Show that a metrizable space is compact if and only if it is sequentially compact.
Munkres, Topology, §28, Theorem 28.2, which adds limit point compactness as a third equivalent condition.
As printed upstream this exercise read “Show that if \(X\) is metrizable, then \(X\) is compact”, which is false: \(\RR\) is metrizable and not compact.
Solution
Let \((X, d)\) be a metric space.
\((\Longrightarrow)\) Compact \(\implies\) Sequentially Compact: Let \((x_n)_{n=1}^\infty\) be a sequence in \(X\).
If the set of values \(S = \{x_n : n \in \mathbb{N}\}\) is finite, then by the Pigeonhole Principle there is a point \(x \in S\) and a subsequence that is identically constant \(x_{n_k} = x\), which converges to \(x\).
If \(S\) is infinite: Suppose towards a contradiction that \((x_n)\) has no convergent subsequence. Then no point \(x \in X\) is an accumulation point of \(S\). Thus, for every \(x \in X\), there exists an open ball \(U_x = B(x, \varepsilon_x)\) containing at most finitely many points of \(S\) (specifically, at most \(x\) itself). The collection \(\{U_x\}_{x \in X}\) is an open cover of the compact space \(X\). By compactness, there exists a finite subcover \(\{U_{x_1}, \ldots, U_{x_m}\}\) covering \(X\). Then \(S \subseteq X \subseteq \bigcup_{i=1}^m U_{x_i}\), which implies that \(S\) is a subset of a finite union of finite sets, hence \(S\) must be finite, a contradiction. Thus \((x_n)\) must have a convergent subsequence.
\((\Longleftarrow)\) Sequentially Compact \(\implies\) Compact: We establish this via Lebesgue Number and Total Boundedness:
Lebesgue’s Number Lemma: Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in A}\) be an open cover of \(X\). Suppose towards a contradiction that there is no Lebesgue number: for every \(n \in \mathbb{N}\), there exists \(x_n \in X\) such that \(B(x_n, 1/n)\) is not contained in any single set \(U_\alpha \in \mathcal{U}\). Since \(X\) is sequentially compact, \((x_n)\) has a subsequence \((x_{n_k})\) converging to some point \(x \in X\). Since \(\mathcal{U}\) covers \(X\), \(x \in U_{\alpha_0}\) for some \(\alpha_0\). Since \(U_{\alpha_0}\) is open, there exists \(r > 0\) such that \(B(x, r) \subseteq U_{\alpha_0}\). Since \(x_{n_k} \to x\), choose \(k\) large enough such that \(d(x_{n_k}, x) < r/2\) and \(1/n_k < r/2\). Then for any \(y \in B(x_{n_k}, 1/n_k)\): \[ d(y, x) \leq d(y, x_{n_k}) + d(x_{n_k}, x) < \frac{1}{n_k} + \frac{r}{2} < \frac{r}{2} + \frac{r}{2} = r. \] Thus \(B(x_{n_k}, 1/n_k) \subseteq B(x, r) \subseteq U_{\alpha_0}\), contradicting the choice of \(x_{n_k}\). Hence, there exists a Lebesgue number \(\delta > 0\) for the cover \(\mathcal{U}\).
Total Boundedness: As proved in standard metric space theory, sequential compactness implies that \(X\) is totally bounded (otherwise an infinite \(\delta\)-separated sequence would have no Cauchy/convergent subsequence). Thus, for the Lebesgue number \(\delta > 0\), \(X\) can be covered by finitely many open balls of radius \(\delta\): \[ X = \bigcup_{i=1}^N B(y_i, \delta). \]
Finite Subcover: By the definition of the Lebesgue number \(\delta\), each ball \(B(y_i, \delta)\) is entirely contained in some open set \(U_{\alpha_i} \in \mathcal{U}\). Therefore: \[ X = \bigcup_{i=1}^N B(y_i, \delta) \subseteq \bigcup_{i=1}^N U_{\alpha_i}. \] Thus \(\{U_{\alpha_1}, \ldots, U_{\alpha_N}\}\) is a finite subcover of \(\mathcal{U}\), proving that \(X\) is compact.