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I have to prove that in any infinite dimension normed linear space we have that the closed unit ball is not compact.

I know that I have to construct a sequence such that $||x_n||=1$ and $|x_m-x_n|\geq \frac{1}{2}$. If I can do this then we have a bounded sequence with no convergent subsequence so that this space is not compact but I have no idea how to actually find such a sequence?

Thanks for any help

hmmmm
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    One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here. – David Mitra Apr 10 '13 at 15:08
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    Won't help with a proof, but to get an intuitive sense of why this is true, I like to think of compact spaces being ones you can't "get lost" in. Like an open interval you could get closer and closer to the edge, or an unbounded one go on forever. In this case, everywhere you turn there's a new dimension to walk in :) – lukemassa Jan 31 '23 at 03:40

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Idea for an indirect proof: Suposse unit ball is compact. Cover it by $\cup_{x\in B_1 (0)} B_{\frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,\dots,n$ such that balls of radius $\frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.

ters
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