BOUNDING AND NONBOUNDING MINIMAL PAIRS IN THEpmt6sbc/preprints/minimal.pdf · BOUNDING AND...

BOUNDING AND NONBOUNDING MINIMAL PAIRS IN THEENUMERATION DEGREES

S. BARRY COOPER, ANGSHENG LI, ANDREA SORBI AND YUE YANG

Abstract. We show that every nonzero ∆0

2e-degree bounds a minimal pair.

On the other hand, there exist Σ0

2e-degrees which bound no minimal pair.

1. Introduction

In a poset 〈P,≤〉 with a least element 0, a pair of elements a, b is said to forma minimal pair if a, b 6= 0 and 0 is the greatest lower bound of a, b. The firstinvestigation of minimal pairs in the enumeration degrees of Σ0

2 sets goes backto Cooper and McEvoy [MC85]. It is well known that there is an order theoreticembedding ι (preserving least element, joins, and the jump operation) of the Turingdegrees into the enumeration degrees, in such a way that the c.e. Turing degreesembed exactly onto the Π0

1 enumeration degrees. Cooper and McEvoy show thatfor every pair of nonzero Π0

1 enumeration degrees a and b in which either side islow we have that

(∀c)[c ≤ a,b ⇒ (∃d ∈ Π01)[c ≤ d ≤ a,b]].

Thus every minimal pair of low c.e. Turing degrees is mapped by ι into a minimalpair of enumeration degrees. As to bounding minimal pairs, Cooper and McEvoyshow that if h is Σ0

2-high (or equivalently high, by [SS99]) and a ≤ h is a nonzerolow enumeration degree, then there exists b ≤ h such that a and b form a minimalpair.

As regards the nonbounding phenomenon, the situation is more complicated.In the Turing degrees, the Lachlan Nonbounding Theorem, [La79], states thatthere is a nonzero c.e. Turing degree that does not bound any minimal pair ofc.e. Turing degrees. However, the image of any nonbounding c.e. Turing degreeunder the embedding ι becomes bounding in the e-degrees, as shown by Arslanov,Cooper and Kalimullin, [ACK01], as follows. For every total enumeration degreea (meaning that a is in the range of ι) there are enumeration degrees b and c suchthat a = b∪ c, and 0e = b∩ c, where 0e denotes the least e-degree (in fact, b andc may be assumed to be quasiminimal, i.e., bounding no total e-degree except for0e). In particular this holds of the Π0

1 enumeration degrees, that is, the images of

1991 Mathematics Subject Classification. 03D30.All four authors were partially supported by NUS Grant No. R-146-000-028-112 (Singapore).

The first and the third authors were partially supported by the INTAS-RFBR project Com-

putability and Models, n. 97-2139, and by EPSRC Research Grant No. GR/M 91419. The secondauthor was partially supported by NSF grant No. 69973048 (CHINA) and NSF major grantNo. 19931020 (CHINA).

1

2 S. BARRY COOPER, ANGSHENG LI, ANDREA SORBI AND YUE YANG

the c.e. Turing degrees under the embedding ι. Thus every Π01 enumeration degree

bounds a minimal pair.

There are other earlier works on bounding and, to a lesser extent, nonboundingphenomena. Copestake [Cop88], in her study of 1-genericity in the enumerationdegrees, had observed that every 2-generic function bounds a minimal pair andthis result does not appear to be true for 1-generic functions. Later Copestake(see [Coo90]) claimed that there is a 1-generic e-degree which bounds no minimalpairs. But her proof did not appear.

In the first part of this paper we show that the above bounding result of Ar-slanov, Cooper and Kalimullin can be extended to the ∆0

2 enumeration degrees.In the subsequent sections, we show that there is a nonzero Σ0

2 enumeration de-gree that does not bound any minimal pair. Furthermore, our construction can beadapted to produce a set 1-generic nonbounding e-degree.

The reader is referred to [So87] for all unexplained computability theoretic no-tations and terminologies. In particular we will refer to an effective enumera-tion Wee∈ω of the c.e. sets, and to computable approximations We,se,s∈ω tothe c.e. sets, satisfying, for all e, s: We,s is finite; We,0 = ∅; We,s ⊆ We,s+1;We =

⋃

s We,s. If Ξe is the enumeration operator (or, simply, e-operator) definedby the c.e. set We then Ξe,s denotes the enumeration operator defined by We,s.If Zss∈ω is a Σ0

2 approximation to a set Z, and Ψs is an approximation to ane-operator Ψ then ΨZs

s not necessarily defines a Σ02-approximation to ΨZ . Starting

from ΨZs

s s∈ω we can however uniformly define a Σ02 approximation ΨZ [s]s∈ω

to ΨZ : For details see [MC85]. It is to these Σ02 approximations that we refer

in the following when dealing with stage approximations of e-operators and Σ02

sets. We denote by 〈u, v〉 the natural number coding the pair of natural numbers(u, v) under the standard coding function; if x = 〈u, v〉, then we let (x)1 = u and(x)2 = v. For every e, we also let ω[e] = 〈e, x〉 : x ∈ ω.

2. Every nonzero ∆02 e-degree is bounding

In this section we show that every nonzero ∆02 e-degree bounds a minimal pair.

Theorem 1. If X is ∆02 and not c.e., then there exists a pair of ∆0

2 sets A0, A1 ≤e

X, whose e-degrees form a minimal pair.

Proof. Let X be a given ∆02 set, with X not c.e., and let Xss∈ω be a ∆0

2-approximation to X.

We build e-operators Θi, with i ∈ 0, 1, such that the sets A0 = ΘX0 and

A1 = ΘX1 satisfy the claim.

2.1. The requirements. We aim to build Θ0 and Θ1 satisfying the followingrequirements, for every i ∈ 0, 1, every c.e. set W , and every pair of e-operatorsΦ0, Φ1:

BOUNDING AND NONBOUNDING IN THE E-DEGREES 3

N iW :ΘX

i 6= W

PΦ0,Φ1:Φ

ΘX0

0 = ΦΘX

1

1 ⇒ ΦΘX

0

0 c.e.

Throughout this proof, we will refer to some effective enumeration (Ψe, Ωe)e∈ω

of all pairs of e-operators. We shall write N ie = N i

W , if W = We, and Pe = PΦ0,Φ1,

if Ψe = Φ0 and Ωe = Φ1.

2.2. The priority ordering. We order the requirements according to the follow-ing priority ordering <:

N00 < P0 < N1

0 < P1 < N01 < P1 < N1

1 < · · ·

so that the e-th requirement Re is

Re =

N0i if e = 4i,

N1i if e = 4i + 2,

Pi if e = 2i + 1.

For an N -requirement N , we let

eN =

2e if N = N0e ,

2e + 1 if N = N1e ;

finally for a P -requirement P we let eP = e if P = Pe.

2.3. The strategies. We briefly describe the strategies to meet the requirements.

The N -strategies. The strategy to satisfy N = N ie is as in Gutteridge’s proof of

the existence of no minimal ∆02 enumeration degree, see e.g. [Co82].

We define the length of agreement function ì(e, s) by

ì(e, s) = µx ≤ s[ΘXi (x)[s] 6= We(x)[s]]

where Θi,s is the finite part of Θi defined at stage s. It is understood that ì(e, s) =s if no such x exists.

To achieve We 6= ΘXi , at stage s we “code” X into the eN -th column of ΘX

i

up to ì(e, s), that is, we enumerate for every j < ì(e, s) the axiom 〈〈eN , j〉, j〉into Θi. Since X ∈ ∆0

2, it follows that lims ì(e, s) always exists, either finite orinfinite. If lims ì(e, s) = ∞, then it easily follows that for all j, j ∈ X if and onlyif 〈eN , j〉 ∈ We, giving that X is c.e., a contradiction. Thus N is satisfied, sincelims ì(e, s) is finite. Furthermore, the action of N is finitary.

The P -strategies. Let us consider a requirement Pe = PΦ0,Φ1. We build a c.e. set

Ze to be equal to ΦΘX

0

0 ∩ ΦΘX

1

1 . The strategy is simply to enumerate all numbers

z ∈ ΦΘX

0

0 ∩ ΦΘX

1

1 into Ze.

Suppose that at some stage s, we enumerate a number z ∈ ΦΘX

0

0 ∩ΦΘX

1

1 into Ze,

but at some later stage t, the number z gets extracted from ΦΘX

0

0 ∩ ΦΘX

1

1 because


Xs 6= Xt. Suppose, say, that z /∈ ΦΘX

1

1 [t]. But at the moment we enumerated zinto Ze, there were axioms 〈z, F0〉 ∈ Φ0 and 〈z, F1〉 ∈ Φ1 such that F0 ⊆ ΘX

0 [s]

and F1 ⊆ ΘX1 [s]. We would like to fix (or put back) z in either Φ

ΘX0

0 or ΦΘX

1

1 asfollows. (The choice of side will depend on the priority, as specified later.) For

example we want to fix z in ΦΘX

0

0 . Then by enumerating the axioms 〈x, ∅〉 ∈ Θ0 for

every x ∈ F0 such that x ∈ ΘX0 [s] \ ΘX

0 [t], we permanently achieve that z ∈ ΦΘX

0

0

(if ΘX0 does not change on the remaining elements of F0, otherwise we have to

repeat the process for these elements). The outcomes of this action are either a

diagonalization z ∈ ΦΘX

0

0 \ ΦΘX

1

1 , or eventually z ∈ ΦΘX

1

1 as well. If all attempts to

diagonalize fail, then under the assumption that ΦΘX

0

0 = ΦΘX

1

1 one easily concludes

that Ze = ΦΘX

0

0 .

Interactions between different strategies. Clearly there is a conflict between thecoding activity of an N -requirement N = N i

k, and a P -strategy Pe = PΦ0,Φ1

enumerating axioms of the form 〈〈eN , j〉, ∅〉 ∈ Θi. The enumeration of such anaxiom may result in Pe injuring N , since we now have 〈eN , j〉 ∈ ΘX

i , but we mighthave j /∈ X.

Observe that the action of an N -strategy is finitary. The injury of the P -requirements which have lower priority than N can be dealt with by initializa-tion. To be more precise, fix a P -requirement: Each N -strategy N having higherpriority than P will only “control” finitely many numbers (those j’s such that〈〈eN , j〉, j〉 ∈ Θi) on its column ω[eN ]. Whenever the value of ΘX

i (〈eN , j〉) changes(due to the fact that the approximation of Xs(j) changes), we initialize the re-quirement Pe, that is, we discard the c.e. set Ze and start over. Therefore, we mayassume that whenever Pe at some stage t sees an axiom of the form 〈x, F 〉 ∈ Θi,with F ∩ ω[eN ] ⊆ ΘX

i [t] then we always have F ∩ ω[eN ] ⊆ ΘXi (where N is of the

form N = N ik and N i

k < Pe).

Thus, we only need to consider the N -strategies below some P -strategy whichhas an infinitary action.

An N -strategy below a P -strategy. Let N = N ik be of lower priority than Pe.

Assume for definiteness that i = 0, the case i = 1 being similar. The conflictbetween N and P may arise when at some stage t there is some z ∈ Ze,t but

z /∈ ΦΘX

0

0 [t] ∩ΦΘX

1

1 [t]. The crucial point here is that Pe need not injure N since forits attempt to diagonalization it may as well choose a finite set F and enumerateaxioms of the form 〈x, ∅〉 ∈ Θ1, for some x ∈ F . This might result in injuring some

N -strategies of the form N = N1k′ , but not N . We just observe that if z ∈ Φ

ΘX0

0 [t]then in fact this action may not yield any direct diagonalization, but in any caseit corrects Ze.

Several N -strategies below a P -strategy. Suppose that Pe < N0k < N1

k′ . If Pe

does not decide to injure N0k , then Pe might have to injure N1

k′ . This happens


when at some stage t there is some z ∈ Ze,t, z /∈ ΦΘX

0

0 [t] ∩ ΦΘX

1

1 [t], and for itsattempt to diagonalization at z, Pe may choose only either side of a pair of finitesets F0, F1 where F0 ⊆ ΘX

0 [s0] or F1 ⊆ ΘX1 [s1] for some s0, s1 < t, and there are

numbers 〈eN0

k, j〉 ∈ F0 such that 〈eN0

k, j〉 /∈ ΘX

0 [t]; thus, in order to avoid injuring

N0k , Pe chooses F1 and injures N1

k′ . But it follows from our discussion above onN -strategies that lims l0(k, s) < ∞, thus there are only finitely many numbers jsuch that 〈〈eN0

k, j〉, j〉 ∈ Θ0. By the fact that X is ∆0

2 there is a stage s2 suchthat for each such j and for all s ≥ s2 we have j ∈ X if and only if j ∈ Xs. Ifz is enumerated for the first time into Ze,s at some stage s ≥ s2, then for every〈eN0

k, j〉 ∈ F0 we have that

〈eN0

k, j〉 ∈ ΘX

i [s] ⇔ (∀v ≥ s)[〈eN0

k, j〉 ∈ ΘX

i [v]],

thus P does not need to enumerate any 〈〈eN0

k, j〉, ∅〉 into Θ0, hence does not injure

N1k′ for the sake of N0

k .

So the extent of the interaction between Pe, N0k , N1

k′ , is finitary being restrictedonly to attempts at diagonalizing with refer to those numbers z which are enu-merated into Ze before stage s2.

2.4. The construction. At the end of each step s, if a parameter has not beenexplicitly defined during step s then it is understood to have retained the samevalue as at the previous stage.

Step 0) For all i ∈ 0, 1 and e ∈ ω, let Θi,0 = Ze,0 = ∅.

Step s + 1) We first look for the highest priority requirement N = Ne, if any, with

e ≤ s such that there is some j with ΘXi [s + 1](〈eN , j〉) 6= ΘX

i [s](〈eN , j〉). If wefind such an N then let e(s + 1) = e, and initialize all Pe′ > N , by discarding theold Ze′ and starting a new c.e. set Ze′ initially set to be ∅ by letting Ze′,s+1 = ∅.If no such N exists, then let e(s + 1) = s + 1.

Next, we look for P -requirements that require attention at s + 1 (as definedbelow) and act accordingly.

We say that a P -requirement Pe with e ≤ e(s + 1) requires attention at stages + 1 if there is z ∈ Ze,s such that either

ΘX0 [v] ¹ F0 6= ΘX

0 [s] ¹ F0

orΘX

1 [v] ¹ F1 6= ΘX1 [s] ¹ F1

where v is the stage at which z was enumerated into Ze (after the last initialization)and F0 and F1 are the least finite sets with 〈z, F0〉 ∈ Φ0,v and 〈z, F1〉 ∈ Φ1,v andF0 ⊆ ΘX

0 [v] and F1 ⊆ ΘX1 [v], respectively.

Action for the P -requirements requiring attention. If Pe requiresattention then act as follows: For each such z ∈ Ze,s, consider the finite sets F0

and F1 defined above. For a finite set F , let e(F ) be the least eN such that thereare some j, i with 〈〈eN , j〉, j〉 ∈ Θi and j ∈ Xv \Xs (informally, N is the highest


priority N -requirement which P can injure because of F ). Among F0 and F1 chooseFi such that e(Fi) is greater. For each j such that 〈k, j〉 ∈ Fi, 〈〈k, j〉, j〉 ∈ Θi

and j ∈ Xv \ Xs, define the axiom 〈〈k, j〉, ∅〉 ∈ Θi by enumerating 〈〈k, j〉, ∅〉 intoΘi,s+1. We say also that Fi injures N through P , if e(Fi) = eN .

Action for the N-requirements. For every e ≤ e(s + 1), and for everyj < `i(e, s) define the axiom 〈〈eN , j〉, j〉 ∈ Θi.

Further action for the P -requirements.

For every P -requirement Pe with e ≤ e(s + 1), act as follows: If there is y suchthat

y ∈ (ΦΘX

0

0 [s] ∩ ΦΘX

1

1 [s]) \ Ze,s

then enumerate the least such y into Ze.

Finally, for every i ∈ 0, 1 let Θi,s+1 consist of Θi,s plus the numbers enumeratedinto Θi at step s + 1. Similarly, for every e, let Ze,s+1 consist of Ze,s plus thenumbers, if any, enumerated into Ze at step s + 1.

2.5. The verification. The verification is based on the following lemmas.

Lemma 2. For each i ∈ 0, 1, ΘXi is ∆0

2.

Proof. This follows from the fact that X is ∆02 and from definition of Θi, as for all

x〈x,D〉 ∈ Θi ⇒ [D = ∅ or D = (x)2].

¤

Lemma 3. For every N = N ie, we have:

(1) the setI ie = 〈eN , j〉 : 〈〈eN , j〉, ∅〉 ∈ Θi

is finite. In fact, the set

H ie = 〈eN , j〉 : (∃D)[〈〈eN , j〉, D〉 ∈ Θi]

is finite.(2) lims `i(e, s) exists and is finite and N i

e is satisfied.

Proof. By induction on r = eN . Suppose that the claim is true of all r′ < r, and letr = eN with N = N i

e. Notice that by induction for every N ′ < N , with N ′ = N i′

e′ ,the set

JN ′ = ΘXi′ ∩ ω[e

N′ ]

is finite.

We begin with observing that by construction, neither any N -requirement withN ′ 6= N nor any P -requirement P with N < P can enumerate any axiom ofthe form 〈〈eN , j〉, ∅〉 ∈ Θi. So we have only to check the effect of higher priorityP -requirements.

Suppose P = Pe′ = PΦ0,Φ1< N , and assume by induction that s0 is a stage such

that at every s > s0:


• no P ′ with P ′ < P < N injures N ie;

• P does not injure any N ′ with P < N ′ < N ;• for N ′ such that P < N ′ < N we have:

(∀x ∈ JN ′)[x ∈ ΘXi′ [s] ⇔ x ∈ ΘX

i′ ].

For every number z, if z is enumerated in Ze′ at any stage t > s0, then z ∈

ΦΘX

0

0 [t]∩ΦΘX

1

1 [t], and therefore there are two finite sets F0, F1 such that Fi ⊆ ΘXi [t]

and neither set injures any N ′ with P < N ′ < N . But the action dictated bythe construction, of putting elements of F1−i back into ΘX

1−i does not injure N .Hence P may injure N only because of the finitely many numbers z′ which areenumerated in Ze′ before stage s0, and can do so at most once for each such z′.We have therefore seen that I i

e is finite.

The finiteness of H ie follows form the proof of (2) below.

Let us now show (2). Since ΘXi is ∆0

2 by Lemma 2, we have that lims ì(e, s)always exists, either finite or infinite; and

lims

ì(e, s) finite ⇔ ΘXi 6= We.

If lims ì(e, s) = ∞, then for all but finitely many numbers j, we have by (1) that〈〈eN , j〉, j〉 ∈ Θi but 〈〈eN , j〉, ∅〉 /∈ Θi . Therefore for all but finitely many j,

j ∈ X ⇔ 〈eN , j〉 ∈ ΘXi ,

giving X c.e., since ΘXi = We, a contradiction. ¤

Lemma 4. For every e, Pe is initialized finitely often and is eventually satisfied.

Proof. Suppose that Pe = PΦ0,Φ1. Since Pe can be initialized only by some N < Pe,

and since for each such N , JN is finite and ΘXi ∈ ∆2, each i ∈ 0, 1, it follows

that Pe is initialized only finitely often.

Assume that ΦΘX

0

0 = ΦΘX

1

1 . We want to show that the set Z = ΦΘX

0

0 = ΦΘX

1

1 isc.e. Let s0 be a stage after which Pe is never initialized. We claim that Z = Ze

where Ze is the set built by Pe after the last initialization, i.e.,

Ze =⋃

s>s0

Ze,s.

Indeed the inclusion ⊆ is trivial, since at almost all stages we enumerate into

Ze the least number in ΦΘX

0

0 ∩ ΦΘX

1

1 that is not already in Ze. This gives Z ⊆ Ze.

As to the other direction, suppose that z is the least number such that z ∈

Ze \ΦΘX

0

0 , and let s1 > s0 be a stage such that z ∈ Ze,s1and for every x < z either

x /∈ Ze, or x ∈ Ze,s1and for every s ≥ s1 we have that x ∈ Φ

ΘX0

0 [s] ∩ Φ01ΘX

1 [s].Then Pe will certainly require attention at some stage t ≥ s1, and the construction

ensures that z will be back into ΦΘX

1

1 or ΦΘX

0

0 permanently, thus either ΦΘX

0

0 6= ΦΘX

1

1

or z ∈ Z, both conclusions contradicting our assumptions. ¤

This concludes the proof of Theorem 1. ¤


In fact, the enumeration operators Θi turn out to be e-operators of a specialform. We recall that an e-operator Φ is called an s-operator if for all x,D if〈x,D〉 ∈ Φ then the cardinality of D is ≤ 1. A set A is s-reducible to a set B(A ≤s B) if A = ΦB for some s-operator Φ. For more on the structure of thes-degrees see e.g. [FHR59], and [Coo90].

Then

Corollary 5. Every nonzero ∆02 s-degree bounds a minimal pair of s-degrees.

Proof. The above proof shows in fact that A0, A1 ≤s X, and for every Z if Z ≤e A0

and Z ≤e A1 then Z is c.e.. ¤

3. Nonbounding minimal pairs

What happens if instead of X ∈ ∆02 we start with a set A which is Σ0

2 but not∆0

2? If one tries to mimic the proof of Theorem 1 starting from a Σ02 set A, the

following scenario may happen, besides other evident immediate difficulties. Letus consider N -requirements N0

k and N1k′ such that P < N0

k < N1k′ . Let P = PΦ0,Φ1

,and for simplicity let N = N0

k and N ′ = N1k′ . When a number j appears in A, the

length of agreement function l0(k, s) increases, the N -strategy adds more axioms inΘ0, hence more elements of the form 〈eN , j〉 enters ΘA

0 . Thus, more elements enter

Z = ΦΘA

0

0 ∩ ΦΘA

1

1 . When j leaves A at some later stage, z ∈ Z leaves ΦΘA

0

0 ∩ ΦΘA

1

1 ,P then must injure N ′ (by adding an axiom 〈〈eN ′ , j′〉, ∅〉 in Θ1) in order not toinjure N . For a ∆0

2 set X, the finiteness of the number of changes on the relevantinitial segment of X comes to rescue: In this case, l0(k, s) is bounded, only finitelymany axioms 〈〈eN , j〉, j〉 ∈ Θ0 are added. For a properly Σ0

2 set A, although wecan try to modify the strategies to overcome some of the difficulties, the damagedone by P on N ′ remains: It could happen that there are infinitely many z’s whichenter Z at some “false” stages and leave at later “true” stages, hence the actionof P at “true” stages would enumerate too many axioms into Θ1, preventing usfrom concluding that ΘA

1 6= Wk′ .

This hints at the possibility that there might be a properly Σ02 set A such

that whatever pair of e-operators Θ0, Θ1 we may consider, there are e-operatorsΦ0, Φ1 such that any successful effort of simultaneously making ΘA

0 is non-c.e. and

Ze = ΦΘA

0

0 = ΦΘA

1

1 c.e. would make ΘA1 c.e., in other words, this hints at the

possibility of constructing a properly Σ02 set whose e-degree does not bound any

minimal pair.

Indeed, in contrast with ∆2 e-degrees, we are able to establish the followingnonbounding theorem for Σ2 e-degrees.

Theorem 6. There is a Σ02 set A, which is not c.e. and for all X,Y ≤e A, either

X is c.e., or Y is c.e., or there is a set D such that D is not c.e. but D ≤e X andD ≤e Y .


3.1. The requirements. We will build a Σ02 set A so as to satisfy the following

requirements, for every c.e. set W and every pair of e-operators Θ0, Θ1:

NW : A 6= W

RΘ0,Θ1: ΘA

0 c.e. ∨ ΘA1 c.e. ∨ (∃Φ0, Φ1)[Φ

ΘA0

0 = ΦΘA

1

1 ∧ (∀ c.e W )[ΦΘA

0

0 6= W ]]

where Φ0 and Φ1 are e-operators built by us.

3.2. The strategies. We briefly outline the strategies.

The obvious starting point is to mimic the Lachlan construction of a nonbound-ing c.e. Turing degree [La79]. In the Turing case the R-requirements take actuallya different form: together with the N -requirements NW ’s, Lachlan’s constructionof a c.e. set A aims at meeting the following requirements for every quadrupleX,Y, Θ0, Θ1, where X,Y are c.e. sets and Θ0, Θ1 are Turing functionals:

RX,Y,Θ0,Θ1: X = ΘA

0 ∧ Y = ΘA1 ⇒ [X computable ∨ Y computable

∨ (∃D, Φ0, Φ1)[D = ΦX0 = ΦY

1 ∧ (∀ c.e W )[D 6= W ]]

where in this case Φ0 and Φ1 are of course Turing functionals, and D is a c.e. set.

In the Turing degree case the strategy for the requirement RX,Y,Θ0,Θ1tests first if

X = ΘA0 and Y = ΘA

1 . If the test is passed, then the job is spread to infinitely manysubstrategies SW , each of which has a diagonal witness d against W . When d isrealized (i.e., d enters W ), SW seeks permission first from X (to get ΦX

0 (d) ↑), thenfrom Y (to get ΦY

1 (d) ↑). A failure of permission makes progress on showing thatX is computable or Y is computable; on the other hand a successful permissionallows us to diagonalize against W , by putting d into D to get D(d) 6= W (d). Wefollow roughly the same plan. However things change considerably when workingin the e-degrees. First of all notice that X and Y do not really enter the playin the e-degree case: Whereas in the Turing case ΘA

0 or ΘA1 may be undefined at

some point, or be both always defined but with ΘA0 or ΘA

1 not being c.e. sets, inthe e-degree case (where we deal with Σ0

2 sets) for every pair of e-operators Θ0 andΘ1 the sets ΘA

0 and ΘA1 are always Σ0

2 sets. A similar remark applies to D, whichmust be explicitly built as a c.e. set in the Turing case, whereas in the context ofthe e-degrees it is of course sufficient to built just Φ0 and Φ1 so that ΦX

0 = ΦY1 .

The symbols X and Y should be therefore regarded in this context only as mereabbreviations of ΘA

0 and ΘA1 respectively, and D as an abbreviation of ΦX

0 , andΦY

1 when equal. All this considered it is however quite useful to think of RΘ0,Θ1as

RX,Y,Θ0,Θ1: X = ΘA

0 ∧ Y = ΘA1 ⇒ [X c.e. ∨ Y c.e. ∨

(∃D, Φ0, Φ1)[D = ΦX0 = ΦY

1 and (∀ c.e W )[D 6= W ]]].

Each such requirement will split into infinitely many subrequirements SWΘ0,Θ1

, where

W is a c.e. set, acting towards making ΦX0 (d) = ΦY

1 (d) 6= W (d), on some witnessd.

In the Turing degrees, the emphasis is on the computability of the set D relativeto X and Y , therefore the use of links and other devices for prompt actions are


essential, whereas in the e-degrees such devices are to some extent less crucial,since we only require the set D to be enumeration reducible to X and Y .

The serious trouble turns out to be “maintaining X- and Y -permission”. Inthe Turing degrees, we can control X- and Y -permission via Θ0 and Θ1 by addingrestraints on A, whereas in the e-degrees the permission may come and go infinitelyoften. To be more precise, let us consider a possible attempt to extract a realizedwitness d from D in order to diagonalize against W . For this we need to extractd from ΦX

0 and ΦY1 . To extract d from ΦX

0 , we need to extract numbers from X(mutatis mutandis, this is similar to the permitting method in the Turing case).It is true that we can extract numbers from A to make some k out of X, but this“permission” could be only temporary since ΘA

0 may have another axiom of theform 〈k,E〉 ∈ Θ0 where E ⊆ A and E does not contain any of the numbers alreadyextracted, thus putting k back into X. To keep the permission, we would needto extract the new finite set E from A as well. As we have no obvious control onthe enumeration of Θ0, in the worst case Θ0 can force us to extract all witnessesappointed by other N -requirements. We have therefore to draw the line betweenextracting numbers out of A for the sake of extracting witnesses from D, andswitching to other strategies. One of the key new ingredients of our constructionis to assign relative priorities (which we will refer to as local priorities) to SW -substrategies and N -strategies, so that SW -substrategies “favor” — so to speak —lower priority N -strategies.

Very briefly, we proceed as follows, relatively to a witness d. We try to find afinite set E without interfering with other higher priority strategies or higher localpriority N -strategies, such that the extraction of E from A guarantees that somek gets out of ΘA

0 . We then arrange that the only Φ0-axiom relevant to d, whichhas been enumerated, is 〈d, k〉 ∈ Φ0. Failure of finding such a finite set E, witha corresponding number k, gives us a global win of RΘ0,Θ1

: X is c.e.. If we areable to find such an E, then we turn to the argument that we can diagonalize ΦX

0

against all c.e. sets. The typical scenario begins with extracting numbers out of Awhich entails that some k goes out of X, and if Θ0 does not enumerate any newaxioms for k then we get X-permission. If Θ0 always responds by enumeratingnew axioms, say of the form 〈k,E ′〉 ∈ Θ0 with E ′ ⊆ A and E ′ does not interfereswith either higher priority requirements or higher local priority N -requirements,we then change our mind: Instead of looking for X-permission, we enumerate kinto a set U and restrain A on E ′. Luckily, Θ0’s activity succeeds in pushing theset E ′ to a safe place, so that higher local priority requirements have room to act.

In a similar way we treat the possible extraction of d from ΦY1 , when Y is

not c.e.. We first arrange that the only Φ1-axiom relevant to d, which has beenenumerated, is 〈d, l〉 ∈ Φ1, for some l. Suppose we have found a finite set Ewhose extraction from A guarantees that k is extracted from X. We then turnto the problem of finding a finite set F whose extraction from A guarantees theextraction from Y of l. Again, failure of finding such a finite set (not interferingwith either higher priority requirements or higher local priority N -requirements),with a corresponding number l, gives us a global win of RΘ0,Θ1

: Y is c.e.. If we


do find such a finite set F then we temporarily keep k ∈ X and l ∈ Y , by puttingE and F back into A. If later d gets realized then we extract E and F from A,getting k /∈ X and l /∈ Y and thus ΦX

0 (d) = ΦY1 (d) 6= W (d).

As in many priority arguments, in the course of the construction we will makeuse of a tree of strategies T , which we will describe in details later. Each nodeδ ∈ T is devoted to act towards satisfying some given requirement. We will saythat δ is an RΘ0,Θ1

-strategy if δ is assigned the requirement RΘ0,Θ1; δ is an NW -

strategy if δ is assigned the requirement NW ; and finally δ is an SWΘ0,Θ1

-strategy if

δ is assigned the requirement SWΘ0,Θ1

.

The N -strategies. The strategy for NW is the usual Friedberg-Muchnick strat-egy:

(1) pick a new a, and keep a ∈ A;(2) wait for a to become realized (i.e., a ∈ W );(3) extract a from A.

The outcomes of this strategy are clear. We say that the outcome is 1 if a nevergets realized, otherwise the outcome is 0.

The R-strategies. The strategy for R = RΘ0,Θ1(= RX,Y,Θ0,Θ1

) is rather com-plicated. Suppose that α is an RΘ0,Θ1

-strategy. The strategy α plays the roleof a mother node and the real jobs are done by its substrategies SW

α . In laterdiscussions, we may drop the indices if there is no confusion.

At α, we set the outcome to be 0 and we split the jobs to α’s S-substrategies.Although the axioms of Φ0 and Φ1 appear to be enumerated by α’s substrategies,strictly speaking we should think of Φ0 and Φ1 as built at α and we should in factbe talking about Φα,0 and Φα,1.

The substrategies SW are located below αˆ0, in fact they form a subtree, andwe try the substrategies SW one by one.

The SW -strategies. Let us consider a substrategy SW = SWX,Y,Θ0,Θ1

of someR-strategy.

An SW -strategy β works as follows. The informal picture of β has two layersof strategies. The upper layer deals with X-permission, and the lower layer workson Y -permission based on the outcome of the upper layer.

The picture we have in mind is the following:


β

∞X 0

∞Y 0 1· · ·

1

∞Y 0 1· · ·

· · · d0

For the sake of simplicity of the tree structure, we compress the two layers intoone:

β

∞X 〈∞Y , 0〉〈0, 0〉〈1, 0〉· · ·

〈∞Y , 1〉〈0, 1〉〈1, 1〉· · ·

d0

There are thus three possible groups of outcomes for β, which are arranged fromleft to right as follows:

∞X <L (T0) <L (T1) <L (T2) <L · · · <L d0.

where (Tk) is the collection of nodes:

〈∞Y , k〉 <L 〈0, k〉 <L 〈1, k〉 <L · · ·

We will be sometimes talking about outcome (Tk) for β if the outcome of β isinside (Tk). Likewise we will be talking about strategy βˆ(Tk), etc..

The intended meanings of the outcomes are as follows.

• The outcome d0 indicates that an unrealized candidate d for the diag-onalization against W is found. This is a Σ2-outcome or essentially aΠ1-outcome.

• For each natural number k, the outcome group (Tk) indicates that we havefound an agitator Ek for k, as described next. It is a Σ2-outcome as far asX is concerned (that is, if we ignore the Π2-outcome 〈∞Y , k〉 ∈ Tk).

We now explain what an agitator set is for a given number. For everye-operator Ψ and any number k, let (with slight modifications we borrowthis notation from [Coo87]):

εΨ(k) = F : F finite and 〈k, F 〉 ∈ Ψ

and let

↑εΨ(k) = E : F finite and (∀F ∈ εΨ(k))[F ∩ E 6= ∅];

thus if some E ∈↑εΨ(k) is restrained out of a set B then k /∈ ΨB.If β has final outcome inside the group (Tk), then k is the least num-

ber such that for some finite set Ek, E ∪ Ek ∈↑ εΘ0(k), where E denotes


the numbers currently kept out of A by higher priority strategies and Ek

interferes with some N -strategies N which have higher local priority thanβˆ(Tk), meaning that each such N has a realized witness a ∈ Ek. (Theprecise description of local priority is discussed later.)

We call Ek a Θ0-agitator of k (or simply an agitator, when Θ0 is under-stood from the context) because we can extract k from ΘA

0 (or put back kinto ΘA

0 ) by extracting Ek from A (or putting back Ek into A, respectively).• The outcome ∞X indicates that for all k ∈ ω, there is an axiom 〈k,E ′〉 ∈ Θ0

such that E ′ does not interfere with any strategy having either higherpriority or N -strategies having higher local priority than βˆ(Tk). Obviouslythis is a Π2-outcome for β. If SW has outcome ∞X , then we will arguethat the set X is c.e. by building a set U at β such that X = U .

Before we discuss the “lower layer”, which deals with Y -permission, let us takea pause and quickly look at the possible conflicts between the action of βˆ(Tk) andthe N -strategies below βˆ∞X . We will explain the notion of “local priority” inthe process.

Let us assume that we have one lower priority N on some node γ extendingβˆ∞X . N has an unrealized witness a. Suppose that at some stage s, we finda number k, such that 〈k,E〉 ∈ Θ0, for some finite E ⊆ A, and a ∈ E. Atthis moment, we would believe that a would remain in A, and put k into U , andhave outcome βˆ∞X . Unfortunately, our belief on a might be misplaced, a couldget realized at some later stage t (or possibly at the very same stage s), so wetake a out of A when N acts. This injures SW since we now have an error in U .To overcome this injury, we temporarily give the diagonalization of D against Whigher priority than N . This is reasonable, as we shall see that failure of correctingU will switch N off the true path. At future stages, when we visit β, from time totime we may put a back into A so that k is back into X, and we can use the samek for an axiom 〈d, k〉 ∈ Φ0, where d is the current witness. So long as Θ0 doesnot provide us with a new axiom, we play with a, the true path will pass throughβˆ(Tk), and the activities of the lower layer will succeed. Notice that this a is acandidate for making A properly Σ2, as we shall see in the next paragraph. IfΘ0 provides us with a “new” axiom 〈k,E ′〉 ∈ Θ0 (i.e., not interfering with higherpriority strategies or with higher local priority N -strategies) with a /∈ E ′, then Uis correct again by restraining A on E ′, we switch back to βˆ∞X and k remainsin X and thus U(k) is correct forever (provided E ′ ⊆ A).

When more than one N -strategies are involved, there are potential dangers thatone k could involve infinitely many neighborhoods. For example, the new E ′

could involve some other witnesses, such as a1 which is a witness for N1; the samecould happen for a different witness a2, then for a3, etc. We have to know whenwe should stop accommodating these ai’s. Notice that the conflict only happensbetween the N ’s below βˆ∞X and the k chosen by the same β. We cannot affordto give absolute higher priority (than SW ) to these N -nodes even if they appearstrictly to the left of βˆ(Tk). The solution is to assign local priority to βˆ(TK)and compare it with the local priority of N . We will allow βˆ(Tk) to injure, say,


N for all N with lower local priority than βˆ(Tk). Whenever k discovers a newaxiom 〈k,E ′〉 ∈ Θ0 with E ′ ⊆ A and E ′ not interfering with either higher priorityor higher local priority strategies, all N with lower local priority than βˆ(Tk) willdiscard their old witnesses which are in E ′. This arrangement will make us winboth N and SW . If N is on the true path, then only finitely many nodes to itsright can force it to discard its witness, because each k could injure N only finitelyoften. Thus the witness a of N eventually does not get discarded anymore, givingto N the possibility to get satisfied.

The above is the upper layer picture of β. We now continue to discuss the lowerlayer of β, which deals with Y -permission.

Each of the outcome group (Tk) itself consists of outcomes which are all basedon the assumption that k has a permanent agitator Ek, which is eventually out ofA.

The intended meaning of the outcomes is as follows.

• The outcome 〈l, k〉 indicates that for this number l there is a “small” εΘ1(l),that is there is a finite set F k

l such that E ∪ Ek ∪ F kl ∈↑ εΘ1(l), with

F kl interfering with some N -strategies, which have higher local priority

than βˆ〈l, k〉. (Again, E denotes the numbers currently kept out of A byhigher priority strategies.) Notice that the conflict now is between the N -strategies below 〈∞Y , k〉 and the activity at nodes βˆ〈l′, k〉. In the luckycase, by taking out Ek and some such F k

l from A, we win SW by successfullyextracting a realized witness d out of D, for which the only relevant axiomsare 〈d, k〉 ∈ Φ0 and 〈d, l〉 ∈ Φ1. We will have outcome 〈l, k〉 forever.

• Of course, most of the cases will be unlucky, that is, either k comes backinto X with an axiom 〈k,E ′〉 ∈ Θ0, i.e., an axiom such that E ′ does notcontain elements currently kept out of A by strategies having higher prioritythan βˆ(Tk) or by N -strategies having higher local priority than βˆ(Tk); orl comes back into Y with a new axiom 〈l, F 〉 ∈ Θ1, i.e., an axiom such thatF does not contain elements currently kept out of A by strategies havinghigher priority than βˆ〈l, k〉, or by N -strategies having higher local prioritythan βˆ〈l, k〉. The former case is handled by the outcome ∞X in the upperlayer. We consider the latter case only. The outcome 〈∞Y , k〉 indicatesthat for all l ∈ ω, there is a neighborhood F which does not interferewith Ek or any strategies having higher priority or any N -strategies havinghigher local priority than 〈l, k〉. Again 〈∞Y , k〉 is a Π2-outcome (modulothe Σ2-outcome (Tk) in the upper layer). If SW has outcome 〈∞Y , k〉, thenwe will argue that the set Y is c.e. by building a set Vk at βˆ〈∞Y , k〉 suchthat Y = Vk.

We now take a closer look at β. The environment of β at stage s consists of thefollowing parameters (not necessarily all are defined at the same time):

• A stage s− when β was visited for the last time, and an outcome o− whichwas the outcome of β at s−. If s is the first stage (after initialization of β)


then s− = s and o− = ∞X , because ∞X indicates that U ⊆ X, which iscertainly the case when U is empty.

• A computable bijection σβ,∞X(k) mapping all numbers k ∈ ω onto the

N -strategies below βˆ∞X . We say that a requirement N has lower localpriority than βˆ(Tk) if N = σβ,∞X

(i) for some i ≥ k. The notion of higherlocal priority than βˆ(Tk) is defined accordingly.

Similarly, for every k, a computable bijection σβ,k(l) mapping all numbersl ∈ ω onto the N -strategies below βˆ〈∞Y , k〉. We say that such an N haslower local priority than βˆ〈l, k〉 if N = σβ,k(I) for some i ≥ l. The notionof higher local priority than βˆ〈l, k〉 is defined accordingly.

• A set U aiming to show that X is c.e.; we intend to make U a subset ofX such that every k in U has an axiom 〈k,E〉 ∈ Θ0 such that E doesnot interfere with either higher priority strategies or higher local priorityN -strategies. Eventually U = X or there is some k as in next item.

• A number k ∈ U \X with a corresponding agitator; this number is due tothe action of higher priority strategies and higher local priority N -strategies(higher than βˆ(Tk)), namely if E denotes the elements currently kept outof A by higher priority strategies, and Ek denotes the elements kept out ofA by higher local priority N -strategies, then E∪Ek ∈↑ εΘ0(k), i.e., the onlyavailable Θ0-axioms for k are of the form 〈k,E ′〉 ∈ Θ0 with E ′∩ (E∪Ek) 6=∅. Notice also that among the elements kept out of A by higher prioritystrategies one must also take into account numbers belonging to possibleagitators used and currently extracted by higher priority S-strategies.

• An agitator Ek for k as above.• For each k, a set Vk aiming to show that Y is c.e.; we intend to make Vk

a subset of Y such that every l in Vk has an axiom 〈l, F 〉 ∈ Θ1 such thatF does not interfere with either higher priority strategies or higher localpriority N -strategies (higher than βˆ〈l, k〉), or the agitator Ek). EventuallyVk = Y , or the is some l as in next item.

• For each k, a number l ∈ Vk \Y with a corresponding agitator; this numberis due to the action of higher priority strategies and higher local priority N -strategies (higher than βˆ〈l, k〉), and the agitator Ek, namely if E denotesthe elements currently kept out of A by higher priority strategies, and F k

l

denotes the elements kept out of A by higher local priority strategies N ,then E ∪ Ek ∪ F k

l ∈↑ εΘ1(l), i.e., the only available Θ1-axioms for l are ofthe form 〈l, F 〉 ∈ Θ1 with F ∩ (E ∪ Ek ∪ F k

l ) 6= ∅.• An agitator F k

l for l as above.• A number d, which is the diagonalization witness.

Of course the above parameters are functions of β, so we should for completenesswrite e.g. Uβ, Vβ,k, etc. Likewise we should talk of Θα,0, Θα,1, Xα, Yα, Φα,0, Φα,1

etc. We shall omit mentioning the node when this is clearly understood from thecontext.

3.3. Coordination between strategies. We now discuss the conflicts betweendifferent strategies and how to solve them. In particular, we will point out the


potential cause for an inequality between ΦX0 (d) and ΦY

1 (d) and how to removethe cause without using any links. Coordination of the strategies also explainswhy there is no injury on the true path (in the typical 0′′′-priority argument fash-ion). It is also worth noting the reason for D to be properly Σ2, if R has theglobal Π3-outcome. Although many ideas will become precise only after the for-mal construction, we believe that the following informal discussion will help readerunderstand the construction.

We begin with the interaction between one S-strategy and one N -strategy, as-suming the corresponding nodes are β (for S) and γ (for N). We might haveincluded the mother node α ⊂ β, at which the e-operators Φ0 and Φ1 are located.However, since every e-operator is total, we may assume that all axioms are enu-merated at S-nodes. It will become clear how the corrections of Φ0 and Φ1 (ifany) should be done. Therefore, we may ignore the node α and the correspondingR-strategy.

Let us briefly set things up as follows:

• β has diagonalization witness d, chooses k ∈ X, l ∈ Y , enumerates〈d, k〉 ∈ Φ0 and 〈d, l〉 ∈ Φ1. There are a Θ0-agitator E = Ek anda Θ1-agitator F = F k

l for k, l respectively. β also has U and Vk. The ac-tions include agitating E and F , i.e., extracting E, F out of A and puttingthem back into A.

• γ has diagonalization witness a. The action consists in extracting a out ofA.

After considering all relative positions of β and γ, many cases are trivially dealtwith by initialization. We list the following less trivial cases.

(1) βˆd0 ⊆ γ or βˆ〈l, k〉 ⊆ γ.The action of N has no impact on S, since N picks its a after S picks

its parameters, such as E and F . If β ever changes its outcome then it hasto make a move to the left, and N gets initialized.

(2) βˆ〈∞Y , k〉 ⊆ γ.The a chosen by N will be bigger than all elements of E. But a could be

in several F ’s (think of the scenario in which β goes through the outcomecycle d0 → 〈l, k〉 → 〈∞Y , k〉 several times). The extraction of a at N willnot create any inequality of ΦX

0 (d) and ΦY1 (d), since d is cancelled when

〈∞Y , k〉 was accessible. Instead, the extraction of a creates errors in Vk,hence contributing to produce some agitator F k

l for some l (maybe someearlier l ∈ Vk).

Then S takes control of a, i.e., S puts a back into A and recycle k, lfor some new d′ by enumerating 〈d′, k〉 ∈ Φ0 and 〈d′, l〉 ∈ Φ1. Whend′ is realized, a is agitated out of A again through agitating F k

l , and wehave outcome 〈l, k〉. Notice again that no inequality arises between theΦi’s. If this outcome ever changes in the future, it means that either F k

l

stops being an agitator for l (at which stage βˆ〈∞Y , k〉 is accessible) or ais initialized (at which stage βˆ∞X is accessible). In summary, a comes inand out of A at most twice for each agitator F k

l .


(3) βˆ∞X ⊆ γ.The case is similar to (2). It will not create any inequalities between the

Φ’s. But it creates errors in U . Since each Vk is initialized, there is no errorin Vk.

As in (2) S takes control of a when it is accessible the next time, i.e.,S puts a back into A and recycles k, together with some l, picks new d′

and enumerates 〈d′, k〉 ∈ Φ0 and 〈d′, l〉 ∈ Φ1. If the true outcome is〈∞Y , k〉 for some fixed k, then N is never accessible, but a remains foreverin the set Ek. Thus a is not in A in a Π2-way. (Notice that this is differentfrom (2), since in (2) a would be in and out finitely many times only.)

We can summarize injury as follows: When an a is extracted at γ, it may injurenodes β above it, and it must be the case that βˆ∞X ⊆ γ or βˆ〈∞Y , k〉 ⊆ γ.When this happens, β takes control of a, and agitates a infinitely often in theformer case, and finitely often in the latter. Before β releases its control, γ isnot accessible. Moreover, the extraction of a at γ will not create any inequalitiesbetween the Φi’s, hence no correction of Φ0 and Φ1 is needed.

We now add one more S-strategy, say S ′ at β′ into play. Again we may ignorethe mother node α′ of β′.

By symmetry, let us only consider the impact of the action of β on the strategyβ′. The impact was made through the number a of which β has taken control. Weselect the most interesting cases below.

(1) β′ˆ〈∞Y , k〉 ⊆ β or β′ˆ∞X ⊆ β.We discuss the former, since latter is the similar. If S ′ sees an agitator F ′

for some l involving a, then S ′ would take control of N instead of S. Whenthe higher priority S ′ takes control of N , S is not accessible. S ′ release itscontrol only when its F ′ has nothing to do with a. Therefore if S agitatesa in the future, it would not affect S ′.

(2) γ <L β′ and βˆ〈∞Y , k〉 ⊆ β′.This turns out to be the trickiest case.The worry is that the d′ chosen at S ′ will be agitated by the action of S

via N .The first observation is that only a realized d′ can have this problem,

for the following reason: When we pick d′ at β′ we have 〈d′, k′〉 ∈ Φα′,0

and 〈d′, l′〉 ∈ Φα′,1, agitators E ′, F ′, and some axiom 〈k′, D′〉 ∈ Φα′,0,with D ∩ E ′ 6= ∅: when we pick d′, the axiom for k′ applies, so E ′ is in A.When β′ is visited, all elements being controlled by higher priorities nodes(such as S) are out of A, thus E ′ must be an agitator whose elementsare only controlled by S ′ itself, namely, E ′ contains realized witnesses ofstrategies N ′’s extending β′ˆ∞X or β′ˆ〈∞Y , k〉. The only possibility for〈k′, D′〉 failing to apply and at the same time S ′ is not initialized is that S ′

acts itself to take E ′ out. This only happens when d′ is realized.After S ′ agitates E ′ (i.e. takes it out of A), Θα′,0 may have an axiom of

the form 〈k′, D〉 with a ∈ D and D \ a ⊆ A (it is possible since we haveno control on the enumeration of Θα′,0). When S agitates a, d′ gets in and


out of ΦX′

α′

α′,0. We cannot discard d′ since S ′ might be on true path. Thisexplains why D is properly Σ2. Also notice that injury on true path willnot help, e.g., in the case R′ < R < S < S′.

Now assume that S agitates a infinitely often. When k′ goes back intoΘA

α′,0 via a new axiom 〈k′, D〉 ∈ Θα′,0 as before, there might be a disagree-

ment d′ ∈ ΦXα′,0 \ ΦY

α′,1. Next time we visit α′ we see that we may need tocorrect d′ at β′. Since β′ may not be accessible any more, we may want tojump to β′ or travel through some kind of link to β′ to do the job.

However, we actually have an alternative way to avoid the occurrenceof the inequality at β′. Notice that the problem arises only when we puta back into A, that is when we give outcome d0 at β. Also notice that ifthis were the true outcome on the true path, then this is finitary! Thus,whenever we give outcome d0 at a node such as β, then by Clause (?) of theconstruction we momentarily put back into A all N -witnesses relative to N -strategies γ ⊇ βˆ〈∞Y , k〉. After all we do not care about them if the truepath goes through βˆd0. Notice that in this case we automatically correctthe disagreement at d′, since l′ does not depend on the agitator chosen byβ, and on other higher priority agitators and realized N -witnesses along β.If d0 is the true outcome, then we have k′ and l′ forever in X and Y , thuswe have Φ′X

0 (d′) = Φ′Y1 (d′) = 1 forever.

When we abandon the outcome d0 at β then we can reestablish thesituation as it was before putting these elements back into A. We reestab-lish things only when we give outcome 〈∞Y , k〉 (or ∞X , according to thecase): Strategies below these outcomes will have the opportunity again ofextracting their witnesses or agitators as needed.

The other cases are similar.

3.4. Construction. First we describe the priority tree T of strategies N and R,and substrategies SW , which we use in the construction.

As we have seen, the only possible outcomes of an N -strategy γ are 0 <L 1, anR-strategy has just one outcome 0, and the outcomes of a substrategy SW are asdescribed before, and ordered as indicated.

As tree of strategies T let us fix some computable subtree of Z<ω, where

Z = 0,∞X , 〈l, k〉, d0 : l ∈ ω ∪ ∞Y , k ∈ ω

such that we have a computable function R assigning requirements to the nodesof T satisfying

• for every infinite path f ∈ [T ], the mapping R assigns requirements in a1-1 fashion onto the nodes of f ;

• if α ∈ T and α is an R-strategy, i.e., α is assigned an R-requirement, thenαˆ0 ∈ T ;

• if α ∈ T and α is an N -strategy, i.e., α is assigned an N -requirement, thenαˆo ∈ T for o ∈ 0, 1;


• if α ∈ T and α is an S-strategy, i.e., α is assigned an S-requirement, thenαˆo ∈ T for o ∈ ∞X , 〈l, k〉, d0 : l ∈ ω ∪ ∞Y , k ∈ ω;

• the empty node ∅ is in T ;• along any infinite path each R-strategy appears before all its S-substrategies;• along any infinite path f , if βˆ∞X ⊂ f or βˆ〈∞Y , k〉 ⊂ f for some k ∈

ω, where β is an S-strategy which is a substrategy of α, then no othersubstrategy of α is assigned to any node extending β in f ;

• T does not contain other nodes.

If α, β ∈ T we say that α <L β (or α is to the left of β) if there is somei < |α|, |β| such that αi 6= βi and αi <L βi for the least such i. We say that α ≤ βif α <L β or α ⊆ β, and α < β if α ≤ β but α 6= β.

The construction is by stages: at stage s we define the current approximationAs to A and the values of several parameters.

At stage 0, set A0 = ω; Φα,0 = Φα,1 = ∅, Uβ = ∅; Vβ,k = ∅, for every α, β ∈ T ,and k ∈ ω. All witnesses (N -witnesses and SW -witnesses) are undefined.

At each stage s, we define the accessible strings of length ≤ s and act accordingly.These strings are all substrings of a (least) string δ of length s. If α is accessibleat s we also say that α is visited at s.

At the end of stage s we initialize the nodes to the right of δ and below δ,namely we reset all parameters of these nodes as follows.

• When we initialize an N -node γ, we discard its current witness a, if any.Notice that by (?) of the construction we permanently enumerate a backinto A.

• When we initialize an R-node α, we start over the e-operators Φα,0 andΦα,1 and all its substrategies get initialized.

• When we initialize an S-node β, we discard its witness d, if any. Whenwe discard a witness d we define 〈d, ∅〉 ∈ Φ0 and 〈d, ∅〉 ∈ Φ1 (if Φi = Φα,i,where α ⊂ β is the R-strategy of which β is an s-substrategy). Also we letUβ = ∅ and Vβ,k = ∅ for all β, k such that δ < βˆ〈∞Y , k〉. If a number xwas restrained at some S-node β, and β is initialized, then x stops beingrestrained.

The string δ is defined inductively. The root of the priority tree is alwaysaccessible. Suppose that we have defined δ ¹ n, with n < s. We define δ(n) anddescribe the actions according to the strategy δ ¹n. By default, if δ ¹n has outcomeo then δ ¹nˆo will be the next accessible string.

Case 1) δ ¹n is an N -strategy γ.

Let us assume that a diagonal witness a for N is available. In the case that a isnot defined (e.g. it was discarded by initialization), simply choose a fresh numbera:

Check if a 6∈ W .

• If yes, then let the outcome be 1.• If no, i.e., a ∈ W , then extract a out of A, and let the outcome be 0.


Case 2) δ ¹n is an R-strategy α.

Let 0 be the outcome.

Case 3 δ ¹n is an SW -strategy β of a strategy RX,Y,Θ0,Θ1.

The action of β depends on the outcome o− which is the one when we previouslyvisited β at stage s−.

In the following we say that an axiom 〈k,E ′〉 ∈ Θ0 is applicable if E ′ does notinterfere with numbers restrained out of A by higher priority strategies, includingpossible agitators of higher S-strategies, or by higher local priority N -strategies.We also say that an axiom 〈l, F ′〉 ∈ Θ1 is applicable if F ′ does not interferewith numbers restrained out of A by higher priority strategies, including possibleagitators of higher priority S-strategies, or by higher local priority N -strategies,or by the agitator Ek (where l has been chosen in correspondence of k as in (3b2)below).

(1) The previous outcome o− is 〈l, k〉 (l, k ∈ ω). It happens if and only if (1c)or (4b) happened at stage s−.

In this case, we have had a realized witness d, and we have extractedfinite sets Ek and F k

l out of A. (Note that this k and l are in U and Vk

respectively already.) Thus at stage s−, d ∈ W \ D.(a) If there is an applicable axiom 〈k,E ′〉 ∈ Θ0, then let the outcome

be ∞X . Discard the diagonal witness d. Initialize all N -strategieslocated at nodes γ ⊇ βˆ∞X which have lower local priorities thanβˆ(Tk). This has the effect of restraining A on E ′. We say that werestrain k in X. (In this case, we made progress on showing thatX = U .)

(b) If case (1a) does not happen and there is an applicable axiom 〈l, F ′〉 ∈Θ1, then let the outcome be 〈∞Y , k〉. Discard the diagonal witness d.Initialize all strategies N located at nodes γ ⊇ βˆ〈∞Y , k〉 which havelower local priorities than βˆ〈l, k〉. This has the effect of restraining Aon F ′. We say that we restrain l in Y . (We made progress on showingthat Y = Vk.) Notice that if 〈∞Y , k〉 is the true outcome, then theelements of Ek are witnesses of the Σ2-ness of A.

(c) If otherwise, then extract the agitators Ek and F kl from A and let the

outcome be 〈l, k〉. (The witness d is still valid. We win the substrategySW , as we still have

ΘA0 (k) = ΘA

1 (l) = 0.)

(2) The previous outcome o− is ∞X . It happens if and only if (1a) or (2a) or(2b1) or the exit clause in (3) happened at stage s−.

Check if there is k ∈ U but k not restrained in X.(a) If no, pick the least k ∈ X \ U (if any), enumerate k into U , and let

∞X be the outcome. If X ⊆ U , then let ∞X be the outcome.(b) If yes, then let k be the least such k:


(b1) If there an applicable axiom 〈k,E ′〉 ∈ Θ0, then act as in (1a).We say that we restrain k in X.

(b2) Otherwise let Ek be a corresponding agitator. The agitator Ek isappointed as follows: Let E be the set of numbers currently keptout by the higher priority strategies (including possible agitatorsextracted by S-strategies), and let Ek be the least set consistingof realized witnesses of N -strategies extending βˆ∞X and havinghigher local priority than βˆ∞X , such that E ∪ Ek ∈↑ εΘ0(k)(of course with respect to the current approximation to Θ0), butk ∈ ΘA

0 if Ek ⊆ A. Such a set Ek exists since k ∈ U and byinitialization has been enumerated at a stage when E is out ofA. Extract Ek from A. Let the outcome be 〈∞Y , k〉.

(3) The previous outcome o− is 〈∞Y , k〉. It happens if and only if (1b) or (2b2)or (3a) happened at stage s−.

Then first check if there is an applicable axiom 〈k,E ′〉 ∈ Θ0. If so thenact as in (1a) (we refer to this as the exit clause of (3)); if no, we processas follows.

Check if there is y ∈ Vk but y not restrained in Y .(a) If no, then pick the least l ∈ Y \ Vk (if any), enumerate l into Vk,

and let 〈∞Y , k〉 be the outcome. If Y ⊆ Vk then let 〈∞Y , k〉 be theoutcome.

(b) If yes then pick the least such l:(b1) If there is a applicable axiom 〈l, F ′〉 ∈ Θ1, then act as in (1b).

We say that we restrain y in Y .(b2) Otherwise let F k

l be a corresponding agitator. The agitatorF k

l is appointed as follows: Let E be the set of numbers cur-rently kept out by the higher priority requirements, and let F k

l

be the least set consisting of consisting of realized witnesses ofN -strategies extending βˆ〈∞Y , k〉 and having higher local pri-ority than βˆ〈∞Y , k〉, such that E∪Ek∪F k

l ∈↑ εΘ1(l) (of coursewith respect to the current approximation to Θ1), but l ∈ ΘA

1 ifEk ∪F k

l ⊆ A. Such a set F kl exists since l ∈ Vk and by initializa-

tion has been enumerated at a stage when both E and Ek areout of A.In this case pick a new witness d. Enumerate 〈d, k〉 into Φ0

and 〈d, l〉 into Φ1. Enumerate Ek and F kl back into A. Let

d0 be the outcome. (Notice that no N -strategy below βˆd0 haswitnesses which interfere with keeping d ∈ ΦX

0 and d ∈ ΦY1 , since

these witnesses by initialization do not belong to either Ek orF k

l .)(4) The previous outcome o− is d0. It happens if and only if (3b2) or (4a)

happened at stage s−.Then check if the diagonal witness d is realized. Notice that we have had

k, l and axioms 〈d, k〉 ∈ Φ0 and 〈d, l〉 ∈ Φ1 selected for this d already.(a) If no, then let the outcome be d0. (D 6= W is achieved trivially.)


(b) If yes, extract the agitators Ek and F kl and let the outcome be 〈l, k〉.

Clause (?) At the end of stage s+1, let As+1 be the whole set of natural numbersω, minus the numbers which have been extracted from A by the accessible stringsat stage s + 1. This in particular has the following effect: when acting for anS-strategy β,

• in (2b2), for each S-node β′ (not necessarily working for the same R) suchthat β′ ⊇ βˆ∞X , we put the agitators of β′ back into A;

• and, in (3b2), for each S-node β′ (not necessarily working for the same R)such that β′ ⊇ βˆ〈∞Y , k〉, we put the agitators of β′ back into A.

(As explained in the section on coordination of strategies this is to prevent anyinequality of Φi’s at β′.)

3.5. Verification. If δ ∈ T , we say that a stage s is δ-true if δ is accessible at s.

Lemma 7 (True path). There is an infinite path f ∈ [T ] such that for every n,

(1) f ¹n eventually is not initialized;(2) there exist infinitely many f ¹n-true stages.

Proof. The proof is by induction on n. For each n we describe f ¹n. For n = 0 wehave that f ¹0 = ∅, thus (1) and (2) are trivially true.

Assume now that the claim is true of n, and let s0 be a stage such that f ¹n isnever initialized at stages s ≥ s0.

First of all the claim is trivial if f ¹n is an R-strategy: In this case, f ¹(n+1) =f ¹nˆ0, since 0 is the only possible outcome. Finally, f ¹(n+1) is never initializedby strategies to its right.

Next, let us consider the case when f ¹n is an S-strategy. Suppose that s ≥ s0

is (f ¹n)-true and let s′ > s be the next (f ¹n)-true stage. Then by construction

• if the outcome of f ¹n at s is d0, and is different from the outcome at s′,then the outcome at s′ is (Tk), for some k.

• if the outcome of f ¹n at s is 〈l, k〉, for some l, k, and the outcome at s′ isnot in (Tk), then the outcome at s′ is ∞X .

This shows that either at almost all (f ¹n)-true stages s ≥ s0 the outcome of f ¹nis d0 (hence f ¹(n + 1) = (f ¹n)ˆd0); or at almost all (f ¹n)-true stages s ≥ s0 theoutcome is inside the group (Tk); or at infinitely many (f ¹n)-true stages we haveoutcome ∞X : In this last case f ¹(n + 1) = (f ¹n)ˆ∞X .

It remains to see what happens when at almost all (f ¹ n)-true stages s ≥ s0,say all (f ¹ n)-true stages s ≥ s1, we have outcome inside one single group (Tk).Suppose again that s ≥ s1 is (f ¹ n)-true and let s′ > s be the next (f ¹ n)-truestage. Then

• if the outcome of f ¹n at s is 〈l, k〉 and is different from the outcome at s′,then the outcome at s′ is 〈∞Y , k〉.


This shows that either there exists l such that at almost all (f ¹ n)-true stagess ≥ s1 the outcome of f ¹ n is 〈l, k〉 (hence f ¹ (n + 1) = f ¹ 〈l, k〉), or infinitelymany times we have outcome 〈∞Y , k〉, in which case f ¹(n + 1) = (f ¹n)ˆ〈∞Y , k〉.

Finally assume that f ¹ n is an N -strategy, say N = NW . Then either f ¹

(n + 1) = (f ¹ n)ˆ1, or f ¹ (n + 1) = (f ¹ n)ˆ0. In either case f ¹ (n + 1) triviallysatisfies (1).

So let us now show that f ¹ (n + 1) can be initialized only finitely often. Afters0 it can no longer be initialized by strings to its left. On the other hand it can beinitialized by S-strategies β ⊂ f ¹(n+1) of higher local priority, more precisely bynodes βˆ(Tk) (with βˆ∞X ⊂ f ¹(n+1)) or βˆ〈l, k〉 (with βˆ〈∞Y , k〉 ⊂ f ¹(n+1)).At stage s we initialize lower priority requirements after acting at an S-strategy βfor the following possible reasons:

(a) Acting in cases (1a), (2b1), or the exit clause of (3) when we give outcome∞X . We initialize all N -requirements γ ⊃ (f ¹n)ˆ∞X with local prioritysmaller than βˆ(Tk).

(b) Acting in cases (1b), (3b1) when we give outcome 〈∞Y , k〉. We initializeall N -requirements γ ⊃ (f ¹ n)ˆ〈∞Y , k〉 with local priority smaller thanβˆ〈l, k〉.

Let us first show that if β is an S-node and βˆ∞X ⊆ f ¹ (n + 1) then for everyk the nodes βˆ(Tk) can initialize lower priority N -requirements located belowβˆ∞X only finitely many times. We show in fact that for every k there are onlyfinitely many stages at which βˆ(Tk) is accessible. For this, notice that thereonly finitely many N -requirements below βˆ∞X having local priority higher thanβˆ(Tk): Let N be the set of these N -requirements. Some of these requirementsN ∈ N may get a permanent witness aN : Let N1 be the subset of N consisting ofsuch requirements, and let N2 = N \ N1. Suppose that there are infinitely many(f ¹ n)-true stages t0 < t1 < t2 < . . ., with s0 < t0, such that at ti the outcomeof β is ∞X and at t−i the outcome was (Tk) (where t−i is the last stage precedingti at which β was visited). Then at ti we have k ∈ ΘA

0 via an applicable axiom〈k,Ei〉 ∈ Θ0. Let t ≥ t0 be a stage such that for every N ∈ N1 the final witness aN

has been already appointed, and is already realized if it ever gets to be realized.Notice that for every N ∈ N2 we have that lims aN = ∞. Let ti > t: Then Ei

does not contain any of the realized witnesses aN for N ∈ N1. It may containsome of the witnesses aN ′ for N ′ ∈ N2 which subsequently may be extracted asthey become realized. But for each such N ′ ∈ N2 we have that lims aN ′ = ∞, andsince any N -witness which is discarded is enumerated back into A there is a stages′ > ti such that again Ei ⊆ A and thus (such some) Ei can be restrained in A,thus k is restrained in X, and at no later stage v > s′ do we have that βˆ〈∞Y , k〉is accessible.

In a similar way one shows that if βˆ〈∞y, k〉 ⊆ f ¹ (n + 1) then for every l,the node βˆ〈l, k〉 (for l ∈ ω) can initialize lower priority N -requirements locatedbelow βˆ〈∞Y , k〉 only finitely many times. As before, we can show in fact that if


βˆ〈∞Y , k〉 ⊆ f ¹(n + 1) then there are only finitely many stages at which βˆ〈l, k〉is accessible.

It follows that f ¹(n + 1) can be initialized in this way only finitely often, sincethere exist only finitely many S-nodes β such that β ⊂ f ¹(n + 1). ¤

Lemma 8. For every pair of e-operators Θ0 and Θ1 the requirement RΘ0,Θ1is

satisfied.

Proof. Let δ be the node on the true path f which works on the R-strategyRΘ0,Θ1

= RX,Y,Θ0,Θ1. We distinguish the following cases based on whether RΘ0,Θ1

has a global Σ3- or Π3-outcome:

• Case 1. Assume that RX,Y,Θ0,Θ1has a global Π3-outcome, namely, for every

S-substrategy β ⊂ f of δ either βˆd0 ⊂ f or there exist l, k ∈ ω such thatβˆ〈l, k〉 ⊂ f .Claim. ΦX

0 = ΦY1 , where Φ0 and Φ1 are the e-operators built by the

strategy δ.Proof of Claim. First of all we notice that we only need to consider thosed’s which have been appointed as witnesses by some S-substrategy β (notnecessarily on the true path) of δ.

Let β be an SW -substrategy of δ and let d be the witness which has beenappointed as a witness by β. We consider all possible locations of β.

If f <L β, then eventually d is discarded by initialization and thus〈d, ∅〉 ∈ Φ0 ∩ Φ1 making trivially ΦX

0 (d) = ΦY1 (d) = 1.

Next, consider the case in which β ⊂ f . We may assume that d is neverdiscarded.

– Suppose that βˆd0 ⊂ f . Let s0 be the least stage after which βˆd0

never gets initialized. Let s1 ≥ s0 be the first stage at which βˆd0

is accessible. By construction, this can only happen in (3b2). Thusat stage s1, we enumerate agitators Ek and F k

l back into A, and enu-merate axioms 〈d, k〉 ∈ Φ0 and 〈d, l〉 ∈ Φ1. By choice of Ek andF k

l , they are disjoint from each other and from the set E which is theset of numbers currently kept out by the higher priority strategies.Hence only β can extract them out of A. By construction again, theextraction by β would result in βˆd0 being initialized, contradictingour choice of s1. Therefore, D(d) = ΦX

0 (d) = ΦY1 (d) = 1[s] for all

stages s > s1. Moreover, we clearly have that d 6∈ W by (4a), thusD(d) = 1 6= 0 = W (d).

– Suppose that βˆ〈l, k〉 ⊂ f . Let s0 be the least stage after which βˆ〈l, k〉never gets initialized. Let s1 ≥ s0 be the first stage at which βˆ〈l, k〉is accessible. By construction, this can only happen in (4b). Thus atstage s1, we extract agitators Ek and F k

l out of A. By choice of Ek

and F kl , we have that k ∈ ΘA

0 and l ∈ ΘA1 if and only if Ek ⊆ A and

F kl ⊆ A respectively (modulo the extractions performed by higher

priority strategies). And this is so at any stages s > s1, otherwiseβˆ〈l, k〉 would be initialized. Therefore, D(d) = ΦX

0 (d) = ΦY1 (d) =


0[s] for all stages s > s1 at which β is accessible, which implies thatD(d) = ΦX

0 (d) = ΦY1 (d) = 0. Moreover, we clearly have that d ∈ W

by condition (4b), thus D(d) = 0 6= 0 = W (d). Notice that we do notrule out the possibility that d reenters D at some stages at which β isnot accessible.

It is left to consider the case when β <L f . Let s0 be the least stage afterwhich β is not accessible. Let s1 ≤ s0 be the last stage at which βˆ〈l, k〉was accessible. Since we had the outcome 〈l, k〉, d must have been realizedat stage s1. It follows from the discussion in the section on coordinationof strategies that the only way a disagreement between the Φi’s at d canarise is that the agitators Ek and F k

l for d are contained in the realizedwitnesses of N -strategies below either ηˆ∞X or ηˆ〈∞Y , k〉 for some η ⊂ βsimultaneously. But by Clause (?) the agitators Ek and F k

l are back to Aby stage s1. Hence ΦX

0 (d) = ΦY1 (d) for all s > s1. This finishes our proof

of Claim.Since β ⊂ f eventually does get a witness d which is never discarded by

Lemma 7, the above argument shows that ΦX0 (d) = ΦY

1 (d) 6= W (d).• Case 2. Assume now that there is a substrategy β such that βˆ∞X ⊂ f ,

which is one of the two global Σ3-outcomes of RΘ0,Θ1. We claim in this

case that either X is finite or X = U , where U is the c.e. set built at β.Suppose X not finite. Suppose that k is the least number such that

X(k) 6= U(k) and assume k ∈ X. Let s0 be a stage after which β isnever initialized, and all k′ < k which will eventually be enumerated intoU have been already enumerated. Since X is infinite there are infinitelymany stages at which we enumerate numbers in U . At the least such stages > s0 we enumerate k in U , contradiction.

Assume now that k ∈ U , but k /∈ X. By initialization there is a stringη, with βˆ∞X ⊆ η ⊂ f such that all non-discarded witnesses of strategieshaving higher local priority than βˆ(Tk) are located at nodes η′ ≤ η.

Let s1 > s0 be a stage such that for all k′ < k either k′ /∈ U or for everys ≥ s1 we have that k′ has been already restrained in X by stage s1, andη is not initialized at s. Let s > s1 be any β-true stage such that s− was(βˆ∞X)-true. Since k 6∈ X, it means that k cannot be restrained at s by(2b1), then we can find a permanent agitator for k, giving that βˆ(Tk) ⊂ f ,contradiction.

• Case 3. The last case to consider is when the above does not happen andthere is a substrategy β such that for some k, βˆ〈∞Y , k〉 ⊂ f . This is thesecond of the two global Σ3-outcomes of RΘ0,Θ1

. We claim in this case thateither Y is finite or Y = Vk where Vk is the c.e. set built at βˆ〈∞Y , k〉.

Assume that Y is infinite, and suppose for a contradiction that Y 6= Vk.Let l be the least number such that Y (l) 6= Vk(l).

Assume first that l ∈ Y . Then there is a β-true stage (following the lastinitialization of Vk and such that all l′ < l which will ever be enumeratedinto Vk have been already enumerated by this stage) at which Vk ⊂ Y ,l ∈ Y \ Vk and we enumerate l ∈ Vk.


Finally assume that l ∈ Vk \ Y . Arguing as for the previous case, lets1 ≥ s0 be a stage such that for all l′ < l either l′ /∈ Vk or for every s ≥ s1 wehave that l′ ∈ Y at s, and all N -strategies γ, such that βˆ〈∞Y , k〉 ⊆ γ < fand γ has higher local priority than βˆ〈l, k〉, are not initialized at s. Lets > s1 be any β-true stage such that s− was (βˆ〈∞Y , k〉)-true. Since l /∈ Y ,we can not restrain l in Y by (3b1), therefore we can find a permanentagitator F k

l for l, giving that βˆ〈l, k〉 ⊂ f , contradiction.

¤

Lemma 9. For every W , the requirement NW is satisfied.

Proof. Let δ ⊂ f be the NW -strategy. By Lemma 7, δ eventually is not initialized.If a is the permanent witness of N , then either a /∈ W and a ∈ A, or a eventuallygets realized. In the latter case a gets extracted infinitely often at almost all δ-truestages, giving that a /∈ A. ¤

Remark 10. If a is any number then a is extracted from A for two possible reasons:either a is a realized witness of some N -strategy, or a belongs to some agitator forsome S-strategy. In the limit a is out of A if and only if a is extracted by somestrategy σ for which there exist infinitely many stages at which σ is accessible andextracts a. For S-strategies this is clear from the construction. For N -strategiesthis is a consequence of Clause (?) of the construction. In either case, if s is a stageat which no σ responsible for the extraction of a is accessible, then a is enumeratedback into A. Of course if σ is on the true path, then in the limit a is out of A ina Π0

2 fashion.

3.6. Other Consequences. The techniques used in the proof of Theorem 6 canbe modified to produce other results. The first one is an immediate consequence.

Corollary 11. There is a nonzero s-degree that does not bound any minimal pairof s-degrees.

Proof. Immediate by the fact that Φ0,α and Φ1,α are s-operators, for every α. Theabove proof shows in fact that for every pair of non c.e. sets X,Y such thatX,Y ≤e A, one can find a non c.e. set D such that D ≤s X,Y . ¤

The nonbounding strategy can also be adapted to produce a set 1-generic e-degree (i.e. an e-degree containing a 1-generic set), hence confirming an old claimof Copestake, quoted in [Coo90, p.73] . Due to the similarity of the proofs, weomit most of the details and only provide the sketch of the crucial ideas.

Theorem 12. There is a set 1-generic nonbounding e-degree.

Recall that the requirements to make A a 1-generic set are, for every c.e. set W :

GW : (∃κ ⊂ cA)[κ ∈ W ∨ (∀λ ⊇ κ)(λ /∈ W )],

where cA denotes the characteristic function of A, κ, λ ∈ 2<ω, and we view W ⊆2<ω as a c.e. subset of finite binary strings.

We will use GW to replace NW in our requirement list.


Let us consider a G-strategy σ on the priority tree T . First of all σ will choose aboundary b which is larger than all numbers controlled by higher priority strategies.σ also guesses that A ¹ b will be eventually equal to a string κσ of length b. Atany stage s, σ looks for a finite extension λ of κσ such that λ ∈ W . If such a λis found, then σ will change a finite portion of A so that λ ⊂ A, hence meet therequirement GW by making λ ∈ A ∩ W . If no such string is ever found, then GW

is satisfied trivially.

To be more precise, let us introduce the following parameters in the environmentof σ:

• E(σ): the set of numbers extracted by strategies τ ⊂ σ.• F (σ): the set of numbers restrained by strategies τ <L σ.• b: a number which is larger than the maximum of E(σ) ∪ F (σ).• κσ: the least string of length b which is compatible with E(σ) and F (σ),

i.e.

x ∈ E(σ) ⇒ x < b and κσ(x) = 0

x ∈ F (σ) ⇒ x < b and κσ(x) = 1.

Observe that by construction if σ is on the true path, then every number inE(σ) (consisting of elements taken out by other G-strategies located along σ, andby those agitators Ek corresponding to some S-node β such that βˆ〈∞Y , k〉 ⊆ σ)will be extracted out of A in a Π0

2 fashion, as analyzed in Remark 10; and everynumber in F (σ) will be restrained in A. Hence κσ will be a true initial segmentof cA. Note the asymmetry, again due the Clause (?), in the definition of E(σ)(i.e. the set of numbers extracted by strategies τ ⊂ σ) and the definition of F (σ)(i.e. the set of numbers restrained by strategies τ <L σ).

The G-strategies. Given a G-strategy σ, the action to satisfy the correspondingrequirement GW , say, is the following:

• wait for some λ ⊇ κσ such that λ ∈ W ;• extract λ−1(0) from A, and enumerate λ−1(1) into A and restrain the ele-

ments of λ−1(1) in A with priority σ; the purpose is to make As+1 look likethe string λ;

• initialize lower priority strategies. When we initialize a G-strategy, wechoose b larger than any number we have seen so far; and release all numbersit restrains.

There two possible outcomes 0, 1 for this strategy:

• outcome 1: wait forever; we win since κσ ⊂ cA, but for no λ ⊇ κσ do wehave λ ∈ W ;

• outcome 0: we win by making λ ⊂ cA, for some λ ∈ W .

Now we compare the actions of G-strategies with those of N -strategies. Noticethat an N -strategy first enumerates a witness, then may extract the witness. AG-strategy, upon acting, enumerates only finitely many numbers, namely the setH+ = x : λ(x) = 1, into A; and extracts finitely many numbers, namely the set


H− = x : λ(x) = 0, out of A. We may treat H+ and H− as blocks, so that theymust enter into or be extracted out of A at the same time.

Then the conflicts between two G-strategies are not serious: whenever the higherpriority one acts, the boundary of the lower priority one is pushed further.

The conflicts between the extracting activity of the G-requirements and the S-requirements are solved in the same way as we did before for N -strategies in theproof of Theorem 6.

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Department of Pure Mathematics, School of Mathematics, University of Leeds,

Leeds, LS2 9JT, U. K.

Institute of Software, Academia Sinica, Beijing, 100080 China.

Dipartimento di Scienze Matematiche ed Informatiche “Roberto Magari”, Via

del Capitano 15, 53100 Siena, Italy.

Department of Mathematics, Faculty of Science, National University of Sin-

gapore, Lower Kent Ridge Road, Singapore 119260.

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