this is related to this question but is simpler, and hopefully is wellknown. There are a number of references that say that the convex hull of a collection of points in a CAT(0) space need not be closed. I was wondering if anyone was aware of an explicit example ?
There are such examples already in Riemannian world! In fact in any generic Riemannian manifold of dimension $\ge3$ convex hull of 3 points in general position is not closed. BUT it is hard to make explicit and generic at the same time :)
If it is closed then there are a lot of geodesics lying in its boundary  that is rare! To see it do the following exercise first: Show that in generic 3dimensional manifold, arbitrary smooth convex surface contains no geodesic. (Here geodesic = geodesic in ambient space.)
To make word "generic" more clear: show that any metric admits $C^\infty$perturbation such that above property holds.
Semisolution: Assume that a geodesic $\gamma$ lies in the boundary of a convex set $K$ with smooth boundary. Let $N(t)$ be the outer normal vector to $K$ at $\gamma(t)$. Note that $N(t)$ is parallel. Further note that from convexity of $K$ we get that for any Jacoby field $J(t)$ such that $$\langle N(t_0),J(t_0)\rangle\le 0\ \text{and}\ \langle N(t_1),J(t_1)\rangle\le 0,$$ we have $$\langle N(t),J(t)\rangle\le 0\ \text{if}\ t_0<t<t_1.$$ Note that this condition does not hold if the curvature tensor on $\gamma$ is generic.
P.S. Roughly it means that convex hulls in Riemannian world are too complicated. But I know one example where it is used, see Kleiner's An isoperimetric comparison theorem. But he is only using that Gauss curvature of nonextremal points on the boundary of convex hulls is zero...
Appendix. (A construction of convex hull.) To construct convex hull you can do the following: start with some set $K_0$ and construct a sequence of sets $K_n$ so that $K_{n+1}$ is a union of all geodesics with ends in $K_n$. The union $W$ of all $K_n$ is convex hull. Now assume it coincides with its closure $\bar w$. In particular if $x\in\partial\bar W$ then $x\in K_n$ for some $n$. I.e. there is a geodesic in $\bar W$ passing through $x$ (if $x\not\in K_0$). From convexity, it is clear that such geodesic lies in $\partial \bar W$...
P.P.S. A more general statement is proved in our paper About every convex set...

$\begingroup$ this is very helpful. However, there are some things that I'm not entirely clear about. * I'm not sure why in your statement about generic 3D manifolds, the geodesic has to be in the ambient space ? * I don't see why this implies the result (I'm not doubting: merely not following) $\endgroup$ – Suresh Venkat Dec 12 '09 at 6:34

$\begingroup$ @Anton, I don't get it. In any manifold, take a point $p$ and a nearby small ball $B$, and connect $p$ to all points of $B$ by geodesic segments. I believe the resulting set is convex if everything is small enough, yet the conical part of the boundary is filled by geodesics. $\endgroup$ – Sergei Ivanov Dec 13 '13 at 17:49

$\begingroup$ @SergeiIvanov, I extended the hint; now it is called "semisolution". I hope it helps. $\endgroup$ – Anton Petrunin Dec 14 '13 at 2:48

$\begingroup$ Thanks, now I understand. Indeed the Euclidean intuition fails. $\endgroup$ – Sergei Ivanov Dec 14 '13 at 20:18
There is a natural example if you don't ask the collection of point to be finite (but where it is closed): take the subset $A_1$ of the real Hilbert space $X=L^2(\mathbb{R})$ consisting of function taking at most one value beside $0$. Then it is easily seen that the geodesic segments between pairs of points in $A_1$ cover the set $A_3$ of function taking at most $3$ values beside $0$. The convex hull of $A_1$ is then the set of functions taking a finite number of values, and is dense in $X$.
For the story, this is a variation on an example I came across in optimal transportation: $X$ was the Wasserstein space of the real line with quadratic cost, which is isometric to the subset of $L^2([0,1])$ consisting of nondecreasing functions.