# Convex sets in Alexandrov spaces

Let $X$ be a compact Alexandrov space with $curv\geq 1$ (and without boundary). Does $X$ always have a nontrivial compact convex subset without boundary?

Definition of a convex subset: $A\subseteq X$ is called convex if for every two points $p ,q\in A$, there exists a minimizing geodesic between them which is completely in $A$.

This is extremally rare, even if $$X$$ is a Riemannian manifold.
If the convex set $$A$$ has interior points, then any boundary point of the subset $$A$$ in $$X$$ lies on the boundary of Alexandrov space $$A$$. So if $$X\ne A$$ then $$\dim A<\dim X$$.
Note that $$A$$ has to be totally geodesic, otherwise an end of geodesic would be a boundary point of $$A$$. Generic Riemannian manifold do not have totally geodesic submanifolds of dimension at least 2.
So, we are left with the case $$\dim A=1$$. In other words, $$A$$ is a closed geodesic in $$X$$. Since $$A$$ is convex, it has to be length minimizing on each half. This is also extremally rare thing --- generic Riemannian manifold does not have such geodesics.
• @Jayq Both types of boundaries (relative to $X$ and intrinsic boundary of $A$) coincide in this case (since $X$ has no boundary). This is indeed requires a proof, it can be done by induction on dimension; you need to show that space of directions $\Sigma_pA$ forms a closed convex set in $\Sigma_pX$. – Anton Petrunin Dec 27 '18 at 18:39
• This just seems to be true if $A$ is open, or what am I understanding wrong here? Take for example an equator as A inside a round sphere. – Bruce Wayne Mar 5 at 17:12