Prof. Girardi, Math 703, Fall 2012 Homework Solutions: Homework 13. in X is Cauchy.


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1 Homework 13 Let (X, d) and (Y, ρ) be metric spaces. Consider a function f : X Y Prove or give a counterexample. f preserves convergent sequences if {x n } n=1 is a convergent sequence in X then {f(x n)} n=1 is a convergent sequence in Y Show that a convergent sequence {x n } n=1 in X is Cauchy Give an example of a metric space (X, d) and a Cauchy sequence {x n } n=1 does not converge. (Enough to quote previous homework problem.) from X that Give an example of a continuous f : X Y that does not preserve Cauchy sequences Show that if f is uniformly continuous, then f preserves Cauchy sequences Prove or give a counterexample. If f preserves Cauchy sequences, then f is uniformly continuous Show the following. f is uniformly continuous if {x n } n=1 and {z n} n=1 are sequences from X s.t. d(x n, z n ) 0, then ρ(f(x n ), f(z n )) 0. Remark 0.1. Let (X, d) and (Y, ρ) be metric spaces. Consider a function f : X Y. (a) Recall [K, Prop. 2.25]. f is continuous at x X if {x n } n=1 X and x n x, then f(x n ) f(x). (b) Definition. f preserves convergent sequences if {x n } n=1 X converges, say x n x, then f(x n ) f(x). (c) So: 1 f is continuous on X f preserves convergent sequences. (d) Definition. f preserves Cauchy sequences if {x n } n=1 X is Cauchy, then {f(x n)} n=1 Y is Cauchy. Remark 0.2. For a dicussion of this problem, visit the below website. Proof. LTGBG This is true. Here comes the proof Let f preserve convergent sequences. Consider a sequence {x n } n=1 from X that converges to some x in X. Then {f(x n )} n=1 converges in Y, specifically to f(x) Assume that if {x n } n=1 is a convergent sequence in X then {f(x n)} n=1 is a convergent sequence in Y. Let {x n } n=1 X be converge sequence, say x n x. We WTS that f(x n ) f(x). 1 Follows directly from (a) and (b). Page 1 of 6
2 Define { x n } n=1 by x 2n 1 x 2n = x n = x for each n N. Then x n x. So by assumption, there is y Y s.t. f( x n ) y. By considering the subsequence {f( x 2n )} n=1 of {f( x n)} n=1 we get that y = lim f( x 2n) = n lim f(x) = f(x). n Thus f(x n ) f(x) Let {x n } n=1 be a convergent sequence in X, say x n x. Fix ɛ > 0. Find N ɛ N s.t. if n N ɛ then d(x n, x) < ɛ 2. Let n, m > N ɛ. Then Thus {x n } n=1 is a Cauchy sequence in X Recall Homework # 8. d(x n, x m ) d(x n, x) + d(x, x m ) < ɛ 2 + ɛ 2 = ɛ. Trench, 8.1, # 25, p Consider the set C[a, b] of continuous real functions on the interval [a, b]. (a) Show that is a norm on C[a, b]. f = b a f(x) dx (b) Show that the sequence {f n } defined by is a Cauchy sequence in (C[a, b], ). f n (x) = (c) Show that (C[a, b], ) is not complete. ( ) x a n b a So we can take the metric space (X, d) to be (C[a, b], d) where d(f, g) := f g and take {f n } n=1 as in (b) above Let (X, d) := ((0, 1), ) and (Y, ρ) := (R, ). Let f(x) := 1/x. Clearly, f : X Y is a continuous function. Consider the sequence {x n } n=1 in X where x n := 1/n. Then {x n } n=1 is Cauchy in X since 0 d(x n, x m ) = 1 n 1 m 1 n + 1 m n,m 0. But {f(x n )} n=1 = {n} n=1 is not Cauchy in Y since ρ(f(x n), f(x n+1 )) = 1 for each n N Let f be uniformly continuous. Consider a Cauchy sequence {x n } n=1 in X. We WTS that {f(x n )} n=1 is a Cauchy sequence in Y. Page 2 of 6
3 Fix ɛ > 0. Since f is uniformly continuous, there exists δ ɛ > 0 such that if x, x X and d(x, x) < δ ɛ, then ρ(f(x), f( x)) < ɛ. (1) Furthermore, since {x n } n is Cauchy, there exists N ɛ N such that if n, m N ɛ, then d(x n, x m ) < δ ɛ. (2) Now let n, m N ɛ. It follows from (1) and (2) that ρ(f(x n ), f(x m )) < ɛ. Therefore, {f(x n )} n is Cauchy in Y Preservation of Cauchy sequences does not imply uniform continuity. Let (X, d) = (Y, ρ) = (R, ). Consider the function f : X Y given by f(x) := x 2. Clearly f is continuous on X. To see that f preserves Cauchy sequences, first recall that a sequence in (R, ) is Cauchy if and only if it converges. Next fix a Cauchy sequence {x n } n in X. Then {x n } n is a convergent sequence in X. By Remark 0.1(c) (or HMWK 13.1), since f is continuous on X, the sequence {f(x n )} n converges in Y. Thus {f(x n )} n is a Cauchy sequence in Y. But To see that f is not uniformly continuous on R, fix δ > 0. Let x = 1 δ and x = 1 δ + δ 17. Then x x = δ 17 < δ. f( x) f(x) = ( x) 2 x 2 = δ > So f cannot be uniformly continuous on R Let f is uniformly continuous. Fix sequences {x n } n=1 and {z n} n=1 from X with the property that d(x n, z n ) 0. We WTS that ρ(f(x n ), f(z n )) 0. Fix ɛ > 0. Since f is uniformly continuous on X, there exists δ ɛ > 0 such that if x, x X and d(x, x) < δ ɛ, then ρ(f(x), f( x)) < ɛ. (3) Since d(x n, z n ) 0, there e exists N δɛ N such that if n N δɛ, then d(x n, z n ) < δ ɛ. (4) Now let n N δɛ. It follows from (3) and (4) that if n > N δɛ then ρ(f(x n ), f(z n )) < ɛ. Thus ρ(f(x n ), f(z n )) Suppose that f is not uniformly continuous. 2 It suffices to show there exist sequences {x n } n=1 and {z n} n=1 in X such that d(x n, z n ) 0 but ρ(f(x n ), f(z n )) 0. The negation of the definition of uniform continuity gives that there exists some ɛ 0 > 0 s.t. for each δ > 0, there exist points x δ, z δ in X s.t. d(x δ, z δ ) < δ but ρ(f(x δ ), f(z δ )) ɛ 0. 2 Do you see why we want to argue by contrapositive? Indeed, we are trying to say P Q and Q deals only with sequences but P is continuous. (5) Page 3 of 6
4 Take this ɛ 0 and consider the sequence {δ n } n=1 where δ n := 1/n. Using (5), find x n, z n X s.t. d(x n, z n ) < 1 n but ρ(f(x n ), f(z n )) ɛ 0. Then lim n d(x n, z n ) = 0. But the sequence {ρ(f(x n ), f(z n ))} n=1 ρ(f(x n ), f(z n )) ɛ 0 for each n N. cannot converge to zero since Homework 14 Let (X, d) be a metric space. Let {x n } n=1 be a Cauchy sequence in X that has a convergent subsequence {x n k } k=1, say {x nk } k=1 converges to x X. Show that {x n} n=1 converges to x. Remark 0.3. Loosely speaking: show that a Cauchy sequence in a metric space that has a convergent subsequence converges (of course, to the element that the convergent subsequence converges to). Proof. LTGBG. Fix ɛ > 0. Since {x n } n=1 is Cauchy, there exists some N ɛ N such that if n, m N ɛ, then d(x n, x m ) < ɛ 2. (6) Since {x nk } k=1 converges to x, there exists some K ɛ N such that if k K ɛ, then d(x nk, x) < ɛ 2. (7) Pick k ɛ such that k ɛ K ɛ and n kɛ N ɛ. (8) Fix n N ɛ. It follows from (6), (7), and (8) that d(x n, x) d(x n, x nkɛ ) + d(x nkɛ, x) < ɛ 2 + ɛ 2 = ɛ. Thus {x n } n=1 converges to x. Homework 15 Let (X, d) be a metric space and A X. Consider the metric space (A, d A ) where d A := d A A Show that if A is complete, then A is closed in X Show that if A is totally bounded, then A is bounded. Page 4 of 6
5 Notation 0.4. Throughout this problem, for x 0 X and a 0 A and ɛ > 0, let N X ɛ (x 0 ) := {x X : d(x, x 0 ) < ɛ} N A ɛ (a 0 ) := {a A: d(a, a 0 ) < ɛ}. Remark 0.5. Let s recall definitions. Let (X, d) and (A, d A ) be as in the statement of Homework A is complete provided each Cauchy sequence in A converges to some element in A. 2. A is totally bounded ɛ > 0 A can be covered by finitely many ɛballs i.e. ɛ > 0 n N and a 1,... a n A s.t. A = n i=1 N ɛ A (a i ) i.e. ɛ > 0 n N and a 1,... a n A s.t. A n i=1 N ɛ X (a i ). 3. A is bounded provided there exists a 0 A and R > 0 s.t. A N A R (a 0). Proof. LTGBG Let A be complete. We WTS that A is closed in X. Fix a sequence {a n } n=1 that converges to some point in X, say lim n a n = x X. Then {a n } n=1 is a Cauchy sequence in X by Homework 13.2 and thus {a n } n=1 is a Cauchy sequence in A. Since A is complete, by def. of complete, {a n} n=1 converges to some element in a A. Thus x = a A. So by [K, Cor. 2.23], A is closed in X Let A be totally bounded. We WTS that A is bounded. Let ɛ = 17. Since A is totally bounded, by definition of totally bounded, there exists n N and a 1,..., a n A such that Pick an a 0 A. Let A = n i=1n A ɛ (a i ). (9) M := max d(a 0, a i ) 1 i n R := M + ɛ. Clearly 0 < R <. It is enough to show that A N A R (a 0 ). (10) Towards this, fix an a A. By (9), there exists i {1,... n} such that a N A ɛ (a i ). Thus Thus a N A R (a 0). So (10) holds. d(a, a 0 ) d(a, a i ) + d(a i, a 0 ) < ɛ + M = R. Page 5 of 6
6 Homework 16 Let U be an open subset of R. Show U can be expressed as a countable union of disjoint open intervals. Proof. Consider an arbitrary x U. Since U is open, there is an ɛ > 0 so that (x ɛ, x + ɛ) U. Thus the sets {a (, x): (a, x] U} and {b (x, ): [x, b) U} each are nonempty and, by the completeness of R, a x := inf{a (, x): (a, x) U} [, x) b x := sup {b (x, ): [x, b) U} (x, ]. Note the following hold. I x := (a x, b x ) is an open (perhaps unbounded) interval. x I x U. I x is the maximal open interval in U containing x. I.e., if J is any open interval in U containing x, then J I x. Thus U equals the union of open intervals, specifically, U = I x. x U Next we shall show that this union can be expressed as a countable union of disjoint open intervals. Claim 1. For any two points x and y in U, I x and I y are either equal or disjoint. Indeed, if I x I y is nonempty, then I x I y is an open interval in U containing both x and y. But I x and I y are the maximal such open intervals containing x and y, respectively. So it must be that I x = I x I y = I y. So Claim 1 holds. Claim 2. The set (so no repeats allowed) I := {I x : x U} is countable. Indeed, for each I I, we may choose some rational number q Q that lies in I. Since the interval in I are distinct by Claim 1, the rational numbers we choose are distinct from one another. As a result, the set of rational numbers we collected is countable. So Claim 2 holds. Thus U = I I I and this union is a countable union of disjoint open intervals. Page 6 of 6
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