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 $A$.



Convex sets in Alexandrov spaces
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.


