Timeline for A characterization of convexity
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2010 at 13:22 | comment | added | Victor Protsak | Benoît: Yes, the removed region is above the parabola, so the remaining set is on the sides. | |
| Jun 22, 2010 at 12:32 | comment | added | Benoît Kloeckner | @Joseph: yes,I guess you're right. @Victor: you mean the region on the sides of the parabola, right? | |
| Jun 22, 2010 at 2:04 | comment | added | Tom Goodwillie | Unboundedness is not the issue. Diego Matessi's example can be modified by intersecting with max(|x|,|y|,z)<1. | |
| Jun 22, 2010 at 0:42 | comment | added | Victor Protsak | As Tom Goodwillie pointed out, the intersection with $x=z+1$ is disconnected (it is the region above a parabola whose vertex is below the boundary of the half-plane), so this doesn't work, either. | |
| Jun 21, 2010 at 22:15 | comment | added | Victor Protsak | Joseph: That's right! It can be fixed by removing a half-cone $\{ (x,y,z): x>0, x^2>y^2+z^2\}$ instead. Then $P\cap X$ is $P_+\setminus Q$, where $P_+$ is a half-plane and $Q$ is empty, a point, a wedge, a region above a parabola or hyperbola, or a an elliptical segment with the chord along the boundary line of $P_+.$ | |
| Jun 21, 2010 at 18:49 | comment | added | Joseph O'Rourke | @Benoit: Would it not then be possible to choose $P$ slanting so that it clips a bit of the base of the cylinder near the origin, and so creating a hole in $P \cap X$? Maybe I misunderstand what you intend. | |
| Jun 21, 2010 at 16:56 | comment | added | Benoît Kloeckner | Nice example, that can be made closed by removing a semi-infinite open cylinder instead of a half-line. | |
| Jun 21, 2010 at 16:47 | history | answered | Diego Matessi | CC BY-SA 2.5 |