Is the boundary of the unit sphere in every normed vector space compact? I wanted to ask whether the boundary of the unit sphere in every normed vector space is compact?
I know that this is true for simple examples, but how is it in general?
 A: No. Take
$$
\ell^2 = \{(x_n) \in \mathbb{R} : \sum |x_n|^2 < \infty\}
$$
with the norm
$$
\|(x_n)\| = \sqrt{\sum |x_n|^2}
$$
And consider the vectors
$$
e_n = (0,0,\ldots, 0, 1,0,\ldots )
$$
with $1$ in the $n^{th}$ spot and $0$ elsewhere.
The sequence $\{e_n\}$ is contained in the unit sphere, and has no convergent subsequence.
In general, the unit sphere is compact iff the space is finite dimensional. One way to prove this fact uses the Riesz' Lemma
A: One can prove quite generally that a (Hausdorff) topological vector space $E$ over the real numbers is locally compact if and only if $E$ is finite-dimensional. This means in particular that the unit ball in any normed vector space $E$ isn't compact if $E$ is infinite-dimensional. It follows that the unit sphere in such $E$ cannot be compact either; if it were, then so would be the unit ball which is a topological cone on the unit sphere. 
Terence Tao has a nice write-up of this general result (originally due to Weil) here. 
Remark: the Hausdorff condition is very weak; for TVS it is equivalent to saying that any two $x, y$ have the same neighborhoods only if $x=y$. 
