After reading about various properties of $V_\alpha$ and how it can be used to model various axioms of Set Theory, Kunen mentions that in $ZFC$, one cannot prove that there is an $\alpha$ such that $V_\alpha \models ZFC$ (assuming $ZFC$ is consistent.) I've seen proofs before that show why certain axioms fail under certain $V_\alpha$'s, but I was intrigued by the strength of this statement.

The Incompleteness theorem is then mentioned to show why this is true by contratiction. Assuming $ZFC \vdash \exists \alpha \thinspace[V_\alpha \models ZFC]$. Then, $ZFC \vdash \mbox{Con}(ZFC)$, contradicting the Second Incompleteness Theorem.

I was suprised to see the instance of a "nicer" contradiction. Assuming $ZFC \vdash \exists \alpha \thinspace[V_\alpha \models ZFC]$, let $\beta$ be least such that $V_\beta \models ZFC$. But then, by our assuption, $V_\beta \models \exists \alpha[V_\alpha \models ZFC]$. Such an $\alpha \in V_\beta$ must be less than $\beta$, which contradicts $\beta$ being the least.

However, to finish the "nicer" contradiction, it must be verified that the statement "$V_\alpha \models ZFC$" is absolute. How would one go about showing that this statement is absolute? Any help/hints would be greatly appreciated.

  • $\begingroup$ @CamiloArosemena I'm not sure if that is right, because the statement $\alpha = \text{rank}(x)$ is absolute to all transitive sets containing $\alpha$ and $x$, but the statement $x = V_\alpha$ (which seems more relevant here) is not. $\endgroup$ – Trevor Wilson Oct 14 '13 at 3:31

Assuming $\mathsf{ZFC} \vdash \exists \alpha\,(V_\alpha \models \mathsf{ZFC})$, let $\beta$ be the least ordinal such that $V_\beta \models \mathsf{ZFC}$ and let $\alpha$ be the unique set such that the model $V_\beta$ satsifies "$\alpha$ is the least ordinal such that $V_\alpha$ satisfies $\mathsf{ZFC}$." Let $X = V_\alpha^{V_\beta}$.

It suffices to show that the following statements, true in $V_\beta$, are absolute between $V_\beta$ and $V$.

  1. $\alpha$ is an ordinal,

  2. $X \models \mathsf{ZFC}$, and

  3. $X = V_\alpha$.

For 1 and 2 we only need to use the fact that $V_\beta$ is a transitive set and that $\Delta_0$ formulas are absolute to transitive sets. For 3, although the property of being a rank initial segment of $V$ is not absolute to all transitive sets, one can show that it is absolute to rank initial segments of $V$. One way to do this is to show by induction on $\xi$ that $(V_\xi)^{V_\beta} = V_\xi$ for all ordinals $\xi < \beta$.

  • $\begingroup$ Thanks for the response. I was able fill in the details as to why "$\alpha$ is an ordinal" is $\Delta_0$. Why does it follow that "$X \models ZFC$" is $\Delta_0$? $\endgroup$ – josh Oct 14 '13 at 17:24
  • $\begingroup$ @josh You're welcome. It's not clear to me whether "$X \models \mathsf{ZFC}$" is $\Delta_0$, but what I meant was that for any particular sentence $\varphi$ the statement "$X \models \varphi$" is a $\Delta_0$ statement about $X$ (all the quantifiers get bounded by $X$.) It now occurs to me that one may need to use the fact that the definition of the set of $\mathsf{ZFC}$ axioms is absolute to $V_\alpha$ and to $V_\beta$, but this is not hard to show. $\endgroup$ – Trevor Wilson Oct 14 '13 at 21:51
  • $\begingroup$ That makes sense. I will look into it. Thanks again! $\endgroup$ – josh Oct 15 '13 at 0:54

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