Baire Category Theorem

Theorem

The Baire Category Theorem states that every complete metric space is a Baire space.

Proof

Let $U_n$ be a countable collection of open sets which are everywhere dense.

We want to show that the intersection $\bigcap U_n$ is (everywhere) dense.

A subset $S$ is dense if every nonempty open subset intersects it. To see this, take any point $x$ in the space, and consider an open neighbourhood of this point. This neighbourhood contains an open set $V$ such that $x\in V$, which intersects $S$ as it is open and nonempty. So every neighbourhood of every point intersects $S$, and so $S$ is dense.

Thus, to show that the intersection is dense, it is sufficient to show that any nonempty open set $W$ has a point $x$ in common with all of the $U_n$.

Since $U_1$ is dense, $W$ intersects $U_1$; thus, there is a point $x_1$ and $r_1 > 0$ such that:

$\overline{B}(x_1, r_1) \subset W \cap U_1$.

where $B(x, r)$ and $\overline{B}(x, r)$ denote an open ball centered at $x$ with radius $r$ and its closure, respectively.

Since $U_n$ are dense, in a recursive manner, we find a pair of sequences $x_n$ and $r_n > 0$ such that:

$\overline{B}(x_n, r_n) \subset B(x_{n-1}, r_{n-1}) \cap U_n$

as well as $r_n < 1/n $.

Since $x_n \in B(x_m, r_m)$ when $n > m$, we have that $x_n$ is a Cauchy Sequence, and $x_n$ converges to some limit $x$ by completeness.

For any $n$, by closedness:

$x \in \overline{B}(x_{n+1}, r_{n+1}) \subset B(x_n, r_n)$.

Hence, $x \in W$ and $x \in U_n$ for all $n$.

$\blacksquare$

Source of Name

This entry was named for René-Louis Baire.

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces