Timeline for Continuously selecting elements from unordered pairs
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 22, 2012 at 23:51 | vote | accept | François G. Dorais | ||
| Oct 19, 2011 at 19:52 | answer | added | KP Hart | timeline score: 6 | |
| Sep 7, 2011 at 21:27 | answer | added | François G. Dorais | timeline score: 4 | |
| Sep 7, 2011 at 3:09 | answer | added | Adam Bjorndahl | timeline score: 5 | |
| Sep 6, 2011 at 16:46 | history | edited | François G. Dorais | CC BY-SA 3.0 |
update
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| Sep 6, 2011 at 12:18 | answer | added | Andreas Blass | timeline score: 13 | |
| Sep 6, 2011 at 5:51 | comment | added | François G. Dorais | I'd call such an $R$ an open tournament on $X$. | |
| Sep 6, 2011 at 4:14 | comment | added | François G. Dorais | The case of $\mathbb{Q}^2$ is actually pretty neat. If I'm not mistaken, it does have a continuous selector, as will any other countable metric space. | |
| Sep 6, 2011 at 4:03 | comment | added | Adam Bjorndahl | I agree my version is basically just a restatement, and so not very satisfying. As long as we're on that track, though, here's another restatement from a slightly different perspective: Suppose there is a binary relation $R$ on X that is total, antisymmetric, and "plays well with the topology", in the sense that, whenever $a \neq b$ and $aRb$, there are open neighbourhoods $U$ and $V$ about $a$ and $b$ such that, for all $x \in U$ and $y \in V$, $xRy$. I think this is also equivalent to the existence of a continuous selector. The usual order on $\mathbb{R}$ is an example of such a relation. | |
| Sep 6, 2011 at 3:53 | comment | added | François G. Dorais | Well, it seems like I managed to both underestimate and overestimate the same argument in the span of a few hours. Anton's argument works fine when $X^2$ is locally connected, but I don't see the general case. Maybe Anton had a bit more in mind? Your decomposition into two symmetric open sets works, but I feel like it's more a restatement of the question than an answer. Then again, maybe there is no better answer than that. | |
| Sep 6, 2011 at 2:52 | comment | added | Adam Bjorndahl | Is it important what the connected components of $X^{2} - \Delta$ look like? I'm worried about an example like $X = \mathbb{Q}^{2}$. Maybe instead of components we want: there is a decomposition of $X^{2} - \Delta$ into disjoint opens $U$ and $V$ such that $(x,y) \in U$ iff $(y,x) \in V$? | |
| Sep 6, 2011 at 0:50 | comment | added | François G. Dorais | Now that you put it in those terms, I think it's me that completely missed the full weight of the necessary condition. Please post your comment as an answer so I can accept it. Follow-up: Is this equivalent to the linear ordering condition? | |
| Sep 5, 2011 at 23:53 | comment | added | Anton Petrunin | After removing diagonal, any point $(x,y)$ should be in a component different from $(y,x)$. It seems that this should be necessary and sufficient condition. Did I miss something? | |
| Sep 5, 2011 at 21:04 | history | edited | François G. Dorais | CC BY-SA 3.0 |
addendum
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| Sep 5, 2011 at 20:17 | history | asked | François G. Dorais | CC BY-SA 3.0 |