Timeline for Is non-existence of the hyperreals consistent with ZF?
Current License: CC BY-SA 3.0
19 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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| Dec 8, 2014 at 13:20 | comment | added | Mikhail Katz | @JoelDavidHamkins, good point, thanks. | |
| Dec 8, 2014 at 13:16 | comment | added | Joel David Hamkins | Transfer plus countable choice proves $\omega_1$-saturation, since transfer gives you the ultrafilter and countable choice gives you the Los theorem, so you can build the ultrapower, which has saturation. | |
| Dec 8, 2014 at 8:32 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
Kanovei-Shelah
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| Dec 7, 2014 at 9:23 | comment | added | Mikhail Katz | @EricWofsey, thanks. Note however that a nonprincipal ultrafilter is not enough to prove transfer, so in principle it is possible that in transfer should imply countable additivity, though your comment about Caratheodory does seem to make this difficult. | |
| Dec 7, 2014 at 9:08 | comment | added | Eric Wofsey | I should perhaps mention that the Caratheodory extension theorem, in the form Goldblatt states it at the end of 16.3, definitely requires countable choice, as it easily implies that $\mathbb{R}$ is not a countable union of countable sets (let $\mathcal{A}$ be finite unions of half-open intervals and $\mu$ be Lebesgue measure on $\mathcal{A}$). | |
| Dec 7, 2014 at 9:04 | comment | added | Eric Wofsey | I'm not any expert on this and I haven't looked through the arguments closely enough to be able to answer your last question, though I will remark that existence of a nonprincipal ultrafilter is indepedent of countable choice and thus it seems unlikely that it implies $\sigma$-additivity of Lebesgue measure. But the key step that Goldblatt leaves out (and which uses countable choice) is section 16.3, where he states without proof the Caratheodory extension theorem. If you read the first paragraph of 16.5, it is clear that this is crucial to his construction of Loeb measure. | |
| Dec 7, 2014 at 8:26 | comment | added | Mikhail Katz | ...countable additivity purely from transfer. Apart from Goldblatt's approach, do you think it is feasible to derive countable additivity of Lebesgue measure using hyperfinite subdivisions and transfer? This seems like an interesting question in reverse mathematics where you may be more of an expert than myself (sorry didn't get a chance to look you up on mathscinet yet ;-) | |
| Dec 7, 2014 at 8:25 | comment | added | Mikhail Katz | @EricWofsey, thanks for your comment. I looked over the proof the other day and did not find any mention of countable choice, but perhaps the proof relies on this implicitly. Goldblatt does provide some proofs so you can't make a blanket statement that he constructs them "without proof". It seems to me that the construction of outer measure should be doable without using choice, but I haven't thought about this seriously in a while. Could you substantiate your claims by more specific references to Goldblatt? Reliance on saturation is certainly a difficulty here if one wants to derive... | |
| Dec 7, 2014 at 7:07 | comment | added | Eric Wofsey | The argument in Goldblatt does not prove countable additivity of Lebesgue measure from the existence of the hyperreals in ZF. Rather, Goldblatt constructs (without proof) both Loeb and Lebesgue measure using the usual outer measure machinery which uses countable choice, and then proves they are equal. In addition, Goldblatt throughout assumes not just existence of a hyperreal field with transfer but also countable saturation. | |
| Dec 6, 2014 at 22:34 | comment | added | Joel David Hamkins | Asaf, my personal view is that we should welcome any mathematically substantive, interesting post that sheds light on the issues of a question. So I don't really agree to have the kind of formal rules that you are advocating. (And my comments here object to your criterion only in so far as you have put it forth as a general rule, apart from the merits of this particular case.) | |
| Dec 6, 2014 at 20:02 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
added 372 characters in body
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| Dec 5, 2014 at 12:41 | comment | added | Asaf Karagila♦ | @Joel: This is a discussion for the meta site, rather than here. But I think that answers should be answers. Of course, a question asking for something, and you want to post an answer showing that certain assumptions are really needed, is some sort of answer. But if you are giving it as an answer, you should at least give some substantial explanation. Here, on the other hand, there is just a remark equivalent to "Hey, by the way, the axiom of choice is needed elsewhere in analysis", and not much longer either. I agree this makes an excellent comment, it makes a bad answer, though. | |
| Dec 5, 2014 at 12:14 | comment | added | Joel David Hamkins | @AsafKaragila I find your criteria to be unnecessarily restrictive. We should welcome any post that makes, as here, a substantive and interesting mathematical contribution bearing on a question. I believe that expert and insightful remarks on the background of a question, in general, make fine answers. | |
| Dec 5, 2014 at 9:39 | comment | added | Asaf Karagila♦ | Yes, and this is a very good comment. Answers should answer the question, not make remarks on its background. | |
| Dec 5, 2014 at 9:34 | comment | added | Mikhail Katz | @Asaf, the philosophical implications are clearly in the background of this question. | |
| Dec 4, 2014 at 18:46 | review | Low quality posts | |||
| Dec 4, 2014 at 20:37 | |||||
| Dec 4, 2014 at 18:17 | comment | added | Asaf Karagila♦ | Seems like a reasonable comment. Not quite an answer to the question at hand. | |
| Dec 4, 2014 at 16:33 | history | answered | Mikhail Katz | CC BY-SA 3.0 |