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?

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?

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?

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?

The answer is very likely to be "no". 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?

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?

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?