Baire category theorem

(Redirected from Baire's theorem)

The Baire category theorem (BCT) is an important tool in general topology and functional analysis. The theorem has two forms, each of which gives sufficient conditions for a topological space to be a Baire space.

The theorem was proved by René-Louis Baire in his 1899 doctoral thesis.

Statement of the theorem

A Baire space is a topological space with the following property: for each countable collection of open dense sets Un, their intersection ∩ Un is dense.

Note that neither of these statements implies the other, since there are complete metric spaces which are not locally compact (the irrational numbers with the metric defined below; also, any Banach space of infinite dimension), and there are locally compact Hausdorff spaces which are not metrizable (for instance, any uncountable product of non-trivial compact Hausdorff spaces is such; also, several function spaces used in Functional Analysis; the uncountable Fort space). See Steen and Seebach in the references below.

This formulation is equivalent to BCT1 and is sometimes more useful in applications. Also: if a non-empty complete metric space is the countable union of closed sets, then one of these closed sets has non-empty interior.

Relation to the axiom of choice

The proofs of BCT1 and BCT2 for arbitrary complete metric spaces require some form of the axiom of choice; and in fact BCT1 is equivalent over ZF to a weak form of the axiom of choice called the axiom of dependent choices.[1]

The restricted form of the Baire category theorem in which the complete metric space is also assumed to be separable is provable in ZF with no additional choice principles.[2] This restricted form applies in particular to the real line, the Baire space ωω, and the Cantor space 2ω.

Uses of the theorem

BCT1 is used in functional analysis to prove the open mapping theorem, the closed graph theorem and the uniform boundedness principle.

BCT1 also shows that every complete metric space with no isolated points is uncountable. (If X is a countable complete metric space with no isolated points, then each singleton {x} in X is nowhere dense, and so X is of first category in itself.) In particular, this proves that the set of all real numbers is uncountable.

BCT1 shows that each of the following is a Baire space:

• The space $\mathbb{R}$ of real numbers
• The irrational numbers, with the metric defined by $d(x,y) = \frac{1}{n+1}$, where $n$ is the first index for which the continued fraction expansions of $x$ and $y$ differ (this is a complete metric space)
• The Cantor set

By BCT2, every finite-dimensional Hausdorff manifold is a Baire space, since it is locally compact and Hausdorff. This is so even for non-paracompact (hence nonmetrizable) manifolds such as the long line.

Proof

The following is a standard proof that a complete pseudometric space $\scriptstyle X$ is a Baire space.

Let $\scriptstyle U_n$ be a countable collection of open dense subsets. We want to show that the intersection $\scriptstyle \bigcap U_n$ is dense. A subset is dense if and only if every nonempty open subset intersects it. Thus, to show that the intersection is dense, it is sufficient to show that any nonempty open set $\scriptstyle W$ in $\scriptstyle X$ has a point $\scriptstyle x$ in common with all of the $\scriptstyle U_n$. Since $\scriptstyle U_1$ is dense, $\scriptstyle W$ intersects $\scriptstyle U_1$; thus, there is a point $\scriptstyle x_1$ and $\scriptstyle 0 \;<\; r_1 \;<\; 1$ such that:

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

where $\scriptstyle B(x, r)$ and $\scriptstyle \overline{B}(x, r)$ denote an open and closed ball, respectively, centered at $\scriptstyle x$ with radius $\scriptstyle r$. Since each $\scriptstyle U_n$ is dense, we can continue recursively to find a pair of sequences $\scriptstyle x_n$ and $\scriptstyle 0 \;<\; r_n \;<\; \frac{1}{n}$ such that:

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

(This step relies on the axiom of choice.) Since $\scriptstyle x_n \;\in\; B(x_m, r_m)$ when $\scriptstyle n \;>\; m$, we have that $\scriptstyle x_n$ is Cauchy, and hence $\scriptstyle x_n$ converges to some limit $x$ by completeness. For any $\scriptstyle n$, by closedness,

$x \in \overline{B}(x_n, r_n).$

Therefore $\scriptstyle x \;\in\; W$ and $\scriptstyle x \;\in\; U_n$ for all $\scriptstyle n$.

See also this blog post [1] by M. Baker for the proof of the theorem using Choquet's game.