q 1-degrees of c.e. sets

Arch. Math. Logic (2012) 51:503–515DOI 10.1007/s00153-012-0278-7 Mathematical Logic

Q1-degrees of c.e. sets

R. Sh. Omanadze · I. O. Chitaia

Received: 16 September 2010 / Accepted: 7 March 2012 / Published online: 17 March 2012© Springer-Verlag 2012

Abstract We show that the Q-degree of a hyperhypersimple set includes an infinitecollection of Q1-degrees linearly ordered under ≤Q1 with order type of the integers andconsisting entirely of hyperhypersimple sets. Also, we prove that the c.e. Q1-degreesare not an upper semilattice. The main result of this paper is that the Q1-degree ofa hemimaximal set contains only one c.e. 1-degree. Analogous results are valid forΠ0

1 s1-degrees.

Keywords Q1-reducibility · s-reducibility · Hyperhypersimple set ·Hemimaximal set

Mathematics Subject Classification 03D25 · 03D30

1 Introduction

Tennenbaum (as quoted by Rogers [12, p. 159]) defined the notion of Q-reducibilityon sets of natural numbers as follows: a set A is Q-reducible to a set B (in symbols:A ≤Q B) if there is a computable function f such that for every x ∈ ω (where ω

denotes the set of natural numbers),

x ∈ A ⇐⇒ W f (x) ⊆ B.

R. Sh. Omanadze · I. O. Chitaia (B)Iv. Javakhishvili Tbilisi State University, 2, University St., Tbilisi 0186, Georgiae-mail: [email protected]

R. Sh. Omanadzee-mail: [email protected]

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504 R. Sh. Omanadze, I. O. Chitaia

In this case, we say that A ≤Q B via f . The relation of Q-reducibility is reflexiveand transitive so that it generates a degree structure on the subsets of ω. It is not hardto show that Q-reducibility is, in general, incomparable with Turing reducibility ≤T .On the computably enumerable (c.e.) sets we have that if A ≤Q B then A ≤T B; theconverse implication does not always hold; this easily follows from the observationthat if A ≤Q B then A is c.e. in B, where A denotes the complement of A.

The notion of Q-reducibility is very natural and important for the Theory of Algo-rithms. With the help of this notion a series of interesting results were obtained. Thisnotion plays a key role in the Marchenkov solution (see [4]) of the well-known Postproblem using Post’s methods. Q-degrees have applications in several fields of theTheory of Algorithms, for instance in the study of word problems and in computationalcomplexity.

If A ≤Q B via f and for all x and y,

x �= y �⇒ W f (x) ∩ W f (y) = ∅,

then we say that A is Q1-reducible to B, denoted by A ≤Q1 B. A c.e. set A isQ-complete (respectively, Q1-complete) if every c.e. set is Q-reducible (respectively,Q1-reducible) to A. Gill and Morris [2] have showed that the two notions are equiva-lent, i.e., a c.e. set A is Q-complete if and only if it is Q1-complete.

In this paper we prove that the Q-degree of a hyperhypersimple set includes aninfinite collection of Q1-degrees linearly ordered under ≤Q1 with order type of theintegers and consisting entirely of hyperhypersimple sets. This is an analogy of a wellknown theorem due to Dekker (see e.g. [12, Theorem 8.XIV]) regarding c.e. 1-degreeswithin a single m-degree. Also, we prove that there exist two c.e. sets having no leastc.e. upper bound in the Q1-reducibility ordering. This result is an analogy of a wellknown theorem due to Young (see e.g. [12, Theorem 10.IV]), showing that the c.e.1-degrees do not form an upper semilattice. The main result of this paper is that ifC , D are hemimaximal sets then C ≡Q1 D if and only if C ≡1 D. A c.e. set H ishemimaximal if there are a maximal set M and disjoint c.e. sets M0, M1 such thatM0 ∪ M1 = M , each Mi , i = 0, 1, is noncomputable, and M0 = H . Similar resultsare valid for Π0

1 s1-degrees.Our notation and terminology are standard, and can be found e.g., in [12] and [14].

2 Relating Q-reducibility and s-reducibility

As is known, see for instance [11], Q-reducibility is strictly related to S-reducibility≤s which is a subreducibility of enumeration reducibility. Namely one can show [2],see also [10] that if B �= ω then

A ≤Q B ⇐⇒ A ≤s B

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Q1-degrees of c.e. sets 505

(where X denotes the complement of a given set X). Thus, if B �= ∅, we have thatA ≤s B if and only if there is a computable function f such that

(∀ x)(x ∈ A ⇐⇒ W f (x) ∩ B �= ∅

).

Therefore, it turns out that ≤Q1 corresponds, via complements of sets, to a subreduc-ibility ≤s1 of ≤s , namely (for B �= ∅), A ≤s1 B if there is a computable function fsuch that for all x , y

x ∈ A ⇐⇒ W f (x) ∩ B �= ∅

and

x �= y �⇒ W f (x) ∩ W f (y) = ∅.

In its version for s1-reducibility, the following result is proved in [11]:

Corollary 1 If A ≤Q1 B, A ∈ �02 and B ∈ Π0

2 then there is a computable functionf such that for all x, y, W f (x) is finite,

x ∈ A ⇐⇒ W f (x) ⊆ B

and

x �= y �⇒ W f (x) ∩ W f (y) = ∅.

3 c.e. Q1-degrees within c.e. Q-degrees

We now look at Q1-degrees inside Q-degrees and prove that the Q-degree of a hyper-hypersimple set includes an infinite collection of Q1-degrees linearly ordered under≤Q1 with order type of the integers and consisting entirely of hyperhypersimple sets.

Definition 1 If A is any noncomputable c.e. set, a nontrivial splitting of A is a pairof disjoint noncomputable c.e. sets A0, A1 such that A = A0 ∪ A1.

Definition 2 A set A is hyperhypersimple if it is a coinfinite c.e. set, and there is nodisjoint weak array, each member of which intersects A, i.e. if there is no computablefunction f such that for all x and y,

1. W f (x) is finite,2. x �= y �⇒ W f (x) ∩ W f (y) = ∅,3. W f (x) ∩ A �= ∅.

The notions of Q-reducibility and Q1-reducibility are different on the class of c.e.sets. Indeed, if A0, A1 is a nontrivial splitting of a hyperhypersimple set A, thenA0 ≤Q A and A1 ≤Q A (see [9]), but A0 �≤Q1 A and A1 �≤Q1 A as is easily seen.

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The following theorem is analogous to a well known theorem by Dekker, see e.g.[12, Theorem 8.XIV], stating that every simple m-degree contains an infinite collectionof 1-degrees, having the same order type as the integers, with each 1-degree consistingentirely of simple sets.

Theorem 1 The Q-degree of a hyperhypersimple set includes an infinite collection ofQ1-degrees linearly ordered under ≤Q1 with order type of the integers and consistingentirely of hyperhypersimple sets.

Proof We begin with

Lemma 1 Let A and B be c.e. sets such that

B = A ∪ {m}

for some m ∈ A. Then

(i) A is hyperhypersimple if and only if B is hyperhypersimple;(ii) if A is hyperhypersimple then

B ≤1 A and A ≤m B and A �≤Q1 B.

Proof of lemma Part (i) immediately follows from Lachlan’s characterization of hy-perhypersimple sets, see [12, Theorem 12.XX].

Let us now show part (ii). The reductions B ≤1 A and A ≤m B are as in the proofof the above mentioned theorem by Dekker on simple m-degrees, see [12, p. 118].

Assume that A ≤Q1 B via f , i.e. for all x and y,

x ∈ A ⇐⇒ W f (x) ⊆ B

and

x �= y �⇒ W f (x) ∩ W f (y) = ∅.

We define, by induction, the values of a computable function g as follows:

Wg(0) = W f (m)\{m},Wg(n+1) =

⋃

i ∈ Wg(n)

i �= m

W f (i).

We now show that for all x and y,

x �= y �⇒ Wg(x) ∩ Wg(y) = ∅.

For all i > 0, Wg(i) ∩ Wg(0) = ∅ by the definition of g.

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Suppose that for all i , 0 ≤ i < n,

Wg(i) ∩ Wg(n) = ∅

and show that for all k, 0 ≤ k ≤ n,

Wg(k) ∩ Wg(n+1) = ∅.

Let for some k, 0 < k ≤ n,

Wg(k) ∩ Wg(n+1) �= ∅,

then( ⋃

i ∈ Wg(k−1)

i �= m

W f (i)

)∩

( ⋃

i ∈ Wg(n)

i �= m

W f (i)

)�= ∅.

Since for all x and y,

x �= y �⇒ W f (x) ∩ W f (y) = ∅,

it follows that

(∃ i)(

i ∈ Wg(n) ∩ Wg(k−1)

),

a contradiction.Then for all x and y,

x �= y �⇒ Wg(x) ∩ Wg(y) = ∅.

Let us show that for all x ,

Wg(x) ∩ B �= ∅.

For n = 0, we have

Wg(0) ∩ B = W f (m) ∩ B �= ∅.

Suppose now that Wg(n) ∩ B �= ∅. Then there is an x ∈ Wg(n) ∩ B and, by thedefinition of g, we have

Wg(n+1) ∩ B ⊇ W f (x) ∩ B �= ∅.

Thus {Wg(x)}x∈ω is a disjoint weak array of finite sets, witnessing that B is nothyperhypersimple, a contradiction.

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Let A be a hyperhypersimple set. Let {a0, a1, . . .} be an infinite subset of A and let{b0, b1, . . .} be an infinite subset of A. Then, by Lemma 1,

. . . , A ∪ {b0, b1}, A ∪ {b0}, A, A\{a0}, A\{a0, a1}, . . .

yields a linear ordering of Q1-degrees having order type of the integers. Moreover, bypart (i), every Q1-degree of the collection consists entirely of hyperhypersimple sets.

��Corollary. The s-degree of a Π0

1 hyperhyperimmune set includes an infinite collectionof s1-degrees linearly ordered under ≤s1 with order type of the integers and consistingentirely of hyperhyperimmune sets.

Proof Immediate by Theorem 1. ��

4 On the partial ordering of the c.e. Q1-degrees

It is well known that the 1-degrees of c.e. sets is not upper semilattice. In this sectionwe have obtained an analogous result for the c.e. Q1-degrees.

Lemma 2 If A and B are c.e. sets and A ≤Q1 B, then there is a computable functionf such that for all x and y,

(1) x ∈ A ⇐⇒ W f (x) ⊆ B,(2) x �= y �⇒ W f (x) ∩ W f (y) = ∅,(3) x ∈ A �⇒ ∣

∣W f (x) ∩ B∣∣ = 1.

Proof Let A ≤Q1 B via g, and by the s − m − n theorem, let f be a computablefunction such that

W f (n) ={

x : x ∈ Wg(n) and (∀ y < x)(y ∈ B)}.

Then f has the desired properties. ��Definition 3 A c.e. set M is maximal (r -maximal) if M is infinite and there is no c.e.(computable) set W such that |W ∩ M| = |W ∩ M| = ∞.

It is well-known (see e.g. [12, Theorem10.IV]) that the c.e. 1-degrees are not anupper semilattice. The c.e. Q1-degrees give another example of a structure of c.e.degrees that is not an upper semilattice.

Theorem 2 There exist two c.e. sets having no c.e. least upper bound on theQ1-reducibility ordering.

Proof Let A and B be two Turing incomparable maximal sets (see [15, Corollary 4.2])and let C = A ⊕ B. Then, as proved in [6], the set C is hyperhypersimple. Assumefor a contradiction that degQ1

(A) and degQ1(B) have least upper bound in the c.e.

Q1-degrees, and let D be a set belonging to this least upper bound. Then D ≤Q1 C ,

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and we can show that D is hyperhypersimple. Indeed, D is infinite, so if D were nothyperhypersimple then there would be a disjoint weak array {Wα(x)}x∈ω such that, forall x , Wα(x) ∩ D �= ∅. In this case, define a computable function β as follows:

Wβ(x) =⋃

i∈Wα(x)

Wγ (i),

where γ is a computable function such that D ≤Q1 C via γ . Then C is not hyperhy-persimple, a contradiction. ��Remark In proving that D is hyperhypersimple, we have in fact shown that if Y ishyperhypersimple, X is c.e. and X ≤Q1 Y , then X is hyperhypersimple, i.e. the hy-perhypersimple degrees are downwards closed in the c.e. degrees under ≤Q1 .

Since A ≤Q1 D and B ≤Q1 D, by Lemma 2 there exist computable functions f ,g such that

x ∈ A �⇒ ∣∣W f (x) ∩ D∣∣ = 1

and

x ∈ B �⇒ ∣∣Wg(x) ∩ D

∣∣ = 1.

Lemma 3 If D ⊆ ⋃x∈A W f (x) or D ⊆ ⋃

x∈B Wg(x), then D is a maximal set.

Proof Let D ⊆ ⋃x∈A W f (x) and E be a c.e. set such that |E ∩ D| = |E ∩ D| = ∞.

Then

F = {x :W f (x) ∩ E �= ∅

}

is c.e. and |F ∩ A| = ∞.If x ∈ F ∩ A then

W f (x) ∩ (D ∩ E) �= ∅ and W f (x) ∩ (E ∩ D) = ∅,

since x ∈ A �⇒ |W f (x) ∩ D| = 1.But

⋃x∈A W f (x) ⊇ D, therefore

⋃

x∈F∩A

W f (x) ⊇ E ∩ D.

E ∩ D is infinite, then F ∩ A is infinite.Hence, we have that |F∩A| = |F∩A| = ∞, but A is a maximal set, a contradiction.

The case D ⊆ ⋃x∈B Wg(x) is similar.

Let D ⊆ ⋃x∈A W f (x) or D ⊆ ⋃

x∈B Wg(x), then by Lemma 3 D is a maximalset. Since degQ(A ⊕ B) is the least upper bound in the Q-degrees of degQ(A) and

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degQ(B), our assumptions on D allow to conclude that D ≡ A⊕B. By [7, Theorem 1],D ≤m A or D ≤m B. Then B ≤Q A or A ≤Q B, a contradiction. Therefore we have

D\⋃

x∈A

W f (x) �= ∅

and

D\⋃

x∈B

Wg(x) �= ∅.

Let a ∈ D\ ⋃x∈A W f (x) and b ∈ D\⋃

x∈B Wg(x).Define a computable function h as follows:

Wh(x) ={

Wg(x) ∪ {b}, if a ∈ Wg(x),

Wg(x), if a �∈ Wg(x).

Then A ≤Q1 D ∪ {a} via f and B ≤Q1 D ∪ {a} via h. But by Lemma 1, D �≤Q1

D ∪ {a}, a contradiction. ��Corollary 2 There exist two Π0

1 sets having no upper bound in the s1-reducibilityordering of Π0

1 sets.

Proof Immediate by Theorem 2. ��

5 c.e. 1-degrees within c.e. Q1-degrees

In this section we prove that the Q1-degree of a hemimaximal set contains only onec.e. 1-degree.

Definition ([1]) A set A is hemimaximal if there are a maximal set M and a nontrivialsplitting M0, M1 of M such that A = M0.

Definition ([13]) A c.e. set A is nowhere simple if for every c.e. set B with B\Ainfinite there is an infinite c.e. set W ⊆ B\A.

It is proved in [8] that if A is a c.e. set, B a nowhere simple set and A ≤Q B, thenA is a nowhere simple set.

Proposition 1 Let A be a hemimaximal set and B be a noncomputable c.e. set suchthat B ≤Q1 A. Then B is a hemimaximal set.

Proof Let A be a hemimaximal set, B be a noncomputable c.e. set such that B ≤Q1 A.By Corollary 1, let f be a computable function such that for all x and y,

1. x ∈ B ⇐⇒ W f (x) ⊆ A,2. x �= y �⇒ W f (x) ∩ W f (y) = ∅,3. W f (x) is finite.

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Let C be a c.e. set such that A ∩ C = ∅ and M = A ∪ C is a maximal set. It isproved in [5, Theorem 8] that if M is maximal and M is the disjoint union of two non-computable c.e. sets M1 and M2, then both M1 and M2 are nowhere simple. ThereforeA is nowhere simple. Consider the set

B1 = {x : W f (x) ∩ C �= ∅

}.

Then B1 is a c.e. set and B1 ∩ B = ∅.Let us show that B1 is not computable. Suppose that B1 is computable. Then B1

is computable and B1\B is infinite: indeed if B1\B were finite, then B would becomputable, a contradiction. Since A is nowhere simple and B ≤Q A, it follows (asalready remarked) that B is nowhere simple too [8]. Then there is an infinite c.e. set Wsuch that W ⊂ B1\B. Let W1 and W2 be noncomputable c.e. sets, W1 ∩ W2 = ∅ andW = W1 ∪ W2. Then

⋃i∈W1

W f (i) ∩ M is infinite and⋃

i∈W2W f (i) ∩ M is infinite

and( ⋃

i∈W1

W f (i)

)∩

( ⋃

i∈W2

W f (i)

)= ∅.

But M is a maximal set, a contradiction.Thus B1 is a noncomputable c.e. set, B ∩ B1 = ∅ and B ∪ B1 is infinite. If B ∪ B1

is not a maximal set, then there is a c.e. set W such that W ∩ (B ∪ B1) is infinite andW ∩(B ∪ B1) is infinite. Then

(⋃x∈W W f (x)

)∩ M is infinite and(⋃

x∈W W f (x)

)∩ Mis infinite, a contradicting maximality of M . ��Question Is there a c.e. Q-degree which contains only one c.e. m-degree?

For Q1-degrees we have the following result.

Theorem 3 If C, D are hemimaximal sets then

C ≡Q1 D ⇐⇒ C ≡1 D.

Proof Let C , D be hemimaximal sets and C ≡Q1 D. Then by Corollary 1 there arecomputable functions f and g such that for all x and y,

(c1) x ∈ C ⇐⇒ W f (x) ⊆ D,(c2) x �= y �⇒ W f (x) ∩ W f (y) = ∅,(c3) W f (x) is finite

and

(d1) x ∈ D ⇐⇒ Wg(x) ⊆ C ,(d2) x �= y �⇒ Wg(x) ∩ Wg(y) = ∅,(d3) Wg(x) is finite.

Let M1, M2 be maximal sets such that M1 = C ∪ C1, M2 = D ∪ D1, C ∩ C1 =D ∩ D1 = ∅.

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Lemma 4 Let M be a maximal set. Then for all partial computable functions ϕ, ifϕ(M) ∩ M is infinite then

{x : x ∈ M and ϕ(x) ∈ M and ϕ(x) �= x

}

is finite.

Proof By the maximality of M it is sufficient to prove the lemma for all computablefunctions f . Indeed, since M is a maximal set, ϕ(M)∩M infinite gives that dom ϕ∩Mis infinite. hence for almost all x ∈ M we have x ∈ dom ϕ. Thus there is a computablefunction coinciding with ϕ for almost all x ∈ M .

Assume that | f (M) ∩ M| = ∞.Kobzev [3] showed that if N is an r -maximal set, then for all computable functions g

∣∣∣{

g(x) : x ∈ N and g(x) ∈ N and g(x) �= x}∣∣∣ < ∞.

By Kobzev’s result applied to our maximal set M and by our assumption on f , itthen follows that

∣∣∣{

f (x) : x ∈ M and f (x) ∈ M and f (x) = x}∣∣∣ = ∞.

Then, trivially,

∣∣∣{

x : x ∈ M and f (x) ∈ M and f (x) = x}∣∣∣ = ∞.

But then it follows that∣∣∣{

x : x ∈ M and f (x) ∈ M and f (x) �= x}∣∣∣ < ∞,

otherwise the c.e. sets

{x : f (x) = x} and {x : f (x) �= x}

would split M into two infinite sets, a contradiction. ��Lemma 5 M1 ⊆∗

⋃x∈M2

Wg(x) and M2 ⊆∗⋃

x∈M1W f (x), where X ⊆∗ Y denotes

that X\Y is finite.

Proof Suppose that

R = M1 ∩( ⋃

x∈M2

Wg(x)

)

be finite, and let

E = {x : x ∈ D1 and Wg(x) ∩ C1 �= ∅

}.

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Then

x ∈ M2 ∪ E ⇐⇒ Wg(x) ∩ (C1 ∪ R) �= ∅,

i.e. M2 ∪ E is a c.e. set. But M2 ∪ D1 = (M2 ∪ E)∪ D1, i.e. M2 ∪ D1 is a c.e. set, a con-tradiction. Therefore R is infinite. If M1\R is infinite, then the c.e. set

⋃x∈ω Wg(x)

split M1 into two infinite sets, a contradiction. The case M2 ⊆∗⋃

x∈M1W f (x) is

similar. ��Corollary 3 M1 ∩ ( ⋃

x∈D1Wg(x)

)and M2 ∩ ( ⋃

x∈C1W f (x)

)are finite sets.

Let us define a partial computable function ϕ as follows. We compute simulta-neously {Wg(i)}i∈ω and {W f ( j)} j∈ω and, for given z, seek the integers x , y (if theyexist) such that

z ∈ Wg(y) and y ∈ W f (x).

If we can find such x and y, then we let ϕ(z) = x .

Lemma 6 ϕ(M1) ∩ M1 is infinite.

Proof It follows from Lemma 5 and Corollary 3 that for almost all x , if x ∈ M1, thenϕ(x) is defined and ϕ(x) ∈ M1. However, for all x , y we have

(m1) Wg(x) ∩ M1 is finite,(m2) x �= y �⇒ Wg(x) ∩ Wg(y) = ∅

and

(n1) W f (x) ∩ M2 is finite,(n2) x �= y �⇒ W f (x) ∩ W f (y) = ∅.

Then it follows from the definition of ϕ that ϕ(M1) ∩ M1 is infinite.

By Lemma 4

{x : x ∈ M1 and ϕ(x) ∈ M1 and ϕ(x) �= x

}

is finite. Therefore for almost all x ∈ M , ϕ(x) converges and it is equal to x , and thus,by definition of ϕ, there exists a (unique) y, such that

x ∈ Wg(y) and y ∈ W f (x).

Then for almost all x we have

x ∈ M1 �⇒ x ∈ Wg(y) and y ∈ W f (x) �⇒ y ∈ M2 ∪ D1.

Let

F = M1\{

x : (∃ y)(y ∈ W f (x) and x ∈ Wg(y)

)}.

Then F is a finite set.

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We define the values of a computable function h which 1-reduces the set C to theset D as follows.

Suppose that h is already defined for all n < x , if n1 �= n2, then h(n1) �= h(n2)

and n ∈ C ⇐⇒ h(n) ∈ D. Let R and R1 be infinite computable subsets of D andD1, respectively. We compute C , C1 ∪ F , {W f ( j)} j∈ω, {Wg(i)}i∈ω simultaneously andstop the procedure the first time any of the following occurs:

Case 1 x is found in C . Then stop the procedure, set

h(x) = min{

n : n ∈ R and n �∈ {h(0), . . . , h(x − 1)

}}.

Case 2 x is found in C1 ∪ F . Then stop the procedure, set

h(x) = min{

n : n ∈ R1 and n �∈ {h(0), . . . , h(x − 1)

}}.

Case 3 x ∈ Wg(y) and y ∈ W f (x). Then if y �∈ {h(0), . . . , h(x − 1)

}, stop the

procedure, set h(x) = y.

If y ∈ {h(0), . . . , h(x − 1)

}, then y �∈ M2. Indeed, let

y ∈ M2 and y ∈ {h(0), . . . , h(x − 1)

}.

Then there is i < x , i ∈ Wg(y) and y ∈ W f (i). Then

W f (i) ∩ W f (x) �= ∅,

a contradiction. Therefore y ∈ D ∪ D1. Compute D and D1 simultaneously. If y ∈ D,then since x ∈ Wg(y) and y ∈ W f (x) we have that x ∈ C : set

h(x) = min{

n : n ∈ R and n �∈ {h(0), . . . , h(x − 1)

}}.

If y ∈ D1, then since

x ∈ Wg(y) and y ∈ W f (x) �⇒ x ∈ C

set

h(x) = min{

n : n ∈ R1 and n �∈ {h(0), . . . , h(x − 1)

}}.

Clearly,

ω = C ∪ C1 ∪ F ∪{

x : (∃ y)(y ∈ W f (x) and x ∈ Wg(y)

)}.

Therefore the three considered cases cover all possibilities. Obviously, C ≤1 Dvia h.

Thus C ≤1 D. The case D ≤1 C can be shown similarly. ��

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From Theorem 3 and Proposition 1 follows the following

Theorem 4 Let a be the Q1-degrees of any hemimaximal set. Then for all c.e. setsA, B ∈ a, we have A ≡1 B.

Corollary 4 There exist a Π01 s1-degree a such that for any Π0

1 sets A, B ∈ a, wehave A ≡1 B.

Proof Immediate by Theorem 4. ��Acknowledgments The first author was partially supported by the Georgian National Science Foundation(Grants # GNSF/ST09_144_3-105, # GRANT/ST09_270_3-105 and # GNSF/CNRS09/561). The authorswould like to thank the anonymous referee for many suggestions and improvements throughout the paper.

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