Timeline for Kunen's use of Countable Transitive Models
Current License: CC BY-SA 2.5
19 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Apr 12, 2015 at 5:24 | comment | added | Transcendental | All is not lost, however. We just work inside $ \mathcal{M} $ to construct a countable and transitive generic extension $ {c^{\mathcal{M}}}[G] $ of $ c^{\mathcal{M}} $, where $ G \in \mathcal{M} $ is a generic filter of some $ E^{\mathcal{M}} $-poset $ \mathbb{P} $ that is $ E^{\mathcal{M}} $-contained in $ c^{\mathcal{M}} $. As $ c^{\mathcal{M}} $ models $ \mathsf{ZFC} $, this can be carried out (only finitely many axioms are needed, but how many, we do not care). We therefore have a model $ {c^{\mathcal{M}}}[G] $ of $ \mathsf{ZFC} $ that models/refutes specific set-theoretic principles. | |
| Apr 11, 2015 at 18:16 | comment | added | Transcendental | Hence, if $ \mathcal{M} $ is a model of $ T $, then both $ \mathcal{M} $ and $ c^{\mathcal{M}} $ are models of $ \mathsf{ZFC} $. Certainly, as $ c^{\mathcal{M}} $ satisfies the Axiom of Regularity, it knows that it is well-founded. However, nothing guarantees that the restriction of the relation $ E^{\mathcal{M}} $ (on $ \mathcal{M} $) to the collection of $ E^{\mathcal{M}} $-members of $ c^{\mathcal{M}} $ is well-founded externally, so we cannot apply the Mostowski Collapsing Lemma. | |
| Apr 11, 2015 at 17:40 | comment | added | Transcendental | Here is another subtlety. Let $ K $ be the signature consisting of a single constant symbol ‘$ c $’ and a single binary-relation symbol ‘$ E $’ to mimic set-membership. Let $ T $ be the first-order theory, based on the signature $ K $, consisting of $ \mathsf{ZFC} $ and also the following sentences: (i) ‘$ c $ is countable and transitive’, (ii) ‘$ \phi^{c} $’ for each $ \phi \in \mathsf{ZFC} $. If $ \mathsf{ZFC} $ is consistent, then by the Reflection Principle, every finite fragment of $ T $ is also consistent. The Compactness Theorem therefore says that $ T $ is consistent as well. | |
| Jun 18, 2011 at 9:46 | answer | added | Monroe Eskew | timeline score: 6 | |
| Mar 3, 2011 at 23:07 | vote | accept | David Fernandez-Breton | ||
| Mar 3, 2011 at 23:04 | comment | added | David Fernandez-Breton | @Stefan: of course, when I wrote $(M,\in)$ I really meant $(M,E)$ where $E$ is to be the interpretation of $\in$ inside the model $M$... now I see, since $M$ is countable, it surely has some subsets $A$ that are not elements of $M$, and one such $A$ could be an infinite $E$-descending sequence, so we cannot apply Mostowski... I guess that comment is like half (or more) of the answer to my question!!! This is an important point that I wasn't aware of before. | |
| Mar 3, 2011 at 20:50 | history | edited | Jason |
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| Mar 3, 2011 at 20:44 | comment | added | Stefan Geschke | David, the problem is that if ZFC is consistent, then the completeness theorem gives you a model of the form $(M,E)$, where $M$ is a set and $E$ is a binary relation on that set. Typically the binary relation $E$ is not the usual $\in$. While $\in$ is wellfounded by the axiom of regularity in the real world, $(M,E)$ satisfies the axiom of regularity, but still $E$ can be illfounded in the real world and $M$ just doesn't know an infinite $E$-descending sequence. In this case $(M,E)$ is not isomorphic to a structure of the form $(X,\in)$. | |
| Mar 3, 2011 at 20:36 | answer | added | Jason | timeline score: 6 | |
| Mar 3, 2011 at 19:29 | comment | added | David Fernandez-Breton | Now that I see oktan's answer, maybe I start understanding the thing... so probably the fact that $(M,\in)$ satisfies the Foundation axiom doesn't necessarily imply that the relation $\in$ is really well-founded!!! (although I still need to assimilate this) | |
| Mar 3, 2011 at 19:21 | comment | added | David Fernandez-Breton | @Michael: I know, Mostowski collapsing lemma requires the relation to be extensional, set-like and well-founded. But if $(M,\in)$ is a set-model of ZFC, then $\in$ satisfy those things because $(M,\in)$ must satisfy, among others, the extensionality and foundation axioms. What's the point that I'm missing? (I'm really sure there's some such point) | |
| Mar 3, 2011 at 18:34 | comment | added | Stefan Geschke | Now, what the way that forcing in presented by Kunen actually shows is that given any fixed finite fragment $\Phi$ of ZFC+$\neg$CH, every model of ZFC contains a model of $\Phi$. It follows that the consistency of ZFC implies the consistency of ZFC+$\neg$CH. | |
| Mar 3, 2011 at 18:26 | comment | added | Stefan Geschke | Carl, what is the problem with assuming the consistency of ZFC? A typical consistency proof shows a statement like "Con(ZFC) implies $\neg$Con(ZFC+CH)". Now such a statement would have to be proved assuming something, typically ZFC itself, since we believe that a true mathematical statement is one that is provable in ZFC. Goedels proof of the consistency of CH does exactly that: Start with a model of ZF and produce from it a model of ZFC+CH. Unfortunately with forcing there are some technical problems with such an approach, and one way around it is to force over c.t.m.s. | |
| Mar 3, 2011 at 12:46 | comment | added | Carl Mummert | It is true that if you're going to study ZFC then you're likely to assume (in your head) that ZFC is consistent. But there are technical problems getting the theory itself to believe it's consistent. If you add Con(ZFC) as an axiom, you get a new theory ZFC'; by the incompleteness theorem ZFC' cannot prove Con(ZFC'), so you can't make models of ZFC' in ZFC'. The easiest solution is to just stick with ZFC, and remember it doesn't prove Con(ZFC). A good way to think about forcing while you learn it is to pretend you're actually working in a model of ZFC' but constructing models of just ZFC. | |
| Mar 3, 2011 at 11:16 | comment | added | Emil Jeřábek | While it is not the most important reason (see oktan's answer), it's perfectly possible that ZFC is consistent, and yet it proves its own inconsistency. | |
| Mar 3, 2011 at 11:08 | answer | added | Stefan Hoffelner | timeline score: 10 | |
| Mar 3, 2011 at 10:34 | comment | added | Michael Greinecker | I don't understand your point. A model of ZFC is just a set with some relation on it. That is not enough to apply the Mostowski collapsing lemma. For this you need a special kind of model and Löwenheim-Skolem doesn't guarantee you the existence of a countable such model. | |
| Mar 3, 2011 at 9:00 | history | asked | David Fernandez-Breton | CC BY-SA 2.5 |