# How to show that quotient space $X/Y$ is complete when $X$ is Banach space, and $Y$ is a closed subspace of $X$?

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.

• Isn't this sometime formulated more briefly by saying that completeness is three space property? See also here. (How this thing is called is perhaps not that relevant for this question, but knowing this terminology might be useful if somebody tries to find this result online, be it for completeness or for other properties of normed spaces.) – Martin Sleziak Sep 23 '16 at 4:32

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

• 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.