Timeline for Is there a universal way to force the Axiom of Choice to be true?
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27 events
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| Mar 27, 2018 at 21:31 | answer | added | Joel David Hamkins | timeline score: 4 | |
| Mar 25, 2018 at 13:43 | history | edited | Oscar Cunningham | CC BY-SA 3.0 |
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| Mar 25, 2018 at 13:34 | comment | added | Oscar Cunningham | @AsafKaragila That's a good observation. If an adjunction existed there would be a unit (or counit) morphism between each model of ZF and the corresponding model of ZFC. Since this can't exist you've answered four parts of my twelve part question! (left or right adjoint $\times$ two kinds of object $\times$ three kinds of morphism) | |
| Mar 25, 2018 at 13:10 | comment | added | Asaf Karagila♦ | Well, the reason I ask is that an elementary embedding does not exist between models with different theories. So no model of ZFC embeds elementary into a model of non-AC, and vice versa. So your inclusion functor would be sort of degenerate. | |
| Mar 25, 2018 at 13:02 | comment | added | Oscar Cunningham | @AsafKaragila I'm afraid I don't know exactly what you mean by "inclusion". From what I've found it seems that logicians sometimes consider various different notions of maps between set theories that are stronger than mere $\in$-homomorphisms but weaker than elementary embeddings. An answer in terms of any of these would be interesting, but from a category theoretic point of view you would want their properties to be strong enough to induce a functor between the corresponding toposes. So the image of a function between sets should be a function between the images of those sets. | |
| Mar 25, 2018 at 12:54 | comment | added | Oscar Cunningham | @AsafKaragila Any model of ZF corresponds to a category whose objects are the sets and whose morphisms are the functions between sets defined in that model. This category is a topos, so a logical functor between models is just a logical functor between the corresponding toposes. I don't know if this has a good characterisations when thinking in terms of the membership relations. | |
| Mar 25, 2018 at 11:24 | comment | added | Asaf Karagila♦ | Just out of curiosity, what are "logical functors" in the set theoretic context? Why isn't just inclusion one of your functors? | |
| Mar 23, 2018 at 17:36 | answer | added | Rachid Atmai | timeline score: 4 | |
| Mar 23, 2018 at 12:14 | comment | added | Asaf Karagila♦ | @Mike: It would be debatable whether or not just SVC means "some choice", because choice can still fail in pretty bad ways while SVC holds. And I don't know about "forcing some choice" either. Of course, you can always add a choice function to some specific family of sets, but it's not really adding "some choice" to the universe, because "some choice" sounds like "a weak choice principle holds" rather than "this specific family of sets admits a choice function". | |
| Mar 23, 2018 at 11:48 | comment | added | Mike Shulman | @AsafKaragila What's wrong with what I said? | |
| Mar 23, 2018 at 11:05 | comment | added | Andrés E. Caicedo | (Changed your notation for $L$ to a more standard term than $L (V) $. The usual meaning of the latter is just $V $.) | |
| Mar 23, 2018 at 11:03 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
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| Mar 23, 2018 at 10:22 | history | edited | Oscar Cunningham | CC BY-SA 3.0 |
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| Mar 23, 2018 at 8:32 | comment | added | Asaf Karagila♦ | @Mike: Well, I wouldn't say it this way. Rather if choice fails because of one set, we can force it back. But if choice fails because of a proper class of sets, then it's impossible to fix. | |
| Mar 22, 2018 at 23:59 | comment | added | Mike Shulman | Thanks Joel and Asaf -- so the answer to my question is that if you already have some choice, then you may be able to force the rest of it, whereas if you don't have any choice then you can force some choice but not all of it. Interesting! | |
| Mar 22, 2018 at 22:51 | comment | added | Asaf Karagila♦ | @Ali: No. I wanted to subsume these results, and hopefully someday this will happen. But I did not do that. I do believe that the framework I developed in my thesis will be useful for that, though. | |
| Mar 22, 2018 at 22:51 | comment | added | Asaf Karagila♦ | @Joel: Yes, that means 'set forcing', although it is possible that even class forcing cannot do it. Gitik's model is a standard example. | |
| Mar 22, 2018 at 21:52 | comment | added | Ali Enayat | @AsafKaragila: Hi Asaf, does the results in your thesis subsume the work done by Morris in his 1970 thesis at the University of Wisconsin? To my knowledge he was the first to show that there are models of ZF that are not extendable to a model of ZFC with the same class of ordinals (cf. problem 14 on p.80 of Jech's book "The Axiom of Choice"). | |
| Mar 22, 2018 at 20:55 | comment | added | Joel David Hamkins | Asaf, you mean "AC can be forced" by set forcing? I can imagine situations where AC cannot be forced by set forcing, but ZFC can be forced by class forcing. | |
| Mar 22, 2018 at 19:30 | comment | added | Asaf Karagila♦ | @Mike: Blass in his paper "injectivity projectivity and the axiom of choice" defined an axiom SVC, small violations of choice, which is equivalent to the statement "AC can be forced". This holds in all symmetric models, or models of the form L(X) or HOD(X) or generally V(X) when V a model of ZFC. But it does not hold in general for models of ZF, of course. Be it via class symmetric extensions of certain kinds, or just by assuming certain axioms hold and/or fail (e.g., in my thesis I show that the failure of KWP is consistent, and it implies the failure of SVC). | |
| Mar 22, 2018 at 17:48 | comment | added | Joel David Hamkins | Concerning the lack of bi-interpretation between set theories, the situation is that the theories are mutually interpretable, so in any model of ZF we can interpret a model of ZFC and vice versa. What is lacking is sufficient uniformity in the interpretations, and this prevents a bi-interpretation, provably so. Since it seems that many universality properties in category theory are aimed at precisely this kind of uniformity, it seems likely to me that the result will correspond to a failure of universality of some kind that is naturally expressed in that category-theoretic language. | |
| Mar 22, 2018 at 16:56 | comment | added | Joel David Hamkins | Does the Gitik model rule out any of the adjoint situations envisioned by the OP? The Gitik model is a ZF model in which every $\aleph$ is singular, and so it has no outer models with the same ordinals that satisfy the axiom of choice. | |
| Mar 22, 2018 at 16:49 | history | edited | Oscar Cunningham | CC BY-SA 3.0 |
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| Mar 22, 2018 at 16:40 | comment | added | Joel David Hamkins | I wonder whether the question of the OP can be answered by means of the fact that ZF and ZFC are not strictly bi-interpretable. Indeed, one can prove that different set theories extending ZF are never bi-interpretable. I wonder what is the topos-theoretic version of this result? | |
| Mar 22, 2018 at 16:38 | comment | added | Joel David Hamkins | @MikeShulman One can force any given set to be well-orderable (or even countable, an extreme case of this), and this means that in many cases, one can force AC. For example, over any model of the form $L(\mathbb{R})$, one can force AC, essentially by adding a generic well-ordering of the reals. If DC holds, one can do this without adding reals. Meanwhile, there are models of ZF, such as the Gitik model where every $\aleph_\alpha$ is singular, where ZFC fails in every extension with the same ordinals, and so in such a case, there can be no forcing extension with ZFC. | |
| Mar 22, 2018 at 16:12 | comment | added | Mike Shulman | Is there really a way to obtain the axiom of choice using forcing? I've never seen such a construction. | |
| Mar 22, 2018 at 15:24 | history | asked | Oscar Cunningham | CC BY-SA 3.0 |