A Minimal Pair of Turing Degrees

Let P(A) be the following property, where A is any infinite set of natural numbers: (∀X)[X ⊆ A ∧ |A − X| = ∞ ⇒ A ≤m X]. Let (R,≤) be the partial ordering of all the r.e. Turing degrees. We propose the study of the order theoretic properties of the substructure (Sm,≤Sm ), where Sm =dfn {a ∈ R: a contains an infinite set A such that P(A) is true}, and ≤Sm is the restriction of ≤ to Sm. In this paper we start by studying the existence of minimal pairs in Sm.


Introduction
For every infinite set A let us formulate the following property P(A): ( where ≤ m denotes the many-one reducibility.Sets with the property (1) are known as m-introimmune sets.This terminology was introduced in (Cintioli & Silvestri, 2003) to denote those sets that fail to be m-introreducible in a strong way, namely, those infinite sets that are not many-one reducible to any of their co-infinite subsets.
In particular, every infinite set A satisfying property (1) does not contain subsets of higher many-one degree.
Let (R, ≤) be the partially ordered structure of all the r.e.Turing-degrees.Let us define S m = dfn {a ∈ R : a contains an infinite set which has the property (1)}.In this paper we propose the study of the partially ordered substructure (S m , ≤ S m ), where ≤ S m denotes the order ≤ restricted to S m .We know that 0 ∈ S m (Cintioli, 2005), that S m contains a low 1 Turing degree (Cintioli, 2011), and that 0 S m (Cintioli & Silvestri, 2003).What other properties hold for (S m , ≤ S m )?In particular, what order theoretic properties hold for (S m , ≤ S m )?For example: 1) is (S m , ≤ S m ) dense?
3) does (S m , ≤ S m ) has minimum element?or minimal elements?and so on.This problematic collapses if it is true that S m = R−{0}: essentially all the order theoretic properties true in (R, ≤) will be inherited by (S m , ≤ S m ) but at most some exception, due to the fact that 0 S m .However, this study could be extended to substructures (S r , ≤ S r ) for other strong reducibilities ≤ r with ≤ r ≤ T , where S r = {a ∈ R : a contains an infinite set which has the property (1) with ≤ r in place of ≤ m }.The reason why ≤ r must be different from ≤ T is that we know by the results contained in (Jockusch,1973) and (Simpson,1978) that S T = ∅.We do not know if S r ∅ for strong reducibilities ≤ r ≤ m .
In this paper we start by studying the existence of minimal pairs in S m .The existence of minimal pairs in S m follows by known results: such pairs are constituted of two high 1 Turing degrees.The class of high 1 Turing degrees is an important class of Turing degrees below 0 .Another important class of Turing degrees below 0 is that one of low 1 Turing degrees.We recall that a Turing degree a < 0 is low 1 if a = 0 .In this paper we provide a minimal pair in S m constituted of two low 1 Turing degrees.

Notations
Throughout the paper we will use some concepts of Computability Theory without define them.We refer to Odifreddi (1999), Rogers (1967) and Soare (1987) for a full exposition on the subject.
Letter N denotes the set of natural numbers.Given two sets A, B ⊆ N, A − B denotes the set theoretic difference of A and B, and |A| denotes the cardinality of A. We fix an acceptable numbering {ϕ e } e≥0 of all the Turing computable unary functions.{W e } e≥0 is the corresponding enumeration of all the recursively enumerable (r.e.) sets.For every e, s ∈ N, W e,s is the finite approximation of W e obtained by performing s steps in the enumeration of W e .For every e ∈ N and every X ⊆ N, ϕ X e : N → N denotes the unary partial function computable by the e-th oracle Turing machine with the aid of the oracle X.For every e, s, x ∈ N and for every oracle X, ϕ X e,s (x) denotes ϕ X e (x) if the e-th oracle Turing machines with oracle X on input x halts in t ≤ s steps, and in this case we write ϕ X e,s (x) ↓; otherwise, we say that ϕ X e,s (x) is undefined.We assume here that if ϕ X e,s (x) ↓ then x, e < s and the elements asked to the oracle X in the computation of ϕ X e,s (x) are less than s.Finally, given two sets A, B ⊆ N, A is many-one reducible to B, in short A ≤ m B, if there exist a recursive function f : N → N such that for every x ∈ N, x ∈ A ⇔ f (x) ∈ B. In this case we say that f m-reduces A to B.

Main Result
We know that (S m , ≤ S m ) has the maximum element 0 because every cohesive co-r.e.set, actually every cohesive set, satisfies the property (1) (Lemma 3.2 below), and we know of the existence of maximal sets that are Σ 0 1 -complete w.r.t. the Turing reducibility ≤ T (Yates, 1965).For what concerns the width of S m we can say at the moment that because every high 1 r.e.Turing degree contains sets which have the property (1), for example the co-maximal sets of the high 1 r.e.Turing degree.The inclusion (2) is proper, because there are low 1 r.e.Turing degrees in S m (Cintioli, 2011), and the present paper.
In this paper we propose the study of the order theoretic properties of the substructure (S m , ≤ S m ).To begin with, we study here the existence of minimal pairs in S m .We recall that two nonrecursive Turing degrees a and b form a minimal pair if for every Turing degree c, c ≤ a and c ≤ b implies c = 0.By the existence of minimal pairs in S m we derive two consequences about the structure (S m , ≤ S m ): (i) there is not minimum, because 0 is not in S m , (ii) (S m , ≤ S m ) is not a lower semi-lattice, because such minimal pair does not have greater lower bound in S m .
We must say that we got (ii) in an unusual way, and even in some sense opposite to the usual way of proving that a certain partially ordered structure of r-degrees is not a lower semi-lattice.In fact, if the structure has minimum, then two r-degrees forming a minimal pair are nontrivial examples of elements having greatest lower bound, and such elements typically do not certify that the structure is not a lower semi-lattice.
Actually, we observe that the existence of minimal pairs in S m is a consequence of known results, mainly of a result contained in Lachlan (1966).Exactly, we are referring to the following theorem: Theorem 3.1 (Lachlan, 1966, cf.Theorem 2, p. 545) There exist maximal r.e.sets A, B, the greatest lower bound of whose degrees is 0.
The Turing degrees of the two sets of Theorem 3.1 form a minimal pair in S m , because every cohesive set possesses property (1), as stated in the following lemma.
Lemma 3.2 (Cintioli, 2005) Let C be a cohesive set.Then, for every X The Turing degrees of the two sets A and B of Theorem 3.1 are high 1 , because of the maximality of A and B (Martin, 1966).We now prove the existence of a minimal pair in S m constituted of two low 1 Turing degrees, and this is our main contribution of this paper.
Theorem 3.3 There are two low 1 Turing degrees in S m which form a minimal pair.
Proof.By the minimal-pair method we will construct two co-immune low 1 r.e.sets A and B with both A and B satisfying property (1), and such that their Turing degrees form a minimal pair.Namely: The two sets A and B will be constructed by infinitely many stages.At every stage s ≥ 0 we will define two finite sets A s and B s , with A 0 = B 0 = ∅, A s ⊆ A s+1 and B s ⊆ B s+1 .Our final sets will be A = s≥0 A s and B = s≥0 B s .We now describe our strategy.

Strategy
The construction of the two sets A and B is perfectly symmetric.It suffices to meet the following requirements, for every e, i, j ∈ N: The definitions of the requirements N 9e+4 and N 9e+5 are justified by the following known lemma.
Furthermore, by an argument of Posner, to satisfy all the requirements {N 9 i, j +8 } i, j≥0 it suffices to satisfy all the requirements {N 9e+8 } e≥0 formulated in a simpler way.Precisely: Lemma 3.5 For every e ≥ 0, let N 9e+8 be the requirement If all the requirements P 9e , P 9e+1 , N 9e+2 , N 9e+3 and N 9e+8 are met for every e ≥ 0, then all the requirements N 9 i, j +8 are met for every i, j ≥ 0.
Proof.From the hypothesis it follows that both A and B are not recursive, and this implies that A B. Without loss of generality assume that A B, and let x 0 ∈ A − B. For the sake of contradiction, assume that there are i 0 and j 0 such that Let e = e(x 0 , i 0 , j 0 ) be such that for every oracle X and every Then clearly ϕ A e = ϕ A i 0 and ϕ B e = ϕ B j 0 .It follows from (3) that ϕ A e = ϕ B e = C with C not recursive, contrary to the hypothesis that N 9e+8 is met.
By the above lemma we can replace each requirement N 9 i, j +8 with the requirement ⇒ C is recursive.Requirements P 9e and P 9e+1 are positive, because to satisfy them we will enumerate numbers into A and B. Requirements N 9e+2 , N 9e+3 , N 9e+4 , N 9e+5 and N 9e+8 are negative, because to satisfy them we will keep numbers out of A and B. Requirements R 9e+6 and R 9e+7 are positive and negative, because to satisfy them we will keep some numbers out of A and B, and we will enumerate some others numbers into A and B.

Actions to Fulfill the Requirements
In this subsection we give a description of the actions to fulfill all the requirements.These actions will be formally stated in the subsequent subsections Requirements requiring attention and Active requirements.We will use a restraint function r : N × N → N with r(n, 0) = −1 for every n ∈ N. To fulfill requirements P 9e and P 9e+1 we do not need to restrains elements, being both P 9e and P 9e+1 positive, thus we conventionally set r(9e, s) = r(9e+1, s) = −1 for every e, s ∈ N. We said above that the construction of the two sets A and B is symmetric.Therefore, except for N 9e+8 , we describe the actions to fulfil requirements relative to one set, say A.
The actions to fulfill P 9e , N 9e+2 and N 9e+4 are the usual.
• For P 9e , we wait for an opportune stage s + 1 ≥ 9e, if it exists, such that − W e,s ∩ A s = ∅, and − there is a number x ∈ W e,s not restrained by requirements of higher priority than P 9e .
Then, we enumerate x into A s+1 .If W e is infinite, then eventually W e A.
• For N 9e+2 we wait for an opportune stage s + 1 ≥ 9e + 2 such that no number x ≥ e is restrained yet, that is r(9e + 2, s) = −1, and we set r(9e + 2, s + 1) = the minimum number x ∈ A s with x ≥ e.If x will not enumerated into A later, then such x will certificate that N 9e+2 is met.
• For N 9e+4 , we wait for an opportune stage s+1 ≥ 9e+4, if it exists, such that ϕ A s e,s (e) is defined and the elements used in the computation of ϕ A s e,s (e) are not restrained yet, that is r(9e + 4, s) = −1.Then it suffices to set the restraint function r(9e + 4, s + 1) = s, by remembering that the computation of ϕ A s e,s (e) ↓ uses only elements less than s.This guarantees that ϕ A e (e) is defined, if no numbers x ≤ s will be enumerated into A after stage s + 1.For what concerns requirement R 9e+6 , we describe first the strategy to meet it.Strategy: to prevent that ϕ e mreduces A to some its subset X with |A − X| = ∞ it is enough to have a number y such that y ∈ A and ϕ e (y) A.
(4) Therefore, we do the following actions: we wait for an opportune stage s + 1 ≥ 9e + 6 such that for some y ≤ s + 1 ϕ e,s (y) ↓ y and y ∈ A s , with ϕ e,s (y) not restrained by requirements of higher priority than R 9e+6 .Then we force ϕ e to be wrong by enumerating ϕ e,s (y) into A s+1 and keeping y out of A by setting r(9e + 6, s + 1) = y.If ϕ e is a potential m-reduction of A to some X ⊆ A with |A − X| = ∞, then the certainty that s and y of (5) exists is a consequence of the immunity of A, as stated in the following lemma.
Lemma 3.6 Let C be any immune set and let X be a subset of C with |C − X| = ∞.Let us suppose that C ≤ m X via a recursive function f .Then, the set with C − X infinite by hypothesis.So f (C − X) cannot be finite, otherwise f −1 ( f (C − X)) would be an infinite r.e.subset of C, contrary to the hypothesis that C is immune.
For the requirement N 9e+8 we employ the minimal-pair method, which is based on the particular definition of the restraint function, in our case the definition of r(9e + 8, s), for every e, s ≥ 0. For a full exposition of the minimalpair method see for example either (Odifreddi, 1999, p. 543 et seq.), or (Soare, 1987, p. 152 et seq.).At every stage s of the construction of the two sets A and B we define the length-agreement function l(e, s), for every e ≥ 0, in the following way: l(e, s) = dfn max{x : (∀y < x)[ϕ A s e,s (y) ↓= ϕ B s e,s (y) ↓]}.
Furthermore, at every stage s we define the restraint function r(9e + 8, s) by induction on e ≥ 0.
Definition 3.7 For every stage s ≥ 0, let us define if s is a 0-expansionary stage, max{t : t < s and t is a 0-expansionary stage}, otherwise, where a stage s is 0-expansionary if either s = 0 or l(0, s) > max{l(0, t) : t < s}.
Given r(9e + 8, s), define r(9(e + 1) + 8, s) as the maximum of: 2) {t : t < s and r(9e + 8, t) < r(9e + 8, s)}, 3) {t : t < s, r(9e + 8, t) = r(9e + 8, s) and stage t is (e + 1)-expansionary, if s is not an (e + 1)-expansionary stage}, where a stage s is (e + 1)-expansionary if either s = 0 or It is possible to prove that for every e ≥ 0 lim inf s→∞ r(9e + 8, s) < ∞. (8) From Definition 3.7 clearly it follows that for every e, s ≥ 0 Therefore, for every e ≥ 0 lim inf and (10) will allows us to prove that all the P-requirements and R-requirements are met.Moreover, the minimalpair method guarantees that to satisfy each requirement N 9e+8 two conditions are sufficient: C1: every P-requirement and R-requirement requires attentions at most finitely often, and C2: at every stage s > 0 at most one side of the equality ϕ A s e,s (y) ↓= ϕ B s e,s (y) ↓ in the definition (7) of the length-agreement function change its values.
We will construct our sets A and B in such a way that both C1 and C2 above are satisfied.

Requirements Requiring Attention
As a direct consequence of the descriptions contained in the previous subsection we make the following formal definitions.We say that: • Requirement P 9e requires attention at stage s + 1 ≥ 9e if W e,s ∩ A s = ∅ and there is x ∈ W e,s with x > max i<9e {r(i, s)}. (11) • Requirement P 9e+1 requires attention at stage s + 1 ≥ 9e + 1 if the condition (11) holds with B and 9e + 1 respectively in place of A and 9e.

Active Requirements
From now on, the letter R will denote any one of the requirements.At every stage s and for every n ≤ s, n 9e + 8 for every e ≥ 0, we define R n active if it requires attention at stage s and it has the highest priority, that is n = min{m : R m requires attention at stage s}.In particular, at every stage s at most one P-requirement or Rrequirement is active, and this feature allows to satisfy condition C2 above.
The negative requirement N 9e+8 is injured at stage s + 1 if a number x < r(9e + 8, s) is enumerated into either A s+1 or B s+1 .
Stage s + 1.Let A s and B s be the sets constructed up to the stage s.See if there is some requirement requiring attention.If not, then do nothing (Note 1).Otherwise, let R n be the active requirement.
• If R n = P 9e for some e, then let x be the minimum for which P 9e requires attention.Set A s+1 = A s ∪ {x} and B s+1 = B s .
• If R n = P 9e+1 for some e, then let x be the minimum for which P 9e+1 requires attention.Set set B s+1 = B s ∪ {x} and A s+1 = A s .
• If R n = N 9e+2 for some e, then let x be the minimum number in A s such that x ≥ e. Set r(9e + 2, s + 1) = x, A s+1 = A s and B s+1 = B s .
• If R n = N 9e+3 for some e, then let x be the minimum number in B s such that x ≥ e. Set r(9e + 3, s + 1) = x, A s+1 = A s and B s+1 = B s .
• If R n = R 9e+6 for some e, then let x ≤ s + 1 be the minimum number via the which R 9e+6 requires attention.Set A s+1 = A s ∪ {ϕ e,s (x)} and set r(9e + 6, s + 1) = x.Set B s+1 = B s .
• If R n = R 9e+7 for some e, then let x ≤ s + 1 be the minimum number via the which R 9e+7 requires attention.Set B s+1 = B s ∪ {ϕ e,s (x)} and set r(9e + 7, s + 1) = x.Set A s+1 = A s .

End construction of A and B.
Let us pose A = s≥0 A s and B = s≥0 B s .We have to prove that every requirement is satisfied.We prove first that each requirement R 9e+i with e ≥ 0 and i ≤ 7 requires attention at most finitely often.From now on, for every set X and every n ∈ N, X|n denotes the set X ∩ {0, 1, . . ., n}.
Lemma 3.8 For every e ≥ 0 and i ≤ 7, requirement R 9e+i requires attention at most finitely often.In particular, for every e ≥ 0 and i ≤ 7, lim s→∞ r(9e + i, s) < ∞.
Proof.The proof is by complete induction on n = 9e + i, for every e ≥ 0 and i ≤ 7. Given n, let s 0 be the minimum stage after which no requirement of higher priority than R n requires attention.We consider only the cases n = 9e, n = 9e + 2, n = 9e + 4 and n = 9e + 6, because the other cases are similar.