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Interlude -- The Baire Category Theorem

21 Sep 2021 - Tags: analysis-qual-prep

We’ve had a few really detailed posts in a row – summarizing the main theorems of some major objects of study in analysis, and putting these theorems in contexts which (I hope) make them feel coherent and memorable. Today’s post is going to be a bit more relaxed. Two of my best friends (Remy and Alyss) are visiting me, so I want a slightly shorter form post to make sure I have time to spend with them. The Baire Category Theorem seems like the perfect topic!

(Future Chris here, they left on Sunday, but I’m only now getting this post up. My qual is tomorrow morning, and I ended up studying for that instead of revising this. I’m still going to write up how I organized my thoughts on Fourier Analysis, but it’ll probably be after I actually take the exam. Wish me luck!)

First, let’s remind ourselves of the (deceptively simple) statement of the Baire Category Theorem. Recall a set is called Meagre if it is a countable union of nowhere dense sets. This forms a $\sigma$-ideal which gives a similar notion of “smallness” to sets of measure $0$1.

Let $X$ be a complete metric space2

Then $X$ is not meagre in itself.

Another way of phrasing the result, which sounds less obvious by expanding some of the complexity trapped in the word “meagre”, is very useful in practice:

Let $X$ be a complete metric space3, and let $U_n$ be a countable family of dense open sets in $X$

Then $\bigcap U_n$ is still dense in $X$4.

This result, which especially in its first form seems obvious, has a surprising number of consequences. I’ll give some applications here, particularly applications which are relevant for qual-related counterexamples, but if you want more you should see here for a long list, and here for what is essentially a “user manual” for proving existence results using the Baire Category Theorem!

The idea behind proving theorems with the BCT is to write think about what properties you want some object to satisfy. Ideally there are only countably many such properties $P_n$, and each $P_n$ is a “dense open condition” in the sense that \(\{ x \mid P_n(x) \}\) is dense and open.

If you find yourself in this scenario, then the BCT promises you can find an element which satisfies all the properties simultaneously!

Contrapositively, if we have countably many meagre conditions, then we can avoid all of them simultaneously!

Let’s see a few examples in action:

Most continuous functions $f : [0,1] \to \mathbb{R}$ are nowhere differentiable.

Here “most” means “a dense $G_\delta$ set”.

$\ulcorner$ We work in $C([0,1])$ with the sup norm. Notice this is a banach space, so we have access to the BCT.

Now we want to characterize those $f$ which are differentiable at a point. A standard trick is to turn \(\{ f \mid f' \text{ exists} \}\) into the countable pieces \(\bigcup \{ f \mid |f'| \leq n \}\). Of course, $f’$ is still hard to talk about in general. So we use this trick again (well, the reciprocal of it), and replace the $\lim_{h \to 0}$ in the definition of $f’$ by the countably many “$h \leq \frac{1}{k}$”s.

In full, we want to look at the sets

\[D_{n,k} \triangleq \left \{ f \ \middle | \ \exists x . \left \lvert \frac{f(x+h) - f(x)}{h} \right \rvert \leq n \text{ for every } 0 \lt |h| \lt \frac{1}{k} \right \}\]

As a quick exercise, you should verify that if $f$ is differentiable at a point, then $f \in D_{n,k}$ for some $n$ and $k$.

A (slightly tedious) argument shows that each $D_{n,k}$ is closed and nowhere dense, thus meagre. So $\bigcup_{n,k} D_{n,k}$ is meagre (and thus so is the set of functions differentiable at even one point).

This tells us that most continuous functions are not differentiable anywhere. $\lrcorner$

This was supposed to be a shorter, lower effort post, and I definitely succeeded in that regard, haha. I was originally going to also prove that if $f : S^1 \to \mathbb{C}$ is merely continuous, we can’t guarantee pointwise convergence. Indeed, “most functions” diverge at “many points” in the sense that for any countable dense set of points in $S^1$, we can find comeagrely many functions in $C(S^1)$ whose fourier series diverge at every point in that set.

I know there is a BCT argument, but I’m feeling too lazy to come up with it myself, and all the places I checked proved this by using the uniform boundedness principle5 (which does indirectly use BCT, but I would like to be a bit more explicit in this post).

If someone happens to have an argument on hand, I’d love to hear about it! Until then, the qual is tomorrow, so wish me luck! Next time will be about fourier analysis, and it’ll be the last post in the series. See you there ^_^.

  1. Though which sets are classified as “small” can differ! There are subsets of the unit interval which have lebesgue measure $1$ despite being meagre.

    The precise differences between the $\sigma$-ideals of meagre and nullsets is a topic of some classical interest in descriptive set theory. If we assume $\mathsf{CH}$, then there is an involution on $\mathbb{R}$ which swaps meagre and null sets. See here, for instance. 

  2. Or a locally compact hausdorff space 

  3. Again, or a locally compact hausdorff space 

  4. If you’ve never seen this proven, it’s worth doing it yourself at least once! It’s not hard, and will help you gain some comfort with the definition of meagre. 

  5. See here, for instance.