Chapter 3 Exercise #20 on p. 82

Transcription

1 19 views Chapter 3 Exercise #20 on p. 82 You may start your answer as follows. ε Let ε > 0. Since { p n } is Cauchy, there exists N 1 such that d ( p n, p m ) < for n, m N. 2 Remark. In coming up with an answer, one has the choice of whether to start with "{ p n } is Cauchy" or "some subsequence converges". I believe it is better to start with "{ p n } is Cauchy" as above. 1/1

2 17 views Chapter 3 Exercise #21 on p. 82 One approach is as follows. For each n choose. x n E n Using E n E n+1 and lim n diam E n = 0, show that { } is a Cauchy sequence. x n Show that its limit point (why does this exist?), call it x0, is in E n=1 E n. Suppose that E contains two distinct points. Obtain a contradiction. 1/1

3 13 views Chapter 3 Exercise #22 on p. 82 Since G 1 is nonempty and open, there exists x 1 X and r 1 > 0 such that E 1 B r1 ( x 1 ) satisfies E 1 G 1. r Show that there exists x 2 X and r (0, 1 2 ) such that E ( ) satisfies. 2 2 B r2 x 2 E 2 E 1 Explain how to continue, preferably by induction. Define E = E n. n=1 Explain how to use Exercise /1

4 12 views Chapter 3 Exercise #23 on p. 82 In the hint Rudin gives, which is very helpful, observe that one can switch n and m. Explain what inequality this gives for the absolute value in the next display. 1/1

5 11 views Chapter 3 Exercise #24 on p. 82 Let be a metric space and let be the set of Cauchy sequences. That is, a sequence p n is in C if and only if p is Cauchy. In other words, a point p in C is the same as a Cauchy sequence in X. X C p = { } { } p n Define the equivalence relation on by: Two Cauchy sequences and are equivalent, written, if. (a) Show that is an equivalence relation on, i.e., the relation is reflexive, symmetric, and transitive. Hint for transitivity: Suppose and. Use the triangle inequality. If, then (why?). Given, the equivalence class of is, i.e, the set of all Cauchy sequences equivalent to. Let be the set of equivalence classes of Cauchy sequences. That is, X for some Given, define, where p n and q = { } Q. (b) (i) Show that the function, given by the formula above, is well defined. That is, the right side is nonnegative and does not depend on the choices of p P and q Q. (ii) Suppose. Show that P = Q. (iii) Show that the triangle inequality for follows from the triangle inequality for d. Remark on (b)(i). Suppose p and q, Q, where and. We need to show that. By hypothesis, and. Use the triangle inequality. (c) Show that is complete. Let be a Cauchy sequence in. (Note that, where each is itself a Cauchy sequence. So P i is a Cauchy sequence of Cauchy sequences.) Show that converges to some. That is, there exists such that as. Hint: For each, we may write, where is a Cauchy sequence. Define, where p 1 p 2 p 3. That is, is the equivalence class of the Cauchy sequence whose n th term is the same as the n th term of. (i) Show that p is a Cauchy sequence, so that P X. C p = { p n } q = { q n } p q lim n d( p n, q n ) = 0 C p q q r lim n d( p n, r n ) 0 p r p C p [p] {q C q p} p X = {P P = [p] p C}. P, Q X Δ(P, Q) = lim n d( p n, q n ) p = { } P Δ(P, Q) = 0 Δ : X X R lim n d( p n, q n ) Δ p, P q p = { p n } q = { q n} lim n d( p n, q n ) = lim n d( p n, q n) lim n d( p n, p n) = 0 lim n d( q n, q n) = 0 X { P i } X { P i } = { P1, P2, P3, } P i { } { P i } P X P X Δ(, P) 0 (ii) Show that converges to. That is,. Note that, for each, P i lim n p i p n. Define by, where P p for the constant sequence. P i i Z + P i = [ p i ] p i = { p i (1), p i (2), p i (3), } P = [p] p = { (1), (2), (3), } P p p n a n i { P i } P Δ( P i, P) 0 i Δ(, P) = d( (n), (n)) φ : X X φ(p) = P p [p] p = {p, p, p, } (d) Show that for any, we have. Why does this imply that is an injection (a.k.a., one to one)? 1/2

6 (d) Show that for any, we have P p P q. Why does this imply that φ is an injection (a.k.a., one to one)? φ(x) (e) Show that is dense in X. p, q X Δ(, ) = d(p, q) Hint: Let P = [p] X, where p = { p n }. Define p n = { p n, p n, p n, } the constant sequence. Let P n = [ p n ] X Show that lim n Δ( P n, P) = 0. Updated 13 hours ago by Bennett Chow Resolved Unresolved Anonymous 16 hours ago For (c), what does it mean by saying a sequence of equivalent classes converges to a specific equivalent class? Bennett Chow 13 hours ago I've edited the post to clarify this. See the third line of (c). I corrected the hint for (e). I've added a few other edits. 2/2