Pseudojump inversion in special r. b. Π^0_1 classes
aa r X i v : . [ m a t h . L O ] F e b Pseudojump inversion in special r. b. Π classes Abstract
The Jump Inversion Theorem says that for every real A ≥ T ′ there is areal B such that A ≡ T B ′ ≡ T B ⊕ ′ . A known refinement of this theoremsays that we can choose B to be a member of any special Π subclassof { , } N . We now consider the possibility of analogous refinements oftwo other well-known theorems: the Join Theorem – for all reals A and Z such that A ≥ T Z ⊕ ′ and Z > T
0, there is a real B such that A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z – and the Pseudojump Inversion Theorem– for all reals A ≥ T ′ and every e ∈ N , there is a real B such that A ≡ T B ⊕ W Be ≡ T B ⊕ ′ . We show that in these theorems, B can befound in some special Π subclasses of { , } N but not in others. ontents Π classes of positive measure 126 Π classes constructed by priority arguments 13 We begin with a well-known theorem.
Theorem 1.1 (Jump Inversion Theorem, due to Friedberg, see [11, Theorem13.3.IX]) . Given a real A ≥ T ′ we can find a real B such that A ≡ T B ′ ≡ T B ⊕ ′ .Here 0 ′ denotes the Halting Problem, B ′ denotes the Turing jump of B ,and ≡ T denotes Turing equivalence. There are two important related theorems,which read as follows. Theorem 1.2 (Join Theorem, due to Posner and Robinson [10, Theorem 1],see also [3, Theorem 2.1]) . Given reals A and Z such that A ≥ T Z ⊕ ′ and Z > T
0, we can find a real B such that A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z . Theorem 1.3 (Pseudojump Inversion Theorem, due to Jockusch and Shore [3,Theorem 2.1]) . Given a real A ≥ T ′ and an integer e ∈ N , we can find a real B such that A ≡ T J e ( B ) ≡ T B ⊕ ′ .Here J e denotes the e th pseudojump operator, defined by J e ( B ) = B ⊕ W Be where h W Be | e ∈ N i is a fixed standard recursive enumeration of the B -recursively enumerable subsets of N .Our results in this paper concern special Π classes. Following [4] we definea special Π class to be a nonempty Π subset of N N with no recursive elements.Here N N is the Baire space, but as in [4, 5] we focus on special Π subclassesof { , } N , the Cantor space. On the other hand, because of [13, Theorem4.10] our results concerning special Π subclasses of { , } N will also apply to recursively bounded special Π subclasses of N N . Following [5] we use “r. b.” asan abbreviation for “recursively bounded.”We now recall another known theorem, which says that we can find the B for Theorem 1.1 in any special Π subclass of the Cantor space.2 heorem 1.4 (due to Jockusch and Soare [5, just after the proof of Theorem2.1]) . Let P be a special Π subclass of { , } N . Given a real A ≥ T ′ , we canfind a real B ∈ P such that A ≡ T B ′ ≡ T B ⊕ ′ .This refinement of Theorem 1.1 suggests a question as to the existence ornonexistence of analogous refinements of Theorems 1.2 and 1.3. In order tostate our results concerning this question, we make the following definition. Definition 1.5.
Let P ⊆ N N be a subclass of the Baire space.1. P has the Join Property if for all reals A and Z such that A ≥ T Z ⊕ ′ and Z > T
0, there exists B ∈ P such that A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z .2. P has the Pseudojump Inversion Property if for all reals A ≥ T ′ and all e ∈ N , there exists B ∈ P such that A ≡ T J e ( B ) ≡ T B ⊕ ′ .The question that we are asking is, which special Π subclasses of the Cantorspace { , } N have one or both of the properties in Definition 1.5? The purposeof this paper is to present some partial results in this direction, as follows. LetCPA be the special Π subclass of { , } N consisting of all complete, consistentextensions of Peano Arithmetic. We show that CPA has both the Join Prop-erty and the Pseudojump Inversion Property. More generally, we show that anyspecial Π class which is Turing degree isomorphic to CPA has both of theseproperties. For example, this holds for the Π class CZF consisting of all com-plete, consistent extensions of Zermelo-Fraenkel Set Theory (assuming that CZFis nonempty), and for the Π class DNR consisting of all { , } -valued diago-nally nonrecursive functions. On the other hand, let P be a special Π subclassof { , } N which is of positive measure. Citing a theorem of Nies and an obser-vation of Patrick Lutz, we note that P need not have the Join Property, but P and indeed any special Π class which is Turing degree isomorphic to P hasthe Pseudojump Inversion Property. Finally, we construct a special Π subclassof { , } N which has neither the Join Property nor the Pseudojump InversionProperty. We do not know whether there exists a special Π subclass of { , } N which has the Join Property but not the Pseudojump Inversion Property.Here is an outline of the paper. In § § § classes, and we apply these results to show that any Π class which is Turingdegree isomorphic to CPA has the Join Property and the Pseudojump InversionProperty. In § classes consisting of Martin-L¨of randomreals have the Pseudojump Inversion Property but not the Join Property. In § subclass Q ⊆ { , } N which has neither the Join Property nor the Pseudojump Inversion Property. Remark 1.6.
The lattice E w of all Muchnik degrees of nonempty Π subclassesof { , } N is of great interest; see for instance the references in [19]. Therefore, itis natural to ask about Muchnik degrees in E w corresponding to the properties3n Definition 1.5. Our results in this paper shed some light on this question.First, it is known [13, Theorem 6.8] that any Π subclass of { , } N which isMuchnik equivalent to CPA – i.e., of Muchnik degree , where denotes thetop degree in E w – is Turing degree isomorphic to CPA. Therefore, our resultsin § classes of Muchnik degree have both of theproperties in Definition 1.5. Second, it is easy to see that each degree in E w contains a Π subclass of { , } N which includes CPA and hence again has bothproperties. On the other hand, by § r ∈ E w containing a special Π subclass of { , } N which has one property but not theother. Moreover, in § E w containing aspecial Π subclass of { , } N which does not have either property.In the remainder of this section, we fix some notation and terminology.We write f : ⊆ A → B to mean that f is a partial function from A to B ,i.e., a function with domain ⊆ A and range ⊆ B . For a ∈ A we write f ( a ) ↓ andsay that f ( a ) converges or f ( a ) is defined , if a ∈ the domain of f . Otherwisewe write f ( a ) ↑ and say that f ( a ) diverges or f ( a ) is undefined . For a ∈ A and f, g : ⊆ A → B , we write f ( a ) ≃ g ( a ) to mean that either ( f ( a ) ↓ and g ( a ) ↓ and f ( a ) = g ( a )) or ( f ( a ) ↑ and g ( a ) ↑ ). We write f ( a ) ↓ = b to mean that f ( a ) ↓ and f ( a ) = b .The set of all natural numbers is N = { , , , . . . } . For any set A let A N denote the set of all sequences X : N → A . Thus N N and { , } N are the Bairespace and the
Cantor space respectively. The points X ∈ N N or X ∈ { , } N are sometimes called reals . We sometimes identify { , } N with the powersetof N in the usual manner. For X, Y ∈ A N the join X ⊕ Y ∈ A N is given by( X ⊕ Y )(2 n ) = X ( n ), ( X ⊕ Y )(2 n + 1) = Y ( n ), and for P, Q ⊆ A N we write P × Q = { X ⊕ Y | X ∈ P and Y ∈ Q } . Note in particular that P × Q ⊆ N N whenever P, Q ⊆ N N , and P × Q ⊆ { , } N whenever P, Q ⊆ { , } N .For any set A let A ∗ be the set of strings , i.e., finite sequences, of elementsof A . For a , . . . , a m − ∈ A we let h a , . . . , a m − i denote the string σ ∈ A ∗ oflength m given by σ ( i ) = a i for all i < m . In this case we write | σ | = m , andfor each k ≤ m we write σ ↾ k = h a , . . . , a k − i . We also write A m = { σ ∈ A ∗ || σ | = m } and A ≥ m = { σ ∈ A ∗ | | σ | ≥ m } and A Proof of Theorem 1.2. Let A and Z be reals such that A ≥ T Z ⊕ ′ and Z > T B such that A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z . We shallconstruct a sequence of strings σ ⊂ σ ⊂ · · · ⊂ σ j ⊂ σ j +1 ⊂ · · · in N ∗ , and thedesired real B ∈ N N will be obtained as the limit S j σ j of this sequence.Let us regard Z as a subset of N . Since Z is not recursive, at least one of Z and its complement N \ Z is not Σ . And then, since Z is Turing equivalent to N \ Z , we may safely assume that N \ Z is not Σ .Stage j = 0. Let σ = hi .Stage j = 2 n + 1. By induction we have σ n . Define S n = { i ∈ N | ∃ σ ( σ ⊇ σ n a h , . . . , | {z } i , i and ϕ (1) ,σn, | σ | (0) ↓ ) } which is Σ . Since N \ Z is not Σ , there are infinitely many i such that i ∈ Z ⇔ i ∈ S n . Choose the first such i , and use this i to define σ n +1 asfollows. If i ∈ Z , then i ∈ S n so let σ n +1 = the least σ ⊇ σ n a h , . . . , | {z } i , i suchthat ϕ (1) ,σn, | σ | (0) ↓ . If i / ∈ Z , let σ n +1 = σ n a h , . . . , | {z } i , i .Stage j = 2 n + 2. Let σ n +2 = σ n a h A ( n ) i .This completes the definition of σ j for all j ∈ N . We now show that B = S j σ j has the desired properties. We begin with the following observations. See for instance [11, §§ §§ If σ n is known, then the i in stage 2 n + 1 can be found recursively in Z ⊕ ′ (because S n is computable from 0 ′ ), or recursively in B (becausethere is exactly one i such that σ n a h , . . . , | {z } i , i ⊂ B ). And then, usingthis i , σ n +1 can be found recursively in Z (to test whether i ∈ Z ), orrecursively in 0 ′ (to test whether i ∈ S n ). • If σ n +1 is known, then σ n +2 can be found recursively in A (by the defini-tion of σ n +2 ), or recursively in B (because σ n +2 = σ n +1 a h B ( | σ n +1 |i ),hence also recursively in B ⊕ Z or in B ⊕ ′ .Combining these observations, we see that the entire sequence h σ j | j ∈ N i is ≤ T B ⊕ Z , and ≤ T B ⊕ ′ , and ≤ T A (using the hypothesis Z ⊕ ′ ≤ T A ). Wealso have: • B ⊕ Z ≤ T A , because B = S j σ j ≤ T A and by hypothesis Z ≤ T A . • B ′ ≤ T h σ j | j ∈ N i , because n ∈ B ′ ⇔ ϕ (1) ,σ n +1 n, | σ n +1 | (0) ↓ . • A ≤ T h σ j | j ∈ N i , because A ( n ) = σ n +2 ( | σ n +1 | ) for all n .Thus A ≤ T h σ j | j ∈ N i ≤ T B ⊕ Z ≤ T A , and B ′ ≤ T h σ j | j ∈ N i ≤ T B ⊕ ′ ≤ T B ′ , hence A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z , Q.E.D.We now turn to Pseudojump Inversion, Theorem 1.3. Proof of Theorem 1.3. Fix e ∈ N , and let A be a real such that A ≥ T ′ . It willsuffice to find a real B such that A ≡ T J e ( B ) ≡ T B ⊕ ′ . We shall construct asequence of strings σ ⊆ σ ⊆ · · · ⊆ σ j ⊆ σ j +1 ⊆ · · · in N ∗ , and the desired real B ∈ N N will be obtained as the limit S j σ j of this sequence.Stage 0. Let σ = hi .Stage 2 n + 1. Let σ n +1 = the least string σ ⊇ σ n for which n ∈ W σe, | σ | ifsuch a σ exists. Otherwise let σ n +1 = σ n .Stage 2 n + 2. Let σ n +2 = σ n +1 a h A ( n ) i .We now show that B = S j σ j has the desired properties. We begin with thefollowing observations. • If σ n is known, then in stage 2 n + 1 the question of whether σ exists canbe answered recursively in 0 ′ (because the question is Σ ), or recursivelyin J e ( B ) (because σ exists if and only if n ∈ W Be ), hence also recursively in A (because by hypothesis 0 ′ ≤ T A ), or recursively in B ⊕ ′ . And once theexistence or nonexistence of σ is known, σ n +1 can be found recursively. • If σ n +1 is known, then σ n +2 can be found recursively in A (by the defini-tion of σ n +2 ), or recursively in B (because σ n +2 = σ n +1 a h B ( | σ n +1 |i ),hence also recursively in J e ( B ) (because B ≤ T J e ( B )), or in B ⊕ ′ .Combining these observations, we see that the entire sequence h σ j | j ∈ N i is ≤ T J e ( B ), and ≤ T B ⊕ ′ , and ≤ T A . We also have:6 B ⊕ ′ ≤ T A , because B = S j σ j ≤ T A and by hypothesis 0 ′ ≤ T A . • J e ( B ) ≤ T h σ j | j ∈ N i , because n ∈ W Be if and only if n ∈ W σ n +1 e, | σ n +1 | . • A ≤ T h σ j | j ∈ N i , because A ( n ) = σ n +2 ( | σ n +1 | ) for all n .Thus A ≤ T h σ j | j ∈ N i ≤ T B ⊕ ′ ≤ T A and J e ( B ) ≤ T h σ j | j ∈ N i ≤ T J e ( B ),hence A ≡ T J e ( B ) ≡ T B ⊕ ′ , Q.E.D. Let PA denote Peano Arithmetic, and let Sent PA denote the set of sentences ofthe language of PA . An extension of PA is a set T ⊆ Sent PA which includesthe axioms of PA and is closed under logical consequence. For any such T , a completion of T is an extension X of PA which includes T and such that foreach ϕ ∈ Sent PA exactly one of the sentences ϕ and ¬ ϕ belongs to X . Givenan extension T of PA , let Comp T denote the set of all completions of T . ByLindenbaum’s Lemma, Comp T = ∅ if and only if T is consistent.Fix a primitive recursive G¨odel numbering ϕ ϕ ) : Sent PA → N .For convenience we shall assume that PA and N . This induces a one-to-one correspondence between subsets X ofSent PA and subsets of N , namely X 7→ { ϕ ) | ϕ ∈ X } . And of course subsets of N are identified with their characteristic functions in { , } N . Therefore, lettingCPA be the subset of { , } N corresponding to Comp PA , we see that CPA is aspecial Π subclass of { , } N .The purpose of this section is to prove that CPA has the Join Propertyand the Pseudojump Inversion Property. The Join Property will follow easilyfrom Theorem 1.2 plus other known results, but for the Pseudojump InversionProperty we shall use a construction involving the G¨odel-Rosser IncompletenessTheorem [8]. Definition 3.1 ( PA -degrees) . By a PA -degree we mean the Turing degreedeg T ( X ) of some X ∈ CPA. A set S of Turing degrees is said to be upwardlyclosed if for all Turing degrees a and b such that a ≤ b , a ∈ S implies b ∈ S . Lemma 3.2. The set of all PA -degrees is upwardly closed. Proof. This result is originally due to Robert M. Solovay. For a proof see [2,Theorem 2.21.3] or [13, Corollary 6.6]. Lemma 3.3. Let P ⊆ { , } N be a nonempty Π class such that { deg T ( X ) | X ∈ P } is upwardly closed. Then P has the Join Property. Proof. Let A and Z be reals such that A ≥ T Z ⊕ ′ and Z > T 0. By Theorem1.2 there exists a real C such that A ≡ T C ′ ≡ T C ⊕ ′ ≡ T C ⊕ Z . By the LowBasis Theorem [5, Theorem 2.1] relativized to C , there exists B ∈ P such that C ′ ≡ T ( B ⊕ C ) ′ ≡ T B ⊕ C ⊕ ′ . Since B ≤ T B ⊕ C , we can find B ∈ P such that B ≡ T B ⊕ C , hence C ′ ≡ T B ′ ≡ T B ⊕ ′ . We now have C ′ ≡ T ⊕ Z ≤ T B ⊕ Z ≤ T C ′ ⊕ Z ≡ T C ′ , hence A ≡ T C ′ ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z ,Q.E.D. Theorem 3.4. CPA has the Join Property. Proof. This is immediate from Lemmas 3.2 and 3.3.Next we shall prove that CPA has the Pseudojump Inversion Property. Theproof will be presented in terms of completions of recursively axiomatizabletheories. An extension T of PA is said to be recursively axiomatizable if thereis a recursive set of sentences S ⊆ Sent PA such that T is the closure of S underlogical consequence. In this case Comp T is clearly a Π subclass of CPA. Theconverse also holds: Lemma 3.5. We have an effective one-to-one correspondence T Comp T between recursively axiomatizable extensions of PA and Π subclasses of CPA. Proof. Given a Π class P ⊆ CPA, we shall exhibit a recursively axiomatizableextension T of PA such that Comp T = P . Namely, let T = { ϕ ∈ Sent PA | X ( ϕ )) = 1 for all X ∈ P } . To see that T is closed under logical consequence,it suffices to note that each X ∈ P belongs to CPA and is therefore closedunder logical consequence. We also have PA ⊆ T , because P ⊆ CPA. MoreoverComp T = P by Lindenbaum’s Lemma. From the effective compactness of P we see that T is recursively enumerable. Hence T is recursively axiomatizable,and from an index of P as a Π class we can compute an index for a recursiveaxiomatization of T . This completes the proof. Remark 3.6. The idea of Lemma 3.5 applies much more generally. Given anylanguage L and any L -theory T , an extension of T is any set e T ⊆ Sent L whichincludes T and is closed under logical consequence. Regarding Comp T as aclosed set in the product space { , } Sent L , we have a one-to-one correspondence e T Comp e T between extensions of T and closed subsets of Comp T .Recall from § h P i | i ∈ N i is a fixed standard recursive enumeration ofthe Π subclasses of { , } N . Lemma 3.7 (splitting property) . There is a primitive recursive function s : N × { , } → N such that for all i ∈ N , if P i is a nonempty Π subclass of CPAthen P s ( i, and P s ( i, are nonempty disjoint Π subclasses of P i . Proof. Lemma 3.5 tell us that, given i ∈ N , we can effectively find a recursivelyaxiomatizable extension T of PA such that Comp T corresponds to P i ∩ CPA.We then apply the G¨odel-Rosser Incompleteness Theorem [8] to effectively finda sentence ψ ∈ Sent PA such that if T is consistent, then neither ψ nor ¬ ψ belongs to T . Let T (respectively T ) be the closure of T ∪ { ψ } (respectively T ∪{¬ ψ } ) under logical consequence. Clearly Comp T and Comp T are disjoint,and they are nonempty if Comp T is nonempty. Apply Lemma 3.5 to effectivelyfind s ( i, , s ( i, ∈ N such that P s ( i, and P s ( i, correspond to Comp T andComp T respectively. This completes the proof.8e shall now use this splitting property to redo Theorem 1.3 within CPA. Theorem 3.8. CPA has the Pseudojump Inversion Property. Proof. Fix e ∈ N , and let A ∈ { , } N be a real such that A ≥ T ′ . It willsuffice to find a real B ∈ CPA such that A ≡ T J e ( B ) ≡ T B ⊕ ′ . To find B ,we shall construct a sequence of Π classes Q ⊇ Q ⊇ · · · ⊇ Q j ⊇ Q j +1 ⊇ · · · ,by induction on j starting with Q = CPA ⊆ { , } N . The intersection T j Q j will be nonempty, and we shall have B ∈ T j Q j . At the same time, we shallconstruct a function f : N → N such that Q j = P f ( j ) for all j .Fix a primitive recursive splitting function s as in Lemma 3.7.Stage 0. Fix f (0) ∈ N such that P f (0) = Q = CPA.Stage 2 n + 1. By induction we have Q n = P f (2 n ) . If P f (2 n ) ∩ { X | n / ∈ J e ( X ) } = ∅ let Q n +1 = P f (2 n ) and f (2 n + 1) = f (2 n ). Otherwise, let Q n +1 = P f (2 n ) ∩ { X | n / ∈ J e ( X ) } and choose f (2 n + 1) so that P f (2 n +1) = P f (2 n ) ∩ { X | n / ∈ J e ( X ) } . This f (2 n + 1) is found primitive recursively from f (2 n ) and n and e .Stage 2 n + 2. By induction we have Q n +1 = P f (2 n +1) . Let f (2 n + 2) = s ( f (2 n + 1) , A ( n )) and Q n +2 = P s ( f (2 n +1) ,A ( n )) , where s is our splitting func-tion. This f (2 n + 2) is found primitive recursively from f (2 n + 1) and A ( n ).This completes the construction.By construction each Q j is nonempty, so by compactness of { , } N theirintersection T j Q j is nonempty. It remains to show that B ∈ T j Q j has thedesired properties. We begin with the following observations. • If f (2 n ) is known then f (2 n +1) can be found recursively in 0 ′ (by checkingwhether the Π class P f (2 n ) ∩ { X | n / ∈ J e ( X ) } is empty or not), orrecursively in J e ( B ) (by checking whether n ∈ J e ( B ) or not). • If f (2 n + 1) is known then f (2 n + 2) can be found recursively in B (byfinding the k ∈ { , } such that B / ∈ P s ( f (2 n +1) ,k ) ), or recursively in A (byevaluating s ( f (2 n + 1) , A ( n ))).Combining these observations, we see that f is ≤ T A ⊕ ′ ≡ T A , and ≤ T J e ( B ),and ≤ T B ⊕ ′ . Conversely, we also have: • A ≤ T f , because A ( n ) = i if and only if f (2 n + 2) = s ( f (2 n + 1) , i ). • J e ( B ) ≤ T f , because n ∈ J e ( B ) if and only if f (2 n + 1) = f (2 n ). • B ⊕ ′ ≤ T f , because B ≤ T J e ( B ) ≤ T f and 0 ′ ≤ T A ≤ T f .Thus f ≡ T A ≡ T J e ( B ) ≡ T B ⊕ ′ and the proof is complete. Remark 3.9. We have used the Splitting Lemma 3.7 to prove that CPA hasthe Pseudojump Inversion Property. Similarly, it would be possible to use theSplittng Lemma to prove directly that CPA has the Join Property. This directproof would be in contrast to our shorter but indirect proof in Theorem 3.4above. Note however that a version of the Splitting Lemma is used in our proofof Solovay’s Lemma 3.2; see [13, § 6] and [14, § Turing degree isomorphism Definition 4.1 (Turing degree isomorphism) . Following [4] we say that P, Q ⊆ N N are Turing degree isomorphic if { deg T ( X ) | X ∈ P } = { deg T ( X ) | X ∈ Q } .The purpose of this section is to prove that the Join Property and the Pseu-dojump Inversion Property hold for all Π subclasses of { , } N which are Turingdegree isomorphic to CPA. We also obtain some more general-looking results.We begin with the Join Property. Theorem 4.2. Let P ⊆ { , } N be a Π class. If P is Turing degree isomorphicto CPA, then P has the Join Property. Proof. Let A and Z be reals such that A ≥ T Z ⊕ ′ and Z > T 0. By Theorem3.4 let B ∈ CPA be such that A ≡ T B ′ ≡ T B ⊕ ′ ≡ T B ⊕ Z . Since P isTuring degree isomorphic to CPA, let C ∈ P be such that B ≡ T C . Then A ′ ≡ T C ′ ≡ T C ⊕ ′ ≡ T C ⊕ Z . Thus P has the Join Property, Q.E.D. Remark 4.3. More generally, for P, Q ⊆ N N we say that P is Turing degreeembeddable into Q if { deg T ( X ) | X ∈ P } ⊆ { deg T ( X ) | X ∈ Q } . The proofof Theorem 4.2 shows that if P has the Join Property and is Turing degreeembeddable into Q , then Q has the Join Property.We now turn to Pseudojump Inversion. Unfortunately, we cannot simplyimitate the proof of Theorem 4.2. This is because pseudojump operators arenot invariant under Turing equivalence, i.e., X ≡ T Y typically does not imply J e ( X ) ≡ T J e ( Y ). Consequently, we do not know whether the PseudojumpInversion Property is invariant under Turing degree isomorphism of special Π subclasses of { , } N . However, we shall settle some interesting special cases ofthis question. As a first step, consider the following notion, which is a specialcase of Turing degree isomorphism. Definition 4.4 (recursive homeomorphism) . For P, Q ⊆ N N , a recursive home-omorphism of P onto Q is a one-to-one onto mapping Φ : P → Q such that bothΦ and its inverse Φ − : Q → P are the restrictions of partial recursive function-als to P and Q respectively. We say that P and Q are recursively homeomorphic if there exists a recursive homeomorphism of P onto Q . Lemma 4.5. Let P, Q ⊆ N N be recursively homeomorphic. If P has the Pseu-dojump Inversion Property, then so does Q . Proof. Assume that P has the Pseudojump Inversion Property. We shall provethat Q has the Pseudojump Inversion Property. Given e ∈ N and A ≥ T ′ , itwill suffice to find C ∈ Q such that A ≡ T J e ( C ) ≡ T C ⊕ ′ .Let Φ : P → Q be a recursive homeomorphism. Let i ∈ N be such that for all X ∈ P and all n ∈ N we have ϕ Xi ( n ) ≃ ϕ Φ( X ) e ( n ). Then for all X ∈ P we have W Xi = { n ∈ N | ϕ Xi ( n ) ↓} = { n ∈ N | ϕ Φ( X ) e ( n ) ↓} = W Φ( X ) e . Since P has thePseudojump Inversion Property, let B ∈ P such that A ≡ T J i ( B ) ≡ T B ⊕ ′ . Let C = Φ( B ). Because Φ is a recursive homeomorphism, we have B ≡ T Φ( B ) = C ,10ence J e ( C ) = C ⊕ W Ce = C ⊕ W Φ( B ) e = C ⊕ W Bi ≡ T B ⊕ W Bi = J i ( B ), hence A ≡ T J e ( C ) ≡ T C ⊕ ′ , Q.E.D.As a bridge from Turing degree isomorphism to recursive homeomorphism,we have the following lemma. Lemma 4.6. Let P, Q ⊆ { , } N be nonempty Π classes. If P is Turing degreeembeddable into Q , then there exist nonempty Π subclasses e P ⊆ P and e Q ⊆ Q such that e P and e Q are recursively homeomorphic. Proof. We shall draw on some facts about hyperimmune-freeness and truth-tablereducibility . For this background, see [11, §§ § X ∈ P be hyperimmune-free. Since P is Turing degreeembeddable into Q , let Y ∈ Q be such that X ≡ T Y . Then Y is alsohyperimmune-free, and by hyperimmune-freeness there exist truth-table func-tionals Φ , Ψ : { , } N → { , } N such that Φ( X ) = Y and Ψ( Y ) = X . Let e P = { X ∈ P | Φ( X ) ∈ Q and Ψ(Φ( X )) = X } . Then e P is a Π subclass of P ,and it is nonempty because it contains X . Moreover e Q = { Φ( X ) | X ∈ e P } isalso a Π subclasss of Q , and we have a recursive homeomorphism Φ ↾ e P of e P onto e Q . Lemma 4.7. Any nonempty Π subclass of CPA is recursively homeomorphicto CPA. Proof. See [14, § Theorem 4.8. Let P ⊆ { , } N be a Π class. If P is Turing degree isomorphicto CPA, then P has the Pseudojump Inversion Property. Proof. Let P be Turing degree isomorphic to CPA. By Lemma 4.6 there arenonempty recursively homeomorphic Π classes e P ⊆ P and ] CPA ⊆ CPA. ByLemma 4.7 ] CPA is recursively homeomorphic to CPA. By Theorem 3.8 CPAhas the Pseudojump Inversion Property, so by Lemma 4.5 ] CPA and hence e P have the Pseudojump Inversion Property. But then, since e P ⊆ P , it followsthat P has the Pseudojump Inversion Property, Q.E.D.Next we prove a more general-looking result. Recall from [13, 16, 17, 18, 19]that a nonempty Π class P ⊆ { , } N is said to be Muchnik complete if everynonempty Π class Q ⊆ { , } N is Muchnik reducible to P , i.e., for all X ∈ P there exists Y ∈ Q such that Y ≤ T X . Lemma 4.9. A Π class P ⊆ { , } N is Muchnik complete if and only if it isTuring degree isomorphic to CPA. Moreover, for such a P the set of Turingdegrees { deg T ( X ) | X ∈ P } is upwardly closed. Proof. See [13, §§ heorem 4.10. Let P ⊆ { , } N be a nonempty Π class. If P is Muchnikcomplete, then P has the Join Property and the Pseudojump Inversion Property. Proof. This is immediate from Lemma 4.9 and Theorems 4.2 and 4.8.An even more general-looking result reads as follows. Theorem 4.11. Let P ⊆ { , } N be a nonempty Π class such that { deg T ( Y ) | Y ∈ P } is upwardly closed. Then any P ⊆ N N which includes P has the JoinProperty and the Pseudojump Inversion Property. Proof. By Lemma 4.9 CPA is Turing degree embeddable into P . Our resultthen follows by Remark 4.3 and Theorems 4.2 and 4.8. Π classes of positive measure For σ ∈ { , } ∗ we write J σ K = { X ∈ { , } N | σ ⊂ X } . Let µ be the fair coinmeasure on { , } N , defined by letting µ ( J σ K ) = 2 −| σ | for all σ ∈ { , } ∗ . This µ is a Borel probability measure on { , } N . A Π class P ⊆ { , } N is said tobe of positive measure if µ ( P ) > 0. In this section we note that such a P musthave the Pseudojump Inversion Property but need not have the Join Property.We also obtain the same results for Π subclasses of { , } N which are Turingdegree isomorphic to such a P .To prove these results we need some basic facts about Martin-L¨of random-ness and LR -reducibility . We cite our semi-expository papers [13, 15, 17, 18]but one can also consult the treatises of Downey and Hirschfeldt [2] and Nies[9]. Let MLR = { X ∈ { , } N | X is Martin-L¨of random } . Lemma 5.1. A Π subclass of { , } N is of positive measure if and only if itincludes a nonempty Π subclass of MLR. Proof. This is because MLR ⊆ { , } N is a Σ class of full measure. See forinstance [13, § Lemma 5.2 (due to Nies [9]) . Let P ⊆ { , } N be a Π class of positive measure.Then P has the Pseudojump Inversion Property. Proof. See [9, Theorem 6.3.9] or a simpler proof in [15, Theorem 5.1]. Theorem 5.3. Let P ⊆ { , } N be a Π class of positive measure, and let Q ⊆ { , } N be a Π class which is Turing degree isomorphic to P . Then Q hasthe Pseudojump Inversion Property. Proof. By Lemma 5.1 let P be a nonempty Π subclass of P ∩ MLR. Then P is Turing degree embeddable in Q , so by Lemma 4.6 we can find nonempty Π classes e P ⊆ P and e Q ⊆ Q which are recursively homeomorphic. By Lemma5.1 e P is of positive measure, so by Lemma 5.2 e P has the Pseudojump Inver-sion Property. It then follows by Lemma 4.5 that e Q and hence Q have thePseudojump Inversion Property. 12elativizing the notion of Martin-L¨of randomness, for any real Y we writeMLR Y = { X ∈ { , } N | X is Martin-L¨of random relative to Y } . If MLR Y ⊆ MLR Z we say that Z is LR -reducible to Y , abbreviated as Z ≤ LR Y . Intuitivelythis means that Y has at least as much “derandomizing power” as Z . Clearly Z ≤ T Y implies Z ≤ LR Y , but the converse does not hold: Lemma 5.4. There exists a nonrecursive real Z such that Z ≤ LR Proof. See [15, Theorem 6.1] or [2, 9].On the other hand, we have: Lemma 5.5. If Z ≤ LR Y then Z ′ ≤ T Z ⊕ Y ′ . Proof. See [15, Theorem 8.8] or [2, 9]. Lemma 5.6 (Lutz [6]) . If Z ≤ LR X ⊕ Z ≤ LR X for all X ∈ MLR. Proof. Assume Z ≤ LR X ∈ MLR. We must prove that MLR X ⊆ MLR X ⊕ Z . Given X ∈ MLR X , it follows by Van Lambalgen’s Theorem ([15,Theorem 3.6], see also [2, Corollary 6.9.3]) that X ⊕ X ∈ MLR. But then X ⊕ X ∈ MLR Z , so by Van Lambalgen’s Theorem relative to Z we have X ∈ MLR X ⊕ Z , Q.E.D. Theorem 5.7 (Lutz [6]) . MLR does not have the Join Property. Proof. By Lemma 5.4 let Z > T Z ≤ LR 0. For any X ∈ MLRwe have X ⊕ Z ≤ LR X by Lemma 5.6, hence ( X ⊕ Z ) ′ ≤ T X ′ ⊕ Z by Lemma5.5. If MLR had the Join Property, there would be an X ∈ MLR such that X ′ ≡ T X ⊕ Z , hence ( X ⊕ Z ) ′ ≤ T X ⊕ Z , a contradiction. Theorem 5.8 (Lutz [6]) . There is a Π class P ⊆ { , } N of positive measurewhich does not have the Join Property. Proof. This is immediate from Theorem 5.7 and Lemma 5.1. Π classes constructed by priority arguments In this section we construct a special Π class Q ⊆ { , } N which has neitherthe Join Property nor the Pseudojump Inversion Property. In addition, this Π class Q has some other interesting features, which we also discuss.We begin by stating a technical theorem which embodies our constructionof Q . Recall from § h P e i e ∈ N is a fixed enumeration of all Π classes. Theorem 6.1. There is a nonempty perfect Π class Q ⊆ { , } N with thefollowing properties. For all e, n ∈ N there exists m ∈ N such that for allpairwise distinct τ , . . . , τ n ∈ { , } ∗ with | τ | = · · · = | τ n | ≥ m , the Π class A topological space is said to be perfect if it has no isolated points, i.e., there is no openset consisting of exactly one point. In particular, a set P ⊆ { , } N is perfect if and only ifthere is no τ ∈ { , } ∗ such that P ∩ J τ K = { X } for some X ∈ { , } N . Q ∩ J τ K ) × · · · × ( Q ∩ J τ n K )is either disjoint from P e or included in P e . Moreover, this m can be computedfrom e using an oracle for 0 ′ .Before proving Theorem 6.1, we spell out some of its consequences which areof more general interest. The next theorem summarizes these features of Q . Theorem 6.2. Let Q ⊆ { , } N be a nonempty Π class as in Theorem 6.1.1. Q is thin , i.e., for every Π subclass P of Q , the complement Q \ P is againa Π subclass of Q . More generally, for any finite sequence Q , . . . , Q n ofpairwise disjoint Π subclasses of Q , the Π class Q × · · · × Q n is thin.2. Q is special , i.e., no X ∈ Q is recursive. More generally, the Turing degreesof members of Q are independent , i.e., no X ∈ Q is ≤ T the join of anyfinitely many other members of Q .3. Every X ∈ Q is of minimal truth-table degree. Consequently, everyhyperimmune-free X ∈ Q is of minimal Turing degree.4. Every finite sequence X , . . . , X n ∈ Q is generalized low , i.e., ( X ⊕ · · · ⊕ X n ) ′ ≡ T X ⊕ · · · ⊕ X n ⊕ ′ . Proof. We prove parts 1 through 4 in that order.1. To see that Q is thin, let P be a Π subclass of Q . Fix e ∈ N such that P = P e , and let m be as in Theorem 6.1. Then for each τ ∈ { , } m the Π class Q ∩ J τ K is either disjoint from P or included in P . Hence Q \ P = S τ ( Q ∩ J τ K )where the union is taken over all τ ∈ { , } m such that Q ∩ J τ K is disjoint from P . Thus Q \ P is is a union of finitely many Π classes, so it too is a Π class.For the generalization, let P be a Π subclass of Q × · · · × Q n where Q , . . . , Q n are pairwise disjoint Π subclasses of Q . Fix e ∈ N such that P = P e , and let m be as in Theorem 6.1. We then have( Q × · · · × Q n ) \ P = S τ ,...,τ n (( Q ∩ J τ K ) × · · · × ( Q n ∩ J τ n K ))where the union is taken over all pairwise distinct sequences τ , . . . , τ n ∈ { , } m such that ( Q ∩ J τ K ) ×· · ·× ( Q n ∩ J τ n K ) is disjoint from P . Thus ( Q ×· · ·× Q n ) \ P is a union of finitely many Π classes, so it too is a Π class, Q.E.D.2. Assume for a contradiction that X ∈ Q is recursive. Fix e ∈ N suchthat P e = { X } , let m ∈ N be as in Theorem 6.1, and let τ = X ↾ m . Clearly X ∈ Q ∩ J τ K , hence Q ∩ J τ K is not disjoint from P e , hence Q ∩ J τ K is included in P e , hence X is the unique member of Q ∩ J τ K . Thus X is an isolated point of Q , contradicting the fact that Q is perfect.For the generalization, it will suffice to show that X (cid:2) T X ⊕ · · · ⊕ X n for allpairwise distinct X, X , . . . , X n ∈ Q . Let Ψ : ⊆{ , } N → { , } N be a partialrecursive functional, and assume for a contradiction that X = Ψ( X ⊕ · · ·⊕ X n ).Let e ∈ N be such that The hypothesis of pairwise disjointness is essential. For instance, if P = { X ⊕ X | X ∈ Q } then ( Q × Q ) \ P is not Π , so Q × Q is not thin. e = { Y ⊕ Y ⊕ · · · ⊕ Y n | ∀ i ∀ j (if Ψ( Y ⊕ · · · ⊕ Y n )( i ) ↓ = j then Y ( i ) = j ) } .Let m be as in Theorem 6.1 and sufficiently large so that τ = X ↾ m , τ = X ↾ m ,. . . , τ n = X n ↾ m are pairwise distinct. The Π class( Q ∩ J τ K ) × ( Q ∩ J τ K ) × · · · × ( Q ∩ J τ n K )contains X ⊕ X ⊕ · · · ⊕ X n and is therefore not disjoint from P e , so it isincluded in P e . In particular, for all Y ∈ Q ∩ J τ K and all i ∈ N we haveΨ( X ⊕ · · · ⊕ X n )( i ) ↓ = X ( i ) = Y ( i ), hence X = Y . Thus Q ∩ J τ K = { X } , soagain X is an isolated point of Q , contradicting the fact that Q is perfect.3. Let X ∈ Q be given. We have already seen that X is not recursive.To prove that X is of miminal truth-table degree, it remains to show that X ≤ tt Ψ( X ) for any truth-table functional Ψ such that Ψ( X ) is not recursive.Given such a functional Ψ, let e be such that P e = { X ⊕ X ∈ Q × Q | Ψ( X ) = Ψ( X ) } .By Theorem 6.1 let m be such that for all τ ∈ { , } ≥ m the Π class ( Q ∩ J τ a h i K ) × ( Q ∩ J τ a h i K ) is either disjoint from P e or included in P e . If it isdisjoint from P e , let us say that τ is splitting .Case 1: For all sufficiently large τ ⊂ X , τ is splitting. Then for all sufficientlylarge τ ⊂ X and all X ∈ Q ∩ J τ a h i K and X ∈ Q ∩ J τ a h i K , we have Ψ( X ) =Ψ( X ). In particular we have Ψ( X ) = Ψ( Y ) for all Y ∈ Q ∩ J τ K such that X = Y . From this it follows that X is truth-table reducible to Ψ( X ).Case 2: There are arbitrarily large τ ⊂ X such that τ is not splitting. Let τ ⊂ X be non-splitting with | τ | ≥ m . Then Q ∩ J τ a h i K and Q ∩ J τ a h i K are nonempty, and for all X ∈ Q ∩ J τ a h i K and X ∈ Q ∩ J τ a h i K we haveΨ( X ) = Ψ( X ). In particular, letting i, j ∈ { , } be such that X ∈ Q ∩ J τ a h i i K and i + j = 1, we have Ψ( X ) = Ψ( Y ) for all Y ∈ Q ∩ J τ a h j i K . From this itfollows that Ψ( X ) is recursive, Q.E.D.A general property of hyperimmune-free reals X is that the Turing degrees ≤ deg T ( X ) are the same as the truth-table degrees ≤ deg tt ( X ). Since every X ∈ Q is of minimal truth-table degree, it follows that every hyperimmune-free X ∈ Q is of minimal Turing degree.4. Up until now we have not used the “moreover” clause of Theorem 6.1,but now we shall use it. We may safely assume that X , . . . , X n ∈ Q arepairwise distinct. Recall that we have defined the Turing jump X ′ ∈ { , } N of X ∈ { , } N to be (the characteristic function of) the set { e ∈ N | X / ∈ P e } . To compute ( X ⊕ · · · ⊕ X n ) ′ we proceed as follows. First we use ouroracle for X ⊕ · · · ⊕ X n to find l ∈ N such that X ↾ l , . . . , X n ↾ l are pairwisedistinct. Then, given e ∈ N , we use our oracle for 0 ′ to compute m ≥ l as inTheorem 6.1. Letting τ i = X i ↾ m for i = 1 , . . . , n , we know that the Π class( Q ∩ J τ K ) × · · · × ( Q ∩ J τ n K ) is either disjoint from P e or included in P e . Usingour oracle for 0 ′ again, we can decide whether this Π class is disjoint from P e or not. The answer to this question tells us whether e ∈ ( X ⊕ · · · ⊕ X n ) ′ ornot. Thus ( X ⊕ · · · ⊕ X n ) ′ is computable from X ⊕ · · · ⊕ X n ⊕ ′ , Q.E.D.15 emark 6.3. A plausible generalization of part 3 of Theorem 6.2 would saythat for every pairwise distinct finite sequence X , . . . , X n ∈ Q , the tt-degrees ≤ deg tt ( X ⊕ · · · ⊕ X n ) should form a lattice isomorphic to the powerset of { , . . . , n } . From this it would follow that the same holds for Turing degrees,provided X ⊕ · · · ⊕ X n is hyperimmune-free. Theorem 6.4. The special Π class Q ⊆ { , } N of Theorem 6.1 has neitherthe Join Property nor the Pseudojump Inversion Property. Proof. We rely mainly on part 4 of Theorem 6.2. To see that the Join Propertyfails for Q , fix Z ∈ Q . Then Z > T B ∈ Q we have ( B ⊕ Z ) ′ ≡ T B ⊕ Z ⊕ ′ , hence 0 ′ (cid:2) T B ⊕ Z . To see that Pseudojump Inversion fails for Q ,consider a pseudojump operator J e with the property that X < T J e ( X ) and( J e ( X )) ′ ≡ T X ′ for all X ∈ N N . Then for all B ∈ Q we have ( J e ( B )) ′ ≡ T B ′ ≡ T B ⊕ ′ , hence 0 ′ (cid:2) T J e ( B ).The rest of this section is devoted to the proof of Theorem 6.1. We use apriority construction in the vein of Martin/Pour-El [7] and Jockusch/Soare [5,Theorem 4.7].Our construction will be presented in terms of treemaps. A treemap is amapping T : { , } ∗ → { , } ∗ such that T ( σ a h i i ) ⊇ T ( σ ) a h i i for all σ ∈ { , } ∗ and all i ∈ { , } . Note that for any treemap T and σ, ρ ∈ { , } ∗ we have σ ⊂ ρ if and only if T ( σ ) ⊂ T ( ρ ). Note also that there is a one-to-one correspondencebetween treemaps T and nonempty perfect closed subsets of { , } N , given by T [ T ] = { T ( X ) | X ∈ { , } N } where T ( X ) = S n ∈ N T ( X ↾ n ).Recall that any Π class P ⊆ { , } N is closed. Therefore, if P is alsononempty and perfect, there is a unique treemap T P such that P = [ T P ]. Lemma 6.5. Let P ⊆ { , } N be a nonempty perfect Π class. Then, thetreemap T = T P corresponding to P is computable using 0 ′ as an oracle. Proof. Given τ ∈ { , } ∗ we can use our oracle for 0 ′ to decide whether P ∩ J τ K is empty or not. Thus, if already know T ( σ ) for some σ ∈ { , } ∗ , we can thencompute T ( σ a h i i ) for i ∈ { , } , because T ( σ a h i i ) is the smallest τ ⊇ T ( σ ) a h i i such that P ∩ J τ a h j i K is nonempty for all j ∈ { , } . Thus T ≤ T ′ , Q.E.D.In our proof of Theorem 6.1, the treemap T = T Q corresponding to Q willbe uniform , in the sense that | T ( σ ) | will depend only on | σ | . This featureof T will be for convenience only, but T will also have another key property,which reads as follows. For all e, l, n ∈ N with e ≤ l and n ≤ l and allpairwise distinct σ , . . . , σ n ∈ { , } l , either J T ( σ ) K × · · · × J T ( σ n ) K ∩ P e = ∅ or( Q ∩ J T ( σ ) K ) × · · · × ( Q ∩ J T ( σ n ) K ) ⊆ P e . Thus by Lemma 6.5 the “moreover”clause of Theorem 6.1 will hold with m = m e = | T ( σ ) | for all σ ∈ { , } e . The existence of such pseudojump operators is well known [20, § VII.1]. One way toobtain such an operator is to combine part 4 of Theorem 6.2 with the R. E. Basis Theorem[4, Theorem 3] to get an e ∈ N such that 0 < T W e and ( W e ) ′ ≡ T ′ . The desired operator J e is then obtained by uniform relativization to an arbitrary Turing oracle X . roof of Theorem 6.1. We shall construct T = T Q as the limit T = lim s T s ofa recursive sequence of recursive uniform treemaps T s , s ∈ N . This will be apointwise limit as s → ∞ , in the sense that for each σ ∈ { , } ∗ we shall have T ( σ ) = T s ( σ ) for all sufficiently large s . Also, the treemaps T s will be nested , inthe sense that for all s ∈ N and all σ ∈ { , } ∗ there will be a ρ ∈ { , } ∗ suchthat T s +1 ( σ ) = T s ( ρ ). From this it follows that [ T s +1 ] ⊆ [ T s ] for all s , and that[ T ] = T s [ T s ]. Thus Q = [ T ] will be a nonempty perfect Π subclass of { , } N .In presenting our construction of T = lim s T s , we shall use the notation P e,s = { X ∈ { , } N | ϕ (1) ,X ↾ se,s (0) ↑} .Note that P e,s = S τ J τ K where the union is taken over a finite subset of { , } s .Note also that P e,s +1 ⊆ P e,s for all s ∈ N , and that P e = { X ∈ { , } N | ϕ (1) ,Xe (0) ↑} = T s P e,s .We now offer a preliminary account of the construction and proof. To each e, l, n ∈ N with e ≤ l and n ≤ l and each pairwise distinct sequence σ , . . . , σ n ∈{ , } l , we associate a requirement R ( e, σ , . . . , σ n ) at level l . Intuitively, thepurpose of this requirement is to insure that ( J T ( σ ) K × · · · × J T ( σ n ) K ) ∩ P e = ∅ “if possible.” The strategy here will be to attempt to arrange that ( J T s ( σ ) K ×· · · × J T s ( σ n ) K ) ∩ P e,s = ∅ for all sufficiently large s , “if possible.” We shall arguethat if this attempt fails, then ( Q ∩ J T s ( σ ) K ) × · · · × ( Q ∩ J T s ( σ n ) K ) ⊆ P e,s forall sufficiently large s , hence ( Q ∩ J T ( σ ) K ) × · · · × ( Q ∩ J T ( σ n ) K ) ⊆ P e .We now give the detailed construction of T s for all s ∈ N . Begin by fixing arecursive linear ordering of all of the requirements, called the priority ordering .Arrange this ordering so that for each l ∈ N , all requirements at level l areof lower priority than all requirements at level < l . The idea here is thatrequirements at level l can be “injured” by requirements at level < l but willnever be “injured” by requirements at level ≥ l . Note that for each l there areonly finitely many requirements at level ≤ l . We proceed by induction on s .Stage 0. Let T ( ν ) = ν for all ν ∈ { , } ∗ . Clearly T is a uniform treemap.Stage s + 1. Assume inductively that we have defined a uniform treemap T s .Our task at this stage is to define T s +1 . A requirement R ( e, σ , . . . , σ n ) at level l is said to be requesting attention at stage s if ( J T s ( σ ) K × · · · × J T s ( σ n ) K ) ∩ P e,s = ∅ but there exist k ≤ s and ρ , . . . , ρ n ∈ { , } k such that ρ i ⊃ σ i for all i = 1 , . . . , n and ( J T s ( ρ ) K × · · · × J T s ( ρ n ) K ) ∩ P e,s = ∅ . If no requirements arerequesting attention, do nothing, i.e., let T s +1 ( ν ) = T s ( ν ) for all ν ∈ { , } ∗ .Otherwise, let R s be the requirement of highest priority which is requestingattention. For this requirement only, choose k > l ≥ e and ρ , . . . , ρ n as aboveand define T s +1 as follows. 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