Shortest paths in Alexandrov spaces

Let $X$ be an Alexandrov space with curvature bounded from below (if necessary, $X$ might be assumed to be finite dimensional or even compact).

Question 1. Is it true that every point of $X$ has a neighborhood $U$ such that any two points from $U$ can be connected by at most one shortest path (which does not have to be contained in $U$)?

Question 2. In the previous question, can one choose $U$ to be geodesically convex, i.e. for any two points from $U$ any shortest path between them (if it is not unique) is contained in $U$?

A reference would be helpful.

Remark. Of course, if $X$ is a smooth Riemannian manifold then the answers to both questions are 'yes'.

• I am afraid, the vertex $p$ of the cone $X$ over the circle of length less than $2\pi$ has no such neighborhood $U$. – valeri Apr 1 '15 at 9:50
• @valeri why don't you make this an answer ? – Thomas Richard Apr 1 '15 at 11:30
• @Thomas Richard May be Semyon Alesker means that the curvature is bounded from above? Then the answer might be different. – valeri Apr 1 '15 at 11:37
• @valeri Thank you! I think this is the final answer. – orbits Apr 1 '15 at 12:07
• @valeri is right that an upper bound on the curvature implies a positive answer to question 1, because one can construct triangles of "arbitrary fatness" from non-unique shortest paths between two points. – D. Kelleher Apr 1 '15 at 18:03