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?
On
Yes.
The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.
In a general, infinite-dimensional normed space (e.g., over the real scalars) complete in its norm (i.e., an infinite-dimensional Banach space), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.
On
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.
Let $c_{0}$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_{0}$ is given by:
$$ \|(a_{1},a_{1},a_{3},\ldots)\|:=\sup_{k\in\mathbb{N}}|a_{k}|. $$
Let $A$ denote the set of all vectors in $c_{0}$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $x\in A$, $B_{1/2}(x):=\{y\in c_{0}:\|x-y\|<1/2\}$. Then, $\{B_{1/2}(x):x\in A\}$ is an open cover of $A$. To see that it has no finite subcover, let, for each $k\in\mathbb{N}$, $e_{k}$ denote the sequence $(0,\ldots,0,1,0,\ldots)$, where the $1$ is in position $k$. Since $\|e_{k}-e_{\ell}\|=1$ if $k\not=\ell$, it follows that no two such vectors can belong to the same $B_{1/2}(x)$ for any $x$.
Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.