Timeline for Forcing the existence of a weakly inaccessible cardinal in some strong set theory
Current License: CC BY-SA 4.0
25 events
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| Sep 16, 2022 at 5:19 | history | edited | user21820 | CC BY-SA 4.0 |
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| Jun 2, 2018 at 14:26 | comment | added | Thomas Benjamin | @NoahSchweber: I should have said, "the 'model' of $ZFC$ + $I$ in question". Sorry. | |
| Jun 2, 2018 at 14:14 | vote | accept | Thomas Benjamin | ||
| Jun 1, 2018 at 22:22 | comment | added | Noah Schweber | But this is not special; one merely assumes the consistency of ZFC, then infers that ZFC has a model, in order to study models of ZFC. Etc., etc. | |
| Jun 1, 2018 at 21:25 | comment | added | Noah Schweber | @ThomasBenjamin Yes - or if you prefer, one proves conditional results ("If ZFC+I is consistent, then ..."). Incidentally, the word "the" is inappropriate: if ZFC+I has a single model, it has many very different models. | |
| Jun 1, 2018 at 19:21 | comment | added | Thomas Benjamin | @NoahSchweber: So then does one merely assume the consistency of $ZFC$ + $I$, then (via the completeness theorem) infer $ZFC$ + $I$ has a model in order to study the 'model' of $ZFC$ + $I$? | |
| Jun 1, 2018 at 19:09 | comment | added | Noah Schweber | And in terms of the computability-theoretic complexity of models of ZFC+I - which is a totally different question - the answer is also boring: like any other recursively axiomatizable theory, if it has any models at all it has models of PA degree, and in fact it's easy to show that ZFC+I has models only of PA degree. But this isn't odd: tons of theories have this property, including ZFC itself (and much weaker). | |
| Jun 1, 2018 at 19:07 | comment | added | Noah Schweber | @ThomasBenjamin Exactly the consistency of ZFC+I. That is precisely what Godel's completeness theorem says. You're not going to get it for cheaper. | |
| Jun 1, 2018 at 19:06 | comment | added | Thomas Benjamin | @NoahSchweber: Indeed. What hypotheses would one need to produce a model of $ZFC$ + $I$? | |
| Jun 1, 2018 at 19:03 | comment | added | Noah Schweber | @ThomasBenjamin From what hypotheses? Without that context, the question is meaningless. Can one in fact produce a model of ZFC? Can one in fact produce a model of ATR$_0$? Can one in fact produce a model of PA? | |
| Jun 1, 2018 at 19:02 | comment | added | Thomas Benjamin | (cont.) does $ZFC$ + "There is an inaccessible cardinal". Can one in fact produce a model of $ZFC$ + $I$? If so, how? | |
| Jun 1, 2018 at 19:00 | comment | added | Noah Schweber | A true statement: "ZFC+Con(ZFC) cannot prove that ZFC+I has a model." A false statement: "From this, we can deduce that ZFC+I has no models." A true statement: "From this, we can deduce that there are models of ZFC+Con(ZFC) in which ZFC+I has no models." | |
| Jun 1, 2018 at 18:59 | comment | added | Noah Schweber | @ThomasBenjamin Jech is speaking within ZFC: ZFC cannot prove the consistency of ZFC + I, even with the added assumption of Con(ZFC). However, this in no way implies (or even suggests) that ZFC+I is inconsistent. I really don't know what you're doing here. | |
| Jun 1, 2018 at 18:57 | comment | added | Thomas Benjamin | @NoahSchweber: Of course there are other ways to build models rather than forcing. Consider, however, Jech's Theorem 12.12[b] in Set Theory: Third Millenium Edition ", "Moreover, it cannot be shown that the existence of inaccessible cardinals is consistent with $ZFC$.", where "consistent with $ZFC$" means (for Jech) "if $ZFC$ is consistent, then so is $ZFC$ + $I$ where $I$ is the statement 'there is an inaccessible cardinal' ". If one translates Jech's consistency statements into the language of models (via Goel's completeness theorem), one has "if $ZFC$ has a model, then so | |
| May 31, 2018 at 17:51 | comment | added | Noah Schweber | (Sorry, I screwed up a negative: the answer to "Would I be incorrect in ..." is yes.) | |
| May 31, 2018 at 17:50 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| May 31, 2018 at 16:34 | comment | added | Noah Schweber | Precisely: if ZFC + "There is an inaccessible cardinal" is consistent, then it has some model $M$; however, this model is not a forcing extension of any model of ZFC + "There is no inaccessible cardinal." There's no tension here, since in general why should we expect a model of a theory $T_0$ to automatically have to be a forcing extension of a model of a different theory $T_1$? | |
| May 31, 2018 at 16:31 | comment | added | Noah Schweber | @ThomasBenjamin "Would I be incorrect in inferring from this answer that there can be no models of ZFC + "There exists an inaccessible cardinal"?" Are you sure you meant what you intended? The answer is absolutely no - the key point being that forcing is not the only way to build models. For your last sentence, I think your "existence" should be "nonexistence," and the answer is: precisely the inconsistency of ZFC + "there exists an inaccessible cardinal." This is an instance of the completeness theorem. | |
| May 31, 2018 at 16:00 | comment | added | Thomas Benjamin | @NoahSchweber: Thanks for this very interesting answer! Would I be incorrect in inferring from this answer that there can be no models of $ZFC$ + "There exists an inaccessible cardinal"? If this inference would, in fact, be incorrect, what assumptions would be necessary to infer the existence of a model of $ZFC$ + "there exists an inaccessible cardinal (other than assumptions of inconsistency)? | |
| May 15, 2018 at 0:35 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| May 14, 2018 at 22:30 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| May 14, 2018 at 22:29 | comment | added | Noah Schweber | @AndreasBlass A fair point, I slipped up there. Fixed! | |
| May 14, 2018 at 21:33 | comment | added | Andreas Blass | I think that (*) (and your reformulation of it) should begin with "Assuming ZFC + Con(ZFC) is consistent, ...." I don't see how to get this result assuming only the consistency of ZFC. | |
| May 14, 2018 at 19:23 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| May 14, 2018 at 18:46 | history | answered | Noah Schweber | CC BY-SA 4.0 |