Timeline for Special infinitary relations and ultrafilters
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 31, 2019 at 16:39 | comment | added | user140765 | @DanielLitt geometric Langlands also was a moving target some time ago (maybe it even is now, I am not an expert). Like Beilinson-Drinfeld's "best hope" was clarified by Arinkin-Gaitsgory's singular support condition. I think I understand the idea of your comment, but conjecture being a moving target is not necessarily a bad thing (if you understand for what reasons it ought to be moving). | |
| S Jan 28, 2015 at 18:10 | history | suggested | Loreno Heer | CC BY-SA 3.0 |
fix latex did not display
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| Jan 28, 2015 at 17:25 | review | Suggested edits | |||
| S Jan 28, 2015 at 18:10 | |||||
| Apr 9, 2011 at 19:59 | vote | accept | porton | ||
| Apr 9, 2011 at 6:18 | comment | added | porton | @Emil Jeřábek: I made a modified version of my conjecture: mathoverflow.net/questions/61118/… | |
| Apr 8, 2011 at 17:25 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
fix formatting
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| Apr 8, 2011 at 17:20 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
expand
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| Apr 8, 2011 at 15:10 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
make sure the restricted product is nonempty
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| Apr 7, 2011 at 17:46 | history | edited | Emil Jeřábek | CC BY-SA 2.5 |
fix
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| Apr 7, 2011 at 17:39 | comment | added | porton | @Emil Jeřábek: Oh, sorry, it's my error, I was uncareful. I will delete my wrong comment where I blame you for an error. | |
| Apr 7, 2011 at 17:36 | comment | added | Emil Jeřábek |
Some attempts on further clarification. In case it’s not obvious, I’m implicitly using the following observation: if $a_i$ is an ultrafilter, then the condition “$\forall A\in a_i\,A\cap R_i\ne\varnothing$” from the definition of $\prod a$ simplifies to “$R_i\in a_i$”. The point of the counterexample is that any $R$ such that $(\prod^Sa)R$ agrees with $S$ on all but finitely many coordinates, whereas for a product $\prod b$ of ultrafilters you can find $R$ such that $R_i\in b_i$ is distinct from $S_i$ for every $i$, so that $(\prod b)R$ but not $(\prod^Sa)R$.
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| Apr 7, 2011 at 17:30 | comment | added | Daniel Litt | @porton: Your conjecture seems to be a moving target. It couldn't hurt to work on it for yourself for a bit. | |
| Apr 7, 2011 at 17:13 | comment | added | porton | I see I do not fully understand your counter-example and also that your result contradicts with my intuition (thus my intuition being wrong). By these reason I will stop my writing (mathematics21.org/algebraic-general-topology.html) now and go to learn. I have purchased the book "The Theory Of Ultrafilters" by W. W. Comfort, S. Negrepontis but have not yet fully learned it. Now it's the time. | |
| Apr 7, 2011 at 16:30 | history | edited | Emil Jeřábek | CC BY-SA 2.5 |
correction
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| Apr 7, 2011 at 15:28 | history | edited | Emil Jeřábek | CC BY-SA 2.5 |
clarify
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| Apr 7, 2011 at 15:21 | history | answered | Emil Jeřábek | CC BY-SA 2.5 |