A space \(X\) is locally metrizable if each point \(x\) of \(X\) has a neighborhood that is metrizable in the subspace topology. Show that a compact Hausdorff space \(X\) is metrizable if it is locally metrizable. [Hint: Show that \(X\) is a finite union of open subspaces, each of which has a countable basis.]
Solution
Since \(X\) is compact Hausdorff, it is regular. For each \(x\in X\), choose a metrizable neighborhood \(N_x\). Choose an open \(U_x\) with \(x\in U_x\subset N_x\), and by regularity choose an open \(V_x\) with \[ x\in V_x\subset\overline{V_x}\subset U_x. \] The closed subspace \(\overline{V_x}\) is compact and metrizable, hence second countable. Therefore the open subspace \(V_x\) has a countable basis.
Compactness of \(X\) gives finitely many such open sets \[ V_{x_1},\dots,V_{x_m} \] covering \(X\). The union of countable bases for these finitely many open subspaces is a countable basis for \(X\). Since compact Hausdorff spaces are regular, the Urysohn metrization theorem now implies \(X\) is metrizable.