I am trying to do an exercise in an introduction to functional analysis course:
1) Let $X$ be a normed space and $\{x_{n}\}_{n=1}^{\infty}\subseteq X$.
Prove that $X$ is a banach space iff every abselutly convergent series is convergent (that is $\sum_{n=1}^{\infty}||x_{n}||<\infty\implies\sum_{n=1}^{\infty}x_{n}$ exist)
2) Use part $1$ to shoe that if $X$ is banach space and $M$ is a closed subspace then $X/M$ is a banach space relative to the norm $$ ||[x]||=\inf\{||x-y||,y\in M\} $$
I have managed to do the first part of the exercise, but not the second one.
A start is to assume $$ \sum_{n=1}^{\infty}||[x_{n}]||<\infty $$
and to try and show that $$ \sum_{n=1}^{\infty}x_{n} $$
exist.
By definition I know that $$ \sum_{n=1}^{\infty}\inf\{||x_{n}-y||,y\in M\}=L<\infty $$
hence for every $\epsilon>0$ there is a sequence $\{y_{n}\}_{n=1}^{\infty}\subseteq M$ s.t $$ \sum_{n=1}^{\infty}||x_{n}-y_{n}||<L+\epsilon $$
I really don't see how to continue. Can someone please help me out and point me to the right direction ?
Let $\pi \colon X \to X/M$ be the canonical projection. Then we note that the norm on $X/M$ is given by
$$\lVert \eta\rVert_{X/M} = \inf \{ \lVert \xi\rVert_X : \xi \in \pi^{-1}(\eta)\}.$$
So if we have a sequence $(\eta_n)$ in $X/M$ with $\sum\limits_{n=0}^\infty \lVert\eta_n\rVert_{X/M} < \infty$, we can, for each $n\in\mathbb{N}$, choose a $\xi_n \in \pi^{-1}(\eta_n)$ with $\lVert \xi_n\rVert_X < \lVert\eta_n\rVert_{X/M} + 2^{-n}$, and that implies that
$$\sum_{n=0}^\infty \lVert\xi_n\rVert_X < \sum_{n=0}^\infty \left(\lVert\eta_n\rVert_{X/M} + 2^{-n}\right) = \sum_{n=0}^\infty \lVert\eta_n\rVert_{X/M} + 2 < \infty.$$
So since $X$ is a Banach space, we know that
$$x = \sum_{n=0}^\infty \xi_n$$
exists. Let $y = \pi(x)$. Then, by the continuity of $\pi$, we have
$$y = \pi(x) = \pi\left(\lim_{N\to\infty}\sum_{n=0}^N \xi_n\right) = \lim_{N\to\infty} \sum_{n=0}^N \pi(\xi_n) = \lim_{N\to\infty} \sum_{n=0}^N \eta_n,$$
and hence $\sum\limits_{n=0}^\infty \eta_n$ exists (and equals $y$). So $X/M$ satisfies the criterion, hence is a Banach space.