Baire Category Theorem

[[concept]]

Baire Category Theorem

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}).

^statement

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 ${\bigcup }_n{C}_n=M$ and all $C_n$ are nowhere dense.

$M$ that does not converge

we’ll show that there is a sequence in

Since $M$ contains an open ball but $C_1$ does not, we have $M\ne C_1$. Thus there is some $p_1\in M\setminus C_1$. Since $C_1$ is closed but $M\setminus C_1$ is open, this implies that $\exists {ϵ}_1>0$ s.t. $B(p_1,{ϵ}_1)\cap C_1=\varnothing$.

Now, $B\left(p_1,\frac{{ϵ}_1}{3}\right)\not\subset C_2$ means there exists some $p_2\in M\setminus C_2$. And since $C_2$ is closed, there exists $0<{ϵ}_2<\frac{{ϵ}_1}{3}$ such that $B(p_2,{ϵ}_2)\cap C_2=\varnothing$

Now, suppose there we have found $k$ such points. Then for each $j=1,\dots ,k$ we have $p_j\not\subset C_j$ with${ϵ}_j$ such that $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

Since $B\left(p_k,\frac{{ϵ}_k}{3}\right)\not\subset C_{k+1}$, there exists some $p_{k+1}\in B\left(p_k,\frac{{ϵ}_k}{3}\right)$ such that $p_{k+1}\notin C_{k+1}$

Then there exists ${ϵ}_{k+1}<\frac{{ϵ}_k}{3}$ such that $B(p_{k+1},{ϵ}_{k+1})\cap C_{k+1}=\varnothing$

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 $k$,

1. $p_j\in B\left(p_{j-1},\frac{{ϵ}_{j-1}}{3}\right)$
2. $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

This sequence $\{p_k{\}}_k$ is Cauchy because for all $k,\ell \in \mathbb{N}$, we have

d(p_{k+1}, p_{k+\ell}) &\leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p_{k+2}) + \dots + d(p_{k+\ell-1}, p_{k+\ell}) \\ &< \frac{\epsilon_{k}}{3} + \frac{\epsilon_{k+1}}{3} + \dots+\frac{\epsilon_{k+\ell-1}}{3} \\ &< \frac{\epsilon_{1}}{3^k}+\frac{\epsilon_{1}}{3^{k+1}}+\dots+\frac{\epsilon_{1}}{3^{k+\ell}} \\ &< \epsilon_{1} \sum_{n=k}^\infty \frac{1}{3^k} \\ &= \frac{\epsilon_{1}}{3^k}\left( \frac{1}{1-\frac{1}{3}} \right) \\ &= \frac{\epsilon_{1}}{2} 3^{-k+1} \end{align}$$ Since $M$ is complete, there exists some $p \in M$ such that $p_{k} \to p$ as $k \to \infty$. Now, for all $k \in \mathbb{N}$, we have $$\begin{align} d(p_{k+1}, p_{k+\ell+1}) &< \epsilon_{k+1} \left[ \frac{1}{3} + \frac{1}{3^2} + \dots+\frac{1}{3^\ell} \right] \\ &< \epsilon_{k+1} \sum_{n=1}^\infty 3^{-n} \\ &= \frac{\epsilon_{k+1}}{2} \\ \implies \lim_{ \ell \to \infty } d(p_{k+1}, p_{l+\ell+1}) &=d(p_{k+1}, p) \\ &\leq \frac{\epsilon_{k+1}}{2} \\ &< \frac{\epsilon_{k}}{6} \\ \end{align}$$ So as $\ell\to \infty$, we have $$d(p_{k+1}, p)$$ $$\implies d(p_{k}, p) \leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p) \leq \frac{1}{3}\epsilon + \frac{1}{6}\epsilon_{k} < \epsilon_{k}$$ ie, $p \in B(p_{k}, \epsilon_{k}) = B_{k}$ *for all* $k$. And each of these balls has $B_{k} \cap C_{k} = \varnothing$. Thus $p$ is not in any of the $C_{k}$ ie $p \not\in\bigcup_{k}C_{k} = M$ $$\implies \impliedby$$ $$\tag*{$\blacksquare$}$$

References

functional

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