Computability and Logic

294 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), NA G ∗ |= S.
But this means A* is arithmetical, contrary to Lemma 23.8.
Problems
23.1 Use 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.
23.2 Show 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.
23.3 Show that if p ⊩ ∼∼∼B, then p ⊩ ∼B.
23.4 Given an example of a sentence B such that the set of even numbers FORCES
neither B nor ∼B.
23.5 Show that the set of pairs (i, j) such that j codes a sentence of L G and i codes
a condition that forces that sentence is not arithmetical.
23.6 Where 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)?
23.7 Show that the only arithmetical subsets of a generic set are its finite subsets.
23.8 Show that if A is generic, then {A} is not arithmetical.
23.9 Show that {A : A is generic} is not arithmetical.
23.10 Show that every generic set contains infinitely many prime numbers.
23.11 Show that the class of generic sets is nonenumerable.
23.12 A 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.