Timeline for Choice vs. countable choice
Current License: CC BY-SA 2.5
23 events
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| Feb 26, 2018 at 17:55 | comment | added | Jules | Isn't it the case that if there constructively exists an element in each set then there constructively exists an element in the product? The constructive existence already chose an element of each set for us (namely, the one that was constructed). | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 4, 2010 at 5:09 | answer | added | Andreas Blass | timeline score: 5 | |
| Apr 30, 2010 at 11:03 | vote | accept | G. Rodrigues | ||
| Apr 30, 2010 at 7:28 | answer | added | Pete L. Clark | timeline score: 5 | |
| Apr 29, 2010 at 22:46 | comment | added | François G. Dorais | This question now has a meta thread - tea.mathoverflow.net/discussion/374/choice-vs-countable-choice | |
| Apr 29, 2010 at 22:05 | comment | added | Aaron Bergman | "I mean, how can the product of a family of non-empty sets fail to be non-empty?" Could't we say, instead, that products over uncountable indexing sets aren't guaranteed to exist? | |
| Apr 29, 2010 at 21:49 | comment | added | Harry Gindi | I don't understand how someone can reasonably say "I'm a constructivist and that's that." As I noted in my post in the thread you posted, constructive mathematics can easily be embedded within nonconstructive mathematics using topos theory and universes. I can understand preferring to do constructive mathematics (the results hold in all toposes!), but isn't it a little bit extreme to take a philosophical position that nonconstructive mathematics is in some way "wrong"? Just a question I had for a dyed in the wool constructivist. | |
| Apr 29, 2010 at 21:28 | answer | added | Andrej Bauer | timeline score: 46 | |
| Apr 29, 2010 at 19:54 | answer | added | David E Speyer | timeline score: 18 | |
| Apr 29, 2010 at 19:19 | answer | added | Pace Nielsen | timeline score: 18 | |
| Apr 29, 2010 at 19:04 | history | edited | G. Rodrigues | CC BY-SA 2.5 |
Some rewriting and expansion of the original post to address some of the comments.; added 92 characters in body
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| Apr 29, 2010 at 18:06 | answer | added | Noam Zeilberger | timeline score: 13 | |
| Apr 29, 2010 at 17:48 | comment | added | Peter LeFanu Lumsdaine | @Niel: this depends on the interpretation of quantifiers you/they have in mind. In some constructive frameworks (eg Martin-Löf Type Theory), all quantifiers have this strong constructive interpretation, and so AC is indeed a theorem. In others (eg the forms of higher type theory that are valid in a topos), this isn't the case; roughly, to show something is non-empty I have to have give a method to construct of some witness, but I don't have to ensure that it will construct the same witness every time I try it, so this can't be used to give the function required for AC. | |
| Apr 29, 2010 at 17:18 | comment | added | Tom Church | @John Goodrick: Pete Clark writes "isn't it only an extremely weak version of AC that is needed here? Something like countable choice? As Kaplansky says in his Set Theory and Metric Spaces, really countable choice should go without comment in any place where non-axiomatic set theory is being discussed. I think that 99.9% of practicing mathematicians would regard countable choice as simply being 'true'." here: mathoverflow.net/questions/22927/… His comment has four up-votes, so at least 5 mathematicians feel similarly. | |
| Apr 29, 2010 at 16:50 | comment | added | Theo Johnson-Freyd | I think this question should be fairly heavily edited. As it is, it is argumentative and subjective, and does not contribute more to the discussion of AC than what is already available on the other questions on the subject, including the linked question. I vote to close. | |
| Apr 29, 2010 at 16:04 | answer | added | Emerton | timeline score: 5 | |
| Apr 29, 2010 at 15:27 | comment | added | John Goodrick | Your question is very unclear to me (although I tried to give some kind of answer below). To whom is countable choice more "unproblematic" than full AC -- to some people you've talked with (I've never met such a person...)? To you? Have you actually done a survey on how mathematicians feel about this? Also, I can't imagine a reasonable and correct way to formalize the statement: "AC allows us to prove more things than the Axiom of Determinacy." | |
| Apr 29, 2010 at 15:22 | answer | added | John Goodrick | timeline score: 6 | |
| Apr 29, 2010 at 15:20 | comment | added | Niel de Beaudrap | Another remark: to a constructivist, isn't the Axiom of Choice a tautology? The infinite product can only be expressed if there is a constructible function which identifies which sets S_n form the factors of the product; and these sets can only be certified to be all non-empty if there is a function mapping each index n in the product to an element of S_n. So, in order for the product to constructibly satisfy the condition of the AC --- having a (constructible) family of (constructivly demonstratably) non-empty sets --- it seems that one must first exhibit a choice function. | |
| Apr 29, 2010 at 15:17 | comment | added | Niel de Beaudrap | Also note that just because physicists "use" the AC, does not mean that they need the full power of the AC. In a more extreme example, grade-schoolers do not require the AC in order to be able to obtain a well-ordering on the cardinal numbers that they are likely to use. | |
| Apr 29, 2010 at 15:02 | comment | added | Qiaochu Yuan | The axiom of determinacy (en.wikipedia.org/wiki/Axiom_of_determinacy) is also "obviously true," and they're inconsistent. I don't see the point of making such statements. In both cases we are extending intuition from the finite to the infinite, and there's no reason this should always be valid. | |
| Apr 29, 2010 at 14:56 | history | asked | G. Rodrigues | CC BY-SA 2.5 |