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I'm studying model theory and i'extremly confused. I know that ZFC has many models (for example the combination : class of all sets + relation $\in$). And if a theory has a model, then it is consistent. But people still doubt about the consistency of ZFC, how can it be possible ? I mean what are the reasons that prevent people from believing that ZFC is consistent while it has already a model ?

I have read this thread but I don't understand very well, and I don't know how to rebound that question, so I decided to create a new one here.

Ref. How can ZFC be inconsistent if it has a model?

Thank you all!

Mauro ALLEGRANZA
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InTheSearchForKnowledge
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  • In order to prove consistency by providing a model we have to descrive the model. For the consistency of set theory A we have to use a different set theory B to describe (in a precise mathematical way) the model of A. Thus, IF we have doubt about the consistency of A, we may also have doubts about the consistency of B. – Mauro ALLEGRANZA Apr 08 '24 at 07:56
  • Conclusion: IF we do not "believe" in e.g. ZF (etc) it is quite useless to use a "stronger" theory to provide a model for ZF. The construction of models for theory is useful fro the mathematical study of theories, not for "unbelievers". – Mauro ALLEGRANZA Apr 08 '24 at 08:01
  • The naive set theory seemed to be absolutely sound. The Russel paradox occured out of nowhere. This is a warning for all people thinking that the consistency of ZFC is beyond any doubt. There is no guarantee that ZFC is consistent therefore it is resaonable to have a slight doubt. But an inconsistency , if detected once , won't break math. We just have to adjust the set theory again and hope again that this time, it is finally consistent. – Peter Apr 08 '24 at 08:13
  • @MauroALLEGRANZA: So in this case, I think that even the combination (Class of all sets, $\in$) is not sure to be a model of ZFC ? Why do people claim that it is a model of ZFC ? Can you please explain it to me ? Many thanks Mauro. – InTheSearchForKnowledge Apr 08 '24 at 08:14
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    Relevant: Does anyone still seriously doubt the consistency of $\mathsf {ZFC}$? – Mauro ALLEGRANZA Apr 08 '24 at 08:23
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    Why can't we prove consistency of $\mathsf {ZFC}$ like we can for $\mathsf {PA}$? – Mauro ALLEGRANZA Apr 08 '24 at 08:24
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    Why is establishing absolute consistency of $\mathsf {ZFC}$ impossible? – Mauro ALLEGRANZA Apr 08 '24 at 08:26
  • Gentzen's proof of the consistency of PA also requires trust in trasnfinite objects and methods. Since I am far away from being an expert in this , it would be interesting to be clarified in which sense this proof is "more convincing". In other words : Has Gentzen proven 100% that PA is consistent ? – Peter Apr 08 '24 at 08:31
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    @Peter - Gentzen's proof does not use models... having said that, some well-known mathematicians has been suspicious; Weyl: "Gentzen proved the consistency of arithmetic, i.e. induction up to the ordinal $\omega$, by means of induction up to $\epsilon_0$." Tarski is similarly reported to have commented that knowledge of the Gentzen proof increased his confidence in the consistency of arithmetic “by an epsilon”. – Mauro ALLEGRANZA Apr 08 '24 at 08:43
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    @InTheSearchForKnowledge Well, believing the class of all sets is a well-defined object is practically the same as believing ZFC is consistent. After all, the class of all sets is not some physical existence that can be physically observed and verified to satisfy ZFC axioms. It is simply assumed to exist, and assumed to satisfy ZFC. And of course if you’re assuming all that, then you’re just assuming ZFC is consistent. And if someone does not believe ZFC is consistent, they certainly will not believe this class of all sets actually exists (or has the properties we usually ascribe to it). – David Gao Apr 08 '24 at 08:49
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    @InTheSearchForKnowledge Now, there are ways to formalize the claim that the class of all sets exists and is a model of ZFC, thereby “proving” ZFC is consistent, but that requires theories even stronger than ZFC, like MK. Of course, if someone does not even believe ZFC is consistent, there’s no reason for them to believe MK is consistent. And if they do not believe MK is consistent, they certainly won’t believe a “proof” that ZFC is consistent that’s done in MK. – David Gao Apr 08 '24 at 08:50
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    @InTheSearchForKnowledge At the beginning of the last century there were hopes that perhaps one can use a weaker theory to prove the consistency of stronger theories. And if the weaker theory used is non-controversial enough, then the consistency of the stronger theories will be in less doubt. But Gödel’s incompleteness theorems basically say that is impossible, that any method formalizable in ZFC cannot prove ZFC is consistent (provided that ZFC is actually consistent), and since basically all mathematics is formalized in ZFC, this makes proving consistency of ZFC practically impossible. – David Gao Apr 08 '24 at 08:56
  • Can ZF prove that a model of ZF exists? I.e. Is there a set model. Obviously, in the theory Cons(ZF) + ZF, ZF is consistant. The interesting question is, can ZF prove its own consistancy? Gödel's work on incompleteness confirms that it cannot. That there exists a model, depends on what Theory one is working with, yes, if we are working with a stronger theory than ZF, we can have a model of ZF, but then how do we know our stronger theory is consistant?? – Michael Carey Apr 08 '24 at 14:15

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