Completing the Compact Space

Previously, we have seen that compact metric spaces are complete. Is it true that complete metric spaces are compact? Generally, ‘No’ is right even if ‘Yes’ feels right.

Example 1. The metric space $(\mathcal C([0, 1]), d_{\infty})$ of continuous real-valued functions on $[0, 1]$ is complete but not compact.

Proof. The sequence $\{ f_n \} \subseteq \mathcal C([0, 1])$ defined by $f_n(x) := x^n$ has no convergent subsequence. Thus, $\mathcal C([0, 1])$ is not sequentially compact, and thus not compact.

I believe we contemplated this example before, but now this raises a new question: what other feature does a complete metric space require in order to be compact? Well, minimally it has to be sequentially compact. Let’s first investigate in this direction.

Let $(K, d)$ be a complete metric space and $\{ x_n \}$ be any bounded sequence in $K$ , from which we need to extract a convergent subsequence. Since $K$ is complete, every Cauchy sequence in $K$ converges in $K$ . Thus, we just need to extract a Cauchy subsequence from $\{ x_n \}$ .

Now, the result is obvious if $\{ x_n \}$ contains only finitely many distinct terms, so assume that it contains infinitely many terms. Suppose for any $\epsilon > 0$ , there exist finitely many points $y_1, \dots, y_{N_\epsilon}$ such that $\{ B_d( y_i, \epsilon )\}_{i = 1}^{N_\epsilon}$ covers $K$ . Then there exists some index $i$ such that $B_d( y_i, \epsilon )$ contains infinitely many terms in $\{ x_n \}$ . This means that choosing small $\epsilon > 0$ allows us to extract a subsequence contained in $B_d( y_i, \epsilon )$ . This tail would cause the subsequence to be Cauchy, yielding our desired conclusion. We formalize it as follows.

Definition 1. A metric space $(K, d)$ is totally bounded if for any $r > 0$ , there exists finitely many points $y_1, \dots , y_{N_r}$ such that $\{ B_d( y_i, r )\}_{i = 1}^{N_r}$ covers $K$ .

Theorem 1. If $(K, d)$ is complete and totally bounded, then $K$ is sequentially compact.

Proof. Fix any bounded sequence $\mathbf x_0 := \{ x_n \}$ in $K$ . Define $n_0 := 1$ . For each $k \in \mathbb N$ , define $\epsilon_k := 1/2^N$ .

We proceed by induction. For each $k$ , use the total boundedness of $K$ to find $y_{k+1} \in K$ such that $B_d(y_{k+1}, \epsilon_{k+1})$ contains infinitely many terms from $\mathbf x_k$ . Define the subsequence $\mathbf x_{k+1} := \mathbf x_k \cap B_d( y_{k+1}, \epsilon_k )$ . By the well-ordering principle, define $n_k := \min \{i \in \mathbb N : x_i \in B_d( y_{k+1}, \epsilon_{k+1})\}$ .

Then, we obtain the subsequence $\{ x_{n_k} \} \subseteq K$. To check that this subsequence is Cauchy, fix $\epsilon > 0$ . Then there exists $N \in \mathbb N$ such that $\epsilon_N = 1/2^N < \epsilon$ . By construction, for $m > k > N+1$ ,l

$d(x_{n_m}, x_{n_k}) \leq d(x_{n_m}, y_{N+1}) + d(y_{N+1}, x_{n_k}) < 2 \epsilon_{N+1} = \epsilon_N < \epsilon,$

so that the subsequence is Cauchy. Since $K$ is complete, this subsequence converges. Therefore $K$ is sequentially compact.

But does sequential compactness imply compactness?

Theorem 2. Let $(K, d)$ be a metric space. The following are equivalent:

$K$ is compact.
$K$ is sequentially compact.
$K$ is complete and totally bounded.

Proof. We just need to prove that sequential compactness implies compactness. Suppose $(K, d)$ is sequentially compact.

Let $\{ V_\alpha \}$ be any open cover for $K$ . We claim that there exists $\delta > 0$ such that for any $x, y \in K$ , $d(x, y) < \delta \Rightarrow \{x, y\} \subseteq V_\alpha$ for some $\alpha$ . Otherwise, for any $n \in \mathbb N$ , there exists $x_n, y_n \in K$ such that $d(x_n, y_n) < 1/n$ and yet $\{x_n , y_n\} \notin V_\alpha$ for any $\alpha$ . By sequential compactness, assuming $\{x_n\}$ and $\{ y_n \}$ converge to $x, y$ respectively. Then $d(x, y) \leq d(x, x_n) + d(x_n, y_n) + d(y_n, y)$ can be made arbitrarily small so that $x = y$ , and there exists $V_\alpha$ that contains $\{x_N, y_N\}$ for large $N$ , a contradiction.

Therefore, let $\delta > 0$ be such a number. We claim that there exist finitely many points $x_1, \dots, x_n$ such that $\{ B_d(x_i, \delta) \}_{i = 1}^n$ covers $K$ . Suppose otherwise. Choose $x_1 \in K$ and inductively choose $x_{k+1} \in C_k := K \backslash \bigcup_{i = 1}^k B_d(x_i, \delta) \neq \emptyset$ . Since $K$ is sequentially compact, we can pass to a subsequence and assume $x_n$ converges to some $x \in K$ . Then there exists sufficiently large $N$ such that $d(x_{N+1}, x_N) \leq d(x_{N+1}, x) + d(x, x_N) < \delta/2 + \delta/2 = \delta$ , so that $x_{N+1} \in C_N \cap B_d(x_N, \delta) = \emptyset$ , a contradiction.

In particular, for any $x \in B_d(x_i, \delta)$ , $d(x, x_i) < \delta$ . Hence, $B_d(x_i, \delta) \subseteq V_{\alpha_i}$ for some $\alpha_i$ . Thus, $\{ V_{\alpha_i} \}_{i = 1}^n$ forms a finite subcover for $K$ .

This equivalence yields the famous Heine-Borel theorem that characterizes compact sets in $\mathbb R^n$ . But to properly discuss that, we will need to study compact spaces with more attention.

—Joel Kindiak, 9 Apr 25, 1730H

KindiakMath

Completing the Compact Space

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Completing the Compact Space

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