0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/802565/… $\endgroup$ – Asaf Karagila Oct 19 '15 at 22:35
  • 3
    $\begingroup$ 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 $A$. $\endgroup$ – Andrés E. Caicedo Oct 19 '15 at 23:09
  • $\begingroup$ Ah, I didn't see that post. I guess I should have searched under a different title. Thanks @AsafKaragila. $\endgroup$ – Jmaff Oct 19 '15 at 23:13
1
$\begingroup$

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 $(\omega,E)$.

$\endgroup$
  • 1
    $\begingroup$ 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 $f \in V_{\kappa}$. $\endgroup$ – BrianO Oct 19 '15 at 20:41

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.