Notes on compact metric spaces

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$Math 3C Homework 3 Solutions$

4 CLASS NOTES FOR APRIL 14, 2000 For instance, R is not bounded (it can’t be enclosed inside a ball) and it is not totally bounded either. The set [0, 1] is both bounded and totally bounded (for any ε > 0 one can cover [0, 1] by a finite number of ε-balls). Now consider the integers Z with the discrete (or “teleport”) metric 1 if x = y d(x, y) = 0 if x = y In other words, all the integers are distance 1 apart from each other under this metric. In this case, Z is actually bounded (e.g. it is contained in B(0, 2)) but it is not totally bounded (it is not possible to cover Z by balls of radius 1/2). Exercise 3.4. Show that every totally bounded set is bounded. Proof (Optional) Let X be a totally bounded set. Then it can be covered by a finite number of balls of radius 1 (for instance). Let’s call these balls B(x1, 1), . . . , B(xn, 1). Let R denote the radius R = 1 + max 1≤i≤n d(xi, x1). From the triangle inequality we see that B(xi, 1) ⊂ B(x1, R) for all 1 ≤ i ≤ n. Since the B(xi, 1) cover X, we thus see that and hence X is bounded. X ⊂ B(x1, R) Theorem 3.5. Let (X, d) be a metric space The following statements are equivalent: • (i) X is compact. • (ii) Every sequence in X has at least subsequence which converges in X. • (iii) X is complete and totally bounded. The properties (ii) and (iii) are a bit more intuitive to work with than (i). For instance, from (ii) we can see why (0, 1) is not compact, and it is plausible that [0, 1] is compact. The proof of this theorem is somewhat complicated. There are three steps: showing that (i) implies (ii); showing that (ii) implies (iii); and showing that (iii) implies (i). Of these, the first two are fairly straightforward. The last one is nastier and will be left to the written notes. 4. Proof of (i) =⇒ (ii). Let X be a compact set, and suppose that x1, x2, . . . is a sequence in X. We have to find a subsequence of the xi which converges in X.
There are two cases: CLASS NOTES FOR APRIL 14, 2000 5 Case 1: There exists a point x ∈ X such that every open ball centered at x contained infinitely many elements of the sequence x1, x2, . . . . Then we would be done, for the following reason. Since the ball B(x, 1) contains infinitely many elements of the sequence, we can pick an element xn1 in B(x, 1). Then, since B(x, 1/2) contains infinitely many elements of the sequence, it must contain at least one element xn2 with n2 > n1. Repeating this, we can find an element xn3 ∈ B(x, 1/3) with n3 > n2. More generally, we can choose a subsequence xn1, xn2, . . . such that nj+1 > nj and xnj ∈ B(x, 1/j) for all j = 1, 2, 3, . . .. Then it is clear that this sequence is convergent to x, and we are done. Case 2: For every point x ∈ X there exists an open ball centered at x which contains at most finitely many elements of the sequence x1, x2, . . .. For every x ∈ X, let Bx be an open ball centered at x which contains at most finitely many elements of the sequence xn. Then the collection of all these balls {Bx}x∈X is an open cover of X, because every x ∈ X is contained in at least one ball in the collection, namely the ball Bx centered at x. However, this open cover cannot have a finite sub-cover, because every ball only contains a finite number of elements of the sequence, and there are an infinite number of elements of the sequence to cover. This contradicts compactness. 5. Proof of (ii) =⇒ (iii). Let X satisfy property (ii), i.e. every sequence in X has a convergent subsequence. We need to show that X is complete and totally bounded. Let’s first show that X is complete. Let x1, x2, . . . be a Cauchy sequence in X. We need to show that this sequence converges. By (ii), we can find a subsequence xn1, xn2, xn3, . . . of x1, x2, . . . which converges to some point in X, say x. This is almost what we want, but we need to show that the entire sequence converges to x, not just a subsequence. I’ll leave this as an exercise: Exercise 5.1. If xn is a Cauchy sequence, and a sub-sequence xnj of xn converges to x, then xn itself converges to x. Proof (Optional) We need to show that for every ε > 0 there exists N > 0 such that d(xn, x) < ε for all n > N. Well, because xnj converges to x, we can find a J > 0 such that d(xnj, x) < ε/2 for all j > J. Also, since xn is Cauchy, we can find an N > 0 such that d(xn, xm) < ε/2
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