The length of inductive iterated full satisfaction classes

by kamerynwilliams

This will be a talk at the CUNY Models of Peano Arithmetic seminar on Wednesday, October 25.

Last week, I gave an application—due to Krajewski—of iterated full satisfaction classes. As part of that we saw that if M \models \mathsf{PA} is resplendent then M admits an iterated full satisfaction class of any length. But in general, these need not be inductive, as even a fragment of induction in the language with the satisfaction class is enough to prove \mathrm{Con}(\mathsf{PA}). This week, we will consider inductive iterated full satisfaction classes.

It is clear that if a resplendent model satisfies the right theory then it admits an inductive iterated full satisfaction class of length n. But that does not imply than any given inductive iterated full satisfaction class of length < n over the model can be extended to one of length n. This is the main result I will talk about: there are models of arithmetic which admit inductive iterated full satisfaction classes S,S' of length n where S can be extended to an inductive iterated full satisfaction class of length n+1 while S' cannot. A slight strengthening of this result will let us draw conclusions about the tree of inductive iterated full satisfaction classes over, say, countable models of arithmetic. Namely, under appropriate hypotheses, for each levels n \le m in the tree there are continuum many nodes on level n in the tree which can be extended to level m but no further.