Difference between revisions of "Nonstandard satisfaction classes."
(Created page with " == Fullness of $M$ == A satisfaction class for a model $M$ is $e$-full, if $S$ decides all If for every $\Sigma_e$ sentence (in the sense of $M$) $\varphi$ with parameters eit...") |
|||
| Line 1: | Line 1: | ||
| − | |||
== Fullness of $M$ == | == Fullness of $M$ == | ||
| Line 6: | Line 5: | ||
Fullness of $M$, ${\rm Full}(M)\}$ is the set of those $e\in M$ M for which $M$ has an $e$-full inductive partial satisfaction class. | Fullness of $M$, ${\rm Full}(M)\}$ is the set of those $e\in M$ M for which $M$ has an $e$-full inductive partial satisfaction class. | ||
| − | It is easy to see that ${\rm Full}(M)$ is a cut of $M$ and, if ${\rm Full}(M)>\omega$ then $M$ is recursively saturated. Also if $M$ is countable and $ | + | It is easy to see that ${\rm Full}(M)$ is a cut of $M$ and, if ${\rm Full}(M)>\omega$ then $M$ is recursively saturated. Also if $M$ is countable and ${\rm Full}(M)$ contains an element greater than all definable elements of $M$, then ${\rm Full}(M)=M$.\end{enumerate} \end{prop} |
Kaufmann and Schmerl showed that there are completions $T$ of $PA$, such that for every $M\models T$, ${\rm Full}(M)$ contains no definable nonstandard elements. | Kaufmann and Schmerl showed that there are completions $T$ of $PA$, such that for every $M\models T$, ${\rm Full}(M)$ contains no definable nonstandard elements. | ||
Revision as of 11:28, 18 January 2013
Fullness of $M$
A satisfaction class for a model $M$ is $e$-full, if $S$ decides all If for every $\Sigma_e$ sentence (in the sense of $M$) $\varphi$ with parameters either $\varphi\in S$ or $\lnot\varphi\in S$.
Fullness of $M$, ${\rm Full}(M)\}$ is the set of those $e\in M$ M for which $M$ has an $e$-full inductive partial satisfaction class.
It is easy to see that ${\rm Full}(M)$ is a cut of $M$ and, if ${\rm Full}(M)>\omega$ then $M$ is recursively saturated. Also if $M$ is countable and ${\rm Full}(M)$ contains an element greater than all definable elements of $M$, then ${\rm Full}(M)=M$.\end{enumerate} \end{prop}
Kaufmann and Schmerl showed that there are completions $T$ of $PA$, such that for every $M\models T$, ${\rm Full}(M)$ contains no definable nonstandard elements.
Problem: Suppose ${\rm Full}(M)=M$, does $M$ have a full inductive satisfaction class?
Reference: Kaufmann, Matt; Schmerl, James H. Remarks on weak notions of saturation in models of Peano arithmetic. J. Symbolic Logic 52 (1987), no. 1, 129–148.
