÐ\ÑÞ

Proof We want to prove that the max or min of two continuous real-valued functions is continuous. But this follows immediately from the formulas

0ÐBÑ < 1ÐBÑ l0ÐBÑ1ÐBÑl ( 0 ” 1ÑÐBÑ œ #< #

0ÐBÑ < 1ÐBÑ l0ÐBÑ1ÐBÑl Ð0•1ÑÐBÑœ# # ñ

292 Definition 3.2 A space \ is called completely regular if whenever J is a closed set and +ÂJß there exists a function 0−GÐ\Ñ such that 0Ð+Ñœ! and 0lJœ"Þ

Informally, “completely regular” means that “+ and J can be separated by a continuous real-valued function.”

Note i) The definition requires that 0lJœ" , in other words, that J©0" ÒÖ"×ÓÞ However, these two sets might not be equal.

ii) If there is such a function 0 , there is also a continuous 1À\ÄÒ!ß"Ó such that 1Ð+Ñœ! and 1lJœ" . For example, we could use 1œÐ0”!Ñ•" which, by Lemma 3.1, is continuous.

iii) Suppose 1À\ÄÒ!ß"Ó is continuous and 1Ð+Ñœ!1lJœ"Þ , The particular values !ß" in the definition are not important.: they could be any real numbers <+=Þ Ð If we choose a homeomorphism 9 ÀÒ!ß"ÓÄÒ<ß=Ó , thenit must be true that either 9 Ð!Ñœ< , 9 Ð"Ñœ= or 9 Ð!Ñœ= , 9Ð"Ñœ<why?. Then 2œ‰1À\ÄÒ<ß=Ó 9 and ß depending on how you chose 9 ß2Ð+Ñœ< and 2lJ œ , or vice-versa. )

Putting these observations together, we see Definition 3.2 is equivalent to:

Definition 3.2w A space \ is called completely regular if whenever J is a closed set, +ÂJß and <ß= are real numbers with <+=ß then there exists a continuous function 0À\ÄÒ<ß=Ó for which 0Ð+Ñœ< and 0lJœ=Þ

In one way, the definition of “completely regular space” is very different the definitions for the other separation properties: the definition isn't “internal”because an “external” space, ‘ is an integral part of the definition. While it is possible to contrive a purely internal definition for “completely regular,” the definition is complicated and seems completely unnatural: it simply imposes some very unintuitive conditions to force the existence of enough functions in GÐ\Ñ .

Example 3.3 Suppose Ð\ß .Ñ is a pseudometric space with a closed subset J and +ÂJÞ Then .ÐBßJÑ 0ÐBÑœ.Ð+ßJÑ is continuous, 0Ð+Ñœ" and 0lJœ! . So Ð\ß.Ñ is completely regular, but if . is not a metric, then this space is not even X! .

Definition 3.4 A completely regular X" -space \ is called a Tychonoff space (or X " -space). $ #

Theorem 3.5 X" ÊX$ ÐÊX # ÊX " ÊXÑ ! $ # Proof Suppose J is a closed set in \ not containing +\X . If is " , we can choose 0−GÐ\Ñ with $ # "" " " 0Ð+Ñœ!and 0lJœ"Þ Then Yœ0 ÒÐ∞ß# ÑÓ and Zœ0 ÒÐß∞ÑÓ # are disjoint open sets with +−YßJ©ZÞ Therefore \ is regular. Since \X\XÞñ is " , is $

Hewitt's example of a X$ space on which every continuous real-valued function is constant is more than enough to show that a X$ space may not be X " (the example, in Ann. Math., 47(1946) 503-50 9, is $ # rather complicated. ). For that purpose, it is a little easier but still nontrivial X\ to find a $ space containing two points :ß; such that for all 0−GÐ\Ñß0Ð:Ñœ0Ð;ÑÞ Then : and Ö;× cannot be separated by a function from GÐ\Ñ so \ is not T" Þ (See D.J. Thomas, A regular space, not $ # completely regular, American Mathematical Monthly, 76(1969), 181-182). The space \ can then be used to construct an infinite X$ space L (simpler than Hewitt's example) on which every continuous

293 real-valued function is constant (see Gantner, A regular space on which every continuous real-valued function is constant, American Mathematical Monthly, 78(1971), 52.) Although we will not present these constructions here, we will occasionally refer to L in comments later in this section.

Note: We have not yet shown that X% Ê X " : this is true (as the notation suggests), but it is not at all $ # easy to prove: try it! This result is in Corollary 5.3.

Tychonoff spaces continue the pattern of good behavior that we saw in preceding separation axioms, and they will also turn out to be a rich class of spaces to study.

Theorem 3.7 a) A subspace of a completely regular (XÑ" space is completely regular ( XÑ" . $ # $ # b) Suppose \œ \Ág\α . is completely regular ( XÑ" iff each \ α is #α−E $ # completely regular (X" Ñ . $ #

Proof Suppose +ÂJ©E©\ , where \ is completely regularand J is a closed set in E . Pick a closed set O in \ such that O∩EœJ and an 0−GÐ\Ñ such that 0Ð+Ñœ! and 0lOœ" . Then 1œ0lE−GÐEÑß1Ð+Ñœ! and 1lJœ"Þ Therefore E is completely regular.

If gÁ\œ\#α−E α is completely regular, then each \ α is homeomorphic to a subspace of \\so each α is completely regular. Conversely, suppose each \α is completely regular and that J is a closed set in \ not containing + . There is a basic open set Y such that

+−Yœ +Yα" ßÞÞÞY α 8  ©\J

For each 3œ"ßÞÞÞß8 we can pick a continuous function 0À\ÄÒ!ß"Óαα33 with 0Ð+Ñœ! αα 33 and

0α3| Ð\ α 3 Y α 3 Ñœ"Þ Define 0À\ÄÒ!ß"Ó by

0ÐBÑœ max ÖÐ0α3 ‰1 α 3 ÑÐBÑÀ3œ"ßÞÞÞß8ל max Ö0α3 ÐB α 3 ÑÀ3œ"ßÞÞÞß8×

Then 0 is continuous and 0Ð+Ñœ max Ö0Ð+ÑÀ3œ"ßÞÞÞß8ל!Þα3 α 3 If B−J , then for some 3ß

Bα3 ÂY α 3 and 0ÐBÑœ" αα 33 , so 0ÐBÑœ" . Therefore 0lJœ" and \ is completely regular. . Since the X" property is both hereditary and productive, the statements in a) and b) also hold for X" Þ ñ $ #

7 Corollary 3.8 For any cardinal 7 , the “cube” Ò!ß"Ó and all its subspaces are X " . $ #

Since a Tychonoff space \ is defined using functions in GÐ\Ñ , we expect thatthese functions will have a close relationship to the topology on \ . We want to explore that connection.

Definition 3.9 Suppose 0−GÐ\Ñ . Then ^Ð0Ñœ0" ÒÖ!×ÓœÖB−\À0ÐBÑœ!× is called the zero set of 0 . If Eœ^Ð0Ñ for some 0−GÐ\Ñ , we call E a zero set in \ . The complement of a zero set in \ is called a cozero set: coz Ð0Ñœ\^Ð0ÑœÖB−\À0ÐBÑÁ!×Þ

∞ A zero set ^Ð0Ñ\ in is closed because 0 is continuous. In addition, ^Ð0ÑœS+8œ" 8 , where " S8 œÖB−\Àl0ÐBÑl+8 ×. Each S8 is open. Therefore a zero set is always a closed K $ -set. Taking complements shows that cozÐ0Ñ is always an open J5 -set in \ .

294 For 0−GÐ\Ñ , let 1œÐ"”0Ñ•"−GÐ\ÑÞ‡ Then ^Ð0Ñœ^Ð1Ñ . Therefore GÐ\Ñ and GÐ\ч produce the same zero sets in \ (and therefore also the same cozero sets).

Example 3.10

1) A closed set J in a pseudometric space Ð\ß.Ñ is a zero set: Jœ^Ð0Ñ , where 0ÐBÑœ.ÐBßJÑÞ

2) In general, a closed set in \ might not be a zero set  in fact, a closed set in \ might not even be a K$ set.

Suppose \ is uncountable and :−\ . Define a topology on \ by letting UB œÖÖB×× be a neighborhood at each point BÁ: and letting U: œÖFÀ:−F and \F is countable × be the neighborhood base at : . ( Check that the conditions of the Neighborhood Base Theorem III.5.2 are satisfied Þ )

All points in \Ö:× are isolated and \ is clearly XÞ" In fact, \X is % .

If EF and are disjoint closed sets in \ , then one of them (say EÑ satisfies E©\Ö:×, so E isclopen. We then have open sets YœE and Zœ\EªF , so \ is normal.

We do not know yet that XÊX% " in general, but it's easy to see that this space \ is also X " . $ # $ #

If J is a closed set not containing B , then either J©\Ö:×ÖBש\Ö:× or . So one of the sets J or ÖB× is clopen and he characteristic function of that clopen set is continuous and works to show that \ is completely regular.

The set Ö:× is closed but Ö:× is not a K$ set in \Ö:× , so is not a zero set in \ .

∞ Suppose :−SS+8œ" 8 where 8 is open. For each 8 , there is a basic neighborhood F 8 of : such that :−F©Sß88 so \S©\F 8 8 is countable. Therefore ∞ ∞ \+8œ" Sœ8 - 8œ" Ð\SÑ 8 is countable. Since \ is uncountable, we conclude ∞ that Ö:×Á+8œ" SÞ8

Even when J is both closed and a K$ set, J might not be a zero set. We will see examples later.

For purely technical purposes, it is convenient to notice that zero sets and cozero sets can be described in a many different forms. For example, if 0 − GÐ\Ñ , then we can see that each set in the left column is a zero set by choosing a suitable 1 − GÐ\Ñ À

^œÖBÀ 0ÐBÑœ< × œ^Ð1Ñß where 1ÐBÑœ0ÐBÑ< ^œÖBÀ0ÐBÑ 0× œ^Ð1Ñ , where 1ÐBÑœ 0ÐBÑl0ÐBÑl ^œÖBÀ0ÐBÑŸ 0× œ^Ð1Ñ , where 1ÐBÑœ 0ÐBÑ

295 On the other hand, if 1−GÐ\Ñ , we can write ^Ð1Ñ in any of the forms listed above by choosing an appropriate function 0 − GÐ\Ñ À

^Ð1ÑœÖBÀ0ÐBÑœ< × where 0ÐBÑœ1ÐBÑ<< ^Ð1ÑœÖBÀ0ÐBÑ 0× where 0ÐBÑœl1ÐBÑl ^Ð1ÑœÖBÀ0ÐBÑŸ 0 × where 0ÐBÑœl1ÐBÑl ^Ð1ÑœÖBÀ0ÐBÑ r× where 0ÐBÑœ<l1ÐBÑl ^Ð1ÑœÖBÀ0ÐBÑŸr× where 0ÐBÑœ<

Taking complements, we get the corresponding results for cozero sets: if 0 − GÐ\Ñ i) the sets ÖBÀ0ÐBÑÁ<× , ÖBÀ0ÐBÑ +!× , ÖBÀ0ÐBÑ !×ßÖBÀ0ÐBÑ +<× , { BÀ0ÐBÑ< } are cozero sets, and ii) any given cozero set can be written in any one of these forms.

Using the terminology of cozero sets, we can see a nice comparison/contrast between regularity and complete regularity. Suppose BÂJ , where J is closed in \Þ\ If is regular, we can find disjoint open sets YZB−YJ©ZÞ and with and But if \ is completely regular, we can separate BJ and with “special” open sets Y and Z cozero sets! Just choose 0−GÐ\Ñ with 0ÐBÑœ! and 0lJœ" , then " " B−YœÖBÀ0ÐBÑ+# × and J©ZœÖBÀ0ÐBÑ# ×

In fact this observation characterizes completely regular spaces  that is, if a regular space fails to be completely regular, it is because there is a “shortage” of cozero sets because there is a “shortage” of functions in GÐ\Ñ (see Theorem 3.12, below) . For the extreme case of a X$ space L on which the only continuous real valued functions are constant, the only cozero sets are g and L !

The next theorem reveals the connections between cozero sets, GÐ\Ñ and the weak topology on \Þ

‡ Theorem 3.11 For any space Ð\ßÑGÐ\Ñg , and GÐ\Ñ induce the same weak topology g A on \ , and a basefor gA is the collection of all cozero sets in \ .

" Proof A subbase for gA consists of all sets of the form 0ÒYÓ , where Y is open in ‘ and 0−GÐ\ÑÞ Without loss of generality, we can assume the sets Yare subbasic open sets of the form Ð+ß∞Ñ and Ð∞ß,Ñ, so that the sets 0" ÒYÓ have form ÖB−\À0ÐBÑ+× or ÖB−\À0ÐBÑ+,× . But these are cozero sets of \ , and every cozero set in \ has this form. So the cozero sets are a subbase for gA In fact, the cozero sets are actually a base because cozÐ0Ñ∩ coz Ð1Ñœ coz Ð01Ñ :the intersection of two cozero sets is a cozero set.

The same argument, with GÐ\ч replacing GÐ\Ñß shows that the cozero sets of GÐ\ч are a base for the weak topology on \ generated by GÐ\ÑÞ‡ But GÐ\Ñ and GÐ\Ñ ‡ produce the same cozero sets in \, and therefore generate the same weak topology gA on \. ñ

Now we can now see the close connection between \ and GÐ\Ñ in completely regular spaces. For any space Ð\ßÑg the functions in GÐ\Ñ certainly are continuous with respect to g (by definition of GÐ\Ñ ). But is g the smallest topology making this collection of functions continuous? In other

296 words, is g the weak topology on \ generated by GÐ\Ñ ? The next theorem says that is true precisely when \ is completely regular.

Theorem 3.12 For any space Ð\ßg Ñ , the following are equivalent:

a) \ is completely regular b) The cozero sets of \ are a base for the topology on \ (equivalently, the zero sets of \ are a base for the closed sets—meaning that every closed set is an intersection of zero sets) c) \ has the weak topology from GÐ\Ñ (equivalently, from GÐ\ч ) d) GÐ\Ñ (equivalently, GÐ\ч ) separates points from closed sets.

Proof The preceding theorem shows that b) and c) are equivalent.

a)Ê b) Suppose +−S where S is open Þ Let Jœ\SÞ Then we can choose 0−GÐ\Ñ " with 0Ð+Ñœ! and 0lJœ" . Then YœÖBÀ0ÐBÑ+×# is a cozero set for which +−Y©SÞ Therefore the cozero sets are a base for \ .

b)Ê d) Suppose J is a closed set not containing + . By b), we can choose 0−GÐ\Ñ so that +−coz Ð0Ñ©\JÞ Then 0Ð+Ñœ<Á! , so 0Ð+Ñ cl 0ÒJÓœÖ!×Þ Therefore GÐ\Ñ separates points and closed sets.

d)Ê a) Suppose J is a closed set not containing + . There is some 0−GÐ\Ñ for which 0Ð+Ñ Âcl 0ÒJÓÞ Without loss of generality (why? ), we can assume 0Ð+Ñœ!Þ Then, for some % ! , ÐßÑ∩0ÒJÓœg%% , so that for B−J , l0ÐBÑl % Þ Define 1−GÐ\ч by 1ÐBÑœ min Öl0ÐBÑlß×Þ% Then 1Ð+Ñœ! and 1lJœ% , so \ is completely regular.

At each step of the proof, GÐ\Ñ can be replaced by GÐ\ч (check! ) ñ

The following corollary is curious and the proof is a good test of whether one understands the idea of “weak topology.”

Corollary 3.13 Suppose \ is a set and let gY be the weak topology on \ generated by any family of \ functions Y‘g© . Then Ð\ßÑY is completely regular ÞÐ\ß gY )is Tychonoff is Y separates points.

Proof Give \ the topology the weak topology gY generated by Y . Now \ has a topology, so the collection GÐ\Ñ makes sense. Let XA be the weak topology on \ generated by GÐ\ÑÞ The topology gY does make all the functions in GÐ\Ñ continuous, so gA ©Þ g Y On theother hand: Y© GÐ\Ñ by definition of g Y ,and the larger collection of functions GÐ\Ñ generates a (potentially) larger weak topology. Therefore gY © g A Þ Therefore gY œ gA . By Theorem 3.12, Ð\ßÑ g Y is completely regular. ñ

297 Example 3.14

‘ 1) If Y‘œÖ0−À0 is nowhere differentiable × , then the weak topology gY on ‘ generated by Y is completely regular.

2) If L is an infinite X $ space on which every continuous real-valued function is constant ( see the comments at the beginning of this Section 3 ), then the weak topology generated by GÐ\Ñ has for a base the collection of cozero sets {gßL× . So the weak topology generated by GÐ\Ñ is the trivial topology, not the original topology on \ .

Theorem 3.12 leads to a lovely characterization of Tychonoff spaces.

Corollary 3.15 Suppose \ is a Tychonoff space. For each 0 −G‡ Ð\Ñ , we have ‡ ranÐ0Ñ©Ò+ß,0 0 ] œM 0 for some ++,−Þ 0 0 ‘ The evaluation map /À\Ä# ÖMÀ0−GÐ\Ñ×0 is an embedding.

‡ Proof \ is X" , the 0 's are continuous and the collection of 0 's ÐœG Ð\ÑÑ separates points and closed sets. By Corollary VI.4.11, / is an embedding. ñ

‡ 7 Since each M0 is homeomorphic to Ò!ß"Ó , # ÖMÀ0−GÐ\Ñ×0 is homeomorphic to Ò!ß"Óß where 7 œ lG‡ Ð\Ñl . Therefore any Tychonoff space can be embedded in a “cube.” On the other hand (Corollary 3.8) Ò!ß "Ó 7 and all its subspaces are Tychonoff. So we have:

Corollary 3.16 A space \\ is Tychonoff iff is homeomorphic to a subspace of a “cube” Ò!ß "Ó 7 for some cardinal number 7 .

The exponent 7 œ lG* Ð\Ñl in the corollary may not be the smallest exponent possible. As an extreme case, for example, we have -œlGÐÑl‡ ‘ , even though we can embed ‘ in Ò!ß"ÓœÒ!ß"ÓÞ" The following theorem improves the value for 7 in certain cases ( and we proved a similar result for metric spaces Ð\ß .Ñ : see Example VI.4.5. )

Theorem 3.17 Suppose \ is Tychonoff with a base U of cardinality 7 . Then \ can be embedded in Ò!ß"Ó7. In particular, \ can be embedded in Ò!ß"Ó AÐ\Ñ .

Proof Suppose 7 is finite. Since \XÖBלÖFÀF is " , + is a basic open set containing B×Þ Only finitely many such intersections are possible, so \ is finite and therefore discrete. Hence top top \ ©Ò!ß"Ó©Ò!ß"Ó7 Þ Suppose U is a base of cardinal 7 where 7 is infinite. Call a pair ÐYßZÑ−‚U U " distinguished if there exists a continuous 0YßZ À\ÄÒ!ß"Ó with 0 YßZ ÐBÑ+# for all B−Y and 0ÐBÑœ"Y ßZ for all B−\ZÞ Clearly, Y©Z for a distinguished pair ÐYßZÑÞ For each distinguished pair, pick such a function 0YßZ and let Y œÖ0 YßZ ÀÐYßZÑ−‚ U U is distinguished × .

298

We note that if +−Z−U , then there must exist Y− U such that +−YÐYßZÑ and is distinguished Þ To see this, pick an 0À\ÄÒ!ß"Ó so that 0Ð+Ñœ! and 0l\Zœ"Þ Then choose Y− U so that " " +−Y©0 ÒÒ!ß# ÑÓ©ZÞ

We claim that Y separates points and closed sets:

Suppose J is a closed set not containing + . Choose a basic set Z−U with +−Z©\JÞ " There is a distinguished pair ÐYßZÑ with +−Y©Z©\JÞ Then 0Y ßZ Ð+Ñœ<+ # and 0YßZ lJœ", so 0 YßZ Ð+Ñ cl 0 YßZ ÒJÓœÖ"×Þ

By Corollary VI.4.11, / À \ Ä Ò!ß "ÓlY l is an embedding. Since 7 is infinite, lY lŸl U ‚ U lœ7œ7# . ñ

A theorem that states that certain topological properties of a space \ imply that \ is metrizable is called a “metrization theorem.” Typically the hypotheses of a metrization theorem involve that 1) \ has “enough separation” and 2) \ has a “sufficiently nice base.” The following theorem is a simple example.

Corollary 3.18 (“Baby Metrization Theorem”) A second countable Tychonoff space \ is metrizable. top Proof By Theorem 3.17, \©Ò!ß"Ói! . Since Ò!ß"Ó i ! is metrizable, so is \ñ.

In Corollary 3.18, \ turns out to be metrizable and separable (since \ is second countable). On the other hand Ò!ß "Ó i! and all its subspaces are separable metrizable spaces. Thus, the corollary tells us that the separable metrizable spaces (topologically) are precisely the second countable Tychonoff spaces.

299 Exercises

E7. Prove that if \ is a countable Tychonoff space, then there is a neighborhood base of clopen sets at each point. (Such a space \ is sometimes called zero-dimensional.)

E8. Prove that in any space \ , a countable union of cozero sets is a cozero set or, equivalently, that a countable intersection of zero sets is a zero set.

E9. Prove that the following are equivalent in any Tychonoff space \ :

a) every zero set is open b) every K$ set is open c) for each 0−GÐ\ÑÀ0Ð:Ñœ! if then there is a neighborhood R of : such that 0 ±R ´!

E10. Let 3 À‘ Ä ‘ be the identity map and let

Ð3ÑœÖ0−GБ ÑÀ0œ13 for some 1−GÐ ‘ Ñ }.

Ð3Ñ is called the ideal in GÐÑ‘ generated by the element 3 .

For those who know a bit of algebra: if we definite addition and multiplication of functions pointwise, then GÐÑБ or, more generally, GÐ\ÑÑ is a commutative ring Þ The constant function ! is the zero element in the ring; there is also a unit element, namely the constant function “" .”

a) Prove that Ð3ÑœÖ0−GÐÑÀ0Ð!Ñœ!‘ andthe derivative 0Ð!Ñw exists. ×

b) Exhibit two functions 0ß1 in GÐÑ‘ for which 01−Ð3Ñ yet 0ÂÐ3Ñ and 1ÂÐ3Ñ .

c) Let \ be a Tychonoff space with more than one point. Prove that there are two functions 0ß1 − GÐ\Ñ such that 01œ\! on yet neither 01 nor is identically 0 on \ . Thus, there are functions 0ß1−GÐ\Ñ for which 01œ! although 0Á ! and 1ÁÞ ! In an algebra course, such elements 0 and 1 in the ring GÐ\Ñ are called “zero divisors.”

d) Prove that there are exactly two functions 0−GÐÑ‘ for which 0# œ0 . ( M8GÐ\Ñß the notation 0# ÐBÑ means 0ÐBц0ÐBÑ , not 0Ð0ÐBÑÑ . )

e) Prove that there are exactly c functions 0G in ( ) for which 0œ0# .

An element in GÐ\Ñ that equals its own square is called an idempotent . Part d) shows that GБ Ñ and GÐÑ are not isomorphic rings since they have different numbers of idempotents. Is eitherGБ Ñ or GÐÑ isomorphic to GÐÑ  ?

One classic part of general topology is to explore the relationship between the space \ and the rings GÐ\Ñ and GÐ\ч . For example, if \ is homeomorphic to ] , then GÐ\Ñ is isomorphic to GÐ]Ñ . This necessarily implies (why?) that GÐ\ч is isomorphic to GÐ]Ñ ‡ . The question “when does

300 isomorphism imply homeomorphism?” is more difficult. Another important area of study is how the maximal ideals of the ring GÐ\Ñ are related to the topology of \ . The best introduction to this material is the classic book Rings of Continuous Functions (Gillman-Jerison).

f) Let HБ Ñ be the set of differentiable functions 0 À‘‘ Ä . Are the rings GÐ ‘ Ñ and HÐ ‘ Ñ isomorphic? Hint: An isomorphism between GБ Ñ and HÐ ‘ Ñ preserves cube roots.

E11. Suppose \ is a connected Tychonoff space with more than one point. Prove l\l -Þ .

E12. Let \ be a topological space. Suppose 0ß1−GÐ\Ñ and that ^Ð0Ñ is a neighborhood of ^Ð1Ñ (that is, ^Ð1Ñ© int ^Ð0ÑÞ )

a) Prove that 0 is a multiple of 1GÐ\Ñß in that is, prove there is a function 2−GÐ\Ñ such that 0ÐBÑœ1ÐBÑ2ÐBÑ for all B−\ .

b) Give an example where ^Ð0Ѫ^Ð1Ñ but 0 is not a multiple of 1 in GÐ\Ñ .

E13. Let \ be a Tychonoff space with subspaces JEßJ and where is closed and E is countable. Prove that if J ∩ E œ g , then E is disjoint from some zero set that contains J .

E14. A space \ is called pseudocompact if every continuous 0 À \ Ä ‘ is bounded, that is, if GÐ\Ñ œ G‡ Ð\Ñ (see Definition IV.8.7 ). Consider the following condition (*) on a space \ :

(*) Whenever Z" ªZ # ª ... ªZ 8 ªÞÞÞ is a decreasing sequence of nonempty open sets,

∞ then +8œ" clZ8 Á g .

a) Prove that if \ satisfies (*), then \ is pseudocompact.

b) Prove that if \ is Tychonoff and pseudocompact, then \ satisfies (*).

Note: For Tychonoff spaces, part b) gives an “internal" characterization of pseudocompactness that is, a characterization that makes no explicit reference to ‘ .

.

301 4. Normal and X% -Spaces

We now return to a topic in progress: normal spaces Ðand X% -spaces Ñ . Even though normal spaces are badly behaved in some ways, there are still some very important (and nontrivial) theorems that we can prove. One of these will give “X% Ê X " ” as an immediate corollary. $ #

To begin, the following technical variation on the definition of normality is very useful.

Lemma 4.1 A space \ is normal iff whenever EßS is closed is open and E©S , there exists an open setU with E©Y© cl Y©SÞ

Proof Suppose \S is normal and that is an open set containing the closed set E . Then E and F œ \  S are disjoint closed sets. By normality, there are disjoint open sets Y and Z with E©Y and F©ZÞ Then E©Y© cl Y©\Z©SÞ

Conversely, suppose \ satisfies the stated condition and that EßF are disjoint closed sets.

Then E©Sœ\Fß so there is an open set Y with E©Y© cl Y©\FÞ Let Zœ\ cl Y . YZ and are disjoint closed sets containing EF and respectively, so \ñ is normal.

302 Theorem 4.2 a) A closed subspace of a normal (X% ) space is normal ( X % ). b) A continuous closed image of a normal (X% ) space normal ( X % ).

Proof a) Suppose J is a closed subspace of a normal space \ and let E and F be disjoint closed sets in JEßF . Then are also closed in \ so we can find disjoint open sets YZ\w and w in containing EF and respectively. Then YœY∩JZœZ∩Jw and w are disjoint open sets in J that contain E and F , so J is normal.

b) Suppose \ is normal and that 0À\Ä] is continuous, closed and onto. If EF and are disjoint closed sets in ] , then 0ÒEÓ0ÒFÓ" and " are disjoint closed sets in \ . Pick Yw and Z w disjoint open sets in \ with 0" ÒEÓ©Y w and 0 " ÒFÓ©ZÞ w Then Yœ]0Ò\YÓw and Zœ]0Ò\ZÓw are open sets in ] .

If C−Y , then CÂ0Ò\YÓw . Since 0 is onto, Cœ0ÐBÑ for some B−Y©\ZÞw w Therefore C−0Ò\ZÓw so CÂZ . Hence Y∩ZœgÞ

If C−E , then 0" ÒÖC×Ó©0 " ÒEÓ©Y w , so 0 " ÒÖC×Ó∩Ð\YÑœg w . Therefore CÂ0Ò\YÓw so C−]0Ò\YÓœYw . Therefore E©Y and, similarly, F©Z so ] is normal.

Since the X" property is hereditary and is preserved by closed onto maps, the statements in a) and b) hold for X% as well as normality. ñ

The next theorem gives us more examples of normal (and X% ) spaces.

Theorem 4.3 Every regular Lindelo¨f space \ is normal (and therefore every Lindelo¨f X $ -space is X% ).

Proof Suppose EF and are disjoint closed sets in \ . For each B−E , use regularity to pick an open set YB such that B−Y© B cl Y©\FÞ B Since the Lindelo¨f property is hereditary on closed subsets, a countable number of the YB 's cover E : relabel these as YßYßÞÞÞßYßÞÞÞ" # 8 . For each 8 , we have clY8 ∩ F œ gÞ Similarly, choose a sequence of open sets ZßZßÞÞÞßZßÞÞÞ" # 8 covering F such that clZ∩Eœg8 for each 8 .

∞ ∞ We have that -8œ"YªE8 and - 8œ" ZªF 8 , but these unions may not be disjoint. So we define

‡ ‡ YœYZ" "" cl ZœZY" " cl " ‡ ‡ YœYÐZ∪ZÑ# #"# clcl ZœZÐY∪YÑ# # clcl " # ã ã ‡ ‡ YœYÐ8 8"# clcl Z∪ Z∪ÞÞÞ∪ cl ZÑ 8 Z8 œZÐ 8"# clcl Y∪ Y∪ÞÞÞ∪ cl YÑ 8 ã ã

∞‡ ∞ ‡ Let Yœ-8œ" Y8 and Zœ - 8œ" Z 8 .

‡ If B−E , then BÂZ cl 8 for all 8 . But B−Y 5 for some 5 , so B−Y©YÞ5 Therefore E©Y and, similarly, F ©ZÞ

To complete the proof, we show that Y∩Zœg . Suppose B−Y .

303 ‡ Then B−Y5 for some 5ß so BÂ cl Z∪Z∪ÞÞÞ∪Zß" cl # cl 5 so BÂZ" ∪Z # ∪ÞÞÞ∪Z 5 ‡ ‡ ‡ so BÂZ" ∪Z # ∪ÞÞÞ∪Z 5 ‡ so BÂZ8 for any 8Ÿ5Þ

‡ ‡ Since B−Y5 , then B−YÞ5 So, if 85 , then BÂZ8 œZÐ8" cl Y∪ÞÞÞ∪ cl Y∪ÞÞÞ∪ 5 cl YÑ 8

‡ So BÂZ8 for all 8 , so BÂZ and therefore Y∩ZœgÞñ

Example 4.4 The Sorgenfrey line W is regular because the sets Ò+ß,Ñ form a base of closed neighborhoods at each point + . We proved in Example VI.3.2 that W is Lindelo¨f, so W is normal. Since WX is " , we have that WX is % .

5. Urysohn's Lemma and Tietze's Extension Theorem

We now turn our attention to the issue of “X% Ê X " ”. This is hard to prove because to show that a $ # space \X is " , we need to prove that certain continuous functions exist; but the hypothesis “X % ” gives $ # us no continuous functions to work with. As far as we know at this point, there could even be X% spaces on which every continuous real-valued function is constant! If X% -spaces are going to have a rich supply of continuous real-valued functions, we will have to show that these functions can be “built from scratch” in a X% -space. This will lead us to two of the most well-known classical theorems of general topology.

We begin with the following technical lemma. It gives a way to use a certain collection of open sets ÖYÀ<−U×< to construct a function 0−GÐ\Ñ . The idea in the proof is quite straightforward, but I attribute its elegant presentation (and that of Urysohn's Lemma which follows) primarily to Leonard Gillman and Meyer Jerison.

Lemma 5.1 Suppose \ is any topological space and let U be any dense subset of ‘ Þ Suppose open sets Y©\< have been defined, one for each <−U , in such a way that:

i) \œ-<−U Y< and + <−U Yœg < ii) if <ß=−U and <+= , then cl Y©Y< = .

For B−\ , define 0ÐBÑœ inf Ö<−UÀB−Y×< . Then 0À\Ä ‘ is continuous.

We will use this Lemma only once, with U œ Þ So if you like, there is no harm in assuming that U œ  in the proof.

Proof Suppose B−\Þ By i) we know that B−Y< for some <, so Ö<−UÀB−Y×Ág< . And by ii), we know that BÂY= for some = . For that = ÀB−Y if < , then (by ii) =Ÿ<= , so is a lower bound for Ö<−UÀB−Y×< . Therefore Ö<−UÀB−Y×< has a greatest lower bound, so the definition of 0 makes sense: 0ÐBÑœ inf Ö<−UÀB−Y×−< ‘ .

304 From the definition of 0 , we get that for <ß=−Uß

a) if B− cl Y< , then B−Y = for all =< so 0ÐBÑŸ< b) if 0ÐBÑ+= , then B−Y = .

We want to prove 0 is continuous at each point +−\U . Since is dense in ‘ ,

ÖÒ<ß=ÓÀ<ß=−U and <+0Ð+Ñ+=×

is a neighborhood base at 0Ð+Ñ in ‘ . Therefore it is sufficient to show that whenever < + 0Ð+Ñ + = , then there is a neighborhood Y of + such that 0ÒYÓ©Ò<ß=ÓÞ

Since 0Ð+Ñ+= , we have +−Y= , and 0Ð+Ñ< gives us that +ÂY cl< . Therefore YœYY =cl < is an open neighborhood of + . If D−Y , then D−Y©= cl Yß = so 0ÐDÑŸ= ; and DÂ cl Y< , so DÂY < and 0ÐDÑ <Þ Therefore 0ÒYÓ©Ò<ß=ÓÞ ñ

Our first major theorem about normal spaces is still traditionally referred to as a “lemma” because it was a lemma in the paper where it originally appeared. Its author, Paul Urysohn, died at age 26, on the morning of 17 August 1924, while swimming off the coast of Brittany.

Theorem 5.2 Ð Urysohn's Lemma Ñ A space \ is normal iff whenever EßF are disjoint closed sets in \ , there exists a function 0−GÐ\Ñ with 0lEœ! and 0lFœ"Þ ( When such an 0 exists, we say that E and F are completely separated . )

Note: Notice that if E and F happen to be disjoint zero sets , say Eœ^Ð1Ñ and Fœ^Ð2Ñ , then the conclusion of the theorem is true in any space, without assuming normality : just let 1# ÐBÑ 0ÐBÑœ1# ÐBÑ<2 # ÐBÑ . Then 0 is continuous, 0lEœ! and 0lFœ" .

The conclusion of Urysohn's Lemma only says that E©0" Ð!Ñ and F©0 " Ð"Ñ : equality might not be true. In fact, if Eœ0" Ð!Ñ and Fœ0 " Ð"Ñ , then E and F were zero sets in the beginning, and the hypothesis of normality would have been unnecessary.

This shows again that zero sets are very special closed sets: in any space, disjoint zero sets are completely separated. Put another way: given Urysohn's Lemma, we can conclude that every nonnormal space must contain a closed set that is not a zero set.

Proof The proof of Urysohn's Lemma in one direction is almost trivial. If such a function 0 exists, " " then YœÖBÀ0ÐBÑ+# × and Z œÖBÀ0ÐBÑ # × are disjoint open sets (in fact, cozero sets) containing E and F respectively. It is the other half of Urysohn's Lemma for which Urysohn deserves credit.

Let EF and be disjoint closed sets in a normal space \ . We will define sets open sets Y<−Ñ< ( in \ in such a way that Lemma 5.1 applies. To start, let Yœg< for <+! and Yœ\ < for <" .

305 Enumerate the remaining rationals in  ∩Ò!ß"Ó as <ß<ßÞÞÞß<ßÞÞÞ" # 8 , beginning the listwith <œ"" and

<œ!Þ# We begin by defining YœYœ\FÞ<" " Then use normality to define YÐœYÑÀ<# ! since

E©Yœ\F<" , we can pick Y < # so that

E©Y<# © cl Y < # ©Y < " œ\F

Then ! œ <# + < $ + < " œ " , and we use normality to pick an open set Y < $ so that

E©Y<# © cl Y << #$ ©Y © cl Y << $" ©Y œ\F

We continue by induction. Suppose 8 $ and that we have already defined open sets Y<" ßY < # ßÞÞÞßY < 8 in such a way that whenever <3 +< 4 +< 5 Ð3ß4ß5Ÿ8Ñß then

clY<<3 ©Y 4 © cl Y < 4 ©Y < 5 ЇÑ

We need to define Y<8<" so that Ð‡Ñ holds for 3ß4ß5Ÿ8<"Þ

Since <œ"" and <œ! # , and < 8<" −Ð!ß"Ñß it makes sense to define

<œ5 the largest among <ß<ßÞÞÞß< "#8 that is smaller than <ß 8<" and <œ6 the smallest among <ß<ßÞÞÞß< "#8 that is larger than <Þ 8<"

By the induction hypothesis, we already have clY<5 ©YÞ < 6 Then use normality to pick an open set

Y<8<" so that

clY©Y<<5 8<" © cl Y < 8<" ©Y < 6 .

The Y< 's defined in this way satisfy the conditions of Lemma 5.1, so the function 0 À \ Ä ‘ defined

by 0ÐBÑœ inf Ö<− ÀB−Y×< is continuous. If B−E , then B−YœY<# ! and BÂY < if <+! , so 0ÐBÑœ!Þ If B−F then BÂY" , but B−Yœ\ < for <" , so 0ÐBÑœ"Þ ñ

Once we have the function 0 we can replace it, if we like, by 1œÐ!”0Ñ•" so that E and F are completely separated by a function 1−GÐ\ÑÞ‡ It is also clear that we can modify 1 further to get an 2−GÐ\ч for which 2lEœ+ and 2lFœ, where + and , are any two real numbers.

With Urysohn's Lemma, the proof of the following corollary is obvious.

Corollary 5.3 X% Ê X " . $ #

There is another famous characterization of normal spaces in terms of GÐ\Ñ . It is a result about “extending” continuous real-valued functions defined on closed subspaces.

We begin with the following two lemmas. Lemma 5.4, called the “Weierstrass Q -Test” is a slight generalization of a theorem with the same name in advanced calculus. It can be useful in “piecing together” infinitely many real-valued continuous functions to get a new one. Lemma 5.5 will be used in the proof of Tietze's Extension Theorem (Theorem 5.6).

306 Lemma 5.4 (Weierstrass Q -Test) Let \ be a topological space. Suppose 08 À \ Ä ‘ is ∞ continuous for each 8− and that l0ÐBÑlŸQ8 8 for all B−\ . If  Q+∞8 , then 8œ" ∞ 0ÐBÑœ0ÐBÑ 8 converges (absolutely) for all B and 0À\Ä ‘ is continuous. 8œ"

∞ ∞ ∞ Proof For each B ,  l0ÐBÑlŸ8  Q+∞ 8 , so  0ÐBÑ 8 converges (absolutely) by the Comparison 8œ" 8œ" 8œ" Test. ∞ % Suppose +−\ and % ! . Choose R so that  Q+8% . Each 0 8 is continuous, so for 8œR<" % 8œ"ßÞÞÞßR we can pick a neighborhood Y8 of + such that for B−Yßl0ÐBÑ0Ð+Ñl+ 888 #R . Then R Yœ+8œ" Y8 is a neighborhood of + , and for B−Y we get l0ÐBÑ0Ð+Ñl R ∞ R ∞ œl Ð088 ÐBÑ0 Ð+ÑÑ<  Ð0 88 ÐBÑ0 Ð+ÑÑlŸ  l0 88 ÐBÑ0 Ð+Ñl<  l0 88 ÐBÑ0 Ð+Ñl 8œ" 8œR<" 8œ" 8œR<" R ∞ ∞ % % % Ÿ l0ÐBÑ0Ð+Ñl<88  l0ÐBÑl

Lemma 5.5 Let E be a closed set in a normal space \ and let + be a positive real number. Suppose < < 2 À E Ä Ò  <ß <Ó is continuous. Then there exists a continuous 9 À\Ä [ $ ß $ Ó such that #< l2ÐBÑ9 ÐBÑlŸ$ for each B−E .

< < Proof Let EœÖB−EÀ2ÐBÑŸ" $ × and FœÖB−EÀ2ÐBÑ " $ ×E . is closed, and E" and F " are disjoint closed sets in EEF , so " and " are closed in \ . By Urysohn's Lemma, there exists a << < < continuous function 9À\ÄÒßÓ$$ such that 9 lEœ" $ and 9 lFœ " $ . < < < If B−E" , then <Ÿ2ÐBÑŸ$ and 9 ÐBÑœ $ , so l2ÐBÑ 9 ÐBÑlŸl<Ð$ Ñl #< #< œ$ à and similarly if B−Fßl2ÐBÑ" 9 ÐBÑlŸ$ Þ If B−EÐE∪FÑ" " , then 2ÐBÑ and 9 ÐBÑ are << #< both in Ò$ ß $ Ó so l2ÐBÑ9 ÐBÑlŸ $ . ñ

Theorem 5.6 (Tietze's Extension Theorem) A space \ is normal iff whenever E is a closed set in \ and 0−GÐEÑ , then there exists a function 1−GÐ\Ñ such that 1lEœ0Þ

Note: if E is a closed subset of \ œ ‘ , then it is quite easy to prove the theorem. In that case, ‘  E is open and can be written as a countable union of disjoint open intervals Mß where each MœÐ+ß,Ñ or Ð∞ß,Ñ or Ð+ß∞Ñ (see Theorem II.3.4 ). For each of these intervals M , the endpoints are in E , where 0 is already defined. If MœÐ+ß,Ñ then extend the definition of 0 over M by using a straight line segment to join Ð+ß0Ð+ÑÑ and Ð,ß0Ð,ÑÑ on the graph of 0 . If MœÐ+ß∞ÑÞ then extend the graph of 0 over M using a horizontal right ray at height 0Ð+Ñà if MœÐ∞ß,Ñßthen extend the graph of 0 over M using a horizontal left ray at height 0Ð,ÑÞ

As with Urysohn's Lemma, half of the proof is easy. The significant part of theorem is proving the existence of the extension 1 when \ is normal.

307 Proof (É ) Suppose EF and are disjoint closed sets in \EF . and are clopen in the subspace E∪F so the function 0ÀE∪FÄÒ!ß"Ó defined by 0lEœ! and 0lFœ" is continuous. Since E∪F is closed in \ , there is a function 1−GÐ\Ñ such that 1lÐE∪FÑœ0Þ Then " " YœÖBÀ1ÐBÑ+# ×and Z œÖBÀ1ÐBÑ # × are disjoint open sets (cozero sets, in fact) that contain E and F respectively. Therefore \ is normal.

(Ê ) The idea is to find a sequence of functions 13 − GÐ\Ñ such that 8 8 l0ÐBÑ1ÐBÑlÄ!8Ä∞ 3 as for each B − E Ðwhere 0 is defined ). The sums  1ÐBÑ3 are 3œ" 3œ" defined on all of \ and as 8 Ä ∞ we can think of them as giving better and better approximations to 8 ∞ the extension 1 that we want. Then we can let 1ÐBÑœlim  1ÐBÑœ1ÐBÑÞ3  3 The details follow. 8Ä∞ 3œ" 3œ" We proceed in three steps, but the heart of the argument is in Case I.

Case I Suppose 0 is continuous and that 0ÀEÄÒ"ß"Ó . We claim there is a continuous function 1À\ÄÒ"ß"Ó with 1lEœ0Þ

" " Using Lemma 5.5 (with 2œ0ß<œ"Ñ we get a function 1œ" 9 À\ÄÒßÓ$ $ such that # # # for B−El0ÐBÑ1ÐBÑlŸ , "$ . Therefore 01 " ÀEÄÒ$ ß $ ÓÞ

# # # Using Lemma 5.5 again (with 2œ01ß<œÑ" $ , we get a function 1À\ÄÒßÓ# * * such %# # % % that for B−Eßl0ÐBÑ1ÐBÑ1ÐBÑlŸ"# * œÐ $ Ñ. So 0Ð1" <1ÑÀEÄÒ # * ß * Ó

% Using Lemma 5.5 again (with 2œ01" 1ß<œ # * Ñ , we get a function %% ) # $ 1$ À\ÄÒ#( ß #( Ó such that for B−Eßl0ÐBÑ1ÐBÑ1ÐBÑ1ÐBÑlŸ" # $ #( œÐÑ $ . ) ) So 0Ð1" <1 # <1ÑÀEÄÒ $ #( ß #( ÓÞ

We continue, using induction, to find for each 3 a continuous function 8 #3" # 3" 8 1À\ÄÒ3 $3 ß $ 3 Ó such that l0ÐBÑ 1ÐBÑlŸÐ#Î$Ñ3 for B−E . 3œ"

∞ ∞ ∞ #3" Since l1ÐBÑlŸ3 $3 +∞ß the series 1ÐBÑœ  0ÐBÑ3 converges (absolutely) for every 3œ" 3œ" 3œ" ∞ B−\, and 1 is continuous by the Weierstrass Q -Test. Since l1ÐBÑlœl1ÐBÑl 3 3œ" ∞ ∞ #3" Ÿ l1ÐBÑlŸ3  $3 œ"ßwe have 1À\ÄÒ"ß"ÓÞ 3œ" 3œ" 8 # 8 Finally, for B−Eßl0ÐBÑ1ÐBÑlœlim l0ÐBÑ 1ÐBÑlŸ3 lim Ð$ Ñ œ!, so 1lEœ0 and 8Ä∞3œ" 8Ä∞ the proof for Step I is complete.

Case II Suppose 0 ÀEÄÐ"ß"Ñ is continuous. We claim there is a continuous function 1À\ÄÐ"ß"Ñ with 1lEœ0Þ

Since 0 ÀEÄÐ"ß"Ñ©Ò"ß"Ó , we can apply Case I to find a continuous function JÀ\ÄÒ"ß"Ówith JlEœ0Þ To get 1 , we merely make a slight modification to J to get a 1 that still extends 0 but where 1 has all its values in Ð"ß"ÑÞ

308 Let FœÖB−\ÀJÐBÑœ„"×E . and F are disjoint closed sets in \ , so by Urysohn's Lemma there is a continuous 2À\ÄÒ"ß"Ó such that 2lFœ! and 2lEœ"Þ If we let 1ÐBÑœJÐBÑ2ÐBÑ , then 1À\ÄÐ"ß"Ñ and 1lEœ0 , completing the proof of Case II.

Case III (the full theorem) Suppose 0 À E Ä ‘ is continuous. We claim there is a continuous function 1À\Ä‘ with 1lEœ0Þ

Let 2À‘ ÄÐ"ß"Ñ be a homeomorphism. Then 2‰0ÀEÄÐ"ß"Ñ and, by Step II, there is a continuous JÀ\ÄÐ"ß"Ñ with JlEœ2‰0Þ

Let 1œ2" ‰JÀ\Ä‘ . Then for B−E we have 1ÐBÑœ2" ÐJÐBÑÑ œ2" ÐÐ2‰0ÑÑÐBÑœ0ÐBÑÞ ñ

It is easy to see that GÐ\ч can replace GÐ\Ñ in the statement of Tietze's Extension Theorem.

Example 5.7 We now know enough about normality to see some of its bad behavior. The Sorgenfrey line W is normal (Example 4.4 ) but the Sorgenfrey plane W ‚ W is not normal.

To see this, let Hœ‚ß  a countable dense set in W‚W . Every continuous real-valued function on W‚W is completely determined by its values on H . ( See Theorem II.5.12. The theorem is stated for the case of functions defined on a pseudometric space, but the proof is written in a way that applies just as well to functions with any space \ as domain. ) Therefore the mapping GÐW‚WÑÄGÐHÑ given by 0È0lH is one-to-one, so lGÐW‚WÑlŸlGÐHÑŸl‘H lœ- i ! œ-Þ

EœÖÐBßCÑ−W‚W ÀB

309 The comments following the statement of Urysohn's Lemma imply that W ‚ W must contain closed sets that are not zero sets. A completely similar argument “counting continuous real-valued functions” shows that the Moore plane > (Example III.5.6) is not normal: use that > is separable and the B -axis in > is an uncountable closed discrete subspace.

Questions about the normality of products are difficult. For example, it was an open question for a long time whether the product of a normal space \ with such a nice, well-behaved space as Ò!ß "Ó must be normal. In the 1950's, Dowker proved that \‚Ò!ß"Ó is normal iff \ is normal and “countably paracompact.”

However, this result was unsatisfying because no one knew whether a normal space was automatically “countably paracompact.” In the 1960's, Mary Ellen Rudin constructed a normal space \ which was not countably paracompact. But this example was still unsatisfying because the construction assumed the existence of a space called a “Souslin line” and whether a Souslin line exists cannot be decided in the ZFC set theory! In other words, the space \ she constructed required adding a new axiom to ZFC.

Things were finally settled in 1971 when Mary Ellen Rudin constructed a “real” example of a normal space \ whose product with Ò!ß"Ó is not normal. By “real,” we mean that \ was constructed in ZFC, with no additional set theoretic assumptions. Among other things, this complicated example made use of the box topology on a product.

Example 5.8 The Sorgenfrey line WXWX is % , so is " and therefore the Sorgenfrey plane W‚W is $ # also XW‚W" . So is an example showing that X" does not imply X % . $ # $ #

Extension theorems such as Tietze's are an important topic in mathematics. In general, an “extension theorem” has the following form:

E©\ and 0ÀEÄF , then there is a function 1À\ÄF such that 1lEœ0 .

For example, in algebra one might ask: if E is a subgroup of \ and 0ÀEÄF is an isomorphism, can 0 be extended to a homomorphism 1À\ÄF ?

If we let 3ÀEÄ\ be the injection 3Ð+Ñœ+ , then the condition “ 1lEœ0 ”can be rewritten as 1 ‰ 3 œ 0 . In the language of algebra, we are asking whether there is a suitable function 1 which “makes the diagram commute.”

310

Specific extension theorems impose conditions on E and \ , and usually we want 1 to share some property of 0 such as continuity. Here are some illustrations, without further details.

1) Extension theorems that generalize of Tietze's Theorem: by putting stronger hypotheses on \ , we can relax the hypotheses on F .

Suppose E is closed in \ and 0ÀEÄF is continuous.

‘ ÝÚ \is normal and F œ (Tietze's Theorem) Ý \ is normal and F œ ‘8 If Û \ is collectionwise normal** and F is a separable Banach s pace* Ý Ü \ is paracompact** and F is a Banach space*

then 0 has a continuous extension 1À\ÄFÞ

The statement that ‘8 can replace ‘ in Tietze's Theorem is easy to prove:

8 If \ is normal and 0ÀEÄ‘ is continuous, write 0ÐBÑœÐ0ÐBÑß0ÐBÑßÞÞÞß0ÐBÑÑ" # 8 where each 03 À E Ä ‘ . By Tietze's Theorem, there exists for each 3 a continuous extension 1À\Ä3‘ with 1lEœ0 33 . If we let 1ÐBÑœÐ1ÐBÑßÞÞÞß1ÐBÑÑ "8 , then 1À\Ä‘8 and 1lEœ0 . In other words, we separately extend the coordinate functions in order to extend 0 . And in this example, 8 could even be an infinite cardinal.

* A normed linear space is a vector space Z with a norm l@lœ ( “absolute value”) that defines the “length” of each vector. Of course, a norm must satisfy certain axioms for example, l@"# <@ lŸl@ " l

** Roughly, a “collectionwise normal” space is one in which certain infinite collections of disjoint closed sets can be enclosed in disjoint open sets. We will not give definitions for “collectionwise normal” (or the stronger condition, “paracompactness”) here, but is true that

Ú metric Û or Ê paracompactÊÊ collectionwise normal normal Ü compact X#

Therefore, in the theorems cited above, a continuous map 0 defined on a closed subset of a metric space (or, compact X# space) and valued in a Banach space F can be continuously extended a function 1 À \ Ä F .

311 2) The Hahn-Banach Theorem is another example, taken from functional analysis, of an extension theorem. A special case states:

Suppose Q is a linear subspace of a real normed linear space\ and that 0 À Q Ä ‘ is linear and satisfies 0ÐBÑŸllBll for all B−QÞ There there is a linear JÀ\Ä ‘ such that JlQœ0 and JÐBÑŸllBll for all B−\Þ

3) Homotopy is usually not discussed in terms of extension theorems, but extensions are really at the heart of the idea.

Let 01ÀÒ!ß"ÓÄ\ , be continuous and suppose that 0Ð!Ñœ1Ð!ÑœB ! and 0Ð"Ñœ1Ð"ÑœBÞ" Then 0 and 1 are paths in \ that start at B! and end at BÞ " Let F be the boundary of the square Ò!ß"Ó©#‘ # and define JÀFÄ\ by

JÐBß!Ñœ0ÐBÑ JÐBß"Ñœ1ÐBÑ JÐ!ß>ÑœB! JÐ"ß>ÑœB "

Thus J0 agrees with on the bottom edge of F1 and with on the top edge. J is constant ÐœBÑ! on the left edge of F and constant ÐœBÑ" on the right edge of F . We ask whether J can be extended to a continuous map defined on the whole square, L ÀÒ!ß"Ó# Ä\Þ

If such an extension L does exist, then we have

For each >−Ò!ß"Óß restrict L to the line segment at height > to define 0ÐBÑœLÐBß>Ñ> . Then for each >−Ò!ß"Ó0 , > is also a path in \ from BB ! to " . As > varies from !" to , we can think of the 0> 's as a family of paths in \ that continuously deform 0œ00œ1Þ! into "

The continuous extension L (if it exists) is called a homotopy between 0 and 1 with fixed endpoints , and we say that the paths 0 and 1 are homotopic with fixed endpoints .

312 In the space \ on the left, below, it seems intuitively clear that 0 can be continuously deformed (with endpoints held fixed) into 1  in other words, that L exists.

However in the space ] pictured on the right, 0 and 1 together form a loop that surrounds a “hole” in ] , and it seems intuitively clear that the path 0 cannot be continuously deformed into the path 1 inside the space ] that is, the extension L does not exist.

In some sense, homotopy can be used to detect the presence of certain “holes” in a space, and is one important part of algebraic topology.

The next theorem shows us where compact Hausdorff spaces stand in the discussion of separation properties.

Theorem 5.9 A compact X# space \ is X % .

Proof \ is Lindelo¨f, and a regular Lindelo¨f space is normal (Theorem 4.3). Therefore it is sufficient to show that \ is regular. Suppose J is a closed set in \ and BÂJÞ For each C−J we can pick disjoint open sets YCC and Z with B−Y C and C−Z C . J is compact so a finite number of the Z C 's 8 8 cover J say ZC" ßZ C # ßÞÞÞßZ C 8 Þ Then B−+3œ" YœYJ©C3 , - 3œ" ZœZ C 3 , and YßZ are disjoint open sets. ñ

Therefore, our results line up as:

Ú compact X# (*) compact metric ____ " ÊÛ ÊXÊX%$# ÊXÊXÊXÊX $#"! Ü metric

In particular, Urysohn's Lemma and Tietze's Extension Theorem hold in metric spaces and in compact X# spaces .

313 Notice that

i) the space Ð!ß"ÑX is % but not compact X #

ii) the Sorgenfrey line W is X% (see example 5.4) but not metrizable. Ð If W were metrizable, then W‚WA9?6. be metrizable and therefore X% which is false À see Example 5.7 ).

- iii) Ò!ß"Ó is compact X # (assuming, for now, the Tychonoff Product Theorem VI.3.10 ) but not metrizable (why? )

iv) Ð!ß "Ñ is metrizable but not compact.

Combining these observations with earlier examples, we see that none of the implications in (*) is reversible.

Example 5.10 (See Example 5.7 ) The Sorgenfrey plane W‚W is X" , so W‚W can be embedded $ # 7 7 in a cube Ò!ß"Ó and Ò!ß"Ó is compact X# (assuming the Tychonoff Product Theorem ). Since W ‚ W is not normal, we see now that a normal space can have nonnormal subspaces. This example, admittedly, is not terribly satisfying since we can't visualize how W‚W “sits” inside Ò!ß"Ó 7 . In Chapter VIII (Example 8.10), we will look at an example of a X% space in which it's easy to “see” why a certain subspace isn't normal.

6. Some Metrization Results

Now we have enough information to completely characterize separable metric spaces topologically.

Theorem 6.1 (Urysohn's Metrization Theorem) A second countable X$ -space is metrizable. Note: We proved a similar metrization theorem in Corollary 3.18, but there the separation hypothesis was X" rather than X $ . $ #

Proof \ is second countable so \ is Lindelo¨f, and Theorem 4.3 tells us that a Lindelo¨f XXÞ$-space is % Therefore \X is " . So by Corollary 3.18, \ is metrizable. ñ $ #

Because a separable metrizable space is second countable and X$ , we have a complete characterization: \ is a separable metrizable space iff \ is a second countable X$ -space .So, with hindsight, we now see that the hypothesis “X " ” in Corollary 3.18 was unnecessarily strong. In fact, $ # we see that X$ and X " are equivalent in a space that is second countable. $ #

Further developments in metrization theory hinged on work of Arthur H. Stone in the late 1940's in particular, his result that metric spaces have a property called “paracompactness.” This led quickly to a complete characterization of metrizable spaces that came roughly a quarter century after Urysohn's work. We state this characterization here without a proof.

314 A family of sets U in Ð\ßÑg is called locally finite if each point B−\ has a neighborhood R that has nonempty intersection witth only finitely many sets in U . The family U is called 5 -locally finite if we can write Uœ -8−  U8 where each subfamily U 8 is locally finite.

Theorem 6.2 (The Bing-Smirnov-Nagata Metrization Theorem) Ð\ßÑg is metrizable iff \X is $ and has a 5 -locally finite base U .

Note: If \ is second countable, a countable base Uœ ÖS" ßS # ßÞÞÞßS 8 ßÞÞÞ× is 5 -locally finite because we can write U œß- U8 where U 8 œÖS×Þ 8 Therefore this Metrization Theorem includes Urysohn's Metrization Theorem as a special case.

The Bing-Smirnov-Nagata Theorem has the typical form of most metrization theorems: \ is metrizable iff “\ has enough separation” and “ \ has a nice enough base.”

315 Exercises

E15. Let Ð\ß.Ñ be a metric space and W©\ . Prove that if each continuous 0ÀWÄ ‘ extends to a continuous 1À\Ä‘ , then W is closed. ( The converse, of course, follows from Tietze's Extension Theorem.)

E16. Urysohn's Lemma says that in a X% -space disjoint closed sets are completely separated. Part a) shows that this is also true in a Tychonoff space if one of the closed sets is compact.

a) Suppose \ is Tychonoff and JßO©\ where J is closed, O is compact. \ and J∩Oœg. Prove that there is an 0−GÐ\Ñ such that 0±Oœ! and 0±Jœ" . ( This is another example of the rule of thumb that “compact spaces act like finite spaces.” If necessary, try proving the result first for a finite set OÞ )

b) Suppose \ is Tychonoff and that :−Y , where Y is open in \Þ Prove Ö:×K is a $ set in \ iff there exists a continuous function 0À\ÄÒ!ß"Ó such that 0" Ð"ÑœÖ:× and 0l\Yœ! .

E17. Suppose ] is a Hausdorff space. Define BµC] in iff there does not exist a continuous function 0À\ÄÒ!ß"Ó such that 0ÐBÁ0ÐCÑ ) . Prove or disprove: ]ε is a Tychonoff space.

E18. Prove that a Hausdorff space \ is normal iff for each finite open cover h œÖY" , ... , Y× 8 of \ , 8 there exist continuous functions 0À\ÄÒ!ß"ÓÐ3œ"ßÞÞÞ8Ñ3 , such that 3œ" 0ÐBÑ3 =1 for each B−\ and such that, for each 3 , 03 ±\Y 3 ´! . ( Such a set of functions is called a partition of unity subordinate to the finite cover h.)

Hint ÐÊÑ First build a new open cover i œÖ× V" ,...,V 8 that “shrinks” h in the sense that, Z©Z©Y3 cl 3 3 for each 3. To begin the construction, let Jœ\YÞ"-3" 3 Pick an open Z" so that J©Z© "" cl Z©YÞ "" Then ÖZßYßÞÞÞßY× "#8 still covers \Þ Continue by looking at Jœ\ÐZ∪# "-3# YÑ 3 and defining Z # so that ÖZßZßYßÞÞÞßY× "#$8 is still a cover and Z©### cl Z©YÞ Continue in this way to replace the U3 's one by one. Then use Urysohn's lemma to getfunctions 13 which can then be used to define the 0 3 's. .

E19. Suppose \ is a compact, countable Hausdorff space. Prove that \ is completely metrizable.

Hint: 1) For each pair of points BÁB\8 7 in pick disjoint open sets Y7ß8 and Z 7ß8 containing these points. Consider the collection of all finite intersections of such sets. 2) Or: Since \ is, countable, every singleton Ö:×K is a $ set. Use regularity to find ∞ a descending sequence of open sets Z8 containing : such that +8œ" cl ZœÖ:×Þ8 Prove that the Z8 's are a neighborhood base at : .

316 E20. A space\ is called completely normal if every subspace of \ is normal. ( For example, every metric space is completely normal).

a) Prove that \ is completely normal if and only if the following condition holds:

whenever EßF©\ and (cl E∩FÑ∪ÐE∩ cl FÑœg (that is, each of EßF is disjoint from the closure of the other), then there exist disjoint open set= Y and Z with E©Y and F©ZÞ

b) Recall that the “scattered line” (Exercise IIIE.10 ) consist of the set \ œ ‘ with the topology g œÖY∪ZÀY is open in the usual topology on ‘ and Z©×  . Prove that the scattered line is completely normal and therefore X% Þ

E21. A X" -space \ is called perfectly normal if whenever E and F are disjoint nonempty closed sets in \ , there isan 0−GÐ\Ñ with 0" Ð!ÑœE and 0 " Ð"ÑœF .

a) Prove that every metric space Ð\ß .Ñ is perfectly normal.

b) Prove that \ is perfectly normal iff \X is % and every closed set in \K is a $ -set. Note: Example 3.10 shows a X% . space \ that is not perfectly normal.

c) Show that the scattered line (see Exercise E20 ) is not perfectly normal, even though every singleton set Ö:× is a K $ -set.

d) Show that the scattered line is X% Þ Hint: Use the fact that ‘ , with the usual topology, is normal. Nothing deeper than Urysohn's Lemma is required but the problem is a bit tricky.

E22. Prove that a X$ space Ð\ßÑg has a locally finite base U iff g is the discrete topology. (Compare to Theorem 6.2. )

317 Chapter VII Review Explain why each statement is true, or provide a counterexample.

1. Suppose Ð\ßg Ñ is a topological space and let g A be the weak topology on \ generated by GÐ\ÑÞ Then g© g A Þ

2. If \ is regular and B−ÖC× cl , then C−ÖB× cl .

3. Every separable Tychonoff space can be embedded in Ò!ß "Ó i! .

4. If 0−GÐÑß we say 1 is a square root of 01−GÐÑ if  and 1œ0# . If a function f in GÐÑ  has more than one square root, then it has - square roots.

5. In a Tychonoff space, every closed set is an intersection of zero sets.

6. A subspace of a separable space need not be separable, but every subspace of the Sorgenfrey line is separable.

7. Suppose  has the cofinite topology. If E is closed in  , then every 0−GÐEÑ can be extended to a function 1−GÐ ÑÞ

8. For 8œ"ß#ßÞÞÞ , let 0À8‘‘ Ä be given by 0ÐBÑœB<8 8 and let g be the weak topology on ‘ i! generated by the 08 's. Then the evaluation map /ÀÄ‘ ‘ given by /ÐBÑÐ8Ñœ0ÐBÑ8 is an embedding.

9. Let G be the set of points in the Cantor set with the subspace topology from the Sorgenfrey line W . Every continuous function 0 À G Ä ‘ can be extended to a continuous function 1ÀWÄ‘ Þ

10. If Ð\ß.Ñ is a metric space, then \ is homeomorphic to a dense subspace of some compact Hausdorff space.

11. Suppose J©ÞJ‘# is closed iff J is a zero set.

12. Suppose O is a compact subset of the Hausdorff space \‚] . Let Eœ1\ ÒOÓ . Then E is XÞ %

13. Every space is the continuous image of a metrizable space.

14. Let Y‘œÖ0−‘ À0 is not continuous × . The weak topology on ‘ generated by Y is the discrete topology.

55. A compact X# space is metrizable if and only if it is second countable.

16. Suppose JO and are disjoint subsets of a Tychonoff space \ , where J is closed and O is compact. There are disjoint cozero sets Y and Z with J©Y and O©ZÞ

7 17. Every X% space is homeomorphic to a subspace of some cube Ò!ß"Ó Þ

18.. Suppose \X is a % -space with nonempty pairwise disjoint closed subspaces J" ß ÞÞÞß J 8 . There is an 0−GÐ\Ñ such that 0lJœ33 for all 3œ"ßÞÞÞß8Þ

318