Show that every compact metrizable space has a countable basis.
Solution
Let \(d\) be a metric inducing the topology on the compact space \(X\).
1. For every positive integer \(n\), there is a finite set \(F_n\subseteq X\) such that\[ X=\bigcup_{c\in F_n}B(c,1/n). \]
The family \[ \{B(x,1/n):x\in X\} \] is an open cover of \(X\). Compactness gives a finite subcover. Let \(F_n\) be the finite set of centers occurring in such a subcover.
2. The collection\[ \mathcal B = \{B(c,1/n):n\in\mathbb N_{>0},\ c\in F_n\} \]is countable.
1. Hence \(\mathcal B\) is a countable union of finite collections and is therefore countable.
3. The collection \(\mathcal B\) is a basis for the topology of \(X\).
Let \(U\subseteq X\) be open and let \(x\in U\). Because \(d\) induces the topology, choose \(\varepsilon>0\) such that \[ B(x,\varepsilon)\subseteq U. \] Choose \(n\) large enough that \[ \frac{2}{n}<\varepsilon. \] By 1 there is \(c\in F_n\) with \[ x\in B(c,1/n). \] If \(y\in B(c,1/n)\), then \[ d(y,x) \le d(y,c)+d(c,x) <\frac1n+\frac1n =\frac2n <\varepsilon. \] Thus \[ x\in B(c,1/n)\subseteq B(x,\varepsilon)\subseteq U. \] So every point of every open set lies in a member of \(\mathcal B\) contained in that open set, which is exactly the basis criterion.
4. Therefore every compact metrizable space has a countable basis.
By 2, \(\mathcal B\) is countable, and by 3 it is a basis.