Let $M$ be a metric space, $A$ a subset of $M$. The reach (defined by Federer) of $A$ in $M$ is the largest $r_0\ge 0$ such that if $x\in M$ and the $d(x, A)< r_0$, then $A$ contains a unique point nearest to $x$. My question is whether a convex subset in Alexandrov space always has positive reach? In Riemannian manifold there is a convex radius which shows the positivity of reach. (We will assume the subset to be compact).
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The answer is "no". Say consider the doubling $M$ of plane region $F=\{y\ge x^2\}$. $M$ is known to be an Alexandrov space. Let $A\subset M$ is the doubling of the intersection of $F$ with the disc with center $(0,1)$ passing through $(1,1)$. Note that $A$ is convex but all the points on the parabola have two minimizing geodesics to $A$ (one in each of to copies of $F$ in $M$). So $A$ has zero reach. 

