# Can we construct a smaller model of ZFC than transitive minimal model?

Let's say there is countable transitive minimal model of ZFC. Then can we construct a smaller nonstandard model of ZFC? How is this possible?

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I don't see how this question is remotely "elementary set theory". I also removed the model theory tag because this pertains more to classical logic (i.e. completeness/incompleteness) rather than model theoretic methods like ultraproducts, prime models, and so on. – Asaf Karagila Jun 28 '13 at 0:27

Note that transitive models are $\omega$-models, i.e. they agree with the universe on the natural numbers. Since the universe has a model of $\sf ZFC$ it has a proof of the number theoretical statement $\text{Con}(\sf ZFC)$. Therefore if $M$ is any transitive model of $\sf ZFC$, and in particular the minimal one, then $M\models\text{Con}(\sf ZFC)$.
If $L_\alpha$ is the least transitive model then $L_\alpha\models\text{Con}(\sf ZFC)$, so by the completeness theorem $L_\alpha$ knows about a model of $\sf ZFC$, but at the same time this model cannot be transitive (by minimality) so it has to be non-standard.
(I couldn't find the link, but I expanded on this a while ago): By $\Sigma^1_1$ absoluteness, any transitive model of set theory (in particular, the minimal transitive model) contains $\omega$-models of set theory, perhaps ill-founded (definitely ill-founded, in the minimal transitive model). By the completeness theorem, any $\omega$-model contains models of set theory (in some cases, only models that are not $\omega$-models). – Andrés E. Caicedo Jun 28 '13 at 2:55
In fact, we will get a model of $\mathsf{ZFC}+\mathrm{Con}(\mathsf{ZFC}+\mathrm{Con}(\mathsf{ZFC}))$, etc. The moral is that there is a long way to go between the minimal transitive model and any model of $\mathsf{ZFC}+\lnot\mathrm{Con}(\mathsf{ZFC})$, which is the only kind that can genuinely claim to be minimal. But then we can go (a bit) further still: – Andrés E. Caicedo Jun 28 '13 at 2:58