Timeline for Is a certain subset of the disc a convex set?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Dec 7, 2015 at 19:52 | comment | added | Igor Rivin | @LoicTeyssier Actually, the argument for non-convexity uses very much that this is the $1/3$ Cantor set (you can show that there is a triangle in the middle of the circle which is not contained in the chordal set), but this argument disappears for $1/N$ Cantor sets for other values of $N.$ | |
| Dec 7, 2015 at 19:48 | comment | added | Loïc Teyssier | Indeed. I'd be surprised to learn the situation is that different, though, but I already was proved wrong once on that topic today so I'm not going to make bets ;) | |
| Dec 7, 2015 at 19:47 | comment | added | Igor Rivin | @LoïcTeyssier But, of course, it might be convex for other Cantor sets... | |
| Dec 7, 2015 at 19:47 | comment | added | Igor Rivin | @LoicTeyssier You are right! I did not read the question carefully. | |
| Dec 7, 2015 at 19:45 | comment | added | Loïc Teyssier | If I understand correctly the statement of the OP in the link you provide, it seems that the set $A$ (which is $Chords(D)$ in the linked post) obtained from the standard Cantor set (the result of removing a fixed ratio of intervals) is never convex, although it happens that its measure be positive (that of $C$ is zero). | |
| Dec 7, 2015 at 19:35 | history | answered | Igor Rivin | CC BY-SA 3.0 |