Baire Spaces

A topological space is said to be a Baire space, if any of the following equivalent conditions holds:

  • The intersection of countably many dense open subsets is still dense.
  • The union of countably many closed subsets with empty interior has empty interior.

Note that a set is dense in a topological space if and only if its complement has empty interior.

Any sufficient condition for a topological space to be a Baire space constitutes a Baire Category Theorem, of which there are several. Here we give one that is commonly used in functional analysis.

Baire Category Theorem

Complete metric spaces are Baire spaces.

Proof: Let XX be a metric space with complete metric dd. Suppose that XX is not a Baire space. Then there is a countable collection {Un}\braces{U_n} of dense open subsets of XX such that the intersection U:=UnU := \bigcap U_n is not dense in XX.

In a metric space, any nonempty open set contains an open ball. It is also true, that any nonempty open set contains a closed ball, since B(y,δ1)B(y,δ2)\overline{B(y,\delta_1)} \subset B(y,\delta_2) if δ1<δ2\delta_1 < \delta_2.

We construct a sequence (Bn)(B_n) of open balls Bn:=B(xn,ϵn)B_n := B(x_n,\epsilon_n) satisfying

Bn+1BnUnϵn<1nnN,\overline{B_{n+1}} \subset B_n \cap U_n \quad \epsilon_n < \tfrac{1}{n} \quad \forall n \in \NN,

as follows: By hypothesis, the interior of XUX \setminus U is not empty (otherwise UU would be dense in XX), so we may choose an open ball B1B_1 with ϵ1<1\epsilon_1 < 1 such that B1XU\overline{B_1} \subset X \setminus U. Given BnB_n, the set BnUnB_n \cap U_n is nonempty, because UnU_n is dense in XX, and it is open, because BnB_n and UnU_n are open. This allows us to choose an open ball Bn+1B_{n+1} as desired.

Note that by construction BmBnB_m \subset B_n for mnm \ge n, thus d(xm,xn)<ϵn<1nd(x_m,x_n) < \epsilon_n < \tfrac{1}{n}. Therefore, the sequence (xn)(x_n) is Cauchy and has a limit point xx by completeness. In the limit mm \to \infty, we obtain d(x,xn)ϵnd(x,x_n) \le \epsilon_n (strictness is lost), hence xBnx \in \overline{B_n} for all nn. This shows that xUnx \in U_n for all nn, that is xUx \in U. On the other hand, xB1XUx \in \overline{B_1} \subset X \setminus U, in contradiction to the preceding statement. \square\enspace