Let \(X\) be a compact Hausdorff space that is the union of the closed subspaces \(X_1\) and \(X_2\). If \(X_1\) and \(X_2\) are metrizable, show that \(X\) is metrizable. [Hint: Construct a countable collection \(\mathcal{A}\) of open sets of \(X\) whose intersections with \(X_i\) form a basis for \(X_i\), for \(i = 1, 2\). Assume \(X_1 - X_2\) and \(X_2 - X_1\) belong to \(\mathcal{A}\). Let \(\mathcal{B}\) consist of finite intersections of elements of \(\mathcal{A}\).]
Solution
Since each \(X_i\) is closed in compact \(X\), each \(X_i\) is compact. Being metrizable, each \(X_i\) therefore has a countable basis \(\mathcal B_i\).
For every \(B\in\mathcal B_i\), choose an open set \(\widetilde B\subset X\) such that \[ \widetilde B\cap X_i=B. \] Let \(\mathcal A\) be the countable collection of all these chosen lifts, together with the two open sets \[ X\setminus X_1,\qquad X\setminus X_2. \] Let \(\mathcal C\) be the collection of all finite intersections of members of \(\mathcal A\). It is countable.
We show \(\mathcal C\) is a basis for \(X\). Let \(x\in U\) with \(U\subset X\) open. For each \(i=1,2\):
- if \(x\in X_i\), choose \(B_i\in\mathcal B_i\) with \[ x\in B_i\subset U\cap X_i, \] and let \(\widetilde B_i\in\mathcal A\) be its chosen lift;
- if \(x\notin X_i\), use the member \(X\setminus X_i\in\mathcal A\).
The intersection \(C\) of the resulting one or two members of \(\mathcal A\) contains \(x\). For each \(i\), \[ C\cap X_i\subset U\cap X_i. \] Since \(X=X_1\cup X_2\), this implies \(C\subset U\). Hence \(\mathcal C\) is a countable basis for \(X\).
By the preceding compact-Hausdorff metrization criterion, \(X\) is metrizable.