Models of Arithmetic and Upper Bounds for Arithmetic Sets

Models of Arithmetic and Upper Bounds for Arithmetic Sets Alistair H. Lachlan and Robert I. Soare Abstract

We settle a question in the literature about degrees of models of true arithmetic and upper bounds for the arithmetic sets. We prove that there is a model of true arithmetic whose degree is not a uniform upper bound for the arithmetic sets. The proof involves two forcing constructions.

1 Introduction

For A  ! de ne the jump of A to be A0 = fhx; yi : x 2 WyA g. The arithmetic sets are the members of the jump ideal A = fX : (9k)[X T ; k ]g. A degree a is a uniform upper bound for the arithmetic sets (uub for ar) if there is an enumeration E of A (possibly with repetitions) such that E is of degree  a. A tree is a subset T  2
[T ] = ff : f 2 2! & (8n)[f  n 2 T ]g: For X; Y  ! de ne the join X  Y to be f2n : n 2 X g[f2n +1 : n 2 Y g: A Scott set is a nonempty set S  P (!) such that for all X; Y  !, The rst author was supported by NSERC (Canada) Grant A3040, and the second author by National Science Foundation Grant DMS 91-06714. 

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(i) X 2 S & Y 2 S =) X  Y 2 S , (ii) X 2 S & Y T X =) Y 2 S ; (iii) If X 2 S codes an in nite tree T  2
r(M) = fX : X is representable in Mg: Scott [7] proved that: if M j= PA then r(M) is a Scott set; and that for any countable Scott set S there is a nonstandard model M of PA such that r(M) = S . From now on we consider nonstandard models M of PA with universe !. The formulas (with parameters from !) will be identi ed with their Godel numbers. For any model M the atomic diagram D(M) will be viewed as a subset of !. The degree of M is the Turing degree of D(M). It is known that (1) r(M) = ffn 2 ! : M j= pM n j ag : a 2 ! g: th where pM n denotes the n prime in M. This set is called the Scott set of M. Let TA denote true arithmetic, i.e. the set of sentences in the language of PA which are true in the standard model N of PA. It is known that if S is a countable Scott set such that ; n 2 S for all n 2 ! then there exists a nonstandard model M of TA such that r(M) = S . The converse is immediate. Let S be a set of subsets of !. An enumeration of S is a binary relation R on ! such that S = fR n : n 2 !g; where R n denotes fx 2 ! : (x; n) 2 Rg: We say that n is the index of E n in E . Note that by (1) for any countable nonstandard model M of PA we have an enumeration E of the Scott set r(M) such that deg(E )  deg(M). We say that E is an eective enumeration of a Scott set if there are recursive functions f , g , and h which witness the eectiveness of the three conditions for being a Scott set, namely: (i) E f x;y = E x  E y ; (ii) if X = figE then X = E g i;y ; and (iii) if E x is an in nite tree T then E h x 2 [T ]: Solovay ([8]) proved that the degrees of nonstandard models of TA are precisely the degrees of eective enumerations of Scott sets containing the ( )

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arithmetic sets. Also the degrees of models of a particular Scott set S are the degrees of eective presentations of S .

Theorem 1.1 (Solovay) The following conditions are equivalent: (i) d is the degree of a nonstandard model of TA; (ii) for some S which is a Scott set containing the arithmetic sets there is an eective enumeration E of S such that deg(E )  d. Solovay wrote out his proof in [8] but never published it. For a published proof of Solovay's result see Knight [1]. It follows that the degrees of models of TA are closed upwards, although this had been proved by D. Marker before Solovay's theorem. After Solovay's theorem Macintyre and Marker ([5]) studied degrees of recursively saturated models of arithmetic and as a corollary proved that in condition (ii) above \eective enumeration" can be replaced simply by \enumeration".

Theorem 1.2 (Marker) If E is an enumeration of a Scott set S , then there is an eective enumeration E 0 of S such that E 0 T E: Hence, the degrees of nonstandard models of TA are the degrees of enumerations of Scott sets containing the arithmetic sets.

Thus, when we wish to build a model of TA in x2 it suces by these results to build an enumeration E of a Scott set which contains the arithmetic sets. Then deg(E ) will be the degree of a model of TA. As further notation, for strings ,  2 2
2 Uniform Upper Bounds

From the results in x1 it is immediately obvious that any degree which is a uniform upper bound for the arithmetic sets is a degree of a nonstandard model of TA. Lerman ([4]) raised the question of whether the converse was true. In this section we refute the converse by proving: 3

Theorem 2.1 There is a model of true arithmetic whose degree is not a uniform upper bound for the arithmetic sets.

The proof will be given at the end of the section following three lemmas. Here is a high level outline of the proof. Let G  ! be a Cohen generic subset of !. Let E  !  ! be a generic enumeration of the sets arithmetic in G. Then if d = deg(E ), it follows from Marker and Macintyre's re nement of Solovay's theorem that d is the degree of a model of true arithmetic. We shall prove that there is no enumeration of the arithmetic sets which is recursive in d. Consider the language L obtained from the language of arithmetic by adjoining a unary relation symbol G. Here G represents an undetermined subset of ! while the symbols of arithmetic are given their standard interpretations. We use Cohen forcing (2
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Lemma 2.2 For all A 2 A, G  A0 T (G  A)0: Hence, B is a jump ideal and no element of B bounds every element of A. Proof. It is clear that for all x; y 2 ! and A 2 A the following set is dense,

fp 2 2
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which is  in G  A. Thus (G  A)0 T G  A0 which suces to prove that B is a jump ideal. Suppose that A 2 A and that B = G  A bounds every element of A (and hence every element of B). Then 1

B 0 = (G  A)0 T G  A0 T B; a contradiction.

De nition 2.3 Let B  ! be in nite and T  2
( +1)

Proof. In the language of arithmetic for each n 2 ! de ne a unary relation symbol Jn whose interpretation is ; n : Let T be a tree satisfying the hypothesis of the lemma and T = figG; : Consider the sentence: ( )

(k )

h

i

 : \figGJ is a tree and Jk 2 figGJ ": k

+1

k

Since G is generic,  is forced by some p  G. Clearly, for each x 2 !, fq 2 2
( )

 : fhx; j i : x  ng ?! 2; and for all  2 2
 2 dom(figjqq;j ) & [figjqq;j ( ) = 1 =)  ?  ]: (k )

(k )

If such a triple is found, then enumerate in Vj all x  n such that (hx; j i) = 1: Otherwise, Vj = ;: By the recursion theorem we can x j such that Vj = Wj; . We consider two cases. Case 1. The triple hq; ; ni is not found for j . We claim that T = G fig ; is free on ! j = fhx; j i : x 2 !g: Given h 2 2! de ne the tree Th = f 2 T :  6? hg. It suces to prove that Th is in nite since by the (k )

(k )

[j ]

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compactness of 2! we can choose any path f 2 [Th]. Suppose not. Then there exist n and  : fhx; j i : x  ng ?! 2 such that  is not compatible with any node on T of length m, where m = 1+ max(dom()): Since p forces \figGJ is total", there exists q; p  q  G; which forces \figGJ ( ) #" for all  2 2m . So there exists q  G such that  2 dom(figjqq;j ) for all such  . Hence, a suitable triple hq; ; ni is found. This contradiction veri es the claim that T is free on ! j . Case 2. Otherwise. Let hq; ; ni be the triple found for j . Consider a generic set G0 such that q  G0. Let m denote max(dom()) + 1. Since G0  p and p , figG ; is a tree and ; k 2 [figG ; ]. By choice of hq; ; ni, k

k

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(8 2 2m) figG ; ( ) # & [figG ; ( ) = 1 =)  ?  ] :

(2)

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Fix  = ; k  m: Then figG ; ( ) = 1 since ; k is on the tree. Since Wj; = Vj ; we have   ; k : So  and  are compatible, contrary to (2). 0

( +1)

(k )

(k )

( +1)

( +1)

Lemma 2.5 There exists an enumeration E of B whose degree is not a uniform upper bound for the arithmetic sets.

Proof. We use a notion of forcing (E ; ); where

= fp : (9n 2 !)[p 2 2!n & (8i < n)(9X 2 B)(8x)[p(hx; ii) = X (x)]]g; and  is inclusion. Consider the language L obtained from the language of arithmetic by adjoining two binary relation symbols E and R. Here E represents an undetermined binary relation on ! while the symbols of arithmetic are given their standard interpretation and the interpretation of R is a xed enumeration R of the arithmetic sets. We use the forcing (E ; ) to force rst-order L -sentences about E. The binary relation E represented by E is thought of as consisting of rows E n , n 2 !, and a particular forcing condition xes E n for some n. E

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Let E be an E -generic binary relation on !. E also stands for the characteristic function of the relation so that, when p 2 E , p  E means that the characteristic function of the relation E extends p. Note that we can choose E so that B is included in the class enumerated by E because we can alternate forcing L -sentences with including members of B as rows of E . Towards a contradiction suppose that figE is an enumeration of the arithmetic sets. There exists p 2 E such that p  E and p where is the sentence: \figE is total & figE and R enumerate the same family of sets". Fix k such that p T G  ; k : There exist q 2 E and m; n 2 ! such that p  q, R n = ; k , and q forces h i \ (8x) [R(x; n) =) figE (x; m) = 1] & [:R(x; n) =) figE (x; m) = 0] ": 2

( )

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Intuitively, this says that q forces the m-th row of the enumeration figE to be ; k . Since ; k 6T G  ; k we have p  q. Let S = dom(q) ? dom(p) and fsn : n < !g be a recursive enumeration of S . For any  2 2
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( +1)

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we de ne a subtree U of U by:  2 U if for all j 2 ! there exists    with  2 U and ( )  h j and j j  j . The empty string is in U by de nition of V . Claim 1. U has no terminal nodes. Proof. Consider  2 U . For each j < ! we have j   such that jj j  j and (j )  h j . Clearly, either  b 0 or  b 1 is extended by j for in nitely many j . Therefore, one of  b 0 and  b 1 is in U . Claim 2. Let f 2 [U ], F : !  ! ?! 2 extend p [ s(f ), and figF be a total function from !  ! into 2. Then 1

1

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figF

[m]

= (f ) = h:

Proof. The rst equation is clear from the de nition of . Suppose that (f ; x) 6= h(x). Choose  2 2
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References [1] J. F. Knight, Degrees of models with prescribed Scott set, in: Proceedings of the U.S.-Israel Workshop on Model Theory in Mathematical 8

[2] [3] [4] [5] [6] [7] [8]

Logic: Classi cation Theory, Chicago, December 15-19, 1985, Lecture Notes in Mathematics No. 1292, Ed. John Baldwin, Springer Verlag, Berlin, Heidelberg, New York, 1987, 182{191. J. F. Knight, A metatheorem for constructions by nitely many workers, Journal of Symbolic Logic 55 1990, 787{804. J. Knight, A. H. Lachlan, and R. I. Soare, Two theorems on degrees of models of arithmetic, Journal of Symbolic Logic 49 (1984), 425{436. M. Lerman, Upper bounds for the arithmetical degrees, Annals of Pure and Applied Logic 29 (1985), 225{253. A. Macintyre and D. Marker, Degrees of recursively saturated models, Trans. Amer. Math. Soc. 282 (1984), 539{554. D. Marker, Degrees of models of true arithmetic, Proc. of the Herbrand Symposium: Logic Colloquium, 1981, ed. J. Stern, North-Holland, Amsterdam, 233{242. D. Scott, Algebras of sets binumerable in complete extensions of arithmetic, in: Recursive Function Theory: Proc. of Sympos. in Pure Math., vol. 5, Amer. Math. Soc., Providence, 1961, 117{121. R. M. Solovay, Degrees of models of true arithmetic, preliminary version, unpublished manuscript, 1983.

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