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Let $X$ be a normed vector space.

  1. Show that if a subsequence of a Cauchy sequence converges, then the whole sequence converges.
  2. Use the part 1 to show that $S = \{x\in X : \|x\| = 1\}$ is complete if and only if $X$ is Banach.

I have found the first part in a book, but I do not how to prove part 2. Thanx regards

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  • $\begingroup$ Is question 2 clear? $\endgroup$ – MBM Mar 30 '13 at 18:08
  • $\begingroup$ jejeje nop, thanx for the correction :) $\endgroup$ – user70195 Mar 30 '13 at 18:29
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2) One direction is straightforward, the other one requires more work.

If $X$ is a Banach space, it is complete. Now $S$ is closed in $X$, so it is complete.

Conversely, assume that $S$ is complete. Take $(x_n)$ a Cauchy sequence in $X$. If $\lim x_n=0$, we are done proving that $(x_n)$ converges in $X$. So assume that $(x_n)$ does not converge to $0$. This means that there exists $\epsilon>0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_k}\|\geq \epsilon$ for all $k$. Then write...

EDIT Thanks to @Jesper for pointing out a mistake...

$$ \frac{x_{n_k}}{\|x_{n_k}\|}-\frac{x_{n_l}}{\|x_{n_l}\|}=\frac{x_{n_k}-x_{n_l}}{\|x_{n_k}\|}+(\|x_{n_l}\|-\|x_{n_k}\|)\frac{x_{n_l}}{\|x_{n_l}\| \| x_{n_k}\|}. $$ It follows, applying triangular, reverse triangular inequalities and the fact that $\|x_{n_k}\|\geq \epsilon$ in the preceding equality that: $$ \left\| \frac{x_{n_k}}{\|x_{n_k}\|}-\frac{x_{n_l}}{\|x_{n_l}\|}\right\|\leq \frac{2}{\epsilon}\|x_{n_k}-x_{n_l} \|. $$ So the sequence $\frac{x_{n_k}}{\|x_{n_k}\|}$ is Cauchy in $S$. Therefore it converges to some $y$ in $S$.

Now note that $\|x_n\|$ is a Cauchy sequence again by the reverse triangular inequality as $|\|x_n\|-\|x_m\|| \le \|x_n - x_m\|$ and $(x_n)$ is Cauchy. Since $\mathbb{R}$ is complete, it follows that $\|x_{n_k}\|$ converges to $M$. Whence $x_{n_k}=\|x_{n_k}\|\cdot\frac{x_{n_k}}{\|x_{n_k}\|}$ converges to $My$. It only remains to apply 1) to see that $(x_n)$ converges. Hence $X$ is complete, i.e $X$ is a Banach space.

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    $\begingroup$ Woooo, Amazing, thanks so much julien!! $\endgroup$ – user70195 Mar 30 '13 at 19:19
  • $\begingroup$ @ShanaKugimiya Actually, my first argument was correct, and I then I removed an essential part for an obscure reason..I've put it back: we need another extraction to get convergence of $\|x_{n_k}\|$ simultaneously. $\endgroup$ – Julien Mar 30 '13 at 19:28
  • $\begingroup$ Thanks for that julien! :) $\endgroup$ – FASCH Mar 30 '13 at 19:31
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    $\begingroup$ I have a question. If $x_n \not\rightarrow 0$ , then $$\exists\; \varepsilon >0\; \mbox{ such that }\; \forall\; N\in\mathbb{N},\; \|x_n - 0\| = \|x_n\| > \varepsilon,\;\;\forall\; n \geq N.$$ So, as is for all $N\in\mathbb{N}$, why do you considered a subsequence and not the whole sequence? $\endgroup$ – FASCH Mar 30 '13 at 20:15
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    $\begingroup$ @FASCH No. The correct negation is: there exists $\epsilon>0$ such that for all $N$ there exists $n\geq N$ such that $\|x_n\|\geq \epsilon$. This yields an extraction, not the whole sequence. Take $x_n=1+(-1)^n$ for instance to see why your claim is false. And whoever upvoted your comment is advised to consider this example too... $\endgroup$ – Julien Mar 30 '13 at 20:19

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