Set of points with a unique closest point in a compact set

Question

Let $K\subset\mathbb{R}^n$ be any compact set. Let $\operatorname{Unp}(K)$ be the set of points in $$ \operatorname{Unp}(K)=\{x\in\mathbb{R}^n\setminus K:\, \exists ! y\in K \ \ |x-y|=d(x,K)\}. $$ Here are some properties.

1. The distance function is Lipschitz and hence differnetiable a.e. (Rademacher's theorem). If the distance function is differentiable at $x$, then $x\in \operatorname{Unp}(K)$. For a proof, see https://mathoverflow.net/a/299066/121665. Therefore almost all points of $\mathbb{R}^n\setminus K$ belong to $\operatorname{Unp}(K)$.

2. For every $x\in \mathbb{R}^n\setminus K$, there is $y\in K$ 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 $\mathbb{R}^n\setminus (\operatorname{Unp}(K)\cup K)$ 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.

Answer 1

I think the following is a counterexample in $\mathbb{R}^2$. Consider the curve whose polar coordinates expression is $r = \sum_{n=1}^\infty \frac{1}{a_n} \sin(2\pi a_n\theta)$, 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 ...

Answer 2

There are many counterexamples as the following result of Zamfirescu [1] shows.

Theorem. For most of the compact sets $K\subset\mathbb{R}^n$, $\operatorname{Unp}(K)$ 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.