Baire's Theorem and Applications

The Baire Category Theorem

Lemma 1: Let $(X,\tau )$ be a topological space and let $A\subseteq X$. If $A$ is a nowhere dense set then for every $U\in \tau$ there exists a $B\subseteq U$ such that $A\cap {\bar {B}}=\emptyset$.

Theorem 1 (The Baire Category Theorem): Every complete metric space is of the second category.

• Proof: Let $X$ be a complete metric space. Then every Cauchy sequence $(x_{n})_{n=1}^{\infty }$ of elements from $X$ converges in $X$. Suppose that $X$ is of the first category. Then there exists a countable collection of nowhere dense sets $A_{1},A_{2},...\subset X$ such that:

${\begin{aligned}\quad X=\bigcup _{i=1}^{\infty }A_{i}\end{aligned}}$

• Let $U\subset X$. For each nowhere dense set $A_{i}$, $i\in \{1,2,...\}$ there exists a set $B_{i}\subset U$ such that $A_{i}\cap {\bar {B_{i}}}=\emptyset$.

• Let $B_{1}(x_{1},r)\subset U$ be a ball contained in $U$ such that $A_{1}\cap {\bar {B_{1}}}=\emptyset$. Let $B_{2}\left(x_{2},{\frac {r}{2}}\right)\subseteq B_{1}$ be a ball contained in $B_{1}$ whose radius is ${\frac {r}{2}}$ and such that $A_{2}\cap {\bar {B_{2}}}=\emptyset$. Repeat this process. For each $n\in \{2,3,...\}$ let $B_{n}\left(x_{n},{\frac {r}{n}}\right)$ be a ball contained in $B_{n-1}$ whose radius is ${\frac {r}{n}}$ and such that $A_{n}\cap {\bar {B_{n}}}=\emptyset$ and such that $A_{n}\cap {\bar {B_{n}}}=\emptyset$.

• The sequence $(x_{n})_{n=1}^{\infty }$ is Cauchy since as $n\in \mathbb {N}$ gets large, the elements $x_{n}$ are very close. Since $X$ is a complete metric space, we must have that this Cauchy sequence therefore converges to some $p\in X$, i.e., $\lim _{n\to \infty }x_{n}=p$.

• Now notice that $x\in {\bar {B_{n}}}$ for all $n\in \mathbb {N}$ because if not, then there exists an $m\in \mathbb {N}$ such that $x\not \in {\bar {B_{n}}}$ for all $n\geq m$. Hence $({\bar {B_{n}}})^{c}$ is open and so there exists an open ball $X$ such that $x\in B\subset ({\bar {B_{n}}})^{c}$ but then $\lim _{x\to \infty }x_{n}\neq x$ because $x_{m}\not \in B$ for all $m\geq n$.

• Since $x\in {\bar {B_{n}}}$ for all $n\in \mathbb {N}$ then since $A_{n}\cap {\bar {B_{n}}}=\emptyset$ we must have that then $x\not \in A_{n}$