Let $X$ be a real Hilbert Space and $C \subseteq X$. Let $d\_C$ be the infimal distance function to $C$ and $P\_C(x) = C \cap S[x; d\_C(x)]$ be the metric projection. We say $C$ is *proximinal* if $P\_C(x) \neq \emptyset$ for all $x \in X$.


I'm wondering if the intersection of two proximinal subsets of $X$ must be proximinal? To avoid trivial counterexamples, I need their intersection to be non-empty.


The answer is most likely "no", or at least, something that is difficult to prove in the affirmative if true. Asplund proved that, given any non-convex Chebyshev subset $C$ (meaning $|P\_C(x)| = 1$ for all $x$), there exists a closed half-space $H$ such that $X \cap H$ is not proximinal. Therefore, were proximinal sets closed under intersection, this would prove the Chebyshev conjecture: that all Chebyshev subsets are convex.


I was wondering if someone had an explicit example of two proximinal sets whose intersection is not proximinal?



Are the intersection of proximinal sets in a Hilbert Space proximinal?
This answer is strongly inspired by example 3.11 in the book by Bauschke and Combettes.


Let $H = \ell^2$ and consider a sequence $\{\alpha\_n\} \in (1,\infty)$ with $\alpha\_n \searrow 1$. Define
\begin{align}
C &= \{ \alpha\_n \, e\_n \mid n \in \mathbb N\} \cup \{0\}\\
D &= \{ x \in \ell^2 \mid \exists n \in \mathbb N : x\_n \ge 1 \}.
\end{align}
The set $C$ is weakly closed, thus a proximinal set (e.g. Prop. 3.12 in Bauschke&Combettes).


Moreover, it is easy to check that $D$ is proximinal: For $y \in \ell^2$ one can construct a projection as follows: Take $j \in \operatorname{arg\,max}\_j y\_j$. If $y\_j \ge 1$, then $y \in D$. Otherwise, obtain $x$ by replacing the entry $y\_j$ in $y$ by $1$. Then, $x$ is a projection of $y$ onto $D$.


Finally, we have $C \cap D = C \setminus \{0\}$ and this set is not a proximinal set, see example 3.11 in Bauschke and Combettes,
as $0$ has no projection onto $C \setminus \{0\}$.


