# Topological Modules/Banach spaces

Definition (Banach space):

A Banach space is a complete normed space.

Proposition (series criterion for Banach spaces):

Let $X$ be a normed space with norm $\|\cdot \|$ . Then $X$ is a Banach space if and only if

$\sum _{n=1}^{\infty }\|x_{n}\|<\infty$ implies that $\lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$ exists in $X$ ,

whenever $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$ .

Proof: Suppose first that $X$ is a Banach space. Then suppose that $\sum _{n=1}^{\infty }\|x_{n}\|$ converges, where $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$ . Then set $S_{N}:=\sum _{n=1}^{N}x_{n}$ ; we claim that $(S_{N})_{N\in \mathbb {N} }$ is a Cauchy sequence. Indeed, for $M>0$ sufficiently large, we have

$N\geq M\Rightarrow \|S_{M}-S_{N}\|=\left\|\sum _{n=M+1}^{N}x_{n}\right\|\leq \sum _{n=M+1}^{N}\|x_{n}\|\leq \sum _{n=M+1}^{\infty }\|x_{n}\|<\epsilon$ .

Hence, $(S_{N})_{N\in \mathbb {N} }$ also converges, because $X$ is a Banach space.

Now suppose that for all sequences $(x_{n})_{n\in \mathbb {N} }$ the implication

$\sum _{n=1}^{\infty }\|x_{n}\|<\infty \Rightarrow \lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$ holds. Let then $(y_{n})_{n\in \mathbb {N} }$ be a Cauchy sequence in $X$ . By the Cauchy property, choose, for all $k\in \mathbb {N}$ , a number $N_{k}\in \mathbb {N}$ such that $\|y_{m}-y_{n}\|<1/2^{k}$ whenever $m,n\geq N_{k}$ . We may assume that $N_{1}\leq N_{2}\leq \cdots \leq N_{k}\leq \cdots$ , ie. $(N_{k})_{k\in \mathbb {N} }$ is an ascending sequence of natural numbers. Then define $x_{1}:=y_{N_{1}}$ and for $k\geq 2$ set $x_{k}:=y_{N_{k}}-y_{N_{k-1}}$ . Then

$\sum _{j=1}^{k}x_{j}=y_{N_{k}}$ .

Moreover,

$\sum _{j=1}^{k}\|x_{j}\|\leq \|y_{N_{1}}\|+\sum _{j=2}^{k}2^{-(k-1)}$ ,

so that

$\sum _{j=1}^{\infty }\|x_{j}\|$ converges as a monotonely increasing, bounded sequence. By the assumption, the sequence $(S_{k})_{k\in \mathbb {N} }$ converges, where

$S_{k}=\sum _{j=1}^{k}x_{j}=y_{N_{k}}$ .

Thus, $(y_{n})_{n\in \mathbb {N} }$ is a Cauchy sequence that has a convergent subsequence and is hence convergent. $\Box$ 