So if minimal model of ZF exists, it is said that it is countable set by Lowenheim-Skolem. So, is Lowenheim-Skolem saying that for any countable theory with existence of infinite model there exists standard model, respecting normal element relation, that is countable infinite?


No. The existence of standard models is strictly stronger.

It is consistent that there are no standard models, to see this note that the standard models are well-founded, in the sense that there is no infinite decreasing chain of standard models such that $M_{n+1}\in M_n$, simply because $\in$ itself is well-founded and standard models use the real $\in$ for their membership relation.

So there is a minimal standard model. But this model has the standard $\omega$ for its integers, so it cannot possible satisfy $\lnot\text{Con}(\mathsf{ZFC})$, so it must have a model of $\sf ZFC$ inside, but this model cannot be standard.

See also: Transitive ${\sf ZFC}$ model on Cantor's Attic.

  • $\begingroup$ OK, then how can L be minimal and standard both? $\endgroup$ – user61182 Feb 21 '13 at 12:45
  • $\begingroup$ @user81182: $L$ is minimal among class models that contain all the ordinals. The minimal standard model does not contain all the ordinals and is of the form $L_\alpha$ for some countable ordinal $\alpha$. $\endgroup$ – Carl Mummert Feb 21 '13 at 12:48
  • $\begingroup$ @user61182: I'm not sure I understand your question. If there is a standard model then there is a minimal one. Since every model has an inner model which satisfies $V=L$, if $M$ is a minimal standard model then $L^M$ is a minimal standard model which satisfies $V=L$. Since $V=L$ is absolute we have that $L^M$ is actually $L_\eta$ (of the "real $L$") for some $\eta$, and this is the least transitive model in the sense of inclusion as well, not only in the sense of membership. $\endgroup$ – Asaf Karagila Feb 21 '13 at 12:48
  • $\begingroup$ @Carl: You might want to post a more elaborated answer, I am just leaving for office hours and won't have time to do this for a couple of hours. :-) $\endgroup$ – Asaf Karagila Feb 21 '13 at 12:50
  • $\begingroup$ @CarlMummert Oh got that part. I was mistaken. $\endgroup$ – user61182 Feb 21 '13 at 12:58

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