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Let $C$ be a compact set in $\mathbb{R}^n$. If $C$ is convex, it must be homeomorphic to a closed ball. Now suppose that instead of convexity we require the intersection of $C$ with any line to have at most $m$ connected components. What topological restrictions does this condition place on $C$ and its embedding in $\mathbb{R}^n$?

For example, the result at shows that for $n=3$ and $m=3$, the complement of $C$ can't be knotted.

I would be interested in results for specific small cases as well as in the following question: For fixed $n$, does there exist $m$ so that $C$ can come from any "reasonable" isotopy class? (Define reasonable reasonably.)

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The empty set or a singleton are compact convex subsets of $\mathbb R^n$ which are not homeomorphic to a closed ball. – Rasmus Jan 23 '12 at 22:49
@Rasmus A singleton of $\mathbb{R}^n$ is homeomorphic to a closed ball of dimension $0$ and the empty set is homeomorphic to a closed ball of dimension $-1$ (by convention). We have a result: if $U\subseteq \mathbb{R}^n$ is a bounded convex open set, then $\overline{U}$ is homeomorphic to the closed ball in $\mathbb{R}^n$. – Amitesh Datta Jan 24 '12 at 1:33

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