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Baire category theorem for complete metric spaces

Statement

Assume the Axiom of Choice. Let $(X, d)$ be a nonempty complete metric space. If $X = ⋃_{n \geq 1} F_{n}$ for closed subsets $F_{n} \subseteq X$ , then at least one $F_{n}$ has nonempty interior in $X$ .

Statement references

Jean-Pierre Demailly, Complex Analytic and Differential Geometry, Chapter IX §1.B

Facts & Assumptions

Given: The Axiom of Choice; a nonempty complete metric space $(X, d)$ ; and closed sets $(F_{n})_{n \geq 1}$ covering $X$ .

[A1]

The Axiom of Choice permits the countable recursive choices of nested balls used below (The Axiom of Choice).

Proof

1.1givenA1construct

Suppose every $F_{n}$ has empty interior. Choose a nonempty open ball $B_{0} \subseteq X$ . Since $F_{1}$ is closed with empty interior, $B_{0} ∖ F_{1}$ is nonempty and open; choose a closed ball $\overline{B} (x_{1}, r_{1}) \subset B_{0} ∖ F_{1}$ with $0 < r_{1} \leq 2^{- 1}$ .

2.1step 1.1A1construct

Inductively, if $\overline{B} (x_{n}, r_{n})$ has been chosen, its interior $B (x_{n}, r_{n})$ is a nonempty open set. The closed set $F_{n + 1}$ has empty interior, so $B (x_{n}, r_{n}) ∖ F_{n + 1}$ is nonempty and open; choose a closed ball $\overline{B} (x_{n + 1}, r_{n + 1})$ contained there with $0 < r_{n + 1} \leq 2^{- n - 1}$ . Thus the closed balls are nested, and the $n$ th ball avoids $F_{n}$ .

3.1givenstep 2.1A1∎

The nested centers form a Cauchy sequence because for $k \geq n$ one has $x_{k} \in \overline{B} (x_{n}, r_{n})$ and $r_{n} \to 0$ . Completeness gives a limit $x \in X$ ; each closed ball contains every later center and hence contains $x$ . Therefore $x \notin F_{n}$ for every $n$ , contradicting $X = ⋃_{n} F_{n}$ . Thus some $F_{n}$ has nonempty interior.

Depends on

The Axiom of Choice