Problem Set 2: Solutions Math 201A: Fall 2016


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1 Problem Set 2: s 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. (a) If F X is closed and (x n ) is a Cauchy sequence in F, then (x n ) is Cauchy in X and x n 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 (x n ) is a sequence in F, then there is a subsequence (x nk ) that converges to x X since X is compact. Then x F since F is closed, so F is compact. Alternatively, If {G α X : α I} is an open cover of F, then {G α : α 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 (x n ) is a convergent sequence in K with limit x X, then every subsequence of (x n ) converges to x. Since K is compact, some subsequence of (x n ) converges to a limit in K, so x K and K is closed. Suppose that K is not bounded, and let x 1 K. Then for every r > 0 there exists x K such that d(x 1, x) r. Choose a sequence (x n ) in K as follows. Pick x 2 K such that d(x 1, x 2 ) 1. Given {x 1, x 2,..., x n }, pick x n+1 K such that By the triangle inequality, d(x 1, x n+1 ) 1 + max 1 k n d(x 1, x k ). d(x k, x n+1 ) d(x 1, x n+1 ) d(x 1, x k ) 1 for every 1 k n. It follows that d(x m, x n ) 1 for every m n, so (x n ) has no Cauchy subsequences, and therefore no convergent subsequences, so K is not compact. 1
2 Problem 2. Let A be a subset of a metric space X with closure Ā. Define the interior A and boundary A of A by A = {G A : G is open}, A = Ā \ A. (a) Why is A open and A closed? (b) Prove that X \ Ā = (X \ A). (c) Prove that A is closed if and only if A A, and A is open if and only if A A c. (d) If A is open, does it follow that (Ā) = A? (a) A union of open sets is open so A is open, and an intersection of closed sets is closed so Ā and A = Ā (A ) c are closed. (b) Note that x A if and only B ɛ (x) A for some ɛ > 0. If x Āc, then B ɛ (x) Āc for some ɛ > 0 since Āc is open. Since Ā A, we have Āc A c, so B ɛ (x) A c, meaning that x (A c ). It follows that Ā c (A c ). For the reverse inclusion, note that if x (A c ), then there exists ɛ > 0 such that B ɛ (x) A c, so x is not the limit of any sequence in A, meaning that x Āc. It follows that (A c ) Āc, so (A c ) = Āc. (c) If A is closed, then Ā = A so A = A (A ) c A. For the converse, note that since A A Ā, we have Ā = (Ā A ) (Ā (A ) c ) = A A. If A A, then it follows that closed. Ā A, so A = Ā, meaning that A is (d) This is not true in general. For example, define A R 2 by A = { (x, y) : x 2 + y 2 < 1 } \ {(x, 0) : 0 x < 1}. Then A is open, but (Ā) = {(x, y) : x 2 + y 2 < 1} A. 2
3 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. Let (x n ) be a Cauchy sequence in X. Since A is dense in X, we can choose a sequence (a n ) in A such that d(x n, a n ) 0 as n. (For example, choose a n A such that d(x n, a n ) < 1/n.) Given any ɛ > 0, there exists M N such that d(x n, a n ) < ɛ/3 and d(x m, x n ) < ɛ/3 for all m, n > M, so d(a m, a n ) d(a m, x m ) + d(x m, x n ) + d(x n, a n ) < ɛ, which shows that (a n ) is a Cauchy sequence in A. It follows that (a n ) converges to some x X. Given any ɛ > 0, there exists N N such that d(x n, a n ) < ɛ/2 and d(a n, x) < ɛ/2 for all n > N, so d(x n, x) d(x n, a n ) + d(a n, x) < ɛ, which shows that (x n ) converges to x and proves that X is complete. 3
4 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)? A subset of X is compact if and only if it is finite. Every finite set is compact. If F = {x 1, x 2,..., x n } X and {G α X : α I} is an open cover of F, then x k G αk for some α k I, so {G α1, G α2,..., G αn } 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 G x = {x}, then {G x : x F } is an open cover of F with no finite subcover. Alternatively, if F is infinite, then there is a sequence (x n ) in F with x m x n for all m n, so d(x m, x n ) = 1 for m n, and (x n ) 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. 4
5 Problem 5. Let c 0 be the Banach space of real sequences (x n ) such that x n 0 as n with the supnorm (x n ) = sup n N x n. Is the closed unit ball B = {(x n ) c 0 : (x n ) 1} compact? The closed unit ball in c 0 is not compact. For example, let e k = (δ nk ) n=1 δ nk = { 1 if n = k 0 if n k denote the sequence whose kth term is one and whose other terms are zero. Then e k c 0 since lim n δ nk = 0, and e k belongs to the closed unit ball in c 0 since e k = 1. However, e j e k = 1 for every j k, so (e k ) k=1 has no Cauchy or convergent subsequences in c 0. Remark. A similar argument using the Riesz lemma shows that the closed unit ball in any infinitedimensional normed space is not compact in the norm topology. 5
6 Problem 6. A metric (or topological) space X is disconnected if there are nonempty open sets U, V X such that X = U V and U V =. A space is connected if it is not disconnected. A space X is totally disconnected if its only nonempty connected subsets are the singleton sets {x} with x X. (a) Show that the interval [0, 1] is connected (in its standard metric topology). (b) Show that the set Q of rational numbers is totally disconnected. (a) Suppose for contradiction that [0, 1] = U V where U, V are nonempty, disjoint open sets in [0, 1]. We assume that 0 U without loss of generality. Let a = sup {x [0, 1] : [0, x) U}. Since 0 U and U is open, we have [0, ɛ) U for some ɛ > 0, so 0 < a 1. If 0 < b < a, then [0, b) U since, by the definition of the supremum, there exists b < c < a such that [0, c) U. (It also follows that [0, a) = 0<b<a [0, b) U, so the supremum is attained and, in fact, {x [0, 1] : [0, x) U} = [0, a].) If a U, then (a ɛ, a + ɛ) U for some ɛ > 0, but then [0, a + ɛ) U, contradicting the definition of a. On the other hand, if a V, then (a ɛ, a] V for some 0 < ɛ < a, but then [0, b) U for a ɛ < b a, also contradicting the definition of a. (b) Let A Q be any subset of the rational numbers with at least two elements. Choose x, y A with x y. The irrational numbers are dense in R, so there exists z R \ Q such that x < z < y. Let U = (, z) A, V = (z, ) A. Then U, V are open sets in the relative topology on A. Moreover, x U, y V so U, V are nonempty, and U V =, U V = A. It follows that Q is totally disconnected. 6
7 Problem 7. Let T = {G R : G c is countable or G c = R}. (a) Show that T is a topology on R. (b) Let I = (0, 1) with closure Ī = {F I : F is closed} in this topology. Show that Ī = R. (c) Is there a sequence (x n ) such that x n I and x n 2 in this topology? (d) What part of your proof in Problem 5 of Set 1 fails in this example? (a) The collection of closed sets C = {F R : F c T } in this topology is given by C = {F R : F is countable or F = R}. The collection C satisfies the axioms for closed sets in a topological space: (1), R C. (2) The intersection of closed sets is closed, since either every set is R and the intersection is R, or at least one set is countable and the intersection in countable, since any subset of a countable set is countable. (3) A finite union of closed sets is closed, since a finite (or countable) union of countable sets is countable. It follows that T is a topology on R (called the cocountable topology). (b) If F I is closed, then F is uncountable, since (0, 1) is uncountable, so F = R and Ī = R. (c) Let (x n ) n=1 be a sequence in (0, 1). Then U = R \ {x n : n N} is an open neighborhood of 2, but x n / U for any n N, so (x n ) does not converge to 2. A similar argument applies to any x / (0, 1), so the sequential closure of I is Ĩ = (0, 1). (d) If X is a topological space, then a neighborhood base of x X is a collection {U α : α A} of neighborhoods of x such that for every neighborhood U of x there exists α A with U α U. Then x n x if and only if for every α A there exists N N such that x n U α for all n > N. The proof that the sequential closure is equal to the closure fails for the cocountable topology on R because x R does not have a countable neighborhood base. On the other hand, if X is a metric space, then {B 1/n (x) : n N} is a countable neighborhood base of any x X. 7
8 Problem 8. Let S be the set of sequences whose terms are 0 or 1: S = {(s k ) k=1 : s k = 0 or s k = 1}. (a) Use a diagonal argument to show that S is uncountable. (b) Show that S has the same cardinality as the power set P(N) of the natural numbers. (a) Let f : N S. Suppose that f(n) = s n where s n = (s kn ) k=1, and define t = (t k ) k=1 S by { 1 if s kk = 0, t k = 0 if s kk = 1. Then t s k for every k N, since the two sequences have different kth terms. It follows that there is no map from N onto S, so S is uncountably infinite. For A N define the characteristic function χ A : N {0, 1} of A by { 1 if k A, χ A (k) = 0 if k / A. is oneto Then the map f : P(N) S defined by f(a) = (χ A (k)) k=1 one and onto, so P(N) and S have the same cardinality. 8
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