Problem Set 2: Solutions Math 201A Fall 2016 Problem 1 ...

Problem Set 2: Solutions Math 201A: Fall 2016

Problem 1. (a) Prove that a closed subset of a complete metric space is complete. (b) Prove that a closed subset of a compact metric space is compact. (c) Prove that a compact subset of a metric space is closed and bounded.

Solution

? (a) If F X is closed and (xn) is a Cauchy sequence in F , then (xn) is Cauchy in X and xn x for some x X since X is complete. Then x F since F is closed, so F is complete.

? (b) Suppose that F X where F is closed and X is compact. If (xn) is a sequence in F , then there is a subsequence (xnk) that converges to x X since X is compact. Then x F since F is closed, so F is compact. Alternatively, If {G X : I} is an open cover of F , then {G : I} F c is an open cover of X. Since X is compact, there is a finite subcover of X which also covers F , so F is compact.

? (c) Let K X be compact. If (xn) is a convergent sequence in K with limit x X, then every subsequence of (xn) converges to x. Since K is compact, some subsequence of (xn) converges to a limit in K, so x K and K is closed.

? Suppose that K is not bounded, and let x1 K. Then for every r > 0 there exists x K such that d(x1, x) r. Choose a sequence (xn) in K as follows. Pick x2 K such that d(x1, x2) 1. Given {x1, x2, . . . , xn}, pick xn+1 K such that

d(x1, xn+1) 1 + max d(x1, xk). 1kn

By the triangle inequality,

d(xk, xn+1) d(x1, xn+1) - d(x1, xk) 1 for every 1 k n.

It follows that d(xm, xn) 1 for every m = n, so (xn) has no Cauchy subsequences, and therefore no convergent subsequences, so K is not compact.

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Problem 2. Let A be a subset of a metric space X with closure A?. Define the interior A and boundary A of A by

A = {G A : G is open} , A = A? \ A.

(a) Why is A open and A closed? (b) Prove that X \ A? = (X \ A). (c) Prove that A is closed if and only if A A, and A is open if and only if A Ac. (d) If A is open, does it follow that (A?) = A?

Solution ? (a) A union of open sets is open so A is open, and an intersection of closed sets is closed so A? and A = A? (A)c are closed. ? (b) Note that x A if and only B (x) A for some > 0. If x A?c, then B (x) A?c for some > 0 since A?c is open. Since A? A, we have A?c Ac, so B (x) Ac, meaning that x (Ac). It follows that A?c (Ac). For the reverse inclusion, note that if x (Ac), then there exists > 0 such that B (x) Ac, so x is not the limit of any sequence in A, meaning that x A?c. It follows that (Ac) A?c, so (Ac) = A?c. ? (c) If A is closed, then A? = A so A = A(A)c A. For the converse, note that since A A A?, we have

A? = (A? A) (A? (A)c) = A A.

If A A, then it follows that A? A, so A = A?, meaning that A is closed. ? (d) This is not true in general. For example, define A R2 by

A = (x, y) : x2 + y2 < 1 \ {(x, 0) : 0 x < 1} .

Then A is open, but (A?) = {(x, y) : x2 + y2 < 1} = A.

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Problem 3. Let X be a metric space with a dense subset A X such that every Cauchy sequence in A converges in X. Prove that X is complete. Solution

? Let (xn) be a Cauchy sequence in X. Since A is dense in X, we can choose a sequence (an) in A such that d(xn, an) 0 as n . (For example, choose an A such that d(xn, an) < 1/n.)

? Given any > 0, there exists M N such that d(xn, an) < /3 and d(xm, xn) < /3 for all m, n > M , so d(am, an) d(am, xm) + d(xm, xn) + d(xn, an) < , which shows that (an) is a Cauchy sequence in A. It follows that (an) converges to some x X.

? Given any > 0, there exists N N such that d(xn, an) < /2 and d(an, x) < /2 for all n > N , so d(xn, x) d(xn, an) + d(an, x) < , which shows that (xn) converges to x and proves that X is complete.

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Problem 4. Let d : X ? X R be the discrete metric on a set X, 1 if x = y,

d(x, y) = 0 if x = y.

What are the compact subsets of the metric space (X, d)? Solution

? A subset of X is compact if and only if it is finite. ? Every finite set is compact. If F = {x1, x2, . . . , xn} X and

{G X : I} is an open cover of F , then xk Gk for some k I, so

{G1, G2, . . . , Gn} is a finite subcover of F . Alternatively, every sequence in F has a constant subsequence, which converges to a point in F . ? Conversely, if F X is infinite and Gx = {x}, then {Gx : x F } is an open cover of F with no finite subcover. Alternatively, if F is infinite, then there is a sequence (xn) in F with xm = xn for all m = n, so d(xm, xn) = 1 for m = n, and (xn) has no Cauchy or convergent subsequences. Remark. A rough heuristic is that compact sets have many properties in common with finite sets. For example, finite sets have the finite intersection property.

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Problem 5. Let c0 be the Banach space of real sequences (xn) such that xn 0 as n with the sup-norm (xn) = supnN |xn|. Is the closed unit ball

B = {(xn) c0 : (xn) 1}

compact?

Solution ? The closed unit ball in c0 is not compact. ? For example, let

ek = (nk) n=1

1 if n = k nk = 0 if n = k

denote the sequence whose kth term is one and whose other terms are

zero. Then ek c0 since limn nk = 0, and ek belongs to the closed unit ball in c0 since ek = 1. However, ej - ek = 1 for every j = k, so (ek) k=1 has no Cauchy or convergent subsequences in c0.

Remark. A similar argument using the Riesz lemma shows that the closed unit ball in any infinite-dimensional normed space is not compact in the norm topology.

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