# Topological Modules/Banach spaces

Definition (Banach space):

A Banach space is a complete normed space.

Proposition (series criterion for Banach spaces):

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

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

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

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

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

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

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

${\displaystyle \sum _{n=1}^{\infty }\|x_{n}\|<\infty \Rightarrow \lim _{N\to \infty }\sum _{n=1}^{N}x_{n}}$

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

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

Moreover,

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

so that

${\displaystyle \sum _{j=1}^{\infty }\|x_{j}\|}$

converges as a monotonely increasing, bounded sequence. By the assumption, the sequence ${\displaystyle (S_{k})_{k\in \mathbb {N} }}$ converges, where

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

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