Timeline for Base of a cone in a vector space: can one always choose a convex base?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2016 at 7:13 | answer | added | Rabee Tourky | timeline score: 4 | |
| Mar 1, 2015 at 20:36 | vote | accept | polmath | ||
| Feb 28, 2015 at 13:56 | answer | added | Tobias Fritz | timeline score: 9 | |
| Feb 27, 2015 at 18:06 | comment | added | Tobias Fritz | @AlexDegtyarev: that's a good question, but actually the answer is negative. Here's a simple example of a pointed convex cone which does not have a convex base. In $\mathbb{R}^2$, define $(x,y)$ to be in $C$ if $x>0$, or if $x=0$ and $y\geq 0$. (This cone is a simple counterexample also to many other elementary conjectures that one might have about convex cones, by the way.) | |
| Feb 27, 2015 at 17:55 | comment | added | Tobias Fritz | @WillieWong: thanks, that's exactly right. I'm not sure whether a convex base always exists; if it does, then the question is uninteresting. So I assume that the OP knows examples in which a convex base does not exist. In these cases, my argument shows that a non-convex base still exists, for trivial reasons. This shows that what the OP is trying to do is not possible. | |
| Feb 27, 2015 at 15:13 | comment | added | Willie Wong | @AlexDegtyarev: my thoughts too. If $V$ is locally convex we can get the hyperplane from Hahn-Banach; perhaps we can ask for even less. | |
| Feb 27, 2015 at 15:09 | comment | added | Alex Degtyarev | I may be missing something, but cannot you just intersect your cone with an appropriate hyperplane? | |
| Feb 27, 2015 at 15:04 | comment | added | Willie Wong | @OlegEroshkin: I suspect Tobias is addressing the hypothesis "if $C$ has at least one base..." which he points out is always achievable. | |
| Feb 27, 2015 at 14:44 | comment | added | Oleg Eroshkin | @TobiasFritz You are proving that a base exists. But the question is can you choose a convex base. | |
| Feb 27, 2015 at 14:20 | comment | added | Tobias Fritz | Every cone has a base in your sense, and this follows from the axiom of choice. In $C\setminus\{0\}$, put $x\sim y$ whenever $x$ and $y$ are positive scalar multiples of each other. This is an equivalence relation, and the axiom of choice lets you pick one representative of each equivalence class. These representatives form a base in your sense. | |
| Feb 27, 2015 at 13:50 | history | asked | polmath | CC BY-SA 3.0 |