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?
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$\begingroup$ 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 $P$. $\endgroup$– Tracy HallCommented Mar 16, 2011 at 17:53
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$\begingroup$ 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. $\endgroup$– Tapio RajalaCommented Mar 16, 2011 at 20:33
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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.
answered Mar 16, 2011 at 14:48