Baire Category Theorem

Theorem
Let $M = \left({A, d}\right)$ be a complete metric space.

Then $M = \left({A, d}\right)$ 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 $\displaystyle \bigcap U_n$ is everywhere dense.

Let $W \subseteq A$ be a nonempty open set of $M$.

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

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

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

Since $U_n$ are everywhere dense, in a recursive manner, we find a pair of sequences $x_n$ and $\epsilon_n > 0$ such that:
 * $\overline B \left({x_n, \epsilon_n}\right) \subset B\left({x_{n-1}, \epsilon_{n-1} }\right) \cap U_n$

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

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

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

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

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

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

So, by definition, $\left({M, d}\right)$ is a Baire space.