63 lines
3.6 KiB
TeX
63 lines
3.6 KiB
TeX
\section{Baire Spaces}
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\label{section:baire}
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\begin{definition}[Baire Space]
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\label{definition:baire}
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Let $X$ be a topological space, then the following are equivalent:
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\begin{enumerate}
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\item For any $\seq{A_n} \subset 2^X$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
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\item For any $\seq{A_n} \subset 2^X$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n$ has empty interior.
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\item For any $\seq{U_n} \subset 2^X$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
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\end{enumerate}
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If the above holds, then $X$ is a \textbf{Baire space}.
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\end{definition}
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\begin{proof}
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$(1) \Rightarrow (2)$: Let $\seq{A_n} \subset 2^X$ be closed with empty interior, then $\seq{A_n}$ are nowhere dense. Hence $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
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$(2) \Rightarrow (3)$: 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.
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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.
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$(3) \Rightarrow (1)$: 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.
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\end{proof}
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\begin{lemma}
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\label{lemma:baire-condition}
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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:
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\begin{enumerate}
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\item[(a)] For all $n > 1$, $\ol V_n \subset U_n \cap V_{n - 1} \subset U$.
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\item[(b)] $\bigcap_{j \in \natp} \ol V_j$ is non-empty.
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\end{enumerate}
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then $X$ is a Baire space.
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\end{lemma}
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\begin{proof}
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By assumption (a),
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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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\begin{theorem}[Baire Category Theorem]
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\label{theorem:baire}
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Let $X$ be a topological space, then the following are sufficient conditions for $X$ to be Baire:
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\begin{enumerate}
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\item $X$ is completely metrisable.
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\item $X$ is locally compact.
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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 (\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 \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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\label{definition:meagre}
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Let $X$ be a topological space, then $X$ is \textbf{meagre} if there exists $\seq{A_n} \subset 2^X$ nowhere dense such that $X = \bigcup_{n \in \natp}A_n$.
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\end{definition}
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