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!

  • $\begingroup$ You have a small typo; your last $\mathsf{ZFC}$ should be $\mathsf{ZC}$ (it is false otherwise). Also, you may want to mention this is Exercise I.16.8. $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 21:01
  • $\begingroup$ @AndresCaicedo: Thanks for pointing that out. I will fix it. $\endgroup$ – josh Oct 10 '13 at 21:07
  • 1
    $\begingroup$ $\mathsf{ZF}$ is not the same as $\mathsf{ZC}$. $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 21:13
  • $\begingroup$ Ahh. I replaced the typo with another typo. Fixed again. Sorry about that. $\endgroup$ – josh Oct 10 '13 at 21:25


  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.

  • $\begingroup$ Asaf: Here's my attempt for your hint. The text stated a theory a few sections back that said the axiom of infinity is equivalent to statement: $\omega$ exists. Also $\omega \subset V_\omega $, which means that $\omega \in V_{\omega + 1}$. This tells us that the formula "$\omega$ exists" would be true in $V_\beta$ for $\omega < \beta$ and false in $V_\omega$. $\endgroup$ – josh Oct 10 '13 at 21:28
  • $\begingroup$ Yes, that is correct. Let me try and find a way to give another hint for the replacement part. $\endgroup$ – Asaf Karagila Oct 10 '13 at 21:34
  • $\begingroup$ (Interestingly, I got a bit stuck and went to see the solution I submitted some three years ago when I had this question in an exercise to some advance course in set theory, and I was a bit sad to see that I hand waved a bit too much over the replacement part and made some mistakes...) $\endgroup$ – Asaf Karagila Oct 10 '13 at 22:06

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