Let \(X\) be a compact Hausdorff space, let \(\ts{A_n}\) be a countable collection of closed sets of \(X\). Show that if each set \(A_n\) has empty interior in \(X\), then the union \(\bigcup A_n\) has empty interior in \(X\). [Hint: Imitate the proof of Theorem 27.7.]
This is a special case of the Baire category theorem, which we shall study in Chapter 8.
Solution
Suppose toward a contradiction that \(U\subset\bigcup_{n\ge1}A_n\) for some nonempty open set \(U\subset X\). Since compact Hausdorff spaces are regular, we can construct inductively nonempty open sets \(U_n\) such that \[ \overline{U_n}\subset U_{n-1}\setminus A_n \] for \(n\ge1\), starting with \(U_0=U\). Indeed, \(A_n\) has empty interior, so \(U_{n-1}\setminus A_n\) is a nonempty open set; choose a point there and then a nonempty open neighborhood whose closure is still contained in that open set.
The compact sets \[ K_n=\overline{U_n} \] form a nested sequence \[ K_1\supset K_2\supset\cdots \] of nonempty closed subsets of compact \(X\). Hence \[ \bigcap_{n\ge1}K_n\ne\varnothing. \] Choose \(x\) in the intersection. Since \(K_n\subset X\setminus A_n\), we have \(x\notin A_n\) for every \(n\). Also \(K_1\subset U\), so \(x\in U\). This contradicts \(U\subset\bigcup_nA_n\). Therefore \(\bigcup_nA_n\) has empty interior.