Using the Downward Lowenheim-Skolem-Tarski Theorem.

Question

I've been reading about the models of Set Theory in Kunen's most recent Set Theory text, and working on exercises since this is my first time working with Model Theory. There is one exercise that I've been looking at for a while now, but I can't seem to wrap my brain around it.

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

The notation $\beta \approx \gamma$ means that $\beta$ and $\gamma$ have the same size, i.e., there is a bijection from $\beta$ to $\gamma$.

A hint is also provided: Use the Downward Lowenheim-Skolem-Tarski Theorem to get a countable $A$ with $\omega, \omega_1 \in A \preccurlyeq V_\alpha$. Then, let $M$ be the Mostowski collapse of $A$; let $\beta = \mbox{mos}(\omega) = \omega$ and $\gamma = \mbox{mos}(\omega_1)$. Then, $\gamma$ will be a countable ordinal that $M$ ``thinks" is true.

The text states the mentioned Theorem (in $ZFC^-$) as follows: Let $\mathfrak{B}$ be any structure for $\mathcal{L}$. Fix $\kappa$ such that max$(|\mathcal{L}|, \aleph_0) \leq \kappa \leq B$, and fix $S \subset B$ with $|S| \leq \kappa$. Then, there is an $\mathfrak{A} \preccurlyeq \mathfrak{B}$ such that $S \subset A$ and $|A| = \kappa$.

I'm not sure how we can have $\omega, \omega_1 \in A$, or what we would be able to use for $A$. Any help or explanations would be greatly appreciated.

Answer 1

What the Downward-Löwenheim-Skolem theorem tells you is that for the relation $R_\in=\{(a,b)\in V_\alpha\times V_\alpha:a\in b\}$ of $V_\alpha$ you can find some countable $A\subseteq V_\alpha$ such that the elements $\omega,\omega_1$ of the model $V_\alpha$ lie at $A$ and $(A,A^2\cap R_\in)\preceq(V_\alpha,R_\in)$, that is, $(A,\in)\preceq(V_\alpha,\in).$

To complete the rest of the exercise, that is, to show $(\operatorname {mos}(\omega)\simeq\operatorname {mos}(\omega_1))^M$ is false, simply show that if $\varphi(\bar x)$ is a $\Delta_0$-formula then $(\varphi(\bar a))^A$ if and only if $(\varphi(\operatorname {mos}(\bar a))^M$, and as the notion of being an ordinal and a bijection can be described with $\Delta_0$-formulas, you get the result.

Answer 2

The Downward Löwenheim–Skolem Theorem allows you to get a countable elementary substructure of a given structure in a countable language, say $(V_\alpha;\in)$, containing any given countably many elements. To do this, add countably many constant symbols to the language corresponding to the elements you want to make sure your substructure contains.

In this case, to get a countable elementary substructure $A \prec V_\alpha$ with $\omega,\omega_1 \in A$, you can apply the theorem to the expanded structure $(V_\alpha;\in,\omega,\omega_1)$.

EDIT: As Asaf pointed out, $\omega$ and $\omega_1$ are definable (in $V_\alpha$ as well as in $V$) so if $A \prec V_\alpha$ then we automatically get $\omega,\omega_1 \in A$. The argument that I described is only necessary if you want to obtain the result for an arbitrary pair of ordinals $\beta,\gamma < \alpha$.