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.