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How to show that quotient space $X/Y$ is complete when $X$ is Banach space, and $Y$ is a closed subspace of $X$?
Here's my attempt: Given a Cauchy sequence $\{q_n\}_{n \in \mathbb{N}}$ in $X/Y$, each $q_n$ is an equivalence class induced by $Y$, I want to find a representative $x_n$ in $q_n$ so that the induced sequence $\{x_n\}_{n \in \mathbb{N}}$ is also a Cauchy sequence in $X$. But I don't know how to construct such sequence.
Theorem. A normed space $X$ is Banach iff for all $\{x_n:n\in\mathbb{N}\}$ convergence of $\sum_{n=1}^\infty \Vert x_n\Vert$ implies that the series $\sum_{n=1}^\infty x_n$ converges in $X$.
Proof. Let $X$ be a Banach space. Assume that for a given $\{x_n:n\in\mathbb{N}\}$ the series $\sum_{n=1}^\infty\Vert x_n\Vert$ is convergent. Then its partial sums $\left\{\sum_{n=1}^N x_n:N\in\mathbb{N}\right\}$ is a Cauchy sequence. Since $X$ is Banach the last sequence have a limit, i.e. the series $\sum_{n=1}^\infty x_n$ converges in $X$.
On the otherer direction, consider arbitrary Cauchy sequence. Then you can choose subsequence $\{n_k:k\in\mathbb{N}\}$ such that $\Vert x_{n_{k+1}}-x_{n_k}\Vert<2^{-k}$. Then the series $\sum_{k=1}^\infty\Vert x_{n_{k+1}}-x_{n_k}\Vert$ is convergent. By assumption this gives that $\sum_{k=1}^\infty (x_{n_{k+1}}-x_{n_k})$ converges in $X$ to some limit $x$. Since $K$-th partial sum of that series is $x_{n_{K+1}}-x_{n_1}$ we conclude that the series $\{x_{n_k}: k\in\mathbb{N}\}$ converges to $x+x_{n_1}$. Since $\{x_n:n\in\mathbb{N}\}$ is a Cauchy sequence with convergent subsequence $\{x_{n_k}:k\in\mathbb{N}\}$, then it is convergent. Since $\{x_n:n\in\mathbb{N}\}$ is a arbitrary Cauchy sequence, then $X$ is Banach.
Theorem. Let $X$ be Banach space and $Y$ be its closed subspace, then $X/Y$ is Banach.
Proof. Now we proceed to the proof of the main result. For each $x\in X$ denote $\hat{x}:=x+Y\in X/Y$. Consider $\{\hat{x}_n:n\in\mathbb{N}\}$ such that the series $\sum_{n=1}^\infty\Vert\hat{x}_n\Vert$ converges. From definition of the norm in $X/Y$ we have that for each $n\in\mathbb{N}$ there exists $x_n\in \hat{x}_n$ such that $\Vert x_n\Vert\leq 2\Vert\hat{x}_n\Vert$. Since $\sum_{n=1}^\infty\Vert\hat{x}_n\Vert$ converges then the last inequality gives that $\sum_{n=1}^\infty\Vert x_n\Vert$ converges also. Since $X$ is Banach we see that $\sum_{n=1}^\infty x_n$ converges in $X$ to some vector $x\in X$. Then from definiton of the norm in $X/Y$ it follows that $\sum_{n=1}^\infty\hat{x}_n$ converges to $\hat{x}$ in $X/Y$. Since $\{\hat{x}_n:n\in\mathbb{N}\}$ was chosen arbitrary then by previous lemma $X/Y$ is Banach
Here is an alternative to Norbert's argument: Do you know the proof of the open mapping theorem? There one shows (after using Baire's theorem) the following: If $T:X\to Z$ is a continuous linear map between Banach spaces such that $\overline{T(B_X)}$ containes some ball in $Z$, then $T$ is open and (hence surjective). Apply this to the completion $Z$ of $Y/X$ and the quotient map.
