Let \(X\) be a compact Hausdorff space. Show that \(X\) is metrizable if and only if \(X\) has a countable basis.
Solution
If \(X\) has a countable basis, then compact Hausdorffness implies regularity, and the Urysohn metrization theorem gives a compatible metric.
Conversely, suppose \(X\) is compact and metrizable with metric \(d\). For each \(n\ge1\), compactness gives finitely many points \[ x_{n,1},\dots,x_{n,k_n} \] such that the balls \(B(x_{n,j},1/n)\) cover \(X\). Let \[ D=\{x_{n,j}:n\ge1,1\le j\le k_n\}. \] Then \(D\) is countable and dense. The family of balls \[ B(d,q),\qquad d\in D,\ q\in\mathbb Q_{>0}, \] is countable and forms a basis for the metric topology. Thus \(X\) has a countable basis.
Hence a compact Hausdorff space is metrizable exactly when it is second countable.