Throughout, is a consistent extension of .
Definition 1 is a Kaufmann model if it is -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 be countable and recursively saturated, be a -class for with nonstandard, and be undefinable. Then there is nonstandard such that is not definable in .
Proof: Suppose otherwise towards a contradiction. That is, for all nonstandard , . By underspill, there is some standard such that is definable in . But then is definable in , a contradiction.
Corollary 3 Let be countable and recursively saturated and let be undefinable. Then there is countable, recursively saturated such that is not coded in .
Proof: Since is recursively saturated, it has a -class for some nonstandard . By the previous lemma we get that is not definable in for some nonstandard . Take countable . Then is as desired.
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 .
Proof: This proof uses . Fix , a diamond sequence. Shelah has an absoluteness argument  that I don’t yet understand which eliminates this assumption.
Start with countable and recursively saturated. We will build an -chain of recursively saturated, elementary end extensions. Our Kaufmann model will be the union of this chain. There are three cases:
- For limit , .
- For a successor ordinal, is an arbitrary countable, recursively saturated end extension of . Such extension exists by the above corollary.
- The final case, for for limit , depends upon . Because is countable, we may assume its universe is and hence that . We’ve two possibilities. If is definable in , then extend arbitrarily as in the successor of a successor case. If is not definable, then extend to so that isn’t coded in .
By construction, is -like and recursively saturated. Let’s check that it is rather classless. Let be a class of and set . Then is club in ; it is obviously closed, and unboundedness follows from a pigeonhole-type argument. Thus, by there is such that and . is a class of and thus is coded in . By construction, this means that is definable in . By elementarity, is definable in . As was an arbitrary class, is rather classless.
Why care about Kaufmann models?
The reason I care about Kaufmann models is that a Kaufmann model cannot be the of a (nonstandard) model of .
Proposition 5 If , then is not rather classless.
Proof: This is not hard to see. must have a tree with an undefinable branch, essentially by Cantor’s theorem. A branch of a tree is a class. So has an undefinable class.
Every of a model of must be recursively saturated, as proves has a satisfaction class. Additionally, it must also satisfy , the statements which proves are true of . If a countable model of has both these properties, then it is the of some model of . Kaufmann models show that this does not extend to the uncountable case.
Problem 6 Charactize when an uncountable model of is the of some model of . Of course, this is a rather difficult problem.
- Shelah, S. (1978). Models with second-order properties. II. Trees with no undefined branches.