Timeline for An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request
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May 7, 2013 at 20:57 | comment | added | Gerhard Paseman | Tom, I've seen postings by a Butch Malahide on other fora, which may help you get in touch with him/her/it. In this day and age, referring to MathOverflow users by number should be socially acceptable. Gerhard "User 3528. Aliases 3206, 3371..." Paseman, 2013.05.07 | |
May 7, 2013 at 19:32 | comment | added | Tom Leinster | PS to Butch: I'll acknowledge you when I update the paper for which I needed this, arxiv.org/abs/1209.3606. Forgive the impertinence, but is Butch Malahide your real name? Feel free to contact me by email. (I'd contact you myself, but there's no address on your profile.) | |
May 7, 2013 at 19:29 | comment | added | Tom Leinster | Excellent. Thanks very much. For the sake of precision, let me add that they don't quite do the case $n=3$ in the sense described in my question. They do show that if a set $\mathcal{U}$ of subsets of $X$ satisfies the partition condition for all $n\leq 3$, then $\mathcal{U}$ is an ultrafilter. But they seem not to observe that it suffices to require it for $n=3$ (which implies it for $n=0,1,2$). Quite possibly they knew it but just didn't think it was worth mentioning. | |
May 7, 2013 at 19:25 | vote | accept | Tom Leinster | ||
May 7, 2013 at 18:56 | history | edited | user33772 | CC BY-SA 3.0 |
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May 7, 2013 at 7:34 | history | edited | user33772 | CC BY-SA 3.0 |
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May 7, 2013 at 7:27 | comment | added | Gerhard Paseman | Welcome to MathOverflow! Gerhard "Ask Me About System Design" Paseman, 2013.05.07 | |
May 7, 2013 at 7:06 | history | answered | user33772 | CC BY-SA 3.0 |