### Kaufmann models

Throughout, ${T}$ is a consistent extension of ${\mathrm{PA}}$.

Definition 1 ${M \models T}$ is a Kaufmann model if it is ${\omega_1}$-like, recursively saturated, and rather classless.

The purpose of this blog post is to demonstrate that Kaufmann models exist. We will first need to detour through some lemmas.

Lemma 2 Let ${M \models T}$ be countable and recursively saturated, ${S}$ be a ${Q_e}$-class for ${M}$ with ${e}$ nonstandard, and ${X \subseteq M}$ be undefinable. Then there is nonstandard ${e'}$ such that ${X}$ is not definable in ${(M,S \upharpoonright e')}$.

Proof: Suppose otherwise towards a contradiction. That is, for all nonstandard ${e'}$, ${X \in \mathrm{Def}(M, S \upharpoonright e'}$. By underspill, there is some standard ${n}$ such that ${X}$ is definable in ${(M, S \upharpoonright n)}$. But then ${X}$ is definable in ${M}$, a contradiction. $\Box$

Corollary 3 Let ${M \models T}$ be countable and recursively saturated and let ${X \subseteq M}$ be undefinable. Then there is countable, recursively saturated ${N \succ_{\mathrm{end}} M}$ such that ${X}$ is not coded in ${N}$.

Proof: Since ${M}$ is recursively saturated, it has a ${Q_e}$-class ${S}$ for some nonstandard ${e}$. By the previous lemma we get that ${X}$ is not definable in ${(M, S \upharpoonright e')}$ for some nonstandard ${e'}$. Take countable ${(N,T) \succ_{\mathrm{end}} (M, S \upharpoonright e')}$. Then ${N}$ is as desired. $\Box$

Now that we know we can end-extend countable, recursively saturated models and avoid coding a certain undefinable set, we have the tools we need to build a Kaufmann model.

Theorem 4 There exists a Kaufmann model of ${T}$.

Proof: This proof uses ${\diamond}$. Fix ${\langle S_\lambda \rangle}$, a diamond sequence. Shelah has an absoluteness argument [1] that I don’t yet understand which eliminates this assumption.

Start with ${M_0 \models T}$ countable and recursively saturated. We will build an ${\omega_1}$-chain ${\langle M_\alpha \rangle}$ of recursively saturated, elementary end extensions. Our Kaufmann model will be the union of this chain. There are three cases:

• For limit ${\lambda}$, ${M_\lambda = \bigcup_{\alpha < \lambda} M_\alpha}$.
• For ${\alpha}$ a successor ordinal, ${M_{\alpha+1}}$ is an arbitrary countable, recursively saturated end extension of ${M_\alpha}$. Such extension exists by the above corollary.
• The final case, for ${M_{\lambda+1}}$ for limit ${\lambda}$, depends upon ${S_\lambda}$. Because ${M_\lambda}$ is countable, we may assume its universe is ${\lambda}$ and hence that ${S_\lambda \subseteq M_\lambda}$. We’ve two possibilities. If ${S_\lambda}$ is definable in ${M_\lambda}$, then extend arbitrarily as in the successor of a successor case. If ${S_\lambda}$ is not definable, then extend to ${M_{\lambda+1}}$ so that ${S_\lambda}$ isn’t coded in ${M_{\lambda+1}}$.

By construction, ${N = \bigcup_{\alpha < \omega_1} M_\alpha}$ is ${\omega_1}$-like and recursively saturated. Let’s check that it is rather classless. Let ${X}$ be a class of ${N}$ and set ${X_\alpha = X \cap M_\alpha}$. Then ${\{ \lambda : (M_\lambda, X_\lambda) \prec (N, X) \}}$ is club in ${\omega_1}$; it is obviously closed, and unboundedness follows from a pigeonhole-type argument. Thus, by ${\diamond}$ there is ${\lambda}$ such that ${X_\lambda = S_\lambda}$ and ${(M_\lambda, S_\lambda) \prec (N,X)}$. ${X}$ is a class of ${N}$ and thus ${S_\lambda}$ is coded in ${M_{\lambda+1}}$. By construction, this means that ${S_\lambda}$ is definable in ${M_\lambda}$. By elementarity, ${X}$ is definable in ${N}$. As ${X}$ was an arbitrary class, ${N}$ is rather classless. $\Box$

The reason I care about Kaufmann models is that a Kaufmann model cannot be the ${\omega}$ of a (nonstandard) model of ${\mathrm{ZFC}}$.

Proposition 5 If ${\mathcal{M} \models \mathrm{ZFC}}$, then ${M = \omega^\mathcal{M}}$ is not rather classless.

Proof: This is not hard to see. ${M}$ must have a tree with an undefinable branch, essentially by Cantor’s theorem. A branch of a tree is a class. So ${M}$ has an undefinable class. $\Box$

Every ${\omega}$ of a model of ${\mathrm{ZFC}}$ must be recursively saturated, as ${\mathbb{ZFC}}$ proves ${\omega}$ has a satisfaction class. Additionally, it must also satisfy ${\mathrm{Th}(\mathbb{N})^\mathrm{ZFC}}$, the statements which ${\mathrm{ZFC}}$ proves are true of ${\mathbb{N}}$. If a countable model of ${\mathrm{PA}}$ has both these properties, then it is the ${\omega}$ of some model of ${\mathrm{ZFC}}$. Kaufmann models show that this does not extend to the uncountable case.

Problem 6 Charactize when an uncountable model of ${\mathrm{PA}}$ is the ${\omega}$ of some model of ${\mathrm{ZFC}}$. Of course, this is a rather difficult problem.

References

1. Shelah, S. (1978). Models with second-order properties. II. Trees with no undefined branches.