Timeline for Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
Current License: CC BY-SA 2.5
10 events
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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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| Oct 24, 2015 at 17:39 | comment | added | Todd Trimble | As a small additional gloss: the nonemptiness of a product of finite sets $\prod_{u \in A} H_u$ doesn't require the full axiom of choice, it just uses the ultrafilter principle. For any set can be totally ordered by an application of the compactness theorem, which uses just the ultrafilter principle. Applying this to totally order the set $H = \sum_{u \in A} H_u$ (the disjoint union), each $H_u$ inherits a total order by restriction, which is a well-order by finiteness. Letting $h_u$ be the least element in $H_u$, the tuple $(h_u)$ is then an element of the product. | |
| Nov 19, 2010 at 23:05 | comment | added | Qiaochu Yuan | @Andres: thanks for the details; it's more than enough. | |
| Nov 19, 2010 at 22:45 | comment | added | Andrés E. Caicedo | @Qiaochu: Send me an email if you want a copy of the paper. | |
| Nov 19, 2010 at 22:37 | history | edited | Andrés E. Caicedo | CC BY-SA 2.5 |
Added the proof of Banaschewski's theorem
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| Nov 19, 2010 at 9:27 | vote | accept | Qiaochu Yuan | ||
| Nov 19, 2010 at 9:27 | comment | added | Qiaochu Yuan | Thanks, Andres! I would really appreciate any details about Banaschewski's paper. | |
| Nov 19, 2010 at 2:30 | history | edited | Andrés E. Caicedo | CC BY-SA 2.5 |
added 154 characters in body
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| Nov 19, 2010 at 1:36 | comment | added | Willie Wong | Who'd have thunk that just mentioning your name would summon an expert? :) | |
| Nov 19, 2010 at 1:29 | history | answered | Andrés E. Caicedo | CC BY-SA 2.5 |