Baire Category & Nowhere Differentiable Functions (Part One)

The Baire Category Theorem is a powerful result that relates a metric space to its underlying topology. (And sadly no, it has nothing to do with category theory!)  Informally, the theorem says that if you can find a metric with respect to which your topological space is complete, then that space cannot be written as a countable union of nowhere dense sets. In other words, a metric can put a restriction on the topology. Here's the formal statement:

The Baire Category Theorem: Let $X$ be a complete metric space and let $\{O_n\}_{n=1}^{\infty}$ be a collection of open, dense subsets. Then $\bigcap_{n=1}^\infty O_n$ is dense in $X$, that is, $\overline{\bigcap_{n=1}^\infty O_n}=X$.

"Now wait a sec," you might be thinking, "how is this equivalent to your 'informal' statement above?" I'm so glad you asked! Notice that "$\cap_nO_n$ is dense in $X$" implies that that $\cap_nO_n\neq \varnothing$. (Otherwise we could conclude that the closure of the empty set is the whole space, which, assuming $X$ is nonempty, is bogus*.) By De Morgan's law, this is equivalent to saying $\cup_nO_n^c\neq X$ where each $O_n^c$ - being the complement of an open, dense set - is a closed and nowhere dense set. And this is precisely what our informal statement said above.

 A quick word on what it means to be nowhere dense: As you might guess, nowhere density is the exact opposite of density. If a set is nowhere dense, it exists but there's not much to it. Sort of like an atom, which is mostly empty space, or a spider web. Technically, we say a set $A$ is nowhere dense if the interior of its closure is empty, i.e. if $$\text{int}({\overline{A}})=\varnothing.$$ Take, for instance, the set $\{\frac{1}{n}:n\in\mathbb{N}\}$. This is nowhere dense in $\mathbb{R}$ because its closure is $C=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}$ (its one and only limit point is 0) and this has empty interior. To see this, recall that $x$ is an interior point of a set $S$ if you can find an open set containing $x$ which lies entirely in $S$. Well, $C$ has no interior points since any open interval $(a,b)\subset\mathbb{R}$ which contains $\frac{1}{n}$ for some $n\in\mathbb{R}$ contains (uncountably many!) irrational numbers - elements which are not members of $C$. That is, there is no open interval in $\mathbb{R}$ which is a subset of $C$. Hence $C$ is nowhere dense. For similar reasons, the Cantor set is also nowhere dense.

To see the Baire Category Theorem in action, we'll sketch the proof that there exists a continuous function $f:[0,1]\to\mathbb{R}$ which is not differentiable on $[0,1]$. In fact, we'll discover that most continuous functions on $[0,1]$ are NOT differentiable! This may be surprising since most of the continuous functions encountered in the classroom are differentiable. Baire Category tells us that those functions are the exception rather than the rule. To avoid writing an overwhelmingly long post, I'll postpone the proof until next week. In the mean time, I'll leave you with two warm-up exercises:

Warm-up #1

Prove that the irrationals are a dense subset of $\mathbb{R}$.

 Proof. Let $\{r_n\}_{n=1}^\infty$ be an enumeration of the rationals and for each $n$ define $O_n=\mathbb{R}\smallsetminus\{r_n\}$. Then $O_n$ is open (since singleton sets in a metric space are closed) and dense (since the closure of $O_n$ is $O_n\cup\{r_n\}=\mathbb{R}$). Then $\mathbb{R}\smallsetminus \mathbb{Q}=\cap_n O_n$ is the set of all irrationals and, by the Baire Category Theorem, this is dense.

Warm-up #2

Prove that $\mathbb{Q}$ cannot be written as a countable intersection of open sets (i.e. prove $\mathbb{Q}$ is not a G$_\delta$ set).

 Proof. Suppose to the contrary that $\mathbb{Q}=\cap_nO_n$ where each $O_n$ is open. Since $\mathbb{Q}\subset O_n$ for each $n$ and since $\mathbb{Q}$ is dense, we know that each $O_n$ is dense. Thus $\mathbb{Q}$ can be written as a countable intersection of open dense sets.

 Let $\{r_n\}_{n=1}^\infty$ be an enumeration of the rationals. Then we can write the irrationals as $\mathbb{R}\smallsetminus\mathbb{Q}=\cap_n\mathbb{R}\smallsetminus\{r_n\}$ which is also a countable intersection of open dense sets (for the same reason as in #1 above).

 Thus by the Baire Category Theorem, $\varnothing=\mathbb{Q}\cap(\mathbb{R}\smallsetminus\mathbb{Q})$ is dense in $\mathbb{R}$, which is a contradiction.*

Footnotes:

 *The closure of the empty set is itself. This is because it's already closed (since its complement, $X$, is open).