Baire Spaces

A topological space is said to be a Baire space, if any of the following equivalent conditions holds:

• The intersection of countably many dense open subsets is still dense.
• The union of countably many closed subsets with empty interior has empty interior.

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.

Baire Category Theorem

Complete metric spaces are Baire spaces.

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. $\square\enspace$