Chapter VII Separation Axioms

Transcription

1 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 X!, X", X, X, 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: X " Þ It turns out that the X " spaces (also called Tychonoff spaces) are an extremely well-behaved class of spaces with some very nice properties. 2. The Basics Definition 2.1 A topological space \ is called a 1) X! space if, whenever BÁC \, there either exists an open set Y with B Y, CÂY or there exists an open set Z with C Z, B Â Z 2) X" space if, whenever BÁC \, there exists an open set Ywith B YßCÂZ 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 iff whenever B Á C, then ab Á ac 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

2 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 BÂFÐCÑ % or CÂFÐBÑ %. 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. 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 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 œöb \À Bis equivalent to some point Cin S œs. 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 " " and 1 is a homeomorphism. Applied to a X! space, the X! -identification accomplishes nothing. If Ð\ß.Ñis a pseudometric space, the X! -identification is the same as the metric identification discussed in Example VI.5.6 À in that case a œ a iff.ðbßcñ œ!þ B C 5) For 3 œ!ß "ß À if Ð\ß g Ñ is a X 3 space and g ª g is a new topology on \, then Ð\ß g Ñ is also a space. X 3 w X! w 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 ÖSÀ Sis open and 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 X! but not X" : if B C, then S œ ÐBß Ñis 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 : 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

3 a) A subspace of a X3 space is a X3space b) If \œ \ Ág, 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 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 Bbut not C, w then YœY Eis an open set in E containing Bbut not C. Similarly we can find an open set Zin E containing C but not BÞ Therefore Eis a X " space. α α α Suppose \œ E \ is a nonempty X " space. Each \ is homeomorphic to a subspace of \, so each Bα is X " (by part a)). 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 containing B, there exist disjoint open sets Y and Z such that B Y and J Z. 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

4 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. Since cl Yis a neighborhood of B, the closed neighborhoods of Bform 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 œint Oand 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 ÒÐ ß ÑÓand 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 \ 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. 286

5 Caution The 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 B \, let a neighborhood base consist of all sets R of the form U B R œ F ÐBÑ Ða 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, \ is called the slotted plane. Note that F% ÐBÑ U B ( 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 \ 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 Y Þ Using the % in the definition of R, we can choose a point T œ ÐBß!Ñ J with! B % Þ Any basic neighborhood set of Tmust 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. 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 Bß J that need to be separated. Of course it is easy to see that an enlargement of a X topology Ð3 œ!ß "ß Ñ is also X Þ

6 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 T on the B-axis and a point Uabove the B-axis. Therefore is X. > Theorem 2.13 a) A subspace of a regular ( XÑ space is regular ( X ). b) Suppose \œ \ Ág. \ is regular ( X ) iff each \ is regular ( X). α E α α Proof a) Let E \ where \ is regular. Suppose + Eand that Jis a closed set in E that does not w w w contain +. There exists a closed set J in \ such that J E œ JÞ Choose disjoint open sets Y w w w w w w and Z in \ with + Y and Z ªJ ÞThen Y œy Eand Z œz Eare open in E, disjoint, + Y, and J Z. Therefore Eis regular. α α α b) If \œ E\ Ág is regular, then part a) implies that each \ is regular since each \ α is homeomorphic to a subspace of \. Conversely, suppose each \ α is regular and that Y œ Yα" ß ÞÞÞß Yα8 is a basic open set containing B. For each α 3, we can pick an open set Zα3 in \ α such that Bα Z α cl Z α Yα Þ Then B Z œ Zα ß ÞÞÞß Zα cl Z " 8 cl Zα" ß ÞÞÞß cl Zα8 Y Þ ( Why is the last inclusion true? ) Therefore \ is regular. Since X" is hereditary and productive, a) and b) also hold for X 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 Example 2.15 a) Every pseudometric space Ð\ß.Ñ is normal (and therefore every metric space is X Ñ..ÐBßFÑ.ÐBßEÑ.ÐBßFÑ In fact, if E and F are disjoint closed sets, we can define 0ÐBÑ œ. Since the % 288

7 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 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 E and F? b) Let have the right ray topology g œ ÖÐBß Ñà B Ögß. i) Ð g ß 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 Yœgand Zœ\Þ ii) Ð g ß 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. ii 3) Ð g ß Ñ is not a X" space À so X% is a stronger condition than normality. " But when we combine normal X" into X%, we have a property that fits perfectly into the separation hierarchy. Theorem 2.16 X ÊX ÐÊX ÊX ÊXÑ % "! Proof Suppose \ 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. 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 Y. Clearly, Y is a base for the closed sets in \ iff U œösàsœ\ JßJ Y is a base for the open sets in \. For a polynomial T in 8real variables, define the zero set of T as ^ÐT Ñ œ ÖÐB" ß B ß ÞÞÞß B8 Ñ 8 À T ÐB" ß B ß ÞÞÞß B8Ñ 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 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 ^ÐTÑ are rather special geometric objects those surfaces in 8 which can be described by polynomial equations T ÐB ß B ß ÞÞÞß B 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 cl Y cl Z œ g. ( Clearly, T ÊX&Î Ê T Þ) a) Prove that a subspace of a X&Î space is a X&Î space. b) Suppose \œ \ α Ág. Prove that \ is X &Î iff each \ α is 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 : W PÀ U : œ ÖF% Ð:Ñ W À %! if : P: U : œ ÖF% Ð:Ñ ÐW PÑ Ö: À %! ( You may assume that these 's satisfy the axioms for a neighborhood base. ) Prove that S is T but not T. X &Î U : E4. If E \, we can define a topology on \ by g œös \À SªE Ög Decide whether or not Ð\ß g Ñ is normal. 290

9 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 regularà and that if \ is X, the ] is also X Þ E6. a) Suppose \ is finite. Prove that Ð\ßg Ñ is regular iff there is a partition Uof \ that is a base for the topology. For parts b)-e) assume that \ is a X -space (not necessarily finite). b) Suppose E is closed in \Þ Prove that i) Prove that Eœ ÖSÀSis open and E S Þ ii) Define BµC iff BœCor 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 Y 8 such that each Y F Á g and then cl Y cl Y œ g if 8 Á 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 \ is regular then ck O is compact. 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, 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 0 can be defined on a metric space Ð\ß.Ñusing formulas like 0ÐBÑ œ.ðbß+ñ 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 X space L 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 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 continuous and bounded. Lemma 3.1 Suppose 0ß1 GÐ\Ñ. Define real-valued functions 0 1 and 0 1 by Then 0 1and 0 1are in GÐ\ÑÞ ( 0 1ÑÐBÑ œ maxö0ðbñß 1ÐBÑ Ð0 1ÑÐBÑ œ minö0ðbñß 1ÐBÑ 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 Ð0 1ÑÐBÑ œ 0ÐBÑ 1ÐBÑ l0ðbñ 1ÐBÑl 0ÐBÑ 1ÐBÑ l0ðbñ 1ÐBÑl ñ Definition 3.2 A space \ is called completely regular if whenever J is a closed set and B  Jß 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

11 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!ß1lJ œ" and 1À\ÄÒ!ß"Ó, we could compose 1with a linear homeomorphism 9 À Ò!ß "Ó Ä Ò+ß,Ó to get a continuous function 2 œ 9 1 À \ Ä Ò+ß,Ó where 2ÐBÑ œ + and 2lJ œ, (or vice versa). (It must be true that either 9Ð!Ñ œ + and 9Ð"Ñ œ,, or vice versa: why?) 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, B  Jß 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 B and J real-valued function. can be separated by a bounded continuous 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. Example 3.3 Suppose +ÂJ Ð\ß.Ñwhere Ð\ß.Ñis a pseudometric space. Then 0ÐBÑœ.ÐBßJÑ.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 is a closed set in \ not containing B. If \ is X ", we can choose 0 GÐ\Ñwith " " " " 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 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 (DJ. 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 realvalued 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

12 Note: We have not shown that X% Ê X ". 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 ( X " Ñspace is completely regular ( X " Ñ. b) Suppose \œ α E \ α Ág. \ is completely regular (T " Ñ iff each \ αis completely regular ( X Ñ. " Proof Suppose +ÂJ E \, where \ is completely regular and Jis a closed set in E. Pick a closed set Oin \ such that O EœJand an 0 GÐ\Ñsuch that 0Ð+Ñœ! and 0lOœ". Then 1œ0lE GÐEÑß1Ð+Ñœ! and 1lJ œ"þ Therefore Eis completely regular. If gá\œ α E\ α is completely regular, then each \ α is homeomorphic to a subspace of \ so each \ α is completely regular. Conversely, suppose each \ α is completely regular and that J is a closed set in \ not containing +. There is a basic open set Y such that + Y œ Y ß ÞÞÞY \ J α α " 8 For each 3 œ "ß ÞÞÞß 8 we can pick a continuous function 0α3 À \ α3 Ä Ò!ß "Ó with 0α3Ð+ α3ñ œ! and 0 Ð\ Y Ñ œ "Þ Define 0 À \ Ä Ò!ß "Ó by α α α ÐBÑ œ max ÖÐ0 1 ÑÐBÑ À 3 œ "ß ÞÞÞß 8 œ max Ö0 ÐB Ñ À 3 œ "ß ÞÞÞß 8 α α α α Then 0 is continuous and 0Ð+Ñ œ max Ö0α Ð+ α Ñ À 3 œ "ß ÞÞÞß 8 œ!þ If B J, then for some 3ß 3 3 Bα3  Yα3 and 0α3ÐBα3Ñ 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 7 Corollary 3.8 Ò!ß "Ó and all its subspaces are X. " Since a Tychonoff space \ is defined using GÐ\Ñ, 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: 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 8. Each S8 is open. Therefore a zero set is always a closed K -set. Taking complements shows that coz Ð0Ñ is always an open J -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)

13 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 might not be a zero set if fact, a closed set in \ might not even be a K set. Suppose \ is uncountable and : \. Define a topology on \ by letting U B 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 E and F are disjoint closed sets in \, then one of them (say EÑsatisfies E \ Ö: and therefore Eis clopen. We then have open sets Y œe and Z œ \ E ª F that work in the definition of normality. We do not know, at this point that X Ê X ", but we can argue directly that this \ 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 Ö: 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. 8œ" Suppose Ö: œ S8 where S8 is open. For each 8, : F 8 S8 for some F8 U : Þ Therefore \ Ö: œ \ 8œ" S8 œ 8œ" Ð\ S8Ñ Ð\ F8 Ñis a countable set which is impossible since \ is uncountable. 8œ" Even when J is both a closed set and a K set, J 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\Ñ À ^œöbà0ðbñœ< 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 ^Ð1Ñin any of the forms listed above by choosing an appropriate function 0 GÐ\ÑÀ ^Ð1ÑœÖBÀ0ÐBÑœ< where0ðbñœ< 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 Ÿ r where 0ÐBÑ œ < l1ðbñl Taking complements, we see that if 0 GÐ\Ñthen 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 Y and Z with B Y and J ZÞ 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 B and J 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 g and L! 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\ß Ñ, GÐ\Ñand G Ð\Ñ induce the same weak topology on \Þ A base for this weak topology is the collection of all cozero sets in \. g 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 coz Ð0Ñ 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 \ and GÐ\Ñ in completely regular spaces. For any space Ð\ßg Ñ, the functions in GÐ\Ñ are continuous with respect to g (by definition of GÐ\Ñ) ß but is g the smallest topology making this collection of functions continuous? In other words, is g 296

15 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\ß g 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 B S where S is open Þ Let J œ \ SÞ Then we can choose 0 GÐ\Ñ " 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) Ê d) Suppose J 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) Ê a) Suppose J is a closed set not containing B. For some 0 GÐ\Ñwe have 0ÐBÑ Â cl 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 minöl0ðbñlß % 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 g Y be the weak topology on \ generated by any family of \ functions Y. Then Ð\ßg Ñis completely regular. Y Proof Give \ the topology the weak topology gy generated by Y. Now that \ has a topology, the collection GÐ\Ñ makes sense. Let X G be the weak topology on \ generated by GÐ\ÑÞ The topology gy does make all the functions in GÐ\Ñ continuous, so gg gy Þ 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 ggþ Therefore g œ g. By Theorem 3.12, Ð\ßg Ñis completely regular. ñ Y G Y 297

16 Example 3.14 by Y 1) If Y œö0 À0 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 ÖM0À 0 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 the 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 possible. In 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 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 \ 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 0YßZ ÐBÑœ" 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 Uis distinguished. YßZ " YßZ 298

17 We note that if B Z U, then there must exist Y U such that B Y and ÐYßZÑis distinguishedþ 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 ÐYßZÑwith B Y Z \ JÞ Then 0YßZÐBÑœ< and 0YßZlJ œ ", so 0YßZÐBÑ Â cl 0YßZ ÒJ Ó œ Ö" Þ lyl By Corollary VI.4.11, /À\ÄÒ!ß"Ó is an embedding. Since 7 is infinite, lylÿl U U 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 metrizable. top i! i! Proof By the theorem, \ Ò!ß "Ó. Since Ò!ß "Ó is metrizable, so is \. ñ \ is 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, topologically, the separable metrizable spaces 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 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\ÑÀif 0Ð:Ñœ! then there is a neighborhood R of : such that 0±R! E10. Let 3À Ä be the identity map and let Ð3ÑœÖ0 GÐ ÑÀ0œ13for some 1 GÐ Ñ}. For those who know a bit of algebra: GÐ Ñ Ðor, 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. 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 0 on \ yet neither 0 nor 1 is identically 0 on \. ÐSo the ring GÐ\Ñ has zero divisors. Ñ d) Prove that there are exactly two functions 0 GÐ Ñfor which 0 œ0. (Here, 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 which equals its own square is called an idempotent in GÐ\Ñ). Part d) shows that GÐ 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\Ñis isomorphic to GÐ]Ñalso (why?). The question when does isomorphism imply homeomorphism is more difficult. Another important area of study is how 300

19 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 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 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 0 is a multiple of 1 in GÐ\Ñ. Also, 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. Suppose Jß E \, where J is closed and E is countable. Prove that if J Eœg, then Eis disjoint from some zero set containing J. E14. A space \ is called pseudocompact ( Definition IV.8.7 ) if every continuous 0 À \ Ä is bounded, that is, if GÐ\Ñ œ G Ð\ÑÞ Consider the following condition on a topological space \: (*) If Z " ª Z ª... ª Z 8 ª ÞÞÞ is a decreasing sequence of nonempty open sets, then cl Z Á g. 8œ" 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, one that makes no explicit mention of.. 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 nice classical (and nontrivial) theorems we can prove. One of these will have X Ê X as a corollary. % " To begin, the following technical variation on the definition of normality is very useful. Lemma 4.1 A space \ is normal iff whenever Sis open, E closed, and E S, then there exists an open set Ywith E Y cl Y SÞ Proof Suppose \ is normal and that Sis an open set containing the closed set E. Then E 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 \ satisfies the stated condition and that EßF are disjoint closed sets. Then E Sœ\ Fß so there is an open set Ywith 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 sets w w in J. Then EßF are also closed in \ so we can find disjoint open sets Y and Z in \ containing E w w and F respectively. Then Y œ Y 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 Fare " " w disjoint closed sets in ], then 0 ÒEÓ and 0 ÒFÓ are disjoint closed sets in \. Pick Y w and Z " w " w disjoint open sets in \ with 0 ÒEÓ Y and 0 ÒFÓ Z Þ w Then Y œ ] 0Ò\ Y Ó and w Z œ ] 0Ò\ Z Ó are open sets in ]. w w w If C Y, then CÂ0Ò\ Y Ó. Since 0 is onto, Cœ0ÐBÑfor some B Y \ Z Þ Therefore w C 0Ò\ Z Óso CÂZ. Hence Y Z œgþ " " w " w If C E, then 0 ÒÖC Ó 0 ÒEÓ Y, so 0 ÒÖC Ó Ð\ Y Ñ œ g. Therefore w w CÂ0Ò\ Y Óso C ] 0Ò\ Y ÓœY. Therefore E Yand, 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 Lindelof space \ is normal (and therefore every Lindelof 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 Lindelof property is hereditary on closed subsets, a countable number of the YB 's cover E: relabel these as Y" ß Y ß ÞÞÞß Y8ß ÞÞÞ. For each 8, we have cl Y8 F œ gþ Similarly, choose a sequence of open sets Z" ß Z ß ÞÞÞß Z8ß ÞÞÞ covering F such that cl Z E œ gfor 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" cl Y" Y œ Y Ðcl Z" cl Z Ñ Z œ Z Ðcl Y" cl Y Ñ ã ã Y8 œ Y8 Ðcl Z" cl Z ÞÞÞ cl Z8Ñ Z8 œ Z8 Ðcl Y" cl Y ÞÞÞ cl Y8Ñ ã ã Let Yœ Y and Zœ Z. 8œ" 8 8œ" 8 If B E, then BÂcl Z for all 8. But B Y for some 5, so B Y YÞ Therefore E Yand, similarly, F ZÞ To complete the proof, we show that Y Z œg. Suppose B Y. 303

22 Then B Y5 for some 5ß so B  cl Z" cl Z ÞÞÞ cl Z5ß so B  Z" Z ÞÞÞ Z5 so B  Z" Z ÞÞÞ Z5 so BÂZ for any 8Ÿ5Þ " 5 8 Since B Y, then B Y Þ So, if 8 5, then B  Z œ Z Ðcl 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 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 Lindelof, 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 ". Proving this is hard 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, for each < U, in such a way that: < i) \œ Y and < U < < UY< œg ii) if <ß = U and < =, then cl Y Y. < = For B \, define 0ÐBÑœinf Ö< UÀB Y <. Then 0À\Ä is continuous. We will only use the Lemma once, with Uœ in the proof. Uœ Þ So if you like, there is no harm in assuming that 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 0ÐBÑœinf Ö< UÀB Y makes sense. From the definition of 0, we get that for <ß= Uß a) if cl, then for all so B Y B Y = < 0ÐBÑŸ< < = < 304

23 b) if 0ÐBÑ =, then B Y =. We want to prove 0 is continuous at each point + \. Since U 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 +  cl Y<. Therefore Y œ Y= cl Y< is an open neighborhood of +. If D Y, then D Y = cl Yß = so 0ÐDÑŸ= ; and DÂcl Y<, so DÂY< and 0ÐDÑ <Þ Therefore 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 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.) " " Notice: the theorem says that E 0 Ð!Ñ and F 0 Ð"Ñß but equality might not be true. If fact, if " " we had E œ 0 Ð!Ñ and F œ 0 Ð"Ñ, 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 If Eœ^Ð1Ñand Fœ^Ð2Ñare disjoint zero sets, then the function 0ÐBÑœ completely separates E and F. 1ÐBÑ 1ÐBÑ 2ÐBÑ 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 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 < ( < Ñin such a way that the Lemma 5.1 applies. To start, let Y œ g for <! and Y œ \ for < ". < < Enumerate the remaining rationals in Ò!ß "Ó as < " ß < ß ÞÞÞß < 8 ß ÞÞÞ, 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 < < < " 305

24 Then! œ < < < " œ ", and we use normality to pick an open set Y< so that E Y cl Y Y cl Y Y œ \ F < < < < < " We continue by induction. Suppose 8 and that we have already defined open sets Y ß Y ß ÞÞÞß Y in such a way that whenever < 3 < 4 < 5 Ð3ß 4ß 5 Ÿ 8Ñß then cl Y Y cl Y Y Ð Ñ < < < < 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 œ the largest among < " ß < ß ÞÞÞß < 8that is smaller than < 8 " ß and < œ the smallest among < ß < ß ÞÞÞß < 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 cl Y 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 0ÐBÑœinf Ö< ÀB Y < is continuous. If B E, then B Y< œy! and BÂY < if <!, so 0ÐBÑœ!Þ If B Fthen BÂY, but B Y œ\ for < ", so 0ÐBÑœ"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 1further to get an 2 G Ð\Ñfor 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). Lemma 5.4 (Weierstrass Q -Test) Let \ be a topological space. Suppose 0 À \ Ä is 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" 8 306

25 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 "ß ÞÞÞß R 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" and F" are disjoint closed sets in E, and since E is closed, E" and F " are closed in \. By Urysohn's Lemma, there exists < < < < a continuous function 9 À\ÄÒ ß Ósuch 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 9ÐBÑl Ÿ Þ If B E ÐE" F" Ñ, then 2ÐBÑ and 9ÐBÑ 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, then it is quite easy to prove that each 0 GÐEÑ can be extended to a function 1defined on all of. In that case, the open set E can be written as a countable union of disjoint open intervals Mß where each M œð+ß,ñor Ð ß,Ñor Ð+ß Ñ( see Theorem II.3.4). Any endpoints of M 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+ß 0Ð+ÑÑ and Ð,ß 0Ð,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. 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\Ñsuch that 1lÐE FÑœ0Þ Then YœÖBÀ1ÐBÑ " ZœÖBÀ1ÐBÑ " 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\Ñsuch that 3 307

26 8 8 l0ðbñ 1 ÐBÑl Ä! as 8 Ä for each B EÐwhere 0 is defined ). The sums 1 ÐBÑ are defined 3 3 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 œ lim 1 ÐBÑ œ 1 ÐBÑÞ The details follow. We 3 3 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"Ñwe get a function 1" œ9 À\ÄÒ ß Ósuch that for B E, l0ðbñ 1 ÐBÑl Ÿ. Therefore 0 1 À E Ä Ò ß ÓÞ " " Using Lemma 5.5 again (with 2œ0 1ß<œ " Ñ, we get a function 1 À\ÄÒ * ß* Ósuch % % % 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 2œ0 1" 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 0 Ð1 1 1 ÑÀEÄÒ 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 ÐÎÑ for B E œ" 8 3 " Since l1 ÐBÑl Ÿ ß the series 1ÐBÑ œ 0 ÐBÑ converges (absolutely) for every œ" 3œ" 3œ" B \, and 1 is continuous by the Weierstrass Q-Test. Since l1ðbñl œ l 1 ÐBÑl 3 " Ÿ l1ðbñlÿ 3 œ"ßwe have 1À\ÄÒ "ß"ÓÞ 3œ" 3 8Ï3œ" Finally, for B Eß l0ðbñ 1ÐBÑl œ lim l0ðbñ 8 1 ÐBÑl Ÿ lim Ð Ñ œ!, so 1lE œ 0 and the proof for Step I is complete. 8Ä 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 "Ó with 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 ". Eand 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 8 3œ" 3 8Ä 3œ" 3 308

27 Let 2À ÄÐ "ß"Ñbe a homeomorphism. Then 2 0ÀEÄÐ "ß"Ñand, 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 Ewe 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 W is normal ( Example 4.4) but the plane W W is not normal. To see this, let Hœ ß a countable dense set in W W. Every continuous real-valued function on W Wis completely determined by its values on H. ( See Theorem II 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 0È0lHis one-to-one, so lgðw WÑlŸlGÐHÑŸl H lœ- i! œ-þ EœÖÐBßCÑ W W ÀB Cœ" 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 Wmust 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

28 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 spaceñ 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 W is X%, so W is X " and therefore the Sorgenfrey plane W W is also X ". So W W 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\ÄFsuch 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 \, we can relax the hypotheses on F. 310

29 Suppose E is closed in \ and 0 À E Ä F is continuous. If \ is normal and F œ (Tietze's Theorem) \ is normal and F œ 8 \ is collectionwise normal** and Fis a separable Banach space* \ is paracompact** and Fis a Banach space* then 0 has a continuous extension 1 À \ Ä FÞ 8 The statement that can replace in Tietze's Theorem is easy to prove: If \ is normal and 0 À E Ä 8 is continuous, write 0ÐBÑ œ Ð0" ÐBÑß 0 ÐBÑß ÞÞÞß 08ÐBÑÑ where each 03 À E Ä. By Tietze's Theorem, there exists for each 3 a continuous extension 13 À \ Ä with 13lE œ 03. If we let 1ÐBÑ œ Ð1" ÐBÑß ÞÞÞß 18ÐBÑÑ, then 1À\Ä 8 and 1lEœ0. In other words, we separately extend the coordinate functions in order to extend 0. And in this example, 8 could even be an infinite cardinal. * A normed linear space is a vector space Z with a norm l@l ( œ 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@ " A Banach space is a normed linear space which is complete in this metric.. For example, 8, with its usual norm lðb" ß B ß ÞÞÞß B8Ñl œ B B ÞÞÞ B is a separable Banach space. " 8 ** 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 compact X Ê paracompact Ê collectionwise normal Ê normal Therefore, in the theorems cited above, a continuous map 0 defined on a closed subset of a metric space (or, compact X space) and valued in a Banach space F 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 0 is a continuous linear functional defined on a subspace Q 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) is usually not formulated in terms of extension theorems, but extensions are Homotopy really at the heart of the idea. 311

30 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 0and 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 JÐ!ß>Ñ œ B J ÐBß "Ñ œ 1ÐBÑ JÐ"ß>Ñ œ B! " Thus J agrees with 0 on the bottom edge of F and with 1 on the top edge. J is constant ÐœBÑ! on the left edge of Fand 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"Ó Ä\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 0 œ 0 into 0 œ 1Þ >! " The continuous extension L(if it exists) is called a homotopy between 0 and 1 with fixed endpoints, and we say that the paths 0 and 1 are homotopic with fixed endpoints. In the space \ below, it seems intuitively clear that 0 can be continuously deformed (with endpoints held fixed) into 1 in other words, that Lexists. 312