Chapter IV Completeness and Compactness

1. Introduction

In Chapter III we introduced topological spaces as a generalization of pseudometric spaces. This allowed us to see how certain ideas
continuity, for example
can be extended into a setting 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 distinct in topological spaces but equivalent in the special case of pseudometric spaces
for example, second countability and separability.

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 ÐBÑ8 in a pseudometric space Ð\ß.Ñ is called a Cauchy sequence if a !b% N −  such that if 7ß8 R , then .ÐBßBÑ7 8 % .

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, as the next theorem confirms.

Theorem 2.2 If ÐBÑÄB8 in Ð\ß.Ñ , then ÐBÑ 8 is Cauchy.

% Proof Let % ! . Because ÐBÑÄB8 , there is an R− such that .ÐBßBÑ8 # for 8 R . % % So if 7ß8 R , we get .ÐBßBÑŸ.ÐBßBÑ.ÐBßBÑ7 8 7 8 #  # œ%. Therefore ÐBÑ8 is Cauchy. ñ

However, a Cauchy sequence does not always converge. For example, look at the space Ð ß.Ñ where . is the usual metric. Consider a sequence in  that converges to È# in ‘ à for example, we could use Ð;8 Ñ œ Ð"ß "Þ%ß "Þ%"ß "Þ%"%ßÞÞÞ Ñ .

ñ Since Ð;Ñ8 is convergent in ‘ ßÐ;Ñ 8 is a Cauchy sequence in  ( “Cauchy” depends only on . and the numbers ;8 , not whether we are thinking of these ; 8 'sas elements of  or ‘ . ñ But Ð;Ñ8 has no limit in Ðß.Ñ . ( Why? To say that “È#   ” 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.Ñhas a limit in \ .

Example 2.4 In all parts of this example, . is the usual metric on subsets of ‘ .

1) Б ß.Ñ is complete ( for the moment, we assume this as a simple fact from analysis; however we will prove it soon ) but Ð ß.Ñ and Ðß.Ñ  are not complete: completeness is not a hereditary property !

2) If ÐBÑ8 is a Cauchy sequence in Ðß.Ñ , then there is some R such that lB
Bl7 8 1 for 7ß8 R . Because the B8 's are integers, this implies that BœB7 8 for 7ß8 R
that is, ÐB8 Ñ is eventually constant and therefore converges. Therefore Ð ß.Ñ is complete.

w " " w 3) Consider  with the equivalent metric .Ð7ß8Ñœl7
8 l . ( . µ . because both metrics produce the discrete topology on .) w w In Ð ß.Ñ , the sequence ÐBÑœÐ8Ñ8 is Cauchy, and if 5− , then .ÐBß5Ñ8 " " " w œl
lÄ8 5 5 Á!as 8Ä∞ , so ÐBÑÄ58 / . So the space Ðß.Ñ is not complete. Since Ðß.Ñ w has the discrete topology, the map 3Ð8Ñœ8 is a homeomorphism between Ðß.Ñw and the complete space (  ß.ÑÞ So completeness is not a topological property . A homeomorphism takes convergent sequences to convergent sequences and nonconvergent sequences to nonconvergent sequences. For example, Ð8Ñ does not converge in Ðß.Ñ w and Ð3Ð8ÑÑœÐ8Ñ does not converge in Ðß.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 equivalent metric . w does not change the open sets and therefore does not change which sequences converge. But the change might create or destroy Cauchy sequences because the Cauchy property does depend specifically on how distances are measured. Another similar example: we know that any open interval Ð+ß ,Ñ is homeomorphic to ‘ . However, with the usual metric on each space, ‘ is complete and Ð+ß ,Ñ is not complete.

4) In light of the observations in 3) we can ask: if Ð\ß .Ñ is not complete, might there be some equivalent metric .w for which Ð\ß.Ñ w is complete? To take a specific example: could we find some metric .w on  so .µ. w but for which Ðß.Ñ  w is complete? We could also ask this question about Ð ß .ÑÞ Later in this chapter we will see the answer
and perhaps surprisingly, the answers for  and  are different x

w " 5) Ðß.Ñ and ÐÖ8 À8−  ×ß.Ñ are countable discrete spaces, so they are " homeomorphic. But (even better!) the function 0Ð8Ñ œ 8 is actually an isometry between these w " " spaces because . Ð7ß8Ñœl7
8 lœ.Ð0Ð7Ñß0Ð8ÑÑÞ 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. w For example, the sequence ÐBÑœ88 is Cauchy in Ðß.Ñ and the isometry 0 carries " " ÐBÑ8 to the Cauchy sequence Ð0ÐBÑÑœÐÑ8 8 in ÐÖÀ8−×ß.Ñ 8  . It is clear that Ð8Ñ has no limit " in the domain and Ð8 Ñ has no limit in the range. w " 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.

155 Theorem 2.5 If B is a cluster point of a Cauchy sequence ÐBÑÐ\ß.Ñ8 in , then ÐBÑÄB 8 .

% Proof Let % ! . Pick R so that .ÐBßBÑ7 8 # when 7ß8 RÞ Since ÐBÑ8 clusters at B , we can pick a O R so that B−FÐBÑ% . Then for this O and for 8 Rß O # % % .ÐB8 ßBÑŸ.ÐB 8O ßB Ñ.ÐB O ßBÑ#  # œ % so ÐB8 ÑÄB . ñ

Corollary 2.6 In a pseudometric space Ð\ß.Ñ : a Cauchy sequence ÐBÑ8 converges iff ÐBÑ 8 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 ÐBÑ8 in Ð\ß.Ñ is bounded
that is, the set ÖB" ßB # ßÞÞÞßB 8 ßÞÞÞ× has finite diameter.

Proof Pick R so that .ÐB7 , BÑ" 8 when 7ß8 R . Then B−FÐBÑ8 " R for all 8 RÞ Let <œ max Ö"ß.ÐBßBÑß.ÐBßBÑßÞÞÞß.ÐB"R #R R
"R ßBÑ×Þ Therefore, for all 7ß8 we have .ÐB78 ßB ÑŸ.ÐB 8R ßB Ñ.ÐB R7 ßB ÑŸ#<, so diam ÖBßBßÞÞÞßB "#8 ßÞÞÞן#<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 BŸB8 8" for all 8 , and weakly decreasing if B B8 8" for all 8 . A monotone sequence is one that is either weakly increasing or weakly decreasing.

Lemma 2.9 If ÐB8 Ñ is a bounded monotone sequence in ‘ , then ÐB8 Ñ converges.

Proof Suppose ÐBÑ8 is weakly increasing. Since ÐBÑ8 is bounded, ÖBÀ8−× 8  has a least upper bound in ‘ . Let Bœ sup ÖBÀ8−8  × . We claim ÐBÑÄBÞ 8 Let %% ! . Since B
Bß we know that B
% is not an upper bound for ÖBÀ8−×8  . Therefore B
BŸB% R for some RÞ Since ÐBÑ8 is weakly increasing and B is an upper bound for ÖBÀ8−×ß8  it follows that B
BŸBŸB % R 8 for all 8 R . Therefore lB
BlŸ8 % for all 8 Rß so ÐBÑÄB8 . If ÐB8 Ñ is weakly decreasing, then the sequence Ð
B8 Ñ is a weakly increasing, so Ð
BÑÄ+8 for some + ; then ÐBÑÄ
+Þñ 8

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

Proof Call B5 a peak point of ÐBÑB B 858 if for all 8 5 . We consider two cases.

i) If ÐB8 Ñ has only finitely many peak points, then there is a last peak point B8! in ÐBÑ 8 .

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

We continue in this way to pick an increasing subsequence ÐB85 Ñ of ÐBÑ 8 .

156 ii) If ÐB8 Ñ has infinitely many peak points, then list the peak points as a subsequence

B88ß" ßB # ÞÞÞ (where 88ÞÞÞÑ "# . Since each of these points is a peak point, we have

B B85 8 5" for each 5ÐBÑ , so 8 5 is a weakly decreasing subsequence of ÐBÑñ8 .

Theorem 2.11 Ð‘ß .Ñ is complete.

Proof Let ÐBÑ8 be a Cauchy sequence in Ðß.Ñ‘ . By Theorem 2.8, ÐBÑ8 is bounded and by

Lemma 2.10, ÐBÑ8 has a monotone subsequence ÐBÑ8 5 which is, of course, also bounded. By

Lemma 2.9, ÐBÑ85 converges to some point B−‘ . Therefore B is a cluster point of the Cauchy sequence ÐBÑ8 , so ÐBÑÄBÞ 8 ñ

Corollary 2.12 Б8ß .Ñ is complete.

Proof Exercise # (Hint: For 8œ# , the sequence ÐÐBßCÑÑ8 8 in ‘ is Cauchy iff both sequences ÐBÑ8 and ÐCÑ 8 # are Cauchy sequences in ‘ . If ÐBÑÄB8 and ÐCÑÄC 8 , then ÐÐBßCÑÄÐBßCÑ 88 in ‘ . )

3. Subspaces of Complete Spaces

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

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+Ñ8 be a Cauchy sequence in E . Then Ð+Ñ8 is also Cauchy in the complete space Ð\ß.Ñ, so Ð+ÑÄB8 for some B−\Þ But E is closed, so this limit B must be in E , that is, Ð+ÑÄB−E8 . Therefore ÐEß.Ñ is complete.

2) If B−E cl , we can pick a sequence in E such that Ð+ÑÄB8 . Since Ð+Ñ 8 converges, it is a Cauchy sequence in E . But ÐEß.Ñ is complete, so we know Ð+ÑÄ+8 for some + − E © \. Since limits of sequences are unique in metric spaces, we conclude that B œ + . Therefore if B−E cl , 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 . be a metric. Can you give an example of a pseudometric space Ð\ß .Ñ for which 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 gives a different characterization
in terms of the existence of points in certain intersections. This illustrates an important idea: the completeness property for Ð\ß .Ñ can be expressed in different ways, but each characterization somehow asserts the existence of certain points.

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

1) Ð\ß .Ñ is complete 2) Whenever J" ªJ # ª... ªJ 8 ª ... is a decreasing sequence of nonempty ∞ closed sets with diamÐJÑÄ!8 , then +8œ" JœÖB× 8 for some B−\Þ

Proof 1ÑÊÑ 2 For each 5 , pick a point B−J55 . Since diam ÐJÑÄ! 5 , the sequence ÐBÑ 5 is Cauchy soÐBÑÄB5 for some B−\Þ For each 8 , the subsequence ÐB85 ÑœÐB 8"8# ßB ßÞÞÞÑÄB . Since the J 8 's are ∞ decreasing, this subsequence is inside the closed set J8 , so B−J 8 . Therefore B−8œ" J 8 . ∞ + If BßC−8œ" J8 , then .ÐBßCÑŸ diam ÐJÑ 8 for every 8Þ Since diam( JÑÄ!8 , + ∞ this means that .ÐBßCÑœ!Þ Therefore BœC , and so +8œ" JœÖB×8 .

2ÑÊÑ 1 Suppose 1) is false, and let ÐBÑ8 be a nonconvergent Cauchy sequence ÐBÑ8 . We will construct sets J8 which violate condition 2). Without loss of generality, we may assume that all the B8 's are distinct. ( Why? If any value BœC5 were repeated infinitely often, then C would be a cluster point of ÐBÑ8 and we would have ÐBÑÄC 8 . Therefore each term of ÐBÑ8 can occur only finitely often, and we can pick a subsequence from ÐB8 Ñ whose terms are distinct. This subsequence is also a nonconvergent Cauchy sequence. We could now construct the J8 's using the terms of the subsequence. But to keep the notation a bit simpler, we might as well assume all the terms in the original sequence are distinct. ) >2 Let JœÖBßB8 8 8" ßÞÞÞßB 85 ßÞÞÞל the “ 8 tail” of the sequence ÐBÑÁg8 . Then ∞ % each JªJß8 8" and +8œ" JœgÞ 8 For % ! , we can pick R so that .ÐBßBÑ7 8 # if 7ß8 R . Then diam ÐJÑR % and for 8 Rß diam ÐJÑŸ8 diam ÐJÑ R % . So diam ÐJ8 Ñ Ä !Þ We claim that these J8 's are closed
which will contradict 2) and complete the proof.

If some particular J8! were not closed, then there would be a point B− cl J8!
J 8 ! and we could find a sequence of distinct terms in J8! that converges to B . But such a sequence is automatically a subsequence of the original sequence ÐBÑ8 , so B would be a cluster point of ÐB8 Ñ
which is impossible since a nonconvergent Cauchy sequence ÐB8 Ñ can't have a cluster point. ñ

How would Theorem 3.2 be different if . were only a pseudometric? How would the proof change?

Example 3.3

1) In the complete space Ðß.Ñ‘ consider: a) JœÒ8ß∞Ñ8 " b) J8 œÐ!ß8 Ñ " c) J8 œÒ8ß88 Ó

∞ In each case, +8œ" J8 œ g . Do these examples contradict Theorem 3.2? Why not?

" " 2) Let JœÖB−8  ÀÒÈÈ #
8 ŸBŸ #Ó× 8 . The J 8 's satisfy all the ∞ hypotheses in Theorem 3.2), but +8œ" J8 œ g . Does this contradict Theorem 3.2?

158

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

If Ò!ß "Ó were countable, we could list itselements in 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+ß,Ó 8 8 of length less than #8" and excluding B 8" . Then Ò+"" ß, ÓªÒ+ ## ß, ÓªÞÞÞªÒ+ 88 ß, ÓªÞÞÞ and these sets clearly satisfy the conditions in ∞ part 2) of the Cantor Intersection Theorem. But +8œ" Ò+8 ß, 8 Óœg . This is impossible since 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
in it, 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 J+ and J , (centered at +, and ) with J©EJ©EÞ+ int and , int

Proof Let % œ.Ð+ß,Ñ . Since +ß,− int Eß we can choose positive radii %+ and % , so that % FÐ+Ñ©%+int E and FÐ,Ñ© % , int E . In addition, we can choose both %+ and % , less than # Þ w w Let %%%œ min Ö+ ß , ×Þ Then we can use the closed balls J+ œÖBÀ.ÐBß+ÑŸ% × and w J, œÖBÀ.ÐBß,ÑŸ% ×. ñ

Proof of Theorem 3.6 The idea of the proof is to construct, inductively, - 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 JJ!" and , centered at BB !" and , each with " diameter # Þ Stage 2 Since B!!! − int J and B is not isolated, we can pickdistinct points B!! and B !" (both ÁBJ!! ) inint and use Lemma 3.5 to pick disjoint closed balls J!! and J !" (centered at B!! and B !" ) and both ©JÞ int ! We can then shrink the balls, if necessary, so that each has " diameter ## Þ We can repeat similar steps inside JÀ" since B−J " int "" and B is not isolated, we can pickdistinct points B"! and B "" (both ÁB " ) inint J " 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

159 " balls, if necessary, so that each has diameter ## Þ At the end of Stage 2, we have 4 disjoint closed balls: J!! ßJ !" ßJ "! ßJ "" Þ Stage 3 We now repeat the same construction inside each of the 4 sets J!! ßJ !" ßJ "! ßJ "" . For example, we can pickdistinct points B!!! and B !!" (both ÁB !!) inint J !! and use the Lemma 3.5 pick disjoint closed balls JJ!!! and !!" (centered at BB !!! and !!" ) and both © int " JÞ!! 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: J!!! ßJ !!" ßJ !"! ßÞÞÞÞ 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 . ª J !!! ß ª J !! à ß ª J !!" J! .... etc. à ª J !"! ß ª J !" à ª J !"" and ª J "!! ß ª J "! à ß ª J "!" J" .... etc. à ª J ""! ß ª J "" à ª J """

160 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 “split” happens when we choose two new nonempty disjoint closed balls inside the current one, making sure that their diameters keep shrink toward ! .

 In the end, for each binary sequence = œ Ð8" ß8 # ßÞÞÞß8 5 ßÞÞÞÑ − Ö!ß"× we will have a corresponding descending sequence of nonempty closed sets whose diameters Ä ! :

J8" ªJ 88 ## ªJ 888 "#$ ªÞÞÞªJ 88ÞÞÞ8 "#5 ªÞÞÞ

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

JªJ! !" ªJ !"" ªJ !""! ªJ !""!! ªJ !""!!" ªÞÞÞ

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

∞ J∩J8" 88 ## ∩J 888 "#$ ∩ÞÞÞ∩J 88ÞÞÞ8 "#5 ∩ÞÞÞœ+5œ" J 88ÞÞÞ8 "#5 œÖB× =

For two different binary sequences, say >œÐ7"# ß7 ßÞÞÞß7 5 ßÞÞÞÑÁÐ8 "# ß8 ßÞÞÞß8 5 ßÞÞÞÑœ= , there is a smallest 5 for which 7Á85 5 . Then B−J = 88ÞÞÞ88"# 5
"5 and B−J > 88ÞÞÞ87 "# 5
"5 . Since these sets are disjoint, we have BÁB= > . Thus, mapping =ÈB = gives a one-to-one function from Ö!ß"× into \ . We conclude that -œ#i! œlÖ!ß"×lŸl\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 3œ"ßÞÞÞß5 and for each 3 -tuple Ð8ßÞÞÞß8Ñ−Ö!ß"×" 3 we have defined points B 8"... 8 3 and " closed balls J8ÞÞÞ8"3 centered at B 88 "3... , with diam ÐJÑ 8ÞÞÞ8 "3 #3 and so that

for each Ð8ßÞÞÞß8Ñ" 3 , J 8" ªJ 88 "# ªÞÞÞªJ 88ÞÞÞ8 "#3

Induction step : We must construct the sets for stage 5"Þ For each Ð5"Ñ -tuple 5" Ð8ßÞÞÞß8ß8" 5 5" Ñ−Ö!ß"×, we need to define a point B 8" ÞÞÞ8 5 8 5" and a closed ball

J8" ÞÞÞ8 5 8 5" in such a way that the conditions in the induction hypothesis remain true with 5" replacing 5 .

For any Ð8ßÞÞÞß8Ñ" 5 : we have B 8ßßß8"5 − int J 8ÞÞÞ8 "5 . Since B 8ßßß8 "5 is not

isolated, we can pick distinct points B8ÞÞÞ8!"5 and B 8ÞÞÞ8" "5 (both ÁB 8ÞÞÞ8 "5 ) in int J 88 "5...

and use the Lemma 3.5 to pick disjoint closed balls J88!"5... and J 88" "5 ... (centered at

B8ßßß8!" 5 and B 8ÞÞÞ8" " 5 ) and both ©J int 88"... 5 . We can then shrink the balls, if necessary, " so that each has diameter #5  " Þ ñ

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

161 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 - . (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\lŸ- . 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 be the set of all condensation points in \ . Each point +−\
G 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 Lindelo¨f. 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∩GœÖ,× . 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 . but for which Ðß.Ñw is complete
that is, is  “completely metrizable?”

Corollary 3.9  is not completely metrizable.

w Proof Suppose . is a metric on  equivalent to the usual metric . , so that g. œ g . w Þ Then Ð ß .w Ñ is a nonempty metric space and, since .µ.w , the space Ðß.Ñ w has no isolated points w (no set Ö;× is in g. œ g . w 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 statements are equivalent:

a) every Cauchy sequence is eventually constant b) Ð\ß .Ñ is complete and g. is the discrete topology c) 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 ] . Prove that Ð]ß.Ñ is complete.

E3. Suppose that Ð\ß.Ñ is a metric space and that ÐBÑ8 is a Cauchy sequence with only finitely many distinct terms. Prove that ÐB8 Ñ is eventually constant, i.e., that for some 8 −  , B8 œB 8" œÞÞÞ

E4. Let p be a fixed prime number. The :-adic norm | |: is defined on  by:

pk m For !ÁB− , write Bœn for integers 5ß7ß8 , where p does not divide 78 or . ( Of
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) lBCl: œlBl : †lCl : c) lBCl: ŸlBl : lCl : d) lBCl: Ÿ max ÖlBl : ßlCl : × l l: is shares properties a), b) and c) with the ordinary norm (or absolute value), l l on  . But d) is a condition stronger than the usual triangle inequality: that is, d) Ê c)

The :-adic metric .: is defined on  by .ÐBßCÑœlB
Cl : :

Because of d), the metric .: satisfies a strengthened version of the triangle inequality for metric spaces: what is this stronger inequality ? For BßCßD− , .: ÐBßDÑ

Prove or disprove À( ß . : ) is complete.

163 E5. Suppose Ð\ß.Ñ iscomplete and that J "ªJ # ª... ªJ 8 ª ... is a sequence of closed sets in \ for which diam ÐJÑÄ8 0. Prove that if ] is a Hausdorff space and 0À\Ä] is continuous, ∞ ∞ +8œ"0ÒJÓœ0Ò8 + 8œ" JÓ8 .

E6. A metric space Ð\ß.Ñ is called locally complete if every point B hasa neighborhood R B (necessarily closed) which is complete.

a) Give an example of a metric space 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 of the function 0À\Ä\0ÐBÑœBÞ if We say that a topological space \ has the fixed point property if every continuous 0 À \ Ä \ has a fixed point.

The fixed point property is a topological property (this is easy to check; do it! )

Example 4.2

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

2) The interval Ò+ß ,Ó has the fixed point property: a continuous 0 À Ò+ß,Ó Ä Ò+ß,Ó must have a fixed point. Certainly this is true if 0Ð+Ñœ+ or 0Ð,Ñœ, . So assume that 0Ð+Ñ + and 0Ð,Ñ, and define 1ÐBÑœ0ÐBÑ
B . Then 1 is continuous, 1Ð+Ñœ0Ð+Ñ
+ ! and 1Ð,Ñ œ 0Ð,Ñ
,  ! . By the Intermediate Value Theorem (from analysis) there must be a point -−Ð+ß,Ñ where 1Ð-Ñœ0Ð-Ñ
-œ! , that is, 0Ð-Ñœ-Þ

3) The Brouwer Fixed Point Theorem states that H8 œÖB−‘ 8 À.ÐBß!ÑŸ"× 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"H, 8 is homeomorphic to Ò+ß,Ó , so Brouwer's Theorem generalizes part 2) of this example. 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 isa contraction mapping (for short, a contraction ) if there is a constant α−Ð!ß"Ñ such that .Ð0ÐBÑß0ÐCÑÑŸ α .ÐBßCÑ for all BßC−\ .

Notice that a contraction 0 is automatically continuous: for % !œ , choose $ % . Then, for all B−\ß .ÐBßCÑ$ implies .Ð0ÐBÑß0ÐCÑÑ α$%  . In fact, this choice of $ depends only on % and not the point B . When that is true, we say that 0 is uniformly continuous .

The definition of uniform continuity for 0 on \ reads:

a%b$ aB aCÐ.ÐBßCÑ $ Ê.Ð0ÐBÑß0ÐCÑÑ % Ñ

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

a% aB b $ aCÐ.ÐBßCÑ $ Ê.Ð0ÐBÑß0ÐCÑÑ % Ñ

165 Example 4.4

" " # 1) The function 0ÀÒ!ß% ÓÄÒ!ß % Ó given by 0ÐBÑœB is a contraction because for any " # # " BßC−Ò!ß% Ó, we have l0ÐBÑ
0ÐCÑlœlB
ClœlBCllB
ClŸ# lB
Cl .

2) However, the function 0ÀÄ‘ ‘ given by 0ÐBÑœB# is not a contraction. For 0 to be a contraction, there would have to be some α − Ð!ß "Ñ for which

lB
ClŸ# # α lB
Cl for all BßC in ‘ , and this would imply that lBClŸ α for all BÁC (which is false).

We could also argue that 0 is not a contraction by noticing that 0 À‘ Ä ‘ is not uniformly continuous. For example, if we choose %œ# , then no $ ! will satisfy the condition # aBaCÐ.ÐBßCÑ $ Ê.Ð0ÐBÑß0ÐCÑÑ#Ñ . For example, given $ ! , we can let Bœ $ and # $ $ # Cœ$ # . Then l0ÐBÑ
0ÐCÑlœ#% # .

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 g from the complete space g to g 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). 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 !α " and that .Ð0ÐBÑß0ÐCÑÑŸ α .ÐBßCÑ for all BßC−\ . Pick any point B−\! and apply the function 0 repeatedly to define

B" œ0ÐB ! Ñß # B#" œ0ÐB Ñœ0Ð0ÐB ! ÑÑœ0 ÐB ! Ñ ã 8 B8 œ0ÐB 8
" ÑœÞÞÞœ0 ÐBÑ ! ã We claim that this sequence ÐBÑ8 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 .ÐBßBÑ8 7 % for large enough 7ß8Þ Assume 7 8 . Applying the “contraction property” 8 times gives:

8 7 8
" 7
" .ÐB87 ßB Ñ œ.Ð0 ÐBÑß0 !! ÐBÑÑœ.Ð0Ð0 ÐBÑÑß0Ð0 ! ÐBÑÑÑ !

8
" 7
" # 8
# 7
# Ÿα .Ð0 ÐBÑß0!! ÐBÑÑŸ α .Ð0 ÐBÑß0 !! ÐBÑÑ

8 7
8 8 ŸÞÞÞŸ α .ÐBß0! ÐBÑÑœ ! α .ÐBßB !7
8 Ñ

166 Using the triangle inequality and the contraction property again, we get

8 .ÐBßB87 ÑŸα .ÐBßB !7
8 Ñ

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

8 # 7
8
" Ÿα Ð.ÐBßBÑ!" αα .ÐBßBÑ !" .ÐBßBÑÞÞÞ !" α .ÐBßBÑÑ !"

8 # 7
8
"8∞ 5 œα .ÐBßBÑÐ"!" αααα  ÞÞÞ Ñ .ÐBßBÑ! " 5œ! α 8 α .ÐB! ßB " Ñ œ"
α Ä ! as 8 Ä ∞ .

8 α .ÐB! ßB " Ñ So "
α % if 8 some RÞ Then .ÐBßBÑ8 7 % if 7 8 R . Therefore ÐBÑ8 is Cauchy. Since Ð\ß.Ñ is complete ÐBÑÄB8 for some B−\ .

By continuity, Ð0ÐBÑÑÄ0ÐBÑ8 . But also Ð0ÐBÑÑÄB8 because 0ÐBÑœB 88" . Because a sequence in a metric space can have at most one limit. we get that 0ÐBÑ œ B . ( If . were just a pseudometric, B might not actually be a fixed point: we could only say that .ÐBß0ÐBÑÑ œ ! . )

If C is also a fixed point for 0 , then .ÐBßCÑœ.Ð0ÐBÑß0ÐCÑÑŸα .ÐBßCÑ . Since ! α " , this implies that .ÐBßCÑœ! , so BœC
that is, the fixed point B is unique. ( If . were just a pseudometric, 0 could have several fixed points at distance ! from each other.) ñ

Notes about the proof

8 α .ÐB! ßB " Ñ 1) The proof gives us that if 7 8 , then .ÐBßBÑ8 7 "
α for each 8−  . If we 8 α .ÐB! ßB " Ñ fix 8 and let 7Ä∞ , Then ÐBÑÄB7 , so .ÐBßBÑŸ 8 "
α . Thus we have a computable bound on how well B8 approximates the fixed point B .

2) For large 8 we can think of B8 as an “approximate fixed point”
in the sense that 0 doesn't move the point B 8 very much. To be more precise,

.ÐB888 ß0ÐB ÑÑ Ÿ.ÐB ßBÑ.ÐBß0ÐB 88 ÑÑœ.ÐB ßBÑ.Ð0ÐBÑß0ÐB 8 ÑÑ

Ÿ.ÐB8 ßBÑα .ÐBßB 8 ÑœÐ" α Ñ.ÐB 8 ßBÑ

8 α .ÐB! ßB " Ñ Ÿ Ð" α Ñ"
α Ä ! as 8 Ä ∞.

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

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

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

Example 4.7 Let \ be a topological space and GÐ\ч be the set of bounded continuous real- valued functions 0À\Ä‘ , with the metric 3 Ð0ß1Ñœ sup Öl0ÐBÑ
1ÐBÑlÀB−\× . ( Since 0ß 1 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 3Ð0ß1Ñ % implies that 0 is “uniformly close” to 1À

3Ð0ß1Ñ % Ê l0ÐBÑ
1ÐBÑl % at every point B−\ ЇÑ

‡ 3 In the specific case of ÐGÐÑßÑ‘ 3 , the statement that Ð0ÑÄ08 is equivalent the statement (in analysis) that “Ð0Ñ8 converges to 0 uniformly on ‘ .”

Theorem 4.8 ÐG‡ Ð\Ñß3 Ñ is complete.

‡ Proof Suppose that Ð0Ñ8 is a Cauchy sequence in ÐGÐ\ÑßÑ3 : given % ! , there is some R so that 3Ð0ß0Ñ8 7 % whenever 8ß7 R . This implies (see Ð * Ñ , above ) that for each B − \ , l0ÐBÑ
0ÐBÑl8 7 % whenever 8ß7 R . For any fixed B , then, Ð0ÐBÑÑ8 is a Cauchy sequence of real numbers, so Ð0ÐBÑÑÄ8 some <− B ‘ . Define 0À\Ä ‘ by 0ÐBÑœ< B .To complete ‡ 3 the proof, we claim that 0−GÐ\Ñ and that Ð0ÑÄ08

First we argue that0 is bounded. Pick R so that 3 Ð0ß0Ñ"8 7 whenever 8ß7 R . Using 7œR , this gives 3 Ð0ß0Ñ"8 R for 8 R and therefore l0ÐBÑ
0ÐBÑl"8 R for every B − \. Now let 8 Ä ∞ , to get that

for every B l0ÐBÑ
0ÐBÑlŸ"R ,

so, for every B
" Ÿ0ÐBÑ
0R ÐBÑ Ÿ" .

Since 0R is bounded, there is a constant Q such that

QŸ0ÐBÑŸQR for every B .

Adding the last two inequalities shows that 0 is bounded because, for every B

l0ÐBÑl Ÿ Q  " . 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\ч ! 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 % ! and pick R so that 3 Ð0ß0Ñ8 7 # whenever % 7ß8 R. Then l0ÐBÑ
0ÐBÑl8 7 # for every B whenever 8ß7 R . Letting 7Ä∞ , we % get that l0ÐBÑ
0ÐBÑlŸ8 # for all B and all 8 R . Therefore 3 Ð0ß0Ñ8 % if 8 R .

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

l0ÐBÑ
0Ð+Ñl Ÿ l0ÐBÑ
0R ÐBÑll0 R ÐBÑ
0 R Ð+Ñll0 R Ð+Ñ
0Ð+Ñl

3 % Since Ð0ÑÄ08 , we can choose R large enough to make 3 Ð0ß0ÑÞR $ This implies that % % l0ÐBÑ
0ÐBÑlR$and l0Ð+Ñ
0Ð+Ñl R $ . Since 0 R is continuous at + , we can choose a % neighborhood [ of + so that l0ÐBÑ
0Ð+ÑlR R $ for all B−[Þ Then for B−[ we have

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

# Theorem 4.9 (Picard's Theorem) LetÐBßCÑ! ! be an interior point of a closed box H in ‘ , and suppose that 0 À H Ä ‘ is continuous. In addition, suppose that there is a constant Q such that for all ÐBßCÑ"" and ÐBßCÑ "# in H :

l0ÐB"" ßC Ñ
0ÐB "# ßC ÑlŸQlC "#
C l (L) Then i) there exists an interval M œÒB!
+ßB ! +ÓœÖBÀlB
B ! lŸ+×ß and ii) there exists a unique differentiable function 1 À M Ä ‘

C œ1ÐB Ñ such that ! ! for B − M œ 1w ÐBÑ œ 0ÐBß1ÐBÑÑ

In other words, on some interval M centered at B! , there is a unique solution Cœ1ÐBÑ to the initial value problem Cw œ 0ÐBßCÑ Ð‡Ñ œ CÐB! Ñ œ C !

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

1) The “strange” condition (L) says that “0 satisfies a Lipschitz condition in the variable C on H.” For our purposes, it's enough to notice that (L) will be true when the partial derivative 0C exists and is continuous on the closed box H . In that case, we know (from analysis) that |0ŸC | some constant QH on . Then, if we think of 0ÐBßCÑ" as a single-variable function of C and use the ordinary Mean Value Theorem, we get a point D between C" and C # for which

|0ÐB"" ßC Ñ
0ÐB "# ßC Ñlœl0 C" ÐB ßDÑl†lC "#
C l ŸQlC "#
C l

2) Consider a specific example: 0ÐBßCÑœC
B , HœÒ
"ß"Ó‚Ò
"ß"Ó , and ÐBßCÑœÐ!ß!Ñ! !. Since 0 C œ" , the preceding comment tells us that condition (L) is true. Cw œC
B Consider the initial value problem . Picard's Theorem states that there œ CÐ!Ñ œ ! is a unique differentiable function Cœ1ÐBÑ , defined on some interval Ò
+ß+Ó that solves this system: 1Ð!Ñœ! and 1ÐBÑœ0ÐBß1ÐBÑÑÐw that is, Cw œC
BÑ . As we will see, the proofs of Picard's Theorem and the Contraction Mapping Theorem can actually help us to find this

169 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.

Proof (Picard's Theorem) We begin by changing the initial value problem Ð‡Ñ into an equivalent problem that involves an integral equation Ї‡Ñ rather than a differential equation. Let C œ 1ÐBÑ be a continuous function defined on an interval M œÒB!
+ßB ! +Ó :

Suppose 1 satisfies (*): 0 is continuous, so 1ÐBÑœ0ÐBß1ÐBÑÑw is continuous. (Why? ) Therefore the Fundamental Theorem of Calculus tells us for all B − M ,

Bw B 1ÐBÑ
1ÐB! Ñœ 1 Ð>Ñ.>œ 0Ð>, 1Ð>ÑÑ.> , so 'B! ' B !

B 1ÐBÑœC!  0Ð> , 1Ð>ÑÑ.> (**) 'B!

B! Suppose 1 satisfies (**): then 1ÐB! ) œC !  0Ð>1Ð>ÑÑ.>œC , ! . Since 0Ð>ß1Ð>ÑÑ is 'B! continuous, the Fundamental Theorem gives that for all B−M , 1w ÐBÑ = 0ÐBß1ÐBÑÑ . Therefore the function 1ÐBÑ satisfies (*).

Therefore a function continuous 1M on satisfies ÐÑ1 * iff satisfies ÐÑ ** . ( Note that the initial condition Cœ1ÐBÑ! ! is“built into” the single equation Ї‡ÑÞ ) Now we show that a unique solution 1 to (**) exists on some interval MBÞ centered at !

First, we establish some notation:

We are given a constant Q in the Lipschitz condition (L).

Pick a constant O so that l0ÐBßCÑlŸO for all ÐBßCÑ−H . ( We will prove later in this chapter that a continuous real-valued function on a closed box in ‘# must be bounded. For now, we assume that fact from analysis.)

Because ÐBßCÑ−H ! ! int , we can pick a constant + ! so that

i) ÖBÀlB
B! lŸ+ׂÖCÀlC
C ! lŸO+שH , and ii) +Q  "

Let M œÖBÀlB
B! lŸ+לÒB !
+ßB ! +ÓÞ

We consider ÐG‡ ÐMÑß3 Ñ , where 3 is the metric of uniform convergence. By Theorem 4.8, this ‡ space is complete. Let F œÖ1−G ÐMÑÀa>−Mß l1Ð>Ñ
C! lŸO+× .

For an arbitrary 1−GÐMÑ1Ð>ч , might be so large that Ð>ß1Ð>ÑÑÂHÞ But if we look only at the functions in F , we avoid this problem: if 1−F and >−M , then Ð>ß1Ð>ÑÑ−H because of how we chose + . So for 1−F , we know that 0Ð>ß1Ð>ÑÑ is defined for all > − MÞ )

Notice that FÁg since the constant function 1ÐBÑœC! is certainly in F .

170 ‡ F is closed in GÐMÑ2−F : if cl , then there is sequence of functions Ð1ÑF8 in such that Ð1ÑÄ28 in the metric 3 . Therefore 1ÐBÑÄ2ÐBÑ8 for all B−M , and subtracting C ! gives

l18 ÐBÑ
C ! lÄl2ÐBÑ
C ! l .

Since l1ÐBÑ
ClŸO+8 ! for each B−M , then l2ÐBÑ
ClŸO+! for each B−M . Therefore 2−F, so F is closed. So ÐFßÑ3 is a nonempty complete metric space .

For 1−F , define on M a function 2œXÐ1Ñ using the formula

B 2ÐBÑœX () 1 ÐBÑœC!  0Ð>ß1Ð>ÑÑ.> . 'B!

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

B B B l2ÐBÑlŸ||| C!  0Ð>ß1Ð>ÑÑ.> | Ÿ||| C !  0Ð>ß1Ð>ÑÑ.> | Ÿ|| C !!  O.> ŸlClO+ , 'B! ' B! ' B ! so 2 is bounded. Therefore XÀFÄGÐMч . But in fact, even more is true: XÀFÄF , because B aB−M , | XÐ1ÑÐBÑ
C! || œ 2ÐBÑ
C ! lœl 0Ð>ß1Ð>ÑÑ.> | ŸO+ 'B!

Now we claim that X ÀÐFß3 ÑÄÐFß 3 Ñ is a contraction. To see this, we simply compute distances: if 1" and 1−F # , then

3ÐXÐ1ÑXÐ1Ñl"#, œ sup Ö | XÐ1ÑÐBÑ
XÐ1ÑÐBÑ " # | ÀB−M× B œ sup Ö | 0Ð>ß1Ð>ÑÑ
0Ð>ß1Ð>ÑÑ.>" # | ÀB−M× 'B! B Ÿ sup Ö || 0Ð>ß1Ð>ÑÑ
0Ð>ß1Ð>ÑÑ" # | .>| ÀB−M× 'B! B Ÿ sup Ö | 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+Q3 Ð1" ß1 # Ñ 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 , 1ÐBÑœXÐ1ÑÐBÑœC!  0Ð>, 1Ð>ÑÑ.> 'B! which is precisely condition (**). ñ

Example 4.10 If we combine the method in the proof of Picard's Theorem with the numerical estimates in the proof of the Contraction Mapping Theorem, we can get useful information about a specific initial value problem. To illustrate, we consider

Cw œC
B

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

171 We begin by choosing a box H with ÐBßCÑœÐ!ß!Ñ! ! in its interior: we select (rather arbitrarily) the box HœÒ
"ß"Ó‚Ò
"ß"ÓÞ Since l0ÐBßCÑlœlC
BlŸlBllClŸ# on H , we can use Oœ# in the proof. Because l0ÐBßCÑlœ"C throughout H , the Lipschitz condition (L) is satisfied with Q œ "Þ

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

i) ÖBÀ lBlŸ+ׂÖC À lClŸO+œ#+שH and ii) +Q œ+†""

" " " Again rather arbitrarily, we choose +œ# , so that MœÒB
+ßB+ÓœÒ
! ! # ß # Ó . ‡ ‡ " " Then FœÖ1−G ÐMÑÀl1ÐBÑ
!lŸO+לÖ1−G ÐMÑÀ| 1ÐBÑlŸ" 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 1! to be the constant function 1ÐBÑœ!Ò
ßÓ! on # # . ( If for simplicity we use a constant 1!, then 1œ! ! is the “best possible” choice since its graph goes through the point ÐB! ßC ! Ñ œ Ð!ß!ÑÞÑ

According to the proof of the Contraction Mapping Theorem, the sequence of functions 8 Ð18 Ñ œ ÐX Ð1 ! ÑÑ will converge (with respect to the metric 3 ) to 1 , where 1 is a fixed point for X ; 1 will be the solution to our initial value problem. We calculate:

B B B B# 1ÐBÑœC" !  0Ð>ß1Ð>ÑÑ.>œ! ! 0Ð>ß!Ñ.>œ
>.>œ
'B! ' ! ' ! #

B># B > # BB $# 1ÐBÑœXÐ1ÐBÑÑœ!# " '! 0Ð>ß
# Ñ.>œ ' !
#
>.>œ
'#

B>>$# B >> $# BBB %$# 1ÐBÑœXÐ1ÐBÑÑœ$ # '! 0Ð>ß
'#
Ñ.>œ ' !
'#

>.>œ
#%'#

and, in general,

B# B $ B 8" 18 ÐBÑœXÐ1 8
" ÐBÑÑœÞÞÞœ
#x
$x
ÞÞÞ
Ð8"Ñx

The functions 18 converge uniformly to the solution 1 .

In this particular problem, moreover, we are lucky enough to recognize that the functions 18 ÐBÑ ∞B8 ∞ B 8 B are just the partial sums of the series 8œ#
8x œ"B
 8œ! 8x œ"B
/Þ Therefore 1ÐBÑ œ "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 , after the fact, that the solution actually is valid for B 's in the much larger interval ‘ . )

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 " α 3 Ð1! ß1 " Ñ " #8 3Ð1! ß1 " Ñ 3(1ß1ÑŸ8 "
α . In this example α œ+Qœ# , so that 3 ( 1ß1ÑŸ8 " "
# "B"""# " " œ#####8
" sup Öl!
Ð
ÑlÀB−Mל8
" †œ $ 8# . Therefore, on the interval Ò
ßÓ# # , " 18 ÐBÑ is uniformly within distance #8# of the exact solution 1ÐBÑ

172 Finally, recall that our initial choice of 1−F! was arbitrary. Since lBlŸ" sin , we know that sin− F , and we could just as well have chosen 1ÐBÑœ! sin B . Then the functions 1ÐBÑ8 computed as XÐ1Ñœ1ßXÐ1Ñœ1ßÞÞÞ! " # # would be quite different Ðtry computing 1" and 1 # Ñ B but it would still be true that 1ÐBÑÄ1ÐBÑœ"B
/8 uniformly on MÀ this must be same limit 1 because the solution 1 is unique. ñ

The Contraction Mapping Theorem can be used to prove other results
for example, the Implicit Function Theorem. (You can see details 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 |0ÐBÑŸOw | for all BÞ Prove that 0 is a contraction (and therefore has a unique fixed point.)

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

|0ÐBÑ
0ÐCÑllB
Cl for all BÁC (**) for all BÁ C −‘ but such that 0 has no fixed point.

Note: The function 0 is not a contraction mapping. If we allow α œ " in the definition of contraction, your example shows that Contraction Mapping Theorem may not be true
not even if (as above) we “compensate” by using “ ” instead of “ Ÿ ” in the definition.

c) Find an example satisfying (**) where 0 ÀÒ!ß"ÓÄÒ!ß"Ó and 0 is not a contraction ( so compactness + (**) doesn't force 0 to be a contraction )

d) (Edelstein's Theorem) Show that if 0ÀÐ\ß.ÑÄÐ\ß.Ñ satisfies Ї‡Ñ and \ is compact, then 0 has a unique fixed point ++œBB and that lim 8 where ! is any point of \ and 8Ä∞ B œ08 B Hints: 1ÐBÑœ.ÐBß0ÐBÑÑ has a minimum value at some point Bœ+Þ Show that 8 !  + is the unique fixed point of 0 . Without loss of generality, assume all B8 Á + ; then .ÐBß+ÑÄj !Þ8 Prove that jœ!Þ At some point, use the fact that each sequence in \ has a convergent subsequence.

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

a) Suppose that b5 − for which 0 5 is a contraction. Then, by the Contraction Mapping Theorem, 05 has a unique fixed point : . Prove that : is also the unique fixed point for 0. b) Prove that the function cosÀÄ‘ ‘ is not a contraction. c) Prove that for some 5 − , cos5 is a contraction . Ð Hint: the Mean Value Theorem may be helpful . ) d) Pick 5−1œ so that cos5 is a contraction and let : be the unique fixed point of 1 . By a), : is also the unique solution of the equation cosB œ B . 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 | 1Ð!Ñ
:±8 <0.00001. e) For this 8 , use a computer or calculator to evaluate 18 Ð!Ñ . ( This “solves” the equation cosBœB with l Error l 0.00001 .)

E10. Consider the differential equation CœBCw with the initial condition CÐ!Ñœ" . Choose a suitable rectangle H and suitable constants OßQ+ 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,  is a dense subspace of the complete space (‘ß .Ñ . This is a model for the definition of a completion for a metric space. We will focus on the important case of metric spaces, but make a few additional comments along the way about pseudometric spaces where slight adjustments are necessary. µ µµ µ Definition 5.1 Ð\.Ñ, is a completion of Ð\ß.Ñ if: Ð\Ñ,. is complete µ . lÐ\‚\Ñ œ . , and µ \ is a dense subspace of \ . µ Since . œ . on \
, we will simplify and slightly abuse
the notation by also using . to refer µ to the metric on \ . Occasionally, if it's helpful to distinguish between . and the “extended” µ µ metric on \ , we may revert to more precise notation and use a different name, such as . , µ for the extended metric on \ .

Loosely speaking, a completion of Ð\ß .Ñ contains the additional points necessary Ð and no others) to provide limits for Cauchy sequences that fail to converge in Ð\ß .Ñ . In the case of Ð ß .Ñ, the additional points are the irrational numbers, and the resulting completion is Б ß.ÑÞ

If a metric space Ð\ß .Ñ is already complete, what would its completion look like? In that case, µ µ Ð\ß.Ñ is a complete subspace of Ð\.Ñ\ , , so must be closed in \. But \ must also be dense µ µ in \ , so \œ\
that is, a complete metric space is its own completion. µ µ If Ð\ß .Ñ is a complete pseudometric space, then \ might not be closed in \ and \ might µ contain additional points C . But each C − \
\ is the limit of a (Cauchy) sequence ÐBÑ from \ , and ÐBÑ already has a limit B−\ , so .ÐBßCÑœ!Þ Any new points C in µ 8 8 \
\ are “unnecessary additions” because every Cauchy sequence in \ already has a limit in \, and each of these “unnecessary additions” C is at distance ! from a point in \ .

Of course different spaces might have the same completion
for example, Ðß.Ñ‘ 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, Ðß.Ñ  " and ÐÖ8 À8− ×.Ñ , are homeomorphic topological spaces (countable sets, each with the discrete topology). But the completion of Ð ß.Ñ Ð œ itself!) is not homeomorphic to the " " completion ÐÖ!×∪Ö8 À8− ×ß.Ñ of ÐÖ 8 À8−  ×ß.ÑÞ

One way to create a completion : recall that if 0 is an (onto ) isometry between two metric spaces Ð\ß.Ñ and Ð]ß=Ñ , then Ð\ß.Ñ and Ð]ß=Ñ can be regarded as “the same” metric spaceÞ We can think of 0 as just “assigning new names” to the points of \ . If 0ÀÐ\ß.ÑÄÐ]ß=Ñ is an isometry from \ into ] , we can identify Ð\ß.Ñ with Ð0Ò\Óß=Ñœ an “exact metric copy” of Ð\ß.Ñ inside Ð]ß=ÑÞ If Ð]ß=Ñ happens to be complete, then Ð0Ò\Óß=Ñ is dense in the complete space Ð cl] 0Ò\Óß=Ñ . We agree to identify Ð0Ò\Óß=Ñ with Ð\ß.Ñ and call Ð cl] 0Ò\Óß=Ñ a completion of Ð\ß.Ñ even though \ is not literally a subset of cl] 0Ò\Ó . To find a completion of Ð\ß .Ñ, then, it is sufficient to find an isometry 0 from Ð\ß.Ñ into any complete space Ð]ß=Ñ .

175 w" " w Example 5.2 Consider with the metric .Ð8ß7Ñœl8
7 l . Ð ß.Ñ is isometric to " " (ÖÀ8−×ß.Ñ8  where . is the usual metric on 
in fact, the mapping 0Ð8Ñœ 8 is an isometry. So these spaces are “exact metric copies” of the other. " BecauseÐÖ!×∪Ö8 À8− ×ß.Ñ is a completion of its dense subspace " " w (ÖÀ8−×ß.Ñ8  , we can also think of ÐÖ!×∪ÖÀ8−×ß.Ñ8  as a completion of Ðß.Ñ
 even " though  is not literally a subspace of Ö!×∪Ö8 À8− ×Þ

The next theorem tells us every metric space has a completion and, as we will see later, it is essentially unique (so the method for creating it usually doesn't matter).

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

Proof By our earlier comments, it is sufficient to find an isometry of Ð\ß .Ñ into some complete metric space. We will use ÐG‡ Ð\Ñß3 Ñ where, as usual, 3 is the metric of uniform convergence (see Example 4.7 and Theorem 4.8 ) The theorem is trivial if \œgß so we assume that we can pick a point :−\ . For each point +−\ , define a function 9+ À\Ä ‘ using the formula 9 + ÐBÑœ.ÐBß+Ñ
.ÐBß:ÑÞ Each map 9+ is continuous because it is a difference of continuous functions and, for each B − \ , ‡ l9+ ÐBÑlœl.ÐBß+Ñ
.ÐBß:ÑlŸ.Ð+ß:Ñ, so 9+ is bounded. Therefore 9 + −GÐ\ÑÞ ‡ Define FÀÐ\ß.ÑÄÐG Ð\Ñß 3F9 Ñ by Ð+Ñœ+ Þ We complete the proof by showing that F is an isometry. This just involves computing some distances: for any +ß , − \ ,

399Ð+, ß Ñœ sup Öl 9 + ÐBÑ
9 , ÐBÑlÀB−\× œsup ÖlÐ.ÐBß+Ñ
.ÐBß:ÑÑ
Ð.ÐBß,Ñ
.ÐBß:ÑÑl À B − \× œ sup Öl.ÐBß+Ñ
.ÐBß,ÑlÀ B−\×Þ

For every B, l.ÐBß+Ñ
.ÐBß,Ñl Ÿ .Ð+ß,Ñ , and letting Bœ, gives l.ÐBß+Ñ
.ÐBß,Ñlœ.Ð+ß,Ñ . Therefore

3Ð 9+ ß 9 , Ñœ sup Öl.ÐBß+Ñ
.ÐBß,ÑlÀ B−\ל.Ð+ß,ÑÞ ñ

The completion of Ð\ß.Ñ given by this proof is Ð clG‡ Ð\Ñ F Ò\ÓÑ , 3 . The proof is “slick” but it has very little intuitive content. For example, if we apply the proof to Ð ß.Ñ , it is not at all clear that the resulting completion is isometric to Б ß.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. ( To reiterate: In Theorem 5.4, we will see that the method doesn't matter: the completion of Ð\ß .Ñ
no matter what method is used to construct it
always comes out “the same.” We will simply sketch this alternate construction here, and in the discussion it's probably clearer µ µ if we use . to refer to the metric . extended to \ .

We call two Cauchy sequences ÐBÑ8 and ÐCÑ 8 in Ð\ß.Ñ equivalent if .ÐBßCÑÄ!Þ8 8 It is easy to check thatµ is an equivalence relation among Cauchy sequences in Ð\ß .Ñ . Clearly, if ÐB8 Ñ Ä D − \ß then any equivalent Cauchy sequence also converges to D . And if two nonconvergent Cauchy sequences are equivalent, then they are “trying to converge to the same point” but that point is “missing” in \ .

176 µ We denote the equivalence class of a Cauchy sequence ÐBÑ by ÒÐBÑÓ and let \ be the µ 8 8 set of equivalence classes. Define the distance . between two equivalence classes by µ . ÐÒÐB88 ÑÓßÒÐC ÑÓÑœlim .ÐB 88 ßC 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 most tedious part) that µ µ µ Ð\ ß . Ñ is complete: any . -Cauchy sequence of equivalence classes ÐÒÐB ÑÓÑ must µ µ 8 converge to an equivalence class in Ð\ß . Ñ . µ For each B−\ , the sequence ÐBßBßBßÞÞÞÑ is Cauchy, and so ÒÐBßBßBßÞÞÞÑÓ−\ . µ µ The mapping 0 ÀÐ\ß.ÑÄÐ\ß. Ñ given by 0ÐBÑœÒÐBßBßBßÞÞÞÑÓ is an isometry, and it µµ µµ 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. Note: ‘ can also be constructed as a completion of  by using a method called Dedekind cuts in  . However that approach makes use of the ordering  in  ß so we cannot imitate this construction in the general setting of metric spaces where no ordering of elements exists.

The next theorem gives us the good news that it doesn't really matter, in the end, how we construct a completion
because the completion of Ð\ß.Ñ is “essentially” unique: all completions are isometric, and in a “special” way! We state the theorem for metric spaces and in the proof it is clearer to give the metrics on the completions new names
not referring to them as . . ( 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 0ÀÐ]ß=ÑÄÐ^ß>Ñ such that 0l\ is the identity map on \ . ( In other words, not only are the completions ] and ^ isometric, but there is an isometry between them that holds \ fixed. The isometry merely “renames” the new points in the “outgrowth” ]
\Þ )

Proof For each C−] , we can pick a sequence ÐBÑ\8 in which converges to C . Since ÐBÑ8 is convergent, ÐBÑ8 is Cauchy in Ð]ß=Ñ
and therefore also in Ð^ß>Ñ since both = and > agree with .\ on . Because Ð^ß>Ñ is complete, ÐBÑÄ8 some point D−^ , and we can define 0ÐCÑœD . (It is easy to check that 0ÀÐ]ß=ÑÄÐ^ß>Ñ is well-defined
that is, we get the same D no matter which sequence ÐBÑ8 we first chose converging to C .) For B−\ß what is 0ÐBÑ ? We can choose ÐBÑ8 to be the constant sequence BœB8 , and therefore 0ÐBÑœB . So 0l\ is the identity map on \ . w w w We need to verify that 0 is an isometry. Suppose C−] and we choose ÐBÑÄC8 . Then w w w .ÐB8 ßB8 Ñœ=ÐB 8 ßB 8 ÑÄ=ÐCßC Ñ and w w w w .ÐB8 ßB8 Ñœ>ÐB 8 ßB 8 ÑÄ >ÐDßD Ñœ>Ð0ÐCÑß0ÐCÑÑ , so =ÐCßCw Ñ œ >Ð0ÐCÑß0ÐC w ÑÑ ñ

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

177 6. Category

There are many different mathematical ways to compare the “size” of sets, and these methods are used 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, and in analysis there is a further 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 topologically useful idea for comparing the size of certain sets. The most interesting results in this section will be about complete metric spaces, but the basic definitions make sense in any topological space Ð\ßg ÑÞ

Definition 6.1 A subset E of a topological space Ð\ßÑg is called nowhere dense in \ if int\ (cl \ E ) œgÞ (In some books, a nowhere dense set is called rare .)

A set has empty interior iff its complement is dense. Therefore we also can say that E is nowhere dense in \ iff \
cl\ E is dense.

Intuitively, we can think of an open set S as including some “elbow room” around each of its points
if B − S , then all sufficiently near 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 is nowhere dense in \ iff int Jœg\
J iff is dense in \ . In particular, if a singleton set Ö:× is a closed set in \Ö:× , then is nowhere dense unless : is isolated in \Þ

2) Suppose E©ÞE‘ is nowhere dense iff cl E contains no interval Ð+ß,ÑÞ For example, each singleton set Ö<× is nowhere dense in ‘ . In particular, for 8 −  ß Ö8} is nowhere dense in ‘ . But note that Ö8× is not nowhere dense in  , because 8 is isolated in . Whether a set is 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"ש © ‘ Þ

" Ö8 À8− × 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 3) 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 that is is nowhere dense in \E iff cl\ is nowhere dense in \ .

Theorem 6.3 Let Ð\ßÑg be a topological space and F©E©\F . If is nowhere dense in E , then F is nowhere dense in \ .

Proof Suppose not. Then there is a point B− int\\ ÐFÑ© cl cl \ F . Since int \\ ÐFÑ cl is an open set containing B and B− cl\ F , then int \ Ð cl \ FÑ∩FÁg . So gÁ intcl\\ Ð FÑ∩F© intcl \\ Ð FÑ∩E© cl \ F∩Eœ cl. E F But int\\ÐFÑ∩E cl is a nonempty open set in E , so intE ÐFÑÁg
cl E which contradicts the hypothesis that F is nowhere dense in E . ñ

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

Lemma 6.4 In a topological space Ð\ßg Ñ :

1) If H is dense in \ and S is open in \ , then cl ÐS∩HÑœSÞ cl 2) If H is dense in \S and is open in \S∩H , then is dense in S . 3) If H is dense in \S and is open and dense in \S∩H , then is dense in \ . In particular, the intersection of two
and therefore finitely many
denseopen sets is dense. (But this is not true for countable intersections; can you provide an example? )

Proof 1) To show clS©ÐS∩HÑ cl , suppose B−SY cl . If is any open set containing B , then Y∩SÁg. Since H is dense, this implies that gÁÐY∩SÑ∩HœY∩ÐS∩HÑ . Therefore B −cl ÐS ∩ HÑÞ 2) Using part 1), we have clSÐS∩HÑœ cl \ ÐS∩HÑ∩SœÐ cl \S SÑ∩Sœ cl SœSÞ 3) If S is dense, then part 1) gives cl ÐS∩HÑœSœ\ñ cl .

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

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

\
cl ÐE"# ∪EÑœ\
Ð clcl E "# ∪ EÑœÐ\
cl EÑ∩Ð\
" cl. # Ñ

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

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

Definition 6.6 In Ð\ßÑg , a subset E 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 \ . ( Books that use the terminology rare for nowhere dense sets usually use the word meager for first category sets. )

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 is nowhere dense in \E , then is first category in \ .

2) is a first category set in ‘ .

3ÑF©E©\E If and is first category in \F , then is first category in \ . However F may not be first category in E , as the example Ö!שÖ!ß"ש ‘ shows. However, using Theorem 6.3, we can easily prove that if F is first category in E , then F is also first category in \. 4) A countable union of first category sets in \ is first category in \ .

" 5) EœÖ!×∪Ö8 À8− × . E is nowhere dense (and therefore first category) in ‘ ; but Eis second category in E
because any subset of E" that contains is not nowhere dense in E .

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

a) Consider the right-ray topology g‘ œÖgß ×∪ÖÐ+ß∞ÑÀ+− ‘‘ × on . ∞ ‘ is uncountable, but ‘ œ-8œ" J8 , where JœÐ
∞ß8ÓÞ 8 Each J 8 is nowhere dense in ÐßÑ‘g‘ , so isfirst category in ÐßÑ ‘g .

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

7) Let 8 "Þ A straight line is nowhere dense in ‘8 , so a countable union of straight lines in ‘8 is first category in ‘ 8 . ( Can ‘8 be written as a countable union of straight lines? The answer follows immediately from the Baire Category Theorem, proved below. But, in fact, the answer is clear from a simple argument using countable and uncountable sets. ) More generally, when 58 , a countable union of 5 -dimensional linear subspaces of ‘8 is first category in ‘ 8 Þ (Can a countable union of 5 -dimensional linear subspaces œ ‘8? )

It is not always easy to say 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 are the questions equivalent? )

As we observed in Lemma 6.4, the intersection of two (and therefore, finitely many) dense open sets is dense. Sometimes , however, 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 In any topological space Ð\ßg Ñ , the following two statements are equivalent :

1) If E is first category in \\
E , then is dense in \ ∞ 2) For each sequence KKK" , # ,..., 5 ,... of dense open sets, the intersection +5œ" K 5 is also dense.

180 Proof (1ÊKKK 2) Let " , # ,..., 5 ,... be a sequence of dense open sets. Each \
K 5 is closed ∞ ∞ ∞ and nowhere dense, so -5œ"Ð\
KÑ5 is first category. By 1), \
-5œ" Ð\
KÑœ5 + 5œ" K 5 is dense.

∞ (2Ê"ÑE Let be a first category set, say EœR8œ" 5 , where each R 5 is nowhere - ∞ dense in \ . Then each \
RœK cl5 5 is dense so, by 2), 5œ" Ð\
RÑ cl5 is dense. Since ∞ ∞ ∞ + +5œ"Ð\
Rќ\
cl5 - 5œ" cl R©\
5 - 5œ" Rœ\
E 5 , we see that \
E is dense. 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 holds.

Intuitively we can think of a nonempty Baire space as one which is “thick,” at least in some places. A “thin” first category set E can't “fill up” \À in fact, its complement \
E is dense (and therefore nonempty). This fact is the basis for how a Baire space is often used in “applications”: if you want to prove that there is an element in a Baire space \ with a certain propertyT , you can consider EœÖB−\ÀB does not have property T× . If you can show that E is first category in \ , it follows that \
EÁg .

Clearly, a space homeomorphic to a Baire space is a Baire space: being a Baire space is a topological property.

The following theorem tells us a couple of important facts.

Theorem 6.10 Suppose S is an open set in a Baire Space Ð\ßÑÞg

1) If S Á gß 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 \ ; since \
S is closed, it would follow that \
Sœ\, and therefore SœgÞ

2) Suppose S is open in the Baire space \ , and let E be a first category set in S . We ∞ must show that S
E is dense in S . Suppose EœR-5œ" 5 , where R 5 is nowhere dense in S . By Theorem 6.3, R5 is also nowhere dense in \ . Therefore E is first category in \\
E so is dense in \ . Since S is open, S∩Ð\
EÑœS
E is dense in S by part 2) of 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 \ œÐ ∩Ò!ß"ÓÑ∪Ö#× , with its usual topology. If we write ∞ \ œ-5œ" R5, then the isolated point “2” must be in some R5 , so R 5 is not nowhere dense. 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.

181 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 with many Baire spaces.

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

Proof If \œgß then \ is Baire, so we assume \ÁgÞ ∞ Let E œ-5œ" R5 , where R 5 is nowhere dense. We must show that \
E is dense in \. Suppose B−\Þ! 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 intJ!" is not a subset of cl R . Therefore we can choose a point B " in the open set intJ
R!" cl . Pick a closed ball J " , centered at B """!"! , so that B−J©J
R©J int cl . " If necessary, choose J" even smaller so that diam ÐJÑÞ" # ( 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 intJ"# is not a subset of cl R . Therefore we can choose a point B # in the open set intJ
R"# cl . Pick a closed ball J # , centered at B ###"#" , so that B−J©J
R©J int cl . " If necessary, choose J# even smaller so that diam ÐJÑÞ# $

For the induction step: suppose we have defined closed balls J" ßJ # ßÞÞÞßJ 5 centered at points " BßBßÞÞÞßB"#5, with diam ÐJÑ 33  " and so that B−J© 33 int J 3
"
R©J cl 33
" .

182 Since intJÁgBR5 (it contains 5 ) and since 5" is nowhere dense, we know that int J 5 is not a subset of clR5" . Therefore we can choose a point B5" in the open set int J
R 5 cl 5" . Pick a closed ball J5" , centered at B 5" , so that B−J©J
R©J 5" 5" int 5 cl 5" 5 . If necessary, " choose J5" even smaller so that diam ÐJÑ5" 5  # .

By induction, the closed sets J5 are defined for all 5, diam ÐJÑÄ!5 and J! ªJ " ªÞÞÞªJ 5 ªÞÞÞÞ Since Ð\ß.Ñ is complete, the Cantor Intersection Theorem says that ∞ there is a point B−5œ" J©JÞ5! For each 5 "B−J© , 55
"5 int J
R cl , so BÂR cl 5 , + ∞ so BÂRÞ5 Therefore B−\
-5œ" Rœ\
E5 , so B−J∩Ð\
EÑ ! and we are done. ñ

Example 6.13

1) Now we can see another reason why  is not completely metrizable. If Ð  ß.Ñw is complete, then Ð ß .w Ñ is a Baire space. But  , with the usual metric . is not a Baire space. w Therefore g.Á g . w so .µ.Þ/

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 see even more: that every point in \ must be a condensation point. Otherwise, there would be a non-condensation point B! − \ , and there would be some countable (possibly finite) open set SœÖBßBßBßÞÞÞBßÞÞÞ×Þ!"#8 Since each B 8 is non-isolated, the singleton sets ÖB×8 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 a subset EÐ\ßÑ of g is called an J5 set if E can be written as a countable union of closed sets, and that E is a K $ set if it can be written countable intersection of open sets. The complement of a K$ set is an J 5 set and vice-versa. ∞  is not an J5 set in ‘ . To see this, suppose that  œ-8œ" J8 , where the J 8 's are closed in ‘ . Since  is second category in ‘ , one of these J8 's must be not nowhere dense in ‘ . This J8 must therefore contain an open interval Ð+ß,Ñ . But then Ð+ß,Ñ©J8 ©  , which is impossible because there are rational numbers in any interval Ð+ß ,ÑÞ Taking complements, we see that  is not a K$ -set in ‘ .

183 Example 6.14 Suppose 0ÀÒ!ß"ÓÄ‘ is continuous. Then 0 has an antiderivative 0 " , by the Fundamental Theorem of Calculus. Since 0" is differentiable (therefore continuous), it has an antiderivative 0# . Continuing in this way, let 08 denote an antiderivative of 0 8
" . It is a trivial observation that: Ðb5aB0ÐBÑœ!Ñ5 Ê 0ÐBÑœ! for all B −Ò!ß"Ó

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

ÐaBb50ÐBÑœ!ÑÊ0ÐBÑœ! 5 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 05 ÐB ! ÑÁ!Þ

∞ The hypothesis in (*) lets us write Ò!ß"Óœ-5œ" J5 , where J 5 œÖB−Ò!ß"ÓÀ0ÐBÑœ!× 5
" œ0ÒÖ!×Ó5 ; these are closed sets in Ò!ß"Ó because each 05 is continuous. Since Ò!ß"Ó is second

category in itself, one of the sets J5! is not nowhere dense and therefore contains an interval

Ð+ß,Ñ. Since 05! is identically ! on Ð+ß,Ñ0 , is identically ! on Ð+ß,ÑÞ

We can repeat the same argument on any closed subinterval N œ Ò-ß.Ó © Ò!ß"Ó À by letting 1œ0lN and 1œ0lN5 5 , we can conclude that N contains an open interval on which 0 is identically ! . In particular , each closed subinterval NœÒ-ß.Ó contains a point B at which 0ÐBÑœ!. Therefore ÖBÀ0ÐBÑœ!× is dense in Ò!ß"Ó . Since 0 is continuous and 0 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 EßF where E∪FœÒ!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 ŸÞ# 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 ∞ +8œ" MœÖB×Þ8 If B−E , Andy wins; if B−Fß Beth wins. We claim that if E is first category, then Beth can always win .

∞ th Suppose EœR-8œ" 5 , where E 5 is nowhere dense in Ò!ß"ÓÞ Let Andy's 5 choice be N ÐœMÑ#5
". 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∩RÑÁg 5 , then cl N ÐN∩RÑ 5 must contain an nonempty open intervalÐ+ß ,ÑÞ But then Ð+ß,Ñ© clN5N5 ÐN∩RÑ© cl Rœ clÒ!ß"Ó R∩N© 5 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
ÐN∩RÑ clN 5 is nonempty and Beth can make her 5 choice M #5 to be a closed interval Ò+ß,Ó© int N
clN5 ÐN∩RÑ . This implies that M∩Rœg #55 , since

M#5 © int N
cl N5 ÐN∩RÑ© int N
ÐN∩RÑ©N
ÐN∩RÑœN
R 5 55

∞ ∞ Therefore +5œ"M∩5 - 5œ" Rœg 5 and Beth wins!

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 General Topology (S. Willard). Roughly, 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 GÐÒ!ß"ÓÑÞ‡ But the complete space ÐGÐÒ!ß"ÓÑßч 3 is a (“thick”) Baire space. Therefore it is second category in itself so G‡ ÐÒ!ß"ÓÑ
R Á g . In fact, GÐÒ!ß"ÓÑ
R‡ is dense in GÐÒ!ß"ÓÑÞ ‡ Any function in GÐÒ!ß"ÓÑ
R ‡ does the trick.

7. Complete Metrizability

Which metric spaces Ð\ß.Ñ are completely metrizable
that is, when does there exist a metric .w on \ for which Ð\ß.Ñ w is complete and .µ. w ? We have already seen, using the Baire Category Theorem, that  is not completely metrizable, and that certain familiar spaces like " Ö8 À8− × are completely metrizable. In this section, we will answer to this question.

Lemma 7.1 Let Ð\ß.Ñ be a pseudometric space. If 1À\Ä‘ continuous and g ( a ) Á!ß 1 then g À\Ä ‘ is continuous at a .

Proof The proof is a perfect “mimic” of the proof in analysis (where \ œ ‘ or ‘ 8 ).

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 ÐO , d wÑ is complete
that is, S is completely metrizable.

If ÐSß .Ñ is not already complete, it is because there are some nonconvergent Cauchy sequences in S . But those Cauchy sequences do have limits in Ð\ß.Ñ
they converge to points outside S but in FrS . The idea of the proof is to create a new metric .w on S that is equivalent to . , and which “blows up distances” 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.

1 1 1 " Let . be the usual metric on the interval NœÐ
ßÑÞ## The sequence ÐBÑœÐ
Ñ8 #8 is a nonconvergent Cauchy sequence in NÞÀNÄ The function tan‘ is a homeomorphism for which Ð0ÐB8 ÑÑ Ä ∞Þ

.ÐBßCÑœlw tan B
tan Cl defines a new metric on N , and tan ÀÐNß.ÑÄÐß.Ñw ‘ is an isometry because .ÐBßCÑœœlw tan B
tan Clœ.Ð tan Bß tan CÑÞ Since Ðß.Ñ‘ is w w complete, so in ÐNß.ÑÞÐThe sequence ÐBÑ8 is still nonconvergent in ÐNß. Ñ because .µ.àwbut in ÐNß.Ñ w , it is no longer a Cauchy sequence. Ñ

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 ÐBÑ8 when they approach Fr SÞ

185 " Proof of 7.2 Define 0ÐBÑœ.ÐBß \
SÑ . Since S is open, the denominator is never !B−S for so 0ÀSÄ‘ is continuous, by Lemma 7.1. On S , define .ÐBßCÑœ.ÐBßCÑl0ÐBÑ
0ÐCÑlÞw It is easy to check that .w is a metric on S .

Since sequences are sufficient to determine the topology in pseudometric spaces, we can show that .w µ . on S by showing that they produce the same convergent sequences in S . Suppose ÐBÑ8 is a sequence in S and B−SÀ

w w If .ÐBßBÑÄ!8 , then since . . we get .ÐBßBÑÄ!8 .

w If .ÐBßBÑÄ!8 , then l0ÐBÑ
0ÐBÑlÄ! 8 (since 0 is continuous), so .ÐBßBÑÄ!8 .

w w w ÐSß.Ñ is complete À suppose ÐBÑ.8 is a -Cauchy sequence in S . Since . . , the sequence . ÐBÑ8 is also . -Cauchy, so ÐBÑÄ 8 some B−\ . But, in fact, this B must be in SÞ

If B−\
Sß then .ÐBß\
SÑÄ!8 and so 0ÐBÑÄ∞Þ 8 In particular, this means that for every 8 we can find 7 8 for which 0ÐBÑ 0ÐBÑ"Þ7 8 Then w l0ÐBÑ
0ÐBÑl "7 8 , so .ÐBßBÑ " 7 8 . This contradicts the assumption that ÐBÑ8 is . w -Cauchy. w .w . w Therefore ÐBÑÄB−S8 . Since .µ. on SÐBÑÄB−S , 8 and so ÐSß.Ñ is complete. ñ

The next theorem generalizes this result, but the really interesting idea is actually 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 metric space and EK\ is a $ in , then there is a metric 3 onE , with 3 µ.E on , for which ÐEßÑ 3 is complete. In other words: a K $ subset of a complete space is completely metrizable.

∞ w Proof Let EœS + 3 , where S 3 is open in \. . Let 3 be a metric on S3 , equivalent to .S on 3 , 3œ" w w w such that ÐS.Ñ3 , 3 is complete. Let .ÐBßCÑœ 3 min Ö.ÐBßCÑß"×3 . Then .µ.µ.3 3 on S , and w ÐS3 ß. 3 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. )

∞ .3 ÐBßCÑ For BßC−E , define 3 ÐBßCÑœ  23 (the series converges since all .3 ÐBßCÑ Ÿ " ). 3œ" 3 is a metric on E :

Exercise

3 µ . on E :

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

186 . Suppose Ð+ÑÄ+−E©S8 3 . Let % ! . For every 3.µ., 3 on S 3 , ∞ .3 1 % so Ð+ÑÄ+−S8 3 . Pick 5 so that  24  2 . For each 3œ"ß5 ... , , we 4œ5" .3 Ð+ 8 ß+Ñ % can pick R3 so if 8 R 3 then 23 Þ 2 5 Then if 8 R 5 ∞ .Ð+ß+Ñ38 .Ð+ß+Ñ 38 œ max { RßR×" ..., 5 , we get 3 Ð++Ñœ 8 , 23   2 3 3œ" 3œ5" % % 3 Ÿ5† 25 + 2 œ %, so Ð+ÑÄ+−E8 . 3 Conversely, if Ð+ÑÄ+−E8 , then for % ! , we can pick R so that ∞ .3 Ð+ 8 ß+Ñ % 8 RÊ3 Ð+ß+Ñœ8  23  2 Ê.Ð+ß+Ñ " 8 % , so 3œ" ." . Ð+ÑÄ+−SÞ8 "" But .µ. on S "8, so Ð+ÑÄ+ .

ÐEß3 Ñ is complete:

Let Ð+Ñ8 be 3 -Cauchy in E . Let % ! . For any 3−  , we can choose R 3 so that

% if 7ß8 R3 , then 3 Ð+ß+Ñ 8723 , and therefore .Ð++Ñ 387, % .

Therefore Ð+Ñ8 is Cauchy in the complete space ÐSß.Ñ3 3 and there is a point .3 . B−S33 such that Ð+ÑÄB−S 8333 . But .µ. on S 3833, so Ð+ÑÄB−S©\ .

This is true for each 3 . But Ð+Ñ8 can have only one limit +−Ð\ß.Ñ , so we conclude that BœBœ"#3 ... œBœ ... œ+Þ Since +−S 3 for each 3 , we . 3 have +−E , so ( +ÑÄ+−E8 . But .µ3 on E , so ( +ÑÄ+−E8 . Therefore ÐEßÑ3 is complete. ñ

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

Proof œ-;−  Ö;×, so  is an J 5 set in ‘ . So, taking complements, we get that œ+;−  Ð
Ö;×Ñ ‘ is a K$ set in ‘ . 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.Ñww is any metric space and 0ÀÐ\ß.ÑÄÐ 0Ò\Óß.Ñ© ww Ð^ß.Ñww isa ww homeomorphismß then 0Ò\Ó is a K$ set in Ð^ß.Ñ ( therefore we say that “\ is an absolute K$ set among metric spaces” ) µ µ iv) \ is a K$ set in the completion Ð \.Ñ , .

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.Ñw are complete metric spaces with w E©\F©]2 and . Let be a homeomorphism from EF onto . Then there exist KE$ sets in \ and Fw in ] with E©E©E w cl and F©F©F w cl and there exists a homeomorphism 2w from E ww onto F 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)Ê ii) Suppose Ð\ß.Ñ is completely metrizable and let .w be a complete metric on \ equivalent to . . Thenthe identity map 3ÀÐ\ß.ÑÄÐ\ß.Ñw is a homeomorphism and 3Ò\Óœ\ w w is certainly a K$ set in Ð\ß.ÑÞ ( In other words, we can use Ð\ß.Ñ œ Ð]ß. Ñ in Part 2 ).

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.ÑÞw Let 0ÀÐ\ß.ÑÄÐ^ß.Ñ© ww the completion Ð ^ , . Ñ be a homeomorphism (into ^ ). We want to prove that Fœ0Ò\Ó must be a K$ set in ^ .

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

∞w ∞ Since EK]ß is a $ in there are open sets S]8 in such that Eœ+8œ" SœE∩S8 + 8œ" 8 ∞ w w ww w œ+8œ" ÐS∩EÑ8 . Therefore E is also a K$ in EÞ But 2ÀEÄF is a w w ∞ homeomorphism, so 2ÒEÓœ2ÒEÓœF is a K$ set in F . Therefore Fœ+8œ" Z 8 , where each w w µ Z8 isopen in F and, in turn, ZœF∩[8 8 where [ 8 is open in ^ .

wµ w ∞ µ Also, FK^FœY is a $ in , so +8œ" 8 , where the Y 8 's are open in ^ .

188 ∞ ∞w ∞ w Putting all this together, Fœ+8œ" Zœ8 + 8œ" Ð[∩FќР8 + 8œ" [Ñ∩F 8 ∞ ∞ µ œÐ[Ñ∩ÐYÑ8œ"8 8œ" 8 . The last expression shows that FK is a $ set in ^ . But then + + ∞ ∞ ∞ ∞ Fœ^∩Fœ^∩Ð+8œ" [Ñ∩Ð8 + 8œ" YќР8 + 8œ" [∩^Ñ∩Ð 8 + 8œ" Y∩^Ñ 8 is a K$ set in ^ .

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

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

Theorem 7.5 characterizes, for metric spaces, the “absolute K$ sets. This suggests other 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 Ð  ß.Ñ is an example of a Baire metric space that isn't complete.

To finish this section on category we mention, just as a curiosity and without proof, a generalization of an earlier result (Blumberg's Theorem: see Example II.5.8 ). I have never seen it used anywhere .

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 0lH À H Ä ‘ is continuous,

The original theorem of this type was proved for \ œ‘ or ‘ # 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$ set and every open set is an J 5 set (see Exercise II.E.15 ).

a) Find a topological space Ð\ßÑg containing a closed set J that is not a K $ set.

b) Recall that the “scattered line” is the space Б ß g Ñ where

gœÖY∪ZÀY is a usual open set in ‘ and Z©× 

Prove that the scattered line is not metrizable. (Hint: K$ or J 5 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 a Hausdorff space Ð\ßÑg and that \ has no isolated points. Prove that H is nowhere dense in \ .

E13. Let g be the cofinite topology. Prove that Ð\ßÑg 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 Ð ß.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−\ßb a constant QB such that l0ÐBÑlŸQ B for all 0− Y

Prove that there exists a nonempty open set Y and a constant Q (independent of B ) such that l0ÐBÑlŸQ 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 0ÀÐ\ßÑÄÐ]ß.Ñg , where Ð]ß.Ñ is a metric space where . is a metric. For each B − \ , define

=0 ÐBÑœ “the oscillation of 0BœÖ at ” inf diam Ð0ÒRÓÑÀR is a neighborhood of B×

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

1 b) Prove that for 8− , ÖB−\À = 0 ÐBÑ8 × is open in \ .

c) Prove that ÖB−\À0 is continuous at B× is a K$ -set in \ .

Note: Since  is not aK$ -set in ‘ , a continuous function 0 À ‘‘ Ä cannot have as its set of points of continuity. On the other hand,  is a K$ -set in ‘ and, from analysis, you should know an example of a continuous function 0ÀÄ‘ ‘ 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− , 0ÐBÑ0ÐCÑŸ ±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 Ð0ÐBÑÑÄ!8 and, likewise, if ÐCÑ 8 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 0 À‘‘ 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 0ÀÄ‘‘ is continuous and that 0 is continuous at BB−Þ iff  ∞ " Then œÖB− ‘ À =0 ÐBÑœ!ל+8œ" Y 88 , where YœÖB−‘ À = 0 ÐBÑ8 ×Þ

Each Y8 ª , and Y 8 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 ‘
Y 8 consists only of the endpoints of some disjoint open intervals. Therefore ‘
Y 8 is countable.

∞ ∞ It follows that ÖB−À0‘ is not continuous at Bל
‘+8œ" Yœ8 - 8œ" ‘
Y 8 is countable. Therefore 0 is continuous on more than just the points in  .

# E18. For each irrational : , construct an equilateral triangle X : (including its interior) in ‘ with one vertex at Ð:ß !Ñ and its opposite side above and parallel to the B -axis. Prove that 1 -ÖX: À : − × must contain an “open box” of the form Ð+ß,Ñ‚Ð!ß5 Ñ for some +ß,− ‘ and some 5 −  . In the hypothesis, we could weaken “equilateral” to read “ ... ” ? " (Hint: Consider RœÖ:−ÀX5 : has height ×5 . )

191 E19. Suppose 0ÀÄ‘‘ . 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\ßÑg is “absolutely open” if whenever 0ÀÐ\ßÑÄÐ]ßÑg g w 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 powerful topological properties. Formally, it is a stronger version of the Lindelo¨f property (although it preceded the Lindelo¨f property historically).

Compact spaces are relatively 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 one nonempty open set 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, Ö8 À8− × is not compact. " 4) The space \œÖ!×∪Ö8 À8− × is compact: if h is any open cover, and !−Y−h, then the set Y covers \ except for perhaps finitely many points. 5) ‘ is not compact since hœÖÐ
8ß8ÑÀ8−  × 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.

It is sometimes useful to have a characterization of compact spaces stated in terms of closed sets, and we can get one using FIP. As you read the proof of Theorem 8.4, you should see that the characterization is nothing more a restatement of the definition of compactness that uses complements to “convert” to closed sets.

Theorem 8.4 The following are equivalent for any topological space Ð\ßg Ñ

1) \ is compact 2) every family Y of closed sets in \ with the finite intersection property also has + Y Á g .

Proof 1)Ê 2) Suppose \ is compact and that Y is a family of closed sets with FIP. Let 8 hœÖ\
JÀJ−× Y. For any JßJßÞÞÞßJ−" # 8 Y , the FIP tells us that \Á\
+3œ" J 8 œÐ\
J" Ñ∪ÞÞÞ∪Ð\
J 8 ÑÞ In other words, no finite subcollection of h covers \. Since \ is compact, h cannot be a cover of cover \ , that is, \
-h œ + Ö\
YÀY−× h œ+Y Á g . 2)Ê 1) The proof of the converse is similar and left as an exercise. ñ

Theorem 8.5 For any spaceÐ\ , g Ñ

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

193 Proof a) The proof is like the proof that a closed subspace of a Lindelof¨ space is Lindelof. ¨ (Theorem III.3.10) Suppose J is a closed set in a compact space \ , and let h œÖYÀ−E×α α be a cover of J by sets open in J . For each α , pick an open set Zα in \ such that YœZ∩JÞ α α Since J is closed, the collection i œÖ\
J×∪ÖZÀ−E×α α is an open cover of \ .

Therefore we can find Zαα" ßÞÞÞßZ 8 so that Ö\
JßZ αα" ßÞÞÞßZ 8 × covers \ . Then ÖY αα" ßÞÞÞY 8 × covers J , so J is compact.

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

b) Suppose O is a compact set in a Hausdorff space\ . Let C−\
OÞ For each B−O, we can choose disjoint open sets YZ\BB and in where B−YC−Z B and B . Then h œÖYÀB−O×B covers Oß so there are finitely many points BßÞÞÞßB−O" 8 such that

YßÞÞÞßYBB" 8 covers O . Then C−ZœZ∩ÞÞÞ∩ZB" B 8 ©\
O , so \
O is open and 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, suppose l\l " and that . is the trivial pseudometric on \ . Then every 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 \ is homeomorphic to a subspace O of a Hausdorff space ]O] , 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 µ µ µ be closed in \ . Therefore Ð\ß.Ñœ\.ÑÐ , , so Ð\ß.Ñ is complete. ñ (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? )

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

Definition 8.7 Atopological space Ð\ßg Ñ iscalled

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 0 À \ Ä ‘ is bounded ( Check that this is equivalent to saying that every continuous real-valued function on \ assumes both a maximum and a minimum value ).

194 We want to look at the relations between these different varieties of compactness.

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\ßÑXg is a " -space
that is, if all singleton sets ÖB× are closed in \ .

Proof Suppose E is an infinite set in \ . Choose a sequence Ð+Ñ8 of distinct terms in E and let B be a cluster point of Ð+ÑÞ8 Then every neighborhood RB of contains infinitely many +ß8 '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 ÐB8 Ñ be a sequence in\ . We want to show that ÐBÑ8 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 infiniteÞB Let be a limit point of the set E . We claim B is a cluster point for ÐBÑÞ8 Suppose Y is an open set containing B and 8− . Let ZœY
ÖBßÞÞÞßB×∪ÖB×Þ" 8 Since ÖBßÞÞÞßB×" 8 is closed, Z is open and B−Z©YÞ Then Z∩ÐE
ÖB×ÑÁg , 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 8  where the : 8 's are distinct points and :Á 8 5 for all 8ß5−Þ " The idea is to make .Ð:ßBÑœ!8 8 for each 8 so that : 8 is a sort of “double” for 8 . So we define

Ú .Ð:8 ßB 8 Ñœ! "" " " .Ð ß Ñœl
lœ.Ð:ß:8 7 Ñ if 7Á8 Û 87 8 7 " " " if Ü .Ð:ß8 7 Ñœl 8
7 l 7Á8

It is easy to check that . is a pseudometric on \Ð\ß.Ñ . In , every nonempty set E (finite or infinite) has a limit point
B−E because 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 (So, by Lemma 8.8, a X" -space Ð\ßg Ñ is countably compact iff every infinite set in \ has a limit point).

Proof Suppose Ð\ßg Ñ is countably compact and consider any sequence ÐBÑ8 . Let X 8 œ the >2 ∞ “8 tail” of ÐBÑœÖBÀ5 8×Þ8 5 We claim that +8œ" cl XÁgÞ8

Assume not. Then every B is in Ð\
XÑ cl8 for some 8 , so Ö\
XÀ8−× cl 8  is a countable open cover of \ . Since \ is countably compact, there exists an R such that \œÐ\
cl XÑ∪ÞÞÞ∪Ð\
" cl XÑ R . Taking complements gives R gœ+8œ" cl XœX8 cl R , which is impossible.

∞ Let B−+8œ" cl X8 and let R be a neighborhood of B . Then R∩XÁg8 for every 8R , so contains an B5 for arbitrarily large values of 8 . Therefore B is a cluster point of ÐBÑÞ8

195 Conversely, suppose \ is not countably compact. Then \ has a countable open cover 8 ÖYßÞÞÞßYßÞÞÞ×" 8 with no finite subcover. For each 8 , pick a point B−\
8-3œ" Y 3 . Then the sequence ÐB8 Ñ has no cluster point. To see this, pick any B−\ ; we know B is in some Y 8 . This Y8 is a neighborhood of BBÂY and 7 8 for 7 8ñ .

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 are true:

Ú \ is compact or Ê\ is countably compact Ê\ is pseudocompact Û is sequentially compact Ü \ Proof It is clear from the definitions that a compact space is countably compact.

Suppose \ is sequentially compact. Then each sequenceÐB8 Ñ 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 \ is countably compact and that 0 À \ Ä ‘ is continuous. The sets
" Yœ08 ÒÐ
8ß8ÑÓœÖB−\À
80ÐBÑ8× form a countable open cover of \ , so for some R , the sets YßÞÞÞßY"R cover \ . Since Y©Y©ÞÞÞ©Y "#R , this implies that Yœ\ R . So
R0ÐBÑR for all B−\
in other words, 0 is bounded, so \ is pseudocompact. ñ

In general , no other implications hold 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 any 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. Conversely, assume \ is countably compact and that ÐBÑ8 is a sequence in \ . By Theorem 8.10, we know ÐBÑ8 has a cluster point, B . Since \ is first countable, there is a subsequenceÐBÑÄB85 (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 BßBßÞÞÞßB"#8 −\ such that \œFÐBÑ∪ÞÞÞ∪FÐBÑ% " % 8 .

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.

196 Example 8.14

1) Neither ‘ nor  (with the usual metric) is totally bounded since neither can be covered by a finite number of " -balls. 2) A compact pseudometric space Ð\ß.Ñ is totally bounded: for any % ! , we can pick a finite subcover from the open cover h œ ÖF% ÐBÑ À B − \×Þ

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

Proof If \œg , then \ is totally bounded, so assume \ÁgÞÐ\ß.Ñ If 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# so that .ÐBßBÑ "#% (or else FÐBÑ% " would cover \ ). We continue inductively. Suppose we have chosen points BßBßÞÞÞßB" # 8 in \ such that .ÐBßBÑ 3 4 % for each 3Á4ß3ß4œ"ßÞÞÞß8 . Then we can choose a point B8" at distance % from each of B" ß ÞÞÞß B 8
because otherwise the % -balls centered at BßÞÞÞßB" 8 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 , choose a finite number of 8 -balls that cover \ and let H 8 be the (finite) set " that contains the centers of these balls. For any C−\ß.ÐBßCÑ 8 for some B−HÞ8 This ∞ means that HœH-8œ" 8 is dense. Since H is countable, \ is separable. 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 Lindelo¨f (see Theorem III.6.5 ) and a countably compact Lindelo¨f 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 ÐBÑ8 so that all the B8 's are distinct ß and .ÐBßBÑ !7 8 if 7Á8ÐÑwhy? . Then we can find open sets Y8 such that i) B−Y 8887 , ii) Y∩Yœg for 7Á8, and so that iii) diam YÄ!8Ä∞Þ8 as

197 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 ÐB Ñ has no cluster point we can find, for each 8 , a ball FÐBÑw that contains 8 $8 8 w w no other B
7 that is, .ÐBßBÑ 8 7 $8 for all 7Á8 . Let $8 œÎ# $ 8 . Then for

7Á8ßwe have B7 ÂFÐBÑ$8 8 and $ 8 .ÐBßBÑÞ 87 Of course, two of these balls might overlap. To get the Y8 's we want, we shrink these balls (choose smaller radii %8 to replace the $ 8 ) 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 8 step, we get a new ball FÐBÑ%8 8 for which

if 7Á8BÂFÐBÑ , 7%8 8 (because % 88  $ Ñ and

FÐBÑ∩FÐBÑœg%88 % 4 4 for all 48 (because %4  % 8 .ÐBBÑ 4ß 8

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

Then we can let Y8 œF%8 ÐB 8 ÑÞ

8.ÐBß\
Y8 Ñ For each 8, define 0À\Ä8‘ by 0ÐBÑœ 8 . Since B−Y8 8 and \
Y 8 is closed, .ÐB8 ß\
Y 8 Ñ the denominator of 0ÐBÑ8 is not ! . Therefore 0 8 is continuous, and 0ÐBÑœ8Þ 88 Define ∞ 0ÐBÑœ 08 ÐBÑ . ( For any B−\ , this series converges because B is in at most one Y 8 and 8œ" therefore at most one term 08 ÐBÑ Á ! .) Then 0 is unbounded on \ because 0ÐBÑœ0ÐBÑ8 8 8 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+ containing + such that Z∩YÁg + 8 for at most finitely many 8 's.

If .Ð+ßBÑœ!8 for some B 8 , we can simply let ZœY +8 .

198 Suppose .Ð+ßBÑ !8 for all 8 . Since + is not a cluster point of ÐBÑ8 , 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− FÐ+Ñ∩Y% , 8# # 8 # 8 % % then .Ð+ßB ÑŸ.Ð+ßDÑ.ÐDßB Ñ  œ %). Then let Z œF% Ð+ÑÞ 8 8 # # + #

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 Ð0lZÑÒYÓ©[+. But Z + is open in \ß so Y is also open in \ , and 0ÒYÓœÐ0lZÑÒYÓ©[+ . So 0 is continuous at +ñ .

We now explore some properties of compact metric spaces.

Theorem 8.18 Suppose E is a compact subset of a metric space Ð\ß.ÑÞE 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 l0ÐBÑlQ for all B−E . But that means that E©FÐ+ÑQ ! , 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 \ . Every subset of Ð\ß.Ñ is both closed and bounded, but an infinite subspace of \ is not compact. However, the converse to Theorem 8.18 is true in ‘8 Ð with the usual metric Ñ, 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 E is closed and bounded, then E is compact. First consider the case 8œ" . Then E is a closed subspace of some interval M œ Ò+ß ,Ó and it is sufficient to show that M is compact. To do this, consider any sequence ÐB8 Ñ in M and choose a monotone subsequence ÐBÑ85 using Lemma 2.10. Since ÐBÑ8 5 is bounded, we know that ÐBÑÄ85 some point <−M (see Lemma 2.9 ).Since ÐBÑ8 has a convergent subsequence, M is sequentially compact and therefore compact. When 8 " , the proof is similar. We illustrate for 8œE 2. If is a closed bounded set in ‘#, then E is a closed subspace of some closed box of the formM œ Ò+ß,Ó‚Ò-ß.Ó . Therefore it is sufficient to prove that M is compact. To do this, consider any sequence ÐBßCÑ8 8 in M . Since

Ò+ß,Ó is compact, the sequence ÐBÑ8 has a subsequence ÐBÑ 8 5 that converges to some point

B−Ò+ß,Ó . Now consider the subsequence ÐBßCÑM85 8 5 in . Since Ò-ß.Ó is compact, the sequence ÐCÑ has a subsequence ÐCÑ that converges to some point C−Ò-ß.Ó . But in ‘# , 85 8 5 6 Ð?ß@ÑÄÐ?ß@Ñ iff Ð?ÑÄ? and Ð@ÑÄ@ in ‘. So ÐB ßC ÑÄÐBßCÑ−MÞ This shows 88 8 8 8856 5 6 that M is sequentially compact and therefore compact. For 8 # , a similar argument clearly works
the argument 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 particular, for compact metric spaces
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 Let % œ" . Pick points BßÞÞÞßB"8 so that \œFÐBÑ∪ÞÞÞ∪FÐBÑ "" "8 and let Qœmax Ö.ÐBßBÑÀ3ß4œ"ßÞÞÞß8×3 4 . For any pair of points BßC−\ , we have B−FÐBÑ" 3 and C−FÐBÑ" 4 for some 3ß4 , so that

.ÐBßCÑŸ.ÐBßBÑ.ÐBßC3 34 Ñ.ÐC 4 ßCÑ"Q "œQ # .

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

Notice that the converse to the theorem is false: “total boundedness” is a stronger condition than “boundedness.” For example, let .ÐBßCÑœw min Ö"ßlB
Cl× on ‘ . Then diam Ðß.Ñœ" ‘ w so (‘ß.Ñw is bounded. Because .ÐBßCÑœlB
Clw when lB
Cl"ß we have 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 BßÞÞÞßB"8 so that \œFÐBÑ∪ÞÞÞ∪FÐBÑ% " 8 . Of course these # # balls also cover E . But to show that ÐEß.Ñ is totally bounded, the definition requires us to show that we can cover E with a finite number of % -balls centered at points in EÞ Let NœÖ4ÀFÐBÑ∩EÁg×% and, for each 4−N , pick +−FÐBÑ∩EÞ% # 4 4# 4 E We claim that the balls FÐ+Ñ cover E . In fact, if +−E , then +−FÐBÑ% for some 4 ; % 4 # 4 % % and of course + is also in FÐBÑÞ% Then .Ð+ß+ÑŸ.Ð+ßBÑ.ÐBß+Ñ  œ %, so 4 # 4 4444 # # 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 (Ê ) We have already seen that the compact space Ð\ß .Ñ is totally bounded and complete. ( For completeness, here is a fresh argument: let ÐB8 Ñ be a Cauchy sequence in Ð\ß.Ñ. Since \ is countably compact, ÐBÑ8 has a cluster point B\ 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.

Since Ð\ß.Ñ is totally bounded, we can cover \ with a finite number of 1-balls, and one of them
call it F
" must contain B 8 for infinitely many 8 . 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 B 8 for infinitely many 8 .

" Continue inductively in this way. At the induction step, suppose we have already defined 5 -balls FªFªÞÞÞªF" # 5 where each ball contains B8 for infinitely many 8 . Since ÐFß.Ñ5 is totally " bounded, we can cover it with a finite number of 5  " -balls, one of which
call it F©F
5"5must contain B 8 for indefinitely many 8 . This inductively defines an infinite descending sequence of balls FªFªÞÞÞªFªÞÞÞ"#5 where each F 5 contains B 8 for infinitely many values of 8 .

Then we can choose B8" − F "

B−F8# # , with 8 8 #" ã

B85 −F 5 , with 8 8 5 5
" ÞÞÞ 8 " ã

The subsequence ÐBÑ85 is Cauchy since B−F 856 for 6 5 and diam FÄ!Þñ 5

Example 9.4 With the usual metric . :

i)  is complete and not compact (and therefore not totally bounded).

" ii) Ö8 À8− × 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 Ö8 À8−  × are homeomorphic because both are countable discrete spaces. Total boundedness is not a topological property. We remark here, without proof, that for every separable metric space Ð\ß .Ñ , there is an equivalent metric .w such that Ð\ß.Ñ w is totally bounded.

" iv) Ö8 À8− ×∪Ö!× is compact and is therefore both complete and totally bounded.

201 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 ÀÐ\ß.ÑÄÐ]ß.w Ñ is uniformly continuous if

a%$ b aBaC−\ ( .ÐBßCÑ $ Ê.w Ð0ÐBÑß0ÐCÑÑ % Ñ

Clearly, if \œÖB" ßÞÞÞßB 8 × is finite, then every continuous function 0 À Ð\ß.Ñ Ä Ð]ß=Ñ is uniformly continuous. ( For % ! and each 3œ"ßÞÞÞß8 , we pick thethe $ 3 thatworks at B 3 in the definition of continuity. Then $œÖßÞÞÞ× min $" $ 8 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 0ÀÐ\ß.ÑÄÐ]ß.Ñw is continuous, then 0 is uniformly continuous.

Proof If 0 is not uniformly continuous, then for some % ! , “no $ works.” In particular, for " that % , $ œ8 “doesn't work”
so we can find, for each 8 , a pair of points ?@8 and 8 with " w .Ð?ß@Ñ888 but .Ð0Ð?Ñß0Ð@ÑÑ 88 % Þ By compactness, there is a subsequence

Ð?85 Ñ Ä B − \. The corresponding subsequence Ð@ÑÄB85 also because .Ð@ßBÑ8 5 " Ÿ.Ð@88 ß? Ñ.Ð? 8 ßBÑ .ÐB 8 ßBÑÄ!Þ Therefore, by continuity, we should have 55 585 5

Ð0Ð?85 ÑÑÄ0ÐBÑandalso Ð0Ð@ 8 5 ÑÑÄ0ÐBÑÞ Butthisisimpossiblesince w .Ð0Ð?85 Ñß0Ð@ 8 5 ÑÑ % for all 5 . ñ

Uniform continuity is a strong and useful condition. For example, Theorem 9.6 implies that a continuous function 0 À Ò+ß ,Ó Ä ‘ must be uniformly continuous. This is one of the reasons why “continuous functions on closed intervals” are so nice to work with in calculus
the following example is a simple illustration:

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

B! œ+B " ÞÞÞB 3
"3 B ÞÞÞB 8 œ, and let 73 and Q 3 denote the inf and sup of the set 0ÒÒBßBÓÓ3
"3 . Let ? 3 BœB
BÞ 3 3
"

8 8 The sums 733? BŸ  Q 33 ? B 3œ" 3œ" are called the lower and upper sums for 0 associated with this partition.

t is easy to see that the sup of the lower sums (over all possible partitions) is finite: it is , called the lower integral of 0 on Ò+ß,Ó and denoted + 0ÐBÑ.B . Similarly, the inf of all the upper '

sums is finite: it is called the upper integral of 0 on Ò+ß,Ó and denoted , 0ÐBÑ.BÞ It is easy to
'+ , , verify that +0ÐBÑ.BŸ + 0ÐBÑ.BÞ If the two are actually equal , we say 0 is (Riemann) '
'
integrable on Ò+ß,Ó and we write , 0ÐBÑ.Bœ , 0ÐBÑ.Bœ , 0ÐBÑ.BÞ '+ '
+ ' +

202 Uniform continuity is just what we need to prove that a continuous 0 on Ò+ß,Ó isintegrable. If 0 is continuous on Ò+ß ,Ó , then 0 is uniformly continuous. Therefore, for % ! , we can choose % $ ! such that l0ÐBÑ
0ÐCÑl,
+ whenever lB
ClÞ$ Pick a partition of Ò+ß,Ó for which % each ?3BÞ $ Then for each 3Q
7Ÿ, 3 3 ,
+ Þ Therefore
8 8 , , '+0ÐBÑ.B
' + 0ÐBÑ.BŸ Q33? B
 7 33 ? B
3œ" 3œ" 8 8 % 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!ß* Ó∪Ò *$ ß Ó∪Ò $* ß Ó∪Ò * ß"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, *
clearly survives forever and ends up in G .

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!ßÓ$ Òß"Ó$ ÐJÑ! ÐJÑ#

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

" #" #( )" #"* #!( )#& #' Ò!ß#( ÓÒßÓÒß #(* *#( ÓÒßÓ #($ Òß $#( ÓÒßÓ #(* Òß *#( ÓÒß"Ó #( (J!!! ) ( J !!# ) ( J !#! ) ÐJÑ !## ÐJÑÐJÑÐJÑÐJÑ #!! #!# ##! ###

etc. etc.

>2 5 " At the 5 stage the set remaining is the union # closed subintervals ß each with length $5 . We 5 label them J88ÞÞÞ8" # 5 where Ð8ß8ßÞÞÞ8Ñ−Ö!ß#×Þ " # 5

Notice that, for example, JªJ! !# ªJ !#! ªJ !#!! ªJ !#!!# ªÞÞÞ . 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.” So for every sequence  = œ Ð8" ß8 # ßÞÞÞß8 5 ßÞÞÞÑ − Ö!ß#× , we have

J8" ªJ 88 "# ªÞÞÞªJ 88ÞÞÞ8 "#5 ªÞÞÞ

203 Since is complete, the Cantor Intersection Theorem gives that ∞ a one-point Ò!ß "Ó +5œ" J8" 8 # ßÞÞÞ8 5 œ set ÖB×Þ= The Cantor set G is then defined as

 G œÖB= À=−Ö!ß#× ×©Ò!ß"Ó .

Comments :

w w w w  w 1) If =Á=œÐ8ß8ßÞÞÞß8ßÞÞÞÑ−Ö!ß#×" # 5 , we can look at the first 5 for which 8Á85 5 . Then B−Jand B−Jw w . Since these closed intervals are disjoint, we conclude = 8ÞÞÞ8"5
"5 8 = 8ÞÞÞ8 "5
" ß8 5  B= Á B = w . Therefore each sequence = − Ö!ß #× corresponds to a different point in G , so lGl lÖ!ß#×lœ# i! œ-. Since G©Ò!ß"Ó , we conclude lGlœ- .

2) An observation we don't really need: each B − Ò!ß"Ó can be written using a “ternary decimal” ∞ >8 expansion Bœ $8 œ!Þ>>ÞÞÞ>ÞÞÞ"#5three , where each >−Ö!ß"ß#×Þ 5 Just as with base 10 8œ" " decimals, the expansion for a particular B is not always unique: for example, $ œ!Þ"!!!ÞÞÞ three œ!Þ!#####ÞÞÞthree Þ It's easy to check B has two different ternary expansions iff B is the endpoint of one of the deleted “open middle thirds” in the construction. It is also 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 (therefore also complete) metric space.

ii) Every point in G is a limit point of G
that is ßG 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. ) To see this, suppose B= −G , where =œÐ8ß8ßÞÞÞß8ß8 "# 55" ,  " w 8ÞÞÞÑ−Ö!ß#×5# . Given % ! , pick 5 so $5  % . Define 85" ( œ! or # ) so that w 85" Á85" ß and let

w w i ! = œÐ8"# ß8 ßÞÞÞß8 5 ß85" ß8 5 2 ÞÞÞÑ−Ö!ß#× .

" w Then BB= and = are distinct points in G , but both are in J8ßÞÞÞ8"5 and diam Jœ 8ßÞÞÞ8 "5 $5 % . Therefore lB
Bl= =w % , 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 lGlœ-Þ However, we can also see this from the discussion in ii): there are #i! œ - different ways we could redefine the infinite “tail” Ð85" ßÞÞÞ8 56 ßÞÞÞ ) of =
and each version produces a different point in GBÞ at distance% from = From Example 6.13(2), every point in G is a condensation point.

iv) In some ways, G is “big.” For example, the Cantor set has as many points as Ò!ß "Ó . Also, since G is complete, it is second category in itself. But G 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.

204 To see this, suppose MœÐ+ß,Ñ©G . At the 5 th stage of the construction, we must have

M ©exactly one of the J 8" ßÞÞÞ8 5 . (This follows immediately from the definition of an interval.) " Therefore ,
+$5 for every 5 , so +œ, and MœgÞ

v) G is also “small” in another sense. In the construction of G , the open intervals that " " " " $ are deleted have total length # %Ð ÑÞÞÞœ# œ" . In some sense, the “length” of $ˆ * ‰ #( "
$ 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 a “Cantor set” obtained by some other construction is actually the same topologically 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 E is homeomorphic to the Cantor set G .

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. Theorem 10.1 can be used to show that the resulting “generalized Cantor set” is homeomorphic to G , but the total lengths of the deleted open intervals may be different: that is, the result is can be a topological Cantor set but with positive measure. “Measure” is not a topological property.

Here's one additional observation that we don't need, but which you might see in an analysis course. As we noted earlier, G consists of the points B−Ò!ß"Ó for which we can write

∞ >8 Bœ $8 œ!Þ>>ÞÞÞ>ÞÞÞ"#5three , with each > 8 −Ö!ß#×Þ 8œ" ∞ 2,8 We can obviously rewrite this as Bœ $8 , where each ,−Ö!ß"×8 and therefore we can define 8œ" a function1 À G Ä Ò!ß"Ó as follows:

∞ ∞ 2,8 , 8 for B−Gß 1ÐBÑœ1Ð$8 Ñœ  # 8 . 8œ" 8œ" More informally,:

1Ð!Þ#,"#$ #, #, ÞÞÞthree Ñœ!Þ, "#$ , , ÞÞÞ two

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

( $ 1Ð* Ñœ1Ð!Þ#!####ÞÞÞÞthree ) œ!Þ"!""""ÞÞÞ two œ % and ) $ 1Ð* Ñœ1Ð!Þ##!!!!ÞÞÞthree ) œ!Þ""!!!ÞÞÞ two œ %

205 In fact, for +ß,−G , we have 1Ð+Ñœ1Ð,Ñ iff the interval Ð+ß,Ñ is one of the “deleted middle thirds.” It should be clear that 1 is continuous and that if +, in G , then 1Ð+ÑŸ1Ð,Ñ
that is, 1 is weakly increasing. (Is 1 onto? )

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 KÀÒ!ß"ÓÄÒ!ß"Ó in the following simple-minded way:

1ÐBÑif B−G KÐBÑ œ 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 horizontally over each “deleted middle third” Ð+ß,Ñ from Ð+ß1Ð+ÑÑ to Ð,ß1Ð,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 B in a “deleted middle third” Ð+ß,ÑK , is differentiable and Kw ÐBÑ œ !Þ

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 1 works in a similar way for any closed set J © ‘ and any continuous function 1ÀJÄ‘ . 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 form Ð+ß,Ñ , Ð
∞ß,ÑÐ+ß∞Ñ or We can extend 0 to a continuous function JÀÄ‘ ‘ simply by extending the graph of 0 over linearly over each M8 À

if MœÐ+ß,Ñ8 , then let the graph J over M 8 be the straight line segment joining Ð+ß0Ð+Ñß >9 Ð,ß0Ð,ÑÑÞ if MœÐ
∞ß,Ñ8 , then let J have the constant value 0Ð,Ñ98 I 8 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 J is a closed subset of Ð\ß.Ñ , 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
CÀBßC−GלÒ
"ß"Ó. Although G has measure ! , the difference set in this case has measure # !

206 Exercises

E21. Suppose that 0À\Ä] is continuous and onto, where \] and are any topological spaces. Prove If \ is compact, then ] 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 g be the usual topology on Ò!ß "Ó and suppose g" and g # are two other topologies on Ò!ß"Ó such that ggg"#"§§ Prove that ÐÒ!ß"Ó , g Ñ is not Hausdorff and that ÐÑ Ò!ß"Ó , g # is Á Á not 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 topological space \ is pseudocompact iff every continuous 0 À \ Ä ‘ achieves both a maximum and a minimum value.

E24. Suppose 0ÀÐ\ß.ÑÄÐ]ß.ÑÞw 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.Ñ and that E is compact. Prove that .ÐEßFÑ !Þ Is this necessarily true if both E and F are not compact?

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 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.

207 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 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\lœi!, 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 (that is: 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 \ such that g. œ g .

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 repeatedly rotate it by some fixed angle " on W " a

b) Prove that if Ð\ß.Ñ is totally bounded and 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 # -balls; one of these balls must contain 0ÐBÑ8 for infinitely many values of 8 .

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.Ñw is totally bounded for every metric .w µ.. 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 $ ! such that for all BFÐBÑ©Y : $ for some Y− h . ( 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 .ÐEßFÑ !for every pair of disjoint closed sets EßF©\Þ

E34. Suppose E©‘8 (with the usual metric). Prove that E is totally bounded if and only if E is bounded.

208 Chapter IV Review

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

1. Let E be the set of fixed points of a continuous function 0À\Ä\ß where \ is a Hausdorff space. E is closed in \ .

2. Let W denote the set of all Cauchy sequences in  that 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 0 À‘ Ä ‘ for which the Cantor set is the set of fixed points.

5. If \ has the cofinite topology, then every subspace of \ 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 ‘Æ for which is a subbase. Then Б ß g Ñ is a Baire space.

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

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

10. If g is the cofinite topology on g , 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 0À\Ä\ . The set GœÖB−À0‘ 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.

209 17. If S is nonempty and open in Ð\ßÑSg , then is not nowhere dense.

18. A nonempty nowhere dense subset of ‘ must contain an isolated point.

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

20. Suppose F is nonempty subset of  with no isolated points. F 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 E is first category in \ and FF © E , then is first category in E .

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

24. If  has the cofinite topology, then every infinite subset is sequentially compact.

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

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

27. Suppose 0ÀÒ!ß"ÓÄ‘ is continuous. Let I be the subset of the Cantor set G that consists of the endpoints of the open intervals deleted from Ò!ß"Ó in the construction of G . If 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 ^ œ Ò!ß "Ó# have the usual topology. Every nonempty closed subset of ^ is second category in itself.

32. Suppose 0ÀÒ!ß"ÓÄ8‘ 7 and GœÖB−Ò!ß"ÓÀ0 8 is continuous at B× . There is a metric . equivalent to the usual metric on G for which ÐGß.Ñ 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 EK is a $ set in ‘ Þ

210 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 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.Ñw is totally bounded. Then Ðß.Ñ  w cannot be complete.

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

39. Suppose 0ÀÒ!ß"ÓÄ8‘ 7 and GœÖB−Ò!ß"ÓÀ0 8 is continuous at B× . Then there is a metric d on G , equivalent to the usual metric, such that ÐGß .Ñ is totally bounded.

40. Suppose Ð\ßÑg is compact, that 0À\Ä ‘ is a continuous function, and 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 also has no isolated points.

43. Suppose that for each point B−Ð\ß.Ñ , there is an open set YB such that ÐYß.Ñ B is complete (that is, \ is “locally complete” ). Then there is a metric dw µ d on \ such that Ð\ß.w Ñ is complete. µ µ µ 44. For a metric space Ð\ß.Ñ with completion ( \ß.Ñ , it can happen that ±\
\±œi ! . µ µ 45. Let . be the usual metric on  . There is a completion ( ß.Ñ of Ð  ß.Ñ 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 is the Cantor set, then ‘
GJ is an 5 set in ‘ .

50. A subspace E of a metric space Ð\ß.Ñ is compact iff E is closed and bounded.

51. Let UœÖÒ+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. Let . be the usual metric on j# . Then Ðjß.Ñ # is sequentially compact.

53. Let . be the usual metric on j# . Then Ðjß.Ñ # is countably compact.

54. A subspace of a pseudocompact space is pseudocompact.

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

56. Suppose G is the Cantor set and 0ÀGÄG is continuous. Then the graph of 0 (a subspace of G ‚ G , with its usual metric) is totally bounded.

±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 E and F are subspaces of Ð\ßg Ñ 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 8 times. Then for each B−Ò
"ß"Ó , there exists an 8 (perhaps depending on B ) such that cos 8 BœBÞ

62. Suppose .w is any metric on !ß"Ó equivalent to the usual metric. Then ÐÒ!ß"Óß.Ñw 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 dw µ d, Ð\ß.w Ñ is totally bounded.

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

212