Chapter 3 - CBU

3.5. THE CAUCHY CRITERION 65Lemma (3.5.4). Cauchy sequences are bounded.Proof. Let X = (x n ) be Cauchy and ✏ = 19 H 2 N 3 8 n H, |x n x H | < 1.Then|x n | |x H | apple x n x H < 1 =)1 < |x n | |x H | < 1 =) |x n | < |x H | + 1.Let M = max |x 1 |, |x 2 |, . . . , |x H 1 |, |x H | + 1 .Then |x n | apple M 8n 2 N.Theorem (3.5.5 — Cauchy Convergence Criterion). A sequence is convergent() it is Cauchy.Proof. [Yet another ✏ 2 argument.](=)) Lemma 3.5.3((=) Let X = (x n ) be Cauchy. Then X is bounded,so by B-W, X has a convergent subsequence, say X 0 = (x nk ) ! x ? .[To show lim(x n ) = x ? .] Let ✏ > 0 be given.Since (x n ) is Cauchy, 9 H 2 N 3 8 n, m H, |x n x m | < ✏ 2 .Since lim(x nk ) = x ? , 9 K 2 N 3 K H and |x K x ? | < ✏ 2 .⇤But |x n x K | < ✏ 2also. Then, for n H,|x n x ? | = |(x n x K ) + (x K x ? )| apple |(x n x K )| + |(x K x ? )| < ✏ 2 + ✏ 2 = ✏.Thus lim(x n ) = x ? .⇤