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.



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


