Show that if every point \(x\) of \(X\) has a neighborhood that is a Baire space, then \(X\) is a Baire space. [Hint: Use the open set formulation of the Baire condition.]
Solution
1. Reduction to an open cover by Baire open sets:Proof:
1.1. If a point \(x \in X\) has a neighborhood \(N\) that is a Baire space, there exists an open neighborhood \(U_x \subseteq N\) containing \(x\).
1.2. Since every open subspace of a Baire space is a Baire space, \(U_x\) is an open subspace of \(X\) that is a Baire space.
1.3. Thus \(X\) possesses an open covering \(\mathcal{U} = \{U_\alpha\}_{\alpha \in J}\) such that each \(U_\alpha\) is open in \(X\) and is a Baire space in the subspace topology.
2. Open set / nowhere dense formulation of the Baire condition:Proof:
2.1. A topological space \(X\) is a Baire space if and only if for every countable collection \(\{A_n\}_{n=1}^\infty\) of closed subsets of \(X\) each having empty interior (\(\operatorname{Int}_X(A_n) = \varnothing\)), and every non-empty open set \(W \subseteq X\), the difference \(W \setminus \bigcup_{n=1}^\infty A_n\) is non-empty.
3. Verification of the Baire property on \(X\):Proof:
3.1. Let \(\{A_n\}_{n=1}^\infty\) be a sequence of closed subsets of \(X\) with \(\operatorname{Int}_X(A_n) = \varnothing\) for all \(n\), and let \(W \subseteq X\) be an arbitrary non-empty open set.
3.2. Pick a point \(x_0 \in W\).
3.3. Since \(\mathcal{U}\) covers \(X\), choose an open Baire set \(U \in \mathcal{U}\) containing \(x_0\).
3.4. The intersection \(V = W \cap U\) is a non-empty open subset of \(U\).
3.5. For each \(n\), the set \(F_n = A_n \cap U\) is closed in \(U\).
3.6. Because \(U\) is open in \(X\), the interior in \(U\) satisfies:\[\operatorname{Int}_U(F_n) = \operatorname{Int}_X(A_n) \cap U = \varnothing \cap U = \varnothing.\]
3.7. Since \(U\) is a Baire space and \(V\) is a non-empty open set in \(U\), \(V \not\subseteq \bigcup_{n=1}^\infty F_n\).
3.8. Thus there exists a point \(y \in V\) such that \(y \notin \bigcup_{n=1}^\infty F_n\).
3.9. Because \(y \in V \subseteq U\), \(y \notin F_n \implies y \notin A_n\) for all \(n \ge 1\).
3.10. Since \(y \in V \subseteq W\), we have \(y \in W \setminus \bigcup_{n=1}^\infty A_n\).
4. Q.E.D.By step 3, \(W \not\subseteq \bigcup_{n=1}^\infty A_n\) for every nonempty open \(W\), so \(X\) is a Baire space by step
2.