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ARITHMETICAL DEFINABILITY

A-correct p forces S. By Lemma 23.6, there exists a (fully) generic set A* such that p is A*-correct. Since p forces S, by Lemma 23.7 (in its original version), NAG |= S. But this means A* is arithmetical, contrary to Lemma 23.8.

Problems

23.1Use Beth’s definability theorem, Tarski’s theorem on the first-order indefinability of first-order arithmetic truth, and the results of section 23.1 to obtain another proof of the existence of nonstandard models of arithmetic.

23.2Show that for each n the set of (code numbers of) true prenex sentences of the language of arithmetic that contain at most n quantifiers is arithmetical. Show the same with ‘prenex’ omitted.

23.3Show that if p B, then p B.

23.4Given an example of a sentence B such that the set of even numbers FORCES neither B nor B.

23.5Show that the set of pairs (i, j) such that j codes a sentence of LG and i codes a condition that forces that sentence is not arithmetical.

23.6Where would the proof of Addison’s theorem have broken down if we had worked with , & , rather than , , (and made the obvious analogous stipulations in the definition of forcing)?

23.7Show that the only arithmetical subsets of a generic set are its finite subsets.

23.8Show that if A is generic, then { A} is not arithmetical.

23.9Show that { A : A is generic} is not arithmetical.

23.10Show that every generic set contains infinitely many prime numbers.

23.11Show that the class of generic sets is nonenumerable.

23.12A set of natural numbers is said to have density r, where r is a real number, if r is the limit as n goes to infinity of the ratio (number of members of A < n)/n. Show that no generic set has a density.