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So on my homework it says that to prove the unit ball in $\ell^2(\mathbb{N})$ is non-compact, it suffices to find countably many elements $x_n$ of $\ell^2(\mathbb{N})$ with $\lVert x_n\rVert \leq \frac{1}{2}$ such that $\lVert x_n - x_m\rVert \geq \delta$ for some $\delta \in (0, \frac{1}{2})$ and $n \neq m$. Why is that?


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up vote 8 down vote accepted

A sequence satisfying your criteria cannot have a convergent subsequence, since no subsequence of such a sequence can be Cauchy. But, compact metric spaces are sequentially compact.

To find the sequence $(x_n)$, consider the unit vectors in $\ell_2$ (suitably scaled).

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Oh, I forgot about that :P thanks! – badatmath Mar 1 '12 at 18:42

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