On relative enumerability of Turing degrees
On relative enumerability of Turing degrees Archive for Mathematical Logic Volume 38, Number 3, 2000, 145-154
Shamil Ishmukhametov 18 Kremlevskya st. Dept.Math. Kazan Federal University, 42008, Kazan, Russia [email protected] Abstract Let d be a Turing degree, R[d] and Q[d] denote respectively classes of recursively enumerable (r.e.) and all degrees in which d is relatively enumerable. We proved in [ta] that there is a degree d containing differences of r.e.sets (briefly, d.r.e.degree) such that R[d] possess a least element m>0. Now we show the existence of a d.r.e. d such that R[d] has no a least element. We prove also that for any REAdegree d below 00 the class Q[d] cannot have a least element and more generally is not bounded below by a non-zero degree, except in trivial cases.
Introduction The recursively enumerable (r.e.) degrees play an important role in Mathematical Logic, since they are exactly the degrees in which axiomatisible theories lie. Relativising this class to some oracle A we obtain so called A-REA (or a-REA, we replace A by its degree) degrees. The hierarchy of a-REA degrees was defined in Jockusch and Shore [1984]. Clearly, if a degree d is a-REA, then it is b-REA for every b, a0 and there is a degree b0. It is known that d.r.e. degrees form a proper subclass of 2-REA degrees (see Jockusch and Shore [1984]). We proved that for some d R[d] is an upper cone with an r.e. base a>0. We proved also uniting our method with the Cooper and Yi construction of isolated d.r.e. degree [ta] that there is a d.r.e. degree d which is enumerable just in one r.e. degree a below d. We conjectured that this a bounds not only r.e. but all degrees in which d is recursively enumerable. Our conjecture was based on a proposition (see Ishmukhametov [ta1]) that, given a d.r.e. set D, there exists a least degree a in which D is enumerable (in the class of all degrees). Now we refute our conjecture and show that there is no REA-degree d below 00 such that the class Q[d] possess a least element greater than 0. Moreover, for no such d Q[d] is bounded below by a non-zero degree. Namely, we show that, given a REA degree d and a, 0 s0 , l(e, s00 ) > x + 1, then restrain D up to s00 . (4) Wait for a stage s000 , l(e, j, s000 ) > s00 . Remove x from D, enumerate x + 1 into De , and restrain Be up to s000 . Define De (y) for all y which are not witnesses of Ne,j - requirements, j ∈ ω, equal to D(y). Clearly, each Ne,j acts finitely often and restrains a finite interval so the cooperation of different Ne,j can be organized in the usual way. Now we verify the Basic Module. Lemma 1.1 Each Ne,j is met. Proof. Fix k =< e, j > and assume the Lemma for all k 0 < k. Assume,the left part of Ne,j holds, and let s0 be a least stage after which no requirement with higher priority acts. Consider witnesses x and x + 1 of Ne,j defined after stage s0 . Since, lim inf l(e, s) = ∞, each step 1,2, and 3 of the Basic Module occurs. If Ne,j fails then lim inf l(e, j, s) = ∞, and step 4 also occurs. 000 We shall prove, that As s00 6= A s00 . (∗) A⊕WkA Since D(x) = Φm (x) at stages s0 , s00 and s000 , then when we put x A⊕W A into D, and then remove it from there, Φm k (x) subsequently takes values 0,1,0. This means, A ⊕ WkA τx changes between stages s0 and s00 , (τx is the A⊕W A use of computation Φm k (x) at stage s0 ) and after stage s00 returns its old ∗ value because ΦD m τx returns the old value after removing x from D. So there is some z < τx , A ⊕ WkA (z) changes between stages s0 and s00 , and then returns. Since, A is r.e., then WkA (z 0 ) changes for z 0 = (z − 1)/2. If the first value of WkA (z 0 ) was 1, then the change of WkA (z 0 ) is possible only if A τx∗
5 ∗ 0 000 changes. But A τx∗ is equal to ΦD and cannot m τx at stages s and s 0 000 change between stages s and s . So the first value WkA (z 0 ) was 0, and z was enumerated in WkA between stages s0 and s00 . In order to return WkA (z 0 ) to 0 it is necessary to change the oracle A s00 , and (∗) holds. B∗ Since A s00 = Φj e s00 at stage s000 , after A s00 change, we get the B∗ disagreement A s00 6= Φj e s00 , and Ne,j is met. This finishes the proof.
Lemma 1.2 If lim inf l(e, s) = ∞, then De ≡T D. Proof. If x is not a witness of some Ne,j -requirement, then x ∈ D ↔ x ∈ De . Assume, x and x + 1 are witnesses of a requirement Ne,j . Wait for a stage s , at which l(e, s) exceeds x + 1. At stage s0 , either x is put into D and De or it is restrained by a higher priority requirement. In the latter case, both x and x + 1 are not ever put into D and De . Assume the first case occurs. By the construction, if x is ever removed from D, then x + 1 is enumerated into De . Therefore, x 6∈ D ↔ x + 1 ∈ De . The proof of the Theorem follows immediately from the lemmas. 0
2
Studying a class Q[d]
Now we study, for a given ∆02 degree d>0, the class Q[d] of all degrees, less d, in which d is r.e. Our main result concerning such classes we divide into two parts. First we show that for any proper d.r.e degree d and any r.e. a n. Go to the next step. 6. Wait for B σ(n) to change, then return to 1. This establishes the module. Notice that due to the module we change some values C(y) and E(x) such that x is not an element of M in(Ds ). But if later D(n) returns to 0, condition M in(Ds ) ⊆ E s can become invalid. In order to prevent it each time when some n is removed from D we add to E all elements of M in(Ds ). This causes an infinte injury to the basic module. Nevertheless the following lemma ensures a satisfaction of the requirement Nj : B Lemma 2.3 If for every k ≤ n ΦC j (k) = A(k) then Σ (n) ↓= D(n).
Proof. Fix some n ∈ ω. Assume the lemma for all k < n. If n was enumerated in D before the stage when procedure hni is started the proof is clear. Assume n was not in D before a stage s when a triple hxn , s0 , s00 i satisfying conditions 2.1-2.3 of the basic module appears. The restraint on C σ(n) posed at stage s can be destroyed in the following cases: 1. Some requirement of higher priority acts. By inductive assumptions we can assume that this case does not occur after some stage s0 . 2. Restoring condition M in(D) ⊆ E we return some number x to E and remove s(x) from C. This happens if some m is removed from D. Let u be such as in the Basic Module. Clearly, this m was put into D before stage u and s(m) < u ≤ σ(n). Therefore, when the restraint on C z0 is destroyed ΣB (n) becomes undefined, and procedure hni can be initialized. Procedure hni eventually finds a triple hxn , s0 , s00 i such that C s cn = C cn (this is ensured by lemma 2.2 and the way of selection of such triples). At this stage we define ΣB (n) = D(n). If later n is enumerated in D then we get the disagreement A(xn ) 6= ΦC j (xn ). Since the restraint on C cn is not injured, A(n) must be changed and become equal to ΦC j (xn ). Then h(n) B is enumerated in B and Σ (n) becomes undefined. Again, D(n) = ΣB (n). This finishes the proof. Since D 6≤T B, there is a (least) n such that either ΣB (n) ↑, or ΦC j (xn ) ↓6= A(xn ).
9 In the full construction we use a linear ordering of requirements. Each requirement can have two possible outcomes: finite and infinite due to defined or undefined is the value ΣB (n) where n is a least number such that D(n) 6= ΣB (n). The finite case is clear. In the case ΣB (n) ↑ we arrange a construction in such a way that all values ΣB (k), k ≥ n, becomes undefined simultaneosly at B-true stages (i.e such stages s that B s bs = B bs , bs is a number enumerated at stage s in B). This causes a delay of C(y)-changes relating to D(n)-change but again we are able to prove E ≤T D since C(y)-changes happen no later than at the nearest B-true stage after D(n) has changed. We leave details to the reader. This establishes the theorem. Theorem 3 For any ∆02 degrees a,d, a
Shamil Ishmukhametov 18 Kremlevskya st. Dept.Math. Kazan Federal University, 42008, Kazan, Russia [email protected] Abstract Let d be a Turing degree, R[d] and Q[d] denote respectively classes of recursively enumerable (r.e.) and all degrees in which d is relatively enumerable. We proved in [ta] that there is a degree d containing differences of r.e.sets (briefly, d.r.e.degree) such that R[d] possess a least element m>0. Now we show the existence of a d.r.e. d such that R[d] has no a least element. We prove also that for any REAdegree d below 00 the class Q[d] cannot have a least element and more generally is not bounded below by a non-zero degree, except in trivial cases.
Introduction The recursively enumerable (r.e.) degrees play an important role in Mathematical Logic, since they are exactly the degrees in which axiomatisible theories lie. Relativising this class to some oracle A we obtain so called A-REA (or a-REA, we replace A by its degree) degrees. The hierarchy of a-REA degrees was defined in Jockusch and Shore [1984]. Clearly, if a degree d is a-REA, then it is b-REA for every b, a0 and there is a degree b0. It is known that d.r.e. degrees form a proper subclass of 2-REA degrees (see Jockusch and Shore [1984]). We proved that for some d R[d] is an upper cone with an r.e. base a>0. We proved also uniting our method with the Cooper and Yi construction of isolated d.r.e. degree [ta] that there is a d.r.e. degree d which is enumerable just in one r.e. degree a below d. We conjectured that this a bounds not only r.e. but all degrees in which d is recursively enumerable. Our conjecture was based on a proposition (see Ishmukhametov [ta1]) that, given a d.r.e. set D, there exists a least degree a in which D is enumerable (in the class of all degrees). Now we refute our conjecture and show that there is no REA-degree d below 00 such that the class Q[d] possess a least element greater than 0. Moreover, for no such d Q[d] is bounded below by a non-zero degree. Namely, we show that, given a REA degree d and a, 0 s0 , l(e, s00 ) > x + 1, then restrain D up to s00 . (4) Wait for a stage s000 , l(e, j, s000 ) > s00 . Remove x from D, enumerate x + 1 into De , and restrain Be up to s000 . Define De (y) for all y which are not witnesses of Ne,j - requirements, j ∈ ω, equal to D(y). Clearly, each Ne,j acts finitely often and restrains a finite interval so the cooperation of different Ne,j can be organized in the usual way. Now we verify the Basic Module. Lemma 1.1 Each Ne,j is met. Proof. Fix k =< e, j > and assume the Lemma for all k 0 < k. Assume,the left part of Ne,j holds, and let s0 be a least stage after which no requirement with higher priority acts. Consider witnesses x and x + 1 of Ne,j defined after stage s0 . Since, lim inf l(e, s) = ∞, each step 1,2, and 3 of the Basic Module occurs. If Ne,j fails then lim inf l(e, j, s) = ∞, and step 4 also occurs. 000 We shall prove, that As s00 6= A s00 . (∗) A⊕WkA Since D(x) = Φm (x) at stages s0 , s00 and s000 , then when we put x A⊕W A into D, and then remove it from there, Φm k (x) subsequently takes values 0,1,0. This means, A ⊕ WkA τx changes between stages s0 and s00 , (τx is the A⊕W A use of computation Φm k (x) at stage s0 ) and after stage s00 returns its old ∗ value because ΦD m τx returns the old value after removing x from D. So there is some z < τx , A ⊕ WkA (z) changes between stages s0 and s00 , and then returns. Since, A is r.e., then WkA (z 0 ) changes for z 0 = (z − 1)/2. If the first value of WkA (z 0 ) was 1, then the change of WkA (z 0 ) is possible only if A τx∗
5 ∗ 0 000 changes. But A τx∗ is equal to ΦD and cannot m τx at stages s and s 0 000 change between stages s and s . So the first value WkA (z 0 ) was 0, and z was enumerated in WkA between stages s0 and s00 . In order to return WkA (z 0 ) to 0 it is necessary to change the oracle A s00 , and (∗) holds. B∗ Since A s00 = Φj e s00 at stage s000 , after A s00 change, we get the B∗ disagreement A s00 6= Φj e s00 , and Ne,j is met. This finishes the proof.
Lemma 1.2 If lim inf l(e, s) = ∞, then De ≡T D. Proof. If x is not a witness of some Ne,j -requirement, then x ∈ D ↔ x ∈ De . Assume, x and x + 1 are witnesses of a requirement Ne,j . Wait for a stage s , at which l(e, s) exceeds x + 1. At stage s0 , either x is put into D and De or it is restrained by a higher priority requirement. In the latter case, both x and x + 1 are not ever put into D and De . Assume the first case occurs. By the construction, if x is ever removed from D, then x + 1 is enumerated into De . Therefore, x 6∈ D ↔ x + 1 ∈ De . The proof of the Theorem follows immediately from the lemmas. 0
2
Studying a class Q[d]
Now we study, for a given ∆02 degree d>0, the class Q[d] of all degrees, less d, in which d is r.e. Our main result concerning such classes we divide into two parts. First we show that for any proper d.r.e degree d and any r.e. a n. Go to the next step. 6. Wait for B σ(n) to change, then return to 1. This establishes the module. Notice that due to the module we change some values C(y) and E(x) such that x is not an element of M in(Ds ). But if later D(n) returns to 0, condition M in(Ds ) ⊆ E s can become invalid. In order to prevent it each time when some n is removed from D we add to E all elements of M in(Ds ). This causes an infinte injury to the basic module. Nevertheless the following lemma ensures a satisfaction of the requirement Nj : B Lemma 2.3 If for every k ≤ n ΦC j (k) = A(k) then Σ (n) ↓= D(n).
Proof. Fix some n ∈ ω. Assume the lemma for all k < n. If n was enumerated in D before the stage when procedure hni is started the proof is clear. Assume n was not in D before a stage s when a triple hxn , s0 , s00 i satisfying conditions 2.1-2.3 of the basic module appears. The restraint on C σ(n) posed at stage s can be destroyed in the following cases: 1. Some requirement of higher priority acts. By inductive assumptions we can assume that this case does not occur after some stage s0 . 2. Restoring condition M in(D) ⊆ E we return some number x to E and remove s(x) from C. This happens if some m is removed from D. Let u be such as in the Basic Module. Clearly, this m was put into D before stage u and s(m) < u ≤ σ(n). Therefore, when the restraint on C z0 is destroyed ΣB (n) becomes undefined, and procedure hni can be initialized. Procedure hni eventually finds a triple hxn , s0 , s00 i such that C s cn = C cn (this is ensured by lemma 2.2 and the way of selection of such triples). At this stage we define ΣB (n) = D(n). If later n is enumerated in D then we get the disagreement A(xn ) 6= ΦC j (xn ). Since the restraint on C cn is not injured, A(n) must be changed and become equal to ΦC j (xn ). Then h(n) B is enumerated in B and Σ (n) becomes undefined. Again, D(n) = ΣB (n). This finishes the proof. Since D 6≤T B, there is a (least) n such that either ΣB (n) ↑, or ΦC j (xn ) ↓6= A(xn ).
9 In the full construction we use a linear ordering of requirements. Each requirement can have two possible outcomes: finite and infinite due to defined or undefined is the value ΣB (n) where n is a least number such that D(n) 6= ΣB (n). The finite case is clear. In the case ΣB (n) ↑ we arrange a construction in such a way that all values ΣB (k), k ≥ n, becomes undefined simultaneosly at B-true stages (i.e such stages s that B s bs = B bs , bs is a number enumerated at stage s in B). This causes a delay of C(y)-changes relating to D(n)-change but again we are able to prove E ≤T D since C(y)-changes happen no later than at the nearest B-true stage after D(n) has changed. We leave details to the reader. This establishes the theorem. Theorem 3 For any ∆02 degrees a,d, a
