I try to prove the Jech's "Set theory", exercises 12.11:

12.11. If $\kappa$ is an inaccessible cardinal, then $V_\kappa\models \text{there is a countable model of ZFC}$.

My attempt. Since $(V_\kappa,\in)$ is a model of ZFC, by Löwenheim-Skolem theorem theorem there is a countable model $(A,R)$ which elementary equivalent to $(V_\kappa,\in)$. Especially, $(A,R)$ is a model of ZFC. Since $A$ is countable, we can find $E\subset \omega\times\omega$ such that $(A,E)$ is isomorphic to $(\omega,E)$. Since $E\in P(\omega\times\omega)$ and $P(\omega\times\omega)\in V_{\omega+\omega}$, so $(\omega,E)\in V_\kappa$.

We will prove that $V_\kappa\models (\omega,E)\text{ is a countable model of ZFC}.$ Since $\omega^{V_\kappa}=\omega$, $V_\kappa$ satisfies "$(\omega,E)$ is a countable structure". To prove $V_\kappa\models ((\omega,E)\models ZFC)$, we will check that $V_\kappa \models \varphi^{\omega,E}$ holds for each axiom $\varphi$ of ZFC. By induction for $\varphi$, we can prove $(\varphi^{\omega,E})^{V_\kappa}\leftrightarrow \varphi^{\omega,E}$. If $\varphi$ is an axiom of ZFC, then $\varphi^{\omega,E}$ holds because $(\omega,E)$ is a model of ZFC, so $(\varphi^{\omega,E})^{V_\kappa}$ holds for each axiom $\varphi$ of ZFC.

My "proof" is correct? If not, how to improve my "proof"? Thanks for any help.

Add : As I think, last part of this "proof" has an error. But I don't know what is wrong exactly.

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    $\begingroup$ This is almost good enough. What you need to additionally argue is that $V_\kappa$ has no "fake" ZFC axioms, which wouldn't correspond to any standard formula $\varphi$. $\endgroup$ – Miha Habič May 21 '14 at 14:36
  • $\begingroup$ @MihaHabič To prove that $V_\kappa$ has no "fake" axioms of ZFC, it is sufficient that $V_\kappa$ does not have non-standard natural numbers? (Every formula has corresponding Gödel numbering. If there is a non-standard number, then this non-standard number does not have any corresponding standard formula.) But I don't know the precise proof. In addition, I don't know why this proof is needed. $\endgroup$ – Hanul Jeon May 21 '14 at 16:04
  • $\begingroup$ Yes, standardness of the naturals is exactly the issue. The key point is that ZFC is recursive and so ZFC (or PA or even much weaker theories) can correctly determine for every standard number whether or not it is a ZFC axiom. As for why this is necessary, consider that your argument only shows that $V_\kappa$ has a countable structure which satisfies all the actual ZFC axioms. Depending on whether $V_\kappa$ has nonstandard naturals, this may or may not be the same as saying that $V_\kappa$ believes that the structure satisfies ($V_\kappa$'s version of) ZFC. $\endgroup$ – Miha Habič May 21 '14 at 20:28
  • $\begingroup$ Of course the issue pointed out by @Miha cannot happen when talking about $V_\kappa$, or even transitive models, but it can happen when talking about arbitrary models of $\sf ZFC$, in particular consider models of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$. $\endgroup$ – Asaf Karagila May 22 '14 at 7:34
  • $\begingroup$ I'd also add that your proof only showed that for each axiom that $V$ thinks is an axiom of $\sf ZFC$, the axiom holds in $(\omega,E)$. You need to conclude that there is a truth predicate for $(\omega,E)$ and that $V_\kappa$ knows it is a truth predicate for $(\omega,E)$. $\endgroup$ – Asaf Karagila May 22 '14 at 7:35

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