On Kunen's Set Theory, Exercise I.16.7

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).

• $A$ is an elementary submodel. What does this mean? Commented Mar 21, 2021 at 14:44
• To extend Jonathan's comment, remind that how you defined $\alpha$ and $\beta$. Commented Mar 21, 2021 at 17:16

1 Answer

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$$(\omega_1)$$.

• Thanks. My main concern is showing that $\mathrm{mos}$ is an isomorphism as we do not know that $A$ satisfies the Axiom of Extensionality ($A$ need not be transitive). How can we guarantee that? Commented Mar 22, 2021 at 2:11
• @ClementYung: Elementarity? Commented Mar 22, 2021 at 2:16
• @AsafKaragila I see, I think I was confused with some concepts. Thank you. Commented Mar 22, 2021 at 2:18