\section{Baire Spaces}
\label{section:baire}


\begin{definition}[Baire Space]
\label{definition:baire}
    Let $X$ be a topological space, then the following are equivalent:
    \begin{enumerate}
        \item For any $\seq{A_n} \subset 2^X$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
        \item For any $\seq{A_n} \subset 2^X$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n$ has empty interior.
        \item For any $\seq{U_n} \subset 2^X$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
    \end{enumerate}
    If the above holds, then $X$ is a \textbf{Baire space}.
\end{definition}
\begin{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.
\end{proof}

\begin{lemma}
\label{lemma:baire-condition}
    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:
    \begin{enumerate}
        \item[(a)] For all $n > 1$, $\ol V_n \subset U_n \cap V_{n - 1} \subset U$.
        \item[(b)] $\bigcap_{j \in \natp} \ol V_j$ is non-empty.
    \end{enumerate}
    then $X$ is a Baire space.
\end{lemma}
\begin{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.
\end{proof}

\begin{theorem}[Baire Category Theorem]
\label{theorem:baire}
    Let $X$ be a topological space, then the following are sufficient conditions for $X$ to be Baire:
    \begin{enumerate}
        \item $X$ is completely metrisable.
        \item $X$ is locally compact.
    \end{enumerate}
\end{theorem}
\begin{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 (\autoref{definition:uniform-separated}/\autoref{proposition:compact-hausdorff-normal}), 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 \autoref{lemma:baire-condition}, $X$ is a Baire space.
\end{proof}

\begin{definition}[Meagre]
\label{definition:meagre}
    Let $X$ be a topological space, then $X$ is \textbf{meagre} if there exists $\seq{A_n} \subset 2^X$ nowhere dense such that $X = \bigcup_{n \in \natp}A_n$.
\end{definition}
