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Observing that if we have a three-dimensional convex set then by sectioning it with a plane we obtain a convex set, I wondered if the converse is true: given a set whose every section is a convex set, it turns out that it is a convex set? It was trivial.

Then I gave the following definition: a three-dimensional set is paraconvex if every planar section is the union of separated convex sets. Then every convex three dimensional set is paraconvex but.. is the converse true?

Obviously a set composed of separated convex subsets is not convex but paraconvex. Then I repeat my question restricting it for connected sets.

Again, taking a differentiable curve with non-zero torsion, I got a counterexample.

Then I restricted my question to connected sets that are also regular closed (sets that equals the closure of their interior). I didn't succeed in manage this new case.

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What kind of separation? (The convex sets live in complementary halfspaces but may meet on the boundary? They live in disjoint halfspaces? Halfspaces at positive distance from each other?) – user21467 Mar 17 '12 at 17:45
@StevenTaschuk each is disjoint from the closure of the other. – Immanuel Weihnacht Mar 17 '12 at 18:30

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