PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 7, July 1998, Pages 2131–2139 S 0002-9939(98)04314-7

ENUMERATIONS, COUNTABLE STRUCTURES AND TURING DEGREES

STEPHAN WEHNER

(Communicated by Andreas R. Blass)

Abstract. It is proven that there is a family of sets of natural numbers which has enumerations in every Turing degree except for the recursive degree. This implies that there is a countable structure which has representations in all but the recursive degree. Moreover, it is shown that there is such a structure which has a recursively represented elementary extension.

1. Introduction In the following we are concerned with countable structures in a recursive lan- guage. Researchers have investigated how one could measure the intuitive idea of information content of such structures and tried to relate each one of them to a Turing degree [2], [3], [1]. The natural starting point is to look at the collection of representations.LetAbe a structure. If B is an isomorphic structure with universe ω,thenBis called a representation of A (written B A). D(B), its open diagram, can be regarded as a subset of ω so that it has a Turing' degree, and one can look at the collection of degrees deg(D(B)) : B A . A first guess for capturing the complexity of A would be to{ let its degree be' the} least element of this collection, especially in the light of the following theorem [2, Theorem 4.1]: Theorem 1.1 (Knight). Let A be a structure in a relational language. Then ex- actly one of the following holds: (1) For any d > deg(D(A)), there is a representa- tion B of A such that deg(D(B)) = d. (2) There is a finite subset S of the universe of A such that all permutations of the universe which fix S are automorphisms of A. But this idea fails. For example, Richter [3, Theorem 3.3] shows that for any countable order C which has no recursive representation the collection deg(D(B)) : B C has no least element. Therefore more involved concepts have{ been tried to assign' } degrees to structures [1]. Now for the particular problems addressed in this paper. Steffen Lempp asked (unpublished): Does a structure with representations in all non-recursive degrees have a recursive representation? Julia Knight asked some related questions: With 2 a binary relation R ω associate a family of subsets of ω given by R := Rn : ⊆ F { n ω ,whereRn := x :(n, x) R ;saythatRis an enumeration of R. She ∈ } { ∈ } F Received by the editors September 17, 1996 and, in revised form, January 6, 1997. 1991 Mathematics Subject Classification. Primary 03D45. Many thanks go to Julia Knight and Carl Jockusch!

c 1998 American Mathematical Society

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asked (also unpublished): If a family has the feature that for every non-recursive set X, has an enumeration recursiveF in X,does have a recursive enumeration? Similarly,F if for every non-recursive set X, has anF enumeration r.e. in X,does have an r.e. enumeration? F F In the next section we give some positive results on Knight’s questions under extra hypotheses. In Section 3 we prove that the answer to Knight’s questions is negative, by constructing a single suitable family. This implies that Lempp’s question also has a negative answer, as is shown in the last section. The same finding is obtained in [4], but by another approach. We close by discussing the difference. The notation is quite standard and follows [5]. All sets considered are subsets of ω, the set of natural numbers. We call countable collections of subsets of ω families.Letϕbe a G¨odel-numbering of the partial recursive functions (of varying arity) and Wi := Rng(ϕi) be an enumeration of the recursively enumerable sets as usual. Wi,s is the set of numbers enumerated in Wi by stage s. We use recursive bijections , , , , between ω and ω2, ω and ω3, respectively. We also use h· ·i h· · ·i projections (.). and (.). so that, for example, ( a, b ). = a.Let x, A be the set 1 2 h i 1 h i x, a : a A . Similarly, A.2 := (a).2 : a A .WeletA+x= a+x:a A {hand Ai x∈= }b : b + x A . { ∈ } { ∈ } Fix− an effective{ listing∈ Ω} of recursively enumerable enumerations of all families with r.e. enumerations:

(e) Ω := (i, x): i, x We , { h i∈ } (e) (e) and write Ωi for x : i, x We ,thei-th set of the enumeration Ω . Define (e) (e) { h i∈ } (e) := Ωi : i ω to be the family enumerated by Ω . C D : ω{ 2ω denotes∈ } the canonical enumeration of the family of finite sets; write → Dn for the n-th finite set. Then the binary predicates x Dn and x = Dn are recursive. ∈ | |

2. Positive results In this section we give conditions on a family which ensure that the implications of Knight’s two questions hold. These are given in Theorems 2.3 and 2.4 and are due to Julia Knight. Jockusch gave a proof of Theorem 2.4 which also showed the following: There is a property which families with a recursive (r.e.) enumeration share with families that have, for all non-recursive degrees d,anenumerationre- cursive (r.e.) in d. These seem to be the only positive statements possible about such families. What is this property? By a rather straightforward forcing construction it fol- lows that if a set of natural numbers is recursive (r.e.) in all non-recursive degrees, then it is recursive (r.e.). Therefore, the members of a family are recursive (recur- sively enumerable) if the family has, for all non-recursive degrees d, an enumeration recursive (r.e.) in d. Hence such families are fully described by the index set F

Ir( ):= i:( A )(ϕi = χA) F { ∃ ∈F } or, respectively,

Ire( ):= i:( A )(Wi = A) . F { ∃ ∈F }

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Both the index set Ir of a family with a recursive enumeration and the index set 0 Ire of a family with a recursively enumerable enumeration are Σ3 in the arithmetical hierarchy. Here is the coincidence: Theorem 2.1. Let be a family. If, for every non-recursive degree d, has an F 0 F enumeration recursive in d, then its index set Ir( ) is Σ . F 3 Theorem 2.2 (Jockusch). Let be a family. If, for every non-recursive degree d, F 0 has an enumeration recursively enumerable in d, then its index set Ire( ) is Σ . F F 3 Towards giving a sufficient condition under which the implication of her first question holds, Julia Knight defines an extension function for a family to be a (possibly partial) function f :2<ω ω such that if σ 2<ω and there existsF a set → ∈ A such that χA σ,thenϕf(σ) =χA for some such A. We mention two facts: Any∈F family with a recursive⊇ enumeration has a partial recursive extension function, and so does a family containing all finite sets. We prove Theorems 2.1 and 2.2 simultaneously with the following two. Theorem 2.3 (Knight). Let be a family which has, for all non-recursive d,an enumeration recursive in d.IfF has a partial recursive extension function, then has a recursive enumeration. F F Proof (of Theorems 2.3 and 2.1). Let be a family which has, for any non-recur- sive set X, an enumeration recursive in FX. We construct a generic set D, attempting to meet the following requirements and expecting to fail. Below, this failure will be exploited to prove the statements of the two theorems for separately. F D Re: ϕ is not the characteristic function of an enumeration of . e F The set D will be Cohen-generic. The set of forcing conditions is 2<ω and the partial order is given by . We use the old-fashioned notion of a complete forcing ⊆ sequence (c.f.s.), where pn pn, with pn entering the nth dense set in some +1 ⊇ +1 countable collection. D is the set with characteristic function n ω pn. Fix a condition p 2<ω and e ω. We consider the following∈ four possibilities ∈ ∈ S for p and e, showing how extensions of p may force satisfaction of Re in each case. P1. For some q p, n, x ω, q ϕD(n, x) =0,1, or q ϕD(n, x) . ⊇ ∈ ` e ↓6 ` e ↑ We include q in the c.f.s., thereby satisfying Re. P2. For some q p and some n, for all q0 q there exist x and r0,r1 q0 such D⊇ ⊇ ⊇ that ri ϕe (n, x)=i. For each` A the set ∈F 1 D D := r : r q ( x)(r ϕ (n, x) = χA(x)) A { ⊇ ⇒ ∃ ` e 6 } 1 is dense. We add q to the c.f.s. and enter the sets DA. Then the requirement Re is satisfied.

We write q En = A if for all x and all q0 q there is an r q0 such that D ` ⊇ ⊇ r ϕ (n, x) = χA(x). Note that for q and n thereisatmostonesetAsuch that ` e ↓ q En = A. ` P3. For some q p, n ω and B we have q En = B. ⊇ ∈ 6∈ F ` By putting q into the c.f.s., we meet Re. P4. Not P2 but there exist A and q p such that for all q0 q and all n,if ∈F ⊇ ⊇ q0 En=Bthen A = B. ` 6

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Since P2 does not hold, the set

2 D := q0 : q0 q ( B)(q0 En = B) n { ⊇ ⇒ ∃ ` } 2 is dense for each n. By including q in the c.f.s. and entering the sets Dn,we meet Re. If for all p 2<ω and e ω one of the cases P1,...,P4 holds, then the forcing construction∈ yields a generic∈ (and so non-recursive) set D,inwhich has no recursive enumeration, contrary to the assumption on . F So let p 2<ω and e ω be such that none of theF cases P1,...,P4 hold. It ∈ ∈ follows that if q En = A,thenA for all q p, n and A ω. Moreover, all elements of occur` in this way. ∈F ⊇ ⊆ We first completeF the proof of Theorem 2.3. Let f be a partial recursive extension function for .Fixn ωand p q 2<ω.Saythatσ 2<ω has a q-computation F ∈ ⊆D ∈ ∈ if there is an r q such that r ϕe (n, x)=σ(x) for all x dom(σ). We make the following observations:⊇ ` ∈

q En = A and q0 q implies q0 En = A. • The` set σ 2<ω :⊇σ has a q-computation` is r.e. (uniformly in q). • { ∈ } <ω By definition, q En = A if and only if any σ 2 has a q-computation if • ` ∈ and only if σ χA. ⊆ <ω Since p, e do not satisfy P1, for any q0 q there is a σ 2 which has a • ⊇ ∈ q0-computation. The previous two items imply that the predicate q p ( A)(q En = A) • in q and n is r.e. ⊇ ⇒ ∀ 6` Since p, e do not satisfy P3, if q p and q En = A for any A ,thenfor • <ω ⊇ 6` ∈F any σ 2 with a q-computation there is a set A such that σ χA. ∈ ∈F ⊆ We form a recursive enumeration R of , using pairs (q, n) as indices, where q 2<ω with q p and n ω. SinceF there is a recursive bijection between ∈ ⊇ ∈ <ω (n) P ω and ω,whereP = q 2 : q p , this suffices. Let (σq )n ω be an× effective enumeration of {σ ∈ 2<ω : σ⊇has} a q-computation by the second∈ observation. We choose this{ enumeration∈ so that, additionally, for} every σ with (i) a q-computation there are infinitely many i such that σq = σ. Define a partial (m) (aq (n)) recursive function aq by aq(0) := 0 and aq(n +1):=µm > n.σq σq .By the fourth observation, let h : ω 2<ω ω be a recursive function⊇ with range → × (q, n):q pand ( A)(q En = A) . { ⊇ ∀ 6` } R(q,n) is defined by

(aq (i)) 0 i m σq ϕ aq (m) if m is least such that h(m)=(q, n), ∪ f(σq ) R := ≤ ≤ (q,n)  S  (aq(i))  i ω σq otherwise. ∈ Fix q p andSn ω. By definition of the functions h,thereismsatisfying the ⊇ ∈ first case if and only if ( A)(q En = A). If q En = A, then the choice of (n) ∀ 6` ` the enumeration (σq )q 2<ω ,n ω and the function a guarantees R(q,n) = A.If ∈ ∈ q En = A for any A, then by the fact that f is an extension function for ,it follows6` that R . F (q,n) ∈F Since for every A there are q and n such that q En = A, it follows that R is an enumeration of∈F, and Theorem 2.3 is proved. ` F

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We turn to the proof of Theorem 2.1. As mentioned above, if none of the cases holds for p and e,thenwehave

= X:( q p)( n)(q En = X) . F { ∃ ⊇ ∃ ` } The ternary relation (q En = X) (χX = ϕi)(inq,nand i), when restricted ` ∧ 0 0 to i such that ϕi is the characteristic function of a set, is Π so that Ir( )isΣ . 2 F 3 Theorem 2.4 (Knight). Let be a family such that for all non-recursive X, has an enumeration r.e. in XF.If contains all finite sets, then has an r.e.F enumeration. F F Proof of Theorems 2.4 and 2.2. As above, we construct a generic set D, attempting to meet the following requirements and expecting to fail. D Re. W is not an enumeration of . e F We use the same forcing notion as was used for the previous proof. Fix p 2<ω and e ω. We consider the following three cases. ∈ ∈ C1. For some q p and some n ω, for all q0 q, there exist x ω and r0,r1 q0 ⊇ D∈ ⊇ D ∈ ⊇ such that r0 (n, x) We and r1 (n, x) We . For each A ` the set∈ ` 6∈ 1 ∈F D DA := r : r q ( x)[(r (n, x) We x A)or { ⊇ ⇒ ∃ (r` (n, x) ∈ W D ∧ x 6∈ A)] ` 6∈ e ∧ ∈ }

1 is dense. We put q into the c.f.s. and enter the sets DA.RequirementReis satisfied.

We write q En = A if for all x,ifx A, then for all q0 q there is r q0 such that r (n,` x) W D,andifx A then∈q (n, x) W D. ⊇ ⊇ ` ∈ e 6∈ ` 6∈ e C2. Case C1 does not hold, but q En = B for some q p, n ω and B . ` ⊇ ∈ 6∈ F Requirement Re is met by putting q in the c.f.s. C3. Case C1 does not hold, but there exists A such that for all q0 q and ∈F ⊇ all n,ifq0 En =B,thenA=B. Since Case` C1 does not hold,6 the set 2 D := q0 : q0 q ( B)(q0 En = B) n { ⊇ ⇒ ∃ ` } is dense for all n. We meet the requirement Rn by including q in the c.f.s. 2 and entering all sets Dn. If for all p 2<ω and e ω one of the cases C1,C2, or C3 holds, then the forcing construction∈ yields a∈ generic (and hence non-recursive) set D,inwhich has no r.e. enumeration. This contradicts the assumption of both Theorem 2.2 andF Theorem 2.4. So let p 2<ω and e ω be such that none of the cases C1,C2,C3 ∈ ∈ 0 holds. We show that the index set Ire of is Σ . As in the proof of Theorem 2.3, F 3 for all q p, n,andA ω,ifq En =A,thenA , and all members of occur in this⊇ way. Therefore⊆ we have` ∈F F

i : Wi = i:( q p)( n)(q En = Wi) . { ∈F} { ∃ ⊇ ∃ ` } 0 0 The relation “q En = Wi”isΠ2, and so the index set of is Σ3. This completes the proof of Theorem` 2.2. To complete the proof of TheoremF 2.4, note that since includes all finite sets and only contains r.e. sets, by a theorem of Yates [7, TheoremF 8], has an r.e. enumeration. F

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3. A family of finite sets In this section we first define a family which has no r.e. enumeration. Then we show that for every non-recursive set XC there is an enumeration of which is recursive in X, and finally that every non-recursive degree contains an enumerationC of . This corrects an earlier statement in [6, p 187]. In particular, we apologize forC connecting the error with Martin Kummer. Let r be the partial recursive function defined by

(e) r(e):=(µi, x, s . e, x Ω ). . h i h i∈ i,s 1 Informally, the value of r(e) is the first index i to be found such that there is a (e) number of the form e, x Ωi ; if there is no such i then r(e) is not defined. Let the family be definedh byi∈ C := e, A : A is finite,e ω e, ω Ω(e) :r(e) . C {h i ∈ }−{h i∩ r(e) ↓} does not have an r.e. enumeration; for suppose Ω(e0) is an enumeration of .Then C (e0) (e0) C r(e0) is defined, and the set Ω = e, ω Ω is not a member of . r(e0) h i∩ r(e0) C Let X be an arbitrary non-recursive set. To see how to construct an enumeration X X S of such that S T X we use the following lemma. C ≤ X Lemma 3.1. Uniformly in i and recursively in X there is a finite set Ai T X X ≤ such that Wi = A . 6 i Let g be a partial recursive function such that g(e) if and only if r(e) and (e) ↓ ↓ if r(e) then W =(e, ω Ω ). .Lethbe a partial recursive function such ↓ g(e) h i∩ r(e) 2 that h(e, a) if and only if g(e) and if g(e) then Wh(e,a) = Wg(e) a. Define ↓ ↓ ↓ − X X S n,t,e := e, B (n, t, e) , h i h i where X Dn (s0 +A )ifthereiss>t,max(Dn)+1 ∪ h(e,s0) such that gs(e) and W = Dn, BX (n, t, e):= ↓ g(e),s  and s0 is the least such,  Dn otherwise.  By the lemma, SXis recursive in X. We claim that SX is an enumeration of . “ ”. First of all, it follows from the lemma that all sets BX (n, t, e) are finiteC and⊆ so the family enumerated by SX is contained in e, A : A is finite . X (e) {h i X } Suppose r(e) and S n,t,e = e, ω Ωr(e). It follows that B (n, t, e)=Wg(e). ↓ h i h i∩ There has to be a stage s>t,max(Dn) + 1 such that Wg(e),s = Dn, because X otherwise B (n, t, e)=Dn =Wg(e), a contradiction. Let s0 be the least such s. X 6 X X Then B (n, t, e)=Dn (s0+A )=Wg(e),sothatA = Wh(e,s ),a ∪ h(e,s0) h(e,s0) 0 contradiction. “ ”. Let e, n ω such that C = e, Dn . We want to find a number z such ⊇ X ∈ h i∈C that Sz = C.Lettbe such that Wg(e),s = Dn for all s>t. (If there is no such t X 6 X then C .) By definition, B (n, t, e)=Dn,sothatS = C. 6∈ C n,t,e We have constructed an enumeration of which is recursiveh i in X. Simple coding is sufficient to obtain an enumeration T-equivalentC to X:ChooseA, B so that ∈C

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A B = . Define another enumeration of , − 6 ∅ C A if n X, X X X ∈ P2n := Sn ,P2n+1 :=  B if n X.  6∈ X Clearly, P is recursive in X and enumerates .Letx A B.Thenz Xif and only x P X ,sothatXis recursive in PCX . ∈ − ∈ ∈ 2z+1 Proof of Lemma 3.1. Let y,n : ω2 ω be two recursive one-one functions such that their ranges are disjoint and cover→ ω.Let y(s, x)ifx X, α(s, x):= ∈   n(s, x)otherwise, X so that α T X. We construct the set A in stages uniformly in i and recursively ≤  i in X. In the course of the construction, numbers ax may become defined. X Stage 0. Set x := 0. Ai,0 is empty. X Stage s+1. If Wi,s = Ai,s, pass to the next stage. Otherwise enumerate X 6 α(s, x)inAi ,letax := s and increase x by one. End of construction. X The set Ai is (by construction) r.e. in X. It is also recursive in X: By inspection, X X X Ai only contains numbers α(a, b). If y(s, x) Ai ,theny(s, x) Ai,s+1 and the same holds for numbers n(s, x). This together with∈ the choice of y and∈ n is sufficient.

X Claim 1.ThesetAi is finite. X X Suppose Ai is infinite. During the construction of Ai infinitely many numbers X ax are defined. By induction on x, it follows that A = α(aj ,j):j x ,and i,ax { ≤ } therefore Wi,a = α(aj ,j):j

4. Application to Lempp’s question Let be a family. With we associate the following countable structure A in the languageF L =(S, Z, I),F where S is a binary predicate symbol and Z and FI are unary predicate symbols. The universe of A is ω ω. For every A , set Z((A, x, 0)) and S((A, x, n), (A, x, n + 1)). SetF I((F×A, x, n×)) if and only if n ∈FA. Thus, countably many S-chains (A, x, 0), (A, x, 1), (A, x, 2),... are associated∈ with every A , and in every chain I holds of the n-th member if n A. ∈F ∈ Theorem 4.1. Let d be a Turing degree and be a family. Then has an enumeration recursive in d if and only if the structureF A has a representationF recursive in d. F

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Proof. “ ”. Let QX be an enumeration of which is recursive in X d. Define X ⇒ F X X ∈ R to be another X-recursive enumeration of by R n,i := Qi . A representation F h i of A is given by B,whereZB(x, 0 ), SB( x, n , x, n +1), and IB( x, n )if F X h i h i h X i h i and only if x Rn . These predicates are recursive in R , and therefore D(B)is recursive in X∈. “ ”. Let B be a representation of A whose open diagram is recursive in X d. ⇐ F ∈ The set Z := x : ZB(x) is recursive in X. With each x Z we associate the { (n) } ∈ set Sx := n : IB(f (x)) ,wheref(x) is the unique y such that SB(x, y), and (n) { } f (x) denotes the n-fold application of f to x.ThesetsSxare uniformly in x Z ∈ recursive in X, and, by definition of A , = Sx : x Z . This suffices. F F { ∈ } Corollary 4.2 (Slaman [4]). There is a structure which has representations only in the non-recursive degrees. Proof. Apply the theorem to the family defined in the previous section to obtain representations of A below any non-recursiveC T-degree. Apply Theorem 1.1 to obtain representationsC in all non-recursive T-degrees. Slaman remarked (private communication) that the construction given in [4] yields a structure which is not elementarily equivalent to any recursively represented structure. Informally, the reason for this is as follows. Essentially, the construction proceeds in such a way that the final outcome of the actions taken to diagonalize against the recursive representations can be read off the theory of the structure (which is called M): 1 R He writes ([4, Section 2.1]): “We will ensure that either R− (Ti) i = ,or R M 1 ∅ R Ti,

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It suffices to show that for any formula ( x)(φ) in variables x ,... ,xn,if ∃ 1 A =( x)(φ)[x1 := c1,... ,xn := cn] D | ∃ and c1,...cn A , then there is c A such that ∈ C ∈ C A = φ[x := c, x1 := c1,... ,xn := cn]. D | Note that if A =( x)(φ)[x1 := c1,... ,xn := cn], then there is a d in A such that either D | ∃ D

A =φ[x := d, x1 := c1,... ,xn := cn] D | S(d, c ) S(c ,d) ... S(d, cn) S(cn,d), ∧¬ 1 ∧¬ 1 ∧ ∧¬ ∧¬ or

A =φ[x := d, x1 := c1,... ,xn := cn] D | (S(d, c ) S(c ,d) ... S(d, cn) S(cn,d)). ∧ 1 ∨ 1 ∨ ∨ ∨ In the former case, choose c from A outside of the S-chains of A which c1,... ,cn belong to such that it satisfies Z andC I inthesamewayddoes. InC the latter case, d is already part of an S-chain which is contained in A , and so an element of A . C C References [1] Christopher J. Ash, Carl G. Jockusch, jr. and Julia F. Knight; Jumps of orderings,Trans. Amer. Math. Soc., vol. 319, (1990), p. 573 – 599. MR 90j:03081 [2] Julia F. Knight; Degrees Coded in Jumps of Orderings, J. Symbolic Logic, vol. 51, (1986), p. 1034 – 1042. MR 88j:03030 [3] Linda J. Richter; Degrees of Structures, J. Symbolic Logic, vol. 46 (1981), p. 723 – 731. MR 83d:03048 [4] Theodore A. Slaman; Relative to any Nonrecursive Set, Proc. Amer. Math. Soc., vol. 126 (1998), 2117–2122. CMP 97:11 [5] Robert I. Soare; Recursively Enumerable Sets and Degrees, Springer Verlag, Berlin, Heidel- berg, New York, Tokyo, 1987. MR 88m:03003 [6] Stephan Wehner; On Injective Enumerability of Recursively Enumerable Classes of Cofinite Sets, Arch. Math. Logic, vol. 34, (1995), p. 183 – 196. MR 96d:03062 [7] C.E.M. Yates; On the Degrees of Index Sets II, Trans. Amer. Math. Soc., vol. 135 (1969), p. 249 – 266. MR 39:2637

Department of Chemistry, University of British Columbia, Vancouver, British Columbia, Canada V6T 1Z1 E-mail address: [email protected]

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