FIVE MAJOR RESULTS IN ANALYSIS AND TOPOLOGY Aaron ...

5. THE BAIRE CATEGORY THEOREM 31
Lemma. If X − A is closed and nowhere dense in X, then A is open and dense in X.
Proof. Certainly A is open, being the complement of a closed set. Suppose that U is
a nonempty open subset of X such that A ∩ U is not dense in U. Let V = U − A ∩ U. If
V were empty, then U ⊂ A ∩ U and hence U ⊂ A ∩ U, and since A ∩ U ⊂ U, A ∩ U ⊂ U,
which contradicts our assumption that A ∩ U is not dense in U. Thus, V is nonempty.
Then V = U ∩ (X − A ∩ U) is open, and it follows that V ⊂ A ∩ (X − U) ⊂ A. But then
V ⊂ A ∩ V and A ∩ V ⊂ V , which contradicts our assumption that A was nowhere dense.
This completes the proof.
The term ‘category’ with respect to a space refers to its ability to be written as a ‘small’
union of ‘small’ subsets of itself. A space that can be written as the countable union of
nowhere dense subsets of itself is said to be of the first category. Any set that is not of the
first category is said to be of the second category. The following theorem, by René-Louis
Baire, characterizes non-empty complete metric spaces along these lines.
Theorem (Baire Category Theorem 1 ). Let X be a complete metric space, and let {An}
be a countable collection of dense open subsets of X. Then
n An is dense in X.
Proof. Let Br0(x0) be an open ball in X of radius r0 centered at x0 ∈ X. Since A1 is
open, choose x1 ∈ A1∩Br0(x0) and 0 < r1 < 1 such that Br1(x1) ⊂ A1∩Br0(x0). Given xn−1,
we proceed by choosing a point xn ∈ X and 0 < rn < 1
n such that Brn(xn) ⊂ An∩Brn−1(xn−1).
Since xm ∈ Brn(xn) if m ≥ n, {xi} is fundamental, so that the completeness of X guarantees
the existence of a point x ∈ X such that lim
n→∞ xn = x. Since x lies in each of the closed
spheres Brn(xn), x ∈ Br0(x0)
n An, which is what was to be shown.
The following form of the Baire category theorem is equivalent to the one already proved,
but it casts the result in terms of categories.
1 We follow here the same proof provided in [5] and [2].