Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we have an open set S whose boundary is a closed Jordan curve that has a unique tangent at each point. Is it true that for every epsilon there is a P polygon contained in S such that there is a (1+epsilon) scaled copy of P that contains S?

share|improve this question
    
As answered below: No, unless $S$ is star-shaped around some point of its interior. On the other hand, it seems like this should always be possible if you just ask for $S$ to be contained in an $\epsilon$-neighborhood of the region bounded by $P$. –  Tracy Hall Mar 16 '11 at 17:53
    
For the $\epsilon$-neighborhood it should be enough to assume that the set $S \subset \mathbb{R}^n$ is a bounded domain. One (probably overcomplicated) proof could go as follows: Decompose the space with cubes of diameter $\epsilon$ and select a point inside $S$ in each of the cubes that contain $S$. Select one of these points as the base point and connect the others to that with paths inside $S$. The union $U$ of these paths is closed and therefore it has positive distance from $\partial S$. Now build a polygon $P$ from the set $U$ and we are done. –  Tapio Rajala Mar 16 '11 at 20:33

1 Answer 1

up vote 8 down vote accepted

I do not see how scaling could give such a property. Consider an annulus in the plane (remove a part of it to make it a Jordan domain and also make it smooth if you want). Scaling any polygon inside the annulus by $1 + \epsilon$ makes also the "missing ball" inside the annulus larger.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.