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Let $V$ be a normed space in which every bounded sequence has a convergent subsequence. Show that $V$ must be complete. Show further that $V$ must be finite-dimensional.

I've done the first part in showing $V$ is complete so I've got a grasp on the question except on showing that $V$ is finite dimensional.

dfeuer
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Raul
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  • @John unfortunately not. – Raul Nov 29 '13 at 03:07
  • Did you learn the Riesz lemma? –  Nov 29 '13 at 03:07
  • The Riesz lemma says that when you have a closed subspace $E \neq V$ in $V$. Then for all $\theta <1$ there is $y\notin E$, $||y||=1$ and $||y-x||\geq \theta $ for all $x\in E$. If you have this, you can form inductively a bounded sequence which does not converges. –  Nov 29 '13 at 03:14
  • The proof of Riesz lemma is quite easy. You might want to have a look. http://www.math.umn.edu/~garrett/m/fun/riesz_lemma.pdf . It is just a one line proof –  Nov 29 '13 at 03:16
  • @John i havent learnt banach spaces just yet, but i can see the proof of the lemma is trivial. – Raul Nov 29 '13 at 03:19

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If V is infinite-dimensional, then the unit sphere is not compact, so it contains sequences without a convergent subsequence.

Rodney Coleman
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