Timeline for Do choice principles in all generic extensions imply AC in $V$?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2021 at 19:28 | answer | added | Asaf Karagila♦ | timeline score: 6 | |
| Aug 2, 2018 at 14:44 | history | edited | Asaf Karagila♦ |
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| Aug 2, 2018 at 14:42 | answer | added | Asaf Karagila♦ | timeline score: 12 | |
| Aug 1, 2018 at 10:18 | comment | added | Joel David Hamkins | Here is a solution of the dual form: weak choice statement implies strong choice statement holds in some forcing extension. Namely, if ZFC and V=HOD(b), then there is a forcing extension with V=HOD. | |
| Aug 1, 2018 at 10:15 | comment | added | Asaf Karagila♦ | @Joel: Admittedly, at this point I am a bit lost as to where the conversation had gone to. So there is a good chance of misunderstanding on my side. | |
| Aug 1, 2018 at 9:54 | comment | added | Joel David Hamkins | My point was that your comment was incorrect, since the solution doesn't work when you allow parameters, precisely because you can't destroy that form of choice by forcing. | |
| Aug 1, 2018 at 9:52 | comment | added | Asaf Karagila♦ | @Joel: Yes, I know that, it wasn't a question. :) | |
| Aug 1, 2018 at 9:52 | comment | added | Joel David Hamkins | @AsafKaragila No you can't, because being HOD of a parameter is forcing invariant. I once made a blog post about this: jdh.hamkins.org/… | |
| Aug 1, 2018 at 9:48 | comment | added | Asaf Karagila♦ | @Joel: You can modify that to "definable from a parameter"... | |
| Aug 1, 2018 at 9:46 | comment | added | Joel David Hamkins | Here is one solution: if every forcing extension has a definable global well-order, then anything follows. (Because adding a Cohen real destroys definable global AC, and so the hypothesis is false.) | |
| Aug 1, 2018 at 9:22 | comment | added | Asaf Karagila♦ | @Noah: By the way, SVC is "generic AC", which is arguably very much a choice principle. The first KWPs are choice and the selection principle, both considered as choice principles through and through. | |
| Aug 1, 2018 at 6:14 | comment | added | Asaf Karagila♦ | @Noah: Uh, no? If you look closely at Monro's paper he doesn't violate bounded version of DC, but rather strong versions of DC. And to your question, I guess my answer would be no. I would add, though, that proper forcing cannot violate DC. So that's something. | |
| Aug 1, 2018 at 1:50 | comment | added | Noah Schweber | @AsafKaragila A silly question: is DC preserved by set forcing? Monro's paper shows that the answer is negative for limited versions of DC, but I couldn't find an answer for DC itself. Also, do you know of a single example of a choice principle other than AC itself which is preserved by set forcing? (This is necessarily subjective, I know, but I'm curious; for what it's worth, I don't consider SVC, "the Dedekind-finite cardinalities are bounded," etc. to be choice principles.) | |
| Jul 31, 2018 at 23:33 | comment | added | Joel David Hamkins | I am pleased to see such a question, which can be expressed explicitly in modal terms as $\Box\text{AC}^-\to\text{AC}^+$, using the forcing modality, where $\text{AC}^-$ is the weak choice principle and $\text{AC}^+$ is the stronger one. | |
| Jul 31, 2018 at 23:32 | comment | added | Asaf Karagila♦ | Unless, of course, the obvious solution works. But for that one has to sit down and think about it. And it's mighty late here for doing just that now. | |
| Jul 31, 2018 at 23:31 | comment | added | Asaf Karagila♦ | While your questions and conjectures are very interesting, I feel that somehow people are often too focused on these families of choice principles, and that's somehow limiting. PP and KWP (Kinna–Wagner Principles) seem to be much more pertinent to the generic multiverse. While they are not as famous, PP does imply DC (for example), so if it is indeed the case that PP is preserved in generic extensions, then countable choice would be preserved too and it would serve as a counterexample. Or a proof that PP implies full choice. In either case, it's a hard problem to tackle. [...] | |
| Jul 31, 2018 at 23:28 | comment | added | Asaf Karagila♦ | This is a tough question. Looking back after seven years, I can now say that I feel that the main reason that AC is preserved under generic extensions is that AC is stating that surjections admits inverses. But in reality this amounts to injections rather than inverses. So the Partition Principle should be enough for that. The problem, however, is that the relation between full choice and the Partition Principle is one of the toughest nuts to crack in choiceless set theory. [...] | |
| Jul 31, 2018 at 22:19 | comment | added | Elliot Glazer | $X$ is the family of sets I want a choice function on, not $P(X).$ | |
| Jul 31, 2018 at 22:18 | comment | added | Elliot Glazer | I mean set-generic but an answer for class-generic would be nice as well. | |
| Jul 31, 2018 at 21:51 | comment | added | Goldstern | I assume that "generic extension" means "set-generic". | |
| Jul 31, 2018 at 21:50 | comment | added | Goldstern | What do you mean by "collapse the cardinality of a set $X$ to $\omega$ without adding a choice function"? If you add a bijection between $X$ and $\omega$, you certainly add a choice function from $P(X)\setminus \{0\}$ to $X$, | |
| Jul 31, 2018 at 19:52 | comment | added | Morteza Azad | This somehow reminds me of Solovay's theorem stating that for every forcing notion $\mathbb{P}$, if a set $a$ exists in every generic extension by $\mathbb{P}$ it must already exist in the ground model. | |
| Jul 31, 2018 at 18:08 | comment | added | Elliot Glazer | I am including trivial forcing. I don't see how that gives a positive answer. I'm only assuming the generic extensions (including $V$) satisfy weak choice principles. | |
| Jul 31, 2018 at 18:04 | comment | added | Stamatis Dimopoulos | I suppose it should be all non-trivial extensions, since if trivial forcing is included the answer is always yes. In any case, I like the question. | |
| Jul 31, 2018 at 17:58 | history | asked | Elliot Glazer | CC BY-SA 4.0 |