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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.

Martin Sleziak
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Metta World Peace
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    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 04:32

2 Answers2

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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

Integral
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Norbert
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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 03 '13 at 23:17
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    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 04 '13 at 06:23
  • 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 08:52
  • @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
  • @Lukkio. Ok let's try it other way. There exists $n_1$ such that $\Vert x_n-x_{n_1}\Vert<2^{-1}$ for all $n>n_1$. By induction we can show that, for each $k>1$ we can find $n_k>n_{k-1}$ such that $\Vert x_n-x_{n_k}\Vert<2^{-k}$ for all $n\geq n_k$. These $n_k$'s are the desired ones. – Norbert Jul 24 '15 at 14:40
  • And why such $n_k$ are the desired ones? – user8469759 Jul 24 '15 at 15:17
  • Because you can put $n=n_k$ in the inequality $\Vert x_n-x_{n_k}\Vert<2^{-k}$. – Norbert Jul 24 '15 at 15:19
  • Ok it is clear now!!! – user8469759 Jul 24 '15 at 15:22
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    you could use the fact that the quocient map is a bounded linear function to conclude that $\sum x_n + Y = x + Y$ – Victor Ronchim Mar 18 '16 at 12:29
  • beautiful proof! – zxzx179 May 05 '17 at 19:57
  • I am sorry but why it follows from the definition of quotient norm that $\sum_{n=1}^{\infty}[x_{n}]$ converges to $[x]$ in $X/Y$? By definition, we have $$\Bigg|\sum_{n=1}^{\infty}[x_{n}]-[x]\Bigg|{X/Y}=\inf\Bigg{\Bigg|\sum{n=1}^{\infty}(x_{n}-x)-y\Bigg|_{X}, y\in Y\Bigg}$$ but I don't know how to show this is zero (or less than $\epsilon$). – JacobsonRadical Mar 01 '21 at 18:16
  • @JacobsonRadical, your formula is incorrect. It should be as follows $$\left\Vert \sum_{n=1}^\infty [x_n] - [x]\right\Vert=\inf\limits_{y\in Y}\left{\left\Vert \sum_{n=1}^\infty x_n - x -y\right\Vert\right}.$$ As for the proof try to show that $\lim\limits_{n\to\infty} x_n=x$ in $X$ implies $\lim\limits_{n\to\infty} [x_n]=[x]$ in $X/Y$. – Norbert Mar 02 '21 at 08:53
  • @JacobsonRadical, Take a look at inequality $$\Vert x_n-x-y\Vert\leq\Vert x_n-x\Vert+\Vert y\Vert$$ – Norbert Mar 02 '21 at 09:01
  • @Norbert Great. got it. Nice proof. – JacobsonRadical Mar 02 '21 at 09:05
  • @Norbert Can you help me to spot error in this one: Given ${x_n+Y}$ is Cauchy so $ \Vert x_n-x_m\Vert\leq\Vert x_n-x_m-y\Vert+\Vert y\Vert$, on taking inf on $y$ gives ${x_n}$ to be Cauchy and then using $\Vert x_n-x+Y\Vert\leq\Vert x_n-x-0\Vert<\epsilon$, which completes the proof. I have seen everyone using subsequences, why is it so if it is not a wrong solution? – Lalbahadur Sahu Dec 01 '23 at 12:09
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    @LalbahadurSahu For infima the inequality is reversed $$ \inf\limits_{y}(f(y) + g(y)) \geq \inf\limits_{y} f(y) + \inf\limits_{y} g(y) $$ – Norbert Dec 02 '23 at 02:25
  • Thank you for your response @Norbert. I later looked it carefully. – Lalbahadur Sahu Dec 02 '23 at 07:28
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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.

Jochen
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