# The Chang Model, and Measurable Cardinals

I’ve been meaning to write up this post for a week or so now, but I haven’t managed to get around to it until now. In this post, I want to briefly go over another cool result from “The Stationary Tower”. We define the Chang Model to be $L({}^\omega ON)=\bigcup_{\alpha\in ON}L(\mathcal{P}_{\omega_1}(\alpha))$. Such a model was first considered by C. C. Chang, albeit his construction was a bit different. In his paper “Sets Constructible in $L_{\kappa\kappa}$“, Chang considered the result of going about the construction of $L$ in the language $\mathcal{L}_{\kappa\kappa}$ for an infinite regular cardinal $\kappa$, where we allow fewer than $\kappa$ many instances of quantification and logical connectives. In the case of $\kappa=\omega$, the result is simply $L$. In the case where $\kappa=\omega_1$, it turns out the the resulting model, denoted $C^{\omega_1}$ is precisely the model $L({}^\omega ON)$.

Lemma (Chang): $L({}^\omega ON)$ satisfies ZF.

Lemma (Chang)$L({}^\omega ON)$ satisfies its definition inside itself, and every element of $L({}^\omega ON)$ is $\Sigma_2$ definable inside the model from a countable sequence of ordinals.

At this point, I would like to present Kunen’s result that, if there exist uncountably many measurable cardinals, then AC fails in the Chang model. We will follow the proof presented in “The Stationary Tower”. We will need the following technical lemma (which I will be black-boxing for the purpose of this post). Unfortunately, I don’t know who to attribute this to.

Lemma: Given an ordinal $\gamma$, let $A_\gamma$ denote the class of all cardinals $\kappa$ such that there exists a countably complete nonprincipal ultrafilter $\mu$ of completeness $\kappa$ such that $j(\gamma)\neq\gamma$, where $j:V\to M$ is the elementary embedding induced by $\mu$. Then for all ordinals $\gamma$, $A_\gamma$ is finite.

Theorem (Kunen): Assume that there are uncountably many measurable cardinals, then the Axiom of Choice fails in $L({}^\omega ON)$.

Proof: Let $\langle \kappa_\alpha : \alpha<\omega_1\rangle$ denote the first $\omega_1$-many measurable cardinals. For each $\alpha$, let $\mu_\alpha$ be an ultrafilter witnessing measurability of $\kappa_\alpha$, and let $j_\alpha V\to M_\alpha$ be the induced elementary embedding. Let $\lambda=\sup \{\kappa_\alpha : \alpha<\omega_1\}$, and note that $j_\alpha(\lambda)=\lambda$ for each $\alpha<\omega_1$. To see this, simply note that each $\kappa_\alpha$ is a fixed point of every other $j_\beta$ for $\beta<\alpha$ since it is strongly inaccessible. Thus, $\lambda$ is the supremum of fewer than $\kappa_\alpha$ many fixed points of any given $j_\alpha$, and thus $j_\alpha(\lambda)=\lambda$.

Next, note that ${}^\omega\lambda\subseteq L({}^\omega ON)$. We will show that there is no well-ordering of ${}^\omega \lambda$ in $L({}^\omega ON)$. Suppose, by way of contradiction, that there is such a well-ordering, $\pi$ in $L({}^\omega ON)$. Then, $\pi$ is $\Sigma_2$ definable from a countable set of ordinals $s$ in $L({}^\omega ON)$ as a parameter. We may fix a regular cardinal $\gamma>\max\{ rank (\pi), \sup (s), \lambda\}$ such that $j(\gamma)=\gamma$. By making $\gamma$ sufficiently large, and possibly adding to $s$, we may assume that $\pi$ is definable in $V_\gamma\cap L({}^\omega ON)$ from $s$. In particular, we may fix a formula $\phi$ such that

$\pi=\{a\in{}^\omega \lambda \times {}^\omega \lambda : V_\gamma\cap L({}^\omega ON)\models\phi(a,s)\}$.

Let $\langle \beta_i : i<\omega\rangle$ enumerate $s$. For each $\alpha<\omega_1$, $j_{\alpha}(s)\neq s$ if and only if $j_\alpha(\beta_i)\neq\beta_1$ for some $i<\omega$. But, by the above lemma, the collection $\{\alpha : j_\alpha (\beta_i)\neq\beta_i\}$ is finite for each $i<\omega$. Thus, there are only countably many $\alpha$ such that $j_\alpha$ moves $s$. In particular, there is then some $\alpha_0$ such that, for each $\alpha>\alpha_0$, $j_\alpha(s)=s$.

We claim that then $j_\alpha(\pi)=\pi$ for every $\alpha>\alpha_0$. To see this, by applying $j_\alpha$, we have that

$j_\alpha(\pi)=\{a\in j_\alpha({}^\omega \lambda \times {}^\omega \lambda) : j_\alpha(V_\gamma\cap L({}^\omega ON))\models\phi(a,j_\alpha(s))\}$.

Furthermore, we know that $j_\alpha(\{s, \gamma, \lambda\})=\{s,\gamma,\lambda\}$. Next, since ${}^\omega(M_\alpha)\subseteq M_\alpha$, it follows that $(L({}^\omega ON))^{M_\alpha}= L({}^\omega ON)$. Thus

$j_\alpha(V_\gamma\cap L({}^\omega ON))$ $=j_\alpha (V_\gamma)\cap L({}^\omega ON)$ $=(M_\alpha)_\gamma\cap L({}^\omega ON)$ $=V_\gamma\cap L({}^\omega ON)$.

From this, it follows that $j_\alpha(\pi)=\pi$.

Now, let $t=\langle \kappa_{\alpha_0+i+1} : i<\omega\rangle\in{}^\omega \lambda$, and note that since $\pi$ is a well-ordering of ${}^\omega \lambda$, there is an ordinal $\rho$ such that $rank_\pi (t)=\rho$. Again, we can apply the above lemma to see that there are only finitely many members of $t$ with associated elementary embeddings that move $\rho$. So, fix $i_0<\omega$ such that, for each $i\geq i_0$, $j_{\alpha_0+i+1}(\rho)=\rho$. Let $\beta=\alpha_0+i_o+1$. We then have the following:

1. $j_\beta(\pi)=\pi$;
2. $j_\beta(\rho)=\rho$;
3. $rank_\pi (t)=\rho$, so $j_{\beta}(t)=t$.

But, this gives us a contradiction, as $\kappa_\beta$ is the $i_o^{th}$ coordinate of $t$, and the critical point of $j_\beta$, which means that $j_\beta(\kappa_\beta)\neq\kappa_\beta$, and therefore that $j_\beta(t)\neq t$.