Chapter IV Completeness and Compactness

1. Introduction

In Chapter III we introduced topological spaces a generalization of pseudometric spaces. This allowed us to see how certain ideas continuity, for example can be extended into a situation where there is no “distance” between points. Continuity does not really depend on the pseudometric . but only on the topology.

Looking at topological spaces also highlighted the particular properties of pseudometric spaces that were really important for certain purposes. For example, it turned out that first countability is the crucial ingredient for proving that sequences are sufficient to describe a topology, and that Hausdorff property, not the metric . , is what matters to prove that limits of sequences are unique.

We also looked at some properties that are equivalent in pseudometric spaces for example, second countability and separability but are not equivalent in topological spaces.

Most of the earlier definitions and theorems are just basic “tools” for our work. Now we look at some deeper properties of pseudometric spaces and some significant theorems related to them.

2. Complete Pseudometric Spaces

Definition 2.1 A sequence ÐB8 Ñ in a pseudometric space Ð\ß .Ñ is called a Cauchy sequence if a!b% MNß − such that if 7ß8R , then .ÐBßBÑ78 % .

Informally, a sequence ÐB8 Ñ is Cauchy if its terms “get closer and closer to each other.” It should be intuitively clear that this happens if the sequence converges, and the next theorem confirms this.

Theorem 2.2 If ÐB88 Ñ Ä B in Ð\ß .Ñ , then ÐB Ñ is Cauchy.

% Proof Let % ! . Because ÐB88 Ñ Ä B , there is an R − such that .ÐB ß BÑ # for 8  R . %% So if 7ß 8  R , we get .ÐB78 ß B Ñ Ÿ .ÐB 7 ß BÑ  .ÐBß B 8 Ñ ##  œ% . Therefore ÐB 8 Ñ is Cauchy. ñ

A Cauchy sequence does not always converge. For example, look at the space Ðß.Ñ where . is the usual metric. Consider a sequence in ‘ that converges to #à in for example, we could used Ð;888 Ñ œ Ð"ß "Þ%ß "Þ%"ß "Þ%"%ß ÞÞÞ Ñ . Since Ð; Ñ is convergent in ‘ ß Ð; Ñ is a Cauchy sequence in  (“Cauchy” depends only on .; and the numbers 8 , not whether we are thinking of these ;8 's as elements of ‘ or ..)

But Ð;8 Ñ has no limit in  . (Why? To say that “#Â ” is part of the answer. )

154 Definition 2.3 A pseudometric space Ð\ß .Ñ is called complete if every Cauchy sequence in Ð\ß .Ñ has a limit in \ .

Example 2.4 Throughout this example, . denotes the usual metric on subsets of ‘ .

1) Ðß.Ñ‘ is complete. (For the moment, we take this as a simple fact from analysis; however we will prove it soon.) But Ð ß .Ñ and Ð ß .Ñ are not complete À completeness is not hereditary!

2) Let ÐB8 Ñ be a Cauchy sequence in Ð ß .Ñ . Then there is some R such that lB78  B l 1 for 7ß 8  R . But the B 8 's are integers, so this means that B 7 œ B 8 for 7ß 8  R. So every Cauchy sequence is eventually constant and therefore converges. Ð ß .Ñ is complete.

"" 3) Consider  with the metric ."8 Ð7ß 8Ñ œ l78  l . Then the sequence ÐB Ñ œ Ð8Ñ is "" " Cauchy. If 5− , then .ÐBß5Ñœl"885  lÄ 5 Á! as 8Ä∞ , so ÐBÑ 8 does not converge to 5Ðß.Ñ. The space  " is not complete. However Ðß.Ñ " has the discrete topology, so the identity function 0Ð8Ñœ8 is a homeomorphism between Ðß.Ñ" and the complete space ( ß.ÑÞ So completeness is not a topological property. A homeomorphism takes convergent sequences to convergent sequences and non convergent sequences to nonconvergent sequences. For example, Ð8Ñ does not converge in Ð ß ." Ñ and Ð0Ð8ÑÑ does not converge in Ð ß .Ñ . These two homeomorphic metric spaces have exactly the same convergent sequences but they do not have the same Cauchy sequences. Changing a metric .. to a topologically equivalent metric w does not change the open sets and does not change which sequences converge. But the change may create or destroy Cauchy sequences since this property depends on distance measurements. Another similar example: we know that any open interval Ð+ß ,Ñ is homeomorphic to ‘ . But Ð+ß ,Ñ is not complete and ‘ is complete (where both space have the metric . ).

4) In light of the comments in 3) we can ask: if Ð\ß .Ñ is not complete, might there be a different metric .µ.ww for which Ð\ß.Ñ is complete? To be specific: could we find a metric .w on  so . µ.ww but where Ð ß.Ñ is complete? We could also ask this question about Ð  ß.ÑÞ Later in this chapter we will see that the answer is “no” in one case but “yes” in the other.

" 5) Ðß.Ñ" and ÐÖÀ8−×ß.Ñ8 are both countable discrete spaces, so they are " homeomorphic. In fact, the function 0Ð8Ñ œ8 is an isometry between these spaces: ." Ð7ß8Ñ "" œ|78  l œ .Ð0Ð7Ñß 0Ð8ÑÑÞ An isometry is a homeomorphism, but since it preserves distances, an isometry also takes Cauchy sequences to Cauchy sequences and non-Cauchy sequences to non-Cauchy sequences. Therefore if there is an isometry between two pseudometric spaces, then one space is complete iff the other space is complete. For example, the sequence ÐB8" Ñ œ 8 is Cauchy is Ð ß . Ñ and the isometry 0 carries "" ÐB88 Ñ to the Cauchy sequence Ð0ÐB ÑÑ œ Ð88 Ñ in ÐÖ À 8 − ×ß .Ñ . It is clear that Ð8Ñ has no limit " in the domain and ÐÑ8 has no limit in the range. " Ðß.Ñ" and ÐÖÀ8−×ß.Ñ8 “look exactly alike”  not just topologically but as metric spaces. We can think of 0 as simply renaming the points in a distance-preserving way.

Theorem 2.5 If BÐBÑÐ\ß.ÑÐBÑÄB is a cluster point of a Cauchy sequence 88 in , then .

155 % Proof Let %  ! . Pick R so that .ÐB78 ß B Ñ # when 7ß 8  RÞ Since ÐB 8 Ñ clusters at B , we can pick a OR so that B −FÐBÑ% . Then for this O and for 8Rß O # %% .ÐB88OO ß BÑ Ÿ .ÐB ß B Ñ  .ÐB ß BÑ ##  œ % so ÐB8 Ñ Ä B. ñ

Corollary 2.6 A Cauchy sequence ÐB8 Ñ in Ð\ß .Ñ converges iff it has a cluster point.

Corollary 2.7 In a metric space Ð\ß .Ñ , a Cauchy sequence can have at most one cluster point.

Theorem 2.8 A Cauchy sequence ÐB8"#8 Ñ in Ð\ß .Ñ is bounded  that is, the set ÖB ß B ß ÞÞÞß B ß ÞÞÞ× has finite diameter.

Proof Pick R so that .ÐB78 , B Ñ  " when 7ß 8 R . Then B 8 − F "R ÐB Ñ for all 8 RÞ Let < œ max Ö"ß .ÐB"R ß B Ñß .ÐB #R ß B Ñß ÞÞÞß .ÐB R"R ß B Ñ×Þ Therefore, for all 7ß 8 we have .ÐB78 ß B Ñ Ÿ .ÐB 8R ß B Ñ  .ÐB R7 ß B Ñ Ÿ #<, so diam ÖB "# ß B ß ÞÞÞ ß B 8 ß ÞÞÞ× Ÿ #<Þ ñ

Now we will prove that ‘ (with its usual metric . ) is complete. The proof depends on the completeness property (also known as the “least upper bound property”) of ‘ . We start with two lemmas which might be familiar from analysis. A sequence ÐB8 Ñ is called weakly increasing if BŸB88"for all 8 , and weakly decreasing if B B 88" for all 8 . A monotone sequence is one that is either weakly increasing or weakly decreasing.

Lemma 2.9 If ÐB88 Ñ is a bounded monotone sequence in ‘ , then ÐB Ñ converges.

Proof Suppose ÐB888 Ñ is weakly increasing. Since ÐB Ñ is bounded, ÖB À 8 − × has a least upper bound in ‘ . Let Bœ sup ÖB88 À8− × . We claim ÐBÑÄBÞ Let %%! . Since B Bß we know that B % is not an upper bound for ÖB8 À8−  × . Therefore B% BR8 ŸB for some RÞ Since ÐBÑ is weakly increasing and B is an upper bound for ÖB8R88 À8−% ×ß it follows that B B ŸB ŸB for all 8 R . Therefore lB BlŸ % for all8 Rß so ÐBÑÄB8 . If ÐB88 Ñ is weakly decreasing, then Ð  B Ñ is a weakly increasing sequence so ÐBÑÄ+88 for some + ; then ÐBÑÄ+Þñ

Lemma 2.10 Every sequence ÐB8 Ñ in ‘ has a monotone subsequence.

Proof Call BÐBÑB B8 55858 a peak point of if for all . We consider two cases.

i) If ÐB8 Ñ has only finitely many peak points, then there is a last peak point B88! in ÐB Ñ .

Then B8!"!8 is not a peak point for 88Þ Pick an 8 8 . Since B " is not a peak point, there is an 8#" 8 with B 8"# B 8. Since B 8 # is not a peak point, there is an 8 $# 8 with B 8#$ B 8.

We continue in this way to pick an increasing subsequence ÐB885 Ñ of ÐB Ñ . ii) If ÐB8 Ñ has infinitely many peak points, then list the peak points as a subsequence

B88ß"# ß B ÞÞÞ (where 8 "  8 #  ÞÞÞÑ . Since each of these points is a peak point, we have B8855" B for each 5ÐBÑ , so 885 is a weakly decreasing subsequence of ÐBÑñ .

156 Theorem 2.11 Бß.Ñ is complete.

Proof Let ÐB8 Ñ be a Cauchy sequence in Б ß .Ñ . By Theorem 2.8, ÐB8 Ñ is bounded and by the

Lemma 2.10, ÐBÑ888 has a monotone subsequence ÐBÑÞ55 Of course ÐBÑ is also bounded. By

Lemma 2.9, ÐB85 Ñ converges to some point B −‘ . Therefore B is a cluster point of the Cauchy sequence ÐB88 Ñ , so ÐB Ñ Ä BÞ ñ

Corollary 2.12 Б8ß.Ñ is complete.

# Proof Exercise (Hint: For 8 œ # , the sequence ÐÐB88 ß C ÑÑ in ‘ is Cauchy iff the sequences ÐB88 Ñ and ÐC Ñ both are Cauchy sequences in ‘ . If ÐB 8 Ñ Ä B and ÐC 8 Ñ Ä C , then # ÐÐB88 ß C Ñ Ä ÐBß CÑ in ‘ .)

3. Subspaces of Complete Spaces

Theorem 3.1 Let .E©Ð\ß.Ñ be a metric and suppose .

1) If Ð\ß .Ñ is complete and E is closed, then ÐEß .Ñ is complete. 2) If ÐEß .Ñ is complete, then E is closed in Ð\ß .Ñ .

Proof 1) Let Ð+88 Ñ be a Cauchy sequence in E . Then Ð+ Ñ is also Cauchy in the complete space Ð\ß .Ñ, so Ð+8 Ñ Ä B for some B − \Þ But E is closed, so this limit B must be in E , that is, Ð+8 Ñ Ä B − E. Therefore ÐEß .Ñ is complete.

2) If B− cl E , we can pick a sequence in E such that Ð+ÑÄB88 . Since Ð+Ñ converges, it is a Cauchy sequence in EÐEß.Ñ . But is complete, so we know Ð+ÑÄ+8 for some +−E©\. Since limits of sequences are unique in metric spaces, we conclude that Bœ+ . Therefore if B− cl E , then B−E . So E is closed. ñ

Part 1) of the proof is valid when . is a pseudometric, but the proof of part 2) requires that .Ð\ß.Ñ be a metric. Can you provide an example of a pseudometric space for which the part 2) of the theorem is false?

The definition of completeness is stated in terms of the existence of limits for Cauchy sequences. The following theorem characterizes complete spaces differently: in terms of the existence of points in certain intersections. This illustrates an important idea: completeness for Ð\ß .Ñ can be expressed in different ways, but all equivalent characterizations of completeness are statements that guarantee the existence of certain points.

Theorem 3.2 (The Cantor Intersection Theorem) The following are equivalent for a metric space Ð\ß .Ñ

157 1) Ð\ß .Ñ is complete 2) Whenever JªJª"#... ªJª 8 ... is a decreasing sequence of nonempty closed sets with diam , then ∞ for some ÐJ88 Ñ Ä !8œ" J œ ÖB× B − \Þ

Proof 1ÑÊ 2 Ñ For each 5 , pick a point B55 −J . Since diam ÐJÑÄ! 5 , the sequence ÐBÑ 5 is Cauchy soÐB5 Ñ Ä B for some B − \Þ For any 8 , the subsequence ÐB85 Ñ œ ÐB 8" ß B 8# ß ÞÞÞÑ Ä B . Since the J 8 's are decreasing, this subsequence is inside the closed set , so . Therefore ∞ . JB−J88 B−J8œ" 8 If ∞ , then diam for every Since diam( , Bß C −8œ" J88 .ÐBß CÑ Ÿ ÐJ Ñ 8Þ J 8 Ñ Ä ! this means that Therefore , and so ∞ . .ÐBßCÑœ!Þ BœC8œ" J8 œÖB×

2ÑÊ 1 Ñ Suppose 1) is false and let ÐBÑ88be a nonconvergent Cauchy sequence ÐBÑ . Without loss of generality, we may assume that all the B8 's are distinct. (If any value BœC5 were repeated infinitely often, then CÐBÑ would be a cluster point of 8 and we would have ÐB8 Ñ Ä C . So, since each term occurs only finitely often, it is possible to pick a subsequence from ÐB8 Ñ whose terms are all distinct, and this subsequence is also a nonconvergent Cauchy sequence.) We will construct sets J8 which violate condition 2). >2 Let J888"85 œ ÖB ß B ß ÞÞÞß B ß ÞÞÞ× œ the “ 8 tail” of the sequence ÐB 8 Ñ . For each 8 , ∞ % JÁgßJªJ888"8and 8œ" JœgÞ Also, for % ! , we can pick R so that .ÐBßBÑ 78# if  % 7ß 8 R. This means that for 8  R , diam ÐJ8R Ñ Ÿ diam ÐJ Ñ Ÿ#  % and therefore diamÐJ8 Ñ Ä ! . We claim that the J8 's are closed which contradicts 2) and completes the proof.

Suppose that some JB−JJ888! is not closed. Then there is a point cl!! and we can find a sequence of distinct terms in ÖB88"8#!! ß B ß B ! ß ÞÞÞ× œ J 8 ! that converges to B . This sequence is a subsequence ÐBÑÄB8885 of the original sequence ÐBÑ , so B is a cluster point of ÐBÑÞ But this is impossibleÐBÑñ a nonconvergent Cauchy sequence 8 cannot have a cluster point.

How would Theorem 3.2 be different if . were only a pseudometric? How would the proof change? Can you locate where the fact that . is a metric is used in each half of the proof ?

Example 3.3

1) In the complete space Ðß.Ñ‘ consider:

a) JœÒ8ß∞Ñ8 " b) JœÐ!ßÑ8 8 " c) JœÒ8ß8Ó8 8

In each case, ∞ . Why are these examples not inconsistent with the Theorem 3.2? 8œ"Jœg8

2) Let "" . The 's satisfy all the JœÖB−88 ÀÒ #ŸBŸ88 #Ó× J hypotheses in Theorem 3.2), but ∞ . Why is this not inconsistent with Theorem 3.2? 8œ"Jœg8

Example 3.4 Here is a proof, using the Cantor Intersection Theorem, that the closed interval Ò!ß "Ó is uncountable.

158 If Ò!ß "Ó were countable, we could list its elements as a sequence ÐB8 ÑÞ Pick a subinterval " Ò+"" ß , Ó of Ò!ß "Ó with length less than # and excluding B" . Then pick an interval " Ò+## ß , Ó © Ò+ "" ß , Ó of length less than % and excluding B # . Continuing inductively, choose " an interval Ò+8" ß , 8" Ó © Ò+ 8 ß , 8 Ó of length less than #8" and excluding B 8" . Then Ò+"" ß , Ó ª Ò+ ## ß , Ó ª ÞÞÞ ª Ò+ 88 ß , Ó ª ÞÞÞ and these sets clearly satisfy the conditions in part 2) of the Cantor Intersection Theorem. But ∞ . This is impossible since 8œ"Ò+88 ß , Ó œ g Ò!ß "Ó is complete.

The following theorem has some interesting consequences. In addition, the proof is a very nice application of Cantor's Intersection Theorem. The completeness is used to prove the existence of “very many” points in \ .

Theorem 3.6 Suppose Ð\ß .Ñ is a nonempty complete metric space with no isolated points. Then l\l -Þ

To prove Theorem 3.5, we will use the following lemma.

Lemma 3.5 Suppose Ð\ß.Ñ is a metric space that E©\Þ If +ß,− int E and +Á,ß then we can find disjoint closed balls JJ+, and (centered at +,J©EJ©EÞ and ) with + int and , int

Proof Let %%%œ.Ð+ß,Ñ . Since +ß,− int Eß we can choose positive radii +, and so that % FÐ+Ñ©EFÐ,Ñ©E%%+ int and , int . We may choose both %%+, and less than # Þ Let % w w œÖß×Þmin%%+, Then we can use the closed balls JœÖBÀ.ÐBß+ÑŸ× + % and w JœÖBÀ.ÐBß,ÑŸ, % ×. ñ

Proof of Theorem 3.6 The idea of the proof is to inductively construct - different descending sequences of closed sets, each of which satisfies condition 2) in the Cantor Intersection Theorem, and to do this in a way that the intersection of each sequence gives a different point in \ . It then follows that l\l -Þ The idea is simple but the notation gets a bit complicated. First we will give the idea of how the construction is done. The actual details of the induction step are relegated to the end of the proof. Stage 1 \ is nonempty and has no isolated points, so there must exist two points BÁB−\!". Pick disjoint closed balls J ! and J " , centered at B ! and B " , each with " diameter Þ# Stage 2 Since B−!!! int J and B is not isolated, we can pick distinct points B !!!" and B (both ÁB! ) in int J ! and use Lemma 3.5 to pick disjoint closed balls J !! and J !" (centered at BB!! and !" ) and both ©JÞ int ! We can then shrink the balls, if necessary, so that each has " diameter Þ## We can repeat the same steps inside JÀ""""since B− int J and B is not isolated, we can pick distinct points BB"! and "" (both ÁBJ " ) in int " and use the Lemma 3.5 to pick disjoint closed balls JJ"! and "" (centered at BB "! and "" ) and both ©JÞ int " We can then shrink the " balls, if necessary, so that each has diameter Þ## At the end of Stage 2, we have 4 disjoint closed balls: JßJßJßJÞ!! !" "! "" Stage 3 We now repeat the same construction inside each of the 4 sets JßJßJßJ!! !" "! "" . For example, we can pick distinct points BB!!! and !!" (both ÁBJ !! ) in int !! and use the Lemma

159 3.5 pick disjoint closed balls JJ!!! and !!" (centered at BB !!! and !!" ) and both ©JÞ int !! Then " we can shrink the balls, if necessary, so that each has diameter Þ#$ See the figure below.

After Stage 3, we have 8 disjoint closed balls: JßJßJßÞÞÞÞ!!! !!" !"! We have the beginnings of 8 descending sequences of closed balls at this stage. In each sequence, at the 8th “stage,” the sets " have diameter #8 .

J!!! Jª!! J!!" Jª! J!"! Jª!" J!"" and

J"!! Jª"! J"!" Jª" J""! Jª"" J"""

We continue inductively (see the details below ) in this way at a given stage, each descending sequence of closed balls splits into two “disjoint branches.” The sequence “splits” when we choose two new nonempty closed balls inside the current one, making sure that their diameters keep shrinking toward ! .

160  In the end, each binary sequence = œ Ð8"# ß 8 ß ÞÞÞß 8 5 ß ÞÞÞÑ − Ö!ß "× will correspond to a descending sequence of nonempty closed sets whose diameters Ä! :

J8" ª J 8 ## 8 ª J 8 "#$ 8 8 ª ÞÞÞ ª J 8 "# 8 ÞÞÞ85 ª ÞÞÞ

For example, the binary sequence= œ Ð!ß "ß "ß !ß !ß "ÞÞÞÑ − Ö!ß "× corresponds to the descending sequence of closed sets

J! ª J !" ª J !"" ª J !""! ª J !""!! ª J !""!!" ª ÞÞÞ

The Cantor Intersection theorem tells us for each such =B−\ , there is an = such that

J ∩ J ∩ J ∩ ÞÞÞ ∩ J ∩ ÞÞÞ œ∞ J œ ÖB × 8" 8 ## 8 8 "#$ 8 8 8 "# 8 ÞÞÞ8555œ" 8 "# 8 ÞÞÞ8 =

For two different binary sequences, say > œ Ð7"# ß 7 ß ÞÞÞß 7 5 ß ÞÞÞÑ Á Ð8 "# ß 8 ß ÞÞÞß 8 5 ß ÞÞÞÑ œ = , there is a smallest 57Á8B−JB−J for which 5 5 . Then = 8"# 8 ÞÞÞ85" 8 5 and > 8 "# 8 ÞÞÞ8 5" 7 5 . Since these  sets are disjoint, we have B=> Á B . Thus, mapping = È B = gives a "  " function from Ö!ß "× into \ . We conclude that - œ #i! œ lÖ!ß "×l Ÿ l\lÞ

Here are the details of the formal induction step in the proof..

Induction Hypothesis: Suppose we have completed 5 stages  that is, for each 3 œ "ß ÞÞÞß 5 3 and for each 3 -tuple Ð8"3 ß ÞÞÞß 8 Ñ − Ö!ß "× we have defined points B 88"3... and closed balls " JBÐJÑ8"3 ÞÞÞ8centered at 8 "3... 8 , with diam 8 "3 ÞÞÞ8 #3 and so that

for each Ð8" ß ÞÞÞß 8 3 Ñ, J 8""# ª J 8 8 ª ÞÞÞ ª J 8 "#3 8 ÞÞÞ8

Induction step: We must construct the sets for stage 5"Þ For each Ð5"Ñ -tuple 5" Ð8" ß ÞÞÞß 8 5 ß 8 5" Ñ − Ö!ß "×, we need to define a point B 8" ÞÞÞ855" 8 and a closed ball

J8" ÞÞÞ855" 8 in such a way that the conditions in the induction hypothesis remain true with 5" replacing 5 .

For any Ð8" ß ÞÞÞß 8 5 Ñ : we have B 8"" ßßß855 − int J 8 ÞÞÞ8 . Since B 8 " ßßß8 5 is not

isolated, we can pick distinct points BB8"" ÞÞÞ855 ! and 8 ÞÞÞ8 " (both ÁBJ 8 " ÞÞÞ8 5 ) in int 8 "... 8 5

and use the Lemma 3.5 to pick disjoint closed balls JJ88!""...55 and 88" ... (centered at

BB8"" ßßß855 ! and 8 ÞÞÞ8 " ) and both ©J int 8 "... 8 5 . We can then shrink the balls, if necessary, " so that each has diameter Þñ#5

Corollary 3.7 If Ð\ß .Ñ is a nonempty complete separable metric space with no isolated points, thenl\l œ - .

Proof Theorem II.5.21 (using separability) tells us that l\l Ÿ - ; Theorem 3.6 gives us l\l -Þ ñ

The following corollary gives us another variation on the basic result.

Corollary 3.8 If Ð\ß .Ñ is an uncountable complete separable metric space, then l\l œ - .

161 (So we might say that “the Continuum Hypothesis holds among complete separable metric spaces.” )

Proof Since Ð\ß .Ñ is separable, Theorem II.5.21 gives us l\l Ÿ - . If we knew that \ had no isolated points, then Theorem 3.6 (or Corollary 3.7) would complete the proof. But \ might have isolated points, so a little more work is needed. Call a point B−\ a condensation point if every neighborhood of B is uncountable. Let G\+−\G be the set of all condensation points in . Each point has a countable open neighborhood SS . Each point of is also a non-condensation point, so +−S©\G . Therefore \G is open so G is closed, and therefore ÐGß.Ñ is a complete metric space. Since Ð\ß .Ñ is separable (and therefore second countable), \  G is also second countable and therefore Lindelof.¨ Since \G can be covered by countable open sets, it can be covered by countably many of them, so \G is countable: therefore GÁg (in fact, G must be uncountable). Finally, ÐGß .Ñ has no isolated points in G : if , − G were isolated in G , then there would be an open set S in \ with S∩G œÖ,× . Since SÖ,ש\G , S would be countable, which is impossible since is a condensation point. Therefore Corollary 3.7 therefore applies to ÐGß.Ñ , and therefore l\l lGl -Þñ

Why was the idea of a “condensation point” introduced in the example? Will the argument work if, throughout, we replace “condensation point” with “non-isolated point?” If not, precisely where would the proof break down?

The next corollary answers a question we raised earlier: is there a metric . w on  which is equivalent to the usual metric .Ðß.Ñ but for which w is complete that is, is “completely metrizable?”

Corollary 3.9  is not completely metrizable.

w Proof Suppose ..œÞ is a metric on gg equivalent to the usual metric , so that ..w Then Ðß.Ñwww is a nonempty metric space and, since .µ. , the space Ðß.Ñ has no isolated points w (no set Ö;× is in gg.. œw À “isolated point” is a topological notion.). If Ð  ß . Ñ were complete, Theorem 3.6 would imply that  is uncountable. ñ

162 Exercises

E1. Prove that in a metric space Ð\ß .Ñ , the following are equivalent :

a) every Cauchy sequence is eventually constant b) Ð\ß .Ñ is complete and g. is the discrete topology c) for every E©\ß each Cauchy sequence in E converges to a point in E (that is, every subspace of Ð\ß .Ñ is complete).

E2. Suppose that X is a dense subspace of the pseudometric space Ð] ß .Ñ and that every Cauchy sequence in X converges to some point in ]Ð]ß.Ñ . Prove that is complete.

E3. Suppose that Ð\ß .Ñ is a metric space and that ÐB8 Ñ is a Cauchy sequence with only finitely many distinct terms. Prove that ÐB8 Ñ is eventually constant, i.e., that for some 8 −  , B88" œ B œ ÞÞÞ

E4. Let p be a fixed prime number. On the set  of rationals, the :-adic absolute value | |: (or :-adic norm ) is defined by:

pmk For !ÁB− , write Bœn for integers 5ß7ß8 , where pOf does not divide 7 or 8 . ( 5 " course, 5 may be negative. ) Define |Bœ:œ |:::5 . . We define | !œ!Þ |

For all Bß C −  ,

a) lBl:: ! and lBl œ! iff Bœ! b) l BCl::: œ l Bl † l Cl c) lBCl::: ŸlBl lCl d) l B  Cl::::: Ÿ max Öl Bl ß l Cl × Ÿ Bl  l Cl

The : -adic metric .::: is defined on  by . ÐBß CÑ œ lB  Cl

Prove or disprove that (ß.: ) is complete.

E5. Suppose that Ð\ß .Ñ is complete, that Ð] ßg Ñ is a Hausdorff space and that 0 À \ Ä ] is continuous. Suppose that J\"#ªJ ª... ªJ 8 ª ... is a sequence of closed sets in for which diam 0. Prove that∞∞ . ÐJ88 Ñ Ä8œ" 0Ò J8 Ó œ 0Ò 8œ" J Ó

E6. A metric space Ð\ß .Ñ is called locally complete if every point B has a neighborhood RB

163 (necessarily closed) which is complete.

a) Give an example of a metric spce Ð\ß .Ñ that is locally complete but not complete.

b) Prove that if ÐHß .Ñ is a locally complete dense subspace the complete metric space Ð\ß .Ñ, then H is open in \ . Hint: it may be helpful to notice if O is open and D is dense, then cl (S∩H ) = cl(O). This is true in any topological space Ð\ßg Ñ .

E7. Suppose E is an uncountable subset of ‘ with lElœ7-Þ (Of course, there is no such set E if the Continuum Hypothesis is assumed.) Is it possible for E to be closed? Explain.

164 4. The Contraction Mapping Theorem

Definition 4.1 A point B−\ is called a fixed point for the function 0À\Ä\ if 0ÐBÑœBÞ We say that a topological space \0À\Ä\ has the fixed point property if every continuous has a fixed point.

It is easy to check that the fixed point property is a topological property.

Example 4.2

1) A function 0À‘‘ Ä has a fixed point if and only if the graph of 0 intersects the line CœB. If E©‘ , then E is the set of fixed points for some function: for example 0ÐBÑ œ B †;;EE ÐBÑßwhere is the characteristic function of E . Is every set E©‘ the set of fixed points for some continuous function 0À‘‘ Ä ? Are there any restrictions on the cardinality of E ?

2) The interval Ò+ß ,Ó has the fixed point property. To see this, suppose that 0 À Ò+ß ,Ó Ä Ò+ß ,Ó is continuous. If 0Ð+Ñ œ + or 0Ð,Ñ œ , , then 0 has a fixed point so assume that 0Ð+Ñ  + and 0Ð,Ñ  ,Þ Define 1ÐBÑ œ 0ÐBÑB . Then 1Ð+Ñ œ 0Ð+Ñ+  ! , 1Ð,Ñ œ 0Ð,Ñ  ,  ! and 1 is continuous. By the Intermediate Value Theorem (from analysis) there is a - − Ð+ß ,Ñ where 1Ð-Ñ œ 0Ð-Ñ  - œ ! , that is, 0Ð-Ñ œ -Þ

3) The Brouwer Fixed Point Theorem states that H88 œ ÖB −‘ À .ÐBß !Ñ Ÿ "× has the fixed point property. For 8"ß the proof of Brouwer's theorem is rather difficult; the usual proofs use techniques from algebraic topology. For 8œ" , H8 is homeomorphic to Ò+ß,Ó , so Brouwer's Theorem genralizes Part 2. In fact, Brouwer's Theorem can be generalized as the Schauder Fixed Point Theorem: every nonempty compact convex subset O of a Banach space has the fixed point property.

Definition 4.3 0 À Ð\ß .Ñ Ä Ð\ß .Ñ is a contraction mapping (for short, a contraction ) if there is a constant αα− Ð!ß "Ñ such that .Ð0ÐBÑß 0ÐCÑÑ Ÿ .ÐBß CÑ for all Bß C − \ .

Notice that a contraction 0!œ is automatically continuous: for %$% , choose . Then, for all B − \ß .ÐBß CÑ $α$%$% implies .Ð0ÐBÑß 0ÐCÑÑ   . In fact, this choice of depends only on and not the point B0 . In this case we say that is uniformly continuous .

The definition of uniform continuity for 0\ on reads:

a%$ b aB aC Ð.ÐBß CÑ  $ Ê .Ð0ÐBÑß 0ÐCÑÑ  % Ñ

Compare this to the weaker requirement for 0\ to be continuous on :

aB a%$ b aC Ð.ÐBß CÑ  $ Ê .Ð0ÐBÑß 0ÐCÑÑ  % Ñ

Example 4.4

165 "" # 1) The function 0ÀÒ!ßÓÄÒ!ßÓ%% given by 0ÐBÑœB is a contraction because for any " ## " Bß C − Ò!ß% Ó, we have l0ÐBÑ  0ÐCÑl œ lB  C l œ lB  CllB  Cl Ÿ# lB  Cl .

2) But the function 0À‘‘ Ä given by 0ÐBÑœB# is not a contraction. To be a contraction would require some α − Ð!ß "Ñ which

lB##  C l Ÿα lB  Cl holds for all choices of Bß C , which would imply that lB  Cl Ÿα for all B Á C (which is false).

We could also argue that 00ÀÄ is not a contraction by noticing that ‘‘ is not uniformly continuous. For example, if we choose %$œ# , then no ! will satisfy the condition # aB aC Ð.ÐBß CÑ $$ Ê .Ð0ÐBÑß 0ÐCÑÑ  #Ñ . For example, given  ! , we can let B œ $ and # $$# C œ$ #%. Then l0ÐBÑ  0ÐCÑl œ #   # .

Theorem 4.5 (The Contraction Mapping Theorem) If Ð\ß .Ñ is a nonempty complete metric space and 0À\Ä\ is a contraction, then 0 has a unique fixed point. (“Nonempty” is included in the hypothesis because the empty function ggg from the complete space to is a contraction with no fixed point.)

A fixed point provided by the Contraction Mapping Theorem can be useful, as we will soon see in Theorem 4.9 (Picard's Theorem) below. The proof of the Contraction Mapping Theorem is also useful because it shows how to make some handy numerical estimates (see the example following Picard's Theorem.)

Proof Suppose ! αα  " and that .Ð0ÐBÑß 0ÐCÑÑ Ÿ .ÐBß CÑ for all Bß C − \ . Pick any point B−\! and apply the function 0 repeatedly to define

Bœ0ÐBÑß"! # B#" œ 0ÐB Ñ œ 0Ð0ÐB ! ÑÑ œ 0 ÐB ! Ñ ã 8 B88" œ 0ÐB Ñ œ ÞÞÞ œ 0 ÐB ! Ñ ã We claim that this sequence ÐB8 Ñ is Cauchy and that its limit in \ is a fixed point for 0 . (Here you can imagine a “control system” where the initial input is B! and each output becomes the new input in a “feedback loop.” This system approaches a “steady state” where “inputœ output.”)

Suppose %%! . We want to show that .ÐBßB87 Ñ for large enough 7ß8Þ Assume 78 . Applying the “contraction property” 8 times gives:

87 8"7" .ÐB87 ß B Ñ œ .Ð0 ÐB ! Ñß 0 ÐB ! ÑÑ œ .Ð0Ð0 ÐB ! ÑÑß 0Ð0 ÐB ! ÑÑÑ

8" 7" # 8# 7# Ÿαα .Ð0 ÐB!! Ñß 0 ÐB ÑÑ Ÿ .Ð0 ÐB !! Ñß 0 ÐB ÑÑ

8788 Ÿ ÞÞÞ Ÿ αα .ÐB!! ß 0 ÐB ÑÑ œ .ÐB !78 ß B Ñ

By the triangle inequality (and the “contraction property,” again) we get

166 8 .ÐB87 ß B Ñ Ÿα .ÐB !78 ß B Ñ

8 Ÿα Ð.ÐBßBÑ.ÐBßBÑ!" "# .ÐBßBÑ #$ ÞÞÞ.ÐB 78"78 ßB ÑÑ

8#78" Ÿααα Ð.ÐBßBÑ!" .ÐBßBÑ !" .ÐBßBÑÞÞÞ !" α .ÐBßBÑÑ !"

8#78"85∞ œαααααα .ÐB!" ß B Ñ Ð"    ÞÞÞ  Ñ  .ÐB !" ß B Ñ 5œ! 8 α .ÐBßBÑ!" œÄ!8Ä∞"α as .

8 α .ÐBßBÑ!" So "α 8RÞ% if some Then .ÐBßB87 Ñ% if 78R . Therefore ÐBÑ 8 is CauchyßÐ\ß.ÑÐBÑÄBB−\ and since is complete 8 for some .

By continuity, Ð0ÐB8888" ÑÑ Ä 0ÐBÑ . But also Ð0ÐB ÑÑ Ä B because 0ÐB Ñ œ B . Therefore 0ÐBÑ œ B because a sequence in a metric space has at most one limit. (If . were just a pseudometric, we could conclude only that .ÐBß 0ÐBÑÑ œ ! .)

If C is also a fixed point for 0 , then .ÐBß CÑ œ .Ð0ÐBÑß 0ÐCÑÑ Ÿαα .ÐBß CÑ . Since !   " , this implies that .ÐBß CÑ œ ! , so B œ C  that is, the fixed point B is unique. (If . were just a pseudometric, 0!ñ could have several fixed points at distance from each other.)

Notes about the proof

8 α .ÐBßBÑ!" 1) The proof gives us that if 78 , then .ÐBßB87 Ñ"α for each 8− . If we 8 α .ÐBßBÑ!" fix 8 and let 7 Ä ∞ , Then ÐB78 Ñ Ä B , so .ÐB ß BÑ Ÿ "α . Thus we have a computable bound on how well BB8 approximates the fixed point .

2) For large 8B we can think of 8 as an “approximate fixed point”  in the sense that 0B doesn't move the point 8 very much. To be more precise,

.ÐB88 ß 0ÐB ÑÑ Ÿ .ÐB 8 ß BÑ  .ÐBß 0ÐB 8 ÑÑ œ .ÐB 8 ß BÑ  .Ð0ÐBÑß 0ÐB 8 ÑÑ

Ÿ.ÐBßBÑ88αα .ÐBßB ÑœÐ" Ñ.ÐBßBÑ 8

8 α .ÐBßBÑ!" ŸÐ"α Ñ"α Ä! as 8Ä∞.

Example 4.6 Consider the following functions that map the complete space ‘ into itself:

1) 0ÐBÑ œ #B" : 0 is (uniformly) continuous, 0 is not a contraction, and 0 has a unique fixed point: Bœ " 2)1ÐBÑ œ B#  " is not a contraction and has no fixed point. 3) 2ÐBÑ œ B  sin B has infinitely many fixed points.

We are going to use the Contraction Mapping Theorem to prove a fundamental theorem about the existence and uniqueness of a solution to a certain kind of initial value problem in differential

167 equations. We will use the Contraction Mapping Theorem applied to the complete space in the following example.

Example 4.7 Let \GÐ\Ñ be a topological space and ‡ be the set of bounded continuous real- valued functions 0À \ Ä‘3 , with the metric Ð0ß 1Ñ œ sup Öl0ÐBÑ  1ÐBÑlÀ B − \× . (Since 0ß1 are bounded, so is l0  1l and therefore this sup is always a real number. It is easy to check that 3 is a metric.) The metric 3 is called the metric of “uniform convergence” because 3%Ð0ß 1Ñ  implies that 0 is “uniformly close” to 1 À

3%Ð0ß 1Ñ  Ê l0ÐBÑ  1ÐBÑl  % at every point B − \ ЇÑ

‡ 3 In the specific case of ÐG Б3 Ñß Ñ , the statement that Ð08 Ñ Ä 0 is equivalent the statement (in analysis) that “Ð08 Ñ converges to 0 uniformly on ‘ .”

Theorem 4.8 ÐG‡ Ð\Ñß3 Ñ is complete.

‡ Proof Suppose that Ð08 Ñ is a Cauchy sequence in ÐG Ð\Ñß3% Ñ : given  ! , there is some R so that 3%Ð087 ß 0 Ñ  whenever 8ß 7  R . This implies (see ÐÑ* , above ) that for any fixed B − \ , l087 ÐBÑß 0 ÐBÑl % whenever 8ß 7  R . Therefore Ð0 8 ÐBÑÑ is a Cauchy sequence of real numbers, so Ð08B ÐBÑÑ Ä some < −‘‘ . We define 0ÐBÑ œ < B . Then 0 À \ Ä Þ To complete the ‡ 3 proof, we claim that 0−GÐ\Ñ and that Ð0ÑÄ08

First we argue that0 is bounded. Pick R so that 3 Ð087 ß0 Ñ" whenever 8ß7 R . Using 7œR , this gives 3 Ð0ß0Ñ"8R for 8 R and therefore l0ÐBÑ0ÐBÑl" 8 R for every B−\. If we now let 8Ä∞ , we conclude that

l0ÐBÑ  0R ÐBÑl Ÿ " , that is "Ÿ0ÐBÑ0R ÐBÑŸ" for every B .

Since 0QR is bounded, there is a constant such that

Q Ÿ0R ÐBÑŸQ for every B.

Adding the last two inequalities gives

l0ÐBÑl Ÿ Q  " for every B , so 0 is bounded. 3 We now claim that Ð08 Ñ Ä 0 . (Technically, this is an abuse of notation because 3 is defined on GÐ\ч‡ and we don't yet know that 0 is in GÐ\Ñ ! However, the same definition for 3 makes sense on any collection of bounded real-valued functions, continuous or not. We are using this % “extended definition” of 3 here.) Let %3! and pick R so that Ð0ß087 Ñ # whenever % 7ß 8 R. Then l087 ÐBÑ  0 ÐBÑl # for every B whenever 8ß 7 R . Letting 7 Ä ∞ , we % get that l08 ÐBÑ  0ÐBÑl Ÿ# for all B and all 8 R . Therefore 3% Ð08 ß 0Ñ  if 8 R .

Finally, we show that 0+−\!B−\R− is continuous at every point and % . For and we have

168 l0ÐBÑ  0Ð+Ñl Ÿ l0ÐBÑ  0RRRR ÐBÑl  l0 ÐBÑ  0 Ð+Ñl  l0 Ð+Ñ  0Ð+Ñl

3 % Since Ð08 Ñ Ä 0 , we can choose an R large enough that 3 Ð0R ß 0Ñ $ Þ This implies that %% l0ÐBÑ  0RR ÐBÑl $$and l0 Ð+Ñ  0Ð+Ñl  . Since 0 R is continuous at + , we can choose a % neighborhood [ of + so that l0RR ÐBÑ0 Ð+Ñl$ for all B−[Þ So, if B−[ then we have

%%% l0ÐBÑ  0Ð+Ñl $$$   œ% , so 0 is continuous at + . ñ

Theorem 4.9 (Picard's Theorem) Let 0ÀHÄ‘‘ be continuous, where H is a closed box in # and ÐB!! ß C Ñ − int H . Suppose that there exists a constant Q such that for all ÐB"" ß C Ñ and ÐB "# ß C Ñ in H l0ÐBßCÑ0ÐBßCÑlŸQlCCl"" "# " # (L) Then i) there exists an interval M œÒB!! +ßB +ÓœÖBÀlBBlŸ+×ß ! and ii) there exists a unique differentiable function 1ÀMÄ‘

Cœ1ÐBÑ!! such that w for B−M  1 ÐBÑ œ 0ÐBß 1ÐBÑÑ

In other words, on some interval MB centered at ! , there is a unique solution Cœ1ÐBÑ to the initial value problem

Cœ0ÐBßCÑw Ð‡Ñ  CÐB!! Ñ œ C

Before beginning the proof, we want to make some comments about the hypotheses.

1) Condition (L) says that “0 satisfies a Lipschitz condition in the variable C on H. ” This may seem a bit obscure. For our purposes, it's enough to notice that (L) is true if the partial derivative 0HC exists and is continuous on the box . In that case, we know (from analysis) that |0ŸC" | some constant Q on H . Thinking of 0ÐBßCÑ as a single-variable function of C and using the ordinary Mean Value Theorem gives us a point DCC between "# and for which

|0ÐB"" ß C Ñ  0ÐB "# ß C Ñl œ l0 C" ÐB ß DÑl † lC "  C # l Ÿ QlC "  C # l

2) As a specific example, suppose 0ÐBßCÑœCB , that HœÒ"ß"Ó‚Ò"ß"Ó , and that ÐB!! ß C Ñ œ Ð!ß !Ñ . Since 0 C œ " , the preceding comment tells us that condition (L) is true. CœCBw Consider the initial value problem . Picard's Theorem states that there  CÐ!Ñ œ ! is a unique differentiable function C œ 1ÐBÑ , defined on some interval Ò  +ß +Ó that satisfies this system: 1Ð!Ñ œ ! and 1ww ÐBÑ œ 0ÐBß 1ÐBÑÑ Ð that is, C œ C  BÑ . Moreover, the proofs of Picard's Theorem and the Contraction Mapping Theorem can actually help us to actually find this solution. Of course, once a solution is actually found, a direct check might show that the solution is actually valid on an interval larger than the interval M that comes up in the proof.

169 Proof (Picard's Theorem) We begin by transforming the initial value problem Ð‡Ñ into an equivalent problem that involves an integral equation Ї‡Ñ rather than a differential equation. Let Cœ1ÐBÑ be a continuous function defined on an interval M œÒB+ßB+Ó!! :

Suppose 1 satisfies (*): 0 is continuous, so 1w ÐBÑ œ 0ÐBß 1ÐBÑÑ is continuous. (Why? ) Therefore the Fundamental Theorem of Calculus tells us for all B−M ,

BBw 1ÐBÑ  1ÐB! Ñ œ 1 Ð>Ñ .> œ 0Ð> , 1Ð>ÑÑ .> , so BB!!

B 1ÐBÑ œ C!  0Ð> , 1Ð>ÑÑ .> (**) B!

B! Suppose 1 satisfies (**): then 1ÐB!! ) œ C  0Ð> , 1Ð>ÑÑ .> œ C ! . Since 0Ð>ß 1Ð>ÑÑ is B! continuous, the Fundamental Theorem gives that for all B − M , 1w ÐBÑ = 0ÐBß 1ÐBÑÑ . Therefore the function 1ÐBÑ satisfies (*).

Therefore a function continuous 1M on satisfies ÐÑ * iff it satisfies ÐÑ ** . (Note that the initial condition C!! œ 1ÐB Ñ is “built into” the single equation Ї‡ÑÞ ) We will find a 1 that satisfies ÐÑ** .

Now we establish some notation:

We are given a constant Q in the Lipschitz condition.

Pick a constant O so that l0ÐBß CÑl Ÿ O for all ÐBß CÑ − H . (We will prove later in this chapter that a continuous function on a closed box in ‘# must be bounded. For now, we assume that as a fact from analysis.)

Because ÐB !! ß C Ñ − int H , we can pick a constant +  ! so that

i) ÖBÀlBBlŸ+ׂÖCÀlCClŸO+שH!! , and ii) +Q  "

Let M œÖBÀlBB!!! lŸ+לÒB +ßB +ÓÞ

We consider ÐG‡ ÐMÑß33 Ñ , where is the metric of uniform convergence. By Theorem 4.8, this ‡ space is complete. Let FœÖ1−G ÐMÑÀa>−Mßl1Ð>ÑClŸO+× ! .

For an arbitrary 1 − G‡ ÐMÑ , 1Ð>Ñ might be so large that Ð>ß 1Ð>ÑÑ Â HÞ If we look only at functions in F , we avoid this problem: if 1 − F and > − M , then Ð>ß 1Ð>ÑÑ − H because of how we chose + . So for 1 − F , we know that 0Ð>ß 1Ð>ÑÑ is defined.)

Notice that FÁg since the constant function 1ÐBÑ´C! is certainly in F .

‡ F is closed in GÐMÑ : if 2− cl F , then there is sequence of functions Ð1Ñ8 in F such that Ð188 Ñ Ä 2 in the metric 3 . Therefore 1 ÐBÑ Ä 2ÐBÑ for all B − M , and subtracting C ! we get

l18! ÐBÑ  C l Ä l2ÐBÑ  C ! l .

170 Since l18! ÐBÑ  C l Ÿ O+ for each B − M , then l2ÐBÑ  C ! l Ÿ O+ for each B − M . Therefore 2−F, so F is closed. Therefore ÐFß3 Ñ is a nonempty complete metric space .

For 1−F , define a function 2œXÐ1Ñ by

B 2ÐBÑ œ X () 1 ÐBÑ œ C!  0Ð>ß 1Ð>ÑÑ .> . B!

Notice that 2B−M is continuous (in fact, differentiable) and that, for every ,

BBB l2ÐBÑl Ÿ||| C!!!!  0Ð>ß 1Ð>ÑÑ .> | Ÿ|| C  | 0Ð>ß 1Ð>ÑÑ | .> Ÿ|| C  O .> Ÿ lC l  O+ , BBB!!! so 2 is bounded. Therefore X ÀFÄG‡ ÐMÑ . But in fact, even more is true: XÀFÄF , because B aB − M , | X Ð1ÑÐBÑ  C!! | œ | 2ÐBÑ  C l œ l 0Ð>ß 1Ð>ÑÑ .> | Ÿ O+ B!

We now claim that XÀÐFßÑÄÐFßÑ33 is a contraction. To see this, we simply compute distances: if 11−F"# and , then

3ÐX Ð1"# Ñ, X Ð1 Ñl œsup Ö | X Ð1 " ÑÐBÑ  X Ð1 # ÑÐBÑ | À B − M× œ sup Ö | B 0Ð>ß 1 Ð>ÑÑ  0Ð>ß 1 Ð>ÑÑ .> | À B − M× B! "# B Ÿ sup Ö || 0Ð>ß 1"# Ð>ÑÑ  0Ð>ß 1 Ð>ÑÑ | .>| À B − M× B! ŸÖ sup | B Ql1Ð>Ñ1Ð>Ñ.>ÀB−M× | | (from Condition L ) B! "# B Ÿ sup| Ö Q3 Ð1ß1Ñ.>ÀB−M×"# | B! œ sup ÖQ 3 Ð1ß1ѱBBlÀB−M×"# ! œ+QÐ1ß1Ñ3 "# œα3 Ð1ß1Ñ"# , where α œ+Q 1.

Then the Contraction Mapping Theorem gives us a unique function 1−F for which 1œXÐ1Ñ . From the definition of X , that simply means:

B aB − M , 1ÐBÑ œ X Ð1ÑÐBÑ œ C!  0Ð>, 1Ð>ÑÑ.> B! which is precisely condition (**). ñ

Example 4.10 The method in the proof of Picard's Theorem and numerical estimates in the proof of the Contraction Mapping Theorem can be used to get useful information about a specific initial value problem. To illustrate, we consider

CœCBw

 CÐ!Ñ œ ! and find a solution that is valid on some interval containing 0.

We begin by choosing a box H with ÐB!! ß C Ñ œ Ð!ß !Ñ in its interior. Rather arbitrarily, we select the box HœÒ"ß"Ó‚Ò"ß"ÓÞ Since l0ÐBßCÑlœlCBlŸlBllClŸ# on H , we can use Oœ# in the proof.

Because |0ÐBßCÑlœ"C throughout H , the Lipschitz condition (L) is satisfied with Qœ"Þ

171

Following the proof of Picard's Theorem, we now choose a constant + so that

i) ÖB À lBl Ÿ +× ‚ ÖC À lCl Ÿ O+ œ #+× © H and ii) +Qœ+†""

""" Again rather arbitrarily, we choose +œ### , so that M œÒB!! +ßB +ÓœÒ ß Ó . ‡‡"" Then F œ Ö1 − G ÐMÑ À l1ÐBÑ  !l Ÿ O+× œ Ö1 − G ÐMÑ À | 1ÐBÑl Ÿ " for all B − Ò ## ß Ó× .

Finally, we choose any function 1−F! ; to make things as simple as possible, we might as well "" choose 11ÐBÑœ!ÒßÓ!! to be the constant function on ## .

According to the proof of the Contraction Mapping Theorem, the sequence of functions 8 3 Ð18! Ñ œ ÐX Ð1 ÑÑ Ä 1, where 1 is a fixed point for X , and 1 is the solution to our initial value problem. We calculate:

BBBB# 1"! ÐBÑ œ C  0Ð>ß 1 ! Ð>ÑÑ .> œ !  0Ð>ß !Ñ .> œ  > .> œ  B!!! #

## $# 1 ÐBÑ œ X Ð1 ÐBÑÑ œ ! BB 0Ð>ß >>BB Ñ .> œ   > .> œ   #" !!## '#

$# $# % $ # 1 ÐBÑ œ X Ð1 ÐBÑÑ œBB 0Ð>ß >>  Ñ .> œ  >>   > .> œ  BBB   $#!!'# '# #%' # and, in general,

BB#$ B 8" 188" ÐBÑ œ X Ð1 ÐBÑÑ œ ÞÞÞ œ #x  $x  ÞÞÞ  Ð8"Ñx

The functions 118 converge (uniformly) to the solution .

In this particular problem, we are lucky enough to recognize that the functions 1ÐBÑ8 are just the 88 partial sums of the series ∞∞BB œ"B œ"B/B Þ Therefore 8œ#8x 8œ! 8x 1ÐBÑœ"B/B is a solution of our initial value problem, and we know it is valid for all "" B−MœÒ## ß Ó. (You can check by substitution that the solution is correct and that it actually is a solution that works for all B−‘.)

Even if we couldn't recognize a neat formula for the limit 1ÐBÑ , we could still make some useful approximations. From the proof of the Contraction Mapping Theorem, we know that 8 " α3Ð1!" ß1 Ñ " #8 3Ð1!" ß1 Ñ 3α3(1ß1ÑŸ8 "α . In this example œ+Qœ# , so that ( 1ß1ÑŸ8 " " # "B# """ œ#####8"sup Öl!Ð ÑlÀB−Mל 8" † $ œ 8# . Therefore 18 ÐBÑ is uniformly within """ distance ###8# of 1ÐBÑ on the interval Ò  ß Ó .

Finally, recall that our initial choice of 1−F! was arbitrary. Since l sin BlŸ" , we know that sin− F , and we could just as well have chosen 1!8 ÐBÑ œ sin B . Then the functions 1 ÐBÑ computed as XÐ1Ñœ1ßXÐ1Ñœ1ßÞÞÞ!"## would be quite different Ðtry computing 11"# and Ñ B but it would still be true that 1ÐBÑÄ1ÐBÑœ"B/8 uniformly on MÀ the same limit 1 because the solution 1ñ is unique.

172 The Contraction Mapping Theorem can be used to prove other results for example, the Implicit Function Theorem. (You can see details, for example, in Topology , by James Dugundji. )

173 Exercises

E8. +Ñ Suppose 0 À‘‘ Ä is differentiable and that there is a constant O  1 such that |0w ÐBÑ | Ÿ O for all BÞ Prove that 0 is a contraction (and therefore has a unique fixed point.)

b) Give an example of a continuous function 0À‘‘ Ä such that

|0ÐBÑ0ÐCÑl  lBCl for all B−ÁC ‘ but such that 0 has no fixed point.

Note: The function 0œ" is not a contraction mapping. If we allow α in the definition of contraction, then the Contraction Mapping Theorem is not be true not even if we “compensate” by using “Ÿ ” instead of “ ” in the definition.

E9. Let 0À Ð\ß .Ñ Ä Ð\ß .Ñ , where Ð\ß .Ñ is a nonempty complete metric space. Let 0 5 denote the “kth iterate of 0 ” that is, 0 composed with itself 5 times.

a) Suppose that b5 − for which 0 5 is a contraction. Then, by the Contraction Mapping Theorem, 0::05 has a unique fixed point . Prove that is also the unique fixed point for .

b) Prove that the function cosÀÄ‘‘ is not a contraction.

c) Prove that cos5 is a contraction for some 5− . ÐH3 nt: the Mean Value Theorem may be helpful.)

d) Let 5− be such that 1œ cos5 is a contraction and let : be the unique fixed point of 1:. By a), is also the unique solution of the equation cos BœB . Start with 0 as a “first approximation” for : and use the technique in the proof of the Contraction Mapping Theorem to find an 8− so that | 18 Ð!Ñ:± < 0.00001.

e) For this 81Ð!Ñ , use a computer or calculator to evaluate 8 . (This “solves” the equation cosBœB with l Error l 0.00001.)

E10. Consider the differential equation CœBCw with the initial condition CÐ!Ñœ" . Choose a suitable rectangle HOßQ+ and suitable constants and as in the proof of Picard's Theorem. Use the technique in the proof of the contraction mapping theorem to find a solution for the initial value problem. Identify the interval M in the proof. Is the solution you found actually valid on an interval larger than M ?

174 5. Completions

The set of rationals  (with the usual metric . ) is not complete. However, the rationals are a dense subspace of the complete space (‘ß.Ñ . This is a model for the definition of a “completion” for a metric space. µ µµ Definition 5.1 Ð\, . Ñ is called a completion of Ð\ß.Ñ if: . | Ð\ ‚ \Ñ œ . , \ is a dense µ µ µ subspace of \Ð\Ñ , and ,. is complete.

Loosely speaking, a completion of Ð\ß .Ñ adds the additional points needed Ð and no others) to provide limits for the Cauchy sequences that fail to converge in Ð\ß.Ñ . In the case of Ð ß.Ñ , we can think of the “additional points” as the irrational numbers, and the resulting completion is Ðß.ÑÞ‘

If a metric space Ð\ß .Ñ is complete, what would a completion look like? Since Ð\ß .Ñ is a µµµ complete subspace of Ð\ , . Ñ , \ must be closed in \ . But \ must also be dense. So µ \œ\ that is, a complete metric space is its own completion. (If Ð\ß .Ñ is a complete µµ pseudometric space, then \C− might not be closed in \\\; but each is at distance ! from some B−\ß so if ÐBÑÄC−]88 , then ÐBÑ already converges to B−\ and “adding the point C ” was unnecessary.

Of course different spaces may have the same completionÐß.Ñ for example, ‘ is a completion for both Ð ß .Ñ and Ð ß .ÑÞ

Notice that a completion of Ð\ß.Ñ depends on . , not just the topology g. Þ For example, Ð ß.Ñ " and ÐÖ8 À 8 − × , .Ñ are homeomorphic topological spaces (both have the discrete topology). " But the completion of Ð ß .Ñ Ð itself!) is not homeomorphic to ÐÖ!× ∪ Ö8 À 8 − ×ß .Ñ , which is a " completion of ÐÖ8 À 8 − ×ß .ÑÞ

Recall that if there is an (onto ) isometry 0Ð\ß.ÑÐ]ß=Ñ between two metric spaces and , then Ð\ß .Ñ and Ð] ß =Ñ can be regarded as “the same metric space”: we can think of 0 as just “assigning new names” to the points of \0ÀÐ\ß.ÑÄÐ]ß=Ñ . If is an isometry from \ into ] , then we can identify Ð\ß.Ñ with Ð0Ò\Óß=Ñ  an “exact metric copy” of Ð\ß.Ñ inside Ð]ß=ÑÞ So if Ð]ß=Ñ happens to be complete, then Ð0Ò\Óß=Ñ is dense in the complete space Ð cl] 0Ò\Óß=Ñ . If we identify Ð0Ò\Óß =Ñ with Ð\ß .Ñ , then we can call Ð cl] 0Ò\Óß =Ñ a completion of Ð\ß .Ñ even though \0Ò\ÓÐ\ß.Ñ is not literally a subset of cl] . Therefore, to find a completion of , it is sufficient to find an isometry 0Ð\ß.Ñ from into some complete space Ð]ß.Ñw .

ww"" Example 5.2 Consider  with the metric .Ð8ß7Ñœl87  l . Ð ß.Ñ is isometric to " " (ÖÀ8−×ß.Ñ8  where . is the usual metric on ( 0Ð8Ñœ8 is an isometry ). These two spaces " are “metrically identical.”. Therefore ÐÖ!× ∪ Ö8 À 8 − ×ß .Ñ contains a dense isometric copy of w " Ð ß . Ñ. The closure of that “copy” is Ö!× ∪ Ö8 À 8 − × , and we can view " w ÐÖ!×∪Ö8 À8− ×ß.Ñ as a completion of Ð ß. Ñ  even though is not literally a subspace " of Ö!× ∪ Ö8 À 8 − ×Þ

175 The next theorem tells us every metric space has a completion and, as we will see later, it is “essentially” unique.

Theorem 5.3 Every metric space Ð\ß .Ñ has a completion.

Proof Our previous comments show that it is sufficient to produce an isometry of Ð\ß .Ñ into some complete metric space: the complete space we will use is ÐG‡ Ð\Ñß33 Ñ where, as usual, is the metric of uniform convergence. The theorem is trivial if \œgß so we assume that we can pick a point :−\ . For any point + − \ , define 9‘9++ À \ Ä by ÐBÑ œ .ÐBß +Ñ  .ÐBß :ÑÞ The map 9 + is a difference of continuous functions so 9+ is continuous. For each B−\ , ‡ l999+++ ÐBÑl œ l.ÐBß +Ñ  .ÐBß :Ñl Ÿ .Ð+ß :Ñ, so is bounded. Therefore − G Ð\ÑÞ ‡ Define F3F9À Ð\ß.Ñ Ä ÐG Ð\Ñß Ñ by Ð+Ñ œ+ Þ We complete the proof by showing that F is an isometry. This just involves computing some distances: for any +ß , − \ ,

39Ð+, ß 9 Ñ œ sup Öl 9 + ÐBÑ  9 , ÐBÑl À B − \× œsup ÖlÐ.ÐBß +Ñ  .ÐBß :ÑÑ  Ð.ÐBß ,Ñ  .ÐBß :ÑÑl À B − \× œÖl.ÐBß+Ñ.ÐBß,ÑlÀB−\×Þ sup

For any B, l.ÐBß +Ñ  .ÐBß ,Ñl Ÿ .Ð+ß ,Ñ . And letting B œ , gives l.ÐBß +Ñ  .ÐBß ,Ñl œ .Ð+ß ,Ñ . Therefore

39Ð+, ß 9 Ñ œ sup Öl.ÐBß +Ñ  .ÐBß ,Ñl À B − \× œ .Ð+ß ,ÑÞ ñ

The completion of Ð\ß .Ñ given in this proof is Ð clGÐ\ч F3 Ò\Ó , Ñ . The proof is “slick” but it seems to lack any intuitive content. For example, if we applied the proof to Ðß.Ñ , it would not at all clear that the resulting completion is isometric to Ðß.Ñ‘ (as we would expect). In fact, there is another more intuitive way to construct a completion of Ð\ß .Ñ , but verifying all the details is much more tedious. We will simply sketch this alternate construction here.

Call two Cauchy sequences ÐB88 Ñ and ÐC Ñ in Ð\ß .Ñ equivalent if .ÐB 88 ß C Ñ Ä !Þ It is easy to check thatµÐ\ß.Ñ is an equivalence relation among Cauchy sequences in . Clearly, if ÐB8 Ñ Ä D − \ß then any equivalent Cauchy sequence also converges to D . If two nonconvergent Cauchy sequences equivalent, then they are “trying to converge to the same point” but the necessary point is “missing” in \ . µ We denote the equivalence class of a Cauchy sequence ÐB Ñ by ÒÐB ÑÓ and let \ be the µ 88 set of equivalence classes. Define the distance . between two equivalence classes by µ . ÐÒÐB88 ÑÓß ÒÐC ÑÓÑ œlim .ÐB 88 ß C ÑÞ (Why must this limit exist? ) It is easy to check that µ 8Ä∞ . does not depend on the choice of representative sequences from the equivalence µ ~ classes, and that .\ is a metric on . One then checks (this is only “tricky” part) that µ µµ Ð\ ß . Ñ is complete: any . -Cauchy sequence of equivalence classes Ð ÒÐB8 ÑÓ Ñ must µ µ converge to an equivalence class in Ð\ ß . Ñ . µ For each B − \ , the sequence ÐBß Bß Bß ÞÞÞÑ is Cauchy, and so ÒÐBß Bß Bß ÞÞÞÑÓ − \ .

176 µ µ The mapping 0 À Ð\ß .Ñ Ä Ð\ ß . Ñ given by 0ÐBÑ œ ÒÐBß Bß Bß ÞÞÞÑÓ is an isometry, and it µµµµ is easy to check that 0Ò\Ó is dense in Ð\ ß . Ñ  so that Ð\ ß . Ñ is a completion for Ð\ß .ÑÞ

This method is one of the standard ways to construct the real numbers from the rationals: ‘ is defined as the set of these equivalence classes of Cauchy sequences of rational numbers. The real numbers can also be constructed as a completion for the rationals by using a method called “Dedekind cuts.” However that approach also uses the order structure in  and therefore we cannot mimic it in the general setting of metric spaces.

The next theorem gives us the good news that, in the end, it doesn't really matter how we construct a completionÐ\ß.Ñ because the completion of is “essentially” unique: all completions are isometric (and, in a “special” way!). We state the theorem for metric spaces. (Are there modifications of the statement and proof to handle the case where . is merely a pseudometric? )

Theorem 5.4 The completion of Ð\ß .Ñ is unique in the following sense: if Ð] ß =Ñ and Ð^ß >Ñ are both complete metric spaces containing \ as a dense subspace, then there is an (onto) isometry 0 À Ð] ß =Ñ Ä Ð^ß >Ñsuch that 0l\ is the identity map on \ . (In other words, not only are two completions of Ð\ß .Ñ isometric, but there is an isometry between them that holds \ fixed. The isometry merely “renames” the new points in the “outgrowth” ]\Þ )

Proof Suppose C−]Þ Since \ is dense in ] , we can pick a sequence ÐBÑ8 in \ which converges to C . Since ÐB8 Ñ is convergent, it is Cauchy in Ð\ß .Ñ , and since >l\ ‚ \ œ . , ÐB8 Ñ is also a Cauchy sequence in the complete space Ð^ß >Ñ . Therefore ÐB8 Ñ Ä some point D − ^ . We define 0ÐCÑ œ D . Then 0 À Ð] ß =Ñ Ä Ð^ß >ÑÞ (It is easy to check that 0 is well-defined that is, we get the same DÐBÑ no matter which of the possibly many sequences 8 we first choose converging to C.) In particular, if B−\ß we can choose ÐBÑ88to be the constant sequence B œB ; then 0ÐBÑ œ B, so that 0l\ is the identity map on \ . www We need to verify that 0C−]ÐBÑÄC is an isometry, Suppose and we choose 8 . We then have that

www .ÐB88 ß B88 Ñ œ =ÐB ß B Ñ Ä =ÐCß C Ñ and www w .ÐB88 ß B88 Ñ œ >ÐB ß B Ñ Ä >ÐDß D Ñ œ >Ð0ÐCÑß 0ÐC ÑÑ , so =ÐCß Cww Ñ œ >Ð0ÐCÑß 0ÐC ÑÑ ñ

According to Theorem 5.4, the space Ðß.Ñ‘ no matter how we construct it  is the completion of the space Ðß.Ñ .

177 6. Category

In mathematics there are many different ways to compare the “size” of sets. These different methods are useful for different purposes. One of the simplest ways is to say that one set is “bigger” if it has “more points” than another set that is, by comparing their cardinal numbers.

In a totally different spirit, we might call one subset of ‘# “bigger” than another if it has a larger area. In analysis, there is a generalization of area. A certain collection `‘ of subsets of # contains sets that are called measurable , and for each set W−`. a nonnegative real number ÐWÑ is assigned. ..ÐWÑ is called the “measure of W ” and we can think of ÐWÑ as a kind of “generalized area.” A set with a larger measure is “bigger.”

In this section, we will look at a third completely different and useful topological idea for comparing the “size” of certain sets. The most interesting results in this section are be about complete metric spaces, but the basic definitions make sense in any topological space Ð\ßg ÑÞ

Definition 6.1 A subset EÐ\ßÑ of the topological space g is called nowhere dense in \ if int\\ (clEœgÞ ) (In some books, a nowhere dense set is called rare .)

A set has empty interior iff its complement is dense. Therefore E\ is nowhere dense in iff int\\ (clEœg ) iff \ cl \ E is dense.

Intuitively, we can think of an open set S as including some “elbow room” around each of its pointsB−S if , then all sufficiently nearby points are also in S . Then we think of a nowhere dense set as being “skinny” so skinny that not only does it contain no “elbow room” around any of its points, but even its closure contains no “elbow room” around any of its points.

Example 6.2

1) A closed set J\Jœg\J\ is nowhere dense in iff int iff is dense in . In particular, if a singleton set Ö:× is a closed set in \ , then Ö:× is nowhere dense unless : is isolated in \Þ

2) Suppose E©‘ Þ E is nowhere dense iff cl E contains no interval of positive length. For example, each singleton Ö<× is nowhere dense in ‘ . In particular, for 8 − ß Ö8 } is nowhere dense in ‘ . But notice that Ö8× is not nowhere dense in  , since 8 is isolated in  . The property of “nowhere dense” is relative to the space in which a set “lives.” If F©E©\ , then int\\ Ð cl FÑÑ© int \\ Ð cl EÑÑ . Therefore if E is nowhere dense in \F, then is also nowhere dense in \ . However, the set F might not be nowhere dense in E . For example, consider Ö"× ©‘ © Þ

" ÖÀ8−×8 ‘ is nowhere dense in .

 and are not nowhere dense in ‘ (the awkward “double negative” in English is one reason why some authors prefer to use the term “rare” for “nowhere dense.”

178 4) Since cl\\\Eœ cl Ð cl EÑ , the set on the left side has empty interior iff the set on the right side has empty interiorßE\E\ that is is nowhere dense in iff cl\ is nowhere dense in .

Theorem 6.3 SupposeF©E©Ð\ßg Ñ . If F is nowhere dense in E , then F is nowhere dense in \.

Proof If not, then there is a point B− int\\ Ð cl FÑ© cl \ F . Since int \\ Ð cl FÑ is an open set containing BB−F and cl\\\ , then gÁÐFÑ∩F int cl . So gÁÐFÑ∩F©ÐFÑ∩E©F∩EœFintcl\\ int \\ cl cl \ clE . Since int\\ÐFÑ∩E cl is a nonempty open set in E , we conclude that intEE ÐFÑÁg cl , which contradicts the hypothesis that FEñ is nowhere dense in .

The following technical results are sometimes useful for handling manipulations involving open sets and dense sets.

Lemma 6.4 In any space Ð\ßg Ñ

1) If S open in \ and H is dense in \ , then cl S œ cl ÐS ∩ HÑÞ 2) If SH\S∩HS is open and is dense in , then is dense in 3) If SH\S∩H\ is open and dense, and is dense in , then is dense in . In particular, the intersection of two and therefore finitely many dense open sets is dense. (This result is not true for countable intersections; can you provide an example? )

Proof 1) We need to show that clS © cl ÐS ∩ HÑÞ Suppose B − cl S and let Y be an open set containing B . Then Y∩SÁg , so, since H is dense, we must have gÁÐY∩SÑ∩H œY ∩ÐS∩HÑ. Therefore B− cl ÐS∩HÑÞ 2) Using part 1), we have clS\ÐS∩HÑœ cl ÐS∩HÑ∩SœÐ cl \S SÑ∩Sœ cl SœSÞ 3) If SÐS∩HÑœSœ\ is dense, then part i) gives cl cl .

Theorem 6.5 A finite union of nowhere dense sets in Ð\ßg Ñ is nowhere dense in \ .

Proof We will prove the result for the union of two nowhere dense sets. The general case follows using a simple induction. If EE"# and are nowhere dense in \ , then

\  cl ÐE"# ∪ E Ñ œ \  Ð cl E " ∪ cl E # Ñ œ Ð\  cl E " Ñ ∩ Ð\  cl # Ñ .

Since the last two sets are open and dense, Lemma 6.4(3) gives that \ cl ÐE"# ∪EÑ is dense. Therefore E∪E"# is nowhere dense. ñ

Theorem 6.5 is false for infinite unions : for each ;−‘ , Ö;× is nowhere dense in , but is not nowhere dense in . ;−Ö;× œ ‘

Definition 6.6 Suppose E©Ð\ßg ÑÞE is called a first category set in \ if E can be written as a countable union of sets that are nowhere dense in X. If E\ is not first category in , we say that E is second category in \ . (In some books, a first category set is called meager .)

179 If we think of a nowhere dense set in \ as “skinny,” then a first category set is a bit larger merely “thin.”

Example 6.7

"Ñ If E\E\ is nowhere dense in , then is first category in .

2)‘ is a first category set in .

3Ñ If F©E©\ and E is first category in \ , then F is first category in \ . However F may not be first category in E , as the example Ö!× © Ö!ß "× © ‘ shows. However, using Theorem 6.3, we can easily prove that if FEF is first category in , then is also first category in \.

4) A countable union of first category sets in \\ is first category in .

" 5) E œ Ö!× ∪ Ö8 À 8 − × is second category in E  since any subset of E containing " is not nowhere dense in EE . But is first category (in fact, nowhere dense) in ‘ .

6) Cardinality and category are totally independent ways to talk about the “size” of a set.

a) Consider the right-ray topology g‘ œ Ögß × ∪ ÖÐ+ß ∞Ñ À + − ‘‘ × on . is uncountable, but ∞ , where Each is nowhere ‘‘œJ8œ" 88 JœÐ∞ß8ÓÞJ 8 dense in Ðß‘g Ñ , so ‘ is first category in Ðß ‘g Ñ .

b) On the other hand,  is countable but (with the usual topology) is second category in .

7) For 8" , a straight line is nowhere dense in ‘8 , so a countable union of straight lines in ‘‘88 is first category in . (Aside: it follows from the Baire Category Theorem, below, that the union of countably many straight lines cannot equal ‘8 . But nothing that fancy is needed: can you give an argument that is based just on countable and uncountable sets to show why a countable union of lines cannot œ ‘8? Can you extend your argument to show why ‘$ is not a countable union of planes? Ñ More generally, if 58 , a countable union of 5 -dimensional linear subspaces of ‘8 is first category in ‘8Þ (Can a countable union of 5œ -dimensional linear subspaces ‘8 ? )

It is not always easy to see what the “category” of a set is. Is  a first or second category set in ‘‘? For that matter, is first or second category in ‘ ? (If you know the answer to either of these questions, you also know the answer to the other: why? )

As we observed in Lemma 6.4, the intersection of two (and therefore, finitely many) dense open sets is dense. However, in certain cases, the intersection of a countable collection of dense open sets is dense. The following theorem discusses this condition and leads us to a definition.

Theorem 6.8 The following are equivalent in any space Ð\ßg Ñ :

1) If E\\E\ is first category in , then is dense in 2) If , ,..., ,... is a sequence of dense open sets, then ∞ is dense. KK"# K 55œ" K 5

180 Proof (1ÊKKK 2) Let "# , ,..., 5 ,... be a sequence of dense open sets. Each \K 5 is closed and nowhere dense, so ∞∞∞ is first category. By 1), 5œ"Ð\  K555 Ñ \  5œ" Ð\  K Ñ œ 5œ" K is dense. (2 Let be a first category set, say ∞ , where each is nowhere Ê"Ñ E Eœ8œ" R55 R dense. Then each cl is dense so, by 2), ∞ cl is dense. Since \ R55 œK5œ" Ð\ R 5 Ñ ∞∞∞cl cl , we see that is dense. 5œ"Ð\ R555 Ñœ\ 5œ" R ©\ 5œ" R œ\E \E ñ

Definition 6.9 A space Ð\ßg Ñ is called a Baire space if one (therefore both) of the conditions in Theorem 6.8 hold.

Intuitively we can think of a Baire space as one which is “thick”E a “thin” first category set can't “fill up”\\E . In fact is not merely nonempty but actually dense in \ . In fact, this is often the way a Baire space is used in “applications”: if you want to prove that there is an element in a Baire space having a certain propertyTEœÖB−\ÀB , you can consider does not have property T× . If you can then show that E is first category in \ , it follows that \ E Á g .

The following theorem tells us a couple of important things.

Theorem 6.10 Suppose SÐ\ßÑÞ is an open set in a Baire Space g

1) If SÁgß then S is second category in \ . (.)In particular, a nonempty Baire space \ is second category in itself

2) S is a Baire space.

Proof 1) If S\ were first category in , then (by definition of a Baire space) \S would be dense in \\S ; since is closed, it would follow that \Sœ\ , and therefore SœgÞ

2) Suppose S\E is open in the Baire space , and let be a first category set in S . We must show that is dense in . Suppose ∞ , where is nowhere dense in . SE S Eœ5œ" R55 R S By Theorem 6.3, R\E\\E5 is also nowhere dense in . Therefore is first category in so is dense in \ . Since S is open, S∩Ð\EÑœSE is dense in S by part 2) of the Lemma 6.4. ñ

Example 6.11 By Theorem 6.10.1, a nonempty Baire space is second category in itself. The converse is false. To see this, let \ œ Ð ∩ Ò!ß "ÓÑ ∪ Ö#× , with its usual topology. If we write ∞ , then the isolated point “2” must be in some , so is not nowhere dense. \œ5œ" R555 R R Therefore \\ is second category in itself. If were Baire, then, by Theorem 6.10, the open subspace  ∩ Ò!ß "Ó would also be Baire and therefore second category in itself. However this is false since  ∩ Ò!ß "Ó is a countable union of (nowhere dense) singleton sets.

To make use of properties of Baire spaces, it would be helpful to have a “large supply” of Baire spaces. The next theorem provides us what we want.

181 Theorem 6.12 (The Baire Category Theorem) A complete metric space Ð\ß .Ñ is a Baire space. (and therefore, by 6.10, a nonempty complete metric space is second category in itself.)

Proof If \œgß then \ is Baire, so we assume \ÁgÞ Let ∞ , where is nowhere dense. We must show that is dense in Eœ5œ" R55 R \E \B−\Þ. Suppose !! The closed balls of the form ÖBÀ.ÐBßBÑŸ×% form a neighborhood base at BJ!! . Let be such a closed ball, centered at B ! with radius % !Þ We will be done if we can show J∩Ð\EÑÁgÞ! We do that by using the Cantor Intersection Theorem. Since intJÁg!!" (it contains B ) and since R is nowhere dense, we know that intJR!" is not a subset of cl . Therefore we can choose a point B " in the open set intJ!" cl R . Pick a closed ball J " , centered at B " , so that B−J© ""!"! int JR©J cl . " If necessary, choose JÐJÑÞ"" even smaller so that diam # (This is the first step of an inductive construction. We could now move to the induction step, but actually include “step two” to be sure the process is clear. ) Since intJÁg""# (it contains B ) and since R is nowhere dense, we know that intJR"# is not a subset of cl . Therefore we can choose a point B # in the open set intJ"# cl R . Pick a closed ball J # , centered at B # , so that B−J© ##"#" int JR©J cl . " If necessary, choose JÐJÑÞ## even smaller so that diam $

For the induction step: suppose we have defined closed balls J"# ß J ß ÞÞÞß J 5 centered at points " B" ß B # ß ÞÞÞß B 5, with diam ÐJ 3 Ñ 3" and so that B 3 − J 3 © int J 3"  cl R 3 © J 3" . Since intJÁg555" (it contains B ) and since R is nowhere dense, we know that int J 5 is not a subset of clRBJR5" . Therefore we can choose a point 5" in the open set int 5 cl 5" . Pick a closed ball JBB−J©JR©J5" , centered at 5" , so that 5" 5" int 5 cl 5" 5 . If necessary, " choose JÐJÑ5" even smaller so that diam 5" 5# .

182 By induction, the closed sets J5ÐJÑÄ!55 are defined for all , diam and J!" ª J ª ÞÞÞ ª J 5 ª ÞÞÞ Þ Since Ð\ß .Ñ is complete, the Cantor Intersection Theorem says that there is a point ∞ For each , int cl , so cl , so B−5œ" J5! ©JÞ 5 " B−J 5 © J 5"5  R B R 5 Therefore ∞ , so and we are done. BÂR55! Þ B−\5œ" R œ\E B−J ∩Ð\EÑ ñ

Example 6.13

1) Now we can see another reason why  is not completely metrizable. Suppose . w is a metric on  that is equivalent to the usual metric .Ðß.Ñ and for which w is complete. By the Baire Category Theorem, Ðß.Ñ w must be second category in itself. But this is false since the space is topologically the same as the usual space of rationals.

2) Suppose Ð\ß .Ñ is a nonempty complete metric space without isolated points. We proved in Theorem 3.6 that l\l - . We can now that, in addition, that each point of \ must be a condensation point. Otherwise, there would be a non-condensation point B−\! , and there would be some countable open set S œ ÖB!"# ß B ß B ß ÞÞÞB 8 ß ÞÞÞ×Þ Since each B 8 is non-isolated, the singleton sets ÖB8 ×are nowhere dense in Ð\ß.Ñ , so S is first category in Ð\ß.Ñ . Since Ð\ß.Ñ is Baire, this would mean that the closed set \S is dense  which is impossible. Notice that this example illustrates again that  cannot be completely metrizable: if its topology were produced by a some complete metric, then each point in  would have to be a condensation point.

3) The set ‘‘‘ is a second category set in : if not, we could write œ∪ , so that would be a first category set in ‘  which contradicts the Baire Category Theorem. (Is  second category in  ?Ñ

4) Recall that E©Ð\ßg Ñ is called an J5 set if E can be written as a countable union of closed sets, and that EK is a $ set if it can be written countable intersection of open sets. The complement of a KJ$5 set is an set and vice-versa. is not an set in . To see this, suppose that ∞ , where the 's are ‘JœJJ5 8œ" 88 closed in ‘ . Since is second category in ‘ , one of these J8 's must be not nowhere dense in ‘ . This JÐ+ß,ÑÐ+ß,Ñ©J©8 must therefore contain an open interval . But then 8  , which is impossible because there are also rational numbers in Ð+ß ,ÑÞ Taking complements, we see that ‘ is not a K$ -set in .

183 Example 6.14 Suppose 0ÀÒ!ß"ÓÄ‘ is continuous. By the Fundamental Theorem of Calculus, 000 has an antiderivative "" . Since is differentiable (therefore continuous), it has an antiderivative 000#88" . Continuing in this way, let denote an antiderivative of . It is a trivial observation that: Ðb5 aB 05 ÐBÑ œ !Ñ Ê 0ÐBÑ œ ! for all B − Ò!ß "Ó

We will use the Baire Category Theorem to prove a less obvious fact:

ÐaB b5 05 ÐBÑ œ !Ñ Ê 0ÐBÑ œ ! for all B in Ò!ß "Ó (*)

In other words, if 0ÐBÑ is not identically 0 on Ò!ß "Ó , then there must exist a point B! − Ò!ß "Ó such that every antiderivative 0ÐBÑÁ!Þ5!

∞ By hypothesis, we can write Ò!ß"Óœ5œ" J55 , where J œÖB−Ò!ß"ÓÀ0ÐBÑœ!× 5 "  œ 05 ÒÖ!×Ó. Since each 055 is continuous, J is closed in Ò!ß "Ó . Since Ò!ß "Ó is second category in itself, one of the sets JÐ+ß,Ñ5! is not nowhere dense and therefore must contain an interval .

Since 0!Ð+ß,Ñ0!Ð+ß,ÑÞ5! is identically on , is identically on

We can repeat the same argument on any closed subinterval N œ Ò-ß .Ó © Ò!ß "Ó À letting 1œ0lN and 155 œ0 lN , we can conclude that N contains an open interval on which 0 is identically !NœÒ-ß.ÓB . In particular , each closed subinterval contains a point at which 0ÐBÑ œ !. Therefore ÖB À 0ÐBÑ œ !× is dense in Ò!ß "Ó . Since 0 is continuous and is ! on a dense set, 0! must be everywhere in Ò!ß"ÓÐ .See Theorem II.5.12 and Exercise III.E.16. )

Example 6.15 (The Banach-Mazur Game) The equipment for the game consists of two disjoint sets Eß F for which E ∪ F œ Ò!ß "ÓÞ Set E belongs to Andy and set F belongs to Beth. Andy gets the first “move” and selects a closed interval M©Ò!ß"Ó" . Beth then chooses a closed interval " M©M#", where M # has length Ÿ# Þ Andy then selects a closed interval M©M$# , where M $ has " length Ÿ $ . They continue back-and-forth in this way forever. (Of course, to finish in a finite time they must make their choices faster and faster.) When all is done, they look at ∞ If , Andy wins; if Beth wins. We claim that if is first category, 8œ"M8 œ ÖB×Þ B − E B − Fß E then Beth can always win.

Suppose ∞ , where is nowhere dense in Let Andy's th choice be E œ8œ" R55 E Ò!ß "ÓÞ 5 N Ð œ M#5" Ñ. Of course, N ∩ R 5 is nowhere dense in Ò!ß "Ó ; but actually N ∩ R 5 is also nowhere dense in the interval N :

If intNN clÐN ∩ R 5 Ñ Á g , then cl N ÐN ∩ R 5 Ñ must contain an nonempty open intervalÐ+ß ,ÑÞ But then Ð+ß ,Ñ © clN5N55 ÐN ∩ R Ñ © cl R œ clÒ!ß"Ó R ∩ N © cl Ò!ß"Ó R 5 , which contradicts the fact that R5 is nowhere dense in Ò!ß "Ó .

>2 Then the open set intN clN5 ÐN∩RÑ is nonempty and Beth can make her 5 choice M #5 to be a closed interval Ò+ß,Ó© int N  clN ÐN∩R 5 Ñ . This implies that M #5 ∩R 5 œg , since

M#5 © int N cl N ÐN∩RÑ© 5 int NÐN∩RÑ©NÐN∩RÑœNR 5 5 5

Therefore ∞∞ and Beth wins! 5œ"M∩55 5œ" Rœg

184 Example 6.16 The Baire Category Theorem can also be used to prove the existence of a continuous function on Ò!ß "Ó that is nowhere differentiable. The details can be found, for example, in Willard's General Topology . In outline, one looks at the space G‡ ÐÒ!ß "ÓÑ with the uniform metric 3 , and argues that the set R of functions which have a derivative at one or more points is a (“thin”) first category set in G‡‡ ÐÒ!ß "ÓÑÞ But the complete space ÐG ÐÒ!ß "ÓÑß3 Ñ is a (“thick”) Baire space. Therefore it is second category in itself so G‡ ÐÒ!ß "ÓÑ  R Á g . In fact, G‡‡ ÐÒ!ß "ÓÑ  R is dense in G ÐÒ!ß "ÓÑÞ Any function in G ‡ ÐÒ!ß "ÓÑ  R does the trick.

7. Complete Metrizability

Which metrizable spaces Ð\ß .Ñ are completely metrizable  that is, when does there exist a metric .\www on with .µ. and with Ð\ß.Ñ complete? We have already seen that  is not completely metrizable (using the Baire Category Theorem), and that certain familiar spaces like " ÖÀ8−×8  are completely metrizable. In this section, we will give an answer to this question.

Lemma 7.1 If Ð\ß .Ñ is a pseudometric space and 1 À \ Ä‘ continuous and ga ( ) Á 0. Then 1 g À\Ä‘ is continuous at a .

Proof The proof is a perfect “mimic” of the standard proof in advanced calculus of the same result for \œ‘ .

Our first theorem tells us that certain subspaces of complete spaces are completely metrizable.

Theorem 7.2 If Ð\ß .Ñ is complete and S is open in \ , then there is a metric d w µ. on S such that ÐSOd ,wÑ is complete that is, is completely metrizable.

If S. is not already complete with metric , it is because there are some nonconvergent Cauchy sequences in S\ . However those Cauchy sequences do have limits in they converge to points outside SS but in Fr . The idea of the proof is to create a new metric .S.w on equivalent of but which “blows up” near the boundary of S : in other words, the new metric destroys the “Cauchyness” of the nonconvergent Cauchy sequences.

Here is a concrete (but slightly simpler) example to illustrate the idea.

11 1 " Let . be the usual metric on the interval N œ Ð ## ß ÑÞ The sequence ÐB8 Ñ œ Ð #  8 Ñ is a nonconvergent Cauchy sequence in NÞ The function tan À N Ä‘ is a homeomorphism for which Ð0ÐB8 ÑÑ Ä ∞Þ

.ÐBßCÑœlw tan B tan Cl defines a new metric on N , and tan ÀÐNß.ÑÄÐw ‘ ß.Ñ is an isometry because .w ÐBß CÑ œ œ l tan B  tan Cl œ .Ð tan Bß tan CÑÞ Since Б ß .Ñ is w w complete, so in ÐN ß . ÑÞ ÐThe sequence ÐB8 Ñ is still nonconvergent in ÐN ß . Ñ because ww .µ.àbut ÐBÑ8 is no longer a Cauchy sequence in ÐNß.ÑÑ

In the proof of Theorem 7.2, we will not have a homeomorphism like “tan” to use. Instead, we define a continuous 0ÀÐSß.ÑÄ‘ and use 0 to create a new metric .w on S that destroys Cauchy sequences ÐB8 Ñ when they approach Fr SÞ

185 " Proof of 7.2 Define 0ÐBÑ œ.ÐBß \  SÑ . Since S is open, the denominator is never ! for B − S so 0 À S Ä‘ is continuous by Lemma 7.1. On S , define .w ÐBß CÑ œ .ÐBß CÑ  l0ÐBÑ  0ÐCÑlÞ It is easy to check that .Sw is a metric on .

Since sequences are sufficient to determine the topology in metric spaces, we show that .µ.w on SSÐBÑ by showing that they produce the same convergent sequences in . Suppose 8 is a sequence in SB−SÀ and

ww If .ÐBßBÑÄ!88 , then since . . we get .ÐBßBÑÄ! .

w . If . ÐB88 ß BÑ Ä ! , then l0ÐB Ñ  0ÐBÑl Ä ! (by continuity) so . ÐB 8 ß BÑ Ä ! .

www ÐSß . Ñ is complete À suppose ÐB88 Ñ is a . -Cauchy sequence in S . Since . .ß ÐB Ñ is also . .ÐBÑÄB−\-Cauchy so 8 some . But we claim that this BSÞ must be in

If B−\Sß then .ÐBß\SÑÄ!88 and so 0ÐBÑÄ∞Þ In particular, this means that for every 8 we can find 78 for which 0ÐB78 Ñ0ÐBÑ"Þ Then w l0ÐB78 Ñ  0ÐB Ñl  " , so . ÐB 78 ß B Ñ  " . This contradicts the assumption that ÐB 8 Ñ is . w-Cauchy. w ..ww Therefore ÐB88 Ñ Ä B − S . Since . µ . on S , ÐB Ñ Ä B − S and so ÐSß . Ñ is complete. ñ

The next theorem generalizes this result. In fact, the interesting “idea” is in Theorem 7.2. The proof of Theorem 7.3 just uses Theorem 7.2 repeatedly to “patch together” a more general result.

Theorem 7.3 (Alexandroff) If Ð\ß .Ñ is complete and E is a K$ in \ , then there is a metric 3 98 A, with 33µ. on E and such that ÐEß Ñ is complete. In other words: a K$ subset of a complete space is completely metrizable.

∞ w Proof Let Eœ S33 , where S is open in \ . Let .3 be a metric on S 3 , equivalent to . on S 3 , 3œ" www such that ÐS33 , .333 Ñ is complete. Let . ÐBß CÑ œ min Ö. ÐBß CÑß "× . Then . 3 µ . µ . on S , and w ÐS33 ß . Ñ is complete. (Note that ..3 and 3 are identical for distances smaller than one: this is why they produce the same convergent sequences and the same Cauchy sequences. )

∞ .ÐBßCÑ3 For Bß C − E , define 3 ÐBß CÑ œ23 (the series converges since all .3 ÐBß CÑ Ÿ " ). 3œ" 3 is a metric on E :

Exercise

3 µ. on E :

We show that 3 and .EÞ produce the same convergent sequences in

. Suppose Ð+83 Ñ Ä + − E © S . Let %  !.

186 ∞ 1 % Choose 53.µ.S so that 224 . For every , 33 on , so 4œ5" .3 Ð+Ñ83 Ä +−S . Therefore, for 3œ"ß ... , 5 , we can pick R 3 so that 8R 3 .Ð+ß+Ñ38 % Ê 223 5 . Then if 8RœRßR× max {"5 ..., , we get 5∞ .Ð+ß+Ñ38 .Ð+ß+Ñ 38 %% 3 3%Ð+8 , +Ñ œ 222233  Ÿ 5 †5 + œ , so Ð+8 Ñ Ä + − E . 3œ"3œ5" 3 Conversely, if Ð+8 Ñ Ä + − E , then for %  ! , we can pick R so that ∞ .Ð+ß+Ñ38 % 8RÊ3% Ð+ß+Ñœ8"8 223  Ê .Ð+ß+Ñ , so 3œ" ." . Ð+8"""8 Ñ Ä + − S Þ But . µ . on S , so Ð+ Ñ Ä + .

ÐEß3 Ñ is complete:

Let Ð+8 Ñ be 3% -Cauchy in E . Let  ! and let 3 − . Choose R3 so that

% if 7ß8R387 , then 3% Ð+ß+Ñ23 , and therefore .Ð++Ñ 387, .

Therefore Ð+833 Ñ is Cauchy in the complete space ÐS ß . Ñ and there is a point .3 . B−S33 such that Ð+ÑÄB−S 8 33 . But .µ. 3 on S 3 , so Ð+ÑÄB−S©\ 8 33 .

This is true for each 3Ð+Ñ . But 8 can have only one limit +−Ð\ß.Ñ , so we conclude that BœBœ"# ... œBœ 3 ... œ+Þ Since +−S 3 for each 3 , we . 3 have +−E , so ( +88 ÑÄ+−E . But .µ3 on E , so ( + ÑÄ+−E . Therefore ÐEß3 Ñ is complete. ñ

Corollary 7.4  is completely metrizable (and therefore is a Baire space, so is second category in ).

Proof , so is an set in . Taking complements, we get that ‘œÖ;×J;− 5 is a set in . ‘œÐÖ;×ÑK;− $ ‘ ñ

In fact, a sort of “converse” to Alexandroff's Theorem is also true.

Theorem 7.5 For any metric space Ð\ß .Ñ , the following are equivalent:

i) Ð\ß .Ñ is completely metrizable w ii) Ð\ß .Ñ is homeomorphic to a G$ set in some complete metric space Ð] ß . Ñ iii) IfÐ^ß .ww Ñ is any metric space and 0ÀÐ\ß.ÑÄÐ 0Ò\Ó ß . ww Ñ © Ð^ß.ww Ñ is a ww homeomorphismß0Ò\ÓK then is a $ set in Ð^ß.Ñ (so we say that “\K is an absolute set among metric spaces” ) $µ µ iv) \\.Ñ is a K$ set in the completion Ð , .

Various parts of Theorem 7.5 are due to Mazurkiewicz (1916) & Alexandroff (1924).

187 The proof of one of the implications in Theorem 7.5 requires a technical result whose proof we will omit. (See, for example Willard, General Topology )

Theorem 7.6 (Lavrentiev) Suppose Ð\ß .Ñ and Ð] ß .w Ñ are complete metric spaces with E © \ w and F©] . Let 2 be a homeomorphism from E onto F . Then there exist K$ sets E in \ and F]ww in with E©E©EF©F©F cl and w cl and there exists a homeomorphism 2EFwww from onto such that 2lEœ2Þ w (Loosely stated: a homeomorphism between subsets of two complete metric spaces can always be “extended” to a homeomorphism between K$ sets.)

Assuming Lavrentiev's Theorem, we now prove Theorem 7.5.

Proof i)ÊÐ\ß.Ñ ii) Suppose is completely metrizable and let .w be a complete metric on \ equivalent to . . Then the identity map 3 À Ð\ß.Ñ Ä Ð\ß.w Ñ is a homeomorphism and 3Ò\Ó œ \ w is certainly a KÐ\ß.ÑÞ$ set in

ii)Ê iii) Suppose we have a homeomorphism 1ÀÐ\ß.ÑÄ1Ò\ÓœE , where E is a K$ µ µ ww set in a complete space Ð] ß .www ÑÞ Let 0 À Ð\ß .Ñ Ä Ð^ß . Ñ © the completion Ð ^ , . Ñ be a homeomorphism (into ^Fœ0Ò\ÓK^ ). We want to prove that must be a $ set in .

We have that 2œ01" ÀEÄF is a homeomorphism. Using Lavrentiev's Theorem, we get an extension of 2 to a homeomorphism 2ww ÀE ÄF w where E ww , F are K sets with E©E w ©] µ $ and F©Fw ©^ (see the figure ).

Since is a in there are open sets in such that ∞ EK]ß$ S]88 EœS8œ" ww∞∞ . Therefore is also a in wwww But is a œE ∩8œ" S88 œ 8œ" ÐS∩EÑ E K$ EÞ 2 ÀE ÄF homeomorphism, so ww is a set in . Therefore ∞ , where is 2ÒEÓœ2ÒEÓœF K$ F Fœ8œ" Z88 Z ww µ open in FZœF∩[[^ and, in turn, 888 where is open in . µµ Also, ww is a in , so ∞ , where the 's are open in . FK^FœY$ 8œ" 88 Y ^

188 ∞∞ w Putting all this together, Fœ8œ" Z88 œ 8œ" Ð[ ∩FÑ ∞∞∞w  œÐ [888 Ñ∩F œÐ [ Ñ∩Ð Y Ñ. The last expression shows that F is a K$ set in µ 8œ" 8œ" 8œ" . But then ∞∞∞ ∞ ^ Fœ^∩Fœ^∩Ð8œ" [88 Ñ∩Ð 8œ" Y ќР8œ" [ 8 ∩^Ñ∩Ð  8œ" Y 8 ∩^Ñ is a K^$ set in .

iii)Ê\Ð^ß.Ñ iv) iii) states that any “homeomorphic copy” of in a metric space ww must ww µ µ be a K$ in ^ . Letting Ð^ß . Ñ œ Ð\ , . Ñ , it follows that \ must be a K$ set in the completion µ µ Ð\,. . Ñ

iv)ÊK i) This follows from Alexandroff's Theorem 7.3: a $ set in a complete space is completely metrizable. ñ

Theorem 7.5 raises the questions: what spaces are “absolutely open”? what spaces are “absolutely closed?” Of course in each case a satisfactory answer might involve some kind of qualification. For example, “among Hausdorff spaces, a space \ is absolutely closed iff ... ”

Example 7.7 Since ‘ is not a K$ in (an earlier consequence of the Baire Category Theorem), Theorem 7.5 gives us yet another reason why  is not completely metrizable.

Example 7.8 Since  is completely metrizable, it follows that is a Baire space. If . is the usual metric on  , then Ðß.Ñ is an example of a Baire metric space that isn't complete.

To finish this section on category, we mention, without proof and just as a curiosity, a generalization of an earlier result (Blumberg's Theorem: see Example II.5.8 ).

Theorem 7.9 Suppose Ð\ß .Ñ is a Baire metric space. For every 0 À \ Ä ‘ , there exists a dense subset H of \ such that 0lHÀHÄ‘ is continuous,

The original theorem of this type was proved for \œ‘‘ or## Þ \œ ‘ Þ (Blumberg, New properties of all real-valued functions, Transactions of the American Mathematical Society, 24(1922) 113-128).

The more general result stated in Theorem 7.9 is (essentially) due to Bradford and Goffman (Metric spaces in which Blumberg's Theorem holds , Proceedings of the American Mathematical Society 11(1960), 667-670)

189 Exercises

E11. In a pseudometric space Ð\ß .Ñ , every closed set is a K$5 set and every open set is an J set (see Exercise II.E.15 ).

a) Find a space Ð\ßg Ñ which contains a closed set J that is not a K$ set.

b) Recall that the “scattered line” is the space Ðß‘g Ñ where

g‘œÖY∪Z ÀY is a usual open set in and Z © ×

Prove that the scattered line is not metrizable. (Hint: KJ$5 or sets are relevant. )

E12. a) Prove that if O is open in Ð\ßg Ñ , then Fr ÐSÑ is nowhere dense.

b) Suppose that D is a discrete subspace of the Hausdorff space Ð\ßg Ñ and that \ has no isolated points. Prove that H\ is nowhere dense in .

E13. Let gg be the confinite topology. Prove that Ð\ß Ñ is a Baire space if and only if \ is either finite or uncountable.

E14. Suppose . is any metric on  which is equivalent to the usual metric on . Prove that the ~ ~ completion Ðß.Ñ is uncountable.

E15. Suppose that Ð\ß .Ñ is a nonempty complete metric space and that Y is a family of continuous functions from \ to ‘ with the following property:

aB − \ß b a constant QBB such that l0ÐBÑl Ÿ Q for all 0 − Y

Prove that there exists a nonempty open set YQ and a constant (independent of B ) such that l0ÐBÑl Ÿ Q for all B − Y and all 0 − Y .

This result is called the Uniform Boundedness Principle.

Hint: Let I5 œÖB−\Àl0ÐBÑlŸ5 for all 0−Y × . Use the Baire Category Theorem.

190 E16. Suppose 0À Ð\ßg Ñ Ä Ð] ß .Ñ , where Ð] ß .Ñ is a metric space where . is a metric. For each B−\, define

=0 ÐBÑ œ “the oscillation of 0 at B ” œ inf {diam Ð0ÒRÓÑ À R is a neighborhood of B×

a) Prove that 0+Ð+Ñœ! is continuous at if and only if =0 .

1 b) Prove that for 8−= , ÖB−\À0 ÐBÑ8 × is open in \ .

c) Prove that ÖB − \ À 0 is continuous at B× is a K$ -set in \ .

Note: Since ‘ is not aK0ÀÄ$ -set in , a continuous function ‘‘ cannot have as its set of points of continuity. On the other hand, ‘ is a K$ -set in and, from analysis, you should probably know an example of a continuous function 0À‘‘ Ä that is continuous at each :−  and discontinuous at each point ;− .

d) Prove that there cannot exist a function 0À‘ ÄÐ!ß∞Ñ such that for all B− and all C −, 0ÐBÑ0ÐCÑ Ÿ ± BC ± Þ Hint for d): Suppose 0 exists.

i) First prove that if (B8 ) is a sequence of rationals converging to an irrational, then Ð0ÐB88 ÑÑ Ä ! and, likewise, if ÐC Ñ is a sequence of irrationals converging to a rational, then Ð0ÐC8 ÑÑ Ä ! .

ii) Define a new function 1À‘‘ Ä by 1±  œ! and 1ÐCÑœ0ÐCÑ for C irrational. Examine the set of points where 1 is continuous. )

E17. There is no continuous function 0À‘‘ Ä for which the set of points of continuity is  . The reason (see problem E16 ) ultimately depends on the Baire Category Theorem. Find the error in the following more elementary “proof”:

Suppose 0À‘‘ Ä is continuous and that 0 is continuous at B iff B−  Þ Then ‘=œÖB− À ÐBÑœ!ל∞ Y, where Y œÖB− ‘= À ÐBÑ" ×Þ 08808œ" 8

Each Yª88 , and Y can be written as a countable union of disjoint open intervals. If Ð+ß ,Ñ and Ð-ß .Ñ are consecutive intervals in Y8 , then , œ -  or else there would be a rational in Ð,ß -Ñ . Therefore ‘  Y8 consists only of the endpoints of some disjoint open intervals. Therefore ‘ Y8 is countable.

It follows that is not continuous at ∞∞ is ÖB −‘‘‘ À 0 B× œ 8œ" Y88 œ 8œ"  Y countable. Therefore 0 is continuous on more than just the points in  .

E18. Suppose that for each irrational :X , we construct an equilateral triangle : (including its interior) in ‘# with one vertex at Ð:ß !Ñ and its opposite side above and parallel to the B -axis. Prove that ÖX À : − × must contain an “open box” of the form Ð+ß ,Ñ ‚ Ð!ß1 Ñ for some  : 5 +ß , −‘ and some 5 − . In the hypothesis, we could weaken “equilateral” to read “ ... ” ? " ()Hint: Consider RœÖ:−ÀX5: has height 5 × .

191 E19. Suppose 0À‘‘ Ä . Prove that if 0 is discontinuous at every irrational :−  , then there must exist an interval MœÐ+ß,Ñ such that 0 is discontinuous at every point in MÞ

E20. Ð\ßggg Ñ is “absolutely open” if whenever 0 À Ð\ß Ñ Ä Ð] ßw Ñ is a homeomorphism (into), then 0Ò\Ó must be open in ] . Find all absolutely open spaces.

192 8. Compactness

Compactness is one of the most important topological properties. Formally, it is a strengthening of the Lindelof¨ property.

Compact spaces are often particularly simple to work with because of the “rule of thumb” that “compact spaces often act like finite spaces.”

Definition 8.1 A topological space Ð\ßg Ñ is called compact if every open cover of \ has a finite subcover.

Example 8.2

1) A finite space is compact. 2) Any set \ with the cofinite topology is compact because any nonempty open set, alone, covers \ except for perhaps finitely many points. 3) An infinite discrete space is not compact because the cover consisting of all " singleton sets has no finite subcover. In particular, ÖÀ8−×8  is not compact. " 4) The space \ œ Ö!× ∪ Ö8 À 8 −h × is compact: if is any open cover, and !−Y −h, then the set Y covers \ except for perhaps finitely many points. 5) ‘h is not compact since œ ÖÐ  8ß 8Ñ À 8 − × has no finite subcover.

Definition 8.3 A family Y of sets has the finite intersection property (FIP) if every finite subfamily of Y has nonempty intersection.

Sometimes it is useful to have a characterization of compact spaces in terms of closed sets, and we can do this using FIP. As you read the proof, you should see that Theorem 8.4 is hardly any more than a reformulation, using complements, of the definition of compactness in terms of closed sets.

Theorem 8.4 Ð\ßgY Ñ is compact iff whenever is a family of closed sets with the finite intersection property, then .  Y Ág Proof Suppose \ is compact and that Y is a family of closed sets with FIP. Let . For any , the FIP tells us that 8 hYœ Ö\  J À J − × J"# ß J ß ÞÞÞß J 8 − Y \ Á \ 3œ" J 8 œÐ\JÑ∪ÞÞÞ∪Ð\J"8 ÑÞIn other words, no finite subcollection of h covers \ . Therefore itself cannot cover , that is, . hhhY\\œÖ\YÀY−לÁg  The proof of the converse is similar. ñ

Theorem 8.5 For any spaceÐ\ ,g Ñ

a) If \J\J is compact and is closed in , then is compact. b) If \O is a Hausdorff space and is a compact subset, then O\ is closed in .

Proof a) The proof is like the proof that a closed subspace of a Lindelof¨¨ space is Lindelof. (Theorem III.3.10) Suppose J\œÖYÀ−E× is a closed set in a compact space , and let hαα be a cover of JZ\YœZ∩JÞ by sets open in J . For each α , pick an open set ααα in such that Since J is closed, the collection iα œ Ö\  J × ∪ ÖZα À − E× is an open cover of \ .

193 Therefore we can find Zαα"8 ß ÞÞÞß Z so that Ö\  J ß Z αα "8 ß ÞÞÞß Z × covers \ . Then ÖY αα "8 ß ÞÞÞY × covers JJ , so is compact.

Note: the definition of compactness requires that we look at a cover of J by sets open in J , but it should be clear that it is equivalent to look at a cover of JÞ using sets open in \

b) Suppose O\C−\OÞ is a compact set in a Hausdorff space . Let For each B−O, we can choose disjoint open sets YBB and Z in \ with B−Y B and C−Z B . Since h œ ÖYB"8BB À B − O× covers Oß there are finitely many points B ß ÞÞÞß B − O so that Y"8 ß ÞÞÞß Y covers O . Then C−Z œZBB"8 ∩ÞÞÞ∩Z ©\O . Therefore \O is open, so O is closed. ñ

Notes 1) Reread the proof of part b) assuming O is a finite set. This highlights how “compactness” has been used in place of “finiteness” and illustrates the rule of thumb that compact spaces often behave like finite spaces. 2) The “Hausdorff” hypothesis cannot be omitted in part b). For example, let \ be a set with l\l  " and let . be the trivial pseudometric on \ . Then each singleton set ÖB× is compact but not closed. 3) Compactness is a clearly a topological property, so part b) implies that if a compact space \O]O] is homeomorphic to a subspace in a Hausdorff space , then is closed in . So we can say that a compact space is “absolutely closed” among Hausdorff spaces “it's closed wherever you put it.” In fact, the converse is also true among Hausdorff spaces a space which is absolutely closed is compact but we do not have the machinery to prove that now.

Corollary 8.6 A compact metric space Ð\ß .Ñ is complete. µ µ Proof Ð\ß .Ñ is a dense subspace of its completion Ð \ , . Ñ . But Ð\ß .Ñ is compact, so \ must µµµ be closed in \Ð\ß.Ñœ\.ÑÐ\ß.Ññ . Therefore Ð , , so is complete. (Note: this proof doesn't work for pseudometric spaces (why?). But is Corollary 8.6 still true for pseudometric spaces? Would a proof using the Cantor Intersection Theorem work in that case?)

You may have seen a different definition of compactness in some other book. In fact there are several different “kinds of compactness” and, in general they are not equivalent . But we will see that they are equivalent in certain spacesÞ for example, in ‘8

Definition 8.7 A topological space Ð\ßg Ñ is called

sequentially compact if every sequence in \\ has a convergent subsequence in countably compact if every countable open cover of \ has a finite subcover (therefore “Lindelof¨ + countably compactœ compact” ) pseudocompact if every continuous 0À\Ä‘ is bounded (You should check that this is equivalent to saying that every continuous real -valued function on \ assumes both a maximum and a minimum value).

We want to look at the relations between these “compactness-like” properties.

194 Lemma 8.8 If every sequence in Ð\ßg Ñ has a cluster point, then every infinite set in \ has a limit point. The converse is true if Ð\ßg Ñ is a X" -space  that is, if all singleton sets ÖB× are closed in \ .

Proof Suppose E\ is an infinite set in . Choose a sequence Ð+ÑE8 of distinct terms in and let BÐ+ÑÞRB+ß be a cluster point of 8 Then every neighborhood of contains infinitely many 8 's so R∩ÐEÖB×ÑÁgÞ Therefore B is a limit point of E . Suppose \X is a " -space in which every infinite set has a limit point. Let ÐBÑ8 be a sequence in\ÐBÑ . We want to show that 8 has a cluster point. Without loss of generality, we may assume that the terms of the sequence are distinct (why?), so that EœÖB8 À8− × is infiniteÞ Let B be a limit point of the set E . We claim B is a cluster point for ÐB8 ÑÞ Suppose Y is an open set containing B and 8 − . Let Z œ Y  ÖB"8 ß ÞÞÞß B × ∪ ÖB×Þ Since ÖB"8 ß ÞÞÞß B × is closed, Z is open and B − Z © Y Þ Then Z ∩ ÐE  ÖB×Ñ Á g , so Y∩ÐEÖB×ÑÁgÞTherefore Y contains a term B5 for some 58ß so B is a cluster point of ÐB8 ÑÞ ñ

Example 8.9 The “X" ” hypothesis in the second part of Lemma 8.8 cannot be dropped. Let " " \œÖÀ8−×∪Ö:À8−×8 88 where the : 's are distinct points and :Á 85 for all 8ß5−Þ  " The idea is to make .Ð:88 ß B Ñ œ ! for each 8 so that : 8 is a sort of “double” for 8 . So we define

.Ð:88 ß B Ñ œ !  "" " " .Ð ß Ñœl  lœ.Ð:ß:87 Ñ if 7Á8  87 8 7 .Ð: ß""" Ñ œ l  l if 7 Á 8  8 787 It is easy to check that .\Ð\ß.ÑE is a pseudometric on . In , every nonempty set (finite or " infinite) has a limit pointB−E if , then its “double” is a limit point of E . But the sequence ÐÑ8 has no cluster point in Ð\ß .ÑÞ

Theorem 8.10 Ð\ßg Ñ is countably compact iff every sequence in \ has a cluster point (Then by Lemma 8.8, a XÐ\ßÑ" -space g is countably compact iff every infinite set in \ has a limit point).

Proof Suppose Ð\ßg Ñ is countably compact and consider any sequence ÐB88 Ñ . Let X œ “the >2 tail of ” We claim that ∞ cl 8 ÐBÑ85 œÖB À5 8×Þ8œ" X 8 ÁgÞ

Assume not. Then every B is in Ð\ cl X88 Ñ for some 8 , so Ö\ cl X À8− × is an open cover of \\ . Since is countably compact, there exists an R such that \ œ Ð\  cl X"R Ñ ∪ ÞÞÞ ∪ Ð\  cl X Ñ . Taking complements gives R cl cl , which is impossible. gœ8œ" X8R œ X Let ∞ cl and let be a neighborhood of . Then for every , so B−8œ" X88 R B R∩X Ág 8 R contains an B8BÐBÑÞ5 for arbitrarily large values of . Therefore is a cluster point of 8

Conversely, suppose \\ is not countably compact. Then has a countable open cover with no finite subcover. For each , pick a point 8 . Then the ÖY"8 ß ÞÞÞß Y ß ÞÞÞ× 8 B83 − \ 3œ" Y sequence ÐB8 Ñ has no cluster point: for any B − \ , we know B is in some Y88 . Y is a neighborhood of BBÂY78ñ and 78 for .

195 The next theorem tells us some connections between the different types of compactness.

Theorem 8.11 In any space Ð\ßg Ñ , the following implications hold:

 \ is compact or Ê\ is countably compact Ê\ is pseudocompact  \ is sequentially compact  Proof It is clear from the definitions that a compact space is countably compact.

Suppose \ÐBÑ\ is sequentially compact. Then each sequence8 in has a subsequence that converges to some point B−\ . Then B is a cluster point of ÐBÑ8 . From Lemma 8.8 we conclude that \ is countably compact.

Suppose \0À\Ä is countably compact and that ‘ is continuous. The sets " Y8 œ 0 ÒÐ  8ß 8ÑÓ œ ÖB − \ À  8  0ÐBÑ  8× form a countable open cover of \ . So Y"R ß ÞÞÞß Y cover \ for some R . Since Y "# © Y © ÞÞÞ © Y R , this implies that Y R œ \ . So R 0ÐBÑR for all B−\ that is, 0 is bounded. Therefore \ is pseudocompact. ñ

In general, no other implications hold in among these four types of compactness, but we do not have the machinery to provide counterexamples now. ( See Corollary 8.5 in Chapter VIII and Example 6.5 in Chapter X ). We will prove, however, that they are all equivalent in a pseudometric space Ð\ß .ÑÞ Much of the proof is developed in the following sequence of lemmas  some of which have intrinsic interest of their own.

Lemma 8.12 If \\ is first countable, then is sequentially compact iff \ is countably compact. (So, in particular, sequential and countable compactness are equivalent in pseudometric spaces.)

Proof We know from Theorem 8.11 that if \\ is sequentially compact, then is countably compact. So assume \ÐBÑ\ is countably compact and that 8 is a sequence in . By Theorem 8.10, we know ÐB8 Ñ has a cluster point, B . Since \ is first countable, there is a subsequenceÐB85 Ñ Ä B (see Theorem III.10.6 ). Therefore \ is sequentially compact. ñ

Definition 8.13 A pseudometric space Ð\ß .Ñ is called totally bounded if ß for each %  !ß \ can be covered by a finite number of %% -balls. If \Ág , this is the same as saying that for ! , there exist B"# ß B ß ÞÞÞß B 8 − \ such that \ œ F%% ÐB " Ñ ∪ ÞÞÞ ∪ F ÐB 8 Ñ .

More informally, a totally bounded pseudometric space is one that can be kept under full surveillance using a finite number of policemen with an arbitrary degree of nearsightedness.

Example 8.14

1) Neither ‘ nor (with the usual metric) is totally bounded since neither can be covered by a finite number of " -balls.

196 2) A compact pseudometric space Ð\ß .Ñ is totally bounded: for any %  ! , we can pick a finite subcover from the open cover h œÖFÐBÑÀB−\×Þ%

Lemma 8.15 If Ð\ß .Ñ is countably compact, then Ð\ß .Ñ is totally bounded.

Proof If \œg , \ is totally bounded, so assume \ÁgÞ If Ð\ß.Ñ is not totally bounded, then for some %%! , no finite collection of -balls can cover \ . Choose any point B" −\ . Then we can choose a point B.ÐBßBÑ FÐBÑ\#"#" so that % (or else % would cover ). We continue inductively. Suppose we have chosen points B"# ß B ß ÞÞÞß B 8 in \ such that .ÐB34 ß B Ñ % for each 3 Á 4ß 3ß 4 œ "ß ÞÞÞß 8 . Then we can choose a point B8" at distance % from each of B"8 ß ÞÞÞß B  because otherwise the % -balls centered at B "8 ß ÞÞÞß B would cover \ . The sequence ÐB8 Ñ chosen in this way cannot have a cluster point because, for any B , FÐBÑ% contains at most one B . Therefore \ is not countably compact. ñ # 8

Lemma 8.16 A totally bounded pseudometric space Ð\ß .Ñ is separable.

" Proof For each 8\H , choose a finite number of 8 -balls that cover and let 8 be the (finite) set " consisting of the centers of these balls. For any C−\ ß.ÐBßCÑ8 for some B−H8 Þ This means that ∞ is dense. Since is countable, is separable. Hœ8œ" H8 H \ ñ

Theorem 8.17 In a pseudometric space Ð\ß .Ñ , the properties of compactness, countable compactness, sequential compactness and pseudocompactness are equivalent.

Proof We already have the implications from Theorem 8.11. If Ð\ß .Ñ is countably compact, then Ð\ß .Ñ is totally bounded (see Lemma 8.15 ) and therefore separable (see Lemma 8.16 ). But a separable pseudometric space is Lindelof¨ (see Theorem III.6.5) and a countably compact Lindelof¨ space is compact. Therefore compactness and countable compactness are equivalent in Ð\ß .ÑÞ We observed in Lemma 8.12 that countable compactness and sequential compactness are equivalent in Ð\ß .Ñ . Since a countably compact space is pseudocompact, we complete the proof by showing that if Ð\ß .Ñ is not countably compact, then Ð\ß .Ñ is not pseudocompact. This is the only part of the proof that takes some work. (A bit of the maneuvering in the proof is necessary because . is a pseudometric. If . is actually a metric, some minor simplifications are possible. ) If Ð\ß .Ñ is not countably compact, we can choose a sequence ÐB8 Ñ with no cluster point (see Theorem 8.10 ). In fact, we can choose this ÐB8 Ñ so that all the B8 's are distinct ß and .ÐB78 ß B Ñ  ! if 7 Á 8 Ðwhy? Ñ . We can then find open sets Y 8 such that i) B 8 − Y 8 , ii) Y∩Yœg87 for 7Á8 , and so that iii) diam YÄ! 8 as 8Ä∞Þ

Here is a sketch for finding the Y8 's; the details are left to check as an exercise. First check that for any +ß, −\ and positive reals <"# ß< À if < " < # .Ð+ß,Ñ then

FÐ+Ñ<<"# and FÐ,Ñ are disjoint  in fact, they have disjoint closures.

Since ÐB8 Ñ has no cluster point we can find, for each 8 , a ball Fw ÐB8 Ñ that contains $8 ww no other B787 that is, .ÐBßBÑ $$$88 for all 7Á8 . Let 8 œ Î# . Then for

7 Á 8ßwe have B78887 Â F$8 ÐB Ñ and $  .ÐB ß B ÑÞ Of course, two of these balls

197 might overlap. To get the Y8 's we want, we shrink these balls (choose smaller radii %$88 to replace the ) to eliminate any overlap. We define the % 8 's inductively:

Let %$""œ .ÐBßBÑÞ "# Pick %# ! so that

%%"# .ÐBßBÑ "# and %$##

Since %$""œ .ÐBßBÑ "$ and %$## .ÐBßBÑß #$we can

pick %$ ! so that %%"$ .ÐBßBÑ "$ and %%#$ .ÐBßBÑ #$ and %$$$

th Continue in this way. At the 8FÐBÑ step, we get a new ball %8 8 for which

if 7Á8 , B78 ÂF%8 ÐBÑ (because %$ 88  Ñ and

FÐBÑ∩FÐBÑœg%%8484 for all 48 (because %% 484ß8  .ÐBBÑ

Finally, when we choose %8 at each step, we can also add the condition that " %88ÐFÐBÑÑÄ!8 , so that diam%8 .

Then we can let YœFÐBÑÞ88%8

8.ÐBß\Y8 Ñ For each 80À\Ä0ÐBÑœ , define 88‘ by . Since B−Y\Y 888 and is closed, .ÐB88 ß\ Y Ñ the denominator of 0ÐBÑ8888 is not ! . Therefore 0 is continuous, and 0ÐBÑœ8Þ Define ∞ 0ÐBÑ œ 08 ÐBÑ. (For any B−\ , this series converges because B is in at most one Y8 and 8œ" therefore at most one term 0ÐBÑÁ!8 .) Then 0\0ÐBÑœ0ÐBÑ is unbounded on because 888 œ8. So we are done if we can show that 0 is continuous at each in \Þ

Let +−\ . We claim that there is an open set Z++8 containing + such that Z ∩Y Ág for at most finitely many 8 's.

If .Ð+ßB88 Ñœ! for some B , we can simply let Z +8 œY .

Suppose .Ð+ß B88 Ñ  ! for all 8 . Since + is not a cluster point of ÐB Ñ , we can choose %%! so that FÐ+Ñ% contains none of the B8 's. For this , choose R so that if 8Rß % then diam Y . Then FÐ+Ñ∩Yœg% for 8R (if 8R and D− F% Ð+Ñ∩Y , 88# # # 8 %% then .Ð+ß B Ñ Ÿ .Ð+ß DÑ  .ÐDß B Ñ   œ %). Then let ZœFÐ+ÑÞ% 88## + #

Then 0lZ8+ is identically ! for all but finitely many 8 . Therefore 0lZ+ is really only a finite sum of continuous functions, so 0lZÀZÄ++ ‘ is continuous. Suppose [ is any neighborhood of 0Ð+Ñ. Then there is an open set Y in Z+++ containing + for which Ð0lZ ÑÒY Ó © [ . But Z is open in \ß so Y is also open in \ , and 0ÒYÓ œ Ð0lZ+ ÑÒYÓ © [ . So 0 is continuous at + . ñ

198 We now explore some properties of compact metric spaces.

Theorem 8.18 Suppose EÐ\ß.ÑÞE is a compact subset of a metric space Then is closed and bounded (that is, E has finite diameter).

Proof A compact subset of the Hausdorff space must be closed (see Theorem 8.5b ). If Eœg , then E is certainly bounded. If EÁg , choose a point +! −E and let 0ÀEÄ‘ by 0Ð+Ñœ.Ð+ß+Ñ! . Then 0 is continuous and 0 is bounded (since \ is pseudocompact): say l0ÐBÑl  Q for all B − E . But that means that E © FQ! Ð+ Ñ , so E is a bounded set . ñ

Caution: the converse of Theorem 8.18 is false. For example, suppose . is the discrete unit metric on an infinite set \Ð\ß.Ñ . Every subset of is both closed and bounded, but an infinite subspace of \ is not compact. However, the converse to Theorem 8.18 is true in ‘8 , as you may remember from analysis.

Theorem 8.19 Suppose E©‘8 . The E is compact iff E is closed and bounded.

Proof Because of Theorem 8.18, we only need to prove that if EE is closed and bounded, then is compact. First consider the case 8œ" . Then E is a closed subspace of some interval MœÒ+ß,Ó and it is sufficient to show that M is compact. To do this, let ÐBÑ8 be a sequence in M and choose (see Lemma 2.10 ) a monotone subsequence ÐB8855 Ñ . Since ÐB Ñ is bounded, we know that ÐB885 Ñ Ä some point < − M (see Lemma 2.9 ). Since ÐB Ñ has a convergent subsequence, M is sequentially compact and therefore compact. When 8" , the proof is similar. We illustrate for 8œ 2. If E is a closed bounded set in ‘#, then EMœÒ+ß,Ó‚Ò-ß.Ó is a closed subspace of some closed box of the form . Therefore it is sufficient to prove that MÐBßCÑMÒ+ß,Ó is compact. To do this, choose a sequence 88 in . Since is compact, the sequence ÐB88 Ñ has a subsequence ÐB5 Ñ that converges to some point B − Ò+ß ,Ó .

Now consider the subsequence ÐB8855 ßC Ñ in M . Since Ò-ß.Ó is compact, the sequence ÐC 85 Ñ has a subsequence ÐC Ñ that converges to some point C − Ò-ß .Ó . But in ‘# , Ð? ß @ Ñ Ä Ð?ß @Ñ iff 856 88 Ð? Ñ Ä ? and Ð@ Ñ Ä @ in ‘ . So ÐB ß C Ñ Ä ÐBß CÑ − MÞ This shows that M is sequentially 88 885566 compact and therefore compact. For 8# , a similar argument clearly works  it just involves “taking subsequences” 8 times. ñ

199 9. Compactness and Completeness

We already know that a compact metric space Ð\ß .Ñ is complete. Therefore all the “big” theorems that we proved for complete spaces are true in a compact metric space Ð\ß .Ñ  for example Cantor's Intersection Theorem, the Contraction Mapping Theorem and the Baire Category Theorem (which implies that a nonempty compact metric space is second category in itself).

Of course, a complete space Ð\ß .Ñ need not be compact: ‘ , for example. We want to see the exact relationship between compactness and completeness.

We begin with a couple of preliminary results.

Theorem 9.1 If Ð\ß .Ñ is totally bounded, then Ð\ß .Ñ is bounded.

Proof For % œ " , pick points B" ß ÞÞÞß B 8 so that \ œ F "" ÐB Ñ ∪ ÞÞÞ ∪ F "8 ÐB ÑÞ Let Q œmax Ö.ÐB34 ß B Ñ À 3ß 4 œ "ß ÞÞÞß 8× . If Bß C − \ , then we have B − F "3 ÐB Ñ and C − F"4 ÐB Ñ for some 3ß 4 , so that

.ÐBßCÑŸ.ÐBßBÑ.ÐBßC3344 Ñ.ÐC ßCÑ"Q "œQ #.

Therefore Ð\ß .Ñ has finite diameter  that is ß Ð\ß .Ñ is bounded. ñ

Notice that the converse to the theorem is false: “total boundedness” is stronger than “boundedness.” For example, let .ww ÐBß CÑ œ min Ö"ß lB  Cl× on ‘‘ . Then diam Ð ß . Ñ œ " so (‘ß.ww Ñ is bounded. Because . and the usual metric . agree for distances smaller that " , . w F" ÐBÑœÐB"ßB"Ñ. Since ‘ cannot be covered by a finite number of these " -balls, (‘ß.w Ñ is not totally bounded.

Theorem 9.2 If Ð\ß .Ñ is totally bounded and E © \ß then ÐEß .Ñ totally bounded: that is, a subspace of a totally bounded space is totally bounded.

Proof Let %  ! and choose B"8 ß ÞÞÞß B so that \ œ F% ÐB " Ñ ∪ ÞÞÞ ∪ F% ÐB 8 Ñ . Of course these # # balls also cover EÐEß.Ñ . But to show that is totally bounded, we need to show that we can cover EÞ with a finite number of % -balls centered at points in E Let N œ Ö4 À F% ÐB Ñ ∩ E Á g× and, for each 4 − N , pick + − F% ÐB Ñ ∩ EÞ # 444# E We claim that the balls FÐ+Ñ cover E . In fact, if +−E , then +−FÐBÑ% for some 4 ; % 44# %% and for this 4 , we also have + − F% ÐB ÑÞ Then .Ð+ß + Ñ Ÿ .Ð+ß B Ñ  .ÐB ß + Ñ   œ % , so 44# 4 444## E +−F% Ð+Ñ4 . ñ

The next theorem gives the exact connection between compactness and completeness. It is curious because it states that “compactness” (a topological property) is the “sum” of two nontopological properties.

200 Theorem 9.3 Ð\ß .Ñ is compact iff Ð\ß .Ñ is complete and totally bounded.

Proof (ÊÐ\ß.Ñ ) We have already seen that the compact space is totally bounded and complete. (For completeness, here is a fresh argument: let ÐB8 Ñ be a Cauchy sequence in Ð\ß .Ñ . Since \ÐBÑB\ is countably compact, 8 has a cluster point in . But a Cauchy sequence must converge to a cluster point. So Ð\ß .Ñ is complete.)

ÐÉÑ Let ÐBÑ8 be a sequence in Ð\ß.Ñ . We will show that ÐBÑ8 has a Cauchy subsequence which (by completeness) must converge. That means Ð\ß .Ñ is sequentially compact and therefore compact. Without loss of generality, we can assume that the terms of the sequence ÐB8 Ñ are distinct (why? ).

Since Ð\ß .Ñ is totally bounded, we can cover \ with a finite number of 1-balls, and one of themF call it "8" must contain infinitely many B 's. Since ÐFß.Ñ is totally bounded (see " Theorem 9.2) we can cover F" with a finite number of # -balls and one of them call it F©F#"must contain infinitely many B 8 's.

" Continue inductively in this way. At the induction step, suppose we have 5 -balls FªFªÞÞÞªF"# 5 where each ball contains infinitely many B 8 's. Since ÐFß.Ñ 5 is totally " bounded, we can cover it with a finite number of 5" -balls, one of which call it F©F5" 5must contain infinitely many B 8 's. This inductively defines an infinite descending sequence of balls F"# ª F ª ÞÞÞ ª F 5 ª ÞÞÞ where each F 5 contains B 8 for infinitely many values of 8 .

Then we can choose B−F8""

B−F8## , with 88 #" ã

B855 −F , with 8 55"" 8 ÞÞÞ8 ã

The subsequence ÐB885556 Ñ is Cauchy since B − F for 6 5 and diam F Ä !Þ ñ

Example 9.4 With the usual metric . :

i)  is complete and not compact (and therefore not totally bounded). " ii) ÖÀ8−×8  is totally bounded, because it is a subspace of the totally bounded space Ò!ß "Ó . But it is not compact (and therefore not complete). " iii)  and ÖÀ8−×8 are homeomorphic because both are countable discrete spaces. Total boundedness is not a topological property. " iv) Ö8 À 8 − × ∪ Ö!× is compact and is therefore both complete and totally bounded.

During the earlier discussion of the Contraction Mapping Theorem, we defined uniform continuity. For convenience, the definition is repeated here.

Definition 9.5 A function 0 À Ð\ß .Ñ Ä Ð] ß .w Ñ is uniformly continuous if

a%$ b aB aC − \ ( .ÐBß CÑ  $ Ê .w Ð0ÐBÑß 0ÐCÑÑ  % Ñ

201 Clearly, if \ œ ÖB"8 ß ÞÞÞß B × is finite, then every continuous function 0 À Ð\ß .Ñ Ä Ð] ß =Ñ is uniformly continuous. (For %$ ! and each 3 œ "ß ÞÞÞß 8 , we pick the the 33 that works at B in the definition of continuity. Then $$$œ min Ö"8 ß ÞÞÞ × works “uniformly” across the space \ .)

The next theorem generalizes this observation to compact metric spaces Ð\ß .Ñ  and illustrates once more “rule of thumb” that “compact spaces act like finite spaces.”

Theorem 9.6 If Ð\ß .Ñ is compact and 0 À Ð\ß .Ñ Ä Ð] ß .w Ñ is continuous, then 0 is uniformly continuous.

Proof Suppose 0! is not uniformly continuous. Then for some %$ , “no works”. In particular, " for that %$ , œ8 “doesn't work” therefore we can find, for each 8?@ , a pair of points 88 and " w with .Ð?88 ß @ Ñ 8 but . Ð0Ð? 8 Ñß 0Ð@ 8 ÑÑ % Þ Since \ is compact, there is a subsequence

Ð?85 Ñ Ä B − \. It follows that the corresponding subsequence Ð@85 Ñ Ä B also because " .Ð@8888 ß BÑ Ÿ .Ð@ ß ? Ñ  .Ð? ß BÑ   .ÐB 8 ß BÑ Ä !Þ Therefore, by continuity, we should 555585 5 haveÐ0Ð?8855 ÑÑ Ä 0ÐBÑ and also Ð0Ð@ ÑÑ Ä 0ÐBÑÞ But this is impossible since w . Ð0Ð?8855 Ñß 0Ð@ ÑÑ % for all 5 . ñ

Uniform continuity is a strong and useful condition. For example, the preceding theorem implies that a continuous function 0ÀÒ+ß,ÓÄ‘ must be uniformly continuous. This is one of the reasons why in calculus “continuous functions on closed intervals” are so nice to work with. The following example is a simple illustration of how uniform continuity might be used in a simple analysis setting.

Example 9.7 Suppose 0 À Ò+ß ,Ó Ä‘ be bounded. Choose any partition of Ò+ß ,Ó

B!" œ+B ÞÞÞB 3"38 B ÞÞÞB œ, and let Q3 and 7 3 denote the sup and inf of the set 0ÒÒB 3" ßBÓÓ 3 . Let ? 3 BœB 3 B 3" Þ The sums 88 7BŸQB33?? 33 3œ" 3œ" are called the lower and upper sums for 0 associated with this partition.

It is easy to see that the sup of the lower sums (over all possible partitions) is finite: it is called , the lower integral of 0Ò+ß,Ó on and denoted + 0ÐBÑ.B . Similarly, the inf of all the upper sums is   finite: it is called the upper integral of 0Ò+ß,Ó on and denoted , 0ÐBÑ.BÞ It is easy to verify that  + ,, ++0ÐBÑ .B Ÿ 0ÐBÑ .BÞIf the two are actually equal , we say 0 is (Riemann) integrable on Ò+ß ,Ó   and we write ,,0ÐBÑ.B œ 0ÐBÑ.B œ , 0ÐBÑ.BÞ +++

Uniform continuity is just what we need to prove that a continuous 0Ò+ß,Ó on is integrable. If 0 is continuous on Ò+ß,Ó , then 0 is uniformly continuous. Therefore, for %$ ! , we can choose ! % such that l0ÐBÑ  0ÐCÑl ,+ whenever lB  Cl $ Þ Pick a partition of Ò+ß ,Ó for which each % ?$333B Þ Then for each 3 , Q 7 Ÿ,+ Þ Therefore

202  88 ,, ++0ÐBÑ.B 0ÐBÑ.B Ÿ Q33?? B 7 33 B  3œ" 3œ" 88 % œ ÐQ7Ñ333?%? B 3 Bœ,+ Ð,+Ñœ % 3œ" 3œ"  Since % ! was arbitrary, we conclude that,, 0ÐBÑ.Bœ 0ÐBÑ.B . ++

10. The Cantor Set

The Cantor set is an example of a compact subspace of ‘ with a surprising combination of properties. Informally, we can construct the Cantor set G as follows. Begin with the closed interval Ò!ß "Ó and delete the open “middle third.” The remainder is the union of 2 disjoint closed "# intervals: Ò!ß$$ Ó ∪ Ò ß "Ó . At the second stage of the construction, we then delete the open middle thirds of each interval leaving a remainder which is the union of 2# disjoint closed intervals: "#"#() Ò!ßÓ∪ÒßÓ∪ÒßÓ∪Òß"Ó**$$**. We repeat this process of deleting the middle thirds “forever.” The Cantor set G is the set of survivors the points that are never discarded.

Clearly there are some survivors: for example, the endpoint of a deleted middle third  for " example, * G clearly survives forever and ends up in .

To make this whole process precise, we name the closed subintervals that remain after each stage. Each interval contains the # new remaining intervals below it:

"# Ò!ß$$ Ó Ò ß "Ó ÐJ!# Ñ ÐJ Ñ

"#"#() Ò!ßÓ**$$** ÒßÓ ÒßÓ Òß"Ó ÐJ!! Ñ ÐJ !# Ñ ÐJ #! Ñ ÐJ ## Ñ

" # " # ( ) " # "* #! ( ) #& #' Ò!ßÓÒßÓÒßÓÒßÓ#( #( * * #( #( $ ÒßÓÒßÓÒßÓÒß"Ó $ #( #( * * #( #( (J!!! ) ( J !!# ) ( J !#! ) ÐJÑ !## ÐJÑÐJÑÐJÑÐJÑ #!! #!# ##! ###

etc. etc.

>2 5 " At the 5#ß stage the set remaining is the union closed subintervals each with length $5 . We 5 label them J8"# 8 ÞÞÞ85 where Ð8 " ß 8 # ß ÞÞÞ8 5 Ñ − Ö!ß #× Þ

Notice that, for example, J! ª J !# ª J !#! ª J !#!! ª J !#!!# ª ÞÞÞ . As we go down the chain “toward” the Cantor set, each new !# or in the subscript indicates whether the next set down is “the remaining left interval” or “the remaining right subinterval.”

 For each sequence = œ Ð8"# ß 8 ß ÞÞÞß 8 5 ß ÞÞÞÑ − Ö!ß #× , we have

J8""# ª J 8 8 ª ÞÞÞ ª J 8 "# 8 ÞÞÞ85 ª ÞÞÞ

Since Ò!ß "Ó is complete, the Cantor Intersection Theorem gives that ∞ J contains a 5œ" 8"# 8 ßÞÞÞ85 single point of Ò!ß "Ó , and we call that point B= . The Cantor set G is then defined by

203  G œ ÖB= À = − Ö!ß #× × © Ò!ß "Ó.

wwww  w If = Á = œ Ð8"# ß 8 ß ÞÞÞß 8 5 ß ÞÞÞÑ − Ö!ß #× , we can look at the first 5 for which 85 Á 85 . Then and w w . Since these closed intervals are disjoint, we conclude B−J= 8" ÞÞÞ85" 8 5 B = −J8 " ÞÞÞ85" ß85  B== Á Bw . Therefore each sequence = − Ö!ß #× corresponds to a different point in G , so lGl lÖ!ß #× l œ #i! œ -. Since G © Ò!ß "Ó , we conclude lGl œ - .

∞ >8 Each B − Ò!ß "Ó can be written using a “ternary decimal” expansion B œ$8 œ !Þ>"# > ÞÞÞ> 5 ÞÞÞthree , 8œ" where each >5 − Ö!ß "ß #×Þ Just as with base 10 decimals, the expansion for a particular B may not "" be unique: for example, $$œ !Þ"!!!ÞÞÞthree but also œ !Þ!#####ÞÞÞ three Þ It's easy to check that an B−Ò!ß"Ó has two different ternary expansions iff Bis the endpoint of one of the deleted “open middle thirds” in the construction. Therefore it is easy to see that a point is in G iff it has a "" ternary expansion involving only 0's and 2's. For example, $#−G and ÂGÞ

What are some of the properties of G ?

i) Ò!ß "Ó  G is the union of the deleted open “middle third” intervals, so Ò!ß "Ó  G is open in Ò!ß "Ó . Therefore G is closed in Ò!ß "Ó , so G is a compact metric space (and therefore complete in the usual metric).

ii) Every point in GGßG is a limit point of that is has no isolated points. (Note: such a space is sometimes called dense-in-itself . This is an awkward but well-established term; it is awkward because it means something different from the obvious fact that every space \ is a dense subset of itself.)  Suppose B="#55"5# − G , where = œ Ð8 ß 8 ß ÞÞÞß 8 ß 8 , 8 ÞÞÞÑ − Ö!ß #× . Given %  ! , " ww pick 5 so $5 % . Define 8œ!#8Á8ß5" ( or ) so that 5" 5" and let

ww i! = œ Ð8"# ß 8 ß ÞÞÞß 8 5 ß 85" ß 8 52 ÞÞÞÑ − Ö!ß #× .

" w Then BB= and = are distinct points in G , but both are in J 8"" ßÞÞÞ855 and diam Jœ 8 ßÞÞÞ8 $5 % . Therefore lB==  Bw l % , so B = is not isolated in G .

iii) With the usual metric .G , is a nonempty complete metric space with no isolated points. It follows from Theorem 3.6 that lGl - and, since G © Ò!ß "Ó , then lGl œ -Þ However, we can also see this from the discussion in ii): there are #œ-i! different ways we could redefine the infinite “tail” Ð85" ß ÞÞÞ8 56 ß ÞÞÞ ) of =  and each version produces a different point in G at distanceBÞ% from =

iv) In some ways, G is “big.” For example, the Cantor set has as many points as Ò!ß "Ó . Also, since GGG is complete with the usual metric, is second category in itself. But is “small” in other ways. For one thing, G is nowhere dense in Ò!ß "Ó (and therefore in ‘ Ñ . Since G is closed, this simply means that G contains no nonempty open interval. To see this, suppose MœÐ+ß,Ñ©G . At the 5th stage of the construction, we must have

M©exactly one of the J8" ßÞÞÞ85 . (This follows immediately from the definition of an interval.) " Therefore ,+$5 for every 5 , so +œ, and MœgÞ

204 v) GG is also “small” in another sense. In the construction of , the open intervals we " "" " $ delete have total length $*# %Ð #( ÑÞÞÞœ# œ" . In some sense, the “length” of what  " $ remains is 0 ! This is made precise in measure theory, where we say that Ò!ß "Ó  G has measure 1 and that G has measure 0 . Measure is yet another way, differing from both cardinality and category, to measure the “size” of a set.

The following theorem states that the topological properties of G which we have discussed actually characterize the Cantor set. The theorem can be used to prove that the “generalized Cantor set” resulting from any of these modified constructions is topologically the same as G .

Theorem 10.1 Suppose E is a nonempty compact subset of ‘ which is dense-in-itself and contains no nonempty open interval. Then EG is homeomorphic to the Cantor set .

Proof The proof is omitted. (See Willard, General Topology .)

It is possible to modify the construction of G by changing “middle thirds” to, say, “middle fifthsß ” or even by deleting “middle thirds” at stage one, “middle fifths” at stage two, etc. The resulting “generalized Cantor set” is homeomorphic to G (using Theorem 10.1) but the total lengths of the deleted open intervals may be different: that is, the result is can be a topological Cantor set with positive measure. “Measure” is not a topological property.

As we noted earlier, G consists of those points B − Ò!ß "Ó for which we can write

∞ >8 B œ$8 œ !Þ>"# > ÞÞÞ> 5 ÞÞÞthree , with each > 8 − Ö!ß #×Þ 8œ"

∞ 2,8 We can obviously rewrite this as B œ$8 , where each ,8 − Ö!ß "× and therefore we can define 8œ" a function1ÀGÄÒ!ß"Ó as follows:

∞∞ 2,,88 for B−Gß1ÐBÑœ1Ð$#88 Ñœ . 8œ" 8œ" More informally,:

1Ð!Þ #,"#$ #, #, ÞÞÞthree Ñ œ !Þ, "#$ , , ÞÞÞ two

This mapping 1 is not one-to-one . For example,

($ 1Ð*% Ñ œ 1Ð!Þ#!####ÞÞÞÞthree ) œ !Þ"!""""ÞÞÞ two œ and )$ 1Ð*% Ñ œ 1Ð!Þ##!!!!ÞÞÞthree ) œ !Þ""!!!ÞÞÞ two œ

In fact, for +ß , − G , we have 1Ð+Ñ œ 1Ð,Ñ iff the interval Ð+ß ,Ñ is one of the “deleted middle thirds.” It should be clear that 1+,G1Ð+ÑŸ1Ð,Ñ is continuous and that if in , then that is, 1 is weakly increasing. (Is 1 onto? )

205 The function 1 is not defined on Ò!ß "Ó  G . But Ò!ß "Ó  G is, by construction, a union of disjoint open intervals. We can therefore extend the definition of 1 to a mapping K À Ò!ß "Ó Ä Ò!ß "Ó in the following simple-minded way:

1ÐBÑif B − G KÐBÑ œ  1Ð+Ñif B − Ð+ß ,Ñ , where Ð+ß ,Ñ is a “deleted middle third”

Since we know 1Ð+Ñ œ 1Ð,Ñ , this amounts to extending the graph of 1 over each “deleted middle third” Ð+ß ,Ñ by using a horizontal line segment from Ð+ß 1Ð+ÑÑ to Ð,ß 1Ð,ÑÑÞ The result is the graph of the continuous function K .

K is sometimes called the Cantor-Lebesgue function. It satisfies:

i) KÐ!Ñ œ !ß KÐ"Ñ œ " ii) K is continuous iii) K is (weakly) increasing iv) at any point BÐ+ß,ÑK in a “deleted middle third” , is differentiable and KÐBÑœ!Þw

Recall that G! has “measure .” Therefore we could say “ KÐBÑœ!w almost everywhere” even though K! rises monotonically from to 1!

Note: The technique that we used to extend 1J© works in a similar way for any closed set ‘ and any continuous function 1ÀJ Ä‘‘ . Since J is open in ‘ , we know that we can write ‘ J as the union of a countable collection of disjoint open intervals M8 which have for Ð+ß,Ñ , Ð  ∞ß ,Ñ or Ð+ß ∞Ñ We can extend 0 to a continuous function J À‘‘ Ä simply by extending the graph of 0MÀ over linearly over each 8

if MœÐ+ß,Ñ88 , then let the graph J over M be the straight line segment joining Ð+ß 0Ð+Ñß >9 Ð,ß 0Ð,ÑÑÞ if MœÐ∞ß,Ñ8 , then let J have the constant value 0Ð,98 I8 if MœÐ+ß∞Ñ8 , then let J have the constant value 0Ð+ÑMß on 8

There is a much more general theorem that implies that whenever JÐ\ß.Ñ is a closed subset of , then each 0 − GÐJÑ and be extended to a function J − GÐ\Ñ . In the particular case \ œ ‘ , proving this was easy because ‘ is ordered and we completely understand the structure of the open sets in ‘ ..

Another curious property of G , mentioned without proof, is that its “difference set” ÖB  C À Bß C − G× œ Ò  "ß "Ó. Although G has measure ! , the difference set in this case has measure #!

206 Exercises

E21. Suppose \0À\Ä]] is compact and that is continuous and onto. Prove that is compact. (“A continuous image of a compact space is compact. ”)

E22. a) Suppose that \] is compact and is Hausdorff. Let 0À\Ä] be a continuous bijection. Prove that 0 is a homeomorphism.

b) Let ggg be the usual topology on Ò!ß "Ó and suppose "# and are two other topologies on Ò!ß "Ó such that ggg"#§§ Prove that ÐÒ!ß "Ó , g " Ñ is not Hausdorff and that Ð Ò!ß "Ó , g # Ñ is not ÁÁ compact. (Hint: Consider the identity map 3 À Ò!ß "Ó Ä Ò!ß "ÓÞ )

c) Is part b) true if Ò!ß "Ó is replaced by an arbitrary compact Hausdorff space Ð\ßg Ñ ?

E23. Prove that a nonempty space Ð\ßg‘‘ Ñ is pseudocompact iff every continuous 0 À Ä achieves both a maximum and a minimum value.

E24. Suppose 0 À Ð\ß .Ñ Ä Ð] ß .w ÑÞ Prove that 0 is continuous iff 0lO is continuous for every compact set O©\ .

E25. Suppose that EF and are nonempty disjoint closed sets in Ð\ß.ÑE and that is compact. Prove that .ÐEß FÑ  !Þ

E26. Let \] and be topological spaces.

a) Prove that \ is compact iff every open cover by basic open sets has a finite subcover.

b) Suppose \‚] is compact. Prove that if \ß] Ág , then \ and ] are compact. (By induction, a similar statement applies to any finite product.)

c) Prove that if \] and are compact, then \‚] is compact. (By induction, a similar statement applies to any finite product.) (Hint: for any B−\ßÖB× ‚ ] is homeomorphic to ] . Part a) is also relevant. )

d) Point out explicitly why the proof in c) cannot be altered to prove that a product of two countably compact spaces is countably compact. (An example of a countably compact space \ for which \‚\ is not even pseudocompact is given in Chapter X, Example 6.8.)

Note: In fact, an arbitrary product of compact spaces is compact. This is the “Tychonoff Product Theorem” which we will prove later. It is true that the product of a countably compact space and a compact space is countably compact. You might trying proving this fact. A very similar proof shows that a product of a

207 compact space and a Lindelof¨¨ space is Lindelof. That proof does not generalize, however, to show that a product of two Lindelof¨¨ spaces is Lindelof. Can you see why?

E27. Prove that if Ð\ß .Ñ is a compact metric space and l \ l œ i! , then \ has infinitely many isolated points.

E28. Suppose \] is any topological space and that is compact. Prove that the projection map 1\ À\‚] Ä\ is closed. (A projection “parallel to a compact factor” is closed. )

E29. Let \ be an uncountable set with the discrete topology g . Prove that there does not exist a totally bounded metric d on \œ such that gg. .

E30. a) Give an example of a metric space Ð\ß .Ñ which is totally bounded and an isometry from Ð\ß .Ñ into Ð\ß .Ñ which is not onto. Hint: on the circleW:" , start with a point and keep rotating it around the boundary by increments of some angle " .

b) Prove that if Ð\ß .Ñ is totally bounded and if 0 is an isometry from Ð\ß .Ñ into Ð\ß .Ñ , then 0Ò\Ó must be dense in Ð\ß.Ñ . % Hint: Given Bß%  !ß cover \ with finitely many # spheres; if the sequence Bß 0ÐBÑß 0Ð0ÐBÑÑ , ÞÞÞ consists of distinct terms, there must be infinitely many of them in one sphere.

c) Show that a compact metric space cannot be isometric to a proper subspace of itself. Hint: you might use part b).

d) Prove that if each of two compact metric spaces is isometric to a subspace of the other, then the two spaces are isometric to each other. Note: Part d) is an analogue of the Cantor-Schroeder-Bernstein Theorem for compact metric spaces.

E31. Suppose Ð\ß .Ñ is a metric space and that Ð\ß .w Ñ is totally bounded for every metric .µ.w . Must Ð\ß.Ñ be compact?

E32. Let h be an open cover of the compact metric space Ð\ß .Ñ . Prove that there exists a constant $h! such that for all B : FÐBÑ©Y$ for some Y − . (The number $ is called a Lebesgue number for h .)

E33. Suppose Ð\ß .Ñ is a metric space with no isolated points. Prove that \ is compact iff .ÐEß FÑ  !for every pair of disjoint closed sets Eß F © \Þ

208 Chapter IV Review

Explain why each statement is true, or provide a counterexample.

1. Let E0À\Ä\ß\ denote the set of fixed points of a continuous function where is a Hausdorff space. E\ is closed in .

2. Let W denote the set of all Cauchy sequences in ‘ which converge to a point in . Then lW ± œ-.

3. For a metric space Ð\ß .Ñ , if every continuous function 0 À \ Ä ‘ assumes a minimum value, then every infinite set in \ has a limit point.

4. There exists a continuous function 0À‘‘ Ä for which the Cantor set is the set of fixed points of 0 .

5. If \ has the cofinite topology, then every subspace is pseudocompact.

6. Let Ò!ß "Ó have the subspace topology from the Sorgenfrey line. Then Ò!ß "Ó is countably compact.

7. Let Æ‘œÖE© ÀE is first category in ‘g × , and let be the topology on ‘ for which Æ is a subbase. Then Ðß‘g Ñ is a Baire space.

8. Every sequence in Ò!ß "Ó# has a Cauchy subsequence.

9. If EÐ\ß.ÑE is a dense subspace of and every Cauchy sequence in converges to some point in \Ð\ß.Ñ , then is complete.

10. If gg is the cofinite topology on , then (ß ) has the fixed point property.

11. Every sequence in Ò!ß "Ñ has a Cauchy subsequence.

1 12. Let ^œÐ!ßÑ∩G3 , where G is the Cantor set. Then ^ is completely metrizable.

13. The Sorgenfrey plane is countably compact.

14. If every sequence in the metric space Ð\ß .Ñ has a convergent subsequence, then every continuous real valued function on \ must have a minimum value.

15. SupposeÐ\ß .Ñ is complete and 0 À \ Ä \ . The set G œ ÖB −‘ À 0 is continuous at B× is second category in itself.

16. In the space GÐÒ!ß "ÓÑ , with the metric 3 of uniform convergence, the subset of all polynomials is first category.

17. A nonempty open set in Ð\ßg Ñ cannot be nowhere dense.

209 18. Every nonempty nowhere dense subset of ‘ contains an isolated point.

19. There are exactly - nowhere dense subsets of ‘ .

20. Suppose FF is nonempty subset of  with no isolated points. cannot be completely metrizable.

21. Suppose Ð\ß .Ñ is a countable complete metric space. If E © \ , then ÐEß .Ñ may not be complete, but E is completely metrizable.

22. If EFFE is first category in \©E and , then is first category in .

23. There are -. different metrics on  , each equivalent to the usual metric, for which Ðß.Ñ is complete.

24. In  , with the cofinite topology, every infinite subset is sequentially compact.

##`0 `0 25. Suppose 0ÀÒ!ß"Ó Ä‘ and EœÖ:− ( !ß" ) : `B and `C both exist at : }. Then E , with its usual metric, is totally bounded.

26. A space Ð\ßg Ñ is a Baire space iff \ is second category in itself.

27. Let IG denote the subset of the Cantor set consisting of the endpoints of the open intervals deleted from Ò!ß "Ó in the construction of G . Suppose 0À Ò!ß "Ó Ä ‘ is continuous and that 0ÒIÓ © Ò!ß !Þ"Ó. Then 0ÒGÓ © Ò!ß !Þ"Ó .

28. Let . be a metric on the irrationals  which is equivalent to the usual metric and such that Ðß.Ñ is complete. Then Ðß.Ñ must be totally bounded.

29. Suppose Ð\ß .Ñ is a nonempty complete metric space and E is a closed subspace such that ÐEß .Ñ is totally bounded. Then every sequence in E has a cluster point.

30. There is a metric d on Ò!ß "Ó , equivalent to the usual metric, such that ÐÒ!ß "Óß .Ñ is not complete.

31. Let ^œÒ!ß"Ó# have the usual topology. Every nonempty closed subset of ^ is second category in itself.

32. Suppose 0 À Ò!ß "Ó87 Ä‘ and G œ ÖB − Ò!ß "Ó 8 À 0 is continuous at B× . There is a metric . equivalent to the usual metric on GÐGß.Ñ for which is complete.

33. If the continuum hypothesis (CH) is true, then ‘ cannot be written as the union of fewer than c nowhere dense sets.

34. Suppose E is a complete subspace of Ò!ß "ÓÞ Then E is a K$ set in ‘ Þ

35. There are 2- different subsets of ‘ none of which contains an interval of positive length.

36. Suppose Ð\ß .Ñ is a nonempty complete metric space and E is a closed subspace such that

210 ÐEß .Ñ is totally bounded. Then every sequence in E has a cluster point.

37. Let . w be a metric on the irrationals  which is equivalent to the usual metric and for which Ðß.Ñww is totally bounded. Then Ðß.Ñ cannot be complete.

38. The closure of a discrete subspace of ‘ may be uncountable.

39. Suppose 0 À Ò!ß "Ó87 Ä‘ and G œ ÖB − Ò!ß "Ó 8 À 0 is continuous at B× . Then there is a metric d on GÐGß.Ñ , equivalent to the usual metric, such that is totally bounded.

40. Suppose Ð\ßg‘ Ñ is compact and 0 À \ Ä is a continuous function such that 0ÐBÑ  ! for all B−\Þ Then there is an %% ! such that 0ÐBÑ for all B−\Þ

41. If every point of Ð\ß .Ñ is isolated, then Ð\ß .Ñ is complete.

42. If Ð\ß .Ñ has no isolated points, then its completion has no isolated points.

43. Suppose that for each point B − Ð\ß .Ñ , there is an open set YBB such that ÐY ß .Ñ is complete. Then there is a metric ddwwµ\ on such that Ð\ß.Ñ is complete. µµµ 44. For a metric space Ð\ß.Ñ with completion ( \ß. Ñ , it can happen that ±\ \ ± œi! . µ µ 45. Let .ß.ÑÐß.Ñ be the usual metric on  . There is a completion ( of for which µ llœiÞ  !

46. A subspace of ‘ is bounded iff it is totally bounded.

47. A countable metric space must have at least one isolated point.

48. The intersection of a sequence of dense open subsets in  must be dense in .

49. If GGJ is the Cantor set, then ‘‘ is an 5 set in .

50. A subspace EÐ\ß.ÑE of a metric space is compact iff is closed and bounded.

51. Let U‘œÖÒ+ß,Ñ∩Ò!ß"ÓÀ+ß,− ß +, × be the base for a topology g on Ò!ß"Ó . The spaceÐÒ!ß "Óßg Ñ is compact.

52. Ðj## ß .Ñ , where . is the usual metric on j , is sequentially compact.

53. A subspace of a pseudocompact space is pseudocompact.

54. Ðj## ß .Ñ , where . is the usual metric on j , is countably compact.

55. An uncountable closed set in ‘ must contain an interval of positive length.

56. Suppose G0ÀGÄG0 is the Cantor set and is continuous. Then the graph of (a subspace of G‚G , with its usual metric) is totally bounded.

211 ±BC± 57. Let G denote the Cantor set ©Ò!ß"Ó , with the metric .ÐBßCÑœ"±BC± . Then ÐGß.Ñ is totally bounded.

58. A discrete subspace of ‘‘ must be closed in

59. A subspace of ‘ which is discrete in its relative topology must be countable.

60. If EF and are subspaces of Ð\ßÑg and each is discrete in its subspace topology, then E∪F is discrete in the subspace topology.

61. Let cos8 denote the composition of cos with itself 8B−Ò"ß"Ó times. Then for each , there exists an 8BBœBÞ (perhaps depending on ) such that cos8

62. Suppose . is any metric on equivalent to the usual metric. Then ÐÒ!ß "Óß .Ñ is totally bounded.

63. Let WW"#" denote the unit circle in ‘ . is homeomorphic to a subspace of the Cantor set G .

64. There is a metric space Ð\ß .Ñ satisfying the following condition: for every metric dd,w µ Ð\ß .w Ñ is totally bounded.

65. Ò!ß "Ó# , with the subspace topology from the Sorgenfrey plane, is compact.

212