Timeline for An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request
Current License: CC BY-SA 3.0
18 events
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| May 7, 2013 at 19:25 | vote | accept | Tom Leinster | ||
| May 7, 2013 at 14:20 | comment | added | YCor | @Pete: write $X=X_1\sqcup X_2\sqcup X_3$ where $X_1=X$ and $X_2=X_3=\emptyset$. For a unique $i$ we have $X_i\in\mathcal{U}$, so necessarily $i=1$ (otherwise uniqueness would fail). So $\emptyset\notin\mathcal{U}$. | |
| May 7, 2013 at 7:06 | answer | added | user33772 | timeline score: 13 | |
| Jul 9, 2011 at 17:49 | history | edited | Tom Leinster |
Added ultrafilter tag
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| Jul 4, 2011 at 23:01 | comment | added | Pete L. Clark | @Tom: great. I never win those contests anyway. :) | |
| Jul 4, 2011 at 22:15 | history | edited | Tom Leinster | CC BY-SA 3.0 |
deleted 7 characters in body
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| Jul 4, 2011 at 22:08 | history | edited | Tom Leinster | CC BY-SA 3.0 |
Clarified word "partition"
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| Jul 4, 2011 at 22:05 | comment | added | Tom Leinster | Hi Pete. Yup, it's just as Todd said. Sorry, I didn't know there was a usage of the word "partition" that forbade the empty set; I'll clarify that in the question. And yes, when I said it was short I had in mind that you get to skip the definition of filter. (But I'm not really bothered about what's shorter than what; this isn't supposed to be a who's-got-the-shortest-one contest.) | |
| Jul 4, 2011 at 20:14 | comment | added | Todd Trimble | (I hope Tom doesn't mind if I answer; the two of us have been talking about this a bit recently.) The empty set is being allowed as a member of the "partition"; more properly, we are really considering properties of functions $X \to [n]$ where $[n]$ is an $n$-element set. | |
| Jul 4, 2011 at 19:25 | comment | added | Pete L. Clark | I don't see how the proposed definition rules out the possibility that $\mathcal{U}$ contains the empty set. Or are you allowing the empty set as a member of your partition? (This is not standard, I believe.) Anyway, is there any advantage to this definition? (You claim it's shorter. I guess it is if you you don't also want to define filters, but that is usually not the case.) | |
| Jul 4, 2011 at 16:51 | comment | added | Qiaochu Yuan | @Tom: yes, that passage, although you're right that I don't assert the converse. | |
| Jul 4, 2011 at 16:42 | answer | added | Gerald Edgar | timeline score: -2 | |
| Jul 4, 2011 at 15:36 | comment | added | Tom Leinster | By the way, Qiaochu, I don't think you need to plant your tongue in your cheek. In my opinion, blog posts (and MO answers) should be taken absolutely seriously when it comes to attributing priority. On the other hand, I'd bet a large amount that this characterization of ultrafilters was found well before either of us was born. | |
| Jul 4, 2011 at 15:34 | comment | added | Tom Leinster | I like your post, Qiaochu. Thanks for pointing it out. Let me probe a bit, though: where would you say that you implicitly give this definition? I'm guessing that you're thinking of the last two paras before the heading "Non-principal ultrafilters". I understood that passage as saying that any ultrafilter satisfies the property in the definition I mentioned, but is it also asserting the converse? I didn't get that impression, though maybe it could be read in different ways. | |
| Jul 4, 2011 at 14:39 | comment | added | Kevin O'Bryant | @Qiaochu: Since this definition is equivalent to the usual, it is "implicitly" everywhere that ultrafilters are discussed. | |
| Jul 4, 2011 at 14:06 | answer | added | Greg Graviton | timeline score: 4 | |
| Jul 4, 2011 at 14:04 | comment | added | Qiaochu Yuan | Does my blog count as part of the literature (he said, tongue firmly planted in cheek)? I implicitly give this definition at qchu.wordpress.com/2010/12/09/ultrafilters-in-topology . | |
| Jul 4, 2011 at 13:42 | history | asked | Tom Leinster | CC BY-SA 3.0 |