Timeline for Do operations generate well-ordered sets only?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 23, 2013 at 18:38 | vote | accept | Włodzimierz Holsztyński | ||
| May 23, 2013 at 14:29 | answer | added | Ramiro de la Vega | timeline score: 4 | |
| May 23, 2013 at 13:30 | answer | added | Emil Jeřábek | timeline score: 10 | |
| May 23, 2013 at 5:33 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
3rd version
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| May 23, 2013 at 5:27 | comment | added | Włodzimierz Holsztyński | Thank you, @Joseph--I am persistently and stubbornly blind. I'll modify my question again. I hope that this time it will be finally fine :-) | |
| May 23, 2013 at 1:54 | comment | added | Joseph Van Name | You will need a bit more than being non-decreasing in each variable since $\lfloor Log(x+1)\rfloor$ is non-decreasing in each variable. | |
| May 23, 2013 at 1:48 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
a name
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| May 23, 2013 at 1:44 | comment | added | Włodzimierz Holsztyński | @Joseph: you're right! I need to reformulate my question. I still believe that under very general assumptions one still not get qa decreasing sequence. | |
| May 23, 2013 at 1:39 | comment | added | Joseph Van Name | Can't you just take $f(x,y)=\lfloor Log(x+1)\rfloor$? | |
| May 23, 2013 at 1:22 | history | asked | Włodzimierz Holsztyński | CC BY-SA 3.0 |