Baire Category Theorem

[[concept]]

Let $M$ be a {as||complete} metric space and let ${C_{n}}_{n}$ be a {as||collection of closed subsets} of $M$ such that $$M = \bigcup_{n \in N}C_{n}$$Then {ha||at least one} of the $C_{n}$ {ha||contain an open ball}
{ha||$$B(x,r) = { y \in M : d(x,y) <r }$$}
(ie, {ha||at least one} $C_{n}$ has {ha||an interior point}).

Note

A set that does not contain an interior point is called nowhere dense. Sometimes when we apply this theorem, we do not need $C_{n}$ to be closed. Then the result is that their closures must contain and open ball. ie, we cannot have all $C_{n}$ be all nowhere dense.

Note

We can use this theorem to prove that there exists a continuous function that is nowhere differentiable

Proof (by contradiction)

Suppose BWOC that there is some collection of closed subsets of $M$ such that $⋃_{n} C_{n} = M$ and all $C_{n}$ are nowhere dense.

Note

we'll show that there is a sequence in $M$ that does not converge

Since $M$ contains an open ball but $C_{1}$ does not, we have $M \neq C_{1}$ . Thus there is some $p_{1} \in M ∖ C_{1}$ . Since $C_{1}$ is closed but $M ∖ C_{1}$ is open, this implies that $\exists ϵ_{1} > 0$ s.t. $B (p_{1}, ϵ_{1}) \cap C_{1} = \emptyset$ .

Now, $B (p_{1}, \frac{ϵ_{1}}{3}) ⊄ C_{2}$ means there exists some $p_{2} \in M ∖ C_{2}$ . And since $C_{2}$ is closed, there exists $0 < ϵ_{2} < \frac{ϵ_{1}}{3}$ such that $B (p_{2}, ϵ_{2}) \cap C_{2} = \emptyset$

Now, suppose there we have found $k$ such points. Then for each $j = 1, \dots, k$ we have $p_{j} ⊄ C_{j}$ with $ϵ_{j}$ such that $B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

Since $B (p_{k}, \frac{ϵ_{k}}{3}) ⊄ C_{k + 1}$ , there exists some $p_{k + 1} \in B (p_{k}, \frac{ϵ_{k}}{3})$ such that $p_{k + 1} \notin C_{k + 1}$

Then there exists $ϵ_{k + 1} < \frac{ϵ_{k}}{3}$ such that $B (p_{k + 1}, ϵ_{k + 1}) \cap C_{k + 1} = \emptyset$

Thus, by indiction, we have found a sequence of points ${p_{k}}_{k} \subset M$ and $ϵ_{k} \in (0, ϵ_{1})$ such that for all $k$ ,

$p_{j} \in B (p_{j - 1}, \frac{ϵ_{j - 1}}{3})$
$B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

This sequence ${p_{k}}_{k}$ is Cauchy because for all $k, ℓ \in N$ , we have

\begin{aligned} d (p_{k + 1}, p_{k + ℓ}) & \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p_{k + 2}) + \dots + d (p_{k + ℓ - 1}, p_{k + ℓ}) \\ < \frac{ϵ_{k}}{3} + \frac{ϵ_{k + 1}}{3} + \dots + \frac{ϵ_{k + ℓ - 1}}{3} \\ < \frac{ϵ_{1}}{3^{k}} + \frac{ϵ_{1}}{3^{k + 1}} + \dots + \frac{ϵ_{1}}{3^{k + ℓ}} \\ < ϵ_{1} \sum_{n = k}^{\infty} \frac{1}{3^{k}} \\ = \frac{ϵ_{1}}{3^{k}} (\frac{1}{1 - \frac{1}{3}}) \\ = \frac{ϵ_{1}}{2} 3^{- k + 1} \end{aligned}

Since $M$ is complete, there exists some $p \in M$ such that $p_{k} \to p$ as $k \to \infty$ .

Now, for all $k \in N$ , we have

\begin{aligned} d (p_{k + 1}, p_{k + ℓ + 1}) & < ϵ_{k + 1} [\frac{1}{3} + \frac{1}{3^{2}} + \dots + \frac{1}{3^{ℓ}}] \\ < ϵ_{k + 1} \sum_{n = 1}^{\infty} 3^{- n} \\ = \frac{ϵ_{k + 1}}{2} \\ ⟹ lim_{ℓ \to \infty} d (p_{k + 1}, p_{l + ℓ + 1}) & = d (p_{k + 1}, p) \\ \leq \frac{ϵ_{k + 1}}{2} \\ < \frac{ϵ_{k}}{6} \end{aligned}

So as $ℓ \to \infty$ , we have

d (p_{k + 1}, p)

⟹ d (p_{k}, p) \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p) \leq \frac{1}{3} ϵ + \frac{1}{6} ϵ_{k} < ϵ_{k}

ie, $p \in B (p_{k}, ϵ_{k}) = B_{k}$ for all $k$ . And each of these balls has $B_{k} \cap C_{k} = \emptyset$ . Thus $p$ is not in any of the $C_{k}$ ie $p \notin ⋃_{k} C_{k} = M$

⟹ ⟸

\begin{matrix} ◼ \end{matrix}

References

#flashcards/math/functional

Mentions

File	Last Modified
Functional Analysis Lecture 3	2025-11-09
Functional Analysis Lecture 4	2025-11-09
open mapping theorem	2025-11-09
uniform boundedness theorem	2025-11-14

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