Difference between revisions of "Cuts in recursively saturated models"

Revision as of 10:22, 18 January 2013

Free cuts

A cut $I$ in a model $M\models PA$ is free if for all $a, b\in I$ if $(M,a)\equiv (M,b)$, then $(M,I,a)\equiv (M,I,b)$. There are free elementary cuts in every countable recursively saturated model of $PA$, and generic cuts of Kaye and Tin Lok Wong are free.

Problem: Let $M\models PA$ be countable and recursively saturated. Does $M$ have a free elementary cut $I$ such that the pair $(M,I)$ is recursively saturated?

References:

Kossak, Roman Four problems concerning recursively saturated models of arithmetic. Special Issue: Models of arithmetic. Notre Dame J. Formal Logic 36 (1995), no. 4, 519–530.

Kaye, Richard; Wong, Tin Lok Truth in generic cuts. Ann. Pure Appl. Logic 161 (2010), no. 8, 987–1005.

Omitting theories of subsets

Suppose $M$ is countable recursively saturated and $X$ is an undefinable subset of $M$. Is there a countable recursively saturated $N$ such that $N$ is an elementary end extension of $M$, and if $Y \subseteq M$ is coded in $N$, then $(M,Y) \not\equiv (M,X)$?

The answer if `yes' is either $(M,X)\not\models PA^*$ or ${\rm Th}(M,X)\notin {\rm SSy}(M)$.

This problem is listed in Kossak Roman; Schmerl, James H. The structure of models of Peano arithmetic, but, unfortunately, with many typos.