Math Online's 1500th Page - 7:56 PM CST on September 28th, 2015

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:

(1)

\begin{align} \quad X = \bigcup_{i=1}^{\infty} A_i \end{align}

• 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 \in