A standard model of ZF need not be transitive, of course, and Joel David Hamkins' answer to Large cardinal axioms and Grothendieck universes gives Tarski sets as an interesting example.

I should clarify since the terminology is not entirely standard. By a standard model I mean what Wikipedia does, and what Joel David Hamkins does in his answer to Standard model of ZFC: the set membership relation $E$ of the model is the actual set membership relation $\in$, restricted to sets in $M$.

Does existence of a standard model imply existence of a transitive one? Existence of a Tarski set does imply existence of a Grothendieck universe. But am I right to suspect that if there is a standard model, then the ordinals in the minimal model are not a true initial segment of the ordinals, so the minimal model is not transitive, and we can cut down to where that minimal model is the only standard one? If I understand Guest289 correctly, his argument shows my guess is wrong, since a minimal model is transitive. Do I understand that correctly? Or does this come back to unclarity abut what is a standard model?

Anyway, does existence of a transitive model have higher consistency strength than existence of a standard model? I would not be surprised if the axiom of choice lays a role here but I do not know if it does.


What precisely do you mean by a standard model? An $\omega$-model? (That is, a model whose set of natural numbers is isomorphic to $\omega$.) Or a $\beta$-model? (That is, a model whose ordinals are well-ordered.)

If the latter, the Mostowski collapse theorem tells us any such model is isomorphic in a unique way to a unique transitive model. If the former, the existence of an $\omega$-model has consistency strength strictly weaker than the existence of a transitive one. (This follows, for instance, from absoluteness considerations: Any transitive model of $\mathsf{ZF}$ has as elements some $\omega$-models.)

It is true that there are $\beta$-models whose ordinals do not form an initial segment of the true ordinals, even if the membership relation of the model is true membership. For instance, if $V_\alpha$ is a model of set theory, consider any of its countable elementary substructures.

Note that models whose membership relation is true membership are $\beta$-models, so they are isomorphic, via the collapse, to transitive models. The point of the Mostowski collapse is that there is a natural rank that we can assign to the elements of a standard model $M$, so that if $M\models x\in y$ then the rank of $x$ is strictly smaller than the rank of $y$. (If the membership relation of $M$ is true membership, we can use as rank the true set-theoretic rank of the sets in question.) A straightforward transfinite recursion then allows us to replace the elements of $M$ by copies that form a transitive model, simply by defining $\pi:M\to V$ by $\pi(y)=\{\pi(x)∣M\models x\in y\}$. Choice plays no role here.

  • $\begingroup$ In my experience, the two most common meanings of "standard model" are "model with standard $\in$-relation" and "transitive model with the standard $\in$-relation." In particular it's stronger than $\beta$-model, but equivalent up to isomorphism. $\endgroup$ – Andreas Blass Dec 27 '14 at 16:06
  • $\begingroup$ @Andreas Merry Christmas, Andreas! Yes, I agree. (I personally feel the term should mean that the model is transitive.) But I feel that the extra generality of addressing $\beta$-models rather than just models with true membership is useful, as these models appear for instance in fine structure (via mastercodes). $\endgroup$ – Andrés E. Caicedo Dec 27 '14 at 16:31
  • $\begingroup$ @AndresCaicedo Compared to proof theory, I tend to think of set theory as having all settled terminology. I have clarified in the question. $\endgroup$ – Colin McLarty Dec 27 '14 at 16:51

Every standard model is well founded so you can take its transitive collapse to end up with a transitive model.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.