Show that every locally compact Hausdorff space is a Baire space.
Solution
1. Reduction:Let \(X\) be a locally compact Hausdorff space, and let \(\{U_n\}_{n=1}^\infty\) be a countable collection of open dense subsets of \(X\).Let \(W \subseteq X\) be an arbitrary non-empty open set.It suffices to show that \(W \cap \left( \bigcap_{n=1}^\infty U_n \right) \neq \varnothing\).
2. Regularity and compact neighborhood property:In a locally compact Hausdorff space, for every point \(x\) and every open neighborhood \(V\) of \(x\), there exists an open neighborhood \(U\) of \(x\) such that \(\overline{U}\) is compact and \(\overline{U} \subseteq V\).
3. Inductive construction of nested compact closures:We inductively construct a sequence of non-empty open sets \(\{V_n\}_{n=1}^\infty\) such that \(\overline{V_1} \subseteq W \cap U_1\), and for each \(k \ge 2\), \(\overline{V_k}\) is compact with \(\overline{V_k} \subseteq V_{k-1} \cap U_k\).Proof:
3.1. Base step (\(k=1\)):Since \(U_1\) is dense in \(X\), the open set \(W \cap U_1\) is non-empty. Pick \(x_1 \in W \cap U_1\). By 2, choose a non-empty open set \(V_1\) such that \(\overline{V_1}\) is compact and \(\overline{V_1} \subseteq W \cap U_1\).
3.2. Inductive step (\(k \ge 2\)):Assume \(V_{k-1}\) is a non-empty open set. Since \(U_k\) is dense in \(X\), \(V_{k-1} \cap U_k\) is a non-empty open set. Pick \(x_k \in V_{k-1} \cap U_k\). By 2, choose a non-empty open set \(V_k\) such that \(\overline{V_k}\) is compact and \(\overline{V_k} \subseteq V_{k-1} \cap U_k\).
3.3. In particular, \(\overline{V_k} \subseteq V_{k-1} \subseteq \overline{V_{k-1}}\) for all \(k \ge 2\).
4. Non-empty intersection via compactness:Proof:
4.1. The sequence \(\{\overline{V_n}\}_{n=1}^\infty\) is a nested sequence of non-empty closed subsets of the compact space \(\overline{V_1}\):\[\overline{V_1} \supseteq \overline{V_2} \supseteq \overline{V_3} \supseteq \cdots\]
4.2. By the finite intersection property of compact spaces, the total intersection is non-empty:\[\bigcap_{n=1}^\infty \overline{V_n} \neq \varnothing.\]
4.3. Choose an element \(x \in \bigcap_{n=1}^\infty \overline{V_n}\).
4.4. For every \(n \ge 1\), \(x \in \overline{V_n} \subseteq U_n\).
4.5. Furthermore, \(x \in \overline{V_1} \subseteq W\).
4.6. Thus \(x \in W \cap \left( \bigcap_{n=1}^\infty U_n \right)\).
5. Q.E.D.By step 4, \(W \cap \left( \bigcap_{n=1}^\infty U_n \right)\) is non-empty for every non-empty open set \(W\), so \(\bigcap_{n=1}^\infty U_n\) is dense in \(X\). Hence \(X\) is a Baire space.