X. Thus, we can set K = Y \ G. 1

Transcription

1 Solutions to Homework #3. 1. Given a metric space (X, d), a point x X and ε > 0, define B ε (x) = {y X : d(y, x) ε}, called the closed ball of radius ε centered at x. (a) Prove that B ε (x) is always a closed subset of X. (b) Deduce from (a) that N ε (x) B ε (x), that is, the closure of the open ball of radius ε centered at x is contained in the respective closed ball. (c) Is it always true that N ε (x) = B ε (x)? Prove or give a counterexample. Solution: (a) By definition, it suffices to show that X \ B ε (x) is open. Take any z X \ B ε (x). Then d(z, x) > ε, so if set δ = d(z, x) ε, then δ > 0. Now take any y N δ (z). Then d(z, x) d(y, x)+d(z, y) < d(y, x)+δ, whence d(y, x) > d(z, x) δ = d(z, x) (d(z, x) ε) = ε. Hence y X\B ε (x). Thus, we showed that for any z X \ B ε (x), there is δ > 0 such that N δ (z) X \ B ε (x), so by definition, X \ B ε (x) is open. (b) By Theorem 5.1(c) from class, any closed subset of X which contains N ε (x) must also contain N ε (x). By (a), B ε (x) is a closed subset (which obviously contains N ε (x)), so N ε (x) B ε (x). (c) No. For instance, take any set X with X 2 and let d be the discrete metric on X (d(x, y) = 1 if x y and d(x, y) = 0 if x = y). Then for any x X we have N 1 (x) = B 1/2 (x) = {x} (one element set consisting of the point x itself). Thus, the set {x} is both open and closed since it is simultaneously an open ball and a closed ball (in fact, it is easy to show that any subset of X is both open and closed). In particular, N 1 (x) = N 1 (x) = {x}. On the other hand, the closed ball B 1 (x) is the entire space X, so B 1 (x) N 1 (x) since X Let X be a metric space, and let Z Y be subsets of X. Prove that Z is closed as a subset of Y Z = Y K for some closed subset K of X. Deduce that if Z is closed in X, then Z is closed in Y. Note: The analogous result with closed sets replaced by open sets was proved in class (Lemma 5.2) and also appears as Theorem 2.30 in Rudin. Solution: Assume that Z is closed in Y. Then Y \ Z is open in Y, so by Theorem 2.30, there exists G open in X such that Y \ Z = Y G. Then Z = Y \ (Y \ Z) = Y \ (Y G) = Y (Y \ G), and Y \ G is closed in X. Thus, we can set K = Y \ G. 1

2 2 Assume that Z = Y K where K is closed in X. Then, by a similar computation Y \ Z = Y G where G = X \ K. Then G is open in X, so by Theorem 2.30, Y \ Z is open in Y. Finally, since Z Y, we can always write Z = Y Z, so if Z is closed in X, applying the above criterion, we deduce that Z is closed in Y. 3. Let (X, d) be a non-empty metric space and S a subset of X. Prove that the following three conditions are equivalent (as defined in class, S is called bounded if it satisfies either of those conditions): (i) There exists x X and R R such that S N R (x). (ii) For any x X there exists R R such that S N R (x). (iii) The set {d(s, t) : s, t S} is bounded above as a subset of R. Solution: (ii) (i) This is clear since X. (iii) (ii) Let K be an upper bound for {d(s, t) : s, t S}. Take any x X. If S =, there is nothing to prove. If S, fix some s 0 S and let R = K d(s 0, x). d(s, s 0 ) + d(s 0, x) K + d(s 0, x) < R, so S N R (x). Then for any s S we have d(s, x) (i) (iii) Suppose that S N R (x) for some x X and M R. Then for any s, t S we have d(s, x) < R and d(x, t) < R, so by triangle inequality d(s, t) < 2R. So the set {d(s, t) : s, t S} is bounded above by 2R. 4. Let K = { 1 n : n N} {0}. Prove that K is compact in two different ways: (i) by showing that K is closed and bounded as a subset of R (ii) directly from definition of compactness Solution: (i) Clearly K is bounded since K [0, 1] ( 1, 1) = N 1 (0). Also note that R \ K = (, 0) (1, + ) 1 ( n + 1, 1 n ). Thus, R \ K is a union of open intervals, hence R \ K is open and therefore K is closed. (ii) Let {U α } be an open cover of K (where we consider K as a subset of R). Since 0 K, one of these sets, call it U γ, must contain 0. Since U γ is open, there exists ε > 0 such that ( ε, ε) U γ. Now choose N N such that N > 1 ε (so that 1 N < ε). Then ( ε, ε) contains 0 and 1 n for all n N, so K U γ S where S = {1, 1 2,..., 1 N } (the important thing is that S is finite). Now for each integer 1 n N we can find α n such that 1 n U α n. Then K U γ N n=1 U α n, so we found a finite subcover of our original cover. n=1

3 3 5. Let X be a metric space. Prove that X is compact X satisfies the following property: (*) Let {K α } be any collection of closed subsets of X such that for any finite subcollection K α1,..., K αn, the intersection K α1... K αn is non-empty. Then the intersection of all sets in {K α } is non-empty. Solution: The above property (*) can be reformulated as follows (via contrapositive): (**) Let {K α } be any collection of closed subsets of X such that the intersection of all sets in {K α } is empty. Then there is a finite subcollection K α1,..., K αn such that the intersection K α1... K αn is empty. We will show that X is compact X satisfies (**). Suppose that X is compact, and let {K α } be any collection of closed subsets of X such that the intersection of all sets in {K α } is empty. Define U α = X \K α. Then each U α is open, and the fact that K α = implies that U α = X, so {U α } is an open cover of X. Since X is compact, there exist α 1,..., α n such that X = n U α i. Then n K α i = n (X \ U α i ) =, as desired. Now assume that (**) holds. Take any open cover {U α } of X, and define K α = X \ U α. Then each K α is closed in X, and since U α = X, we have K α =. By (**), we deduce that there exist α 1,..., α n such that n K α i =. Then X = n U α i, so {U αi } n is a finite subcover of {U α}. Thus we proved that X is compact. 6. Let X be a set, and let d 1 and d 2 be two different metrics on X. Given x X and ε > 0, define Nε 1 (x) = {y X : d 1 (y, x) < ε}, the ε-neighborhood of x with respect to d 1, and similarly define Nε 2 (x) = {y X : d 2 (y, x) < ε}. We will say that d 1 and d 2 are topologically equivalent if a subset S of X is open with respect to d 1 it is open with respect to d 2. (a) Prove that d 1 and d 2 are topologically equivalent if and only if for every ε > 0 and every x X there exist δ 1, δ 2 > 0 (depending on both ε and x) such that Nδ 1 1 (x) Nε 2 (x) and Nδ 2 2 (x) Nε 1 (x). (b) Suppose that there exist real numbers A, B > 0 such that d 1 (x, y) Ad 2 (x, y) and d 2 (x, y) Bd 1 (x, y) for all x, y X. Use (a) to prove that d 1 and d 2 are topologically equivalent. (c) Now use (b) to prove that the Euclidean and Manhattan metrics on R n are topologically equivalent.

4 4 Solution: To simplify the terminology further, we will say that a set S is d 1 -open if S is open with respect to d 1 (and similarly for d 2 ). (a) Assume that d 1 and d 2 are topologically equivalent. Take any ε > 0 and x X. We know that the set N 2 ε (x) is d 2 -open, hence by assumption it is also d 1 -open. By definition, the latter means that for any y N 2 ε (x) there exists δ > 0 such that N 1 δ (y) N 2 ε (x). In particular, this is true for y = x, so there exists some δ 1 > 0 such that N 1 δ 1 (x) N 2 ε (x). Similarly, there must exist δ 2 > 0 such that N 2 δ 2 (x) N 1 ε (x). Now assume that for every ε > 0 and every x X there exist δ 1, δ 2 > 0 (depending on both ε and x) such that N 1 δ 1 (x) N 2 ε (x) and N 2 δ 2 (x) N 1 ε (x). Let U be any d 1 -open set, and take any x U. Then there must exist ε > 0 such that N 1 ε (x) U. By assumption, there exists δ 2 > 0 such that N 2 δ 2 (x) N 1 ε (x), so N 2 δ 2 (x) U. This proves that U is d 2 -open. Similarly, we argue that any d 2 -open set must be d 1 -open, whence d 1 and d 2 are topologically equivalent. (b) Take any ε > 0 and x X, and let δ 1 = ε B. We claim that N 1 δ 1 (x) N 2 ε (x). Indeed, if y N 1 δ 1 (x), then d 1 (y, x) < δ 1 = ε B, whence d 2(y, x) Bd 1 (y, x) < ε, so y N 2 ε (x). Similarly, the inclusion N 2 δ 2 (x) N 1 ε (x) always holds for δ 2 = ε A. Thus, we obtained the desired inclusion between open balls with respect to d 1 and d 2, so by (a), d 1 and d 2 are topologically equivalent. (c) Let us denote the Euclidean metric by d E and the Manhattan metric by d M. We claim that for any x, y R n we have d E (x, y) d M (x, y) n d E (x, y). ( ) By (b), this would imply that d M and d E are equivalent. Recall that if x = (x 1,..., x n ) and y = (y 1,..., y n ), then d M (x, y) = n a i and d E (x, y) = n a2 i where a i = x i y i. Thus, we need to show that for any non-negative real numbers a 1,..., a n, the following inequalities hold: n a 2 i n a i n n The first inequality is easy: indeed, ( n a i) 2 = n a2 i + 2 i<j a ia j n a2 i, and taking square roots of both sides, we get the result. The second inequality n a i n n a2 i is a special case of Cauchy-Schwartz inequality (set b i = 1 for each i in Theorem 1.35 in Rudin). a 2 i

5 Definition: Let (X, d) be a metric space and ε > 0. A subset S of X is called an ε-net if for any x X there exists s S such that d(x, s) < ε. In other words, S is an ε-net if X is the union of open balls of radius ε centered at elements of S. 7. Let S be a subset of a metric space (X, d). Prove that the following are equivalent: (i) The closure of S is the entire X; (ii) U S for any non-empty open subset U of X; (iii) S is an ε-net for every ε > 0. The subset S is called dense (in X) if it satisfies these equivalent conditions. Solution: As suggested in the hint, we will prove that negations of (i), (ii) and (iii) are equivalent to each other. We start by explicitly formulating the negations, denoted below by (a), (b) and (c) (a) The closure of S is strictly contained in X, that is, there exists x X which is NOT a contact point of S (b) There exists a non-empty open subset U of X such that U S = (c) There exists ε > 0 such that S is not an ε-net. Now we prove that (a), (b) and (c) are equivalent (a) (b) If x X is not a contact point of S, there exists ε > 0 such that N ε (x) S =. This means that (b) holds with U = N ε (x) (since N ε (x) is open and non-empty). (b) (a) Take any x U (such x exists since U is non-empty). Since U is also open, there exists ε > 0 such that N ε (x) U. Since U S =, we must have N ε (x) S =, so x is not a contact point of S. (a) (c) As in implication (a) (b), there exists x X and ε > 0 such that N ε (x) S =. This means that d(s, x) ε for all s S, so S is not an ε-net. (c) (a) Analogous to the proof of (a) (c). 5