# Convex Sets and Nearest Neighbors

For a set $$S \subseteq \mathbb{R}^n$$ and a point $$x \in \mathbb{R}^n$$, let $$c_S(x)$$ be the point $$s \in S$$ that minimizes $$\|x-s\|$$ if such a point exists and is unique. It is known that $$c(x) = s$$ exists and is unique if $$S$$ is closed and convex. My question is: is this "if and only if"? More precisely: Are the following statements equivalent:

1. $$S$$ is closed and convex.
2. $$c_S(x)$$ is well-defined for all $$x \in \mathbb{R}^n$$.

If $$S$$ is not closed, then there are points that do not have a nearest neighbor in $$S$$. If $$S$$ is closed, then, for every $$x \in \mathbb{R}^n$$, there exist points that are nearest neighbors to $$x$$. Thus, the question boils down to the following: Let $$S \subseteq \mathbb{R}^n$$ be closed. Then the following two statements are equivalent:

1. $$S$$ is convex.
2. For every $$x \in \mathbb{R}^n$$, there is a unique element $$s \in S$$ that minimizes $$\|x-s\|$$.

Is this true?

This is the celebrated Chebyshev problem. The answer is positive in $$\mathbb{R}^n$$, and still open in the Hilbert space.
A nonempty set $S$ in a normed linear space $X$ is called a Chebyshev set if for each $u \in X$ there is exactly one nearest point in $S$ to $u$ (i.e., for a Chebyshev set, nearest points always exist, and they are unique).
Theorem. A nonempty set in the Euclidean space $R^n$ is Chebyshev if and only if it is closed and convex.