TITLE: Compactness in normed vector spaces.
QUESTION [5 upvotes]: Let $(X,\Vert\cdot\Vert)$ be a normed $\mathbb{K}$-vector space and $A \subset X$ be closed and bounded. My problem is how to determine whether $A$ is compact? 
I know that a compact subset is always closed and bounded. And that in $\mathbb{R}$ the converse holds due to Bolzano–Weierstraß. 
But I think I am missing something. Do there exist subsets in normed vector spaces which are closed and bounded but not compact?

REPLY [2 votes]: Example.  The unit ball $B$ in Hilbert space $l^2$.  It is closed and bounded, but not compact.  Let the orthonormal unit vectors be $e_n, n=1,2,3,\dots$.  Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence.  This is because all the distances are $\|e_n-e_m\| = \sqrt{2}$ so in fact no subsequence is even a Cauchy sequence.