Introductory Analysis 1 Fall 2009 Homework 4 Solutions to Exercises 1 3

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1 Introductory Analysis 1 Fall 2009 Homework 4 Solutions to Exercises 1 3 Note: This homework consists of a lot of very simple exercises, things you should do on your own. A minimum part of it will be due Monday, October 5, Let X be a set and let d 1, d 2 be two distance functions for X. We say they are equivalent iff they define the same family of open sets. That is, d 1 d 2 if and only if whenever a subset of X is open with respect to d 1 it is open with respect to d 2 and whenever a subset of X is open with respect to d 2 it is open with respect to d 1. Prove: (a) Two distance functions for a set X are equivalent if and only if they both have the same convergent sequences. That is, if the functions are d 1, d 2, then a sequence converges with respect to d 1 if and only if it converges with respect to d 2. (b) A metric space is said to be discrete if and only if its metric is equivalent to the discrete metric. Prove that a metric space is discrete if and only if it has no accumulation points (i.e., all points are isolated). (c) Prove that Z, as a subspace of R, is discrete. Solution. First a comment. Sometimes equivalence between metrics is defined by saying that d 1 d 2 iff there exist positive constants a, b such that ad 1 (x, y) d 2 (x, y) bd 1 (x, y). This is NOT the way I defined it, and it is NOT equivalent to my definition. That is, if two metrics are equivalent in this new definition, they are also equivalent in mine. But the converse is not true. A counterexample is provided by the two metrics of Exercise 4, which are equivalent in my definition, but not in the new one. Another (more common) misconception was to assume that because the metrics have the same open sets they also have the same open balls. If you consider the three following three metrics in R 2 : d 1 ((x 1, x 2 ), (y 1, y 2 )) = x 1 y 1 + x 2 y 2, d 2 ((x 1, x 2 ), (y 1, y 2 )) = (x 1 y 1 ) 2 + (x 2 y 2 ) 2, d ((x 1, x 2 ), (y 1, y 2 )) = max { x 1 y 1, x 2 y 2 } ( the metrics of exercise 3 restricted to the case of n = 2, X 1 = X 2 = R), they are equivalent. But their open balls are quite different. The open balls of the first metric are squares with sides parallel to the axes, the second metric has discs, the third one again squares but with sides forming a 45 degrees angle with the axes. Proof. (a) Since convergence of a sequence can be defined exclusively in terms of open sets, this is fairly clear. The only subtle point is assuming that d 1, d 2 have the same convergent sequences, do these sequences also have to have the same limit? The answer is yes. Assume a sequence {p n } converges to p with respect to d 1, to q with respect to d 2. Now

2 consider the sequence {p 1, p, p 2, p 3,, p,...}. This sequence converges to p with respect to d 1, thus it has to converge with respect to d 2. The sequence has a subsequence (the odd terms) converging to q with respect to d 2 ; the subsequence of even terms is constant, all terms are p, thus converges to p. It follows that p = q. (b) Since the metric of a discrete space is equivalent to the discrete metric, all singleton sets, hence all subsets are open. If {x} is open, there has to exist r > 0 such that B(x, r) {x}, thus (necessarily) B(x, r) = {x}. Obviously B(x, r) {x} = {x}, so B(x, r) contains no point of X, hence is not an accumulation point of X. Conversely, if X has no accumulation points, then given x X there must exist a ball centered at x intersecting X in no other point than x. This is another way of saying that {x} is an open ball, hence open. Hence all subsets of X are open and X is discrete. (c) {n} = B(n, 1 2 ) for all n Z. 2. Let X be a metric space and let Y X; consider Y as a metric space with the distance function of X (restricted to Y ). Prove: (a) A subset A of Y is open in Y (i.e., is an open subset of the metric space Y ) if and only if A = U Y for some open subset U of X. (b) A subset A of Y is closed in Y (i.e., is an closed subset of the metric space Y ) if and only if A = F Y for some closed subset F of X. (c) Let A be a subset of Y. The closure of A in Y is the intersection of the closure of A in X with Y. Solution. It can be useful to introduce the notation B X (p, r) for an open ball of X, B Y (p, r) for an open ball of Y. That is, with d the distance function of X (which is the same as that of Y ), B X (p, r) = {q X : d(q, p) < r} if p X, B Y (p, r) = {q Y : d(q, p) < r} if p Y. It is obvious that if p Y, then B Y (p, r) = B X (p, r) Y. I ll also introduce the notation ĀY for the closure of a subset A of Y in Y, reserving Ā for its closure in X. Here are the required proofs. Proof. (a) Assume first that A is open in Y. For every p A there is r p > 0 such that B Y (p, r p ) A. It is then immediate that A = B Y (p, r p ). Let U = B X (p, r p ). As a union of open balls of X, U is open in X. Moreover, U Y = B X (p, r p ) Y = (B X (p, r p ) Y ) = B Y (p, r p ) = A. 2

3 Conversely, assume A = U Y where U is open in X. Let p A. Then p U and there exists r > 0 such that B X (p, r) U. Then B Y (p, r) = B X (p, r) Y U Y = A. It follows that A is open in Y. (b) This is an immediate consequence of the previous item: If A is closed in Y, then Y \A is open in Y, hence Y \A = U Y for some open subset of X. It follows that A = Y F where F = X\U is closed in X. Conversely, if A = Y F where F is closed in X, then Y \A = (Y \A) U where U = X\F. All this is very easy to check. Thus U is open in X, hence Y \A is open in Y, therefore A is closed in Y. (c) Since Ā Y is a closed subset of Y containing A, it follows that Ā Y Ā Y. Let F be closed subset of Y such that A F. There is G closed in Y such that F = G Y. Then A G Y G; since G is a losed subset of X containing A we get Ā G, hence Ā Y F. It follows that Ā Y is a subset of all closed subsets of Y containing A, thus is contained in ĀY. The result follows. 3. Let X 1,..., X n be metric spaces with distance functions d 1,..., d n respectively. Let X = X 1 X n = {p = (p 1,..., p n ) : p i X i, i = 1,..., n}. Consider the following three functions from X X to R. In the definition, p = (p 1,..., p n ), q = (q 1,..., q n ) X. δ 1 (p, q) = δ 2 (p, q) = n d i (p i, q i ). ( n ) 1/2 d i (p i, q i ) 2. δ (p, q) = max{d i (p i, q i ) : i = 1,..., n}. (a) Prove all three metrics for X are equivalent and a sequence {p m } m=1, where p m = (p m1,..., p mn ) converges to p = (p 1,..., p n ) if and only if each sequence {p mi } m=1 converges in X i for i = 1,..., n. (b) Prove that all three metrics for X have the same Cauchy sequences. (Equivalent metrics do NOT necessarily have the same Cauchy sequences so this is not an immediate consequence of the previous point. It is however essentially the same proof as one uses for the previous point.) (c) Prove that X is complete in any one of the three metrics if and only if it is complete in the others, and this happens if and only if each one of the spaces X 1,..., X n is complete. (d) Explain how all this applies to R n. In particular, show that R n is complete. Solution. It will be convenient to prove first the following lemma: 3

4 Lemma 1 Let X be a set and let d 1, d 2 be distance functions on X. Assume there exist real positive constants a, b such that a d 1 (x, y) d 2 (x, y) b d 1 (x, y) for all x, y X. Then d 1 d 2 in the sense of the definition of Problem 1. Moreover both distance functions have the same Cauchy sequences. Proof. The easiest thing to show is that both metrics have the same convergent sequences and the same Cauchy sequences. That s almost obvious, I think (therefore I am). Assume {p n } is a sequence in X. Assume first it converges with respect to d 2, say to p. Let ɛ > 0 be given. Then ɛ/a > 0 and there is N such that d 2 (p n, p) < ɛ/a whenever n N. If n N, then d 1 (p n, p) < a d 2 (p n, p) < a ɛ a = ɛ. The sequence converges with respect to p 1. Assume next the sequence is Cauchy with respect to d 2. We essentially repeat the same argument. Let ɛ > 0 be given. Then ɛ/a > 0 and there is N such that d 2 (p n, p m ) < ɛ/a whenever n, m N. If n, m N, then d 1 (p n, p m ) < a d 2 (p n, p m ) < a ɛ a = ɛ and the sequence is Cauchy with respect to d 1. To prove this we only used that p 1 (1/a)p 2. Since p 2 b p 1, switching the roles of p 1, p 2, replacing 1/a by b we see that if a sequence converges with respect to p 1, or is Cauchy with respect to p 1, then it converges or is Cauchy, respectively, with respect to d 2. The rest is easy. But there is a minor detail that perhaps has to be covered. Let us define a stronger equivalence relation for two metrics d 1, d 2 in a set X by: d 1 = d2 iff there exist real positive constants a, b such that a d 1 (x, y) d 2 (x, y) b d 1 (x, y) for all x, y X. Lemma 1 proves (in part) that d 1 = d2 implies that d 1 d 2. It will be convenient to PROVE (what an inconvenient word!) that this is actually an equivalence relation. That d = d for all metrics d in X is of course obvious; just take a = b = 1. If d 1 = d2 one has then one also has a d 1 (x, y) d 2 (x, y) b d 1 (x, y) for all x, y X, 1 b d 2(x, y) d 1 (x, y) 1 a d 2(x, y) for all x, y X. It follows that d 2 = d1. Finally, if d 1 = d2 and d 2 = d3, then there exist positive constants a, b, α, β such that a d 1 (x, y) d 2 (x, y) b d 1 (x, y), α d 2 (x, y) d 3 (x, y) β d 2 (x, y), for all x, y X then one also has thus d 1 = d3. a a d 1 (x, y) d 3 (x, y) b β d 1 (x, y) 4

5 Now to the exercise! We can apply the Cauchy-Schwarz inequality to get for x = vex1n, y = (y 1,..., y n ) X = X 1 X n : δ 1 (x, y) = n x i y i = = n δ 2 (x, y). ( n n ) 1/2 ( n ) 1/2 (1 x i y i ) 1 2 (x i y i ) 2 (I was not expecting anybody to use this approach, though quite a few of you did; the exercise can be done without using Cauchy-Schwarz). On the other hand it is elementary that if a, b, c... are positive numbers, then a 2 + b 2 + c 2 + (a + b + c + ) 2 ; it follows that δ 2 (x, y) δ 1 (x, y) for all x, y X. We proved: 1 n δ 1 (x, y) δ 2 (x, y) δ 1 (x, y), x, y X thus d 1 = δ2. It is quite easy to see that δ (x, y) δ 1 (x, y) nδ (x, y), x, y X proving δ 1 = δ3 Thus the three metrics are equivalent in the strong sense hence, by Lemma 1 share the same convergent sequences, and the same Cauchy sequences. In particular if the space X is complete with respect to one of these metrics, it is complete with respect to the other two. To see that a sequence converges, or is Cauchy in any of these metrics if and only if the component sequences converge, or are Cauchy, it suffices to prove this holds (for example) for δ. Let x m = (x m1, ldots, x mn ) X for m N. I ll just do convergence; the proof for the Cauchy condition is quite similar. Assume first that each one of the component sequences converge, say lim m x mi = x i for i = 1,..., n. Let x = (x 1,..., x n ). Let ɛ > 0 be given. There exist then N 1,..., N n such that if m N i then d i (x mi, x i ) < ɛ for i = 1,..., n. Let N = max(n 1,..., N n ). If m N, then m N i for i = 1,..., n, hence d i (x mi, x i ) < ɛ for i = 1,..., n, hence δ (x m, x) = max 1 i n d i (x mi, x i ) < ɛ, proving the sequence {x m } converges to x. Conversely, assuming the sequence {x m } converges to x = (x 1,..., x n ), assume 1lej n. Due to the convergence of the sequence {x m } to x, given ɛ > 0 there is N such that m N implies δ (x + m, x) < ɛ. If m N, then d j (x mj, x j ) δ (x m, x) < ɛ. It follows tat lim m x mj = x j ; j = 1,..., n. Assume now X is complete. At this stage we don t really need to mention y specific metric; they are equivalent. Any one of them will do. To see each one of X 1,... X n is complete, let X j be one of these factor spaces. A small subtlety appears here: The result in question is false if any of the spaces X 1,..., X n is empty. This is a sort of silly situation, but if any one of X 1,..., X n is empty, then X is empty and complete since all of its non-existent Cauchy sequences surely converge to some non-existent limit. So to avoid nonsense we assume X i for i = 1,..., n. Back to X j. Suppose that {x jm } is a Cauchy sequence in X j. Select y i X i for i j (ere is where non-emptyness comes in!) and define the sequence {x m } in X by letting {x mj } be the given sequence in X j and {x mi } the 5

6 constant sequence x mi = y i for all m N if i j. The sequence {x m } has the property that all its components sequences are Cauchy; the j-th component by hypothesis, the other components by construction (they re constant sequences). Thus {x m } is Cauchy in X, because X is complete it converges, thus by what we have already shown all of its component sequences converges, in particular the j-th one, proving X j is complete. Conversely, if X 1,..., X n are complete, given a Cauchy sequence in X, all of its component sequences are Cauchy, thus converge, thus the original sequence converges. This proves X is complete. The relevancy of all this to R n is that if we take X 1 = X 2 = = X n = R with d i (x, y) = x y for i = 1,..., n, then δ 2 is the ordinary, standard metric for R n and we proved (among oter thigs) that R n is complete. 4. Let d be the usual metric in N; d(n, m) = n m for all n, m N. Let d 2 : N N R be defined by d(n, m) = 1 n 1 m. Prove that d 1 d 2 (for example, show both are discrete) but (N, d) is complete while (N, d 2 ) is not complete. 5. Let (X, d) be a metric space, assume it is not complete. Let X be the set of all Cauchy sequences of points in X; that is, p X means p = {p n } n=1 where {p n } is a Cauchy sequence in X. We define an equivalence relation in X by p q if and only if lim n d(p n, q n ) = 0. (Of course, {d(p n, q n )} is a sequence of non-negative real numbers.) Prove this is indeed an equivalence relation. Let ˆX = X / be the set of all equivalence classes. We will denote the elements of ˆX by [p], p X ; [p] = [q] if and only if p q. Prove: If p, q X, then the sequence of real numbers {d(p n, q n )} is a Cauchy sequence in R, hence converges. Moreover, If p p, q q then lim n d(p n, q n ) = lim n d(p n, q n). This allows us to define ˆd : ˆX ˆX R by ˆd([p], [q]) = limn infty d(p n, q n ). Prove that ( ˆX, ˆd) is a complete metric space. Here the notation may get a bit fuzzy, so one should be careful. If p X, define p X by p n = p for all n N. Next define a map φ : X ˆX by φ(p) = [ p]. Show this map is one-to-one and preserves the metric; i.e., ˆd([ p], [ q]) = d(p, q). This allows one to identify X with the metric space φ(x). Show that φ(x) is a dense subset of ˆX. The space hatx is the completion of X. One can show (and perhaps we will) that it is unique up to metric isomorphism. 6