# Exercise in Kunen (IV.10)

I am reading Kunen's book on set theory and am puzzled by Exercise IV.10, which is:

Find a sentence $\phi$ such that for any $\beta$, if $\phi$ is absolute for $R(\beta)$, then $\beta=\omega_\beta$. Then, find a formula $\psi(x)$, such that for any non-0 transitive $M$, if $\psi(x)$ is absolute for $M$, then $M=R(\beta)$ for some $\beta$ such that $\beta=\omega_\beta$. Hint. $\phi$ will be enough of ZF to guarantee that $\forall\alpha(\omega_\alpha\hbox{ exists})$. $\psi(x)$ can be "$\phi \wedge (x\hbox{ is an }R(\alpha))$".

(Here, $R(\alpha)$ is the set of sets of rank $<\alpha$, which is also called $V_{\alpha}$.)

Finding $\phi$ is no problem, but I don't see how the provided hint for $\psi(x)$ can work. If we use it, then given absoluteness, we have $R(\beta)\in M \Rightarrow R(\beta)=R(\alpha)^M$ for some $\alpha\in M$, but this says nothing about the possibility $R(\beta)\notin M$.

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I’ve not thought about absoluteness since I took Ken’s course almost 40 years ago, so I may be missing something obvious, but can’t you argue that if $\xi\in\mathbf{ON}^{\mathbf{M}}$, then $\Big(\psi\big(R(\xi)^{\mathbf{M}}\big)\Big)^{\mathbf{M}}$, so $\psi\big(R(\xi)^{\mathbf{M}}\big)$ by absoluteness of $\psi$ for $\mathbf{M}$, and hence $R(\xi)^{\mathbf{M}}=R(\alpha)$ for some $\alpha\in\mathbf{ON}$? – Brian M. Scott Jan 12 '12 at 11:10
Yes, taking $x:=R(\alpha)^M$ is the trick. (Also, $\phi$ must contain enough ZF to show that any set is in some $R(\alpha)$.) – Polichinelle Jan 14 '12 at 0:51

Argue that if $\xi\in\mathbf{ON}^{\mathbf{M}}$, then $\left(\psi\big(R(\xi)^{\mathbf{M}}\big)\right)^{\mathbf{M}}$, so $\psi\big( R(\xi)^{\mathbf{M}}\big)$ by absoluteness of $\psi$ for $\mathbf{M}$, and hence $R(\xi)^{\mathbf{M}}=R(\alpha)$ for some $\alpha\in\mathbf{ON}$. As you say, this will work if $\phi$ contains enough of $\mathbf{Z}$ to show that every set is in some $R(\alpha)$.