### Kaufmann models

#### by kamerynwilliams

Throughout, is a consistent extension of .

Definition 1is aKaufmann modelif 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 2Let 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 3Let 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 4There exists a Kaufmann model of .

*Proof:* This proof uses . Fix , a diamond sequence. Shelah has an absoluteness argument [1] 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 5If , 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 6Charactize when an uncountable model of is the of some model of . Of course, this is a rather difficult problem.

** References **

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