Show that every compact metrizable space \(X\) has a countable basis. [Hint: Let \(\mathcal{A}_n\) be a finite covering of \(X\) by \(1/n\)-balls.]
Solution
Goal: Prove that every compact metrizable space \(X\) is second-countable (admits a countable topological basis).
1. Construction of candidate basis \(\mathcal{B}\):Proof:
1.1. Let \(d\) be a compatible metric on \(X\).
1.2. For each positive integer \(n \in \mathbb{Z}_+\), the family of open balls \(\{B_d(x, 1/n)\}_{x \in X}\) is an open cover of \(X\).
1.3. Because \(X\) is compact, there exists a finite subcover:\[\mathcal{A}_n = \left\{ B_d\left(x_{n, 1}, \frac{1}{n}\right), \dots, B_d\left(x_{n, k_n}, \frac{1}{n}\right) \right\}.\]
1.4. Define \(\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{A}_n\).
1.5. As a countable union of finite sets, \(\mathcal{B}\) is a countable collection of open subsets of \(X\).
2. Verification that \(\mathcal{B}\) is a basis for the topology of \(X\):Proof:
2.1. Let \(U \subseteq X\) be an open set and let \(x \in U\).
2.2. Since \(U\) is metric-open, there exists \(\varepsilon > 0\) such that the metric ball \(B_d(x, \varepsilon) \subseteq U\).
2.3. Choose an integer \(n \in \mathbb{Z}_+\) sufficiently large such that \(\frac{1}{n} < \frac{\varepsilon}{2}\).
2.4. Because \(\mathcal{A}_n\) covers \(X\), there exists an index \(i \in \{1, \dots, k_n\}\) such that \(x \in B_d(x_{n, i}, \frac{1}{n})\).
2.5. We claim that \(B_d(x_{n, i}, \frac{1}{n}) \subseteq B_d(x, \varepsilon)\):- For any \(y \in B_d(x_{n, i}, \frac{1}{n})\), the triangle inequality gives:\[d(y, x) \le d(y, x_{n, i}) + d(x_{n, i}, x) < \frac{1}{n} + \frac{1}{n} = \frac{2}{n} < \varepsilon.\]- Thus \(y \in B_d(x, \varepsilon) \subseteq U\).
2.6. Hence \(x \in B_d(x_{n, i}, \frac{1}{n}) \subseteq U\), with \(B_d(x_{n, i}, \frac{1}{n}) \in \mathcal{B}\).
3. Conclusion:\(\mathcal{B}\) is a countable basis for \(X\), so \(X\) is second-countable. Q.E.D.