$\ell^2$ denotes the set of all real square summable sequence with standard $\ell^2$ norm,

$S=\{x\in \ell^2: \|x\|<1\}$ is compact? interior of $S$ is compact?closure of $S$ is compact?

$\ell^2$ is an infinite dimensional Banach Space , $\bar S$ is its closed unit ball which must not be compact by known standard theorem. is that okay?

But I don't know about other answers. Thank you for help.

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    $\begingroup$ It is known that the unit closed ball of an infinite-dimensional normed vector space is not compact. As a corollary, any compact set in such spaces has an empty interior. $\endgroup$ – Seirios Sep 12 '13 at 7:48
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    $\begingroup$ Well first of all the open unit ball in a (non-zero) normed space is never closed (normed spaces are path-connected), so $\{x \in \ell^2 \colon \lVert x\rVert < 1\}$ is not compact. $\endgroup$ – kahen Sep 12 '13 at 7:55
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    $\begingroup$ If $e_i$ denotes the vector that is $1$ in the $i$th coordinate and $0$ elsewhere, what is the distance between $e_i/2$ and $e_j/2$ for $i\ne j$? Could $\{e_i/2\}$ have a convergent subsequence? $\endgroup$ – David Mitra Sep 12 '13 at 9:01

It seems the following (I don’t checked the following claims, you should do it yourself using known theorems).

Put $S’=\{x\in \ell^2: \|x\|\le 1\}$. We have two topologies on $\ell_2$: the strong and the weak.

In the strong topology, $\overline {S}=S’$, $\operatorname{int} S’=S$, and no one of these sets is compact (see the first comments to your question).

In the weak topology, $\overline {S}=S’$ too (because the set $S'$ is convex), $\operatorname{int} S’=\operatorname{int} S=\varnothing$ (because the space $\ell_2$ is infinitely dimensional, so each its weakly open subset is strongly unbounded), and the set $S'$ is compact because the space $\ell_2$ endowed with the weak topology is naturally isomorphic to its dual space endowed with $*$-weak topology, and we can apply Banach-Alaoglu Theorem.

PS. See also first sentences from here.


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