Symmetry of the derivative (normed/frechet).

src/topology/main/baire.tex

@@ -6,13 +6,38 @@
 \label{definition:baire}
     Let $X$ be a topological space, then the following are equivalent:
     \begin{enumerate}
-        \item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
-        \item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
-        \item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat^+}A_n = X$, there exists $N \in \nat^+$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
-        \item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
+        \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}
@@ -22,3 +47,15 @@
         \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 (\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 $\text{diam}(V_n) \le 1/n$.
+
+    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 \ref{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}