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.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s