Chapter 11: Chapter 11

Problem 326

Define a Cauchy sequence. Prove: a) In any metric space \(\mathrm{X}\), every convergent sequence is a Cauchy sequence. b) Suppose \(\left\\{p_{n}\right\\}\) is a Cauchy sequence in a compact metric space \(\mathrm{X}\), then \(\left\\{\mathrm{p}_{\mathrm{n}}\right\\}\) converges to some point of \(\mathrm{X}\)

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A Cauchy sequence is a sequence \(\left\\{p_n\right\\}\) in a metric space \(X\) with distance function \(d\) such that, for any given \(\epsilon > 0\), there exists a natural number \(N\) with \(d\left(p_m, p_n\right) < \epsilon\) for all \(m,n \ge N\). (a) Every convergent sequence in a metric space is a Cauchy sequence, as proven using the triangle inequality and taking \(N = \max\{N_1, N_2\}\) for appropriate choices of \(N_1\) and \(N_2\). (b) If \(\left\\{p_n\right\\}\) is a Cauchy sequence in a compact metric space \(X\), then there exists a point \(p \in X\) such that \(\lim_{n \to \infty} p_n = p\), as proven by showing the existence of a convergent subsequence and applying the triangle inequality again.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of a Cauchy Sequence

A sequence \(\left\\{p_n\right\\}\) in a metric space \(X\) with distance function \(d\) is called a Cauchy sequence if, for any given \(\epsilon > 0\), there exists a natural number \(N\) such that for any \(m,n \ge N\), we have \(d\left(p_m, p_n\right) < \epsilon\).

Step 2: Proof of (a): Every convergent sequence is a Cauchy sequence

Let \(\left\\{p_n\right\\}\) be a convergent sequence in a metric space \(X\) with distance function \(d\). Let \(p\) be the limit of this convergent sequence. That is, \(\lim_{n \to \infty} p_n = p\). Given \(\epsilon > 0\), we need to show that there exists an \(N\) such that if \(m, n \ge N\), then \(d\left(p_m, p_n\right) < \epsilon\). Since \(\lim_{n \to \infty} p_n = p\), for \(\frac{\epsilon}{2} > 0\), there exists an \(N_1\) such that for all \(n \ge N_1\), \(d\left(p_n, p\right) < \frac{\epsilon}{2}\). Similarly, there exists an \(N_2\) such that for all \(m \ge N_2\), \(d\left(p_m, p\right) < \frac{\epsilon}{2}\). Now, let \(N = \max\{N_1, N_2\}\). If \(m, n \ge N\), then by the triangle inequality, we have \[d\left(p_m, p_n\right) \le d\left(p_m, p\right) + d\left(p, p_n\right) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.\] Therefore, the sequence \(\left\\{p_n\right\\}\) is a Cauchy sequence.

Step 3: Proof of (b): A Cauchy sequence in a compact metric space converges

Suppose \(\left\\{p_n\right\\}\) is a Cauchy sequence in a compact metric space \(X\) with distance function \(d\). We need to prove that there exists a point \(p \in X\) such that \(\lim_{n \to \infty} p_n = p\). Since \(X\) is compact, every sequence in \(X\) has a convergent subsequence. So, there exists a subsequence \(\left\\{p_{n_k}\right\\}\) of \(\left\\{p_n\right\\}\) which converges to a point \(p \in X\). Now, given \(\epsilon > 0\), since \(\left\\{p_n\right\\}\) is a Cauchy sequence, there exists an \(N_1\) such that for any \(m, n \ge N_1\), \(d\left(p_m, p_n\right) < \frac{\epsilon}{2}\). Also, since \(\lim_{k \to \infty} p_{n_k} = p\), there exists an \(N_2\) such that for any \(k \ge N_2\), \(d\left(p_{n_k}, p\right) < \frac{\epsilon}{2}\). Let \(N = \max\{N_1, N_2\}\). For any \(n \ge N\), we have: 1. By the definition of a Cauchy sequence, \(d(p_n, p_{n_N}) < \frac{\epsilon}{2}\). 2. Since \(n_N \ge N \ge N_2\), by the convergence of the subsequence, \(d(p_{n_N}, p) < \frac{\epsilon}{2}\). Now, using the triangle inequality, we have: \[d(p_n, p) \le d(p_n, p_{n_N}) + d(p_{n_N}, p) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.\] Thus, \(\lim_{n \to \infty} p_n = p\), and the Cauchy sequence \(\left\\{p_n\right\\}\) converges to a point of \(X\).

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Chapter 11: Chapter 11

Short Answer

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Step 1: Definition of a Cauchy Sequence

Step 2: Proof of (a): Every convergent sequence is a Cauchy sequence

Step 3: Proof of (b): A Cauchy sequence in a compact metric space converges

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