Let \(X\) equal the countable union \(\bigcup B_n\). Show that if \(X\) is a nonempty Baire space, at least one of the sets \(\overline{B}_n\) has a nonempty interior.
Solution
If every \(\overline{B_n}\) had empty interior, then the closed sets \(\overline{B_n}\) would all be nowhere dense. But \[ X=\bigcup_nB_n\subset\bigcup_n\overline{B_n}=X, \] so \(X\) would be a countable union of nowhere dense sets. This contradicts the Baire property for the nonempty open set \(X\) itself. Hence some \(\overline{B_n}\) has nonempty interior.