1
$\begingroup$

The question of whether the minimal transitive model $M$ of $\sf ZFC$ have a model of each consistent extension $\sf T$ of $\sf ZF$, is answered to the negative, basically because $M$ is countable and we have $2^\omega$ many theories $\sf T$ extending $\sf ZF$.

Now I have three responses to that:

Can $M$ have a model of each consistent parameter free definable theory $\sf T$ that extends $\sf ZF$, among its elements? I know that this is true for all such theories if they are elements of $M$ (see answer)! Here the question includes even those definable outside of $M$?

Can every first order theory that consistently extend $\sf ZF$ by finitely many axioms, have a model in $M$?

Can every theory that consistently extends $\sf ZF$, have a model in $\mathcal P(M)$?

$\endgroup$

1 Answer 1

2
$\begingroup$

The answer to question 1 is no, because the theory of the minimal transitive model itself is parameter-free definable as that theory, and in the other answer I explained why this theory is not an element of the minimal model. More generally, one cannot in general make much of a conclusion about an object from it being definable, becauase in the light of the universal definition, any object can in principle be made definable in a forcing extension.

The answer to the second question is yes. Because the minimal transitive model of ZFC is transitive, it has the true $\omega$ and thus it has the true ZFC and all finite extensions, and it agrees on consistency statements since it has all the same proofs that we do.

I don't understand the third question (or your remarks about subsets on the other post). I'm not sure about what kind of models you intend---do you intend submodels of the minimal model $\langle L_\alpha,\in\rangle$, that is, with the same $\in$ relation? All such models would be well-founded and thus would have transitive collapses either to a set in $L_\alpha$ or to $L_\alpha$ itself.

$\endgroup$
7
  • $\begingroup$ Thanks! Yes, the third one is about submodels with the same $\in$ relation . I like the second possibility, so there are models of "$\sf ZFC$+ there exists a transitive model of $\sf ZFC$" that are elements of $M$, but of course those would not be transitive since $M$ is minimal. Also, regarding $\sf GBC$ it is finitely axiomatized, so all consistent finite extentions of it would have models in the minimal transitive model of $\sf GBC$, is that correct? $\endgroup$ Commented Jan 2, 2023 at 15:45
  • $\begingroup$ Yes, that is correct. $\endgroup$ Commented Jan 2, 2023 at 16:05
  • $\begingroup$ Thank you very much! $\endgroup$ Commented Jan 2, 2023 at 16:58
  • 1
    $\begingroup$ The theory of the minimal transitive model of ZFC is complete! So it is not recursively enumerable, and so cannot be recursively axiomatized. Can it be the case that all recursively axiomatized first order theories extending $\sf ZFC $ have models (albeit non-standard) in the minimal transitive model of $\sf ZFC$? I've just came to know that there are countably many such theories, and I guess that all such theories can be conservatively extended into finite extensions of $\sf GB$. $\endgroup$ Commented Jan 4, 2023 at 8:49
  • 1
    $\begingroup$ Yes, because the programs for enumerating those theories are in the minimal model, and it can enumerate them, and it will agree that they are consistent (if they are) and thus build models for them by the Henkin construction. The same argument applies to arithmetically definable theories, not just c.e. For example, you could use the halting problem as an oracle or whatever, and this would still be absolute to the minimal model. $\endgroup$ Commented Jan 4, 2023 at 12:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .