Chapter VII Separation Axioms
1. Introduction
“Separation” refers here to whether or not objects like points or disjoint closed sets can be enclosed in disjoint open sets; “separation properties” have nothing to do with the idea of “separated sets” that appeared in our discussion of connectedness in Chapter 5 in spite of the similarity of terminology.. 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 “more separation” is added to spaces, they 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 older literature. Therefore the standard separation axioms were historically named XXXX!"#$ , , , , and X % , each stronger than its predecessors in the list. Once these were common terminology, another separation axiom was discovered to be useful and “interpolated” into the list: XÞ"" 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 Y with B−Y , CÂY or there exists an open set ZC−ZBÂZ with ,
2) X" space if, whenever BÁC−\ , there exists an open set Y with B−YßCÂZ and there exists an open set ZBÂYßC−Z with
3) XBÁC−\Y# space (or, Hausdorff space) if, whenever , there exist disjoint open sets and Z\ in such that B−YC−Z and .
It is immediately clear from the definitions that XÊXÊXÞ#"!
Example 2.2
1) \X is a !BC space iff whenever BÁC , then aa Á that is, different points in \ have different neighborhood systems.
2) If l\l " and \ has the trivial topology, then \ is not a X! space.
3) A pseudometric space Ð\ß .Ñ is metric iff Ð\ß .Ñ is a X! space.
283 Clearly, a metric space is XÐ\ß.ÑXBÁCÞ!! . On the other hand, suppose is and that Then for some %%! either BÂFÐCÑ%% or CÂFÐBÑ . Either way, .ÐBßCÑ , so . is a metric.
4) In any topological space \BµCœ we can define an equivalence relation iffaaBC . and let 1 À \ Ä ] œ \Î µ by 1ÐBÑ œ ÒBÓÞ 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 1Ò1ÒSÓÓ" is open in \ . But 1Ò1ÒSÓÓ " œ ÖB − \ À 1ÐBÑ − 1ÒSÓ× œ ÖB − \ Àfor some C − S , 1ÐBÑ œ 1ÐCÑ× œÖB−\À B is equivalent to some point C in SלS .
If ÒBÓ Á ÒCÓ − ] , then B is not equivalent to C , so there is an open set S © \ with (say) B − S and C Â S . Since 1 is open, 1ÒSÓ is open in ] and ÒBÓ − 1ÒSÓ . Moreover, ÒCÓ Â 1ÒSÓ or else C would be equivalent to some point of S implying C − SÞ
]X is called the X! -identification of \ . This identification turns any space into a ! space by identifying points that have identical neighborhoods. If \X is a ! space to begin with, then 1""1 is and is a homeomorphism. Applied to a XX!! space, the -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.5.6Àœ.ÐBßCÑœ!Þ in that caseaaBC iff
ww 5) For 3 œ !ß "ß # À if Ð\ßggg Ñ is a X3 space and ª is a new topology on \ , then Ð\ß g Ñ is also a X3 space.
Example 2.3
1) (Exercise) It is easy to check that a space \X is a" space
iff for each B−\ , ÖB× is closed iff for each , is open and B−\ ÖBל ÖSÀ S B−S×
2) A finite X" space is discrete.
3) Sierpinski space \ œ Ö!ß "× with topology g œ Ögß Ö"×ß Ö!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 XXBC−SœÐBß∞Ñ!" but not : if , then 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 XX"# but not : in an infinite cofinite space, any two nonempty open sets have nonempty intersection.
These separation properties are very well-behaved with respect to subspaces and products.
Theorem 2.4 For 3 œ !ß "ß # :
284 a) A subspace of a XX33 space is a space b) If , then is a space iff each is a space. \œα−E \αα Ág \ X33 \ X Proof All of the proofs are easy. We consider only the case 3œ" , leaving the other cases as an exercise.
w Suppose +Á,−E©\ , where \ is a X" space. If Y is an open set in \ containing B but not C , then YœYw ∩E is an open set in E containing B but not C. Similarly we can find an open set Z in E containing CBÞEX but not Therefore is a " space.
Suppose is a nonempty space. Each is homeomorphic to a subspace of , so \œα−E \αα X" \ \ each BXαααα is "" (by part a)). Conversely, suppose each \X is and that BÁC−\ . Then BÁC for some α . Pick an open set Y\αα in containing B α but not C α . Then YœY α is an open set in \ containing BC but not . Similarly, we find an open set Z\CB in containing but not . Therefore \ is a Xñ" space.
Exercise 2.5 Is a continuous image of a XX33 space necessarily a 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 \JBÂJß is called regular if whenever is a closed set and containing BYZB−YJ©Z , there exist disjoint open sets and such that and .
There are some easy variations on the definition of “regular” that are useful to recognize.
Theorem 2.7 The following are equivalent for any space \ :
285 i) \ is regular
ii) if SB is an open set containing , 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−SÞJœ\S Letting , 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−B−SœR iii) If aB , then int . By ii), we can find an open set Y so that B−Y©cl Y©S . Since cl Y is a neighborhood of B , the closed neighborhoods of B form a neighborhood base at B .
iii)ÊJ i) Suppose is closed and BÂJ . By ii), there is a closed neighborhood OB of such that B−O©\J . We can choose Y œ int O and Z œ\O to complete the proof that \ is regular. ñ
Example 2.8 Every pseudometric space Ð\ß .Ñ is regular. Suppose +  J and J is closed. We have a continuous function 0ÐBÑ œ .ÐBß J Ñ for which 0Ð+Ñ œ - ! and 0lJ œ !Þ This gives us disjoint "-- " open sets with +−Y œ0 ÒÐß∞ÑÓ## and J ©Z œ0 ÒÐ∞ß ÑÓ . 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! .
To bring things into line, we make the following definition.
Definition 2.9 A topological space \X\ is called a $" 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 .
286 Caution The 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 XX!$ space must be (so it would have been equivalent to use “X! ” instead of “XX"$ ” in the definition of “ ”).
# Example 2.11 XÊ#$ÎX . We will put a new topology on the set \œ‘ . At each point B−\ , let a neighborhood base UB consist of all sets R of the form
R œ F% ÐBÑ Ð a finite number of straight lines through BÑ ∪ ÖB× for some % !Þ
(Check that the conditions in the Neighborhood Base Theorem III.5.2 are satisfied.) With the resulting topology, \FÐBÑ− is called the slotted plane . Note that % UB (because “! ” is a finite number ), so # each FÐBÑ% is among the basic neighborhoods in U‘B so the slotted plane topology on is contains the usual Euclidean topology. It follows that \X is # .
The set J œ ÖÐBß !Ñ À B Á !× œ “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 © Y ÞUsing the % in the definition of R , we can choose a point T œ ÐBß !Ñ − J with !B% ÞAny basic neighborhood set of T 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$ ).
Note: The usual topology in ‘# is regular. T his example shows that an “enlargement” of a regular (or XX$$ ) topology may not be regular (or ). 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. Of course it is easy to see that an “enlargement” of a X33 topology Ð3 œ !ß "ß #Ñ is also X Þ
287 Example 2.12 The Moore plane > (Example III.5.6) is clearly XÞ# In fact, at each point, there is a neighborhood base of closed neighborhoods. The figure illustrates this for a point TB on the -axis and a point UB above the -axis. Therefore > is X$ .
Theorem 2.13 a) A subspace of a regular (XÑ$$ space is regular ( X ). b) Suppose . is regular ( ) iff each is regular ( ). \œα−E \αα Ág \ X$$ \ X Proof a) Let E©\ where \ is regular. Suppose +−E and that J is a closed set in E that does not contain +JJ∩EœJÞY . There exists a closed set ww in \ such that Choose disjoint open sets w and Zwwwwww in \ with +−Y and Z ªJ Þ Then Y œY ∩E and Z œZ ∩E are open in E , disjoint, +−Y, and J ©Z . Therefore E is regular.
b) If is regular, then part a) implies that each is regular since each \œα−E \αα Ág \ \\\α is homeomorphic to a subspace of . Conversely, suppose each α is regular and that
Y œ Yαα"8 ß ÞÞÞß Y is a basic open set containing B . For each α3 , we can pick an open set Zα3 in
\ααααα33333such that B − Z © cl Z © Y Þ Then B − Z œ Z αα "8 ß ÞÞÞß Z © cl Z
© cl Zαα"8 ß ÞÞÞß cl Z © Y Þ (Why is the last inclusion true? ) Therefore \ is regular.
Since “XXñ" ” is hereditary and productive, a) and b) also hold for “$ ”
The obvious “next step up” in separation is the following:
Definition 2.14 A topological space \EßF is called normal if, whenever are disjoint closed sets in \YZ\E©YF©ZÞ\X\, there exist disjoint open sets , in with and is called a % space if is normal and XÞ1
Example 2.15 a) Every pseudometric space Ð\ß .Ñ is normal (and therefore every metric space is X% Ñ . .ÐBßFÑ In fact, if EF and are disjoint closed sets, we can define 0ÐBÑœ.ÐBßEÑ .ÐBßFÑ . Since the
288 " denominator cannot be !0 , is continuous and 0lEœ!0lFœ"Þ , The open sets YœÖBÀ0ÐBÑ# × " and ZœÖBÀ0ÐBÑ# × are disjoint and contain E and F respectively. Therefore \ is normal.
Note: the argument given is slick and clean. Can you show Ð\ß .Ñ is normal by constructing directly a pair of open sets separating EF and ?
b) Let ‘g‘‘ have the right ray topology œ ÖÐBß ∞Ñà B − × ∪ Ögß × .
i)Ðß‘g Ñ is normal À the only possible pair of disjoint closed sets is g and \ and we can separate these using the disjoint open sets Yœg and Zœ\Þ
ii) Ðß‘g Ñ is not regular: for example " is not in the closed set JœÐ∞ß!Ó , but every open set that contains J"Ê also contains . So “normal Î regular.”
ii3Ðß ) ‘g Ñ is not a X"% space À so X is a stronger condition than normality.
But when we combine “normalX"% ” into X , we have a property that fits perfectly into the separation hierarchy.
Theorem 2.16 X%$ ÊX ÐÊX #"! ÊX ÊXÑ
Proof Suppose \X is % . If J is a closed set not containing B , then ÖB×J and are disjoint closed sets. By normality, we can find disjoint open sets separating ÖB× and J . It follows that \ is regular and therefore Xñ$ .
.
289 Exercises
E1. A topological space \ is called a door space if every subset is either open or closed. Prove that a X# space is a door space if and only if it has at most one non-isolated point.
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 YY . Clearly, is a base for the closed sets in iff UYœÖSÀSœ\JßJ− × is a base for the open sets in \ .
For a polynomialT8 in real variables, define the zero set of T as
8 ^ÐT Ñ œ ÖÐB"# ß B ß ÞÞÞß B 8 Ñ −‘ À T ÐB "# ß B ß ÞÞÞß B 8 Ñ œ !×
a) Prove that Ö^ÐT Ñ À T 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 XX"# but not .
c) Prove that the Zariski topology on ‘ is the cofinite topology, but that if 8" , the Zariski topology on ‘8 is not the cofinite topologyÞ
Note: The Zariski topology arises in studying algebraic geometry. After all, the sets ^ÐTÑ are rather special geometric objects—those “surfaces” in ‘8 which can be described by polynomial equations T ÐB"# ß B ß ÞÞÞß B 8 Ñ œ !.
E3. A space \X is a &Î# space if, whenever BÁC−\, there exist open sets YZ and such that BYZœgXÊÞ−Y, C−Z and cl ∩ cl. ( Clearly, TT$ Ê &Î# # ) a) Prove that a subspace of a XX&Î# space is a &Î# space. b) Suppose . Prove that is iff each is . \œ \αα Ág \ X&Î# \ X &Î# c) Let W œ ÖÐBß CÑ −‘# À C !× and P œ ÖÐBß !Ñ − W À C œ !×Þ Define a topology on W with the following neighborhood bases:
if :œÖFÐ:Ñ∩WÀ!×−WPÀ U%: % if : − P: U%: œ ÖF% Ð:Ñ ∩ ÐW PÑ ∪ Ö:× À !× ()You may assume that these U: 's satisfy the axioms for a neighborhood base.
Prove that ST is X$&Î# but not T .
E4. If E©\ , we can define a topology on \ by
g œÖS©\À SªE×∪Ög×
Decide whether or not Ð\ßg Ñ is normal.
290 E5. A function 0À\Ä] is called perfect if 0 is continuous, closed, onto, and, for each C−]ß " 0]ÐCÑ is compact. Prove that if \ is regular and 0 is perfect, then is regular à and that if \ is X$ , the ]XÞ is also $
E6. a) Suppose \Ð\ßÑ\ is finite . Prove that gU is regular iff there is a partition of that is a base for the topology.
For parts b)-e) assume that \X is a $ -space (not necessarily finite).
b) Suppose E\Þ is closed in Prove that
i) Prove that is open and Eœ ÖSÀS E©S×Þ ii) Define BµC iff BœC or BßC−EÞ Prove that the quotient space \ε is Hausdorff.
c) Suppose F\ is an infinite subset of . Prove that there exists a sequence of open sets Y8 such that each Y∩FÁg887 and then cl Y ∩ cl Y œg8Á7 if .
d) Suppose each point C] in a space has a neighborhood Z such that cl Z is regular. Prove that ] is regular.
e) Give an example to show that a compact subset O\ of a space need not be closed, but show that if \O is regular then ck is compact.
291 3. Completely Regular Spaces and Tychonoff Spaces
The XX$ property is well-behaved. For example, we saw in Theorem 2.13 that the $ property is hereditary and productive. However, it is not a strong enough property to give us really nice theorems.
For example, it's very useful in studying a space \ if there are a lot of (nonconstant) continuous real valued functions on \ available. Remember how many times we have used the fact that continuous real-valued functions 0Ð\ß.Ñ0ÐBÑœ.ÐBß+Ñ can be defined on a metric space using formulas like or 0ÐBÑ œ .ÐBß J Ñ; when l\l " , we get many nonconstant real functions defined on Ð\ß .ÑÞ But a X$ space can be very deficient in continuous real-valued functions in 1946, Hewett gave an example of a infinite XL$ space on which the only continuous real-valued functions are the constant functions.
On the other hand, the X% property is strong enough to give some really nice theorems (for example, see Theorems 5.2 and 5.6 later in this chapter). But X% spaces turn out to also have 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 not even finitely productive. Examples of this bad behavior are a little hard to get right now, but they will appear rather naturally later.
These observations lead us (out of historical order) to look at a class of spaces with separation somewhere “between XX$% and .” 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\Ñ œ Ö0 − \‘ À 0 is continuous × œ the collection of continuous real-valued functions on \ G‡ Ð\Ñ œ Ö0 − GÐ\Ñ À 0 is continuous and bounded × .
Lemma 3.1 Suppose 0ß1 − GÐ\Ñ . Define real-valued functions 0 ”1 and 0 •1 by
(0 ” 1ÑÐBÑ œ max Ö0ÐBÑß 1ÐBÑ× Ð0 • 1ÑÐBÑ œ min Ö0ÐBÑß 1ÐBÑ×
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
0ÐBÑ1ÐBÑ l0ÐBÑ1ÐBÑl (0 ” 1ÑÐBÑ œ ## 0ÐBÑ1ÐBÑ l0ÐBÑ1ÐBÑl Ð0 • 1ÑÐBÑ œ## ñ
Definition 3.2 A space \JBÂJß is called completely regular if whenever is a closed set and there exists a function 0 − GÐ\Ñ such that 0ÐBÑ œ ! and 0lJ œ "Þ
Note i) The definition requires only that 0lJ œ " , in other words, that 0" ÒÖ"×Ó ª J Þ However, the two sets might not be equal.
292 ii) If there is such a function 0 , there is also a continuous 1 À \ Ä Ò!ß "Ó such that 1ÐBÑ œ ! and 1lJ œ " . For example, we could use 1 œ Ð0 ” !Ñ • " which, by Lemma 3.1, is continuous.
iii) If 1ÐBÑœ!ß1lJœ" and 1À\ÄÒ!ß"Ó , we could compose 1 with a linear homeomorphism 99À Ò!ß "Ó Ä Ò+ß ,Ó to get a continuous function 2 œ ‰ 1 À \ Ä Ò+ß ,Ó where 2ÐBÑ œ + and 2lJ œ , (or vice versa). (It must be true that either 99Ð!Ñ œ + and Ð"Ñ œ , , or vice versa: why?)
Putting these observations together, we see Definition 3.2 is equivalent to:
Definition 3.2w A space \JBÂJß is called completely regular if whenever is a closed set, and +Á, are real numbers, then there exists a continuous function 0À\Ä‘ for which 0ÐBÑœ+ and 0lJ œ ,, and for all Bß + Ÿ 0ÐBÑ Ÿ ,Þ
Informally, “completely regular” means that “BJ and can be separated by a bounded continuous real-valued function.”
The definition of “Tychonoff space” is quite different from the definitions of the other separation properties since the definition isn't “internal” it makes use of an “outside” topological space, ‘ , as part of its definition. Although it is possible to contrive a purely internal definition of “Tychonoff space,” the definition is complicated and seems completely unnatural: it simply imposes some fairly unintuitive conditions to force the existence of enough functions in GÐ\Ñ .
.ÐBßJÑ Example 3.3 Suppose +ÂJ ©Ð\ß.Ñ where Ð\ß.Ñ is a pseudometric space. Then 0ÐBÑœ .Ð+ßJ Ñ is continuous, 0Ð+Ñ œ " and 0lJ œ ! . So Ð\ß.Ñ is completely regular (and so a completely regular space might not even be X!ÑÞ
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\B\X0−GÐ\Ñ is a closed set in not containing . If is " , we can choose with $ # """ " 0ÐBÑ œ !and 0lJ œ "Þ Then Y œ 0 ÒÐ ∞ß## ÑÓ and Z œ 0 ÒÐ ß ∞ÑÓ are disjoint open sets with B−YßJ ©ZÞ Therefore \ is regular. Since \ is X"$ , \ is XÞñ
Hewitt's example of a X$ space on which every continuous real-valued function is constant is more than enough to show that a XX$ space may not be " (the example, in Ann. Math., 47(1946) 503-50 9, is $ # rather complicated. ). For that purpose, it is a little easierX\ but still nontrivial to find a $ space containing two points :ß ; such that for all 0 − GÐ\Ñß 0Ð:Ñ œ 0Ð;ÑÞ Then : and Ö;× cannot be separated by a function from GÐ\Ñ so \ is not T" Þ (DJ. Thomas, A regular space, not completely $ # regular, American Mathematical Monthly, 76(1969), 181-182). The space \ can then be used to construct an infinite XL$ space (simpler than Hewitt's example) on which every continuous 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.
293 Note: We have not shown that XÊX% " . This statement is true (as the notation suggests), but it is not $ # at all easy to prove: try it! This result will follow as a Corollary in Section 4.
Tychonoff spaces continue the pattern of “good behavior” that we saw earlier and 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 \αα Ág . \ is completely regular (T" Ñ iff each \ is α−E $ # completely regular (XÑ" . $ #
Proof Suppose +ÂJ ©E©\ , where \ is completely regular and 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 is completely regular, then each is homeomorphic to a subspace of gÁ\œα−E \αα \ \\so each αα is completely regular. Conversely, suppose each \ is completely regular and that J is a closed set in \+ not containing . There is a basic open set Y such that
+ − Y œ Yαα"8 ß ÞÞÞY © \ J
For each 3 œ "ß ÞÞÞß 8 we can pick a continuous function 0αα33 À \ Ä Ò!ß "Ó with 0 αα 33 Ð+ Ñ œ ! and
0αα33| Ð\ Y α 3 Ñ œ "Þ Define 0 À \ Ä Ò!ß "Ó by
0ÐBÑ œ max ÖÐ0αα33 ‰1 ÑÐBÑ À 3 œ "ß ÞÞÞß 8× œ max Ö0 αα 33 ÐB Ñ À 3 œ "ß ÞÞÞß 8×
Then 0 is continuous and 0Ð+Ñ œ max Ö0αα33 Ð+ Ñ À 3 œ "ß ÞÞÞß 8× œ !Þ If B − J , then for some 3ß
Bαα33  Y and 0 αα 33 ÐB Ñ œ " , so 0ÐBÑ œ " . Therefore 0lJ œ " and \ is completely regular. . Since the X" property is both hereditary and productive, the statements in a) and b) also hold for XÞ" ñ $ #
7 Corollary 3.8 Ò!ß "Ó and all its subspaces are X " . $ #
Since a Tychonoff space \GÐ\Ñ is defined using , we expect that continuous real-valued functions have a close relationship to the topology on \. We want to explore that connection.
Definition 3.9 Suppose 0 − GÐ\Ñ . Then ^Ð0Ñ œ 0" ÒÖ!×Ó œ ÖB − \ À 0ÐBÑ œ !× is called the zero set of 0 . If E œ ^Ð0Ñ for some 0 − GÐ\Ñ , we call + a zero set in \ . The complement of a zero set in \ is called a cozero set: coz Ð0Ñ œ \ ^Ð0Ñ œ ÖB − \ À 0ÐBÑ Á !×Þ
∞ A zero set ^Ð0Ñ in \ is closed because 0 is continuous. In addition, ^Ð0Ñ œ8œ" S8 , where " SœÖB−\Àl0ÐBÑl×888 . Each S is open. Therefore a zero set is always a closed K$ -set. Taking complements shows that cozÐ0Ñ is always an open J5 -set in \ .
For 0 − GÐ\Ñ , let 1 œ Ð " ” 0Ñ • " − G‡‡ Ð\ÑÞ Then ^Ð0Ñ œ ^Ð1Ñ . Therefore GÐ\Ñ and G Ð\Ñ produce the same zero in \ (and therefore also the same cozero sets).
294 Example 3.10
1) A closed set J in a pseudometric space Ð\ß.Ñ is a zero set: J œ ^Ð0Ñ , where 0ÐBÑ œ .ÐBßJÑÞ
2) In general, a closed set might not be a zero set\K if fact, a closed set in might not even be a $ set.
Suppose \ is uncountable and : − \ . Define a topology on \ by letting UB œ ÖÖB×× 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 satisfiedÞ )
All points in \Ö:× are isolated and \ is clearly XÞ"% In fact, \ is X .
If EF and are disjoint closed sets in \ , then one of them (say EÑ satisfies E©\Ö:× and therefore E is clopen. We then have open sets Y œE and Zœ\EªF that work in the definition of normality.
We do not know, at this point that XÊX% " , but we can argue directly that this \ is also X" . $$# #
If JBJ©\Ö:×ÖBש\Ö:× is a closed set not containing , then either or . So one of the sets JÖ:× or is clopen. The characteristic function of this clopen set is continuous and works to show that \ is completely regular.
The set Ö:× is a closed set but Ö:× is not a K$ set in \ (so Ö:× is not a zero set in \Ñ .
Suppose ∞ where is open. For each , for some Ö:× œ8œ" S88 S 8 : − F 88 © S Therefore ∞∞ F−8:U Þ \Ö:ל\8œ" Sœ 8 8œ" Ð\SÑ 8 ∞ is a countable set which is impossible since is uncountable. ©Ð\FÑ8œ" 8 \
Even when JKJ is both a closed set and a $ set, might not be a zero set. We will see examples later.
For 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 each set in the left column is a zero set obtained from some other function 1−GÐ\ÑÀ
^œÖBÀ0ÐBÑœ< ×1ÐBÑœ0ÐBÑ< œ^Ð1Ñß where ^ œ ÖB À 0ÐBÑ 0× œ ^Ð1Ñ , where 1ÐBÑ œ 0ÐBÑ l0ÐBÑl ^ œ ÖB À 0ÐBÑ Ÿ 0× œ ^Ð1Ñ , where 1ÐBÑ œ 0ÐBÑ l0ÐBÑl ^ œ ÖB À 0ÐBÑ r× œ ^Ð1Ñ , where 1ÐBÑ œ Ð0ÐBÑ <Ñ l0ÐBÑ 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\ÑÀ ^Ð1ÑœÖBÀ0ÐBÑœ< × where 0ÐBÑœ<1ÐBÑ< ^Ð1Ñ œ ÖB À 0ÐBÑ 0× where 0ÐBÑ œ l1ÐBÑl ^Ð1Ñ œ ÖB À 0ÐBÑ Ÿ 0 × where 0ÐBÑ œ l1ÐBÑl ^Ð1Ñ œ ÖB À 0ÐBÑ r× where 0ÐBÑ œ < l1ÐBÑl ^Ð1Ñ œ ÖB À 0ÐBÑ Ÿr× where 0ÐBÑ œ < l1ÐBÑl Taking complements, we see that if 0−GÐ\Ñ then a set with any one of the forms ÖBÀ0ÐBÑÁ<× , ÖB À 0ÐBÑ !×, ÖB À 0ÐBÑ !×ß ÖB À 0ÐBÑ <× , { B À 0ÐBÑ < } is a cozero set, and that 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 YZ and with B−YJ©ZÞ\ and If is completely regular and we choose 0−GÐ\Ñ with 0ÐBÑ œ ! and 0lJ œ " , then "" B−Y œÖBÀ0ÐBÑ## × and J ©Z œÖBÀ0ÐBÑ × Thus, in a completely regular space we can “separate” BJ and with special disjoint open sets: cozero sets. In fact this observation (according to Theorem 3.12, below) 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 GÐ\Ñ contains “too few” functions. In the extreme case of a X$ space L on which the only continuous real valued functions are constant (see the remarks at the beginning of this section), the only cozero sets are gL and ! The next theorem gives the connections between the cozero sets, GÐ\Ñ and the weak topology on \Þ Theorem 3.11 For any space Ð\ßg Ñ , GÐ\Ñ and G‡ Ð\Ñ induce the same weak topology on \Þ A base for this weak topology is the collection of all cozero sets in \ . Proof A subbase for the weak topology generated by GÐ\Ñ consists of all sets of the form 0" ÒYÓ , where Y is open in ‘ and 0 − GÐ\ÑÞ Without loss of generality, we can assume the sets Y are subbasic open sets of the form Ð+ß∞Ñ and Ð∞ß,Ñ , so the sets 0" ÒYÓ have the form ÖB − \ À 0ÐBÑ +×or ÖB − \ À 0ÐBÑ ,× . But these are cozero sets of \ , and every cozero set in \ has this form. So the cozero sets are a subbase for the weak topology generated by GÐ\Ñ . In fact, the cozero sets are actually a base because cozÐ0Ñ ∩ coz Ð1Ñ œ coz Ð01Ñ : the intersection of two cozero sets is a cozero set. The same argument, with G‡‡ Ð\Ñ replacing GÐ\Ñß shows that the cozero sets of G Ð\Ñ are a base for the weak topology on \ generated by G‡‡ Ð\ÑÞ But GÐ\Ñ and G Ð\Ñ have the same cozero sets in \ , and therefore generate the same weak topology on \ñ . Now we can now see the close connection between \GÐ\Ñ and in completely regular spaces. For any space Ð\ßgg Ñ , the functions in GÐ\Ñ are continuous with respect to (by definition of GÐ\Ñ ) ß but is g the smallest topology making this collection of functions continuous? In other words, is g 296 the weak topology on \GÐ\Ñ generated by ? 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 G‡ Ð\Ñ ) d) GÐ\Ñ (equivalently, G‡ Ð\Ñ ) separates points from closed sets. Proof The preceding theorem shows that b) and c) are equivalent. a)ÊB−SSÞJœ\SÞ0−GÐ\Ñ b) Suppose where is open Let Then we can choose " with 0ÐBÑ œ ! and 0lJ œ " . Then Y œ ÖB À 0ÐBÑ # × is a cozero set for which B − Y © SÞ Therefore the cozero sets are a base for \ . b)ÊJ d) Suppose is a closed set not containing B . By b), we can choose 0−GÐ\Ñ so that B −coz Ð0Ñ © \ J Þ Then 0ÐBÑ œ < Á ! , so 0ÐBÑ Â cl 0ÒJ Ó œ Ö!×Þ Therefore GÐ\Ñ separates points and closed sets. d)ÊJ a) Suppose is a closed set not containing B0−GÐ\Ñ . For some we have 0ÐBÑ Âcl 0ÒJ ÓÞ Without loss of generality (why? ), we can assume 0ÐBÑ œ !Þ Then, for some % ! , Ð %% ß Ñ ∩ 0ÒJ Ó œ g, so that for B − J , l0ÐBÑl % Þ Define 1 − G‡ Ð\Ñ by 1ÐBÑ œ min Öl0ÐBÑlß % ×Þ Then 1ÐBÑ œ ! and 1lJ œ% , so \ is completely regular. At each step of the proof, GÐ\Ñ can be replaced by G‡ Ð\Ñ (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‘©Ð\ßÑ . Then gY is completely regular. Proof Give \\ the topology the weak topology gYY generated by . Now that has a topology, the collection GÐ\Ñ makes sense. Let XG be the weak topology on \ generated by GÐ\ÑÞ The topology gggY does make all the functions in GÐ\Ñ continuous, so G ©Y Þ On the other hand: Yg© GÐ\Ñ by definition of Y , and the larger collection of functions GÐ\Ñ generates a (potentially) larger weak topology. Therefore ggY ©ÞG Therefore ggYYœÐ\ßÑñG . By Theorem 3.12, g is completely regular. 297 Example 3.14 ‘ 1) If Y‘œÖ0− À0 is nowhere differentiable × , then the weak topology g‘Y on generated by Y is completely regular. 2) If LX is an infinite $ 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 {gß L× . 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 \0−GÐ\Ñ is a Tychonoff space. For each ‡ , we have ran ] for some The evaluation map ‡ is Ð0Ñ © Ò+00 ß , œ M 0 + 0 , 0 −‘ Þ / À \ Ä ÖM 0 À 0 − G Ð\Ñ× an embedding. ‡ Proof \X is " , the 0 's are continuous and the collection of 0ÐœGÐ\ÑÑ 's separates points and closed sets. By Corollary VI.4.11, /ñ is an embedding. Since each is homeomorphic to , ‡7 is homeomorphic to where M00 Ò!ß "Ó ÖM À 0 − G Ð\Ñ× Ò!ß "Ó ß 7 œ lG‡ Ð\Ñl. Therefore any Tychonoff space can be embedded in a “cube.” On the other hand (Corollary 3.8) Ò!ß "Ó7 and all its subspaces are Tychonoff. So we have: Corollary 3.16 A space \\ is Tychonoff iff is homeomorphic to a subspace of the cube Ò!ß"Ó7 for some cardinal number 7 . The exponent 7 œ lG* Ð\Ñl in the corollary may not be the smallest possible. In an extreme case, for example, we have - œ lG‡ Б‘ Ñl , even though we can embed in Ò!ß "Ó œ Ò!ß "Ó" Þ However, the following theorem improves the value for 7 in certain cases. (We proved a similar result for metric spaces Ð\ß .Ñ : see Example VI.4.5.) Theorem 3.17 Suppose \7\ is Tychonoff with a base U of cardinality . Then can be embedded in Ò!ß "Ó7AÐ\Ñ. In particular, \ can be embedded in Ò!ß "Ó . Proof Suppose is finite. Since is , is a basic open set containing Only 7\XÖBלÖFÀFB×Þ" finitely many such intersections are possible, so \ is finite and therefore discrete. Hence top top \ © Ò!ß "Ó © Ò!ß "Ó7 Þ Suppose UUU is a base of cardinal 77 where is infinite. Call a pair ÐYßZÑ−‚ " distinguished if there exists a continuous 0À\ÄÒ!ß"Ó0ÐBÑYßZ with YßZ # for all B−Y and 0YßZ ÐBÑœ" for all B−\ZÞ Clearly, Y ©Z for a distinguished pair ÐYßZÑÞ For each distinguished pair, pick such a function 0œÖ0ÀÐYßZÑ−‚×YßZ and let YUU YßZ is distinguished . 298 We note that if B−Z −UU , then there must exist Y − such that B−Y and ÐYßZÑ is distinguished Þ To see this, pick an 0 À \ Ä Ò!ß "Ó so that 0ÐBÑ œ ! and 0l\ Z œ "Þ Then choose Y − U so that " " B − Y © 0 ÒÒ!ß# ÑÓ © Z Þ We claim that Y separates points and closed sets: Suppose J is a closed set not containing B . Choose a basic set Z−U with B−Z©\JÞ " There is a distinguished pair ÐYßZÑ with B−Y ©Z ©\JÞ Then 0YßZ ÐBÑœ< # and 0YßZ lJ œ ", so 0 YßZ ÐBÑ Â cl 0 YßZ ÒJ Ó œ Ö"×Þ By Corollary VI.4.11, /À\ÄÒ!ß"ÓllY is an embedding. Since 7 is infinite, lYUU lŸl ‚ lœ7# œ7. ñ A “metrization theorem” is one that states that certain topological properties of a space \\ imply that is metrizable. Typically the hypotheses of a metrization theorem involve asking 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 the theorem, \ © Ò!ß "Óii!! . Since Ò!ß "Ó 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 "Ói! and all its subspaces are separable metrizable spaces. Thus, the corollary tells us that, topologically, the separable metrizable spaces 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 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\ÑÀ if 0Ð:Ñœ! then there is a neighborhood R of : such that 0±R´! E10. Let 3À‘‘ Ä be the identity map and let Ð3ÑœÖ0−GБ‘ ÑÀ0œ13 for some 1−GÐ Ñ }. For those who know a bit of algebra: GБ Ñ Ð or, more generally, GÐ\Ñ Ñ with addition and multiplication defined pointwise, is a commutative ring with unit. Ð3Ñ is called the ideal generated by the element 3 . a) Prove that Ð3Ñ œ Ö0 − GБ Ñ À 0Ð!Ñ œ ! and the derivative 0w Ð!Ñ exists × . b) Exhibit two functions 0ß1 in GБ Ñ for which 01 − Ð3Ñ yet 0  Ð3Ñ and 1  Ð3Ñ . c) Let \ be a Tychonoff space with more than one point. Prove that there are two functions 0ß1 − GÐ\Ñ such that 01 ´ 0 on \ yet neither 0 nor 1 is identically 0 on \ . ÐÑSo the ring GÐ\Ñ has zero divisors. d) Prove that there are exactly two functions 00ÐBÑ−GБ Ñ for which 0## œ0 . (Here, means 0ÐBÑ † 0ÐBÑ, not 0Ð0ÐBÑÑ . ) e) Prove that there are exactly c functions 0G in ( ) for which 0œ0# . An element which equals its own square is called an idempotent in GÐ\Ñ ). Part d) shows that GБ Ñ and GÐ Ñ are not isomorphic rings since they have different numbers of idempotents. Is either isomorphic to GÐ Ñ? A classic part of general topology is the exploration of the relationship between the space \ and the rings GÐ\Ñ and G‡ Ð\Ñ . For example, if \ and ] are homeomorphic, then GÐ\Ñ is isomorphic to GÐ]Ñ. This necessarily implies that GÐ\ч‡ is isomorphic to GÐ]Ñ also (why?). The question “when does isomorphism imply homeomorphism” is more difficult. Another important area of study is how 300 the maximal ideals of the ring GÐ\Ñ are related to the topology of \ . The best introduction to this material is the classic 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 \l\l -Þ is a connected Tychonoff space with more than one point. Prove . E12. Let \ be a topological space. Suppose 0ß 1 − GÐ\Ñ and that ^Ð0Ñ is a neighborhood of ^Ð1Ñ (that is, ^Ð0Ñ © int ^Ð1ÑÞ ) a) Prove that there is a function 2 − GÐ\Ñ such that 0ÐBÑ œ 1ÐBÑ2ÐBÑ for all B − \ , i.e., that 01GÐ\Ñ is a multiple of in . Also, b) Give an example where ^Ð0Ñ ª ^Ð1Ñ but 0 is not a multiple of 1 in GÐ\Ñ . E13. Let \JßE©\JE be a Tychonoff space. Suppose , where is closed and is countable. Prove that if J∩Eœg , then E is disjoint from some zero set containing J . E14. A space \0À\Ä is called pseudocompact (Definition IV.8.7 ) if every continuous ‘ is bounded, that is, if GÐ\Ñ œ G‡ Ð\ÑÞ Consider the following condition on a topological space \ : (*) If ZªZª"# ... ªZªÞÞÞ 8 is a decreasing sequence of nonempty open sets, then ∞ cl . 8œ" ZÁg8 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, one that makes no explicit mention of ‘ . . 301 4. Normal and X% Spaces We now return to a topic in progress: normal spaces ÐX and % spaces Ñ . Even though normal spaces are badly behaved in some ways, there are still some very nice classical (and nontrivial) theorems we can prove. One of these will have “XÊX% " ” as a corollary. $ # To begin, the following technical variation on the definition of normality is very useful. Lemma 4.1 A space \SEE©S is normal iff whenever is open, closed, and , then there exists an open set Y with E©Y © cl Y ©SÞ Proof Suppose \S is normal and that is an open set containing the closed set EE . Then and Fœ\Sare 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 \EßF satisfies the stated condition and that are disjoint closed sets. Then E©Sœ\Fß so there is an open set Y with E©Y © cl Y ©\FÞ Let Z œ\ cl Y . YZ and are disjoint closed sets containing EF and respectively, so \ is normal. ñ 302 Theorem 4.2 a) A closed subspace of a normal (XX%% ) space is normal ( ). b) A continuous closed image of a normal (XX%% ) space normal ( ). Proof a) Suppose J\EF is a closed subspace of a normal space and let and be disjoint closed sets in JEßF . Then are also closed in \ so we can find disjoint open sets YZ\ww and in containing E and FYœY∩JZœZ∩JJE respectively. Then ww and are disjoint open sets in that contain and FJ , so is normal. b) Suppose \0À\Ä]EF is normal and that is continuous, closed and onto. If and are disjoint closed sets in ] , then 0" ÒEÓ and 0 " ÒFÓ are disjoint closed sets in \ . Pick Y w and Z w disjoint open sets in \ with 0" ÒEÓ © Y w and 0 " ÒFÓ © Z w Þ Then Y œ ] 0Ò\ Y w Ó and Zœ]0Ò\ZÓw are open sets in ] . If C−Y , then CÂ0Ò\Ywww Ó . Since 0 is onto, Cœ0ÐBÑ for some B−Y ©\Z Þ Therefore C−0Ò\Zw Ó so CÂZ . Hence Y∩Z œgÞ If C−E , then 0ÒÖC×Ó©0ÒEÓ©Y" " w , so 0ÒÖC×Ó∩Ð\YÑœg " w . Therefore CÂ0Ò\Yww Ó so C−] 0Ò\Y ÓœY . 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 Xñ% as well as normality. The next theorem gives us more examples of normal (and X% ) spaces. Theorem 4.3 Every regular Lindelof¨¨ space \X is normal (and therefore every Lindelof $ space is X%). Proof Suppose EF and are disjoint closed sets in \. For each B−E , use regularity to pick an open set YB−Y©Y©\FÞBBB such that cl Since the Lindelof¨ property is hereditary on closed subsets, a countable number of the YB"#8 's cover E : relabel these as Y ß Y ß ÞÞÞß Y ß ÞÞÞ . For each 8 , we have clY8 ∩ F œ gÞ Similarly, choose a sequence of open sets Z"# ß Z ß ÞÞÞß Z 8 ß ÞÞÞ covering F such that clZ∩Eœg8 for each 8 . We have that ∞∞ and , but these unions may not be disjoint. So we define 8œ"YªE88 8œ" ZªF ‡‡ YœY"""" cl Z ZœZ " cl Y " ‡‡ Y## œYÐZ∪#"#cl cl ZÑ Z œZÐY∪ #cl "# cl YÑ ãã ‡‡ Y88 œ Y8"# Ðcl Z ∪ cl Z ∪ ÞÞÞ ∪ cl Z 8 Ñ Z œ Z 8 Ðcl Y "# ∪ cl Y ∪ ÞÞÞ ∪ cl Y 8 Ñ ãã Let ∞∞‡‡and . Yœ8œ" Y88 Zœ 8œ" Z ‡ If B−E , then B cl Z85 for all 8 . But B−Y for some 5 , so B−Y5 ©YÞ 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 BÂZ∪Z∪ÞÞÞ∪Zß cl"# cl cl 5 so BÂZ∪Z∪ÞÞÞ∪Z"# 5 ‡‡ ‡ so BÂZ∪Z∪ÞÞÞ∪Z"# 5 ‡ so BÂZ8 for any 8Ÿ5Þ ‡‡ Since B − Y58 , then B − Y58"58 Þ So, if 8 5 , then B  Z œ Z Ð cl Y ∪ ÞÞÞ ∪ cl Y ∪ ÞÞÞ ∪ cl Y Ñ ‡ So BÂZ8 for all 8 , so BÂZ and therefore Y∩Z œgÞ ñ Example 4.4 The Sorgenfrey line WÒ+ß,Ñ is regular because the sets form a base of closed neighborhoods at each point +WW . We proved in Example VI.3.2 that is Lindelof,¨ so 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 “XÊX% " ”. Proving this is hard 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 ÖY< À<−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 \UÞ is any topological space and let be any dense subset of ‘ Suppose open sets Y©\< have been defined, for each <−U , in such a way that: i) and \œ<−U Y<< <−U Y œg ii) if <ß = − Uand < =, then cl Y<= © Y . For B−\ , define 0ÐBÑœ inf Ö<−UÀB−Y×< . Then 0À\Ä‘ is continuous. We will only use the Lemma once, with Uœ Þ So if you like, there is no harm in assuming that Uœ in the proof. Proof First note : 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 = À if B−Y , 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 0ÐBÑœ inf Ö<−UÀB−Y×< makes sense. From the definition of 0<ß=−Uß , we get that for a) if B− cl Y<= , then B−Y for all =< so 0ÐBÑŸ< 304 b) if 0ÐBÑ =, then B − Y=. We want to prove 0+−\U is continuous at each point . 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 D cl Y < , so DÂY < and 0ÐDÑ <ÞTherefore 0ÒY Ó © Ò<ß =ÓÞ ñ Our first major theorem about normal spaces is 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 ÐÑ Urysohn's Lemma A space \EßF is normal iff whenever are disjoint closed sets in \0−GÐ\Ñ0lEœ!0lFœ"Þ, there exists a function with and (When such an 0 exists, we say that EF and are completely separated .) Notice: the theorem says that E©0Ð!ÑF©0Ð"Ñß" and " but equality might not be true. If fact, if we had Eœ0Ð!ÑFœ0Ð"Ñ" and " , then E and F would have been zero sets in the first place. And in that case, normality would not even be necessary because, in any space \ À 1ÐBÑ# If Eœ^Ð1Ñ and Fœ^Ð2Ñ are disjoint zero sets, then the function 0ÐBÑœ 1ÐBÑ2ÐBÑ## completely separates EF and . This shows again that zero sets are very special closed sets: disjoint zero sets are always completely separated. So, 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 ZœÖBÀ0ÐBÑ × are disjoint open sets (in fact, cozero sets) containing EF and 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 such a way that the Lemma 5.1 applies. To start, let Yœg<< for <! and Yœ\ for <" . Enumerate the remaining rationals in ∩ Ò!ß "Ó as <"# ß < ß ÞÞÞß < 8 ß ÞÞÞ , beginning the list with < " œ " and <œ!Þ#<" E©Y<<"# œ\F, we can pick Y so that E©Y©Y©Yœ\F<<<##" cl 305 Then !œ<<<œ"#$" , and we use normality to pick an open set Y <$ so that E©Y©Y©Y©Y©Yœ\F<<<<<##$$" cl cl We continue by induction. Suppose 8 $ and that we have already defined open sets Y<<"# ß Y ß ÞÞÞß Y < 8 in such a way that whenever <345 < < Ð3ß 4ß 5 Ÿ 8Ñß then clY©Y©Y©YЇÑ<<34 cl < 4 <5 We need to define Y<8" so that Ð‡Ñ holds for 3ß 4ß 5 Ÿ 8 "Þ Since <"# œ " and < œ ! , and < 8" − Ð!ß "Ñß it makes sense to define <5"#88" œ the largest among < ß < ß ÞÞÞß < that is smaller than < ß and <6"#88" œ the smallest among < ß < ß ÞÞÞß < that is larger than < Þ By the induction hypothesis, we already have clY©YÞ<<56 Then use normality to pick an open set Y<8" so that clY©Y©Y©Y<<568" cl < 8" < . The Y0À\Ä< 's defined in this way satisfy the conditions of Lemma 5.1, so the function ‘ defined by 0ÐBÑœ inf Ö<− ÀB−Y×< Once we have the function 01œÐ!”0Ñ•"EF we can replace it, if we like, by so that and are completely separated by a function 1−G‡ Ð\ÑÞ It is also clear that we can modify 1 further to get an 2−GÐ\ч for which 2lEœ+2lFœ,+, and where and are any two real numbers. With Urysohn's Lemma, the proof of the following corollary is obvious. Corollary 5.3 XX% Ê " . $ # 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). Lemma 5.4 (Weierstrass Q -Test) Let \0À\Ä be a topological space. Suppose 8 ‘ is ∞ continuous for each 8 − and that l088 ÐBÑl Ÿ Q for all B − \ . If Q 8 ∞ , then 8œ" ∞ 0ÐBÑ œ 08 ÐBÑ converges (absolutely) for all B and 0 À \ Ä ‘ is continuous. 8œ" 306 ∞∞ ∞ Proof For each Bl0ÐBÑlŸQ∞0ÐBÑ , 88 , so 8 converges (absolutely) by the Comparison 8œ" 8œ" 8œ" Test. ∞ % Suppose +−\ and % ! . Choose R so that Q88 % . Each 0 is continuous, so for 8œR" % 8 œ "ß ÞÞÞß R we can pick a neighborhood Y8888 of + such that for B − Y ß l0 ÐBÑ 0 Ð+Ñl #R . Then R is a neighborhood of , and for we get Y œ8œ" Y8 + B − Y l0ÐBÑ 0Ð+Ñl R∞ R∞ œ l Ð088 ÐBÑ 0 Ð+ÑÑ Ð0 88 ÐBÑ 0 Ð+ÑÑl Ÿ l0 88 ÐBÑ 0 Ð+Ñl l0 88 ÐBÑ 0 Ð+Ñl 8œ"8œR" 8œ" 8œR" R∞ ∞ %%% Ÿ l088 ÐBÑ 0 Ð+Ñl l0 8 ÐBÑl l0 8 Ð+Ñl R †#R #Q 8 # # œ% Þ 8œ"8œR" 8œR" Therefore 0+ñ is continuous at . 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 #< l2ÐBÑ 9 ÐBÑl Ÿ$ for each B − E . << Proof Let E"""" œ ÖB − E À 2ÐBÑ Ÿ $$ × and F œ ÖB − E À 2ÐBÑ × . E and F are disjoint closed sets in EEEF\ , and since is closed, "" and are closed in . By Urysohn's Lemma, there exists << < < a continuous function 999À\ÄÒ$$ ß Ó such that lE"" œ $ and lF œ $ . << < If B − E" , then < Ÿ 2ÐBÑ Ÿ $$ and 99 ÐBÑ œ , so l2ÐBÑ ÐBÑl Ÿ l < Ð $ Ñl #< #< œ$$ à and similarly if B − F""" ß l2ÐBÑ 99 ÐBÑl Ÿ Þ If B − E ÐE ∪ F Ñ , then 2ÐBÑ and ÐBÑ are << #< both in Ò $$ ß Ó so l2ÐBÑ 9 ÐBÑl Ÿ $ . ñ Theorem 5.6 (Tietze's Extension Theorem) A space \E is normal iff whenever is a closed set in \ and 0 − GÐEÑ , then there exists a function 1 − GÐ\Ñ such that 1lE œ 0Þ Note: if E0−GÐEÑ is a closed subset of ‘ , then it is quite easy to prove that each can be extended to a function 1E defined on all of ‘‘ . In that case, the open set can be written as a countable union of disjoint open intervals Mß where each M œÐ+ß,Ñ or Ð∞ß,Ñ or Ð+ß∞Ñ (see Theorem II.3.4 ). Any endpoints of ME are in , where 0 is already defined. If MœÐ+ß,Ñ then extend the definition of 0 over Mby using a straight line segment to join Ð+ß 0Ð+ÑÑ and Ð,ß 0Ð,ÑÑ on the graph of 0 . If M œ Ð+ß ∞ÑÞ then extend the graph of 0M over using a horizontal right ray at height 0Ð+ÑàMœÐ∞ß,Ñß if then extend the graph of 0M over 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. Proof (ÉEF ) Suppose and are disjoint closed sets in \EF . and are clopen in the subspace E ∪ F so the function 0 À E ∪ F Ä Ò!ß "Ó defined by 0lE œ ! and 0lF œ " is continuous. Since E∪F is closed in \ , there is a function 1−GÐ\Ñ such that 1lÐE∪FÑœ0Þ Then "" YœÖBÀ1ÐBÑ## ×and ZœÖBÀ1ÐBÑ × are disjoint open sets (cozero sets, in fact) that contain EF and respectively. Therefore \ is normal. (Ê1−GÐ\Ñ ) The idea is to find a sequence of functions 3 such that 307 8 8 l0ÐBÑ 13 ÐBÑl Ä !as 8 Ä ∞for each B−E Ðwhere 0 is defined ). The sums 13 ÐBÑ are defined 3œ"3œ" on all of \ and as 8Ä∞ we can think of them as giving better and better approximations to the 8∞ extension 1 that we want. Then we can let 1ÐBÑ œlim 133 ÐBÑ œ 1 ÐBÑÞ The details follow. We 8Ä∞ 3œ" 3œ" proceed in three steps, but the heart of the argument is in Step I. Step I Suppose 0 À E Ä Ò "ß "Ó is continuous. We claim there is a continuous function 1 À \ Ä Ò "ß "Ó with 1lE œ 0Þ "" Using Lemma 5.5 (with 2œ0ß<œ"Ñ we get a function 1" œ9 À\ÄÒ$$ ß Ó such that ### for B − E , l0ÐBÑ 1"" ÐBÑl Ÿ$$$ . Therefore 0 1 À E Ä Ò ß ÓÞ ### Using Lemma 5.5 again (with 2œ01ß<œ"#$** Ñ , we get a function 1 À\ÄÒ ß Ó such %## %% that for B − Eß l0ÐBÑ 1"# ÐBÑ 1 ÐBÑl Ÿ*$ œ Ð Ñ . So 0 Ð1 "# 1 Ñ À E Ä Ò ** ß Ó %%% Using Lemma 5.5 again (with 2œ01"# 1ß<œ*#(#( Ñ , we get a function 1 $ À\ÄÒ ß Ó )#$ such that for B − Eß l0ÐBÑ 1"#$ ÐBÑ 1 ÐBÑ 1 ÐBÑl Ÿ#( œ Ð $ Ñ . )) So 0Ð1"#$ 1 1 ÑÀEÄÒ#( ß #( ÓÞ We continue, using induction, to find for each 3 a continuous function 8 ##3" 3" 8 133 À \ Ä Ò $$33 ß Ó such that l0ÐBÑ 1 ÐBÑl Ÿ Ð#Î$Ñ for B − E . 3œ" ∞∞ ∞ #3" Since l133 ÐBÑl Ÿ$3 ∞ß the series 1ÐBÑ œ 0 ÐBÑ converges (absolutely) for every 3œ" 3œ" 3œ" ∞ B − \, and 1 is continuous by the Weierstrass Q -Test. Since l1ÐBÑl œ l 13 ÐBÑl 3œ" ∞∞ 3" Ÿ l1ÐBÑlŸ# œ"ßwe have 1À\ÄÒ"ß"ÓÞ 3 $3 3œ" 8Ï3œ" 8 # 8 Finally, for B − Eß l0ÐBÑ 1ÐBÑl œlim l0ÐBÑ 13 ÐBÑl Ÿ lim Ð$ Ñ œ ! , so 1lE œ 0 and 8Ä∞3œ" 8Ä∞ the proof for Step I is complete. Step II Suppose 0 À E Ä Ð "ß "Ñ is continuous. We claim there is a continuous function 1 À \ Ä Ð "ß "Ñ with 1lE œ 0Þ Since 0 À E Ä Ð "ß "Ñ © Ò "ß "Ó , we can apply Case I to find a continuous function J À \ Ä Ò "ß "Ówith J lE œ 0Þ 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 Let FœÖB−\ÀJÐBÑœ „"× . E and F are disjoint closed sets in \ , so by Urysohn's Lemma there is a continuous 2 À \ Ä Ò "ß "Ó such that 2lF œ ! and 2lE œ "Þ If we let 1ÐBÑ œ J ÐBÑ2ÐBÑ , then 1 À \ Ä Ð "ß "Ñ and 1lE œ 0 , completing the proof of Case II. Step III (the full theorem) Suppose 0ÀEÄ‘ is continuous. We claim there is a continuous function 1À\Ä‘ with 1lEœ0Þ 308 Let 2À‘ ÄÐ"ß"Ñ be a homeomorphism. Then 2‰0ÀEÄÐ"ß"Ñ and, by Step II, there is a continuous J À \ Ä Ð "ß "Ñ with J lE œ 2 ‰ 0Þ Let 1œ2" ‰J À\Ä‘ . Then for B−E we have 1ÐBÑœ2 " ÐJÐBÑÑ œ 2" ÐÐ2 ‰ 0ÑÑÐBÑ œ 0ÐBÑÞ ñ It is easy to see that G‡ Ð\Ñ 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 WW‚W is normal (Example 4.4 ) but the plane 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 ‚WÑ Ä GÐHÑ given by 0È0lH is one-to-one, so lGÐW‚WÑlŸlGÐHÑŸl‘Hi lœ-! œ-Þ EœÖÐBßCÑ−W‚W ÀBCœ"× is closed and discrete in the subspace topology, so every function defined on E is continuous, that is, ‘E-- œ GÐEÑ and so lGÐEÑl œ - œ # Þ If W ‚ W were normal , then each 0 − GÐEÑ could be extended (by Tietze's Theorem) to a continuous function in GÐW ‚WÑ . This would mean that lGÐW ‚ WÑl l‘E-- l œ - œ # - . which is false. Therefore normality is not even finitely productive. 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 a closed discrete subspace. 309 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 Ò!ß "Ó Ða very nice, well-behaved space Ñ 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.” Then, in the 1960's, Mary Ellen Rudin constructed a normal space \ which was not countably paracompact. 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, constructing her space \ 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 \ can be constructed in ZFC with no additional set theoretic assumptions. Among other things, this example makes use of the box topology on a product. Example 5.8 The Sorgenfrey line WX is % , so WX is " and therefore the Sorgenfrey plane W‚W is $ # also XW‚W" . So is an example that shows X" does not imply X% . $$# # Extension theorems are an important idea in mathematics. In general, an “extension theorem” has the following form: E©\ and 0ÀEÄF , then there is a function 1À\ÄF such that 1lEœ0 . If we let 3ÀEÄ\ be the injection 3Ð+Ñœ+ , then the condition “ 1lEœ0 ” can be rewritten as 1‰3œ0. In the language of algebra, we are asking whether there is a suitable function 1 which “makes the diagram commute.” 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 the specifics. 1) Extension theorems that generalize of Tietze's Theorem: by putting stronger hypotheses on \F , we can relax the hypotheses on . 310 Suppose E\0ÀEÄF is closed in and is continuous. \Fœis normal and ‘ (Tietze's Theorem) \Fœ is normal and ‘8 If \F is collectionwise normal** and is a separable Banach space* \F is paracompact** and is a Banach space* then 01À\ÄFÞ has a continuous extension The statement that ‘‘8 can replace in Tietze's Theorem is easy to prove: 8 If \ is normal and 0 À E Ä‘ is continuous, write 0ÐBÑ œ Ð0"# ÐBÑß 0 ÐBÑß ÞÞÞß 0 8 ÐBÑÑ where each 0ÀEÄ3 ‘ . By Tietze's Theorem, there exists for each 3 a continuous extension 1333"8 À \ Ä‘ with 1 lE œ 0 . If we let 1ÐBÑ œ Ð1 ÐBÑß ÞÞÞß 1 ÐBÑÑ , then 1À\Ä‘8 and 1lEœ0 . In other words, we separately extend the coordinate functions in order to extend 08 . And in this example, could even be an infinite cardinal. * A normed linear space is a vector space Zl@lœ with a norm ( “absolute value”) that defines the “length” of each vector. Of course, a norm must satisfy certain axioms for example, l@"# @ l Ÿ l@ " l l@ # l . These properties guarantee that a norm can be used to define a metric: .Ð@ß@Ñœl@@lÞ"# " # A Banach space is a normed linear space which is 8 complete in this metric . . For example, ‘ , with its usual norm lÐB"# ß B ß ÞÞÞß B 8 Ñl ## # is a separable Banach space. œ B"# B ÞÞÞB8 ** 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 metric, or ÊÊparacompact collectionwise normalnormal Ê compact X# Therefore, in the theorems cited above, a continuous map 0 defined on a closed subset of a metric space (or, compact XF# space) and valued in a Banach space and be continuously extended a function 1À\ÄF . 2) The Hahn-Banach Theorem is another example, taken from functional analysis, of an extension. Roughly, it states: Suppose 0Q is a continuous linear functional defined on a subspace of a normed linear space \ . For such a map, a “norm” ll0ll can be defined Then 0 can be extended to a continuous linear functional 1 À \ Ä‘ for which ll1l œ ll0llÞ 3) Homotopy is usually not formulated in terms of extension theorems, but extensions are really at the heart of the idea. 311 Let 0 , 1 À Ò!ß "Ó Ä \ be continuous and suppose that 0Ð!Ñ œ 1Ð!Ñ œ B! 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 J À F Ä \ by J ÐBß !Ñ œ 0ÐBÑ J ÐBß "Ñ œ 1ÐBÑ JÐ!ß>Ñœ B!" JÐ"ß>Ñœ B 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, LÀÒ!ß"ÓÄ\Þ# If L does exist, then we have For each > − Ò!ß "Óß restrict L to the line segment at height > to define 0> ÐBÑ œ LÐBß >Ñ . Then for each >−Ò!ß"Ó , 0>!" is also a path in \ from B to B . As > moves from ! to " , we can think of the 0\>!" 's as a family of paths in that continuously deform 0œ00œ1Þ into The continuous extension L01 (if it exists) is called a homotopy between and with fixed endpoints, and we say that the paths 01 and are homotopic with fixed endpoints . In the space \0 below, it seems intuitively clear that can be continuously deformed (with endpoints held fixed) into 1 in other words, that L exists. 312 However in the space ]0 , and 1 together form a loop that surrounds a “hole” in ] , and it seems intuitively clear that the path 01 cannot be continuously deformed into the path within 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\X#% space is . Proof It is sufficient to show that \\ is regular because is Lindelof¨¨ and a regular Lindelof space is normal (Theorem 4.3). Suppose J is a closed set in \ and BÂJÞ For each C−J we can pick disjoint open sets YZCC and with B−YC−Z C and C. JZ is compact so a finite number of the C 's cover Jsay ZßZßÞÞÞßZÞB−YœYJ©ZœZYßZ Then 88 , , and are disjoint CC"# C 8 3œ"CC33 3œ" open sets. ñ Therefore, our results line up as: compact X# " (*) compact metric Ê or ÊXÊX%$# ÊXÊXÊXÊX $#"! metric In particular, Urysohn's Lemma and Tietze's Extension Theorem hold in metric spaces and in compact X# spaces. Notice that 313 i) the space Ð!ß "Ñ is X%# but not compact X ii) the Sorgenfrey line WX is % (see example 5.4) but not metrizable. ÐIf 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 $ # 77 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 it is hard to visualize how W ‚ W “sits” inside Ò!ß "Ó7 . 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 We have enough information now to completely characterize separable metric spaces topologically. Theorem 6.1 (Urysohn's Metrization Theorem) A second countable X$ space is metrizable. ÐNote: earlier we proved a similar metrization theorem” (Corollary 3.18), but the separation hypothesis then was XXÑ" rather than $ . $ # Proof \\ is second countable so is Lindelof,¨¨ and Theorem 4.3 tells us that a Lindelof XXÞ\X$%space is Therefore 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 “XXX" ” in Corollary 3.18 was unnecessarily strong. In fact, we see that $ and " $ # $ # 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 came roughly a quarter century after Urysohn's work. We state this characterization here without a proof. A family of sets U in Ð\ßg Ñ is called locally finite if each point B − \ has a neighborhood R that meets only finitely many sets in UU5 . The family is called -locally finite if we can write where each subfamily is locally finite. UUœ 8− 88 U 314 Theorem 6.2 (The Bing-Smirnov-Nagata Metrization Theorem) Ð\ßg Ñ is metrizable iff \ is X$ and has a 5U -locally finite base . Note: If \ is second countable, a countable base U5 œ ÖS"# ß S ß ÞÞÞß S 8 ß ÞÞÞ × is -locally finite because we can write where Therefore this Metrization Theorem UU œ 888 ß U œ ÖS ×Þ 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 0À W Ä ‘ extends to a continuous 1À \ Ä‘ , then W is closed. (The converse, of course, follows from Tietze's Extension Theorem.) E16. Let \ be a Tychonoff space. a) Suppose 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 :−Y , where Y is open in \ . Prove Ö:× is a G$ set in \ iff there exists a continuous function 0 À \ Ä Ò!ß "Ó such that 0" Ð"Ñ œ Ö:× and 0l\ Y œ ! . E17. Suppose ]BµC] is a Hausdorff space. Define in iff there does not exist a continuous function 0 À \ Ä Ò!ß "Ó such that 0ÐB ) Á 0ÐCÑ . Prove or disprove: ] Î µ is a Tychonoff space. E18. Prove that a Hausdorff space \œÖYY×\ is normal iff for each finite open cover h "8 , ... , of , there exist continuous functions , such that 8 = 1 for each 033 À \ Ä Ò!ß "Ó Ð3 œ "ß ÞÞÞ 8Ñ3œ" 0 ÐBÑ B − \ and such that, for each 30±\Y´! , 33 . (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 ih œÖ V"8 ,...,V × that “shrinks” in the sense that, cl for each . To begin the construction, let Pick an open Z©333 Z©Y 3 Jœ\ "3" YÞ 3 Z""""""#8 so that J © Z © cl Z © Y Þ Then ÖZ ß Y ß ÞÞÞß Y × still covers \Þ Continue by looking at and defining so that is still a J#"3 œ \ ÐZ ∪3# Y Ñ Z #"#$8 ÖZ ß Z ß Y ß ÞÞÞß Y × cover and Z©### cl Z©YÞ Continue in this way to replace the U 3 's one by one. Then use Urysohn's lemma to get functions 1033 which can then be used to define the 's. . E19. Suppose \\ is a compact, countable Hausdorff space. Prove that is completely metrizable. Hint: 1) For each pair of points BÁB87 in \ pick disjoint open sets Y 7ß87ß8 and Z containing these points. Consider the collection of all finite intersections of such sets. 2) Or: Since \Ö:×K is, countable, every singleton is a $ set. Use regularity to find a descending sequence of open sets containing such that ∞ cl Prove that the 's are Z888 :8œ" Z œ Ö:×Þ Z a neighborhood base at : . E20. A space\\ is called completely normal if every subspace of is normal. (For example, every metric space is completely normal). 316 a) Prove that \ is completely normal if and only if the following condition holds: whenever Eß F © \ and each of Eß F is disjoint from the closure of the other (i.e., (clE∩FÑ∪ÐE∩ cl FÑœgÑ , 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∪Z À Y is open in the usual topology on and Z © × . Prove that the scattered line is completely normal and therefore XÞ% E21. A X\" space is called perfectly normal if whenever EF and are disjoint nonempty closed sets in \ , there is an 0 − GÐ\Ñ with 0" Ð!Ñ œ E and 0 " Ð"Ñ œ F . a) Prove that every metric space Ð\ß .Ñ is perfectly normal. b) Prove that \\X\K is perfectly normal iff is % and every closed set in 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 XÞ% 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 XÐ\ßÑ$ space gUg has a locally finite base iff 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\ßgg Ñ is a topological space and let A be the weak topology on \ generated by GÐ\ÑÞ Then gg©ÞA 2. If \B−ÖC×C−ÖB× is regular and cl , then cl . 3. If JJ"# , , and J $ are pairwise disjoint closed sets in the normal space \ , then there exist pairwise disjoint open sets Y"# , Y , and Y $ such that for each 3œ"ß#ß$ , Y 3 ªJ 3 . 4. Let \+ß∞Ñ denote the real numbers with the topology g for which the “right rays” ( form a base. g is the same as the weak topology generated by GÐ\Ñ . 5. If \\ is infinite and has the cofinite topology, then is completely regular. 6. Every separable Tychonoff space can be embedded in Ò!ß "Ói! . 7. If 0 − GÐ Ñß we say 1 is a square root of 0 if 1 − GÐ Ñ and 1# œ 0 . If a function f in GÐ Ñ has more than one square root, then it has - square roots. 8. In a Tychonoff space, every closed set is an intersection of zero sets. 9. A subspace of a separable space need not be separable, but every subspace of the Sorgenfrey line is separable. 1 if BÁ! 10. Let 0À‘‘ Ä be given by 0ÐBÑœ sin lBl and 1À ‘‘ Ä be given by 1ÐBÑœ B . !Bœ! if The weak topology on ‘ induced by the functions 01 and is completely regular. 11. Suppose has the cofinite topology. If E0−GÐEÑ is closed in , then every can be extended to a function 1−GÐ ÑÞ 12. Let \\ be the Sorgenfrey plane. Then i! is second countable. 13. For 8 œ "ß #ß ÞÞÞ , let 088 À‘‘ Ä be given by 0 ÐBÑ œ B 8 and let g be the weak topology on ‘ i! generated by the 08 's. Then the evaluation map / À‘‘ Ä given by /ÐBÑÐ8Ñ œ 08 ÐBÑ is an embedding. 14. Let G be the set of points in the Cantor set with the subspace topology from the Sorgenfrey line W . Every continuous function 0ÀGÄ‘‘ can be extended to a continuous function 1ÀWÄ Þ 15. The product of two Lindelof¨ spaces cannot contain an uncountable closed discrete subset. 16. The Sorgenfrey line W is homeomorphic to an open subspace of the cube Ò!ß "Ó$ . 17. If Ð\ß .Ñ is a metric space, then \ is homeomorphic to a dense subspace of some compact Hausdorff space. 318 18. Suppose J©‘# ÞJ is closed iff J is a zero set. 19. A closed subspace of a product of two normal spaces is normal. 20. Suppose O\‚]EœÒOÓEXÞ is a compact subset of the Hausdorff space . Let 1\% . Then is 21. For any cardinal 7\Ò!ß"Ó , a subspace of 7 is metrizable if and only if \ is second countable. 22. Every space is the continuous image of a metrizable space. 23. Let Y‘œÖ0−‘ À0 is not continuous × . The weak topology on ‘ generated by Y is the discrete topology. 7 24. Every Lindelof¨ X$ space can be embedded in Ò!ß "Ó for some cardinal 7 . 25. A compact X# space is metrizable if and only if it is second countable. 26. If J[ is closed and _ is open in the Sorgenfrey line WJ©[S , and , then there is an open set in W such that J ©S©S©[ . 27. Suppose JO and are disjoint subsets of a Tychonoff space \ , where J is closed and O is compact. There are disjoint cozero sets YZ and with J©YO©ZÞ and 28. A separable metric space with a basis of clopen sets is homeomorphic to a subspace of the Cantor set. 7 29. Every X% space is homeomorphic to a subspace of some cube Ò!ß "Ó Þ 30. If every function in G‡ Ð\Ñ is constant, then every function in GÐ\Ñ must be constant. i! 31. is X%. 32. If 0À\Ä] is continuous and onto and \ is normal, then ] is normal. 33. If WW is the Sorgenfrey line, and let g be the weak topology on generated by the functions in GÐWÑ. Then ÐWßg Ñ is X% . 34. There is a closed set in the Sorgenfrey plane that is not a zero set. 35. Every closed set in is a zero set. 319