NOTES WEEK 13 DAY 2 SCOT ADAMS

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1 NOTES WEEK 13 DAY 2 SCOT ADAMS Recall: Let px, dq be a metric space. Then, for all S Ď X, we have p S is sequentially compact q ñ p S is closed and bounded q. DEFINITION 0.1. Let px, dq be a metric space. proper means: for all S Ď X, we have Then px, dq is p S is closed and bounded q ñ p S is sequentially compact q. So, in a proper metric space, sequentially compact is exactly the same as closed and bounded. Let s make a couple of nonproper metric spaces. Recall that the standadrd metric d R on R is defined by d R px, yq x y. Then define X : p0, 8q. Recall that d X : d R px ˆ Xq. Let d : d X. Then px, dq is a metric space. Let S : p0, 1s. Then S r 1, 1s X X, so, since r 1, 1s is closed in R, we see that S is closed in X. Also S Ď B d p1{2, 1q, so S contained in a ball in px, dq. Then S is bounded in px, dq. On the other hand p1, 1{2, 1{3, 1{4,...q is a sequence in S that does not subconverge in S. So S is not sequentially compact. Therefore, in px, dq, it is not true that every closed bounded set is sequentially compact. That is, px, dq is not proper. Let s make one more nonproper metric space. Let X and d be as in the last paragraph, so X p0, 8q and, for all x, y P X, we have dpx, yq x y. Note that t ÞÑ e t : R Ñ X is a bijection, with inverse x ÞÑ ln x : X Ñ R. Define a metric d 1 on R by d 1 ps, tq dpe s, e t q. Then, d 1 is set up so that, under the mapping t ÞÑ e t : R Ñ X, the metric d 1 on R corresponds to d on X. Geometers would say that pr, d 1 q is isometric to px, dq, i.e., there s a distance-preserving bijection R Ñ X. To a geometer, the metric spaces pr, d 1 q and px, dq look the same. So, since px, dq is nonproper, pr, d 1 q is also nonproper. On the other hand, we will show below that pr, d R q is proper. (This is the 1-dimensional special case of the Heine-Borel Theorem see below.) However, because t ÞÑ e t : R Ñ X is a homeomorphism (continuous Date: December 1, 2016 Printout date: December 6,

2 2 SCOT ADAMS with continuous inverse), it follows that T d 1 T dr. That is, T d 1 is just the standard topology on R. Consequently, pr, d R q and pr, d 1 q look the same to a topologist. However, to a geometer, they re very different because the first is proper, while the second one is not. DEFINITION 0.2. Let px, dq be a metric space, let p P X and let r ą 0. Then Bpp, rq : tx P X dpp, xq ď ru. We assigned HW#50. REMARK 0.3. Let px, dq be a metric space, let S Ď X. Assume that S is bounded in px, dq. Then Cl X S is bounded in px, dq. Proof. Since S is bounded in px, dq, choose B P B d such that S Ď B. Choose p P X and r ą 0 such that B Bpp, rq. Let A : Bpp, r ` 1q. Then C P B d, so it suffices to show that Cl X S Ď A. Since S Ď B Bpp, rq Ď Bpp, rq, we get Cl X S Ď Cl X pbpp, rqq. By HW#50, Bpp, rq is closed in X, so Cl X pbpp, rqq Bpp, rq. Then Cl X S Ď Cl X pbpp, rqq Bpp, rq Ď Bpp, r ` 1q A, as desired. LEMMA 0.4. Let px, dq be a proper metric space, let a P X N. Assume that a is bounded in px, dq. Then a is subconvergent in X. Proof. Let C : Cl X pimra sq. Then a P pimra sq N Ď C N. By assumption a is bounded in px, dq, i.e., imra s is bounded in px, dq. Then, by Remark 0.3, C is bounded in px, dq. So, since C is closed in X and since px, dq is proper, we see that C is sequentially compact. So, since a P C N, it follows that a is subconvergent in C. Then, by an unassigned exercise in Week 13 Day 1, a is subconvergent in X. We assigned HW#50. Unassigned exercise: Let px, dq be a metric space. proper if and only if Show: px, dq Ď X, r p S is bounded q ñ p Cl X S is sequentially compact q s. Recall that a closed subset C of a sequentially compact topological space X is itself sequentially compact, in the relative topology on C inherited from X. (In fact, the proof is not hard: any sequence in C is a sequence in X and must be subconvergent in X, and, as C is closed in X, is subconvergent in C as well.) Also, recall the Bolzano-Weierstrass Theorem, which asserts that r0, 1s is sequentially compact. The next result is called the Heine-Borel Theorem:

3 NOTES WEEK 13 DAY 2 3 Recall, for all n P N, that the standard norm on R n is denoted and that the standard metric on R n is denoted d n. They are defined by: x a x 2 1 ` ` x 2 n and d n px, yq x y. THEOREM 0.5. Let n P N. Then pr n, d n q is proper. Proof. Let a closed bounded subset S of R n be given. We wish to show that S is sequentially compact. Since S is bounded in R n choose B P B n such that S Ď B. Choose p P R n and r ą 0 such that B Bpp, rq. Let I : t1,..., nu. For all j P I, let C j : rp j r, p j ` rs. For all j P I, C j is homeomorphic to r0, 1s, and so, by Bolzano-Weierstrass, must be sequentially compact. Let K : C 1 ˆ ˆ C n. By Theorem 0.8 of Week 13 Day 1, K is sequentially compact. Claim: S Ď K. Proof of claim: We wish to P S, z P K. Let z P S be given. We wish to show that z P K. Since K C 1 ˆ ˆ C n, we wish to P I, z j P C j. Let j P I be given. We wish to show that z j P C j. That is, we wish to show that z j P rp j r, p j ` rs. We have pz j p j q 2 ď pz 1 p 1 q 2 ` ` pz n p n q 2 z p 2 ă r 2. Then 0 ď pz j p j q 2 ă r 2, so a pz j p j q 2 ă? r 2. That is, z j p j ă r. Then p j r ă z j ă p j ` r. Then z j P pp j r, p j ` rq Ď rp j r, p j ` rs, as desired. End of proof of claim. Since S is closed in R n, it follows that S X K is closed in K. From the claim, we see that S XK S. Then S is closed in K. So, since K is sequentially compact, we conclude that S is sequentially compact. Recall HW#47-48: convergent implies Cauchy implies bounded. Recall HW#49: Cauchy and subconvergent implies convergent. DEFINITION 0.6. Let px, dq be a metric space. complete means: for all a P X N, p a is Cauchy q ñ p a is convergent q. Then px, dq is So a complete metric space is one in which convergent is the same as Cauchy. Recall that the standard metric d R on R is defined by d R px, yq x y. Let the set Q of rational numbers be given the metric d Q : d R pq ˆ Qq. Then T dq is the relative topology on Q inherited from R. We argue that pq, d Q q is not complete, as follows: Let a : p3, 3.1, 3.14, 3.141, , ,...q P Q N be the sequence of decimal approximations to π. Then a is convergent in R, hence

4 4 SCOT ADAMS Cauchy in pr, d R q, hence Cauchy in pq, d Q q. However, because π R Q, we see that a is not convergent in Q. The intuition you want to develop here is that pq, d Q q is not such a good metric space because it has many holes or missing points. For example, π is missing from Q and one wants to fill in that hole, in order to make Q into a better metric space. The process of filling in all these missing points is called completion, and once we have finished that process, we end up with pr, d R q. Similarly, for all n P N, if we complete Q n, we get pr n, d n q. So, following this intuition, we should be able to show, for all n P N, that pr n, d n q is complete; however, by Theorem 0.5, we already know that pr n, d n q is proper, and the next theorem show that proper implies complete. THEOREM 0.7. Let px, dq be a proper metric space. Then px, dq is complete. Proof. We wish to show, for all a P X N, that p a is Cauchy q ñ p a is convergent q. Let a P X N be given. We wish to show p a is Cauchy q ñ p a is convergent q. Assume that a is Cauchy. We wish to show that a is convergent. Because a is Cauchy, by HW#48, we see that a is bounded. So, since px, dq is bounded, by Lemma 0.4, we conclude that a is subconvergent. So, since a is Cauchy, by HW#49, we conclude that a is convergent, as desired. After Theorem 0.7, it is natural to ask whether a complete metric space is necessarily proper. The answer is no. For example, let d be the metric on R defined by dpx, yq mintd R px, yq, 0.99u. It s not hard to show that a sequence in R is d-cauchy iff it is d R -Cauchy. It s not hard to show that T d T dr. Then a sequence in R is d-convergent iff it is d R -convergent. Also, since T d T dr, it follows that a sequence in R is d-subconvergent iff it is d R -subconvergent. It s not hard to show that B d p0, 1q R. Let a : p1, 2, 3,...q P R N. Then imra s Ď R B d p0, 1q. Then a is d-bounded. Since a is not d R -subconvergent in R, we see that a is not d-subconvergent in R, So, since a is d-bounded in R, we see, from Lemma 0.4, that pr, dq is nonproper. By Theorem 0.5, the

5 NOTES WEEK 13 DAY 2 5 metric space pr, d R q is proper, so, by Theorem 0.7, the metric space pr, d R q is complete. We conclude that every d R -Cauchy sequence in R is d R -convergent. Then every d-cauchy sequence in R is d-convergent. Thus pr, dq is complete. Then pr, dq is complete, but not proper. There s another example, I would like to describe next, but I want first to talk about the idea of dimension in topology. In linear algebra, dimension means the cardinality of a basis, but vector spaces are much less stretchy than topological spaces. In the context of topology, it s even difficult to say what the word dimension means. Just as an example, consider the question of whether r0, 1s is homeomorphic to r0, 1s ˆ r0, 1s Ď R 2. Let I : r0, 1s and, for all n P N, let I n : I ˆ ˆ I Ď R n. Can we pull and stretch at something onedimensional like I and stretch it out into something two-dimensional, like I 2? Note, by the way that the idea of space-filling curves allows us to create a continuous surjection I Ñą I 2, but such a map, it turns out, is never injective. It turns out that I and I 2 are not homeomorphic, and there s a clever way to see this: Remember that a topological space X is connected if its only clopen sets are H and X. Some people say 0-connected to mean connected. Note that, for any p P I 2, I 2 ztpu is connected. Perhaps we might say that I 2 is 0-connected after any puncture. That s a topological property. Note that I does not have that property; in fact Izt0.5u is not 0-connected. Therefore I and I 2 cannot be homeomorphic. For all n P Nzt1u, the topological space I n is 0-connected after any puncture, and is therefore not homeomorphic to I. We can show that I 2 and I 3 are not homeomorphic, as follows. There is a topological property called 1-connected which sometimes goes by the alternative name of simply connected. It turns out that the topological space I 3 is 1-connected after any puncture, but I 2 ztp0.5, 0.5qu is not 1-connected. Then I 2 and I 3 are not homeomorphic. There are further topological properties called 2-connected, and then 3-connected, and so on. We can use them to n P N, [ ( m n ) implies ( I m is not homeomorphic to I n ) ]. So, even in the stretchy, fluid world of topology, dimension seems to have some kind of meaning. An entire mathematical discipline, called algebraic topology, has sprung up for a variety of reasons, but one of them is to try to get some sense of how to define the dimension of a topological

6 6 SCOT ADAMS space. It turns out that there are many possible definitions, and, for metric spaces, there are even more; it s a very complicated subject. Typically, in college courses, we tend to stick to the finite-dimensional world, and expose students to infinite-dimensional examples once they enter graduate school, when they take a course on functional analysis. In this class, we break from this convention. Define X : I N and define a metric d on X by dpx, yq supt x i y i s.t. i P Nu. Because N is infinite, the metric space px, dq feels infinite-dimensional, and, in any reasonable definition of dimension, it should be. In class, we provided a way to visualize elements of X as sections of an I-bundle over N. Given such a section, it has infinitely many coordinates, one for each element of N. Each coordinate of a section is an element of I. We talked about how to think the 0.1-ball about a point in X, by viewing the point a section, and then drawing the 0.1-neighborhood about each of its coordinates. Define p P X N by: p 1 : p1, 0, 0, 0, 0, 0,...q P X p 2 : p0, 1, 0, 0, 0, 0,...q P X p 3 : p0, 0, 1, 0, 0, 0,...q P X p 4 : p0, 0, 0, 1, 0, 0,...q P X.. For all x, y P X, we have dpx, yq ď 1, so X is bounded, and so any subset of X is bounded. Then imrp s is bounded, i.e., p is a bounded sequence in X. However, for all i, j P N, we have: p i j q ñ p dpp i, p j q 1 q. It follows that p has no Cauchy subsequence, and, since (HW#47) convergent implies Cauchy, we see that p has no convergent subsequence. Thus, by Lemma 0.4, px, dq is nonproper. We will defer the proof that px, dq is complete until next semester. For now, just remember that many of the typical infinite-dimensional metric spaces are complete and nonproper. Let s now retreat from infinite-dimensional topology and geometry, and go back to the one-dimensional topological space R. DEFINITION 0.8. Let S Ď R. Then S has a max or has a maximum means: [ S X rubpsqs H ], i.e., [ Dy P S s.t. S ď y ]. Recall, for any S Ď R, that max S : ELTpS X rubpsqq; then S has a max iff max S.

7 NOTES WEEK 13 DAY 2 7 Recall the following characterization of closure: LEMMA 0.9. Let X be a topological space, let S Ď X and let p P X. Then: p p P Cl X S q ô U of p in X, U X S H q. Proof. Proof of ñ: Assume: p P Cl X S. We wish to U of p in X, we have U X S H. Let a nbd U of p in X be given. We wish to show: U X S H. Assume: U X S H. We aim for a contradiction. Since U X S H, we see that S Ď XzU. Then Cl X S Ď Cl X pxzuq. Since U is open in X, it follows that XzU is closed in X. Then Cl X pxzuq XzU. Then p P Cl X S Ď Cl X pxzuq XzU. We conclude that p R U. However, U is a nbd of p in X, so p P U. Contradiction. End of proof of ñ. Proof of ð: U of p in X, we have U X S H. We wish to show: p P Cl X S. By definition of closure, Cl X S č t C Ď X p C is closed in X q and p S Ď C q u. We therefore wish to show, for all C Ď X, that r p C is closed in X q and p S Ď C q s ñ r p P C s. Let C Ď X be given. We wish to show that r p C is closed in X q and p S Ď C q s ñ r p P C s. Assume: ( C is closed in X ) and ( S Ď C ). We wish to show: p P C. Assume that p R C. We aim for a contradiction. Let U : XzC. Then U is open in X and p P U. Then U is a nbd of p in X. Then, by assumption, we have U X S H. So, since S Ď C, we get U XC H. However, U XC pxzcqxc H. Contradiction. End of proof of ð. LEMMA Let S Ď R. Assume that S is nonempty, closed in R, and bounded above in R. Then S has a max. Proof. We wish to show: Dy P S s.t. S ď y. Let y : sup S. Then y P R and S ď y. We wish to show: y P S. Since S is closed in R, we have Cl R S S. We wish to show: y P Cl R S. Since S is nonempty, choose a P S. Then a P R, so 8 ă a. Also, a P S ď y, so a ď y. Then 8 ă a ď y, so 8 ă y. Since S is bounded above in R, choose z P R s.t. S ď z. Then z P UBpSq, so, since sup S minrubpsqs, we get sup S ď z. Since z P R, we conclude that z ă 8. Then y sup S ď z ă 8, so y ă 8.

8 8 SCOT ADAMS Since y P R and 8 ă y ă 8, we conclude that y P R. By Lemma 0.9, we wish to show, for any nbd U of y in R, that U XS H. Let a nbd U of y in R be given. We wish to show that U X S H. The set tpy ε, y ` εq ε ą 0u is a nbd base of y in R. Choose ε ą 0 such that py ε, y `εq Ď U. Since y sup S, we get both [ S ď y ] and [ NOT(S ď y ε) ]. Since we have [ NOT(S ď y ε) ], choose s P S such that s ą y ε. Since s P S ď y ă y ` ε, we get y ε ă s ă y ` ε. Then s P py ε, y ` εq Ď U. So, as s P S, we conclude that s P U X S. Then U X S H, as desired. We now prove the Maximum Value Theorem (MaxVT): THEOREM Let X be a nonempty, sequentially compact topological space and let f : X Ñ R. Assume that f : X Ñ R is continuous. Then imrfs has a max. Proof. Because domrfs X H, it follows that imrfs H. Because X is sequentially compact and because f : X Ñ R is continuous, it follows, from Lemma 0.7 Week 12 Day 1, that imrfs is sequentially compact. Then, by Lemma 0.8 Week 12 Day 1, we see that imrfs is closed in R and bounded in R. Since imrfs is bounded in R, it follows that imrfs is bounded above in R. Then, by Lemma 0.10, imrfs has a max, as desired.