Baire Category Theorem
Contents
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 $r_n > 0$ 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.
$\blacksquare$
Source of Name
This entry was named for René-Louis Baire.
Sources
- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology ... (previous) ... (next): $\text{I}: \ \S 5$: Complete Metric Spaces