Answer to Question #230848 in Functional Analysis for smi

Question

State and prove Baire' s category theorem.

Answer 1

Definition 1. A topological space X is said to be a Baire space, if for any given countable collection $A_{n}$ of closed sets with empty interior in X, their union $\bigcup\limits_{n=1}^{\infty} A_n$ also has empty interior in X.

In terms of open sets this property can be reformulated as follows:

Definition 2. A topological space X is said to be a Baire space, if for any given countable collection $G_{n}$ of open dense subsets of X, their intersection $\bigcap\limits_{n=1}^{\infty} G_n$ is dense.

Baire' s category theorem. Every complete metric space is a Bair space.

Proof. The set $\bigcap\limits_{n=1}^{\infty} G_n$ is dense iff for every open set U, $\bigcap\limits_{n=1}^{\infty} G_n\cap U\ne \empty$. We will construct a monotone sequence of closed balls $\bar B(x_{k+1},r_{k+1})\subset \bar B(x_{k},r_{k})$ such that $\bar B(x_{k},r_{k})\subset \bigcap\limits_{n=1}^{k} G_n\cap U$ and $r_k<2^{-k}$. Then the sequence $\{x_n\}$ will be a Cauchy sequence. Indeed, for all $\varepsilon>0$ let $N\in\mathbb{N}$ be arbitrary such that $2^{-N}<\varepsilon/2$ and $n,m>N$. Since $x_n\in \bar B(x_{n},r_{n})\in \bar B(x_{N},r_{N})$ and $x_m\in \bar B(x_{m},r_{m})\in \bar B(x_{N},r_{N})$ then $dist(x_n,x_m)<2r_N<2\cdot 2^{-N}<\varepsilon$. Therefore, the sequence $\{x_n\}$ is a Cauchy sequence and has a limit $x\in\bigcap\limits_{n=1}^{\infty}\bar B(x_{n},r_{n})\subset \bigcap\limits_{n=1}^{\infty} G_n\cap U$. Therefore, $\bigcap\limits_{n=1}^{\infty} G_n\cap U\ne \empty$.

A monotone sequence of closed balls $\bar B(x_{k+1},r_{k+1})\subset B(x_{k},r_{k})$ can be constructed by induction, since for every $k\in\mathbb{N}$ the set $G_{k+1}$ is dense and open, and hence, has a non-empty intersection with non-empty open set $B(x_{k},r_{k})$. Therefore, it contains some closed ball $\bar B(x_{k+1},r_{k+1})\subset \bar B(x_{k},r_{k})$.

The proof is completed.