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My question is if in a normed space (maybe complete if necessary), a convex (non empty) set with convex complement has non-empty interior.

I cannot think of a counterexample, but neither how to prove it.

Any ideas?

Eparoh
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1 Answers1

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Take an unbounded functional $\ell$ and consider $$ A = \{ x \in X \mid \ell(x) \le 0 \}. $$ The complement $$ B = \{ x \in X \mid \ell(x) > 0 \} $$ is convex as well and neither of these sets have an nonempty interior.

gerw
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    I don't understand why this has been downvoted! This answer is a bit terse, but it seems correct to me. I believe the missing steps are "any infinite-dimensional normed space admits an unbounded linear functional", and "the kernel of an unbounded functional is dense" - so all the fibres are dense. Hence these two complementary convex subsets are both dense, so each must have empty interior (giving a counterexample in any infinite-dimensional space). These facts are fairly well-known I think, and proofs can be found on this site. – Izaak van Dongen Mar 04 '24 at 21:39