I need to show that $K$ is compact and that $co(K)$ is bounded, but not closed. Let $x_n$ be a sequence in a Hilbert space such that
$\left\Vert x_n \right\Vert=1$ and $ \langle x_n,\ x_m \rangle =0 $, for all $n \neq m$.
Let $ K= \{ x_n/ n : n \in  \mathbb{N} \} \cup \{0\} $.
I need to show that $K$ is compact, $\operatorname{co}(K)$ is bounded, but not closed and finally find all the extreme points of $ \overline{\operatorname{co}(K)} $ .
 A: We can assume without loss of generality that $\mathcal H$ is the separable Hilbert space with $x_n$ as a basis (because all the objects considered lie within the complete span of $x_n$).
$K$ is compact because it is sequentially compact -- any infinite sequence in $K$ tends to $0\in K$.
$co(K)$ is bounded, because the norm of a convex combination of any points is no greater than the largest of their norms (by triangle inequality), and thus because $K$ is bounded, so is $co(K)$.
$co(K)$ is not closed, because for example $\sum x_n2^{-n}/n$ lies in the closure, but is not a convex combination of $x_n/n$ (because any convex combination of $x_n$ is eventually orthogonal to $x_n$). In fact, $cl(co(K))$ is the set of infinite convex combinations of $x_n/n$.
The extreme points of $K'=cl(co(K))$ are precisely the elements of $K$:


*

*First, we can see that $K$ are extreme points of $K'$, beacause for any nontrivial convex combination $x$ of elements of $K'$ we have for some $n$ that $\langle x\vert x_n\rangle\in (0,1/n)$, so $x$ is not any of $K$

*Suppose $x=\sum \alpha_nx_n$ is an extreme point of $K'$.

*Then $\sum n\alpha_n\leq 1$ and $0\leq n\alpha_n\leq 1$.

*If there for some $n\neq m$ we have $\alpha_n,\alpha_m\neq 0$, then choose $\varepsilon>0$ smaller than $n\alpha_n$ and $m\alpha_m$. Put $y_1=x-x_n\varepsilon/n+x_m\varepsilon/m$, $y_2=x+x_n\varepsilon/n-x_m\varepsilon/m$. Then clearly $x=(y_1+y_2)/2$, and because of the way we have chosen $\varepsilon$, we have $y_1,y_2\in K'$, so we have a contradiction.

*Therefore it must be that at most one of $\alpha_n$ is nonzero. If all are zero, then we're done. If there is $\alpha_n\neq 0$, then it is easy to see that $\alpha_n=1/n$ and we're done.

