# $V_{ \kappa}$ ( $\kappa$ inaccessible ) models there is a countable model of ZFC

I think that this statement is very well-known but I am a bit unclear on some of the reasoning.

I am aware that $V_{ \kappa }$ models ZFC when $\kappa$ is an inaccessible cardinal. Therefore by Downward Lowenheim Skolem theorem we know that there is some countable elementary submodel $M$ such that $M \preccurlyeq V_\kappa$, which since ZFC is a collection of sentences , must also model ZFC( right?). Now I read in Jech:

Thus there is $E \subset \omega \times \omega$ such that $\mathfrak{A} = ( \omega, E)$ is a model of ZFC.

One is then to verify that $V_\kappa \models$ ZFC.

If I am not mistaken I think that we wan to take advantage of the fact that $\Delta_0$ formulas are upward absolute which requires that our model be transitive . I think this is why we want such a $\mathfrak{A}$ model as above. However, I do not know why we can assert the existence of such a model. Is it due to the Mostowski collapsing function, $\pi$?

Thank you for any hints/help.

• Oct 19, 2015 at 22:35
• Look: If there is a model of set theory at all ($V_\kappa$ or whatever), there is a countable one, because this shows that $\mathsf{ZFC}$ is a countable theory that is consistent. Once you have a countable model, you have that, for any countably infinite set $A$, there is a countable model with universe $A$. Oct 19, 2015 at 23:09
• Ah, I didn't see that post. I guess I should have searched under a different title. Thanks @AsafKaragila. Oct 19, 2015 at 23:13

Once you have a model $(M,{\in})$ where $M$ is countable, you also have a bijection $f:\omega\to M$ -- because that's what it means for $M$ to be countable. Now let $$E = \{ (a,b)\in\omega\times\omega \mid f(a)\in f(b) \}$$
This makes $(\omega,E)$ isomorphic to $(M,{\in})$. Since $M$ is known to satisfy ZFC, so does $(\omega,E)$.
• You need a little more, though, to show that $V_{\kappa} \models (\mathfrak{A} \text{ is a countable model of ZFC})$. It suffices to show that the bijection $f \in V_{\kappa}$: then $\mathfrak{A} \in V_{\kappa}$, and confirming the rest is routine. But clearly, $\sup_{n<\omega} (rank(f(n)) < \kappa$, so $f \in V_{\kappa}$. Oct 19, 2015 at 20:41