Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 http://www.ics.uci.edu/~eppstein/junkyard/knot-colinear.html 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.)

share|improve this question
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.