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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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up vote 4 down vote accepted

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 $M$).

alt text

So $A$ has zero reach.

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@Anton, It seems that although there are two geodesics but they are actually connecting $p$ to the same point $(1, 1)$ on the boundary of $A$, right? Which by definition is of positive reach. – Ralph Mar 8 '13 at 9:20
@Ralph: no, see the picture. – Anton Petrunin Mar 8 '13 at 15:25
@Anton, I see, thanks – Ralph Mar 8 '13 at 15:58
@Anton, Question: $F$ is doubled to get $M$, are you saying $A$ is a doubled (portion of) disk?, if so $A\nsubseteq M$. – horse with no name Mar 18 '13 at 4:30
@horse, $A$ is not glued along the top portion. – Ralph Mar 20 '13 at 19:30

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