Cleanup
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@@ -36,6 +36,7 @@
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\[
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U \cap \bigcap_{n \in \natp}\ol V_n \subset U \cap \bigcap_{n \in \natp}U_n
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\]
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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.
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\end{proof}
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@@ -48,11 +49,11 @@
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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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 (\ref{definition:uniform-separated}/\ref{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 $\text{diam}(V_n) \le 1/n$.
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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 $\text{diam}(V_n) \le 1/n$.
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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 $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$.
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By \ref{lemma:baire-condition}, $X$ is a Baire space.
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By \autoref{lemma:baire-condition}, $X$ is a Baire space.
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\end{proof}
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\begin{definition}[Meagre]
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