Timeline for Taking a proper class as a model for Set Theory
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2018 at 16:45 | vote | accept | Elie Ben-Shlomo | ||
| Feb 16, 2018 at 10:47 | comment | added | Simon Henry | Thanks ! your blog post really made things clearer to me. I was actually wondering for some time what was the precise technical point that made GBC and KM different regarding the consistency of ZFC. Now I know. | |
| Feb 15, 2018 at 23:09 | comment | added | Joel David Hamkins | Meanwhile, every model of PA does have a definable model of PA plus not Con(PA), since one can use the Henkin model of the left most branch through the tree of attempts to build a complete consistent Henkin theory. | |
| Feb 15, 2018 at 23:07 | comment | added | Joel David Hamkins | You are objecting to something I am not asserting. Your stronger assertion amounts to the same as what I call the scheme of assertions that L satisfies every individual axiom of ZFC. Of course this is stronger than the arithmetic implication that Con(ZF) implies Con( ZFC). I think we agree about this. | |
| Feb 15, 2018 at 22:41 | comment | added | Asaf Karagila♦ | Joel, I disagree with your claim that this is not a stronger result. From a model of PA you get a model of PA+not Con(PA). But there is no "reasonable way" to define that model within the natural numbers. There is no internal way through which PA proves that every finite part of PA is consistent with not Con(PA). The situation with ZF and ZFC is different (although admittedly the situation is different because both AC and its negation are equiconsistent over ZF). | |
| Feb 15, 2018 at 21:34 | comment | added | Joel David Hamkins | But in ZFC or GBC we are only ensured truth predicates for set models, not proper class models. Meanwhile, in Kelley-Morse set theory, we do have first-order truth predicates for the class models, and thus the argument about L does show Con(ZFC) in KM. See jdh.hamkins.org/km-implies-conzfc. | |
| Feb 15, 2018 at 21:34 | comment | added | Joel David Hamkins | @SimonHenry There is a little more to it than that, since after all, GBC is finitely axiomatizable, but we don't say that GBC implies Con(GBC) just because in GBC we can prove that the universe itself is a model of GBC. The point is that the argument that exists-a-model implies consistency requires having a satisfaction class or truth predicate for the model (since one proves by induction on proofs that the proofs are truth-preserving with respect to the satisfaction class, and contradictions are never true with respect to a truth predicate). | |
| Feb 15, 2018 at 21:28 | comment | added | Simon Henry | Just to check that I understand correctly: the key point here is that ZFC is not finitely axiomatizable right ? It means that checking in the meta theory that each axiom is satisfied is different from checking internally that " for all axiom in the recursively enumerated sets of axioms, L satisfies it " ? Or is there already a distinction to be made for a single axiom or a finitely axiomatizable theory ? | |
| Feb 15, 2018 at 18:39 | comment | added | Joel David Hamkins | @ElieBen-Shlomo I don't view Jech's argument as unrigorous, since I interpret him to be saying the same as what I said. | |
| Feb 15, 2018 at 18:38 | comment | added | Joel David Hamkins | @AsafKaragila Well, of course the set models are closely related, in that the ZFC model is the $L$ of the ZF model, so whatever extra you are getting is gotten that way, because of the closeness of the models. I don't agree that your way of talking about it is stronger, since you are talking about a scheme of proofs, and this takes place in the meta-theory. | |
| Feb 15, 2018 at 18:37 | comment | added | Asaf Karagila♦ | @Elie: Probably the old Kunen book; you might to check Felgner's Models of ZF set theory; and since judging by your name there's a good chance you can read Hebrew, you might want to see if Matti Rubin's notes on axiomatic set theory from the early 2000s are still online. | |
| Feb 15, 2018 at 18:28 | comment | added | Elie Ben-Shlomo | So, are you saying that Jech was being slightly unrigorous in his proof of Con(ZF) implies Con(ZFC)? If so, do you have any references to where I can find such a rigorous proof? | |
| Feb 15, 2018 at 18:27 | comment | added | Asaf Karagila♦ | I somewhat disagree with your last paragraph. The point of the proof is that you get a stronger result. You don't just assume there is a model of ZF and produce a model of ZFC. You actually prove a stronger result, that ZF itself proves any finite number of axioms of ZFC hold in L. Of course, that requires a meta-theoretic jump to conclude from Con(ZF) to Con(ZFC), but it is in fact more than saying "Assume there is a set model, produce another set model". | |
| Feb 15, 2018 at 18:20 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |