Chapter VII Separation Properties

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1 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 X! ßX" 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 X!, X", X, X$, and X%, each one stronger than its predecessor in the list. Once these had become common terminology, another separation axiom was discovered to be useful and interpolated into the list: X$ " 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 B Á C \, there either exists an open set Y with B Y, C Â Y or there exists an open set Z with C Z, B Â Z 2) X" -space if, whenever B Á C \, there exists an open set Y with B YßC Â Z and there exists an open set Z with B Â YßC Z 3) X -space (or, Hausdorff space) if, whenever B Á C \, 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 Ê X Ê X Þ "! Example 2.2 1) \ is a X! -space if and only if: whenever B Á C, then ab Á ac that is, different points in \ have different neighborhood systems. 283

2 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.Ñ is X! and that B Á CÞ Then for some %! either B  F% ÐCÑ or C  F% ÐBÑ. Either way,.ðbßcñ %, so. is a metric. 4) In any topological space \ we can define an equivalence relation B µ C iff ab œ ac. Let 1 À \ Ä \Î µ œ ] 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 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 œ ÖB \ À 1ÐBÑ 1ÒSÓ œ ÖB \ À for some C S, 1ÐBÑ œ 1ÐCÑ œ ÖB \ À B is equivalent to some point C in S œ S. If ÒBÓ Á ÒCÓ ], then B is not equivalent to C, so there is an open set S \ with (say) B S and C  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 ] is called the X!- identification of \. This identification turns any space into a X! -space by identifying points that have identical neighborhoods. If \ is a X! -space to begin with, then 1 is oneto-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, a œ a if and only if.ðbßcñ œ!þ w w 5) For 3 œ!ß"ß À if Ð\ßg Ñ is a X 3 space and g ª g is a new topology on \, then Ð\ßg Ñ is also a space. X 3 B C X! 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 \, ÖB œ + ÖS À S is open and B S 2) A finite X " space is discrete. 3) Sierpinski space \ œ Ö!ß" with topology g œ ÖgßÖ" 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 S œ ÐBß Ñ is an open set that contains C and not B; but there is no open set that contains B and not C. 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

3 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 X3-space b) If \ œ \ Á g, then \ is a X -space iff each \ is a X -space. α E α 3 α 3 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 \ is a X" space. If Y is an open set in \ containing B but not C, w then Y œ Y E is an open set in E containing B but not C. Similarly we can find an open set Z in E containing C but not BÞ Therefore E is a X " -space. α α α b) Suppose \ œ E \ is a nonempty X "-space. Each \ is homeomorphic to a subspace of \, so, by part a), each \ α is X" Þ Conversely, suppose each \ α is X" and that B Á C \. Then Bα Á Cα for some α. Pick an open set Y α in \ α containing Bα but not Cα. Then Y œ Yα is an open set in \ containing B but not C. Similarly, we find an open set Z in \ containing C but not B. Therefore \ is a X -space. ñ " Exercise 2.5 Is a continuous image of a X3-space necessarily a X3-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, but with a closed set replacing one of the points. X Definition 2.6 A topological space \ is called regular if whenever J is a closed set and B  Jß there exist disjoint open sets Y and Z such that B Y and J Z. 285

4 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 cly 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 J œ \ 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) Ê iii) If R a B, then B S œ int R. By ii), we can find an open set Y so that B Y cl Y S RÞ 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 B  J. By ii), there is a closed neighborhood O of B 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.ðbßjñ for which 0Ð+Ñ œ -! and 0lJ œ!þ This gives us disjoint " - " - open sets with + Y œ 0 ÒÐ ß ÑÓ and J Z œ 0 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!. 286

5 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 X$ Ê X ( Ê X" Ê X!): suppose \ is X$ and B Á C \. Then J œ ÖC is closed so, by regularity, there are disjoint open sets Y, Z with B Y and C ÖC Z. Caution Terminology varies from book to book. For some authors, the definition of regular includes X" À 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 X ÊÎ X $. We will put a new topology on the set \ œ. At each point : \, let a neighborhood base consist of all sets R of the form U : R œ F Ð:Ñ Ð 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 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 YÞ Using the % in the definition of R, we can choose a point: œ ÐBß!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

6 Note: The usual topology in is regular. This 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 BßJ that need to be separated. By contrast, it is easy to see that an enlargement of a X topology Ð3 œ!ß"ßñ is still X Þ 3 3 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 \ Á g. \ is regular ( X ) iff each \ is regular ( X ). α E α $ α $ Proof a) Let E \ where \ is regular. Suppose + E and that J is a closed set in E that does w w not contain +. There exists a closed set J in \ such that J E œ JÞ Choose disjoint open sets w w w w w w w Y and Z in \ with + Y and J Z Þ Then Y œ Y E andz œ Z E are open in E, disjoint, + Y, and J Z. Therefore E is regular. α α α b) If \ œ E\ Á g is regular, then part a) implies that each \ is regularß because each \ α is homeomorphic to a subspace of \. Conversely, suppose each \ α is regular and that Y œ Yα ßÞÞÞßYα is a basic open set containing B. For each α Z " 8 3, we can pick an open set α in 3 \ α such that Bα Z α cl Z α Yα Þ Then B Z œ Zα ßÞÞÞßZα clz " 8 clzα ßÞÞÞß cl Zα YÞ ( Why is the last inclusion true? ) Therefore \ is regular. " 8 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 Y, Z in \ with E Y and F ZÞ \ is called a X %-space if \ is normal and X1Þ 288

7 Example 2.15 a) Every pseudometric space Ð\ß.Ñ is normal (so every metric space is X Ñ..ÐBßEÑ.ÐBßEÑ.ÐBßFÑ In fact, if E and F are disjoint closed sets, we can define 0ÐBÑ œ. Since the denominator cannot be!, 0 is continuous and 0lE œ!, 0lF œ "Þ The open sets Y œ ÖB À 0ÐBÑ " and Z œ ÖB À 0ÐBÑ are disjoint that 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 directly constructing a pair of disjoint open sets that contain E and F? b) Let have the right ray topology g œ ÖÐBß ÑàB Ögß. Ð ßg 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 Y œ g and Z œ \Þ Also, Ð ßg Ñ is not regular: for example " is not in the closed set J œ Ð ß!Ó, but every open set that contains J also contains ". So normal ÊÎ regular. But Ð ßg Ñ is not X" and therefore not X% Þ % " When we combine normal X" into X%, we have a property that fits perfectly into the separation hierarchy. Theorem 2.16 X Ê X Ð Ê X Ê X Ê X Ñ % $ "! Proof Suppose \ is X%. If J is a closed set not containing B, then ÖB and J are disjoint closed sets. By normality, we can find disjoint open sets separating ÖB and J. It follows that \ is regular and therefore X. ñ $. 289

8 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 U œ ÖS À S œ \ Jß J Y is a base for the open sets in \. For a polynomial T in 8 real variables, define the zero set of T as ^ÐTÑ œ ÖÐB" ßB ßÞÞÞßB8 Ñ 8 À T ÐB" ßB ßÞÞÞßB8Ñ œ! a) Prove that Ö^ÐTÑ À T a polynomial in 8 real variables is the base for the closed sets of a topology (called the Zariski topology) on 8 Þ b) Prove that the Zariski topology on is X but not X. 8 " 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 ^ÐT Ñ are rather special geometric objects those surfaces in 8 which can be described by polynomial equations TÐB ßB ßÞÞÞßB Ñ œ! " 8. E3. A space \ is a X&Î space if, whenever B Á C \, there exist open sets Y and Z such that B Y, C Z and cly cl Z œ g. ( Clearly, T Ê X Ê T Þ) $ &Î a) Prove that a subspace of a X space is a X space. &Î b) Suppose \ œ \ Á g. Prove that \ is X iff each \ is X. &Î α &Î α c) Let W œ ÖÐBßCÑ À C! and P œ ÖÐBßCÑ W À C œ! Þ Define a topology on W with the following neighborhood bases: You may assume that these Prove that is but not. S T T &Î $ if : W P À U : œ ÖF% Ð:Ñ W À %! if : P: U œ ÖF Ð:Ñ ÐW PÑ Ö: À %! : % &Î 's satisfy the axioms for a neighborhood base. U : 290

9 E4. Suppose E \. Define a topology on \ by Decide whether or not g œ ÖS \ À S ª E Ög 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 regularà and that if \ is X $, the ] is also X Þ $ E6. a) Suppose \ is finite. Prove that Ð\ßgÑ 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 J œ + ÖS À S is open and J S Þ ii) Define B µ C iff B œ C or BßC JÞ Prove that the quotient space \Î µ 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 Y8 F Á g and that cly8 cl Y7 œ g whenever 8 Á 7. e) Suppose each point C in a space ] has a neighborhood Z such that cl Z is regular. Prove that ] is regular. 291

10 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 0ÐBÑ œ.ðbß+ñ or 0ÐBÑ œ.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\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 bounded œ 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 Then 0 1 and 0 1 are in GÐ\ÑÞ ( 0 1ÑÐBÑ œ maxö0ðbñß1ðbñ Ð0 1ÑÐBÑ œ minö0ðbñß1ðbñ 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 (0 1ÑÐBÑ œ 0ÐBÑ 1ÐBÑ l0ðbñ 1ÐBÑl Ð0 1ÑÐBÑ œ 0ÐBÑ 1ÐBÑ l0ðbñ 1ÐBÑl ñ 292

11 Definition 3.2 A space \ is called completely regular if whenever J is a closed set and +  Jß there exists a function 0 GÐ\Ñ such that 0Ð+Ñ œ! and 0lJ œ "Þ Informally, completely regular means that + and J function. can be separated by a continuous real-valued Note i) The definition requires that 0lJ œ ", 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 1 À \ Ä Ò!ß"Ó such that 1Ð+Ñ œ! and 1lJ œ ". For example, we could use 1 œ Ð0!Ñ " which, by Lemma 3.1, is continuous. iii) Suppose 1 À \ Ä Ò!ß"Ó is continuous and 1Ð+Ñ œ!, 1lJ œ "Þ The particular values!ß" in the definition are not important.: they could be any real numbers < =Þ ÐIf we choose a homeomorphism 9 À Ò!ß"Ó Ä Ò<ß=Ó, then it must be true that either 9Ð!Ñ œ <, 9Ð"Ñ œ = or 9Ð!Ñ œ =, 9Ð"Ñ œ < why?. Then 2 œ 9 1 À \ Ä Ò<ß=Ó andß depending on how you chose 9ß 2Ð+Ñ œ < and 2lJ œ, or vice-versa. ) Putting these observations together, we see Definition 3.2 is equivalent to: Definition 3.2 w A space \ is called completely regular if whenever J is a closed set, +  Jß and <ß= are real numbers with < =ß then there exists a continuous function 0 À \ Ä Ò<ß=Ó for which 0Ð+Ñ œ < and0lj œ =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 +  JÞ Then.ÐBßJ Ñ.Ð+ßJÑ 0ÐBÑ œ is continuous, 0Ð+Ñ œ " and 0lJ œ!. 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 " Ê X Ð Ê X Ê X Ê X Ñ $ $ "! " $ Proof Suppose J is a closed set in \ not containing +. If \ is X$ ", we can choose 0 GÐ\Ñ with " " " " 0Ð+Ñ œ! and 0lJ œ "Þ Then Y œ 0 ÒÐ ß ÑÓ and Z œ 0 ÒÐ ß ÑÓ are disjoint open sets with + Yß J ZÞ Therefore \ is regular. Since \ is X, \ is X Þ ñ " $ 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) , is rather complicated.). For that purpose, it is a little easier but still nontrivial to find a X $ 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 (See D.J. Thomas, A regular space, not completely regular, American Mathematical Monthly, 76(1969), ). The space \ can then be used to construct an infinite X space L (simpler than Hewitt's example) on which every continuous $ 293

12 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 ( X " Ñ space is completely regular ( X " $ $ Ñ. b) Suppose \ œ α E \ α Á g. \ is completely regular ( X$ " Ñ iff each \ α is completely regular ( X$ " Ñ. Proof Suppose +  J E \, where \ is completely regular and J is a closed set in E. Pick a closed set O in \ such that O E œ J and an 0 GÐ\Ñ such that 0Ð+Ñ œ! and 0lO œ ". Then 1 œ 0lE GÐEÑß 1Ð+Ñ œ! and 1lJ œ "Þ Therefore E is completely regular. If g Á \ 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 + Y œ Y ßÞÞÞY \ J α α " 8 For each 3 œ "ßÞÞÞß8 we can pick a continuous function 0α À \ α Ä Ò!ß"Ó with 0α Ð+ α Ñ œ! and Ð\ Y Ñ œ "Þ Define 0 À \ Ä Ò!ß"Ó by α α α ÐBÑ œ maxöð0 1 ÑÐBÑ À 3 œ "ßÞÞÞß8 œ maxö0 ÐB Ñ À 3 œ "ßÞÞÞß8 α α α α Then 0 is continuous and 0Ð+Ñ œ max Ö0α Ð+ α Ñ À 3 œ "ßÞÞÞß8 œ!þ If B J, then for some 3ß 3 3 Bα  Yα and 0α ÐBα 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 X$ " Þ ñ Corollary 3.8 For any cardinal 7, the cube Ò!ß"Ó 7 and all its subspaces are X. $ " Since a Tychonoff space \ is defined using functions in GÐ\Ñ, we expect that these functions will 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 E a zero set in \. The complement of a zero set in \ is called a cozero set: cozð0ñ œ \ ^Ð0Ñ œ ÖB \ À 0ÐBÑ Á! Þ + 8œ" A zero set ^Ð0Ñ in \ is closed because 0 is continuous. In addition, ^Ð0Ñ œ S8, where " S8 œ ÖB \ À l0ðbñl. Each S8 is open. Therefore a zero set is always a closed K -set. 8 $ Taking complements shows that coz Ð0Ñ is always an open J -set in \

13 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 sets in \ (and therefore also the same cozero sets). Example ) 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 in \ might not be a zero set in fact, a closed set in \ might not even be a set. K $ Suppose \ is uncountable and : \. Define a topology on \ by letting U B œ ÖÖB be a neighborhood at each point B Á : and letting U : œ ÖF À : 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 X" Þ In fact, \ is X%. If E and F are disjoint closed sets in \, then one of them (say EÑ satisfies E \ Ö:, so E is clopen. We then have open sets Y œ E and Z œ \ E ª F, so \ is normal. We do not know yet that X Ê X 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: or ÖB \ Ö:. 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 \, so Ö: is not a zero set in \. + 8œ" Suppose : S where S is open. For each 8, there is a basic neighborhood F of : such that : F 8 S8ß so \ S 8 \ F8 is countable. Therefore \ + S œ - 8œ" 8 8œ" Ð\ S8Ñ is countable. Since \ is uncountable, we conclude that Ö: Á S Þ + 8œ" 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 ^ œ ÖB À 0ÐBÑ œ < œ ^Ð1Ñß where 1ÐBÑ œ 0ÐBÑ < ^ 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 <l ^ œ ÖB À 0ÐBÑ Ÿ r œ ^Ð1Ñ, where 1ÐBÑ œ Ð0ÐBÑ <Ñ l0ðbñ <l 295

14 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 ^Ð1Ñ œ ÖB À 0ÐBÑ œ < where ^Ð1Ñ œ ÖB À 0ÐBÑ 0 where ^Ð1Ñ œ ÖB À 0ÐBÑ Ÿ 0 where ^Ð1Ñ œ ÖB À 0ÐBÑ r where ^Ð1Ñ œ ÖB À 0ÐBÑ Ÿ r where 0ÐBÑ œ 1ÐBÑ < 0ÐBÑ œ l1ðbñl 0ÐBÑ œ l1ðbñl 0ÐBÑ œ < l1ðbñl 0ÐBÑ œ < l1ðbñl Taking complements, we get the corresponding results for cozero sets: if 0 GÐ\Ñ i) the sets ÖB À 0ÐBÑ Á <, ÖB À 0ÐBÑ!, ÖB À 0ÐBÑ! ßÖB À 0ÐBÑ <, { B À 0ÐBÑ < } 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 B  J, where J is closed in \Þ If \ is regular, we can find disjoint open sets Y and Z with B Y and J ZÞ But if \ is completely regular, we can separate B and J with special open sets Y and Z cozero sets! Just choose 0 GÐ\Ñ with 0ÐBÑ œ! and 0lJ œ ", then B Y œ ÖB À 0ÐBÑ " and J Z œ ÖB À 0ÐBÑ " 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\ßg Ñ, GÐ\Ñ and G Ð\Ñ induce the same weak topology ga on \, and a base for g A is the collection of all cozero sets in \. " Proof A subbase for g A 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 that the sets 0 ÒYÓ have 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 forg A In fact, the cozero sets are actually a base because cozð0ñ cozð1ñ 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 produce the same cozero sets in \, and therefore generate the same weak topology on \. ñ g A Now we can now see the close connection between \ and GÐ\Ñ in completely regular spaces. For any space Ð\ßgÑ 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

15 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 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) Ê b) Suppose + S where S is open Þ Let J œ \ SÞ Then we can choose 0 GÐ\Ñ " with 0Ð+Ñ œ! and 0lJ œ ". Then Y œ ÖB À 0ÐBÑ 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+Ñ œ < Á!, so 0Ð+Ñ Â cl 0ÒJÓ œ Ö! 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 Â cl 0ÒJÓÞ Without loss of generality ( why? ), we can assume 0Ð+Ñ œ!þ Then, for some %!, Ð % ß% Ñ 0ÒJÓ œ g, so that for B J, l0ðbñl % Þ Define 1 G Ð\Ñ by 1ÐBÑ œ minöl0ðbñlß % Þ Then 1Ð+Ñ 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 g Y be the weak topology on \ generated by any family of \ functions Y. Then Ð\ßg Ñ is completely regular Þ Ð\ßg ) is Tychonoff is Y separates points. Y Proof Give \ the topology the weak topology gy generated by Y. Now \ has a topology, so the collection GÐ\Ñ makes sense. Let X A be the weak topology on \ generated by GÐ\ÑÞ The topology gy does make all the functions in GÐ\Ñ continuous, so ga gyþ On the other hand: Y GÐ\Ñ by definition of gy, and the larger collection of functions GÐ\Ñ generates a (potentially) larger weak topology. Therefore gy gaþ Therefore g œ g. By Theorem 3.12, Ð\ßg Ñ is completely regular. ñ Y A Y Y 297

16 Example 3.14 by Y 1) If Y œ Ö0 À 0 is nowhere differentiable, then the weak topology g Y on generated 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 { 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 \ is a Tychonoff space. For each 0 G Ð\Ñ, we have ran Ð0Ñ Ò+ 0ß, 0 ] œ M 0 for some + 0, 0 Þ The evaluation map / À \ Ä ÖM0 À 0 G Ð\Ñ is an embedding. Proof \ is X", the 0's are continuous and the collection of 0's Ð œ G Ð\ÑÑ separates points and closed sets. By Corollary VI.4.11, / is an embedding. ñ Since each M0 is homeomorphic to Ò!ß"Ó, 7 ÖM0À0 G Ð\Ñ is homeomorphic to Ò!ß"Ó ß 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 - œ lg Ð Ñl, even though we can embed in Ò!ß"Ó œ Ò!ß"Ó Þ 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 7 AÐ\Ñ Ò!ß"Ó. In particular, \ can be embedded in Ò!ß"Ó. Proof Suppose 7 is finite. Since \ is X, ÖB œ ÖF À F is a basic open set containing B Þ Only finitely many such intersections are possible, so \ is finite and therefore discrete. Hence top top " + 7 \ Ò!ß"Ó Ò!ß"Ó 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 \ Ä Ò!ß"Ó with 0YßZÐBÑ for all B Y and 0Y ßZ ÐBÑ œ " for all B \ ZÞ Clearly, Y Z for a distinguished pair ÐYßZÑÞ For each distinguished pair, pick such a function 0 and let Y œ Ö0 À ÐYßZÑ U U is distinguished. YßZ YßZ 298

17 We note that if + Z U, then there must exist Y U such that + Y and ÐYßZÑ is distinguishedþ To see this, pick an 0 À \ Ä Ò!ß"Ó so that 0Ð+Ñ œ! and 0l\ Z œ "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 0YßZÐ+Ñ œ < and 0YßZlJ œ ", so 0YßZÐ+Ñ Â cl0yßzòjó œ Ö" Þ By Corollary VI.4.11, / À \ Ä Ò!ß"Ó l Y l is an embedding. Since 7 is infinite, lyl Ÿ l U U l œ 7 œ 7. 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 i Proof By Theorem 3.17, \ Ò!ß"Ó! i. 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"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

18 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\Ñ À if 0Ð:Ñ œ! then there is a neighborhood R of : such that 0 ± R! E10. Let 3 À Ä be the identity map and let Ð3Ñ œ Ö0 GÐ Ñ À 0 œ 13 for some 1 GÐ Ñ}. Ð3Ñ is called the ideal in GÐ Ñ generated by the element 3. For those who know a bit of algebra: if we definite addition and multiplication of functions pointwise, then GÐ Ñ Ðor, more generally, GÐ\ÑÑ is a commutative ringþ The constant function! is the zero element in the ring; there is also a unit element, namely the constant function ". w a) Prove that Ð3Ñ œ Ö0 GÐ Ñ À 0Ð!Ñ œ! and the derivative 0 Ð!Ñ exists. b) Exhibit two functions 0ß1 in GÐ 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 œ! on \ yet neither 0 nor 1 is identically 0 on \. Thus, there are functions 0ß1 GÐ\Ñ for which 01 œ! although 0 Á! and 1 Á! Þ 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 GÐ Ñ for which 0 œ 0. ( M8GÐ\Ñß the notation 0 ÐBÑ means 0ÐBÑ 0ÐBÑ, not 0Ð0ÐBÑÑ. ) e) Prove that there are exactly c functions 0 in G( ) for which 0 œ 0. An element in GÐ\Ñ that equals its own square is called an idempotent. Part d) shows that GÐ Ñ and GÐ Ñ are not isomorphic rings since they have different numbers of idempotents. Is either GÐ Ñ or GÐ Ñ isomorphic to GÐ Ñ? One classic part of general topology is to explore the relationship between the space \ and the rings GÐ\Ñ and G Ð\Ñ. For example, if \ is homeomorphic to ], then GÐ\Ñ is isomorphic to GÐ]Ñ. This necessarily implies (why?) that G Ð\Ñ is isomorphic to G Ð]Ñ. The question when does 300

19 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 HÐ Ñ be the set of differentiable functions 0 À Ä. Are the rings GÐ Ñ and HÐ Ñ isomorphic? Hint: An isomorphism between GÐ 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 1 in GÐ\Ñß that is, prove there is a function 2 GÐ\Ñ such that 0ÐBÑ œ 1ÐBÑ2ÐBÑ for all B \. b) Give an example where ^Ð0Ñ ª ^Ð1Ñ but 0 is not a multiple of 1 in GÐ\Ñ. E13. Let \ be a Tychonoff space with subspaces J and Eß where J 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 (s ee Definition IV.8.7). Consider the following condition (*) on a space \: (*) Whenever Z " ª Z ª... ª Z 8 ª ÞÞÞ is a decreasing sequence of nonempty open sets, + 8œ" then cl Z Á g. 8 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

20 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 E is closed ßS is open and E S, there exists an open set U with E Y cl Y SÞ Proof Suppose \ is normal and that S 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 cly \ Z SÞ Conversely, suppose \ satisfies the stated condition and that EßF are disjoint closed sets. Then E S œ \ Fß so there is an open set Y with E Y cl Y \ FÞ Let Z œ \ cl Y. Y and Z are disjoint closed sets containing E and F respectively, so \ is normal. ñ 302

21 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 w w sets in J. Then Eß F are also closed in \ so we can find disjoint open sets Y and Z in \ containing E and F respectively. Then Y œ Y w w J and Z œ Z J are disjoint open sets in J that contain E and F, so J is normal. b) Suppose \ is normal and that 0 À \ Ä ] is continuous, closed and onto. If E and F are " " w w disjoint closed sets in ], then 0 ÒEÓ and 0 ÒFÓ are disjoint closed sets in \. Pick Y and Z disjoint open sets in \ with 0 " ÒEÓ Y w and 0 " ÒFÓ Z w Þ Then Y œ ] 0Ò\ Y w Ó and w Z œ ] 0Ò\ Z Ó are open sets in ]. If C Y, then C  0Ò\ Y w Ó. Since 0 is onto, C œ 0ÐBÑ for some B Y w \ Z w Þ Therefore w C 0Ò\ Z Ó so C  Z. Hence Y Z œ gþ If C E, then 0 " ÒÖC Ó 0 " ÒEÓ Y w, so 0 " ÒÖC Ó Ð\ Y w Ñ œ g. Therefore C  0Ò\ Y w Ó so w C ] 0Ò\ Y Ó œ Y. 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 ) spaces. X % Theorem 4.3 Every regular Lindelöf space \ is normal (and therefore every Lindelöf X $ -space is ). X % Proof Suppose E and F are disjoint closed sets in \. For each B E, use regularity to pick an open set YB such that B Y B cl Y B \ FÞ Since the Lindelöf property is hereditary on closed subsets, a countable number of the YB 's cover E: relabel these as Y" ßY ßÞÞÞßY8ßÞÞÞ. For each 8, we have cl Y8 F œ gþ Similarly, choose a sequence of open sets Z" ßZ ßÞÞÞßZ8ßÞÞÞ covering F such that clz E œ g for each 8. 8 We have that - and - Y ª E Z ª F, but these unions may not be disjoint. So we define 8œ" 8 8œ" 8 Y" œ Y" cl Z" Z" œ Z" cly" Y œ Y ÐclZ" cl Z Ñ Z œ Z ÐclY" cly Ñ ã ã Y8 œ Y8 ÐclZ" clz ÞÞÞ cl Z8Ñ Z8 œ Z8 ÐclY" cly ÞÞÞ cly8ñ ã ã Let Y œ - Y and Z œ - Z. 8œ" 8 8œ" 8 If B E, then B  cl Z8 for all 8. But B Y5 for some 5, so B Y 5 YÞ Therefore E Y and, similarly, F ZÞ To complete the proof, we show that Y Z œ g. Suppose B Y. 303

22 Then B Y5 for some 5ß so B  clz" clz ÞÞÞ clz5ß so B  Z" Z ÞÞÞ Z5 so B  Z" Z 5 so B  Z for any 8 Ÿ 5Þ 8 Since B Y, then B Y Þ So, if 8 5, then B  Z œ Z ÐclY ÞÞÞ cly ÞÞÞ cly Ñ " 5 8 So B  Z8 for all 8, so B  Z and therefore Y Z œ gþ ñ 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 Lindelöf, so W is normal. Since W is X, we have that W is X. " % 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 \ is X " $, 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 \ 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 - Y and + < U < < UY< œ g ii) if <ß= U and < =, then cl Y Y. < = For B \, define 0ÐBÑ œ inf Ö< U À B Y <. Then 0 À \ Ä is continuous. We will use this Lemma only once, with U œ in the proof. U œ Þ So if you like, there is no harm in assuming that Proof Suppose B \Þ By i) we know that B Y < for some <, so Ö< U À B Y < Á g. And by ii), we know that B  Y = for some =. For that = À if B Y <, then (by ii) = Ÿ <, so = is a lower bound for Ö< U À B Y <. Therefore Ö< U À B Y < has a greatest lower bound, so the definition of 0 makes sense: 0ÐBÑ œ inf Ö< U À B Y. < 304

23 From the definition of 0, we get that for <ß= Uß a) if B cl Y<, then B Y = for all = < so 0ÐBÑ Ÿ < b) if 0ÐBÑ =, then B Y =. We want to prove 0 is continuous at each point + \. Since U is dense in, ÖÒ<ß=Ó À <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 +  cl Y<. Therefore Y œ Y= cl Y< is an open neighborhood of +. If D Y, then D Y = cl Y= ß so 0ÐDÑ Ÿ = ; and D  cl Y<, so D  Y< 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 ÐUrysohn's Lemma Ñ A space \ is normal iff whenever EßF are disjoint closed sets in \, there exists a function 0 GÐ\Ñ with 0lE œ! and 0lF œ "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 E œ ^Ð1Ñ and F œ ^Ð2Ñ, then the conclusion of the theorem is true in any space, without assuming normality: just let 1 ÐBÑ 0ÐBÑ œ 1 ÐBÑ 2 ÐBÑ. Then 0 is continuous, 0lE œ! and 0lF œ ". " " 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 E œ 0 " Ð!Ñ and F œ 0 " 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 Y œ ÖB À 0ÐBÑ and Z œ ÖB À 0ÐBÑ 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 E and F 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 Y œ g for <! and Y œ \ for < ". < < < 305

24 Enumerate the remaining rationals in Ò!ß"Ó as < " ß < ßÞÞÞß< 8 ßÞÞÞ, beginning the list with < " œ " and < œ!þ We begin by defining Y< œ Y" œ \ FÞ Then use normality to define Y< Ð œ Y! Ñ À since " E Y œ \ 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 cly Y cly Y œ \ F < < < < < $ $ " We continue by induction. Suppose 8 $ and that we have already defined open sets Y ßY ßÞÞÞßY in such a way that whenever < 3 < 4 < 5 Ð3ß4ß5 Ÿ 8Ñß then cly Y cly Y Ð Ñ < < < < We need to define Y< 8 " so that Ð Ñ holds for 3ß4ß5 Ÿ 8 "Þ Since < " œ " and < œ!, and < 8 " Ð!ß"Ñß it makes sense to define < 5 œ the largest among < " ß< ßÞÞÞß< 8 that is smaller than < 8 " ß and < œ the smallest among < ß< ßÞÞÞß< that is larger than < Þ 6 " 8 8 " < < < " 8 By the induction hypothesis, we already have cl Y < Y< Þ Then use normality to pick an open set 5 6 Y <8 " so that cly Y cl Y Y. < < < < 5 8 " 8 " 6 The Y< 's defined in this way satisfy the conditions of Lemma 5.1, so the function 0 À \ Ä defined by 0ÐBÑ œ infö< À B Y < is continuous. If B E, then B Y< œ Y! and B  Y < if <!, so 0ÐBÑ œ!þ If B F then B  Y, but B Y œ \ for < ", so 0ÐBÑ œ "Þ ñ " < Once we have the function 0 we can replace it, if we like, by 1 œ Ð! 0Ñ " so that E and F are completely separated by a function 1 G Ð\ÑÞ It is also clear that we can modify 1 further to get an 2 G Ð\Ñ for which 2lE œ + and 2lF œ, 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

25 Lemma 5.4 (Weierstrass Q -Test) Let \ be a topological space. Suppose 0 À \ Ä is 8 continuous for each 8 and that l0 ÐBÑl Ÿ Q for all B \. If Q, then œ" 0ÐBÑ œ 08 ÐBÑ converges (absolutely) for all B and 0 À \ Ä is continuous. 8œ" Proof For each B, l0 ÐBÑl Ÿ Q, so 0 ÐBÑ converges (absolutely) by the Comparison Test œ" 8œ" 8œ" Suppose + \ and %!. Choose R so that % Q Each 0 is continuous, so for 8œR " 8 %. 8 8 œ "ßÞÞÞßR we can pick a neighborhood Y 8 of + such that for B Y8ß l08ðbñ 08Ð+Ñl Then R Y œ Y 8 is a neighborhood of +, and for B Y we get l0ðbñ 0Ð+Ñl + 8œ" R R œ l Ð0 ÐBÑ 0 Ð+ÑÑ Ð0 ÐBÑ 0 Ð+ÑÑl Ÿ l0 ÐBÑ 0 Ð+Ñl l0 ÐBÑ 0 Ð+Ñl œ" 8œR " 8œ" 8œR " R % % % R 8 % 8œ" 8œR " 8œR " Ÿ l0 ÐBÑ 0 Ð+Ñl l0 ÐBÑl l0 Ð+Ñl R Q œ Þ Therefore 0 is continuous at +. ñ % R. 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 is closed, and E and F $ $ are disjoint closed sets in E, so E" and F " are closed in \. By Urysohn's Lemma, there exists a < < < < continuous function 9 À \ Ä Ò $ ß $ Ó such that 9lE" œ $ and 9lF" œ $. < < < If B E", then < Ÿ 2ÐBÑ Ÿ and 9ÐBÑ œ, so l2ðbñ 9ÐBÑl Ÿ l < Ð Ñl $ $ $ < < œ à and similarly if B F ß l2ðbñ ÐBÑl Ÿ Þ B E ÐE F Ñ 2ÐBÑ ÐBÑ $ " 9 If $ " ", then and 9 are < < < both in Ò ß Ó so l2ðbñ 9ÐBÑl Ÿ. 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 1lE œ 0Þ 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 M œ Ð+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 M œ Ð+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 M œ Ð+ß ÑÞ then extend the graph of 0 over M using a horizontal right ray at height 0Ð+Ñà if M œ Ð ß,Ñß then 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

26 Proof ( É ) Suppose E and F are disjoint closed sets in \. E and F 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 FÑ œ 0Þ Then Y œ ÖB À 1ÐBÑ " Z œ ÖB À 1ÐBÑ " and 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 1 GÐ\Ñ such that l0ðbñ 1 ÐBÑl Ä! as 8 Ä for each B E Ðwhere 0 is defined ). The sums 1 ÐBÑ are 3 3 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 1ÐBÑ œ lim 1 ÐBÑ œ 1 ÐBÑÞ The details follow Ä 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 0 À E Ä Ò "ß"Ó. We claim there is a continuous function 1 À \ Ä Ò "ß"Ó with 1lE œ 0Þ Using Lemma 5.5 (with 2 œ 0ß < œ "Ñ we get a function 1" œ 9 À \ Ä Ò 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 2 œ 0 1" ß< œ Ñ, we get a function 1 À \ Ä Ò ß Ó such $ * * % % % that for B Eß l0ðbñ 1 ÐBÑ 1 ÐBÑl Ÿ œ Ð Ñ. So0 Ð1 1 Ñ À E Ä Ò ß Ó " " " * $ " * * Using Lemma 5.5 again (with 2 œ 0 1" 1 ß< œ Ñ, we get a function * % % ) $ 1$ À \ Ä Ò ß Ó such that for B Eß l0ðbñ 1" ÐBÑ 1 ÐBÑ 1$ ÐBÑl Ÿ œ Ð Ñ. ( ( ( $ ) ) So 0 Ð1 1 1 Ñ À E Ä Ò ß ÓÞ " $ ( ( We continue, using induction, to find for each 3 a continuous function 3 " 3 " 8 1 À \ Ä Ò 3 ß 3 Ó such that l0ðbñ 1 ÐBÑl Ÿ ÐÎ$Ñ for B E. 3 $ $ 3 3œ" 8 Since 3 " l1 ÐBÑl Ÿ ß the series 1ÐBÑ œ 0 ÐBÑ converges (absolutely) for every 3 $ 3 3 3œ" 3œ" 3œ" B \, and 1 is continuous by the Weierstrass Q-Test. Since l1ðbñl œ l 1 ÐBÑl 3 " Ÿ l1 ÐBÑl Ÿ 3 œ "ß we have 1 À \ Ä Ò "ß"ÓÞ 3 $ 3œ" 3œ" Finally, for B Eß l0ðbñ 1ÐBÑl œ liml0ðbñ 8 1 ÐBÑl Ÿ limð Ñ œ!, so 1lE œ 0 and the proof for Step I is complete. 8Ä Case 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"Ó with JlE 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 8 3œ" % 3 8Ä $ 3œ" 3 308

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