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Please help me in this question Its known that every closed convex subset of a Hilbert space is a Chebyshev set.

But is the converse true ? I.e does every Chebyshev subset of a Hilbert space is convex?..Also it is given a fact that If the Hilbert space is finite dimensional, then the answer is affirmative.

Note that a Chebyshev set is a set having exactly one best approximation.

Thank you in advance

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This is still an open problem.

On the negative side, there is an example of a non-convex Chebyshev set in inner-product spaces, but the one given in a $1987$ paper by G.Johnson, was given in a non-complete inner-product space. On the positive side, it is known that boundedly compact Chebyshev sets in inner-product spaces are convex. (A set $K$ is boundedly compact if each bounded sequence in $K$ has a convergent subsequence, to some limit in $K$.)

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