If we suppose $ZFC + \exists \text { transitive standard model} (M,\in) \text{of } ZFC$ and then we find a model $(N,\in_{N})$ of ZFC that is a proper end elementary extension of M and contain a set $(N',\in_{N'}=\in_{N}\cap(N'\times N'))\in N$ such that $(N',\in_{N'})\approx(M,\in)$, then is a model of $ZFC^{N'}$ , we obtain contradiction with the second Godel theorem?

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    $\begingroup$ No, we do not. Obviously. $\endgroup$ May 22, 2017 at 12:48
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    $\begingroup$ If you are only looking at $\omega $-models, then $ZFC^{N'}$ is just ZFC. Anyway, if this were not the case, it would not make sense to ask for a model of $ZFC^{N'} $, unless the question is to take place inside $N'$ or a related model, but this then needs to be made explicit in the body of the question. $\endgroup$ May 22, 2017 at 13:35
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    $\begingroup$ In any case, you may even have standard ordinals $\alpha <\beta $ such that $(L_\alpha,\in)\prec (L_\beta,\in)\models\mathsf {ZFC} $. I imagine the question you are really after is whether you can prove in $\mathsf{ZFC}+$"there is a transitive model of $\mathsf{ZFC}$" that there is some transitive model $M $ such that there is such an $N $ as indicated. Is this closer to what you are after? $\endgroup$ May 22, 2017 at 13:42
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    $\begingroup$ No, let's not.${}$ $\endgroup$ May 22, 2017 at 13:47
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    $\begingroup$ You should probably edit the question making the new version explicit. $\endgroup$ May 22, 2017 at 14:03

1 Answer 1


We have to be careful here about what theory is proving what statement about what kinds of models.

So let's start small. ZFC does not prove the statement "there is a model of ZFC" (let alone "there is a transitive model of ZFC" or etc.). This does not prevent some models of ZFC from satisfying "there is a model of ZFC": ZFC proves $\varphi$ iff $\varphi$ is true in every model of ZFC, so all that we know is that some models of ZFC satisfy "there is a model of ZFC."

Now suppose $M$ is a model of ZFC+"There is a transitive model of ZFC" - that is, there is some $N\in M$ such that $M$ thinks "$N$ is a transitive model of ZFC." Then we do not know whether, in $M$, $N$ is contained in some larger transitive model (that is, whether there is an $N'\in M$ with $N\subseteq N'$ and $N'$ is a transitive model of ZFC, in the sense of $M$). This is because such an $N'$ satisfies ZFC+"There is a transitive model of ZFC," so if this were always true then the completeness theorem would tell us ZFC+"There is a transitive model of ZFC" proves Con(ZFC+"There is a transitive model of ZFC"), contradicting Godel (that is, assuming ZFC+"There is a transitive model of ZFC" is in fact consistent).

Now if we think about the paragraph above, we'll quickly see that the word "transitive" isn't really relevant. By the same argument, there is (assuming ZFC+Con(ZFC) is consistent) a model $M$ of ZFC+Con(ZFC) containing a model $N$ of ZFC, such that $M$ does not contain any $N'$ containing $N$ and satisfying ZFC. And again, this doesn't prevent other models of ZFC from exhibiting the opposite behavior, e.g. the extreme property "every set is contained in a transitive model of ZFC."

So the answer to your question as posed - "... we obtain contradiction with the second Godel theorem?" is no: the existence of a model of a theory with certain properties doesn't imply that every model of that theory has certain properties, so being able to in some instances build such an extension doesn't contradict Godel at all, even though those models will satisfy strong consistency properties; conversely, given any procedure you can think of for extending models of ZFC, there will be models of ZFC+Con(ZFC) where that procedure doesn't work. The procedure you have in mind will probably work under further assumptions, which of course means that we do not have a contradiction with Godel after all.


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