Lemma 4.21.2. Let $X$ be a topological space. If for any $U \subset X$ open and $\seq{U_n}\subset 2^{X}$ open and dense, there exists $\seq{V_j}\subset 2^{X}$ open such that:
For all $n > 1$, $\ol V_{n} \subset U_{n} \cap V_{n - 1}\subset U$.
$\bigcap_{j \in \natp}\ol V_{j}$ is non-empty.
then $X$ is a Baire space.
Proof. By assumption (a),
\[U \cap \bigcap_{n \in \natp}\ol V_{n} \subset U \cap \bigcap_{n \in \natp}U_{n}\]
Since $\bigcap_{n \in \natp}\ol V_{n} \ne \emptyset$ by assumption (b), $U \cap \bigcap_{n \in \natp}U_{n} \ne \emptyset$, so $\bigcap_{n \in \natp}U_{n}$ is dense.$\square$