Baire Category Theorem

Theorem
Let $M = \struct {A, d}$ be a complete metric space.

Then $M = \struct {A, d}$ is also a Baire space.

Proof
Let $U_n$ be a countable set of open sets of $M$ all of which are everywhere dense.

The strategy of this proof is to show that the intersection $\ds \bigcap U_n$ is everywhere dense.

Let $W \subseteq A$ be a non-empty open set of $M$.

From Open Set Characterization of Denseness, since $U_1$ is everywhere dense:
 * $W \cap U_1 \ne \O$

Thus, there is a point $x_1 \in A$ and $\epsilon_1 \in R_{>0}$ such that:
 * $\map {\overline B} {x_1, \epsilon_1} \subset W \cap U_1$

where $\map B {x, \epsilon}$ and $\map {\overline B} {x, \epsilon}$ denote an open $\epsilon$-ball of $x$ and its closure, respectively.

Recall that the $U_n$ are everywhere dense.

Hence we can use in a recursive manner to find a pair of sequences $\sequence {x_n}$ and $\sequence {\epsilon_n} > 0$ such that:
 * $\map {\overline B} {x_n, \epsilon_n} \subset \map B {x_{n - 1}, \epsilon_{n - 1} } \cap U_n$

as well as $\epsilon_n < 1/n $.

Since $x_n \in \map B {x_m, \epsilon_m}$ when $n > m$, we have that $x_n$ is a Cauchy Sequence.

Thus $x_n$ converges to some limit $x$ by definition of completeness.

For any $n$, by closedness:
 * $x \in \map {\overline B} {x_{n + 1}, \epsilon_{n + 1} } \subset \map B {x_n, \epsilon_n}$

Thus:
 * $\forall n \in \N: x \in W \cap U_n$

From Open Set Characterization of Denseness it follows that $\ds \bigcap U_n$ is (everywhere) dense.

So, by definition, $\struct {M, d}$ is a Baire space.