Metric Spaces, Baire Category Theorem

Baire Category Theorem

Let S be a complete metric space, and let O1 O2 O3 etc be a countable collection of dense open sets in S. Let O be the intersection of all these open sets. We will show that O is nonempty and dense in S. This is the Baire category theorem. (biography)

Let W be an arbitrary nonempty open set in S, and intersect W with O1, having x1 in the intersection. Let S1 be an open ball in the intersection, with radius r1 centered at x1. Remember that the open ball S1 is the set of points less than r1 distance from x1. Let C1 be the closure of S1, and remember that C1 cannot include any points farther than r1 from x1.

Cut r1 in half, if need be, so that C1 lies entirely inside the intersection of O1 and W.

Since O2 is dense, O2 intersects S1. As before, the intersection is a nonempty open set. Let x2 lie in the intersection, and build an open ball S2 about x2 with radius r2, subject to the following criteria.

Shrink r2 so that it is less than half of r1. We want the diameters to approach 0.

Let d2 be the distance from x1 to x2. We know d2 < r1. Shrink r2 so that it is less than r1-d2. This assures us that C2, the closure of S2, lies entirely inside S1.

Shrink r2 if necessary, so that C2 lies in O2.

Proceed by induction on n. At the nth step, Cn is contained in On and in Sn-1.

The sequence of centers x1 x2 x3 etc is cauchy, with centers beyond xn lying in Sn, no more than 2rn apart. Let x be the limit of this cauchy sequence. Since the metric space is complete, x exists.

If x is not in Cn, then x is at least rn distance from xn. However, Sn+1 is bounded away from the edge of Sn, bounded away from x. The rest of the sequence cannot approach x, and that is a contradiction. hence x lies in each Cn.

Since Cn+1 lies in On, x lies in each On. Remember that O is the intersection O1∩O2∩O3∩O4∩… It follows that x lies in O, and O is nonempty. (This assumes our initial metric space S is nonempty, allowing us to choose the nonempty open set W at the outset.)

since x lies in C1 lies in W, x is in W. This means O and W intersect. Since W is arbitrary, O is a dense set.

this theorem also holds for any locally compact hausdorff space. to review, a space is hausdorff if two points can be separated in disjoint open sets, and a space is locally compact if each point is contained in an open set whose closure is compact. The common metric spaces are hausdorff and locally compact, but others are not, so this is not a generalization of the above. We are really proving the theorem again, for a new class of spaces.

If you aren't familiar with compact spaces, you should read through the introduction. In particular, we will need the theorem that says a closed subspace of a compact space is compact. This is presented on the introductory page, so you don't have to go very far.

As before, we are given O1 O2 O3 etc, a countable collection of dense open sets. Let W be an arbitrary open set. Let x1 be a point in W∩O1, and let C1 be the closure of an open set Q1, containing x1, such that C1 is compact. We can do this because S is locally compact.

Intersect Q1 with W and with O1 to give a possibly smaller open set S1 containing x1. The closure of S1 is a closed subspace inside C1, and is compact. Shrink C1 down to the closure of S1.

We're still not where we want to be. We want C1 to lie entirely inside W and O1. So far we only know this about S1, not its closure. We need to use the fact that the space is hausdorff. You might want to review some of the theorems about compact hausdorff spaces.

Let B1 be the boundary of S1, i.e. the points in C1 that are not in S1. This is a closed set C1, intersect the complement of an open set S1, hence B1 is closed. And it is part of C1, which is compact.

If B1 is empty, then C1 is S1, and is entirely inside W and O1, and there's nothing more to do; so assume B1 is nonempty.

Remember that x1 is in S1, and is not part of B1. Also x1 is compact, and a compact set in a hausdorff space is closed, so x1 is closed. These are two distinct closed sets in C1. since C1 is both compact and hausdorff, it is normal. This means x1 and B1 can be separated in disjoint open sets. Let U contain x1 and let V contain B1. The closure of U misses V, and B1, hence the closure of U lies entirely inside S1, which is in O1 and W. Also, the closure of U is a closed set in C1, and is compact. Shrink S1 down to U, and let C1 be its closure. Now C1 is in W and O1, and contains x1, and is compact.

Since O2 is dense, O2 intersects S1. Let x2 lie in the intersection, and find an open set Q2 containing x2 whose closure is compact. Intersect Q2 with O2 and S1 to find S2. Let C2 be the closure of S2, which is compact. Shrink S2 down, if necessary, so that its closure C2 is entirely contained in O2 and S1. This uses the boundary hausdorff technique described above.

Continue this process forever, building a descending chain of closed compact sets Cn. At the nth step, Cn is contained in On and in Cn-1. Suppose this chain has no intersection. The complements form an ascending chain of open sets that cover C1. Since C1 is compact, a finite subcover will do. This means one of the closed sets Cn, and all its descendants, are empty, which is a contradiction. Therefore the chain intersects in at least one point; call it x.

As before, x is in every On, and in W, hence O intersects W, and O is a dense set in S.

This theorem fails when the collection of dense open sets is uncountable. The interval [0,1] is compact, hausdorff, a complete metric space, the nicest space you could think of, and the complement of any point is a dense open set, yet the intersection of these sets is empty.