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---
title: Baire Spaces
parent: General Topology
nav_order: 7
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.
---
# {{ page.title }}
{% definition Baire Space %}
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.
{% enddefinition %}
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 * Baire Category Theorem #1 %}
Every complete metric space is a Baire space.
{% endtheorem %}
{% 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 * Baire Category Theorem #2 %}
Every locally compact Hausdorff space is a Baire space.
{% endtheorem %}
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