A question dealing with conditions for which $V_\alpha$ models $ZFC$

Question

I've been reading through models of Set Theory in Kunen's most recent Set Theory text and practicing exercises. He mentions that $V_\alpha$ can be used to satisfy certain axioms of $ZFC$ when $\alpha$ is strongly inaccessible. Here, $V_\alpha$ is the set of all well-founded sets whose rank is less than $\alpha$. From here, he presents an exercise to show why $V_\alpha$ models $ZFC$ under certain conditions.

$(ZFC^-)$ Assume that $0 < \alpha < \beta$ and $V_\alpha \preccurlyeq V_\beta$. Prove that $V_\alpha \models ZFC$ and hence $V_\beta \models ZFC$. You may use the fact, to be proved later, that $V_\alpha \models ZC$ for any limit $\alpha > \omega$.

From here, he gives a hint to show how to do this. I broke up the hint into three parts.

1.) Show $\alpha$ is a limit, since if $\alpha = S(\gamma)$, then $ ``\preccurlyeq" $ would fail with the formula $ ``S(a) \mbox{ exists}" $.

2.) Show that $\alpha > \omega$.

3.) For the Replacement Axiom in $V_\alpha$, if $A \in V_\alpha$ and if $\forall x \in A \exists ! y \varphi(x,y)$ holds, then $\exists B \forall x \in A \exists y \in B \varphi(x,y)$ must hold in $V_\beta$.

I was able to successfully show the first part by using the definition of $V_\alpha$. Thus, $\alpha$ must be a limit ordinal. I'm stuck on second and third parts. For the second part, my guess is to rule out the case when $\alpha = \omega$ by coming up with a formula that is true in $V_\beta$ but not in $V_\omega$, but I can't think of one. For the third part, I'm not sure how to show it satisfies Replacement.

Any help would be greatly appreciated!

Answer 1

Hints:

1. The axiom of infinity is false in $V_\omega$. Note that since $\alpha<\beta$ we have that $V_\beta\neq V_\omega$. Conclude the same about $\alpha$.

2. For the axiom of replacement, suppose that $\varphi(u,v)$ is a functional formula (I'm omitting the parameters here) on domain $x\in V_\alpha$. We want to show that $y=\{v\in V_\alpha\mid\exists u\in x.\varphi(u,v)\}$ is in $V_\alpha$.

   Next we use the fact that $V_\beta$ is a model of the separation schema (which is a part of $\sf ZC$) to have that $y\in V_\beta$. Conclude that $V_\beta\models\{v\mid\exists u\in x.\varphi(u,v)\}=y$ (by the fact that $\varphi$ is functional on $x$) and use elementarity again.