MathOverflow Asked by Piotr Hajlasz on December 18, 2020
Let $Ksubsetmathbb{R}^n$ be any compact set. Let $operatorname{Unp}(K)$ be the set of points in
$$
operatorname{Unp}(K)={xinmathbb{R}^nsetminus K:, exists ! yin K |x-y|=d(x,K)}.
$$
Here are some properties.
The distance function is Lipschitz and hence differnetiable a.e. (Rademacher’s theorem). If the distance function is differentiable at $x$, then $xin operatorname{Unp}(K)$. For a proof, see https://mathoverflow.net/a/299066/121665.
Therefore almost all points of $mathbb{R}^nsetminus K$ belong to $operatorname{Unp}(K)$.
For every $xin mathbb{R}^nsetminus K$, there is $yin K$ such that $|x-y|=d(x,K)$ (although $y$ is not unique). Then the interior of the segment $xy$ is contained in $operatorname{Unp}(K)$ (triangle inequality).
Thus $operatorname{Unp}(K)$ contains the union of disjoint open segments and the set $mathbb{R}^nsetminus (operatorname{Unp}(K)cup K)$ is of measure zero and is contained in the endpoints of these segments.
Question. Is it true that $operatorname{Unp}(K)$ contains an open set?
The interest in the study of the properties of the distance function on the set $operatorname{Unp}(K)$ is motivated by results of Federer about sets of positive reach.
H. Federer,
Curvature measures.
Trans. Amer. Math. Soc. 93 (1959), 418–491.
I think the following is a counterexample in $mathbb{R}^2$. Consider the curve whose polar coordinates expression is $r = sum_{n=1}^infty frac{1}{a_n} sin(2pi a_ntheta)$, where $a_n = 100^n$, say. Let $K$ be this curve together with all points interior to it.
It seems to me that any point $x$ with a unique closest point $y$ in $K$ has arbitrarily close neighbors whose closest point is nonunique. Because in a small neighborhood of $y$ the curve looks like a little sine wave, and there will be a point $x'$ close to $x$ which is equidistant from two peaks near $y$. As you zoom in more you might need to adjust $x'$ slightly but it will still be close to $x$. Needs some work, but ...
Correct answer by Nik Weaver on December 18, 2020
There are many counterexamples as the following result of Zamfirescu [1] shows.
Theorem. For most of the compact sets $Ksubsetmathbb{R}^n$, $operatorname{Unp}(K)$ has empty interior, meaning that the set of points in $mathbb{R}^n$ without a unique nearest point in $E$ is dense in $mathbb{R}^n$.
Here "most of the compact sets" is understood in the Baire category sense with respect to the Hausdorff metric on the space of compact sets in $mathbb{R}^n$.
[1] T. Zamfirescu, The nearest point mapping is single valued nearly everywhere. Arch. Math. (Basel) 54 (1990), 563–566.
Answered by Piotr Hajlasz on December 18, 2020
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