Timeline for Uniformizing a relation on ordered sets
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2015 at 19:58 | vote | accept | Bjørn Kjos-Hanssen | ||
| Nov 11, 2014 at 7:41 | answer | added | Dominic van der Zypen | timeline score: 0 | |
| Nov 5, 2014 at 19:26 | comment | added | Bjørn Kjos-Hanssen | @FrançoisG.Dorais it would be interesting if you post an answer including how you thought of that | |
| Nov 5, 2014 at 18:41 | comment | added | Joel David Hamkins | But I also like "lower envelope" and "left envelope", since these functions are the envelopes of the collection of all uniformizations (even computable partial uniformizations) of $R$. | |
| Nov 5, 2014 at 18:37 | comment | added | Bjørn Kjos-Hanssen | Thanks @JoelDavidHamkins in general that sounds good (in my case of interest $R$ is upward closed in the product order). | |
| Nov 5, 2014 at 13:02 | comment | added | Joel David Hamkins | I don't know any standard terminology for this, but I would likely refer to $f$ as the lower boundary of $R$ and $g$ as the left boundary. If the order also had supremums, then there would also be a corresponding upper boundary and right boundary. | |
| Nov 5, 2014 at 12:58 | comment | added | Joel David Hamkins | Since he only takes infimums, he only needs lower semi-complete. | |
| Nov 5, 2014 at 8:54 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
deleted 101 characters in body
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| Nov 5, 2014 at 8:43 | comment | added | Dominic van der Zypen | Note that $\mathbb{N}$ is not complete with respect to the usual ordering $\leq$ because $\mathbb{N}$ does not have a supremum. | |
| Nov 5, 2014 at 3:52 | comment | added | François G. Dorais | Antitone Galois connections? (At least in some cases.) | |
| Nov 4, 2014 at 23:40 | history | asked | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |