Timeline for Is there a good approximating polygon for every smooth set?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2011 at 21:04 | vote | accept | domotorp | ||
| Mar 16, 2011 at 20:33 | comment | added | Tapio Rajala | 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. | |
| Mar 16, 2011 at 17:53 | comment | added | Tracy Hall | 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 $P$. | |
| Mar 16, 2011 at 14:48 | answer | added | Tapio Rajala | timeline score: 7 | |
| Mar 16, 2011 at 14:01 | history | asked | domotorp | CC BY-SA 2.5 |