### Select your language

Suggested languages for you:

Americas

Europe

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 $$\mathrm{X}$$

Expert verified
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.
See the step by step solution

## Unlock all solutions

Get unlimited access to millions of textbook solutions with Vaia Premium

Over 22 million students worldwide already upgrade their learning with Vaia!

## 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 $$X$$.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features
• Access our smart AI features to upgrade your learning

## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner