On Kunen's Set Theory, Exercise I.16.7

Question

Exercise I.16.7 of Kunen's Set Theory, is as follows:

($\mathsf{ZFC}^-$) Let $\gamma > \omega_1$ be a limit ordinal. Prove that there is a countable transitive $M$ and ordinals $\alpha,\beta \in M$ such that $M \equiv R(\gamma)$ and $(\alpha \approx \beta)^M$ is false but $(\alpha \approx \beta)^{R(\gamma)}$ is true.

Hint. By the Downward Lowenheim-Skolem-Tarski Theorem, get a countable $A$ with $\omega,\omega_1 \in A \preccurlyeq R(\gamma)$. Then, let $M$ be the Mostowski collapse of $A$; let $\alpha = \mathrm{mos}(\omega) = \omega$, and $\beta = \mathrm{mos}(\omega_1)$. Then $\beta$ will be a countable ordinal that $M$ "thinks" is uncountable.

My attempt so far is as follows:

Let $A$ be the model as described in the hint with the additional requirement that $\omega \subseteq A$, and let $M$ be its Mostoski collapse. Define $\alpha = \mathrm{mos}(\omega) = \omega$ (last equality holds because $\omega \subseteq A$), and $\beta = \mathrm{mos}(\omega_1)$. Both are clearly transitive and well-ordered by $\in$, and hence are ordinals. Since $\beta \subseteq M$ is countable, we have that $(\alpha \approx \beta)^{R(\gamma)}$ is true, as they are both countably infinite ordinals. It remains to show that $(\alpha \approx \beta)^M$ is false.

This is where I get stuck. My first thought is to prove by contradiction, by assuming there exists a bijection between $\beta$ and $\omega$ in $M$, and show that $(\alpha \approx \beta)^{R(\gamma)}$ is true, but I'm not sure how to proceed.

Any help is appreciated (I would also appreciate any comments on my attempt thus far).

Answer 1

Since there's no bijection between $\omega$ and $\omega_1$ (and since being a bijection is an absolute property), $R(\gamma)$ satisfies $\omega\not\approx\omega_1$. So the elementary submodel $A$ satisfies the same thing. Finally, since the Mostowski collapse mos is an isomorphism from $A$ onto $M$, it follows that $M$ satisfies mos$(\omega)\not\approx$ mos$(\omega_1)$.