Definition 4.21.1 (Baire Space). Let $X$ be a topological space, then the following are equivalent:
For any $\seq{A_n}\subset 2^{X}$ nowhere dense, $\bigcup_{n \in \nat^+}A_{n} \subsetneq X$.
For any $\seq{A_n}\subset 2^{X}$ closed with empty interior, $\bigcup_{n \in \nat^+}A_{n}$ has empty interior.
For any $\seq{U_n}\subset 2^{X}$ open and dense, $\bigcap_{n \in \nat^+}U_{n}$ is dense.
If the above holds, then $X$ is a Baire space.
Proof. $(1) \Rightarrow (2)$: Let $\seq{A_n}\subset 2^{X}$ be closed with empty interior, then $\seq{A_n}$ are nowhere dense. Hence $\bigcup_{n \in \nat^+}A_{n} \subsetneq X$.
$(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_{n} = U_{n}^{c}$, then $A_{n}$ is closed. For any $\emptyset U \subset A_{n}$ open, $U \cap U_{n} \ne \emptyset$ by density of $U_{n}$, so $A_{n}$ has empty interior.
Suppose that $\bigcap_{n \in \natp}U_{n}$ is not dense, then there exists $\emptyset \ne V \subset X$ open such that $\bigcup_{n \in \natp}A_{n} \supset V$, which contradicts the fact that $\bigcup_{n \in \natp}A_{n}$ has non-empty interior.
$(3) \Rightarrow (1)$: For each $n \in \natp$, let $U_{n} = A_{n}^{c}$, then $U_{n}$ is open and dense. Since $\bigcap_{n \in \natp}U_{n}$ is dense, $\bigcup_{n \in \natp}A_{n}$ has non-empty interior.$\square$