aa r X i v : . [ m a t h . L O ] J un CODING POWER OF PRODUCT OF PARTITIONS
LU LIU
Abstract.
Given two combinatorial notions P and P , can we encode P via P . In this talk we address the question where P is 3-coloring of integers and P is product of finitely many 2-colorings of integers.We firstly reduce the question to a lemma which asserts that certain Π class of colorings admit two members violating a particular combinatorial con-straint. Then we took a digression to see how complex does the class has tobe so as to maintain the cross constraint.We weaken the two members in the lemma in certain way to address anopen question of Cholak, Dzhafarov, Hirschfeldt and Patey, concerning a sortof Weihrauch degree of stable Ramsey’s theorem for pairs. It turns out theresulted strengthen of the lemma is a basis theorem for Π class with additionalconstraint. We look at several such variants of basis theorem, among themsome are unknown. We end up by introducing some results and questionsconcerning product of infinitely many colorings. Introduction
Given two combinatorial notions P , P , does P encode P . Depending on what“encode” means, the question appear in many fields of computer science. Forexample, consider the error correcting code. The encoding scheme is a function f : k n → k ˆ n such that for every ρ = ρ ∈ k n , the hamming distance between f ( ρ ) , f ( ρ ) is greater than certain value m . In this case, given a piece of infor-mation ρ ∈ k n , a corrupted message ˆ ρ ∈ k ˆ n (which means the hamming distancebetween ˆ ρ, f ( ρ ), namely |{ l : ˆ ρ ( l ) = f ( ρ )( l ) }| , is below a certain value m ), based onˆ ρ we can recover ρ since ρ is the unique string in k n such that its f value is withinhamming distance m to ˆ ρ . In this example, an instance of problem P is a string ρ ∈ k n , and a solution to ρ is ρ itself; an instance of problem P is a string ˆ ρ ∈ k ˆ n ,and a solution to ˆ ρ is a string ˜ ρ ∈ k ˆ n such that the hamming distance between ˜ ρ, ˆ ρ is no greater than m . By P encode P , it means for every instance ρ of P , thereis an instance of P , namely f ( ρ ), such that every solution of f ( ρ ) recovers ρ .In computability theory, we define “recover” by Turing computable. Therefore,the resulted version of this encoding question becomes: whether P is soc-reducibleto P . Where P is soc-reducible (strongly omniscent reducible [15]) to P iff forevery P instance I , there is a P instance I such that every solution to I Turingcomputes a solution to I . Let’s see more example of instance solution framework. • For the problem RT nk (Ramsey’s theorem for n tuples with k colors), a RT nk instance is a coloring C : [ ω ] n → k , a solution is an infinite set G ⊆ ω monochromatic for C , i.e., | C ([ G ] n ) | = 1. Mathematics Subject Classification.
Primary 68Q30 .
Key words and phrases. computability theory, reverse math, Ramsey’s theorem, Muchnickdegree. • For the problem
DNR h ( h -bounded DNR where h is a function from ω to ω ), a DNR h instance is an infinite sequence of integers X ∈ ω ω , a solutionto X is a Y ∈ h ω such that Y ( n ) = X ( n ) for all n ∈ ω . Coding power of ( RT ) <ω . In this paper, we address a particular encodingquestion whether RT is soc-reducible to ( RT ) <ω . Where an ( RT ) <ω instanceis, for some r ∈ ω , an r -tuple of 2-colorings ( C , · · · , C r − ) ∈ (2 ω ) r and a solu-tion to ( C , · · · , C r − ) is an r -tuple of infinite sets ( G , · · · , G r − ) such that G s ismonochromatic for C s for all s < r .Firstly, we prove that RT is not soc-reducible to ( RT ) <ω in Theorem 2.1. Ourapproach to sperate RT from ( RT ) <ω appeals to many soc-reducibility questionswhere the solution of a P , P instance I is an “almost” Π defined combinatorialrelation. Where “almost” Π means, taking RT k as an example: for a RT k instance C , the definition G is solution to C means, (a) G ⊆ C − ( j ) for some j ∈ k ; (b) G is infinite. Part (a) is a Π relation; and part (b) eliminate a small portion of2 ω from the solution space. This approach reduce Theorem 2.1 to a lemma whichasserts that for a Π class Q in the product space P × P (where for each member( I , I ) of P × P , I i is P i instance), if the projection of Q on its first componentis sufficiently large, then there are two members ( X , Y ) , ( X , Y ) of Q such that,roughly speaking, X , X has no common solution while Y , Y has. For Theorem2.1, this lemma is the following. For two RT k instances C , C , we say C , C are almost disjoint if for every j ∈ k , C − ( j ) ∩ C − ( j ) is finite. Let Q ⊆ ω × (2 ω ) r be aΠ class such that for every X ∈ ω , there exists a Y ∈ (2 ω ) r such that ( X, Y ) ∈ Q (in which case we say Q has full projection on 3 ω ). Lemma 1.1.
There exist ( X , Y ) , ( X , Y ) ∈ Q such that: suppose Y i = ( Y i , · · · , Y ir − ) ,we have X , X are almost disjoint and Y s , Y s are not almost disjoint for all s < r . The proof is given in section 2 Lemma 2.2. Lemma 1.1 is interesting in its ownand further rises two questions:(1) How complex does Q has to be in order to maintain the cross constraint,i.e,for every ( X , Y ) , ( X , Y ) ∈ Q, if X , X are almost disjoint , then Y s , Y s are almost disjoint for some s < r. (2) When Q is a Π class, how weak can the two members, violating the con-straint, be. For example, can they be low? Complexity of the cross constraint.
Note that when r = 1, the conclusionfollows obviously and the reason is “finitary”: there are three 3-colorings that aremutually almost disjoint while for every three 2-colorings, two of them are notalmost disjoint.On the other hand, when r >
1, for every finitely many 3-colorings X , · · · , X n − ,there exist some 2-colorings Y , · · · , Y n − such that for every m = m ′ < n , X m , X m ′ being almost disjoint implies that there exists s < r such that Y ms , Y m ′ s are almost disjoint. Even if we let two players play this in a game, the party who ODING POWER OF PRODUCT OF PARTITIONS 3 wants to maintain the constraint has a winning strategy. i.e., The game is an in-finite sequence X , Y , X , Y , · · · where A presents X m then B presents Y m ; Bwins iff for every m = m ′ , X m , X m ′ being almost disjoint implies that there exists s < r such that Y ms , Y m ′ s are almost disjoint. It’s easy to see that B has a win-ning strategy. But does B has a winning strategy without looking at the historyof the game? i.e., Does there exists a function f : 3 ω → (2 ω ) r such that for every X , X being almost disjoint, there exists an s < r such that the s th component of f ( X ) , f ( X ) are almost disjoint. Lemma 2.2 says that such function f , if exists,can not be ∆ . We will further address these questions in section 3. The weakness of the witnesses.
Clearly, weakening the two members in Lemma1.1 give rise to constraint version basis theorem for Π class. For example, theassertion that the two members does not compute a given incomputable Turingdegree, is the constraint version cone avoidance for Π class; the assertion thatthey (together) are low, is the constraint version low basis theorem. Exploringthe combinatorial nature of Q , we prove the constraint version cone avoidance asa demonstration how these basis theorems are proved. We also introduce somesimilar constraint version basis “theorem”, among which some are unknown.Despite interesting in its own, the cross constraint version basis theorem is alsomotivated by a question of Cholak, Dzhafarov, Hirschfeldt and Patey [1]. Let D k denote the problem that an instance is a coloring C : [ ω ] → k that is stable, i.e.,for every n ∈ ω , there exists a j ∈ k such that C ( { n, m } ) = j for almost all m ∈ ω ;a solution to C is an infinite set G monochromatic for C . We say P is computablyreducible to P iff: for every P instance I , I computes a P instance I such thatfor every solution G of I , I ⊕ G compute a solution of I . Question . Is D computably reducible to D × D .By relativization, this question boils down to: is there a ∆ RT instance C such that for every ∆ RT instances C , C , there exists a solution ~G to ( C , C )such that ~G does not compute a solution to C . In another word, this imposerestriction on both the encoding instance and the encoded instance. Therefore, itis neither a strengthen or a weakening of the question whether RT is soc-reducibleto RT × RT . However, we strength the result RT (cid:2) soc ( RT ) <ω by showing,in Theorem 4.1, that there is a ∆ RT instance as a witness, answering question1.2 in negative. It turns out this strengthen can be proved by weakening the twomembers of Lemma 1.1 in certain ways, i.e., to preserve a sort of hyperimmune. Related literature.
We review some literature in computability theory concerning weakness of aproblem. The are mainly two types of these questions, the first one consider instanceof certain complexity; while the second one consider arbitrary instance.In reverse math, it is established that there are five axioms where each is equiv-alent to many natural mathematical theorems (see [17] for an introduction on re-verse math). Among the five axioms, Recursive comprehension (or
RCA for short)is the weakest. For certain type of mathematical theorems (those can be writtenas ∀ X ∃ Y ψ ( X, Y ) where X is seen as an instance and Y is seen as a solution to LU LIU X ) P , P , to prove P does not imply P (over RCA ), it almost always boils downto constructing weak solution of a computable instance of P (see [7] for recentdevelopment in reverse math). For example, Seetapun and Slaman [16] proved RT does not imply RT by showing that every computable instance of RT admits asolution that does not compute a given incomputable degree.In computability randomness theory, there are questions concerning how to ex-tract randomness by certain combinatorial notions P . Since not able to extractrandomness is a sort of weakness, these questions sometimes boils down to con-struct weak solution of P instance of certain complexity. For example, Greenbergand Miller [6] proved that there exists a DNR h that does not compute any 1-randomreal. Kjos-Hanssen asked whether every 1-random real X ∈ ω admit an infinitesubset that does not compute any 1-random real. Thinking of “subset” as a com-binatorial notion where the solution is an infinite subset of the instance, this isa question concerning 1-random instance of “subset” problem. The question isanswered through a sequence of papers [9][10].Sometimes, studying the P instance of certain complexity translates to studyan arbitrary instance of another problem. For example, to study computable RT instance, Seetapun and Slaman [16] (and later Cholak, Jocksuch and Slaman [2])applied so called CJS-decomposition to reduce RT to SRT and COH . It turnsout, any computable
SRT instance is coded by a ∆ RT instance. Meanwhile,almost every avoidance results concerning ∆ RT instance generalize to the strongavoidance version. Here we list a few of other examples studying an instance ofarbitrary complexity and show how these research are related to reverse math. • A well known simple result is that fast-grow-function (hence forth
FGF )admit strong cone avoidance for non hyperarithmetic Turing degree. i.e.,Given a function f ∈ ω ω there exists a function g ≥ f such that g does notcompute a given non hyperarithmetic Turing degree. • A major question in reverse math was whether
SRT implies COH . Thisquestion boils down to constructing certain weak solution of a computable
SRT instance. In [4], Dzhafarov, Patey, Solomon and Westrick considersarbitrary instance of SRT , and studied whether COH is soc-reducible to
SRT . They proved that RT is not soc-reducible to FGF ; actually theyproved that RT k +1 is not soc-reducible to FGF × RT k , thus SRT is notsoc-reducible to COH since
SRT is soc-equivalent to RT × FGF .Chong, Slaman and Yang employ the nonstandard model answering the
SRT vs COH question in negative. Later, Monin and Patey [14] study thejump control of an arbitrary RT instance, this approach finally solve thestandard model version of the question. More recently, Monin and Pateyproved strong hyperarithmetic and arithmetic avoidance of RT . Theirmethod also answers some questions of Wang [19] and generalize the low construction in [2]. • An old question of Sacks asked whether there is a solution to a the uni-versal
DNR instance of minimal degree, which is answered in affirmativeby Kumabe [11] using bushy tree method. Later, people start to wonderwhether there is a real of positive hausdorff dimension that is of minimaldegree. These reals are coded by
DNR h for some computable non decreasingunbounded function h . In [8][13], it is shown that for an arbitrary DNR h ODING POWER OF PRODUCT OF PARTITIONS 5 instance admit minimal degree solution if h is fast enough. Hopefully, re-ducing the fast growing condition of h could finally solve the question. • A major question in reverse math was whether RT implies WKL , where
WKL is the weak K¨ o nig Lemma. In [5], Dzhafarov and Jockusch provedstrong cone avoidance of RT . This is later generalized by author [12] tostrong avoidance of PA degree, answering the question negatively. • Downey, Greenberg, Jockusch and Milans [3] proves that
DNR is not soc-reducible to DNR in a uniform manner, i.e., given a Turing function Ψ,there always exists a DNR instance X such that for every DNR instanceˆ X , there is a solution to ˆ X that does not compute a solution to X . Organization.
In section 2 we prove Lemma 1.1 and demonstrate how RT (cid:2) soc ( RT ) <ω can be proved using Lemma 1.1. Section 3 discuss how complex does theset Q in Lemma 1.1 need to be in order to satisfy the cross constraint. In section 4,we presents some constraint version basis theorem; among them is the preservationof Γ-hyperimmune of a 3-coloring (see section 4.3 for a definition). In section 5, weintroduce some results and questions on product on infinitely many colorings. Notations.
For a sequence of poset ( W , < p ) , · · · , ( W n − , < p n − ), and ( x , · · · , x n − ), ( y , · · · , y n − ) ∈ W × · · · × W n − , we write ( y , · · · , y n − ) ≤ p ( < p respectively)( x , · · · , x n − ) if y m ≤ p m ( < p m respectively) x m for all m < n .We use σ, τ, ρ, ξ to denote strings of finite length. For a string ρ , we let [ ρ ] (cid:22) = { σ : σ (cid:23) ρ } ; similarly, for a set S of strings, let [ S ] (cid:22) = { σ : σ (cid:23) ρ for some ρ ∈ S } .For a tree T , let [ T ] denote the set of infinite paths on T ; for a string ρ , let[ ρ ] = { X ∈ ω ω : X (cid:23) ρ } ; for a finite set V of strings, let [ V ] = { [ ρ ] : ρ ∈ V } . For astring ρ , we use | ρ | to denote the length of ρ . For a set S of strings, let ℓ ( S ) denotethe set of leaves of S (i.e., the set of strings in S having no proper extension in S when S = ∅ ) and we rules ℓ ( ∅ ) = ⊥ .Two k -colorings X , X ∈ k ω are almost disjoint on Z ⊆ ω if for every j ∈ k , X − ( j ) ∩ X − ( j ) ∩ Z is finite; when Z = ω , we simply say X , X are almostdisjoint. 2. Weakness of a product of partitions
The main objective in this section is to address the question whether RT is soc-reducible to ( RT ) r . Let ( RT ) r denote the problem whose instance is an r -tuple( C , · · · , C r − ) of 2-colorings, a solution to ( C , · · · , C r − ) is an r -tuple of infinitesets of integers ( G , · · · , G r − ) such that G s is monochromatic for C s for all s < r .Let ( RT ) <ω denote S r ( RT ) r . Theorem 2.1.
We have RT (cid:2) soc ( RT ) <ω . Moreover, there exists a ∅ ( ω ) -computable RT instance witnessing the conclusion. The rest of this section will prove Theorem 2.1. Let’s firstly briefly discuss theframework of its proof. First, we fix a somewhat complex P instance I and anarbitrary P instance I . The goal is to construct a solution of I that does notcompute any solution of I . The general approach is to construct a sequence ofconditions d , d , · · · : LU LIU • where each condition is essentially a closed set of candidates of the weaksolution we construct, • each requirement is forced by some condition (meaning every member ofthe condition satisfy the requirement).Then we take the common element G of d , d , · · · (which exists by compactness),so it satisfies all requirements. Thus G is the desired weak solution of I notcomputing any solution of I .The main point in this approach is how to extend a condition in order to forcea given requirement. Usually, this turns out to be a uniform encoding question.i.e., whether for every instance of the encoded problem, there is an instance of theencoding problem so that every solution of the encoding instance compute, via agiven algorithm , a solution of the given encoded instance. In particular, in theorem2.1, the uniform encoding question is exactly the uniform version of the originalencoding question (which might not be the case in general). i.e., roughly speaking,it boils down to the following: let C be the complex RT instance;given a tuple of Turing functional { Ψ k } k ∈ r , a tuple of colors { j k } k ∈ r ⊆ , (2.1)a ( RT ) r instance ( C , · · · , C r − ) , there is a solution ~G in color k of ( C , · · · , C r − ) for some k such thatΨ ~G is not a solution of C in color j k . If C is not encoded via { Ψ k } k ∈ r , { j k } k ∈ r in a fashion as (2.1), then we canfinitely extend the initial segment of a condition to force the Turing functional Ψ k to violate C deterministically.On the other hand, if C is encoded via { Ψ k } k ∈ r , { j k } k ∈ r in that way, weobserve the behavior of Ψ k and look at all 3-colorings ˜ C encoded in that way. Itturns out that the encoded ˜ C forms a Π class. Since the very complex C is in it,this class contains a “lot” of 3-colorings. Now we look at the set of pairs ( ˜ C, ˆ C )where ˜ C is uniformly encoded by ˆ C via { Ψ k } k ∈ r , { j k } k ∈ r similarly as (2.1). Weprove in Lemma 2.2 that there are ( ˜ C , ˆ C ) , ( ˜ C , ˆ C ) such that • ˜ C , ˜ C has no common solution in any color j ∈ • ˆ C , ˆ C do have a common solution in some color k ∗ ∈ r .Thus, let ~G be that common solution of ˆ C , ˆ C . We have Ψ ~G k ∗ ⊆ ∗ ( ˜ C ) − ( j k ∗ ) ∩ ( ˜ C ) − ( j k ∗ ), therefore it is a finite set.More specifically, what we need is the following lemma. Let Q ⊆ ω × (2 ω ) r bea Π class that has full projection on 3 ω . Lemma 2.2.
There exist ( X , Y ) , ( X , Y ) ∈ Q such that: suppose Y i = ( Y i , · · · , Y ir − ) ,we have X , X are almost disjoint and Y s , Y s are not almost disjoint for all s < r .Moreover, ( X , Y ) ⊕ ( X , Y ) ≤ T ∅ ′ .Proof. The approach is to build two sequences of pairs ( ρ i , σ i ) ≺ ( ρ i , σ i ) ≺ · · · , i ∈ X i = ∪ t ρ it , Y i = ∪ t σ it = ( Y i , · · · , Y ir − ) are the desiredwitness. ODING POWER OF PRODUCT OF PARTITIONS 7
For ρ ∈ <ω , σ ∈ <ω , ( ρ, σ ) is good iff for every X ∈ [ ρ ], Q ( X, · ) ∩ [ σ ] = ∅ where Q ( X, · ) = { Y ∈ (2 ω ) r : ( X, Y ) ∈ Q } . It’s easy to see thatfor every good pair ( ρ, σ ) , every ˆ ρ (cid:23) ρ, (2.2) there exists a good pair (˜ ρ, ˜ σ ) ≻ (ˆ ρ, σ ) . A condition in this lemma is a 2-tuple (( ρ , σ ) , ( ρ , σ )) such that ( ρ i , σ i ) ∈ <ω × <ω is good for all i ∈
2. We will try to build the desired two sequences by ∅ ′ -computing a sequence of condition which fulfill all necessary requirements. Morespecifically, for each condition (( ρ t , σ t ) , ( ρ t , σ t )) in the sequence, try to extend itto (( ρ t +1 , σ t +1 ) , ( ρ t +1 , σ t +1 )) so that • ( ρ t +1 , ρ t +1 ) does not positively progress on any color j ∈ compared to ( ρ t , ρ t ),i.e., ( ρ t +1 ) − ( j ) ∩ ( ρ t +1 ) − ( j ) = ( ρ t ) − ( j ) ∩ ( ρ t ) − ( j ) for all j ∈ • while for each s < r , for some k ∈
2, ( σ t +1 , σ t +1 ) positively progress oncolor k on the s th component compared to ( σ t , σ t ),i.e., ( σ t +1 ,s ) − ( k ) ∩ ( σ t +1 ,s ) − ( k ) ) ( σ t,s ) − ( k ) ∩ ( σ t,s ) − ( k ) . Where σ it = ( σ it, , · · · , σ it,r − ).If we can keep positively progress on each component, then we are done. We willshow in Claim 2.3 that this is indeed the case.A condition (( ρ , σ ) , ( ρ , σ )) exclude component s if it cannot be extended tomake positive progress on the s th component. More specifically, for every condi-tion ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ )) extending (( ρ , σ ) , ( ρ , σ )), if (ˆ ρ , ˆ ρ ) does not positivelyprogress on color any color j ∈ ρ , ρ ), then (ˆ σ , ˆ σ ) does notpositively progress on any color on the s th component compared to ( σ , σ ). Claim 2.3.
If condition (( ρ , σ ) , ( ρ , σ )) exclude component s , then there existstwo computable Y , Y ∈ ω such that for every i ∈ , every good pair (ˆ ρ i , ˆ σ i ) (cid:23) ( ρ i , σ i ) , we have ˆ σ is ≺ Y i ( in which case we say ( ρ i , σ i ) lock the s th component ) .Proof. It suffices to prove that for every sufficiently large n ∈ ω , there exists k , k ∈ n ) such that for every condition ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ )) extend-ing (( ρ , σ ) , ( ρ , σ )), ˆ σ is ( n ) = k i whenever ˆ σ is ( n ) is defined. Suppose otherwise.Then there exists an i ∈ i = 0) and two good pairs (ˆ ρ, ˆ σ ), (˜ ρ, ˜ σ ) extending( ρ , σ ) such that ˆ σ s ( n ) = ˜ σ s ( n ) are defined. Now extends ( ρ , σ ) to a good pair( ρ ′ , σ ′ ) where σ ′ s ( n ) is defined and neither of ( ρ ′ , ˆ ρ ) , ( ρ ′ , ˜ ρ ) positively progress on anycolor j ∈ ρ , ρ ). But clearly, either ( σ ′ , ˆ σ ) or ( σ ′ , ˜ σ ) will positivelyprogress on some color k ∈ s th component compared to ( σ , σ ) (dependingon σ ′ s ( n ) = ˆ σ s ( n ) or σ ′ s ( n ) = ˜ σ s ( n )). A contradiction to exclude component s . (cid:3) Let I ⊆ r be a maximal set such that there exists a good pair ( ρ, σ ) lock s th component for all s ∈ I . Starting with the condition (( ρ, σ ) , ( ρ, σ )), it is clear tosee that for any condition (( ρ , σ ) , ( ρ , σ )) extending (( ρ, σ ) , ( ρ, σ )) with ( ρ , ρ )not positively progress on any color j ∈ ρ, ρ ), (( ρ , σ ) , ( ρ , σ ))does not exclude any component. For example, if I = r , then there even existscomputable X , X ∈ ω , Y ∈ ω such that ( X , Y ) , ( X , Y ) ∈ Q are as desired,namely X , X are almost disjoint. Thus we will be able to keep making positiveprogress on each component. Thus we are done. LU LIU (cid:3)
Now we are ready to prove Theorem 2.1.
Definition 2.4 (Hyperimmune) . A RT k instance C is hyperimmune relative to D iffor every D -computable array of k -tuple of mutually disjoint finite sets { ( F n, , · · · , F n,k − ) } s ∈ ω with S j
Let C be a ∅ ( ω ) -computable RT instance hyperimmune rel-ative to any arithmetic degree; let r ∈ ω and let ( C , · · · , C r − ) be a ( RT ) r in-stance.The condition we use in this theorem is a tuple ( { ~F k } k ∈ r , ~X ) such that • Each ~F k = ( F k , , · · · , F k ,r − ) is an r -tuple of finite set in color k of( C , · · · , C r − ). i.e., F k ,s ⊆ C − s ( k ( s )) for all s < r . • ~X = ( X , · · · , X r − ) is arithmetic; each X s ⊆ ω is infinite and X s > F k ,s for all k ∈ r , s < r .As in the Mathias forcing, let ( ~F k , ~X ) denote the collection { ( G ′ , · · · , G ′ r − ) : F k ,s ⊆ G ′ s ∧ G ′ s ⊆ F k ,s ∪ X s for all s < r } . A condition ( { ~F k } k ∈ r , ~X ) is seen as a collection of candidates of the solution( G , · · · , G r − ) we construct. Namely: S k ∈ r ( ~F k , ~X ).A condition ( { ~ ˆ F k } k ∈ r , ~ ˆ X ) extends another condition ( { ~F k } k ∈ r , ~X ) (written as( { ~ ˆ F k } k ∈ r , ~ ˆ X ) ⊆ ( { ~F k } k ∈ r , ~X )) if the collection represented by ( { ~ ˆ F k } k ∈ r , ~ ˆ X ) is asubset of that of ( { ~F k } k ∈ r , ~X ). Or in another word, F k ,s ⊆ ˆ F k ,s ⊆ F k ,s ∪ X s , ˆ X s ⊆ X s for all s < r, k ∈ r .Given a tuple of requirements {R k } k ∈ r , a condition ( { ~F k } k ∈ r , ~X ) forces W k ∈ r R k iff there exists a k ∈ r such that for every ( G , · · · , G r − ) ∈ ( ~F k , ~X ), we have( G , · · · , G r − ) satisfies R k .For every Turing functional Ψ, every j ∈
3, let R j Ψ denote the requirement:Ψ ( G , ··· ,G r − ) is not an infinite subset of C − ( j ) . Fix a tuple of requirements {R k } k ∈ r , where for some Turing functional Ψ k , some j k ∈ R k is R j k Ψ k ; fix a condition ( { ~F k } k ∈ r , ~X ). We will show that there existsan extension of ( { ~F k } k ∈ r , ~X ) that forces W k ∈ r R k . In this way, we will havea sequence of conditions d ⊇ d ⊇ · · · ⊇ d t = ( { ~F t k } k ∈ r , ~X t ) ⊇ · · · such thatfor every 2 r -tuple of Turing functionals { Ψ k } k ∈ r and every 2 r -tuple of integers { j k } k ∈ r , W k ∈ r R j k Ψ k is forced by some d t . Then there exists a k ∗ ∈ r suchthat, let G s = ∪ t F t k ∗ ,s for each s < r , we have ( G , · · · , G r − ) satisfies all require-ments R j Ψ . This means ( G , · · · , G r − ) does not compute a solution of C . Clearly( G , · · · , G r − ) is in color k ∗ of ( C , · · · , C r − ). It is automatic that G s is infinitefor all s < r since for every n , there are such Turing functional Ψ that for every m ,Ψ ( G , ··· ,G r − ) ( m ) ↓ = 1 if and only if | G s | > n for all s < r .As we said, it remains to prove the following. Claim 2.5.
There exists an extension of ( { ~F k } k ∈ r , ~X ) , forcing W k ∈ r R k . ODING POWER OF PRODUCT OF PARTITIONS 9
Proof.
For every n , consider the following set Q n of RT instance ˜ C encoded bysome ( RT ) r instance via the condition and the given Turing functionals. That is,for some ( RT ) r instance ( ˆ C , · · · , ˆ C r − ), ( ˆ C , · · · , ˆ C r − ) together with ( { ~F k } k ∈ r , ~X )forces Ψ ( G , ··· ,G r − ) k ∩ ( n, ∞ ) ⊆ ˆ C − ( j k ) for all k ∈ r . More specifically, ˜ C ∈ Q n iff: There exists a ( RT ) r instance ( ˆ C , · · · , ˆ C r − ) such thatfor every k ∈ r , every finite ( ˆ F , · · · , ˆ F r − ) ∈ ( ~F k , ~X )with ( ˆ F \ F k , , · · · , ˆ F r − \ F k ,r − ) in color k of ( ˆ C , · · · , ˆ C r − ) , we have: Ψ ( ˆ F , ··· , ˆ F r − ) k ∩ ( n, ∞ ) ⊆ ˜ C − ( j k ) . Case 1 . For every n , Q n = 3 ω .In this case, since the sequence Q n does not contain sufficiently many 3-colorings,by hyperimmune of C , we show that C will be violated by some finite extension ofthe condition.By compactness argument and since Q n is a Π , ~X class uniformly in n there existsan ~X -computable increasing array of disjoint triple of finite sets { ( E n, , E n, , E n, ) } n ∈ ω ,where each ( E n, , E n, , E n, ) is a partition of ( n, l n ) ∩ ω for some l n > n such that:For every ( RT ) r instance ( ˆ C , · · · , ˆ C r − ) , (2.3) there exists a k ∈ r , a finite ( ˆ F , · · · , ˆ F r − ) ∈ ( ~F k , ~X )with ( ˆ F \ F k , , · · · , ˆ F r − \ F k ,r − ) in color k of ( ˆ C , · · · , ˆ C r − )such that Ψ ( ˆ F , ··· , ˆ F r − ) ∩ ( n, l n ) * E n,j k . Since C is hyperimmune relative to ~X , there exists an n ∗ such that C − ( j ) ∩ ( n, l n ) = E n ∗ ,j for all j ∈
3. Taking ( ˆ C , · · · , ˆ C r − ) in (2.3) to be the given ( C , · · · , C r − ),we have:There exists a k ∗ ∈ r and a finite ( ˆ F , · · · , ˆ F r − ) ∈ ( ~F k ∗ , ~X )with ( ˆ F \ F k ∗ , , · · · , ˆ F r − \ F k ∗ ,r − ) in color k ∗ of ( C , · · · , C r − )such that Ψ ( ˆ F , ··· , ˆ F r − ) ∩ ( n, l n ) * C − ( j k ∗ ).Thus let ( { ~F ∗ k } k ∈ r , ~X ) be such a condition that ~F ∗ k = ~F k if k = k ∗ and ~F ∗ k ∗ =( ˆ F , · · · , ˆ F r − ), then we are done. Case 2 . For some n , Q n = 3 ω .Let Q = { ( ˜ C, ˆ C ) : ˜ C ∈ ω , ˆ C ∈ (2 ω ) r witnesses ˜ C ∈ Q n } . Clearly Q sat-isfies the hypothesis of Lemma 2.2. Apply (the relativized) Lemma 2.2 to get( ˜ C , ˆ C ) , ( ˜ C , ˆ C ) ∈ Q and a k ∗ ∈ r such that • ( ˜ C , ˆ C ) ⊕ ( ˜ C , ˆ C ) ≤ T ( ~X ) ′ ; • ˜ C , ˜ C are almost disjoint, while • ( ˆ C s ) − ( k ∗ ( s )) ∩ ( ˆ C s ) − ( k ∗ ( s )) ∩ X s = X ∗ s is infinite for all s < r .By definition of Q , let ~X ∗ = ( X ∗ , · · · , X ∗ r − ), for every ( G , · · · , G r − ) ∈ ( ~F k ∗ , ~X ∗ ),we have: Ψ ( G , ··· ,G r − ) k ∗ ∩ ( n, ∞ ) ⊆ ( ˜ C ) − ( j k ∗ ) ∩ ( ˜ C ) − ( j k ∗ ) is finite. Thus the condition ( { ~F k } k ∈ r , ~X ∗ ) extends ( { ~F k } k ∈ r , ~X ) and forces W k ∈ r R j k Ψ k . Thus weare done. (cid:3)(cid:3) Complexity of the cross constraint
Let Q ⊆ ω × (2 ω ) r be such that Q has full projection on 3 ω . We wonder howcomplex does Q has to be in order to satisfy the following cross constraint:for every ( X , Y ) , ( X , Y ) ∈ Q, if X , X are almost disjoint , (3.1) then Y s , Y s are almost disjoint for some s < r. It turns out that Q can not be analytic. Proposition 3.1. If Q is Σ , then Q does not satisfy ( ) .Proof. The proof is by constructing a sequence of good pairs (as a sequence ofconditions) as in Lemma 2.2, except that we redefine “good” to adapt to Σ set.We inductively define a forcing notion in spirit of Cohen forcing. For a closedset ˆ Q ⊆ ω ω × ω ω , let ˆ T be the tree defining ˆ Q , i.e., [ ˆ T ] = ˆ Q ; for a ρ, σ ∈ ω <ω , • we write ρ ⊢ ˆ Q ∩ [( ρ, σ )] = ∅ if ( ρ, σ ) / ∈ ˆ T ; • for an ordinal γ , we write ρ ⊢ γ ˆ Q ∩ [( ρ, σ )] = ∅ if the set (cid:8) ˆ ρ (cid:23) ρ : for some ˆ γ < γ, ˆ ρ ⊢ ˆ γ ˆ Q ∩ [(ˆ ρ, σ a n )] = ∅ (cid:9) is dense over ρ for all n ∈ ω . • we write ρ ⊢ ˆ Q ∩ [( ρ, σ )] = ∅ if ρ ⊢ γ ˆ Q ∩ [( ρ, σ )] = ∅ for some ordinal γ . Claim 3.2. If ρ ˆ Q ∩ [( ρ, σ )] = ∅ , then for some n ∈ ω , some ˜ ρ (cid:23) ρ , we have: ˆ ρ ˆ Q ∩ [(ˆ ρ, σ a n )] = ∅ for all ˆ ρ (cid:23) ˜ ρ .Proof. For each n ∈ ω , consider the set A n = { ˆ ρ (cid:23) ρ : ˆ ρ ⊢ ˆ Q ∩ [(ˆ ρ, σ a n )] = ∅} . Since A n is countable, there exists an ordinal γ such that A n = { ˆ ρ (cid:23) ρ : ˆ ρ ⊢ γ ˆ Q ∩ [(ˆ ρ, σ a n )] = ∅} for all n. If A n is dense over ρ for all n , then we have ρ ⊢ ˆ Q ∩ [( ρ, σ )] = ∅ , a contradiction.Therefore, suppose A n is not dense over ρ . Then there exists ˜ ρ (cid:23) ρ such that A n ∩ [˜ ρ ] (cid:22) = ∅ . Thus ˜ ρ, σ a n is the desired object. (cid:3) We think of Q as a subset set of the product space of 3 ω and (2 ω ) r × ω ω . A good triple is a triple ( ρ, σ, τ ) ∈ <ω × <ω × ω <ω such that for every ˆ ρ (cid:23) ρ ,ˆ ρ Q ∩ [(ˆ ρ, σ, τ )] = ∅ .Similarly as (2.2), we have the following:for every good triple ( ρ, σ, τ ) , every ˆ ρ (cid:23) ρ, there exists a good triple (˜ ρ, ˜ σ, ˜ τ ) ≻ (ˆ ρ, σ, τ ) . The rest of the proof proceeds exactly as Lemma 2.2. (cid:3)
ODING POWER OF PRODUCT OF PARTITIONS 11
On the other hand, Johannes Schrz point out that a non principal ultrafilteron ω give rise to such a set Q . Recall that a non principal ultrafilter U on ω is acollection of sets of integers such that: • U is closed under finite intersection (filter), • U is upward closed (filter); • for every Z ⊆ ω , either Z ∈ U or ω \ Z ∈ U (ultrafilter); • it is not of form { Z ⊆ ω : n ∈ Z } for some n ∈ ω (non principal). Proposition 3.3 (Johannes Schrz) . If there exists a non principal ultrafilter on ω ,then there exists a function f : 3 ω → (2 ω ) such that for every two almost disjoint X , X ∈ ω , f ( X ) s , f ( X ) s are almost disjoint for some s < .Proof. Let U be a non principal ultrafilter on ω . For X ∈ ω , let h ( X ) = j iff X − ( j ) ∈ U . Let f ( X ) = ( ∅ , ∅ ) , ( ∅ , ω ) , ( ω, ∅ ) respectively iff h ( X ) = 0 , , ∅ , ∅ ) represents such a ( Y , Y ) ∈ (2 ω ) that Y − (1) = Y − (1) = ∅ ,and similarly for ( ω, ∅ ) , ( ∅ , ω )). Note that if X , X ∈ ω are almost disjoint, then h ( X ) = h ( X ), since U is non principal (which means no finite set is in U ). Butclearly, if h ( X ) = h ( X ), then f ( X ) s , f ( X ) s are almost disjoint for some s < (cid:3) The following result shows that it is consistent with
ZFC that “there exists a Π set Q ⊆ ω × (2 ω ) with full projection on 3 ω satisfying (3.1)”. Proposition 3.4 (Jonathan) . If V = L , then there exists a Π set Q ⊆ ω × (2 ω ) ( a Σ set Q ⊆ ω × (2 ω ) respectively ) with full projection on ω satisfying ( ) .Proof. In L , we can construct a Σ non principal ultrafilter U on ω . Suppose U isthe projection of a Π set V ⊆ ω × ω on the first component. Let Y , Y , Y denote( ∅ , ∅ ) , ( ∅ , ω ) , ( ω, ∅ ) ∈ (2 ω ) respectively. Let Q consists of such ( X, Y ) ∈ ω × (2 ω ) that: for some j < , Y = ( Y j , ˆ Y ) and ( X − ( j ) , ˆ Y − (1)) ∈ V . Clearly Q is Π and has full projection on 3 ω . It’s also clear that Q satisfies(3.1). (cid:3) As demonstrated in Proposition 3.4, a set Q in 3 ω × (2 ω ) is more complex thanone in 3 ω × (2 ω ) . We wonder if it is necessarily more complex. i.e., let ECC ( r )denote the assertion “ there is a set Q ⊆ ω × (2 ω ) r satisfying (3.1)”; let EU ( ω )denote the assertion “there exists a non principal ultrafilter on ω ”. By Proposition3.3, over ZF , EU ( ω ) → ECC ( r ) → ECC ( r −
1) for all r > Question . Are implications (set theoretic) in EU ( ω ) → ECC ( r ) → ECC ( r − Basis theorem for Π class with cross constraint In [1], Cholak, Dzhafarov, Hirschfeldt, Patey asked whether D is computablyreducible to D . This is equivalent (modulo the relativization) to ask whether thereis a ∆ C ∈ ω such that for every two ∆ C , C ∈ ω , thereis a solution ( G , G ) to ( C , C ) such that ( G , G ) does not compute a solutionof C . The main objective of this section is to prove the following improvement ofTheorem 4.1 which address the above question in affirmative. Theorem 4.1.
We have RT (cid:2) soc ( RT ) <ω . Moreover, there exists a ∆ RT instance C witnessing the conclusion. Just like Theorem 2.1 is reduced to Lemma 2.2, Theorem 4.1 is reduced tothe following improvement of Lemma 2.2 where the two members are weakenedin certain ways, i.e., preserving a sort of hyperimmune of the 3-coloring C . Let Q ⊆ ω × (2 ω ) r be a Π class having full projection on 3 ω . Let C ∈ ω be Γ-hyperimmune (see section 4.3 for the definition). Lemma 4.2.
There exist ( X , Y ) , ( X , Y ) ∈ Q such that suppose Y i = ( Y i , · · · , Y ir − ) ,we have X , X are almost disjoint and Y s , Y s are not almost disjoint for all s < r .Moreover, C is Γ -hyperimmune relative to ( X , Y ) ⊕ ( X , Y ) . The proof will be delayed to section 4.5. Clearly, Lemma 4.2 is a cross constraintversion basis theorem for Π class since non empty Π class do admit memberspreserving the Γ-hyperimmune. Of course, Lemma 2.2 can be seen as a crossconstraint version basis theorem for Π class as well and its normal version issimply that every non empty Π class admit a ∆ member.The proof of Lemma 4.2 consists of two separate ideas:(1) Cross constraint version basis theorem for Q (say cone avoidance, seeLemma 4.5);(2) Defining the appropriate hyperimmune notion, namely Γ-hyperimmune,and demonstrate how it is preserved by, say WKL .We illustrate the idea of cross constraint version basis theorem in section 4.2where we prove the cross constraint version cone avoidance (Lemma 4.5). In section4.3 we define the Γ-hyperimmune and prove in section 4.4 Lemma 4.14 that it ispreserved by
WKL . This not only serves as a demonstration of how Γ-hyperimmuneis preserved but is also needed in Lemma 4.2. Combining the two ideas, we proveLemma 4.2 in section 4.5. Before we proceed to Lemma 4.5, we explore somecombinatorial aspect on Q in section 4.1. We end up this part of the section byproving Theorem 4.1 using Lemma 4.2. Proof of Theorem 4.1.
Let C be a ∆ r ∈ ω and r many 2-colorings C , · · · , C r − .We use similar condition ( { ~F k } k ∈ r , ~X ) as in Theorem 2.1 except that insteadof ~X being arithmetic, we require C to be Γ-hyperimmune relative to ~X . Thedefinition of extension, forcing and requirement are identical as Theorem 2.1. Itsuffices to show how to extend a condition to force a requirement. This part followsexactly as Claim 2.5 except that:(1) In case 1, to see C is not uniformly encoded, we take advantage of C beingΓ-hyperimmune relative to ~X (instead of hyperimmune relative to ~X );(2) In case 2, instead of selecting ( ˜ C , ˆ C ) , ( ˜ C , ˆ C ) so that ( ˜ C , ˆ C ) ⊕ ( ˜ C , ˆ C ) ≤ T ( ~X ) ′ , we require that C is Γ-hyperimmune relative to ( ˜ C , ˆ C ) ⊕ ( ˜ C , ˆ C ),promised by Lemma 4.2.The rest of the proof are exactly the same. (cid:3) Combinatorial lemmas.
For ρ , ρ ∈ k <ω , we say ρ , ρ are disjoint iff ρ − ( j ) ∩ ρ − ( j ) = ∅ for all j ∈ k . ODING POWER OF PRODUCT OF PARTITIONS 13
For n, ˆ n ∈ ω , a set A ⊆ k n , an injective function ext : A → k ˆ n , we say ext is a disjoint preserving extension over ( ρ , ρ ) iff • ext ( ρ ) (cid:23) ρ for all ρ ∈ A ; and • for every ˜ ρ , ˜ ρ ∈ n , if (˜ ρ , ˜ ρ ) does not positively progress on any color j ∈ ρ , ρ ), then neither does ( ext (˜ ρ ) , ext (˜ ρ )). Lemma 4.3.
For every k ∈ ω , n ≤ ˆ n ∈ ω , every ρ ∈ k n , ˆ ρ ∈ k ˆ n , there is a disjointpreserving extension over ( ⊥ , ⊥ ) , namely ext : k n → k ˆ n such that ext ( ρ ) = ˆ ρ .Proof. The function ext could simply be a point wise isomorphism. (cid:3)
Let k ∈ ω , and for each i ∈
2, let f i : 3 <ω → P ( k ) be two functions. For a ρ ∈ <ω , a j ∈ k , we say ( ρ, j ) is good for f i iff: for every ˆ ρ (cid:23) ρ , j ∈ f i (ˆ ρ ). Lemma 4.4.
Suppose for every ˆ ρ (cid:23) ρ , f i (ˆ ρ ) ⊆ f i ( ρ ) = ∅ for all i ∈ . Then forevery n , every set A ⊆ n , there exist an ˆ n ∈ ω , a disjoint preserving extensionover ( ⊥ , ⊥ ) ext : A → ˆ n such that for every ρ ∈ A , every j ∈ k , every i ∈ , either ( ext ( ρ ) , j ) is good for f i or j / ∈ f i ( ext ( ρ )) .Proof. Suppose A = { ρ , · · · , ρ n − } . We handle each member one at a time, bythat we mean to extend all members, preserving the mutually disjoint relation,while the handled member ρ m is extended to some ˆ ρ so that (ˆ ρ, j ) is either goodfor f i or (ˆ ρ, j ) / ∈ f i (ˆ ρ ) for all j ∈ k, i ∈ ρ , let ˆ ρ (cid:23) ρ be such that for every i ∈
2, every j ∈ k ,either (ˆ ρ , j ) is good for f i or (ˆ ρ , j ) / ∈ f i (ˆ ρ ) . Such ˆ ρ exists by the hypothesis on f i . Extend each ρ , · · · , ρ n − to ˆ ρ , · · · , ˆ ρ n − ∈ | ˆ ρ | respectively so that for every m , m < n , if ρ m , ρ m are disjoint, then ˆ ρ m , ˆ ρ m are disjoint (such extensionexists by Lemma 4.3).At step 1, repeat what we do in step 0 (with A replaced by { ˆ ρ , · · · , ˆ ρ n − } ) todeal with ˆ ρ .Repeat this procedure until all elements are dealt with. In the end, we get aset { ρ ∗ , · · · , ρ ∗ n − } . Clearly for every m , m < n , if ρ m , ρ m are disjoint, then ρ ∗ m , ρ ∗ m are disjoint since the mutually disjoint relation is preserved throughouteach step.Now consider the function ext ( ρ m ) = ρ ∗ m for all m < n . It’s easy to see that ext satisfies the conclusion. (cid:3) Note that for any n ∈ ω , the function f : 3 <ω ∋ ρ
7→ { σ ∈ n : [( ρ, σ )] ∩ Q = ∅} ∈ P (2 n )satisfies the hypothesis of Lemma 4.4.4.2. Cross constraint version cone avoidance.
Let Q ⊆ ω × (2 ω ) r be a Π class that has full projection on 3 ω ; let ˇ D be an incomputable oracle. Lemma 4.5.
There exists ( X , Y ) , ( X , Y ) ∈ Q such that: suppose Y i = ( Y i , · · · , Y ir − ) ,we have X , X are almost disjoint and Y s , Y s are not almost disjoint for all s < r .Moreover, ( X , Y ) ⊕ ( X , Y ) (cid:3) T ˇ D .Proof. For ρ ∈ <ω , σ ∈ <ω , for a closed set P ⊆ ω × ω , we say ( ρ, σ ) is good for P iff for every X ∈ [ ρ ], P ( X, · ) ∩ [ σ ] = ∅ where P ( X, · ) = { Y ∈ (2 ω ) r : ( X, Y ) ∈ Q } .A condition in this lemma is a tuple (( ρ , σ ) , ( ρ , σ ) , P, D ) such that • D is an oracle that does not compute ˇ D ; • P ⊆ ω × (2 ω ) r is a Π ,D class; • ( ρ i , σ i ) ∈ <ω × <ω is good for P for all i ∈ ρ , σ ) , ( ρ , σ ) , P, D ) is seen as a set of candidates of thedesired (( X , Y ) , ( X , Y )) (also denoted as (( ρ , σ ) , ( ρ , σ ) , P, D )), namely (cid:8) (( ˆ X , ˆ Y ) , ( ˆ X , ˆ Y )) ∈ ( P × P ) ∩ ([( ρ , σ )] × [( ρ , σ )]) :( ˆ X , ˆ X ) does not positively progress on any color j ∈ ρ , ρ ) . (cid:9) A condition ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ ) , ˆ P , ˆ D ) extends a condition (( ρ , σ ) , ( ρ , σ ) , P, D )iff ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ ) , ˆ P , ˆ D ) ⊆ (( ρ , σ ) , ( ρ , σ ) , P, D );or equivalently: • (ˆ ρ , ˆ ρ ) does not positively progress on any color j ∈ ρ , ρ ); • (ˆ ρ i , ˆ σ i ) (cid:23) ( ρ i , σ i ) for each i ∈ P ⊆ P .The requirement we want to satisfy is: for each Turing functional Ψ, R Ψ : Ψ ( X ,Y ) ⊕ ( X ,Y ) = ˇ D. A condition (( ρ , σ ) , ( ρ , σ ) , P, D ) forces R Ψ iff every member (( ˆ X , ˆ Y ) , ( ˆ X , ˆ Y ))in (( ρ , σ ) , ( ρ , σ ) , P, D ) satisfies R Ψ .Fix a Turing functional Ψ, a condition (( ρ , σ ) , ( ρ , σ ) , P, D ), we prove that: Claim 4.6.
There exists an extension of (( ρ , σ ) , ( ρ , σ ) , P, D ) forcing R Ψ .Proof. For simplicity, assume | ρ | = | ρ | , | σ | = | σ | , otherwise extend them to beso; and suppose D = ∅ . Let T P be a computable tree in 3 <ω × <ω defining P sothat for every ρ ∈ <ω , the set { σ ∈ <ω : ( ρ, σ ) ∈ T P } is a pruned tree.For every oracle ˜ D ∈ ω , consider the following set T ˜ D of trees such that T ∈ T ˜ D iff [ T ] is a condition forcing Ψ ( X ,Y ) ⊕ ( X ,Y ) = ˜ D . More specifically, T ∈ T ˜ D iff: • T ⊆ T P and ( ρ i , σ i ) is good for [ T ] for all i ∈ • for every (˜ ρ i , ˜ σ i ) ∈ T ∩ [( ρ i , σ i )] (cid:22) , every n ∈ ω , if (˜ ρ , ˜ ρ ) does not positivelyprogress on any color j ∈ ρ , ρ ), then Ψ (˜ ρ , ˜ σ ) ⊕ (˜ ρ , ˜ σ ) ( n ) ↓→ Ψ (˜ ρ , ˜ σ ) ⊕ (˜ ρ , ˜ σ ) ( n ) = ˜ D ( n ).The key note is that T ˜ D is a Π , ˜ D class uniformly in ˜ D . Case 1.
For every ˜ D ∈ ω , T ˜ D = ∅ .In particular, T ˇ D = ∅ . By compactness, there exists an N ∈ ω , such that forevery tree T ⊆ ≤ N × ≤ N with ℓ ( T ) ⊆ N × N , if T ⊆ T P and for every ρ ∈ N ,(4.1) there exists a σ ∈ N such that ( ρ, σ ) ∈ T , then we have thatthere exist, for each i ∈ , a (˜ ρ i , ˜ σ i ) ∈ ℓ ( T ) ∩ [( ρ i , σ i )] (cid:22) , an m ∈ ω, such that(4.2)(˜ ρ , ˜ ρ ) does not positively progress on any color j ∈ ρ , ρ ) , andΨ (˜ ρ , ˜ σ ) ⊕ (˜ ρ , ˜ σ ) ( m ) ↓6 = ˜ D ( m ) . ODING POWER OF PRODUCT OF PARTITIONS 15
Now we illustrate how to use Lemma 4.4. Let k = 2 N −| σ | ; let f i : 3 <ω →P (2 N −| σ | ) be such that for every ρ ∈ <ω , f i ( ρ ) = { σ ∈ N −| σ | : ( ρ i ρ, σ i σ ) ∈ T P } . Since for each i ∈
2, ( ρ i , σ i ) is good for P , f i satisfies the hypothesis of Lemma 4.4for all i ∈
2. Let A = 3 N −| ρ | and let ext : 3 N −| ρ i | → <ω be as in the conclusionof Lemma 4.4. Let g i : 3 N −| ρ i | → N −| σ i | be such that g i ( ρ ) ∈ f i ( ext ( ρ )) for all ρ ∈ N −| ρ | and all i ∈
2. Which means, by definition of ext ,( ext ( ρ ) , g i ( ρ )) is good for f i for all ρ ∈ N −| ρ | and all i ∈ T ⊆ ≤ N × ≤ N to be a tree generated by the downward closure ofthe following set: B = { ( ρ, σ ) ∈ N × N : for some i ∈ , some ρ ′ , ρ = ρ i ρ ′ and σ = σ i g i ( ρ ′ ) } . First we note that T ⊆ T P and satisfies (4.1). To see this, for each ( ρ, σ ) ∈ B ,suppose ρ = ρ i ρ ′ and σ = σ i g i ( ρ ′ ). By definition of g i , g i ( ρ ′ ) ∈ f i ( ext ( ρ ′ )) ⊆ f i ( ρ ′ ).Therefore, by definition of f i , ( ρ, σ ) ∈ T P .By (4.2), let (˜ ρ i , ˜ σ i ) ∈ B ∩ [( ρ i , σ i )] (cid:22) be such that for some n ∈ ω , (˜ ρ , ˜ ρ ) does notpositively progress on any color j ∈ ρ , ρ ) and Ψ (˜ ρ , ˜ σ ) ⊕ (˜ ρ , ˜ σ ) ( m ) ↓6 =ˇ D ( m ).We now apply the conclusion of Lemma 4.4 on ext to show that there are ˆ ρ i (cid:23) ˜ ρ i such that • (ˆ ρ i , ˜ σ i ) is good for P for all i ∈
2; and • (ˆ ρ , ˆ ρ ) does not positively progress on any color j ∈ ρ , ρ ).By definition of B , suppose ˜ ρ i = ρ i ρ ′ i and ˜ σ i = σ i g i ( ρ ′ i ). It is clear that ρ ′ , ρ ′ aredisjoint since (˜ ρ , ˜ ρ ) does not positively progress on any color j ∈ ρ , ρ ) (and since | ρ | = | ρ | ). By definition of ext (where ext is required to be adisjoint preserving extension over ( ⊥ , ⊥ )), • ext ( ρ ′ i ) (cid:23) ρ ′ i for all i ∈ • ext ( ρ ′ ) , ext ( ρ ′ ) are disjoint and • ( ext ( ρ ′ i ) , g i ( ρ ′ i )) is good for f i .Unfolding the definition of f i , ( ext ( ρ ′ i ) , g i ( ρ ′ i )) is good for f i means for ev-ery ρ ′′ ∈ [ ext ( ρ ′ i )] (cid:22) , g i ( ρ ′ i ) ∈ f i ( ρ ′′ ), i.e., for every ρ ′′ ∈ [ ext ( ρ ′ i )] (cid:22) , ( ρ i ρ ′′ , ˜ σ i ) =( ρ i ρ ′′ , σ i g i ( ρ ′ i )) ∈ T P . Thus, let ˆ ρ i = ρ i ext ( ρ ′ i ) , we have (ˆ ρ i , ˜ σ i ) is good for P . Obviously, ˆ ρ i (cid:23) ρ i for all i ∈
2. On the otherhand, since ext ( ρ ′ ) , ext ( ρ ′ ) are disjoint and | ρ | = | ρ | , therefore (ˆ ρ , ˆ ρ ) does notpositively progress on any color j ∈ ρ , ρ ).In summary, ((ˆ ρ , ˜ σ ) , (ˆ ρ , ˜ σ ) , P, D ) is the desired extension of (( ρ , σ ) , ( ρ , σ ) , P, D )forcing the requirement in a deterministic way. Case 2.
There exists a ˜ D such that T ˜ D = ∅ .Note that the set of ˜ D such that T ˜ D = ∅ consists of a Π class. By cone avoidanceof Π class, there exists a ˜ D , a T ∈ T ˜ D such that ˜ D ⊕ T (cid:3) T ˇ D . Let ˆ P = [ T ]. By defi-nition of T ˜ D , ( ρ i , σ i ) is good for ˆ P . Thus, (( ρ , σ ) , ( ρ , σ ) , ˆ P , ˜ D ⊕ T ) is a condition extending (( ρ , σ ) , ( ρ , σ ) , P, D ). To see that (( ρ , σ ) , ( ρ , σ ) , ˆ P , ˜ D ⊕ T ) forces R Ψ , we note that for every (( ˆ X , ˆ Y ) , ( ˆ X , ˆ Y )) ∈ (( ρ , σ ) , ( ρ , σ ) , ˆ P , ˜ D ⊕ T ), ifΨ ( ˆ X , ˆ Y ) ⊕ ( ˆ X , ˆ Y ) is total, then Ψ ( ˆ X , ˆ Y ) ⊕ ( ˆ X , ˆ Y ) = ˜ D = ˇ D . Thus we are done inthis case. (cid:3) A condition (( ρ , σ ) , ( ρ , σ ) , P, D ) exclude component s if it cannot be ex-tended to make positive progress on the s th component. More specifically, forevery condition ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ ) , ˆ P , ˆ D ) extending (( ρ , σ ) , ( ρ , σ )), (ˆ σ , ˆ σ ) doesnot positively progress on any color on the s th component compared to ( σ , σ ).Given a pair ( ρ, σ ), a closed set P ⊆ ω × (2 ω ) r we say ( ρ, σ ) lock the s th component iff: there exists a Y ∈ ω such that for every (ˆ ρ, ˆ σ ) (cid:23) ( ρ, σ ) that is goodfor P , ˆ σ s ≺ Y where ˆ σ = (ˆ σ , · · · , ˆ σ r − ).Let I ⊆ r be a maximal set such that there exists a D (cid:3) T ˇ D , a Π ,D class P ⊆ Q , a pair ( ρ, σ ) good for P such that ( ρ, σ ) locks the s th component for all s ∈ I . Starting with the condition d = (( ρ, σ ) , ( ρ, σ ) , P, D ), note that for anyextension d of d , d does not exclude any component (see Claim 2.3).Thus combine with Claim 4.6, there is a sequence of conditions d ⊇ d ⊇ · · · such that every requirement is forced by some d t ; moreover, let (( X , Y ) , ( X , Y )) = ∪ t (( ρ t , σ t ) , ( ρ t , σ t )), we have that Y s , Y s are not almost disjoint for all s < r sincewe can make positive progress on each component infinitely often; while X , X are almost disjoint by definition of extension. (cid:3) A notion of hyperimmune.
For two finite trees T , T in ω <ω , we say T isa one step variation of T iff: • either there is a ξ ∈ ℓ ( T ), a finite non empty set B ⊆ ω | ξ | +1 such that T = T ∪ B ; • or there is a non leaf ξ ∈ T , a non empty set B ( T ∩ [ ξ ] (cid:22) ∩ ω | ξ | +1 suchthat T = ( T \ [ ξ ] ≺ ) ∪ B .The key fact in this definition is that the set B in the or case is non empty. Thiswill be used in Lemma 4.7.Given a poset ( W, ≺ p ) with a root, a tree-computation path in W is a sequenceof pairs ( T , φ ) , ( T , φ ) , · · · (finite or infinite) for every u ∈ ω , T u is a finite tree(possibly empty) in ω <ω ; φ u : T u ∪ ℓ ( T u ) → W is a function such that, • T = ∅ ; • for every ˆ ξ, ξ ∈ T u with ˆ ξ ≻ ξ , φ u ( ˆ ξ ) ≻ p φ u ( ξ ); • T u +1 is a one step variation of T u ; moreover, • φ u +1 ↾ T u +1 ∩ T u = φ u ↾ T u +1 ∩ T u . • φ u ( ⊥ ) = the root of W .In our application, the function φ u will be given by a Turing functional we diagonalagainst; a tree in the sequence represent a tree of initial segments. The either case ofone step variation represents the case that for some initial segment, which is the leafof the tree, sufficiently many extension makes the Turing functional’s computationprogress; the or case represents that it is found that some initial segment cannotbe extended to be the next condition.A simple observation is ODING POWER OF PRODUCT OF PARTITIONS 17
Lemma 4.7. If ( W, ≺ p ) is well founded, then there is no infinite tree-computationpath in W .Proof. The key fact here is that in the or case of one step variation, the set B isnon empty. Therefore, suppose ( T , φ ) , · · · is an infinite tree-computation path in W . Then there is an X ∈ ω ω and a sequence of integers u < u < · · · such that X ↾ n ∈ T v for all n, v ≥ u n . Since φ u +1 inherits φ u by definition of tree-computationpath, therefore φ u n +1 ( X ↾ n + 1) ≻ p φ u n ( X ↾ n ). Thus, a contradiction to the wellfoundness of W . (cid:3) Definition 4.8 (The Γ m space) . We inductively define the following set Γ m .(1) Let Γ be the set of partial functions from ω to 3 with finite domain.Although Γ admit a natural partial order, we here define ≺ p on Γ so thatevery two elements of Γ are incomparable provided they have non emptydomain; and the root of Γ is the partial function with empty domain.(2) Suppose we have inductively defined Γ , · · · , Γ m − where each is a posetwith a root. DefineΓ m as the set of finite tree-computation path in Γ m − .The root of Γ m is clearly the singleton ( ∅ , φ ) (by definition of tree-computationpath, φ is the function with domain {⊥} = ℓ ( ∅ ) and φ ( ⊥ ) = the root ofΓ m − ).(3) Since each element of Γ m is a sequence, it makes sense to say one tree-computation path in Γ m − is an initial segment of the other. This give riseto a natural partial order ≺ p m on Γ m where tcp ≺ p m tcp means tcp isan initial segment of tcp .(4) For a c ∈ Γ , we say c is over n if dom ( c ) ⊆ ( n, ∞ ).(5) Suppose we have defined over n for elements in Γ m − , for a tcp = (( T , φ ) , · · · , ( T u − , φ u − )) ∈ Γ m , we say tcp is over n if for every v < u , every ξ ∈ T v ∪ ℓ ( T v ), φ v ( ξ ) is over n .According to Lemma 4.7, by induction, it is direct to see Lemma 4.9.
For every m , Γ m is well founded. Definition 4.10 (Diagonal against) . For a partial function C from ω to 3,(1) We say C diagonal against a c ∈ Γ iff C ↾ dom ( c ) = c .(2) Suppose we have defined what it means for C diagonal against tcp when tcp ∈ Γ m − . For a tcp = (( T , φ ) , · · · , ( T u − , φ u − )) ∈ Γ m , we say C diagonal against tcp iff there exists a ξ ∈ ℓ ( T u − ), such that C diagonalagainst φ u − ( ξ ).Intuitively, C diagonal against tcp means C agree with some partial function ω → T s − . It can be shown by induction that if tcp ∈ Γ m is over n , then there is a partial function C from ω to 3 with a finitedomain dom ( C ) ⊆ ( n, ∞ ) such that C diagonal against tcp . Definition 4.11 (Γ-approximation) . For each m ∈ ω , a Γ m - approximation is afunction f : ω × ω → Γ m such that for every n, s ∈ ω , f ( n, s ) is over n , f ( n, s +1) (cid:23) p m f ( n, s ) and f ( n,
0) = ( ∅ , φ ). A Γ -approximation is a Γ m -approximation forsome m . Note that by Lemma 4.9, for every Γ-approximation f , every n , there is an s sothat the computation of f ( n, · ) converges to f ( n, s ), i.e., f ( n, t ) = f ( n, s ) for all t ≥ s ; denote this f ( n, s ) as f ( n ). Definition 4.12 (Γ-hyperimmune) . A 3-coloring C is Γ -hyperimmune relative toa Turing degree D iff for every D -computable Γ-approximation f , there exists an n ∈ ω such that C diagonal against f ( n ). When D is computable, we simply say C is Γ-hyperimmune.Obviously, Γ-hyperimmune generalizes the idea of hyperimmune and implies hy-perimmune. Due to Lemma 4.9, Lemma 4.13.
There exists a ∆ -coloring C ∈ ω that is Γ -hyperimmune.Proof. Note that we can computably enumerate all computable Γ-approximations f , f , · · · . Suppose we have defined C on [0 , n ] ∩ ω . To diagonal against a Γ m -approximation f , simply ∅ ′ -compute f (ˆ n ) for some ˆ n ≥ n . That is, thanks toLemma 4.9, we can ∅ ′ -compute the s such that f (ˆ n, s ) = f (ˆ n, t ) for all t ≥ s . (cid:3) Preservation of the Γ -hyperimmune for Π class. Fix a Γ-hyperimmune3-coloring C , a non empty Π class Q ⊆ ω . Lemma 4.14.
There exists an X ∈ Q such that C is Γ -hyperimmune relative to X .Proof. The condition we use in this lemma is a pair ( ρ, P ) where ρ ∈ <ω , P isa Π ,D class for some Turing degree D so that C is Γ-hyperimmune relative to D ;and P ⊆ [ ρ ].Each condition ( ρ, P ) is seen as a collection of candidates of the X we construct,namely P .A condition (ˆ ρ, ˆ P ) extends a condition ( ρ, P ) iff ˆ P ⊆ P . A condition ( ρ, P ) forces a requirement R iff for every Y ∈ P , Y satisfies R .The requirement we deal with is, for each Turing functional Ψ, R Ψ : for some n, C diagonal against Ψ X ( n ) . Here we adopt the convention that for each Turing functional Ψ, there exists a m ,such that for every oracle Y , every n, s , Ψ Y ( n, s ) ∈ Γ m .Starting with the condition ( ⊥ , Q ), we will build a sequence of conditions so thateach requirement is satisfied, then the common element of these conditions satisfiesall requirements. Therefore, it suffices to show how to extend a given condition toforce a given requirement. Fix a Turing functional Ψ, a condition (ˇ ρ, P ). Claim 4.15.
There exists an extension of (ˇ ρ, P ) forcing R Ψ .Proof. We adopt the convention that for every ρ , every n , Ψ ρ ( n, s ) is defined iff s ≤ | ρ | ; and we write Ψ ρ ( n ) for Ψ ρ ( n ). Assume that for every oracle Y , every n, s ∈ ω , Ψ Y ( n, s ) ∈ Γ m − ; and Ψ Y ( n,
0) is the root of Γ m − . For notation simplicity,suppose P is a Π ,D class where D = ∅ . Let T P be a pruned, co-c.e. tree so that P = [ T P ].We observe the behavior of Ψ on oracles ρ ∈ T P , and define an Γ m -approximation f as following. Fix an n . Definition 4.16 (Computing f ( n, · )) . We define f ( n, s ) by induction on s . ODING POWER OF PRODUCT OF PARTITIONS 19 (1) By convention, f ( n,
0) is the sequence consisting of ( ∅ , φ ) alone.(2) Suppose we have defined f ( n, s ) = tcp = (( T , φ ) , · · · , ( T u − , φ u − )) ∈ Γ m and together with an injective function ψ : T u − ∪ ℓ ( T u − ) → <ω ∪ {⊥} with ψ ( ⊥ ) = ⊥ so that P ⊆ [ ψ ( ℓ ( T u − ))];for every ξ ∈ T u − , φ u − ( ξ ) = Ψ ψ ( ξ ) ( n ) . for every ˆ ξ, ξ ∈ T u − ∪ ℓ ( T u − ) , if ˆ ξ ≻ ξ, then ψ ( ˆ ξ ) ≻ ψ ( ξ ) . (3) Define f ( n, s + 1) depending on which of the following cases occur:for some ξ ∈ ℓ ( T u − ) , for every ρ ∈ [ ψ ( ξ )] (cid:22) ∩ T P [ s + 1] ∩ s +1 , (4.4) Ψ ρ ( n ) ≻ p m − φ u − ( ξ );for some ξ ∈ T u − ∪ ℓ ( T u − ) , [ ψ ( ξ )] (cid:22) ∩ T P [ s + 1] = ∅ ;(4.5) otherwise . (4.6)(4) • In case of (4.4), suppose [ ψ ( ξ )] (cid:22) ∩ T P [ s + 1] ∩ s +1 = { ρ , · · · , ρ w − } .We add a set B ⊆ ω | ξ | +1 ∩ [ ξ ] (cid:22) of size w to T u − getting T u ; then define ψ on T u by inheriting ψ and 1-1 mapping B to { τ , · · · , τ w − } ; finally,define φ u by inheriting φ u − and φ u ( ˆ ξ ) = Ψ ψ (ˆ ξ ) ( n ) for all ˆ ξ ∈ B . Let f ( n, s + 1) = tcp a ( T u , φ u ). Clearly f ( n, s + 1) is a tree-computationpath in Γ m − since T u is a one step variation of T u − (fulfilling theeither case). The inductive hypothesis on ψ is easily verified. • In case of (4.5), suppose ξ is the minimal string witness (4.5), i.e.,non proper initial segment of ξ is a witness. Let ˆ ξ be the immediatepredecessor of ξ and suppose T u − ∩ [ ˆ ξ ] (cid:22) ∩ ω | ξ | = B . Define T u as( T u − \ [ ˆ ξ ] ≺ ) ∪ B ; let ˆ ψ, φ u inherit ψ, φ u − on T u respectively. Let f ( n, s + 1) = tcp a ( T u , φ u ). By minimality of ξ , P ⊆ [ ψ ( ℓ ( T u ))]. Theother part of inductive hypothesis is easily verified (with respect toˆ ψ, φ u ). • In case of (4.6), let f ( n, s + 1) = f ( n, s ).As we argued in item (4) of the definition of f ( n, · ), we have f is a Γ m -approximation.By compactness, f is computable. Since, C is Γ-hyperimmune, there are n ∗ , s ∗ suchthat • C diagonal against f ( n ∗ , s ∗ ) and • f ( n ∗ , t ) = f ( n ∗ , s ∗ ) for all t ≥ s ∗ .Suppose f ( n ∗ , s ∗ ) = (( T , φ ) , · · · , ( T u − , φ u − )). By definition of diagonal against,there is a ξ ∈ ℓ ( T u − ) such that C diagonal against φ u − ( ξ ). Consider ρ ∗ = ψ ( ξ )and the Π class P ∗ ⊆ P ∩ [ ρ ] on which the computation of Ψ( n ∗ , · ) converges atΨ ρ ∗ ( n ∗ ), that is, Y ∈ P ∗ iff: Y ∈ P ∩ [ ρ ∗ ] and Ψ Y ( n ∗ , s ) = Ψ ρ ∗ ( n ∗ ) for all s ≥ | ρ ∗ | .Note that by our definition of f , since f ( n ∗ , t ) = f ( n ∗ , s ∗ ) for all t ≥ s ∗ , this means(4.4) no longer occurs after step s ∗ . Therefore, P ∗ = ∅ .But φ u − ( ξ ) = Ψ ρ ∗ ( n ∗ ), therefore we have C diagonal against Ψ Y ( n ∗ ) for all Y ∈ P ∗ . Thus ( ρ ∗ , P ∗ ) is the desired extension of (ˇ ρ, P ) forcing R Ψ . (cid:3) (cid:3) Cross constraint version of the Γ -hyperimmune preservation. Fix aΓ-hyperimmune 3-coloring C , a Π class Q ⊆ ω × (2 ω ) r having full projection on3 ω . Lemma 4.2.
There exist ( X , Y ) , ( X , Y ) ∈ Q such that suppose Y i = ( Y i , · · · , Y ir − ) ,we have X , X are almost disjoint and Y s , Y s are not almost disjoint for all s < r .Moreover, C is Γ -hyperimmune relative to ( X , Y ) ⊕ ( X , Y ) .Proof. The definition of condition, extension, forcing are the same as Lemma 4.5.The requirement is as in Lemma 4.14. Following the proof of Lemma 4.5, it sufficesto show that any condition can be extended to force a given requirement. Fix acondition ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ ) , P, D ) and a Turing functional Ψ. Claim 4.17.
There is an extension of ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ ) , P, D ) forcing R Ψ .Proof. For simplicity, assume | ˇ ρ | = | ˇ ρ | , | ˇ σ | = | ˇ σ | , otherwise extend them to beso. In this proof, whenever we write ( ρ, σ ) it automatically implies | ρ | ≥ | σ | . Weadopt the similar convention as in Claim 4.15, i.e., for every ( ρ, σ ) ∈ <ω × <ω ,every n , Ψ ( ρ,σ ) ( n, s ) is defined iff s ≤ | σ | ; we write Ψ ( ρ,σ ) ( n ) for Ψ ( ρ,σ ) ( n, | σ | ); forsome m , for every oracle Y , every n, s ∈ ω , Ψ Y ( n, s ) ∈ Γ m − ; D = ∅ ; T P is apruned, co-c.e. tree so that P = [ T P ].Let T × denote the set of pairs (( ρ , σ ) , ( ρ , σ )) ∈ ( T P ∩ [(ˇ ρ , ˇ σ )] (cid:22) ) × ( T P ∩ [(ˇ ρ , ˇ σ )] (cid:22) ) such that ( ρ , ρ ) does not positively progress on any color j ∈ ρ , ˇ ρ ).The proof generally follows that of Lemma 4.14 except that we need to in-corporate the combinatorics of P as in Lemma 4.9 in the following way. In thecomputation of f ( n ): • The range of function ψ is T × instead of 2 <ω . • Case (4.6) can be interpreted as for some ξ ∈ T u − ∪ ℓ ( T u − ), it is foundthat the initial segment ψ ( ξ ) can not be extended to be the next condition.In this Lemma, this becomes for some ξ ∈ T u − ∪ ℓ ( T u − ), for the initialsegment (( ρ , σ ) , ( ρ , σ )) corresponding to ξ , ( ρ i , σ i ) is not good for P forsome i ∈
2. We will delete such ξ just like in Lemma 4.14. • Case (4.5) can be interpreted as there are “sufficiently” many extensionˆ ρ of ψ ( ξ ) that make the Turing functional Ψ ˆ ρ ( n ) progress (compared toΨ ψ ( ξ ) ( n )). By “sufficient” many, it means: provably, one of them can beextended to be the next condition’s initial segment.We have a similar case in this Lemma except that the definition of “suf-ficient” becomes: provably, one of them can be extended to some goodpairs as in Lemma 4.5 case 1. i.e., by our definition 4.18 of sufficient,Lemma 4.5 case 1 can be rephrased as there are sufficiently many extensions((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) of ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ )) such that Ψ ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) ( n ) = ˜ D ( n )for some n • Since we here have a very different definition of “sufficient” (see definition4.18) than that of Lemma 4.14, the key point is that in case (4.6), howthe “sufficient” of a set A is preserved while deleting certain element from A . What we actually do is to: firstly extends the 3 <ω -component of themembers in A in a disjoint preserving manner so that the non good mem-ber’s component is extended to the witness of its non good; then delete the ODING POWER OF PRODUCT OF PARTITIONS 21 element not in T P × T P . We prove that this action produces a sub-sufficientset (see Claim 4.19). Definition 4.18 (Sufficient) . Given a set A ⊆ (3 <ω × <ω ) × (3 <ω × <ω ), a(( ρ , σ ) , ( ρ , σ )) ∈ (3 <ω × <ω ) × (3 <ω × <ω ) with | ρ | = | ρ | , | σ | = | σ | , we say A is a sufficient over (( ρ , σ ) , ( ρ , σ )) iff there does not exist a tree T such that • T ⊆ T P and ( ρ i , σ i ) is good for [ T ] for all i ∈
2; and • for every (˜ ρ i , ˜ σ i ) ∈ T ∩ [( ρ i , σ i )] (cid:22) , if (˜ ρ , ˜ ρ ) does not positively progress onany color j ∈ ρ , ρ ), then ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) / ∈ A .We say A is sub-sufficient over (( ρ , σ ) , ( ρ , σ )) iff there exists a ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) ∈ A with (˜ ρ i , ˜ σ i ) (cid:23) ( ρ i , σ i ) for all i ∈
2, a ˆ ρ i (cid:23) ˜ ρ i for each i ∈ ρ , ˆ ρ )does not positively progress on any color j ∈ ρ , ρ ) and (ˆ ρ i , ˜ σ i ) isgood for P for all i ∈ D , the set A consisting of ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) such that Ψ ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) ( n ) = ˜ D ( n )for some n is sufficient over the initial segment of the given condition.For every partial function e on 3 <ω , e give rise to a function on (3 <ω × <ω ) × (3 <ω × <ω ), also denoted as e , such that e (( ρ , σ ) , ( ρ , σ )) = (( e ( ρ ) , σ ) , ( e ( ρ ) , σ )).For a set A ⊆ (3 <ω × <ω ) × (3 <ω × <ω ), a ρ ∈ <ω , we say ρ is involved in A iff there exists a (( ρ , σ ) , ( ρ , σ )) ∈ A such that ρ = ρ i for some i ∈
2. Let
Involve ( A ) denote the set of strings ρ ∈ <ω involved in A ; similarly for ( ρ, σ ) isinvolved in A . Claim 4.19.
Suppose ( ρ i , σ i ) is good for P . Suppose A ⊆ (3 N × N ) × (3 N × N ) is sufficient over (( ρ , σ ) , ( ρ , σ )) where | ρ | = | ρ | ∧ | σ | = | σ | . Suppose ext : Involve ( A ) → ˜ N is a disjoint preserving extension over ( ρ , ρ ) . Then the set ext ( A ) ∩ ( T P × T P ) is sub-sufficient over (( ρ , σ ) , ( ρ , σ )) .Proof. By Lemma 4.4 there is a disjoint preserving extension over ( ρ , ρ ), namely˜ ext : ext ( Involve ( A )) → ˆ N for some ˆ N such thatfor every ρ ∈ ext ( Involve ( A )) , every σ ∈ N , (4.7) either ( ˜ ext ( ρ ) , σ ) / ∈ T P or ( ˜ ext ( ρ ) , σ ) is good for P. Consider ˆ ext = ˜ ext ◦ ext. Clearly ˆ ext is a disjoint preserving extension over ( ρ , ρ ) since it is the composi-tion of two disjoint preserving extension over ( ρ , ρ ). We show that for some ˜ τ ∈ ext ( A ) ∩ ( T P ∩ T P ), ˆ τ = ˜ ext (˜ τ ) and ˜ τ witness the sub-sufficient over (( ρ , σ ) , ( ρ , σ ))of ext ( A ) ∩ ( T P ∩ T P ).Consider the set B such that ( ρ, σ ) ∈ B iff: ρ ∈ N ; and when ρ ∈ Involve ( A ), ( ˆ ext ( ρ ) , σ ) is good for P .Since for every i ∈
2, ( ρ i , σ i ) is good for P , we have for every ρ ∈ ˆ N ∩ [ ρ i ] (cid:22) ,there exists a σ ∈ N ∩ [ σ i ] (cid:22) such that ( ρ, σ ) ∈ T P . In particular, for every ρ ∈ Involve ( A ), there exists a σ ∈ N ∩ [ σ i ] (cid:22) such that ( ˆ ext ( ρ ) , σ ) ∈ T P , whichmeans by (4.7), ( ˆ ext ( ρ ) , σ ) is good for P . Therefore,for every i ∈
2, every X ∈ [ ρ i ], there exists a σ ∈ N ∩ [ σ i ] (cid:22) such that ( ρ, σ ) ∈ B and P ( X, · ) ∩ [ σ ] = ∅ . Therefore, we can enlarge B to a tree T ⊆ T P such that • ( ρ i , σ i ) is good for [ T ] for all i ∈
2; and • T ∩ N × N = B .Since A is sufficient over (( ρ , σ ) , ( ρ , σ )) and A ⊆ (3 N × N ) × (3 N × N ),there exists a ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) ∈ A such that • (˜ ρ i , ˜ σ i ) ∈ B and (˜ ρ i , ˜ σ i ) (cid:23) ( ρ i , σ i ) for all i ∈ • (˜ ρ , ˜ ρ ) does not positively progress on any color j ∈ ρ , ρ ).By definition of B and since ˜ ρ , ˜ ρ ∈ Involve ( A ), we have ( ˆ ext (˜ ρ i ) , ˜ σ i ) is good for P ; by definition of ˆ ext , ( ˆ ext (˜ ρ ) , ˆ ext (˜ ρ )) does not positively progress on any color j ∈ ρ , ρ ) and ˆ ext (˜ ρ i ) (cid:23) ext (˜ ρ i ) for all i ∈
2. Also note that forevery i ∈
2, ( ext (˜ ρ i ) , ˜ σ i ) ∈ T P since ( ˆ ext (˜ ρ i ) , ˜ σ i ) ∈ T P while ˆ ext (˜ ρ i ) (cid:23) ext (˜ ρ i ) forall i ∈
2. Thus, (( ext (˜ ρ ) , ˜ σ ) , ( ext (˜ ρ ) , ˜ σ )) ∈ ext ( A ) ∩ ( T P × T P ) witnesses thesub-sufficient of ext ( A ) ∩ ( T P × T P ) and we are done. (cid:3) We observe the behavior of Ψ on oracles in the condition ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ ) , P, D )and define an Γ m -approximation f as following. Fix an n . Definition 4.20 (Computing f ( n, · )) . We define f ( n, s ) by induction on s .(1) By convention, f ( n,
0) is the singleton ( ∅ , φ ).(2) Suppose we have defined f ( n, s ) = tcp = (( T , φ ) , · · · , ( T u − , φ u − )), aninjective function ψ : T u − ∪ ℓ ( T u − ) → T × . Moreover, for each non leaf ξ ∈ T u − , suppose we have assigned: a set A ξ ⊆ T × ∩ (3 N × N ) × (3 N × N ) for some N , N ∈ ω ; a disjoint preserving extension over (ˇ ρ , ˇ ρ ) ext ξ : Involve ( A ξ ) → ˆ N for some ˆ N , such that let ˜ ξ be the immediatepredecessor of ξ , • A ξ is sufficient over ext ˜ ξ ( ψ ( ξ )); • ψ restricted on T u − ∩ ω | ξ | +1 is a one-one function with the range { τ ∈ A ξ : ext ξ ( τ ) ∈ ( T P [ s ] × T P [ s ]) } ; and ψ ( ⊥ ) = ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ )); • for every ˆ ξ, ξ ∈ T u − ∪ ℓ ( T u − ) , if ˆ ξ ≻ ξ, then ψ ( ˆ ξ ) ≻ ext ˜ ξ ( ψ ( ξ )); • for every ⊥ 6 = ξ ∈ T u − ∪ ℓ ( T u − ) , φ u − ( ξ ) = Ψ ext ˜ ξ ( ψ ( ξ )) ( n ).(3) Define f ( n, s +1) depending on which of the following cases occur: for somenon root ξ ∈ T u − ∪ ℓ ( T u − ), suppose ˜ ξ is the immediate predecessor of ξ and ψ ( ξ ) = (( ρ , σ ) , ( ρ , σ )), ξ ∈ ℓ ( T u − ) and it is found at this time that the set(4.8) (cid:8) τ ∈ T × ∩ [ ext ˜ ξ ( ψ ( ξ ))] ≺ : Ψ τ ( n ) ≻ p m − φ u − ( ξ ) (cid:9) is sufficient over ext ˜ ξ ( ψ ( ξ ));for some i ∈ , ( ext ˜ ξ ( ρ i ) , σ i ) is not good for [ T P [ s + 1]];(4.9) otherwise . (4.10)(4) • In case of (4.8), by compactness, let A ξ ⊆ ((3 N × N ) × (3 N × N )) ∩ T × be a witness for the sufficient over ext ˜ ξ ( ψ ( ξ )) of that set.Let ext ξ : Involve ( A ξ ) → N be an identity function. We add a In case ξ = ⊥ , ext ˜ ξ ( ψ ( ξ )) = ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ )) ODING POWER OF PRODUCT OF PARTITIONS 23 set B ⊆ ω | ξ | +1 ∩ [ ξ ] (cid:22) of size | A ξ | to T u − getting T u ; then define ψ on T u by inheriting ψ and 1-1 mapping B to A ξ ; finally, define φ u by inheriting φ u − and φ u ( ˆ ξ ) = Ψ ext ξ ( ψ (ˆ ξ )) ( n ) for all ˆ ξ ∈ B . Let f ( n, s + 1) = tcp a ( T u , φ u ). Clearly f ( n, s + 1) is a tree-computationpath in Γ m − since T u is a one step variation of T u − (fulfilling theeither case). The inductive hypothesis on ψ is easily verified. • In case of (4.5), suppose ξ is the minimal string witness (4.5), i.e., noproper initial segment of ξ is a witness. Without loss of generality,suppose for some ˆ ρ (cid:23) ext ˜ ξ ( ρ ), (ˆ ρ, σ ) / ∈ T P [ s + 1]. We will replace ext ˜ ξ by ˆ ext which extends the value of the ext ˜ ξ so that ˆ ρ is now inthe range of ˆ ext . Therefore, for ˆ ξ ∈ T u − ∩ ω | ξ | , if ext ˜ ξ ( ψ ( ˆ ξ )) involves( ρ , σ ), then ψ ( ˆ ξ ) has no extension in ˆ ext ( A ˜ ξ ) ∩ ( T P [ s +1] × T P [ s +1]).Such ˆ ξ will be deleted from T u − ∩ ω | ξ | . On the other hand, A ˜ ξ remains.More specifically, by Lemma 4.3, there is a disjoint preserving exten-sion ˆ ext : Involve ( A ˜ ξ ) → | ˆ ρ | over (ˇ ρ , ˇ ρ ) such that ˆ ext ( ρ ) = ˆ ρ andˆ ext ( ρ ) (cid:23) ext ˜ ξ ( ρ ) for all ρ ∈ Involve ( A ˜ ξ ). Let B ⊆ T u − ∩ ω | ξ | be theset of ˆ ξ such that ψ ( ˆ ξ ) does not involve ( ρ , σ ). Clearly B is a propersubset of T u − ∩ ω | ξ | since ξ / ∈ B . The point is B = ∅ since by Claim 4.19 ˆ ext ( ψ ( B )) is sub-sufficient.Thus, let T u = ( T u − \ [ ˜ ξ ] ≺ ) ∪ B , we have T u is a one step variationof T u − (fulfilling the or case). Let ˆ ψ ↾ T u = ψ ↾ T u , φ u = φ u − ↾ T u .Let f ( n, s + 1) = tcp a ( T u , φ u ). The other part of inductive hypothesis(with respect to ˆ ψ, ˆ ext ) is easily verified. • In case of (4.6), let f ( n, s + 1) = f ( n, s ).As we argued in item (4) of the definition of f ( n, · ), we have f is a Γ m -approximation.By compactness, f is computable. Since, C is Γ-hyperimmune, there are n ∗ , s ∗ suchthat • C diagonal against f ( n ∗ , s ∗ ) and • f ( n ∗ , t ) = f ( n ∗ , s ∗ ) for all t ≥ s ∗ .Suppose f ( n ∗ , s ∗ ) = (( T , φ ) , · · · , ( T u − , φ u − )). By definition of diagonal against,there is a ξ ∈ ℓ ( T u − ) such that C diagonal against φ u − ( ξ ). Consider ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ )) = ext ˜ ξ ( ψ ( ξ )) where ˜ ξ is the immediate predecessor of ξ ; and the Π class T of trees T such that: • T ⊆ T P ∩ ([(ˆ ρ , ˆ σ )] (cid:22) ∪ [(ˆ ρ , ˆ σ )] (cid:22) ) and (ˆ ρ i , ˆ σ i ) is good for [ T ] for all i ∈ • for every (˜ ρ i , ˜ σ i ) ∈ T ∩ [(ˆ ρ i , ˆ σ i )] (cid:22) , if (˜ ρ , ˜ ρ ) does not positively progress onany color j ∈ ρ , ˆ ρ ), then Ψ ((˜ ρ , ˜ σ ) , (˜ ρ , ˜ σ )) ( n ∗ ) = φ u − ( ξ ).Since case f ( n ∗ , t ) = f ( n ∗ , s ∗ ) for all t ≥ s ∗ , in particular, case (4.8) does notoccur after step s ∗ , therefore the class T is non empty (check the definition ofsufficient). Since T is a Π class, by Lemma 4.14, let ˆ T ∈ T be such that C isΓ-hyperimmune relative to ˆ T . Since ψ ( ξ ) ∈ A ˜ ξ ⊆ T × and ext ˜ ξ is disjoint preserveextension over (ˇ ρ , ˇ ρ ), we have (ˆ ρ , ˆ ρ ) does not positively progress on any color Note that Ψ ˆ ext ( ˆ ψ (ˆ ξ )) ( n ) = Ψ ext ˜ ξ ( ψ (ˆ ξ )) for all ˆ ξ ∈ B . j ∈ ρ , ˇ ρ ). Combine with (ˆ ρ i , ˆ σ i ) being good for [ ˆ T ] for all i ∈ ρ , ˆ σ ) , (ˆ ρ , ˆ σ ) , [ ˆ T ] , ˆ T ) is a condition extending ((ˇ ρ , ˇ σ ) , (ˇ ρ , ˇ σ ) , P, D ).By definition of T , ((ˆ ρ , ˆ σ ) , (ˆ ρ , ˆ σ ) , [ ˆ T ] , ˆ T ) forces the requirement R Ψ . (cid:3) Now the rest of the proof follows exactly as that of 4.5. (cid:3)
Yet some other basis theorem.
Note that if two colorings are almost dis-joint, then they are Turing equivalent. Therefore, if we replace “ Y , Y are notalmost disjoint” in Lemma 4.5 by Y , Y are not Turing equivalent, then we arriveat the following question. Question . Given two incomputable Turing degree D (cid:3) T D , a non empty Π class Q ⊆ ω , does there exist a X ∈ Q such that X (cid:3) T D and D ⊕ X (cid:3) T D ?Clearly the difficulty of question 4.21 is that when forcing some fact about D ⊕ X , the condition we use is usually a Π ,D class. But there is no guarantee that aΠ ,D class contains a member not computing D .5. Product of infinitely many Ramsey’s theorem
Note that ( RT ) ω is capable of encoding any hyperarithematic degree since it iscapable of encoding fast growing function. Proposition 5.1.
For every function f ∈ ω ω , there is a ( RT ) ω instance ( C , C , · · · ) such that every solution of ( C , C , · · · ) compute a function g such that g ≥ f .Proof. Fix f ∈ ω ω . Let C n ∈ ω be such that C − (0) = [0 , f ( n )] ∩ ω . Let( G , G , · · · ) be a solution to ( C , C , · · · ). Clearly G n ⊆ C − n (1) for all n ∈ ω .Thus clearly ( G , G , · · · ) compute a function g ≥ f . (cid:3) On the other hand, strong cone avoidance of non hyperarithmetic degree for( RT ) ω follows from the following Solovay’s theorem [18]. Theorem 5.2 (Solovay) . For every non hyperarithmetic Turing degree D , thereexists an infinite set X such that none of the subset of X computes D . Proposition 5.3.
Given a non hyperarithmetic Turing degree D , every ( RT ) ω instance admit a solution that does not compute D .Proof. Fix a ( RT ) ω instance ( C , C , · · · ). Let X be an infinite set as in Solovay’stheorem. Let G = { n < n < · · · } be an infinite subset of X such that for every t ∈ ω , G ∩ [ n t , ∞ ) is monochromatic for C t . Clearly G computes a solution to( C , C , · · · ) but G does not compute D . (cid:3) We wonder if ( RT ) ω is capable of encoding RT . Question . Is it true that RT ≤ soc ( RT ) ω ? ODING POWER OF PRODUCT OF PARTITIONS 25
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Department of Mathematics, Central South University, City Changsha, Hunan Province,China. 410083
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