Classiication: Mathematics

A set A of nonnegative integers is recursively enumerable (r.e.) if A can be computably listed. It is shown that there is a rst order property, Q(X), de nable in E, the lattice of r.e. sets under inclusion, such that: (1) if A is any r.e. set satisfying Q(A) then A is nonrecursive and Turing incomplete; and (2) there exists an r.e. set A satisfying Q(A). This resolves a long open question stemming from Post's Program of 1944, and it sheds new light on the fundamental problem of the relationship between the algebraic structure of an r.e. set A and the (Turing) degree of information which A encodes. 2 Recursively enumerable (r.e.) sets have been a central topic in mathematical logic, in recursion theory (i.e. computability theory), and in undecidable problems. They are the next most e ective type of set beyond recursive (i.e. computable) sets, and they occur naturally in many branches of mathematics. This together with the existence of nonrecursive r.e. sets has enabled them to play a key role in famous results such as G odel's incompleteness theorem, the unsolvability of Hilbert's tenth problem on Diophantine equations, and the unsolvability of the word problem for nitely presented groups. For sets A;B !, A is Turing reducible to (also called recursive in) B, written A T B, if there is an algorithm for deciding whether x 2 A provided we are given answers to any questions of the form \Is y 2 B?". We write A T B if A T B and B T A. The equivalence class of A under T is the (Turing) degree (degree of unsolvability) of A, written deg(A) = a. In his famous incompleteness theorem paper (1) G odel de ned a set of natural numbers which (in modern terminology) is the complement of the canonical nonrecursive r.e. set K = fe : e 2 Weg, where fWege2! is an acceptable numbering of all r.e. sets. Post (2) noted thatK is (Turing) complete in the sense that We T K for every We. In an attempt to classify the degree of unsolvability of problems in mathematics, Post (2) posed his famous question (Post's Problem) of whether there exist nonrecursive incomplete r.e. sets. (If not then there would be only one unsolvable problem for Diophantine equations, for nitely presented groups, and for many other problems in mathematics and computer science.) Post's Program for resolving his problem was to nd some property of an r.e. set A (not involving relative computability) which guarantees A nonrecursive and incomplete, i.e. ; 000). Martin's result generalized an earlier result by Yates that maximal sets could be complete and therefore no \thinness" property in the style of Post on A = ! A, could guarantee incompleteness. Soare (15) developed a new method for generating automorphisms of E and used it to show that all maximal sets are automorphic, so that no denable property together with maximality ensures incompleteness. Cholak, Downey and Stob (16) used this automorphism method to prove that for every coin nite r.e. A there is a complete r.e. B such that L(A) = L(B), and hence no E-de nable property of A alone could answer Question 1 afrmatively. The automorphism method was widely used to produce new homogeneity properties for r.e. sets (see ref. 5, Chapters XV and XVI), and it was proposed that a negative solution to Post's Program could be obtained by positively settling the following. Question 2 For every nonrecursive r.e. set A does there exist an automorphism of E such that (A) is complete? This seemed quite promising and a number of positive partial results were obtained (see ref. 5, p. 379). However, in 1985 Harrington showed that the automorphism of Question 2, even if it exists, must necessarily be 4 non 03, and so more complicated than the current techniques for building automorphisms would allow. This means that there is a dynamic obstacle (see x1) to a positive answer to Question 2. In 1984 Harrington (see ref. 5, p. 339) initiated the methodology for converting dynamic obstacles into de nable properties. By pressing these ideas further Harrington and Soare now settle Question 1 positively (and therefore Question 2 negatively) in the following which is the main result of this paper. Theorem 1 There is a nonempty E-de nable property Q(A) such that every r.e. set A satisfying Q(A) is nonrecursive and incomplete. (Furthermore, all r.e. sets A satisfying Q(A) are not only incomplete but also deg(A) forms half of a minimal pair of r.e. degrees.) Note that every r.e. setA satisfyingQ (see x1) is a major subset and hence has high degree. A closely related property includes even sets of low degree. We say (in the style of (17)) that set B is hemi-Q, written HQ(B), if there is an r.e. set A satisfying Q(A) such that A can be split into the disjoint union of nonrecursive r.e. sets B and C. Note that if HQ(B) then B T A so B must also be incomplete. Since HQ(B) is E-de nable, any automorphic image of B must also be incomplete. Hence, there is a nonrecursive low r.e. set B which cannot be mapped to a complete set by any automorphism of E. In the direction of pressing progress on Question 2 as far as possible, Harrington and Soare proved a theorem almost complementary to Theorem 1. Theorem 2 If A is any r.e. set of promptly simple degree then A is e ectively automorphic to a complete set, and this automorphism can be found uniformly (in an r.e. index for A and an index for the recursive function witnessing that A has promptly simple degree). See ref. 5 for a de nition of promptly simple sets and degrees and for all other de nitions and notation used here. (Theorem 2 strengthens a result by Downey, Cholak, and Stob (16) which used the much stronger hypothesis that A is a promptly simple set, and unlike Theorem 2, their proof is neither uniform nor e ective.) Harrington and Soare have proved Theorem 20, which asserts that Theorem 2 holds for sets A in a strictly larger class of r.e. degrees called almost promptly simple degrees, and hence the class of non promptly simple (i.e. tardy) degrees,M = R PS (i.e. the degrees of halves of minimal 5 pairs) is not invariant as de ned below. Harrington and Soare also believe that they can characterize exactly the class of degrees containing an r.e. set not automorphic to a complete set. A major open problem has been to determine which subclasses of the r.e. degrees R (particularly which jump classes Hn and Ln and their complements) are invariant. A class C of r.e. degrees is invariant if it is the set of degrees of some class C of r.e. sets which is invariant under automorphisms of E (e.g. if C is E-de nable). For example, the result by Martin above shows that H1 is invariant, and those by Lachlan and Shoen eld above that L2 is invariant, where Hn = na 2 R : a(n) = 0(n+1)o, Ln = na 2 R : a(n) = 0(n)o, and C = R C. For over 15 years these and the trivial classes L0 and H0 have been the only ones for which the answer to invariance was known. In the direction of Theorem 2 and Question 2, Harrington and Soare have proved: Theorem 3 For every nonrecursive r.e. set A there is an r.e. set B which is high (i.e. deg(B 0) = 000) such that A is 03-automorphic to B. An immediate corollary is that for every n > 0, the classes Ln and Hn are noninvariant. This automorphism is not e ective, however, but requires the new tree methodology invented in 1984 by Harrington for generating none ective 03-automorphisms of E, which he used to show that for every r.e. set A, ; , or 0 = and k0 > k, and prevents that strategy from later acting unless the Step 2 restraint imposed by the h ; ki-strategy is rst injured. Fix the least satisfying F ( ). For each k the conjunction of (8), (9), and (10) (with replaced by and measured with respect to all z k) is a 02 condition G( ; k). The least k satisfying G( ; k) gives the \true path" node h ; ki for which the h ; ki-strategy succeeds as in x1.1. (Notice that here unlike x1.1 the h ; ki-strategy may be injured nitely often if it has assigned B-restraint r(s) = u and later some y < k u enters A. In this case the h ; ki-strategy begins anew on a fresh x, but such injury can occur at most k times.) This proves Lemma 1. 1.3 Proof of Lemma 2 To complete the proof of Theorem 1 we need to prove: Lemma 2 (9A)[Q(A)]: Proof. This proof is very similar to the standard proof (see ref. 5, p. 194) that every nonrecursive r.e. set C has a small major subset A (A sm C) to 11 which the reader should now refer. Let C be any nonrecursive r.e. set. (If we choose C simple (maximal) then A will be simple (r-maximal).) To make A m C it su ces to meet for every e the requirement, Pe : C We =) A We: Replace We by Ve = Ss2! Ve;s de ned by x 2 Ve;s () x 2 Ve;s 1 _ [x 2 (We;s Cs) & (8y x)[y 2 We;s[Cs]]: (13) Note that C & Vi = ;, for every i. De ne the e-state (e; x; s) = fi : i e & x 2 Vi;sg, with the usual ordering of e-states. Let C = f(!) for f a 1:1 recursive function, and let ci = f(i): Let Cs As = fds0; ds1; g, in the ordering induced by fc0; c1; g. (Hence, if x = dsi = dtj for t > s, then j i.) The strategy for Pe is as follows. If i e, and j > i is minimal such that (e 1; dsi ; s) = (e 1; dsj ; s), dsi 62 Ve;s, and dsj 2 Ve;s then Pe wants to enumerate into A all the elements fdsk : i k < jg (but subject to the negative restraint by Ni, i e, as described below). Let fBi : i 2 !g and n(Sj; Ŝj) : j 2 !o be an e ective listing of all r.e. sets and all pairs of r.e. sets respectively. Let RED play Di against Bi, Di C, and also construct Ti;j to meet (2) if BLUE satis es (1). Let = hi; ji; and let D , B , S , Ŝ , and T denote Di, Bi, Sj, Ŝj, and Ti;j respectively. For each the conjunction of the matrix of (1) for (B ;D ; S ) with the conditions B C, and S t Ŝ = C is a 02 relation F ( ). Let fZ g 2! be an r.e. array of r.e. sets such that F ( ) holds i jZ j =1. De ne T by x 2 T ;s () x 2 T ;s 1 _ [x 2 Cs & x jZ ;sj]: (14) The negative requirementN onA asserts that if (1) holds for (B ;D ; S ) then (2) holds for (B ; S ; T ). The strategy for N is this. If x 2 T & C, then N restrains x from A until x 2 S t Ŝ . (If the latter never occurs then F ( ) fails so Z and T are nite, and only nitely many such x 2 T \ C are permanently restrained by N .) If x 2 Ŝ then N imposes no further restraint on x. If x 2 S , and while x 2 A, x is enumerated in B n D then N restrains x from both D and A forever (unless some Pe, e < , enumerates x in A). 12 (If N successfully keeps x 2 D \ A then x violates (1) so F ( ) fails and Z and T are nite.) Otherwise, suppose that (3) holds on S A. If x 2 S and some Pe, e , wants to enumerate x in A then N rst enumerates x in D and then restrains x from A until x is enumerated in B at which time N releases x (forever). (If x remains in B forever then x 2 (D B ) \ (S A) so x violates (1) and again T is nite.) Hence, in any of the three cases N permanently restrains at most nitely many elements. We now combine the Pe and N -strategies to give the full construction of A. If x 2 Cs, choose the least e (if any) such that Pe wants to enumerate x in A. We say that Pe controls x at s. If e controls x at s then before enumerating x in A, Pe performs the following steps during stages t s. (Some of these steps may never terminate in which case Pe never enumerates x in A.) First Pe waits until x 2 S t Ŝ for all e. Next let 0; 1; ; n be a listing of all e such that x 2 S and D 6= D for all < , with x 2 S . For each k, 0 k n, we make x pass through the N k -strategy above (also called the N k-gate) in the order N n; : : : ;N 1;N 0. Hence, for example, when x is released by the N 2-gate by being enumerated in B 2, RED then enumerates x 2 D 1 and waits for x to be released by N 1 by being enumerated in B 1. When x is released by the N 0-gate RED enumerates x in A. Notice that if e controls x at s, it is possible for some di erent e0 to control x at some t > s, but in this case e0 < e, so the previous action by Pe before stage t is entirely compatible with the action of Pe0 at stages t0 t. Now for every e, requirement Pe is satis ed because everyN can permanently restrain only nitely many elements. To see that N is satis ed, note that the de nition of Ve in (13) ensures that either C Ve or Ve is nite. Let G be the intersection of those Ve for e < such that Ve is in nite. Then, modulo nite sets, N is satis ed by ~ T = T T G ; whose strategy guarantees that if (1) holds for (B ;D ; S ) then (2) holds for (B ; S ; ~ T ). (Adjust ~ T for the nite set of elements permanently restrained by N or enumerated in A by Pe for < or e < and Ve nite.) This completes the proof of Lemma 2 and hence of Theorem 1. We can prove that if Q(A) holds via C then A sm C. Because of the similarity of the proof of Lemma 2 to the small major subset construction, it is natural to ask whether the Q-like property Q̂(A) : (9C)[A sm C] guarantees 13 A The rst author was supported by National Science Foundation Grant DMS89-10312, and the second author by National Science Foundation Grant DMS88-07389. This research was mostly carried out while the authors were vis-iting the Mathematical Sciences Research Institute in Berkeley, California,during the Special Year in Mathematical Logic, from September 1, 1989through August 24, 1990, partially supported by National Science Founda-tion Grant DMS 85-05550. The authors are grateful to C. G. Jockusch, Jr.and the referees for several suggestions and corrections on the rst draft ofthis paper.15 REFERENCES1. Godel, K. (1931), Monatsh. Math. Phys. 38, 173{198.2. Post, E. L. (1944) Bull. Am. Math. Soc. 50, 284{316.3. Marchenkov, S. S. (1976) Mat. Zametki 20, 473{478.4. Soare, R. I. (1977) J. Symbolic Logic 42, 545{563.5. Soare, R. I. (1987) Recursively Enumerable Sets and Degrees: A Studyof Computable Functions and Computably Generated Sets, (Springer-Verlag, Heidelberg).6. Ambos-Spies, K., and A. Nies, Cappable recursively enumerable de-grees and Post's program, Archive of Math. Logic, in press.7. Friedberg, R. M. (1957) Proc. Natl. Acad. Sci. 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