Timeline for Can the intersection of the boundaries of compact and convex sets be a single element?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2013 at 14:44 | comment | added | Roc Armenter | @TapioRajala I think I have proved that $n=2$ contains at least two points given the conditions above, so I was half-expecting the same conditions would deliver it in higher $n$. Alas, no. | |
| Sep 13, 2013 at 14:34 | comment | added | Roc Armenter | That's excellent, thanks. I still feel there is something non-generic about this but clearly I need to work out what it is. | |
| Sep 13, 2013 at 14:33 | vote | accept | Roc Armenter | ||
| Sep 13, 2013 at 9:17 | comment | added | Tapio Rajala | In the case $n = 2$ the intersection of the two boundaries seems to always contain at least two points. | |
| Sep 13, 2013 at 9:07 | comment | added | Tapio Rajala | @Benoît: You are right. I was not careful enough. I missed $n=n$ (among other things...). | |
| Sep 13, 2013 at 7:46 | comment | added | Benoît Kloeckner | @Tapio: Hum, for $n=2$, it seems to me that one creates another component in the intersection of boundaries. Also, for $n=1$, you have one convex and compact set, whose boundary must have two points. So, this example should answer $n\ge 3$, and only $n=2$ looks unsettled. | |
| Sep 13, 2013 at 5:20 | comment | added | Tapio Rajala | I was just about to post a similar answer. The same construction works for all $n \ge 1$. | |
| Sep 13, 2013 at 5:20 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| Sep 13, 2013 at 5:15 | history | undeleted | Robert Israel | ||
| Sep 13, 2013 at 5:14 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| Sep 13, 2013 at 4:59 | history | deleted | Robert Israel | via Vote | |
| Sep 13, 2013 at 4:37 | history | answered | Robert Israel | CC BY-SA 3.0 |