Kaufmann models

by kamerynwilliams

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

Why care about Kaufmann models?

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.


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