4.21 Baire Spaces
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.$\square$
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),
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.$\square$
Theorem 4.21.3 (Baire Category Theorem). Let $X$ be a topological space, then the following are sufficient conditions for $X$ to be Baire:
$X$ is completely metrisable.
$X$ is locally compact.
Proof. Let $U \subset X$ be open and $\seq{U_n}\subset 2^{X}$ be open and dense. Let $V_{0} = U$. For each $n \in \natp$, by density of $U_{n}$, there exists $x \in U_{n} \cap V_{n - 1}$. Since $X$ is regular (Definition 5.1.17/Proposition 4.16.4), there exists $V_{n}\in \cn^{o}(x)$ such that $x \in V_{n}\subset \ol U_{n}\subset U_{n} \cap V_{n-1}$. If $X$ is locally compact, choose $V_{n}$ to be precompact. If $X$ is completely metrisable, choose $V_{n}$ such that $\text{diam}(V_{n}) \le 1/n$.
Now, if $X$ is locally compact, then by the finite intersection property, $\bigcap_{n \in \natp}\ol{V_n}\ne \emptyset$. If $X$ is completely metrisable, then $\seq{V_n}$ is a Cauchy filter base, and converges to at least one point, so $\bigcap_{n \in \natp}\ol{V_n}\ne \emptyset$.
By Lemma 4.21.2, $X$ is a Baire space.$\square$
Definition 4.21.4 (Meagre). Let $X$ be a topological space, then $X$ is meagre if there exists $\seq{A_n}\subset 2^{X}$ nowhere dense such that $X = \bigcup_{n \in \natp}A_{n}$.