ON Σ1 -STRUCTURAL DIFFERENCES AMONG ERSHOV HIERARCHIES YUE YANG AND LIANG YU Abstract. We show that the structure R of recursively enumerable degrees is not a Σ1 -elementary substructure of Dn , where Dn (n > 1) is the structure of all n-r.e. degrees in Ershov hierarchy.

1. Definitions and notations Putnam [6] is the first one introduced the n-r.e. sets. Definition 1.1. (i) A set A is n-r.e. if there is a recursvie function f : ω × ω 7→ ω so that for each m, – A(m) = lims f (s, m). – |{s|f (s + 1, m) 6= f (s, m)}| ≤ n. • A Turing degree is n-r.e. if it contains an n-r.e. set. We use Dn to denote the collection of n-r.e. degrees. For simplicity, we assume D0 = D1 . For the other recursion notations, please refer to Soare [10]. In this paper, we work in the partial ordering language, L(≤), through out. L(≤) include variables a, b, c, x, y, z, ... and a binary relation ≤ intended to denote a partial order. Atomic formulas are x = y, x < y. Σ0 formulas are built by the following induction definition. • • • • •

Each atomic formula is Σ0 . ¬ψ for some Σ0 formula ψ. ψ1 ∨ ψ2 for two Σ0 formula ψ1 , ψ2 . ψ1 ∧ ψ2 for two Σ0 formula ψ1 , ψ2 ψ1 =⇒ ψ2 for two Σ0 formula ψ1 , ψ2 .

A formula ϕ is Σ1 if it is the form ∃x1 ∃x2 ...∃xn ψ(x1 , x2 , ..., xn ) for some Σ0 formula ψ. For each n ≥ 1, a formula ϕ is Πn if it is the form ¬ψ for some Σn formula ψ and a formula ϕ is Σn+1 if it is the form ∃x1 ∃x2 ...∃xm ψ(x1 , x2 , ..., xm ) for some Πn formula ψ. A sentence is a formula without free variables. 1991 Mathematics Subject Classification. 03D25. Both authors were partially supported by NUS Grant No. R-146-000-078-112 (Singapore). The second author is supported by postdoctoral fellowship from NUS, NSF of China No. 10471060 and No.10420130638. Both authors would like to thank Ted Slaman for many informative discussions. 1

2

YUE YANG AND LIANG YU

Given two structures A(A, ≤A ) and B(B, ≤B ) for L(≤), we say that A(A, ≤A ) is a substructure of B(B, ≤B ), write A(A, ≤A ) ⊆ B(B, ≤B ), if A ⊆ B and the interpretation ≤A is a restriction to A of ≤B . Definition 1.2. For n ≥ 0. Given structures A(A, ≤A ) and B(B, ≤B ) for L(≤). (i) We say that A(A, ≤A ) is a Σn substructure of B(B, ≤B ), write A(A, ≤A ) Σn B(B, ≤B ), if A(A, ≤A ) ⊆ B(B, ≤B ) and for all Σn formulas ϕ(x1 , x2 , ..., xn ) and any a1 , a2 , ..., an ∈ A, A(A, ≤A ) |= ϕ(a1 , a2 , ..., an ) if and only if B(B, ≤B ) |= ϕ(a1 , a2 , ..., an ). (ii) We say that A(A, ≤A ) is Σn -elementary-equivalent to B(B, ≤B ), write A(A, ≤A ) ≡Σn B(B, ≤B ), if for all Σn sentences ϕ, A(A, ≤A ) |= ϕ if and only if B(B, ≤B ) |= ϕ. For more model theoretic facts, please refer to [4]. 2. Background In this paper, we study the mode theoretical properties of ∆02 Turing degrees as a structure D∞ (D∞ , ≤) 1 of L(≤). We are interested various substructure of D∞ (D∞ , ≤ ), particularly, the structures of n-r.e. degrees Dn (Dn , ≤). 2 For simplicity, we use Dp to denote Dp (Dp , ≤) for p ∈ ω ∪ {∞}. Comparing the difference between Dp (Dp , ≤) and Dq (Dq , ≤), where p, q range over ω ∪ {∞}, has a long history beginning with Cooper(1970’s) and Lachlan’s (1968) unpublished work. They constructed a properly 2-r.e, actually n + 1-r.e, degree (i.e, Dn ⊂ Dn+1 ) and proved that each n + 1-r.e. degree bounds an non-recursvie n-r.e. degree. In this paper, we are interested in the relations between D1 and Dp . The first remarkable result related to the difference between Dn and D∞ was obtained by Slaman in 1983. Theorem 2.1 (Slaman). (i) There are r.e. sets A, B and C and a ∆02 set E such that – ∅ 0 ∧ z 6≥ c. Then for each n ≥ 1, Dn 6|= ϕ(a, b, c). If not, then there is an n-r.e. degree f > 0 so that f ≤ a and f ∪ b 6≥ c. But, by Cooper and Lachlan’s result that each n-r.e. degree bounds a r.e. degree, there is a non-recursvie r.e. degree w ≤ f . So g ∪ b 6≥ c. This is impossible by (i).  1Note 2We

we are using D∞ to denote ∆02 Turing degrees. use “1-r.e.” to denote “r.e.”

R IS NOT A Σ1 -ELEMENTARY SUBSTRUCTURE OF Dn

3

Theorem 2.1 says that Dn is different with D∞ as much as possible. Having proved Theorem 2.1, Slaman raised the following conjecture which remained open more than twenty years. Conjecture 2.2 (Slaman [2]). For each n > 1, D1 Σ1 Dn ? Furthermore, Lempp raised the following conjecture. Conjecture 2.3 (Lempp). For all n > m, Dm Σ1 Dn ? Before discussing the two questions above, we briefly survey some recent results related to the elementary difference between Dp ’s. The first baby result is the following “positive” theorem. Theorem 2.4 (Folklore). For any p, q ∈ ω ∪ {∞}, Dp ≡Σ1 Dq . Theorem 2.4 is the only one known positive result. The first difference between Dn and D∞ was obtained by Cooper, Lachlan and Sacks. Theorem 2.5. For all n ≥ 1, Dn 6≡Σ2 D∞ since (i) (Cooper 1971, Lachlan 1960’s) Dn |= ∀x∃y∃z(z ≤ y ≤ x ∧ x 6≤ y ∧ y 6≤ z). (ii) (Sacks [7])D∞ |= ∃x∀y∀z((y ≤ x ∧ z ≤ x ∧ x 6≤ y ∧ x 6≤ z) =⇒ z = y). Due to Theorem 2.8, people do not think that Dn and D∞ share any interesting properties. So it is natural to put more attention to compare Dn with Dm . One of the most important result related the difference between D1 and Dn is the following two theorems. Theorem 2.6 (Sacks[8]). D1 |= ∀x∀y∃z((x ≤ y ∧ y 6≤ x) =⇒ (x ≤ z ≤ y ∧ z 6≤ x ∧ y 6≤ z)). Theorem 2.7 (Cooper, Harrington, Lachlan, Lempp, Soare[2]). For each natural number n > 1, Dn |= ∃x∃y∀z(x 6≤ y ∧ (x ≤ z ≤ y =⇒ z ≤ x ∨ y ≤ z)). So the following results can be obtained. Corollary 2.8 (Cooper, Harrington, Lachlan, Lempp, Soare[2]). For each natural number n > 1, D1 6≡Σ2 Dn . Thus Corollary 2.8 sheds some light on the Conjecture 2.2. A further question is how difference between Dn and Dn+m for n > 1. Downey formulated the following ambitious question which is now known as Downey’s Conjecture. Conjecture 2.9 (Downey [3]). For each n > 1 and k ≥ 0, Dn ≡Σk Dn+m ? Ironically, Downey Conjecture is not that easy as it looks at the first face. Actually, it is very hard. The conjecture remained open more than fifteen years. The difficulty of Conjecture 2.9 lies in the technique used in the locally theory of Dn . Usually one can generalize a (local) result in D2 to Dn without any difficult. Recently, Arslanov, Kalimullin and Lempp announced a negative solution to Conjecture 2.9. They proved the following result.

4

YUE YANG AND LIANG YU

Theorem 2.10 (Arslanov, Kalimullin, Lempp [1]). D2 6≡Σ2 D3 . But the question whether Dn 6≡Σ2 Dn+m is true for very large n, m still remains open. We turn to the discussion about the relations between D1 and Dn . To solve conjecture 2.2, one possible argument is to build an finite array just as Slaman did in Theorem 2.1. However, by the Cooper and Lachlan observation that every nonrecursvie n-r.e. degree bounds a nonrecursvie r.e. degree, we cannot hope that any n-r.e. degree D plays the role of E as in Slaman Theorem. In fact Cooper and Lachlan’s observation is so critical that the second author thought that it should give a positive answer to Slaman’s conjecture. The original work of this paper began with a workshop at IMS of Singapore. During the workshop, Lempp announced the solution to Downey Conjecture and asked this question again. The second author thought that it should be true but Wang, a graduate in Nanjing University, tried to give a negative answer. Both of us failed. After the workshop, the first author began to be interested in the question. After a brief discussion, we arrived at an agreement that Slaman Conjecture may not be true and had a Σ1 formula by hand to refute the conjecture. Soon after we sent an email to Slaman to tell him our ideas, Ted suggested us a more reasonable Σ1 formula based on our formula. With a slight modification, eventually we have the following Σ1 formula ϕ(x1 , x2 , x3 , x4 ). ϕ(x1 , x2 , x3 , x4 ) ≡ ∃d∃g(d ≤ x1 ∧ d 6≤ x4 ∧ g ≥ x2 ∧ g ≥ d ∧ x3 6≤ g). The solution to Conjecture 2.2 follows from the following technique result: Theorem 2.11. There are r.e. sets A, B, C and E and a d.r.e. set D such that (1) D ≤T A and D 6≤T E; (2) C ≤ 6 T B ⊕ D; (3) For all r.e. set W (W ≤T A ⇒ either C ≤T W ⊕ B or W ≤T E). Assuming Theorem 2.11, we can obtain the following result to refute Slaman conjecture: Theorem 2.12. For all n > 1, D1 6Σ1 Dn . Proof. Assume n > 1. Let a, b, c, d, e be just as in Theorem 2.11. Note all of them except d belong to D1 and d belongs to Dn . By Theorem 2.11, just take g = b ∪ d ∈ Dn , Dn |= ϕ(a, b, c, e). However, by Theorem 2.11 again, D1 |= ∀w∀g((w ≤ a ∧ g ≥ w ∧ g ≥ b ∧ w 6≤ e) =⇒ c ≤ g). In other words, D1 |= ¬ϕ(a, b, c, e). 

R IS NOT A Σ1 -ELEMENTARY SUBSTRUCTURE OF Dn

5

3. Some further remarks (1) A reasonable question is what we can talk about the parameters of ϕ in Theorem 2.11. Inspired by Shore and Slaman [9], we conjecture that each high r.e. degree bounds the four parameters as in Theorem 2.11 so that Slaman conjecture fails. But is there a fragment E ⊂ D1 so that E  D2 ? Specifically, is E = {a ∈ D1 |a00 ∈ 000 } a Σ1 -substructure of D2 ? (2) Recently, Wang and Yu [11] proved that each non-principal ideal in D1 is a Σ1 -substructure of D1 . But it is unknown for D2 . A set A ⊆ Dn is said to be definable in Dn if there is a formula ψ so that a ∈ A if and only if Dn |= ψ(a). For D1 , by the recent work due to Nies [5], Yang and Yu [12], there are many definable non-principal ideals in D1 . A natural question is what about D2 ? To construct a non-principal idea in Dn , we just need to take a non-principal idea I in D1 and then build a non-principal idea J = {b|∃a ∈ I(b ≤ a)}. The problem is whether it is definable in Dn . We formulate the following questions which we are very interested in. Question 3.1. For n > 1, is there a non-trivial definable Σ1 -substructure of Dn ? (3) From the discussion above, we have seen that the definable ideals play a critical role in the study of global theory. Although there are some nontrivial definable ideals in D1 . It is unknown whether there are infinitely many definable ones in D1 . For D2 , we don’t even know whether there is a nontrivial definable ideal in it. Wang also recently studied definable filters in D1 . It is unknown whether there is a non-trivial definable filter in D2 . We say that an incomplete degree a ∈ Dn is cappable if there is an incomplete degree b ∈ Dn so that a∪b = 00 . Otherwise, a is said to be non-cappable. A possible candidate of definable filters is the collection of non-cappable degrees in D2 . To show that, it suffices to prove that each non-cappable non-cappable 2-r.e. degree bound a non-cappable r.e. degree. References [1] M. M. Arslanov, I Kailimulin, and S Lempp. On downey’s conjecture. to appear. [2] S. Barry Cooper, Leo Harrington, Alistair H. Lachlan, Steffen Lempp, and Robert I. Soare. The d.r.e. degrees are not dense. Ann. Pure Appl. Logic, 55(2):125–151, 1991. [3] Rod Downey. D.r.e. degrees and the nondiamond theorem. Bull. London Math. Soc., 21(1):43– 50, 1989. [4] Wilfrid Hodges. Model theory, volume 42 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1993. [5] Andr´e Nies. Parameter definability in the recursively enumerable degrees. J. Math. Log., 3(1):37–65, 2003. [6] Hilary Putnam. Trial and error predicates and the solution to a problem of Mostowski. J. Symbolic Logic, 30:49–57, 1965. [7] Gerald E. Sacks. A minimal degree less than 00 . Bull. Amer. Math. Soc., 67:416–419, 1961. [8] Gerald E. Sacks. The recursively enumerable degrees are dense. Ann. of Math. (2), 80:300–312, 1964. [9] Richard A. Shore and Theodore A. Slaman. Working below a high recursively enumerable degree. J. Symbolic Logic, 58(3):824–859, 1993.

6

YUE YANG AND LIANG YU

[10] Robert I. Soare. Recursively enumerable sets and degrees. Springer-Verlag, Berlin, 1987. [11] Wei Wang and Liang Yu. Realizing σ1 formulas in R. unpublished notes. [12] Liang Yu and Yue Yang. On the definable ideal generated by nonbounding c.e. degrees. J. Symbolic Logic, 70(1):252–270, 2005. Department of Mathematics, Faculty of Science, National University of Singapore, Lower Kent Ridge Road, Singapore 119260. E-mail address: [email protected] Department of Mathematics, Faculty of Science, National University of Singapore, Lower Kent Ridge Road, Singapore 117543. E-mail address: [email protected]