From 777f9d3fd8caf56e6bc6999a4b05379307d0733f Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Tue, 12 Sep 2023 07:36:33 +0200
Subject: Initial commit

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+---
+title: Baire Spaces
+parent: General Topology
+nav_order: 1
+description: >
+  A Baire space is a topological space with the property that the intersection
+  of countably many dense open subsets is still dense. One version of the Baire
+  Category Theorem states that complete metric spaces are Baire spaces. We give
+  a self-contained proof of Baire's Category Theorem by contradiction.
+# spellchecker:words
+---
+
+# {{ page.title }}
+
+{: .definition }
+> A topological space is said to be a *Baire space*,
+> if any of the following equivalent conditions holds:
+> {: .mb-0 }
+>
+> - The intersection of countably many dense open subsets is still dense.
+> - The union of countably many closed subsets with empty interior has empty interior.
+> {: .mt-0 .mb-0 }
+
+Note that
+a set is dense in a topological space
+if and only if
+its complement has empty interior.
+
+Any sufficient condition
+for a topological space to be a Baire space
+constitutes a *Baire Category Theorem*,
+of which there are several.
+Here we give one
+that is commonly used in functional analysis.
+
+{: .theorem-title }
+> Baire Category Theorem #1
+> {: #baire-category-theorem }
+>
+> Every complete metric space is a Baire space.
+
+{% proof %}
+Let $X$ be a metric space
+with complete metric $d$.
+Suppose that $X$ is not a Baire space.
+Then there is a countable collection $\braces{U_n}$ of dense open subsets of $X$
+such that the intersection $U := \bigcap U_n$ is not dense in $X$.
+
+In a metric space, any nonempty open set contains an open ball.
+It is also true, that any nonempty open set contains a closed ball,
+since $\overline{B(y,\delta_1)} \subset B(y,\delta_2)$ if $\delta_1 < \delta_2$.
+
+We construct a sequence $(B_n)$ of open balls $B_n := B(x_n,\epsilon_n)$ satisfying
+
+$$
+\overline{B_{n+1}} \subset B_n \cap U_n \quad \epsilon_n < \tfrac{1}{n} \quad \forall n \in \NN,
+$$
+
+as follows: By hypothesis,
+the interior of $X \setminus U$ is not empty (otherwise $U$ would be dense in $X$),
+so we may choose an open ball $B_1$ with $\epsilon_1 < 1$
+such that $\overline{B_1} \subset X \setminus U$.
+Given $B_n$,
+the set $B_n \cap U_n$ is nonempty, because $U_n$ is dense in $X$,
+and it is open, because $B_n$ and $U_n$ are open.
+This allows us to choose an open ball $B_{n+1}$ as desired.
+
+Note that by construction $B_m \subset B_n$ for $m \ge n$,
+thus $d(x_m,x_n) < \epsilon_n < \tfrac{1}{n}$.
+Therefore, the sequence $(x_n)$ is Cauchy
+and has a limit point $x$ by completeness.
+In the limit $m \to \infty$, we obtain $d(x,x_n) \le \epsilon_n$ (strictness is lost),
+hence $x \in \overline{B_n}$ for all $n$.
+This shows that $x \in U_n$ for all $n$, that is $x \in U$.
+On the other hand, $x \in \overline{B_1} \subset X \setminus U$,
+in contradiction to the preceding statement.
+{% endproof %}
+
+{: .theorem-title }
+> Baire Category Theorem #2
+> {: #baire-category-theorem }
+>
+> Every compact Hausdorff space is a Baire space.
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