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Bolzano-Weierstrass theorem holds in any NLS. (Normed Linear Space). Is this statement True?

If it is true how to prove it.

I know how to prove it in $\mathbb R$. I proved it in $\mathbb R$ by using supremum infimum of the set. But in NLS supremum infimum concept do not exist.

Can anyone please help me by giving any hint.

cmi
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  • Are you sure this is true? This question or rather the answers, seem to show the opposite. Or do you mean for you space to be finite-dimensional? – saulspatz Sep 08 '18 at 03:54
  • https://math.stackexchange.com/q/2293062/489079 please check this link – cmi Sep 08 '18 at 04:06
  • This is not true in general normed spaces, unless you specify that the sequence is contained in a compact set. – Rodrigo Dias Sep 08 '18 at 04:12
  • Can you please give a counter example?@rldias – cmi Sep 08 '18 at 04:15
  • The link you posted refers to a finite-dimensional normed vector space. Not all vector spaces are finite-dimensional. The link I posted in an earlier comment has counterexamples. – saulspatz Sep 08 '18 at 04:47
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    Bolzano - Wierstrass holds in a normed linear space iff it is finite dimensional. In other words, every infinite dimensional normed linear space has a closed and bounded set which is not compact. – Kavi Rama Murthy Sep 08 '18 at 12:12

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Consider the space $l_1$ and the sequence $e_n$ (the $n$th unit vector). This is bounded but had no convergent subsequence.

copper.hat
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