Our context is the study of -models of subsystems of second order arithmetic [5, Chapter VIII]. As in [5, Chapter VII], a -model is an -model M such that, for all sentences with parameters from M, is true if and only if . Theorems 1.1 and 1.3 below are an interesting supplement to the results on -models which have been presented in Simpson [5, §§VII.2 and VIII.6].
Let HYP denote the set of hyperarithmetical reals. It is well known that, for any -model M, HYP is properly included in M, and each is definable in M.
Proof.Fix a recursive enumeration Se, , of the sets of reals. If p is a finite subset of , say that meets p if for all . Let be the set of p such that there exists meeting p. Put if and only if . Say that is dense if for all there exists such that . Say that is definable if it is definable over the -model HYP, i.e., arithmetical in the complete subset of . Say that is generic if for every dense definable there exists such that meets p. We can show that for every there exists a generic meeting p. (This is a fusion argument, a la Gandy forcing.) Clearly is a -model. We can also show that, if C is countable and , then there exists a generic such that . Let be the language of second order arithmetic with additional set constants Xn, . Let be a sentence of . We say that p forces , written , if for all generic meeting p, the -model satisfies . It can be shown that, for all generic , the -model satisfies if and only if meets some p such that . If is a permutation of , define an action of on and by and . It is straightforward to show that if and only if . The support of is defined by . Clearly if and , then . We claim that if and are generic, then the -models and satisfy the same L2-sentences. Suppose not. Then for some we have and , for some L2-sentence . Let be a permutation of such that . Since , we have , hence , a contradiction. This proves our claim. Finally, let where is generic. Suppose is definable in M. Let be generic such that has . Let be an L2-formula with X as its only free variable, such that exactly one , and . Then exactly one . Let be such that . Then for each , we have that if and only if and , if and only if and , if and only if . Thus A=A'. Hence . This completes the proof.
We now improve Theorem 1.1 as follows.
Let denote hyperarithmetical reducibility, i.e., if and only if X is hyperarithmetical in Y.
if X is definable from Y, then .
The -model which we shall use to prove Theorem 1.3 is the same as for Theorem 1.1, namely where is generic. In order to see that M has the desired property, we first relativize the proof of Theorem 1.1, as follows. Given Y, let be the set of such that there exists meeting p with X0=Y. (Obviously 0 plays no special role here.) Say that is generic over Y if, for every dense definable from Y over , there exists such that meets p.
Proof.The proof of this lemma is a straightforward relativization to Y of the proof of Theorem 1.1.
Consequently, in order to prove Theorem 1.3, it suffices to prove the following lemma.
Proof.It suffices to show that, for all p forcing is dense in , there exists forcing and meets r). Assume is dense in . Since , it follows that and . Fix and such that . Put meets . Then S' is a set, so let be such that S'=Se. Claim 1: . If not, let be such that . Let be a permutation such that and . Then and , a contradiction. This proves Claim 1. Claim 2: . To see this, let be generic meeting . By Claim 1 we have . Hence , i.e., there exists meeting q' with X0=G'0. Thus meets . This proves Claim 2. Finally, put . Then and and meets q'). This proves our lemma.
The proof of Theorem 1.3 is immediate from Lemmas 1.4 and 1.5.