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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.

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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.

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

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