# The Unapologetic Mathematician

## Baire Sets

Looking over my notes from topology it seems I completely skipped over Baire sets. This was always one of those annoying topics that I never had much use for, partly because I didn’t do point-set topology or analysis. Also, even in my day the usual approach was a very classical and awkward one. Today I’m going to do a much more modern and streamlined one, and I can motivate it better from a measure-theoretic context to boot!

Basically, the idea of a Baire set is one that can’t be filled up by “negligible” sets. We’ve used that term in measure theory to denote a subset of a set of measure zero. But in topology we don’t have a “measure” to work with. Instead, we use the idea of a closed “nowhere dense” set — one for which there is no open set on which it is dense. The original motivation was a set like a boundary of a region; in the context of Jordan content we saw that such a set was negligible.

Clearly such a set has no interior — no open set completely contained inside — and any finite union of them is still nowhere dense. However, if we add up countably infinitely many we might have enough points to be dense on some open set. However, we don’t want to be able to actually fill such an open set. In the measure-theoretic context, this corresponds to the way any countable union of negligible sets is still negligible.

So, let’s be more specific: a “Baire set” is one for which the interior of every countable union of closed, nowhere dense sets is empty. Equivalently, we can characterize Baire sets in complementary terms: every countable intersection of dense open sets is dense. We can also use the contrapositive of the original definition: if a countable union of closed sets has an interior point, then one of the sets must itself have an interior point.

We’re interested in part of the famous “Baire category theorem” — the name is an artifact of the old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space $X$ is a Baire space. Let $\{U_n\}$ be a countable collection of open dense subsets of $X$. We will show that their intersection is dense by showing that any nonempty open set $W$ has some point $x$ — the same point — in common with all the $U_n$.

Okay, Since $U_1$ is dense, then $U_1\cap W$ is nonempty, and it contains a point $x_1$. As the intersection of two open sets, it’s open, and so it contains an open neighborhood of $x_1$ which set can take to be an open metric ball of radius $r_1>0$. But then $B(x_1,r_1)$ is an open set, which will intersect $U_2$. This process will continue, and for every $n$ we will find a point $x_n$ and a radius $r_n$ so that $B(x_n,r_n)\subseteq B(x_{n-1},r_{n-1})\cap U_n$. We can also at each step pick $r_n<\frac{1}{n}$.

And so we come up with a sequence of points $\{x_n\}$. At each step, the ball $B(x_n,r_n)$ contains the whole tail of the sequence past $x_n$, and so all of these points are within $r_n$ of each other. Since $r_n$ gets arbitrarily small, this shows that $x_n$ is Cauchy, and since $X$ is complete, the sequence must converge to a limit $x$. This point $x$ will be in each set $U_n$, since $x\in B(x_n,r_n)\subseteq U_n$, and it’s obviously in $W$, as desired.

The other part of the Baire category theorem says that any locally compact Hausdorff space is a Baire space. In this case the proof proceeds very similarly, but with the finite intersection property for compact spaces standing in for completeness.

August 13, 2010 - Posted by | Point-Set Topology, Topology

1. Should the last word be ‘completeness’?

Also, thanks for the interesting and well-written stuff.

Comment by Greg Simon | August 13, 2010 | Reply

2. Yes, sorry, fixing…

Comment by John Armstrong | August 13, 2010 | Reply

3. I’m glad that you are including this historically interesting topic.

Comment by Jonathan Vos Post | August 15, 2010 | Reply

4. […] union of these closed subsets has an interior point. But since is a complete metric space, it is a Baire space as well. And thus one of the must have an interior point as […]

Pingback by Associated Metric Spaces and Absolutely Continuous Measures I « The Unapologetic Mathematician | August 16, 2010 | Reply

5. […] have measurable sets, but we do have something almost as good. Just as when we discussed Baire spaces, we can use nowhere-denseness as a topological stand-in for negligibility. In fact, we’ll […]

Pingback by Stone’s Representation Theorem III « The Unapologetic Mathematician | August 20, 2010 | Reply