Complete Metrics

In real analysis, we have a rather useful property of $\mathbb R$ .

Theorem 1. Let $\{x_n\}$ be an $\mathbb R$ -sequence. Then $\{x_n\}$ converges in the usual topology if and only if $\{x_n\}$ is Cauchy.

In this case, we call $\mathbb R$ a complete metric space. What other metric spaces are complete?

Definition 1. We say that $K$ is a complete if every Cauchy $K$ -sequence converges in $K$ .

In terms of the metric $d$ , recall that $\{x_n\}$ is Cauchy if for any $\epsilon > 0$ , there exists $N \in \mathbb N$ such that for $m \geq n \geq N$ , $d(x_m, x_n) < \epsilon$ .

Lemma 1. If $K$ is compact, then any $K$ -sequence contains a convergent subsequence (i.e. $K$ is sequentially compact).

Proof. For the direction $(\Rightarrow)$ , suppose $K$ is compact. Let $\{x_n\}$ be any $K$ -sequence. If $\{x_n\}$ is finite, then there exists some $x \in K$ such that $x_k = x$ for infinitely many indices $k$ . Thus, the subsequence $\{x\}$ trivially converges to $x$ .

Suppose instead $X := \{x_n : n \in \mathbb N\}$ is infinite and non-repeating. We claim that there exists $x \in K$ such that for any $k \in \mathbb N$ , $B_d(x,1/k)\backslash \{x\} \cap X \neq \emptyset$ . In this instance, choose $x_{n_k} \in B_d(x, 1/k)$ so that $x_{n_k} \to x$ .

Suppose for a contradiction that for any $x \in K$ , there exists $\epsilon_x > 0$ such that $B_d(x, \epsilon_x) \backslash \{x\} \cap X = \emptyset$ . We first claim that $X$ is closed. For any $x \in K \backslash X$ , find $\epsilon_x > 0$ such that

$B_d(x, \epsilon_x) \backslash \{x\} \cap X = \emptyset.$

In particular, $B_d(x, \epsilon_x) \cap X = \emptyset$ so that $x \in B_d(x, \epsilon_x) \subseteq K \backslash X$ . Therefore, $K \backslash X$ is open and $X$ is closed.

By hypothesis, for any $n \in \mathbb N$ , there exists $k_n \in \mathbb N$ such that $B_d(x_n, 1/k_n) \backslash \{x_n\} \cap X = \emptyset$ . Defining $V_0 := K \backslash X$ and $V_n := B_d(x_n, 1/k_n)$ , the collection $\{V_0\} \cup \{V_n : n \in \mathbb N\}$ forms an open cover for $K$ . Since $K$ is compact, there exists some $N \in \mathbb N$ such that without loss of generality, $\{V_0,V_1,\dots,V_N\}$ covers $K$ .

Hence, there exists $i = 1, \dots, N$ such that $x_{N+1} \in B_d(x_i, 1/k_i) \backslash \{x_i\}$ . Therefore, $B_d(x_i, 1/k_i) \backslash \{x_i\} \cap X \neq \emptyset$ , a a contradiction, as required.

For the direction $(\Leftarrow)$ , suppose $K$ is sequentially compact. Let $\{V_\alpha\}$ be any open cover of $K$ .

Lemma 2. If $K$ is sequentially compact, then it is complete.

Proof. Let $\{x_n\}$ be any Cauchy sequence in $K$ . Since $K$ is sequentially compact, $\{x_n\}$ contains a convergent subsequence $\{x_{n_k}\}$ such that $x_{n_k} \to x$ . We claim that $x_n \to x$ .

Fix $\epsilon > 0$ . Since $\{x_{n_k}\}$ converges, for any $k_1 > 0$ , there exists $N_1 \in \mathbb N$ such that for $k \geq N_1$ , $n_k \geq N_1$ and $d(x_{n_k}, x) < k_1 \cdot \epsilon$ . Since $\{x_n\}$ is Cauchy, for any $k_2 > 0$ , there exists $N_2 \in \mathbb N$ such that for $m \geq n \geq N_2$ , $d(x_m, x_n) < k_2 \cdot \epsilon$ . Therefore, for $n > \max \{N_1,N_2\} =: N_3$ ,

$d(x_n,x) \leq d(x_n,x_{n_{N_3}}) + d(x_{n_{N_3}},x) < (k_1+k_2)\cdot \epsilon.$

Setting $k_1 = k_2 = 1/2$ yields the desired result.

Theorem 2. If $K$ is compact, then it is complete.

Proof. Since $K$ is compact, it is sequentially compact by Lemma 1. Thus, $K$ is complete by Lemma 2.

Theorem 3. Suppose $K$ is a compact metric space. Let $\mathcal C(K) \subseteq \mathcal F(K, \mathbb R)$ denote the collection of continuous real-valued functions equipped with the supremum norm and supremum metric respectively:

$\displaystyle \| f \|_\infty := \sup_{x \in K} |f(x)|,\quad d_\infty(f,g) := \|f - g \|_\infty.$

Then $(\mathcal C(K), d_\infty)$ is a complete metric space.

Proof. We note that convergence in $(\mathcal C(K), d_\infty)$ corresponds to convergence in the supremum norm. Thus, $f_n \to f$ means that $f_n$ converges uniformly to $f$ . Let $\{f_n\}$ be a Cauchy sequence. This implies that for each $x \in K$ ,

$|f_m(x) - f_n(x)| = |(f_m - f_n)(x)| \leq \| f_m - f_n \|_\infty = d_\infty(f_m,f_n).$

Thus, $\{f_n(x)\}$ is Cauchy and converges in $\mathbb R$ to some $\{ f(x) \}$ . We claim that $f_n \to f$ uniformly. Fix $\epsilon > 0$ . Find $N \in \mathbb N$ such that for any $m, n \geq N$ , for any $x \in K$ ,

$\displaystyle |f_m(x) - f_n(x)| < \frac{ \epsilon }2.$

Since $| \cdot |$ is continuous in the usual topology on $\mathbb R$ , taking $m \to \infty$ ,

$\displaystyle |f(x) - f_n(x)| = \lim_{m \to \infty}|f_m(x) - f_n(x)| \leq \lim_{m \to \infty} \frac{ \epsilon }2 = \frac{\epsilon}2.$

Since this inequality holds for all $x \in K$ , $d_\infty (f_n,f) \leq \epsilon/2 < \epsilon$ . Thus, $f_n \to f$ uniformly, as required.

Remarkably, however, $\mathcal C(K)$ is not always compact.

Example 1. Define the sequence $\{f_n\} \subseteq \mathcal C([0,1])$ by $f_n(x) = nx$ . Then $\{f_n(x)\}$ diverges at every $x \in [0, 1]$ pointwise, so that $\{f_n\}$ does not have any convergent subsequence.

When then, is a subset of functions $\mathcal F \subseteq \mathcal C(K)$ compact? The answer to this question is the famous Arzelà-Ascoli theorem, and we will explore this theorem in the future. For now, we want to use completeness to deduce Banach’s fixed-point theorem, which helps us establish the existence of solutions to first-order differential equations.

—Joel Kindiak, 1 Apr 25, 2008H

KindiakMath

Complete Metrics

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Complete Metrics

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