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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.

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up vote 9 down vote accepted

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

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How did you get this"From definition of the norm in $X/Y$ we have that for each $n\in\mathbb{N}$ there eixists $x_n\in \hat{x}_n$ such that $ x_n \leq 2 \hat{x}_n$." I know the proof of this theorem a bit different, that is, we have that for each $n \in \mathbb{N}$ $\| \hat{x}_n \| =\inf_{y \in Y}\|x_n + y\|$, hence there is $y_n \in Y$ such that $$\| x_n + y_n \| \leq \| \hat{x}_n \| + \frac{1}{2^n}$$ (definition of infimum). Therefore, $\sum_n \|x_n + y_n\| < \infty$.But $(x_n + y_n)$ is a sequence in $X$ so $\sum_{n} x_n + y_n$ converges to some $x in X$. Now we use the rest of your proof. – Frank Tessla Mar 3 '13 at 23:17
Just take $\varepsilon=\Vert\hat{x}_n\Vert$ in the expression $\Vert x_n\Vert\leq\Vert \hat{x}_n\Vert+\varepsilon$ which follows from the definition of $\Vert \hat{x}_n\Vert$ – Norbert Mar 4 '13 at 6:23
Hi sorry to bother you, i was looking at the first theorem you prooved. Specifically i'm struggling to understand the implication " $x_n$ is a cauchy then you can pick up a subsequence $x_{n_k}$ such that $|| x_{n_{k+1}} - x_{n_k}|| \leq 2^{-k} $ ". Can you explain formally why? – user8469759 Jul 24 '15 at 8:52
@Lukkio Set $n_1$<1. Since $(x_n)$ is Cauchy sequence, then you can find some $m$ such that $\Vert x_m-x_{n_1}\Vert<2^{-1}$. So set $n_2=m$. Again, since $(x_n)$ is a Cauchy sequence you can find $m'$ such that $\Vert x_{m'}-x_{n_2}\Vert<2^{-2}$. So set $n_2=m'$. And etc. – Norbert Jul 24 '15 at 13:47
Shouldn't you choose, as first step $\epsilon = 2^{-1}$ then get $n_1,n_2$, and as second step choose $\epsilon = 2^{-2}$ choose one more couple and so on...? how do you prove that for every two couple one index could be in common? You said "set $n_1$ " but actually such $n_1$ should depend from the chosen $\epsilon$ isn't? – user8469759 Jul 24 '15 at 14:06

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.

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