- Email: [email protected]
Incomparable prime ideals of recursively enumerable degrees
Annals of Pure and Applied North-Holland
Logic 63 (1993) 39-56
39
Incomparable prime ideals of recursively enumerable degrees William C. Calhoun* Department
of Mathematics,
Kalamazoo
College,
Kalamazoo,
MI 49006, USA
Communicated by A. Nerode Received 15 June 1991 Revised 18 November 1992
Abstract Calhoun, W.C., Incomparable prime and Applied Logic 63 (1993) 39-56. We show that there is a countably degrees. This solves a generalized
ideals of recursively
enumerable
infinite antichain of prime form of Post’s problem.
degrees,
Annals
ideals of recursively
of Pure
enumerable
1. Background Recursion theory may be broadly divided into two parts: degree theory and applied recursion theory. Applied recursion theory includes both effective mathematics and undecidability results such as the unsolvability of Hilbert’s 10th problem. This paper concerns degree theory. Simply put, the aim of degree theory is to understand the degree structures that arise naturally from various notions of relative computability. In particular, our focus here will be (R, c), the Turng degrees of recursively enumerable sets (r.e. degrees) ordered by Turing reducibility. The r.e. degrees are a natural object to study since they are ‘almost computable’ in the sense that they can be effectively generated. Turing reducibility is the most general kind of relative computability. In addition to being a natural object to study, the structure of (R, S) has turned out to be fascinatingly complicated and notoriously difficult to uncover. The main tool for investigating the r.e. degrees is the priority argument. Priority arguments were invented by Friedberg [6] and Muchnik [14] to solve Post’s problem [15]. Friedberg and Muchnik showed that there are incomparable Correspondence to: W.C. Calhoun, Department of Mathematics, Kalamazoo College, Kalamazoo, MI 49006, USA. *This paper is based on a section of the author’s Ph.D. thesis, written at the University of California at Berkeley. The author would like to thank his thesis advisor, Leo Harrington, for his invaluable help. 0168-0072/93/$06.00
@ 1993-
Elsevier
Science
Publishers
B.V. All rights reserved
40
W. C. Calhoun
r.e. degrees, thus affirmatively answering Post’s question of whether there are r.e. degrees strictly between 0, the degree of the recursive sets, and 0’, the degree of the halting problem. The difficulty in a priority argument is to design a construction in which r.e. sets are effectively generated while ensuring that the sets satisfy highly noneffective requirements. For instance, in the FriedbergMuchnik theorem, sets A and B must be constructed so that A and B are recursively enumerable and such that A is incomparable with B. The latter requirement is highly noneffective. In fact, the set {(i, i): w is incomparable with y} has degree 04. Friedberg and Muchnik solved the problem by breaking up the requirement “A is incomparable with B" into a countably infinite list of subrequirements of the form @ # B and @f #A. Since there are countably many subrequirements, they may be prioritized by choosing some effective w-ordering of them. At each stage of the construction an attempt is made to satisfy the strongest priority requirement that is not yet satisfied. Attempts to satisfy a stronger priority requirement may ‘injure’ weaker priority requirements. In Friedberg and Muchnik’s construction, no requirement is injured more than finitely many times, so each requirement is eventually satisfied. This type of construction is called a finite injury priority argument, or a 0’ priority argument, since a 0’ oracle is needed to determine how each requirement is finally satisfied. A more delicate kind of construction, the infinite injury, or 0” priority argument was developed by Sacks [16] and Shoenfield [19]. Sacks [17] used a 0” priority argument to show that (R, S) is dense as a partial ordering. This led Shoenfield [20] to conjecture that (R, =s) is dense as an upper semi-lattice. However this was refuted by Lachlan [lo] and Yates [25] who showed that some pairs of incomparable degrees in (R, S) have infima. In particular they showed that there is a pair a, b such that a A b = 0. Such a pair is called a minimal pair. The existence of minimal pairs, which contradicts Shoenfield’s conjecture, reveals that the structure of (R, S) is more complicated than Shoenfield had predicted. An even more difficult type of construction, the 0”’ priority argument, was invented by Lachlan [12] and used by him to show additional complexity of (R, s). Harrington found ways to simplify Lachlan’s method, which had been Harrington’s methods also have the something of a technical tour-de-force. advantage of giving a unified approach to 0” priority arguments for all IZE w (and even to 0” priority arguments for ordinals a 3 0). We will present a version of Harrington’s approach in Section 5. Harrington and Shelah [7] used a 0”’ priority argument to show that the first-order theory of (R, S) is undecidable. This was later extended by Harrington and Slaman to show that deg(theory of (R, s)) is O”, the degree of the set of true sentences in the language of first-order arithmetic. (Due to the complexity of the proofs, full proofs have not been published. Recently, simplified proofs have been found by Slaman and Woodin [22] and by Ambos-Spies and Shore [2].) Thus the interaction between degree theory and applied recursion theory has gone in both directions. Degree theory has given us a better understanding of the
Incomparable
prime ideals of r.e. degrees
41
relationships between the nonrecursive sets which arise in applied recursion theory. In the other direction, we have undecidability results about the degrees themselves.
2. Algebraic structure Although the results of Harrington and Shelah and of Harrington and Slaman show that the first order theory of (R, =s) is undecidable, and is in fact as complicated as the set of true sentences of first-order arithmetic, recursion theorists continue to try to understand the structure as best we can. Of course, most mathematicians strive to obtain deep, although necessarily incomplete, understanding of undecidable theories. Number theorists, for instance, are undeterred by the fact that the degree of true arithmetic is 0”. So our position is in no way unusual. One approach is to consider the algebraic structure of (R, G). It is easy to see that (R, S) is an upper semi-lattice since if a = deg(A) and b = deg(B) then aU b = deg(A Cl3B) where A @B is the disjoint union of A and B. However, (R, C) is not a lattice. As mentioned above, Lachlan and Yates showed that some pairs of r.e. degrees have an infimum, but others do not. Thus we consider the algebraic structure (R, C, v , A, 0, l), where v is the join operation and A is the meet partial operation. We may define an ideal as in lattice theory. Definition 2.1. An ideal I, is a nonempty subset which is closed downward and under join, that is, if a, b E I then a v b E I and if c c a then c E I. Since the infimum of two degrees may not exist, the appropriate generalization for an upper semi-lattice of the notion of a filter in lattice theory is the strong filter. Definition 2.2. A strong jilter, F, is a nonempty subset which is closed upward and contains a lower bound for each pair from F, that is, if a, b E F then (3cEF)(cSa,b) and if d?=a then dEF. There have been some interesting recent theorems concerning ideals of r.e. degrees. Shore [21] proved a noninversion theorem for jump, which implies that jump inversion fails for ideals, if we take the jump of an ideal to be its image under the jump operator. An investigation of the structure of the set of ideals of r.e. degrees was begun in the author’s Ph.D. thesis [5]. The main result of this paper, and another interesting recent result described below, concern prime ideals of (R, s).
W. C. Calhoun
42
Definition 2.3. A prime ideal is an ideal whose complement
is a strong filter.
Since the structure of (R, S) has proven to be so complicated, it was somewhat surprising when Ambos-Spies, Jockusch, Shore and Soare [l] showed that the well studied subset M of R is a prime ideal. A degree a is said to be half of a minimal pair if there is a degree b such that a, b is a minimal pair. The set M consists of all halves of minimal pairs.
3. Algebraic strengthening
of the Friedberg-Muchnik
theorem
We are now ready to state the main theorem of this paper, a solution to Post’s problem generalized to the set of prime ideals of (R, s). Theorem 3.1. There are incomparable prime ideals of R. (In fact, countably infinite antichain of prime ideals of R.)
there is a
As a corollary we get a strengthening of Friedberg and Muchnik’s theorem showing that there are r.e. degrees which are incomparable in a strong algebraic sense. Definition
3.2. The radical of an r.e. degree ideals containing a.
a is the intersection
of all prime
Corollary 3.3. There are r.e. degrees a and b whose radicals are incomparable. (In fact, there is a countable sequence of r.e. degrees whose radicals form an injinite antichain.) Proof. Let 1 and J be incomparable
prime ideals. Let a E I\J, and b E J\I. The second statement follows in a similar way from Theorem 3.1. 0 To see how this corollary algebraically strengthens the Friedberg-Muchnik theorem, we describe one consequence of the corollary involving the 1-3-1 or MS lattice (see Fig. 1). 1
@ 0 Fig. 1. The 1-3-1 or M, lattice.
Incomparable prime ideals of r.e. degrees
Fig. 2. Two r.e. degrees
with incomparable
radicals.
The 1-3-1 lattice was shown to be embeddable in R by Lachlan [ll]. Since then embeddings of the 1-3-1 have been studied extensively. For our purposes, the important observation is that if f : M5 + R is an embedding and f(0) E I, where I is a prime ideal, then f(l) E I. Thus if a, b are as in the corollary and f(0) c a then f(0) is in the radical of a, and hence f(1) is also in the radical of a. Therefore f(1) j b, since otherwise b is in the radical of a which contradicts the corollary. The same reasoning shows that if fi, f2, . . . , fn are embeddings of MS into R,f,(O) 4 a, and J(O) CA-r(1) for i > 1, then &(l) 3 b. So no ‘stack’ of 1-3-1’s starting below a can lead to a degree greater than or equal to b. (See Fig. 2.) Of course the same statement is true if we reverse the roles of a and b.
4. Notation Our notation will be fairly standard, similar to that in Soare [23]. Light face lower case Roman letters usually represent integers. Light face upper case Roman letters represent sets of integers, usually r.e. sets. Bold face lower case Roman letters represent degrees, usually r.e. degrees. Bold face upper case Roman letters represent sets of r.e. degrees, usually ideals of r.e. degrees. A string is a member of o<“‘. Lower case Greek letters are used for strings and finite unions of strings. We identify sets of integers with their characteristic functions so that it makes sense to say CYc A, for instance. We hope our typographical conventions will prevent a possible confusion here. When we say
W. C. Calhoun
44
a E B we mean the characteristic
function
of B extends
a, but when we say
A c B we mean A is a subset of B. 4.1. For A = U A,, to make the restraint LYon A at stage s means LYc_A, and we require that a c A, for all t > s such that the restraint is in force at t. If we wish the restraint to cease to be in force at some stage t we will say that we drop the restraint at t. Definition
functions) A, B, A 63 B is the disjoint union of A and B pairing function). To make the restraint a: CBp on A CI3B is the restraints a on A and p on B. In general, when we we mean a restraint of the form p = a, Cl3a2 Cl3- . - CBa;, on where the Ai are finitely many of the sets being A,@A,$--.@A,, constructed, and oi is a restraint on Ai. Upper case Greek letters are used for reductions, which we now define. For sets (or partial (coded by an effective the same as to make speak of a restraint p,
Qi is an r.e. set of pairs ((u, p) such that if en LY, # c+, and if (Y,c LX~then /3, c &. For a reduction (au,, P,), (a*, PJ E @ th @ and a set of integers A, 0“ = IJ {/?: (3a)(a, j3) E bi and (YCA}. Notice that ti may be total or partial. That is, its domain may be o or merely an integer. For a possibly partial function Y, we say Y(x) converges, and write Y(x)J, if x is in the domain of W, otherwise we say that Y(X) diverges, and write Y(X)?.
Definition
4.2. A reduction
Definition 4.3. If @(x)l,
the use of this computation
is the least a c A such that
@%)L For a string /3, we write @(/3)1 if /I c @. The use of this computation is the union of the uses for all x in the domain of /3. All r.e. sets (including reductions) are constructed in stages. The finite approximation at stage s is indicated by the subscript S. Thus we have A,, as,, etc. When an expression in parentheses is subscripted, all sets within the parentheses are taken to be subscripted. For example, (@3x)), means @$(x). We assume standard enumerations @Cand W, of the partial recursive functions and r.e. sets. To simplify the notation we will usually write @ and W, taking it as understood that we, in fact, mean not just the partial recursive function or r.e. set, but also a particular index e for @ or W.
5. Priority arguments
We will use a formulation of priority arguments modeled on that of Harrington. We give our own presentation of this method, but the ideas are his. For another presentation see Weinstein [24]. A related framework for priority
Incomparable
prime ideals of r.e. degrees
45
arguments has been proposed by Lempp and Lerman [13]. Also, Knight [8,9] has given a framework for constructions in recursive model theory that is inspired by Harrington’s method as well as work of Ash [3,4]. We assume that the reader is familiar with priority arguments as described in Soare [23], for instance. In a priority argument
Harrington’s
formulation,
many
for some ordinal
levels,
is a construction
CY.We will only treat
that
the case that
involves
(Y is finite.
Q A
level n argument is one which uses levels 0, 1,2, . . . , II. We begin ordering.
by describing
level 1 arguments.
Level 0 is a recursive
of members
of P,. In this paper,
Let (P,,, co) be a recursive
partial
sequence
P;, will consist
of the potential
finite approxima-
tions to the r.e. set (or sets) we wish to construct (which we will represent by strings), ordered by A 2 B iff A c B (as sets, not as strings). We will often call the subscript t of A, a level 0 stage. Level 1 is a recursive in 0’ sequence level 0. We will often call the A, = S,, S,, &, . . . of level 1 strategies controlling subscript s of S, a level 1 stage. We think of stages representing a progression of time and use terminology like ‘wait for * - . ’ to mean ‘at the least stage such that . . . ‘. A Level 1 strategy is a recursive function S : P;“-+ 8,) where 8, is the set of allowed level 1 environments for the construction. A level 1 environment is a nonempty recursive subset of PO, i.e. potential entries for level 0. Particular sets Y,, Z, of allowed level 1 strategies and allowed level 1 environments are chosen for a particular construction in such a way that Yi, %‘, and the inclusion relation on %‘,are recursive. The intention is that if S(& It) = E then S ‘wants’ A, to be chosen so that A, E E. For any A E p0, the level 1 environment EA = {B: B GA} is always allowed. So the environment E*,_, which allows all possible extensions of A,_, to appear is allowed. In fact, since we must have that A,
46
W. C. Calhoun
Proposition 5.1. For all sets A and B, and functions f recursive in A and g recursive in B, there are integers a and b such that c&’ = C$Ca,bjand @,S = cD$~,~). Proof. By the relativized Recursion Theorem
function function
with parameters there is a recursive G(y) such that for all y, @&Bo,j,Yj= CD&,,. Let h be the B-recursive
h(y) = g@(y), Y). Then there is a b such that @&,, = @f. Let a = B(b). Then e @6”= @&b).
= 4&,b,
and
0
The proposition tells us that in constructing level 0 and level 1 we may assume that we have indices for & and A,. Thus on level 1 we can know what happens at any level 0 stage and even use 0’ to see if Ai properties hold on level 0. In building level 1, we will informally say that we ‘ask 0” or simply that we ‘ask’ if some Ai property holds on level 0. This means that the use of 0’ in computing that level 1 stage is sufficiently large to answer this question. We need to use some caution in this regard, however. We cannot assume that & and A, are total. Thus, on level 1 we cannot ask questions about a level 1 stage s for which we have not yet specified S,. (We might get no answer because we wait forever for an answer, and therefore never specify S,.) However, we will construct level 0 so that & is total even if A, is not. Thus, we can safely ask A’: questions about level 0 on level 1 as described above. This type of consideration implies that for any level n argument, questions asked in order to decide what strategy should appear at stage s on level m must be about level m stages
Incomparable prime ideals of r.e. degrees
47
level 1 strategy for every i, and Al respects priority, then for every level 1 stage s there is a level 0 stage t, such that for all level 0 stages t > t,, S, is respected at t. Proof. By the limit lemma (see Soare [23, p. 571) there is a recursive double sequence S,,, such that lim, S,,, = S, for all s. At level 0 stage t let s, 6 t be largest so that S,,, E 9’i for all s G s, and &J/i, It) I> .!$+,,,(A, It) for all s
The fact that this construction always produces a total &, even if A, is partial, means that the & given by the Recursion Theorem will be total, so level 1 can assume level 0 is total, as mentioned above. In practice we will not explicitly describe how we are building level 0. We will describe how we are building level 1, and leave it as understood that level 0 is built as in the proof of the theorem. Occasionally, it is convenient to take some action on a lower level that is not for the purpose of respecting strategies on the level above. It will be clear that we can do this, and still build the lower level so as to eventually respect the strategies on the level above, as in the theorem. We now turn to the general case of level n arguments. Level m + 1, for m < n, is a recursive in Om+’ sequence A,,, of level m + 1 strategies controlling the construction of the sequence A,,, on level m. The definitions of level m + 1 strategy, level m + 1 environment, A,+I respects priority, etc. are obtained by relativizing the definitions for level 1. The relationship between level m + 1 and level m is simply the relativization of the relationship between level 1 and level 0. The partial ordering (P,, 6, ) on level m, corresponding to (P,,, +,), is given by P,,, = Ym, the allowed level m strategies, and s,,, is the ordering of the allowed level m strategies defined by Sam T * (Va E Pz’?,) S(a) 2 T(o). Notice that since the relation E, 1 E, on level m environments is recursive in Om-‘, the relation S, am S, on level m strategies is recursive in 0”. Thus (P,, G,) is a recursive in 0” partial ordering and the relativization to higher levels can be carried out as we have described. Furthermore, the relativization of Theorem 5.2 shows that for any recursive in 0” sequence of level n strategies, A,, which respects priority, there is a recursive in O”-’ sequence A,_, such that each level IZ strategy is eventually respected on level 12- 1. Then, by the definition of the ordering on level n - 1, A,-1 respects priority, so there is a recursive in O”-’ sequence A,,_, which eventually respects each level it - 1 strategy. Repeating this argument we finally get a recursive sequence & whose construction is controlled by level 1, whose construction is
W. C. Calhoun
0”
s, \
0’
Tt \
Fig. 3. A level 2 argument
controlled by level 2, and so on. (In fact we build levels 0, 1,2, . . . , n simultaneously.) Thus, in any level n argument, the construction of level 0 is ultimately controlled by level n, but as it gets larger this control becomes increasingly indirect (and the arguments, therefore, become increasingly delicate). We mention one particular example of a use of this framework which comes up in many priority arguments. We often wish to have a level 2 strategy S (see Fig. 3) build a reduction @ so that cP’ = B, for some sets A and B. To do so, S first requires that a strategy T appear on level 1 (we will say that S ‘plays’ T on level 1). T ensures that if ((u, 6) E CDand (Yc A then p c B. To ensure this, T does not allow ((u, p) to enter @ at level 0 stage s unless (YG A, and /3 G B,. Then, if at some later stage t, /I $ B,, T requires that a number enter A so that cy+ A. Thus, T insures that the reduction @ is ‘correct’. In addition, S must ensure that CC@ is total. We will say that S ‘protects’ @, meaning that S plays a level 2 environment to ensure that no strategy T’ may appear after T on level 1 which restrains pairs ((u, 0) from entering @ at any level 0 stage s without an ‘excuse’. An excuse is a II’: fact C$about level 0 which is ‘known’ to be true by S. (That is, the use of 0” in constructing level 2 up to S is enough to determine that # is true.) With the excuse $, T’ may restrain all new pairs from entering @. We will say in this case that @ has been ‘made finite’. However, at the first level 0 stage t at which a new witness to the truth of $ appears, the excuse ceases to be valid and T’ must drop its restraint on @. Thus, Qi can only be made finite temporarily. If Qi is protected, @ can be made total by placing strategies T,, T2, . . . on level 1 such that T ensures that @(x)i for all x < i.
6. Proof of the main theorem In this section we prove Theorem 3.1. The proof is a level 4 priority argument in which we build the ideals Ir, 12, . . . with complementary, strong filters, such that A’ E Ii t, i #j. For each i, and each and r.e. sets A’, A’, . . . F,, F,, . . . > r.e. set W,, we must decide if We E Ii or We E F;. ‘Injury’ occurs when an attempt to put an r.e. set We into Ii fails, and switches to an attempt to put We into Fi. In
Incomparable
prime ideals of r.e. degrees
49
the next section, we will discuss a slight modification of the construction which allows the argument to be done on level 3. The prime ideals built by the level 3 construction are A,. This is a better bound on the complexity of the prime ideals we build than we have for the level 4 construction, which we now describe. The argument dovetails arguments for building 1; and Fj for each i. Since the same construction is used for each i, and the interaction between the constructions of the different ideals is slight, we will simplify the notation by describing the argument as if we were only concerned with one fixed i. At the end of the proof we will comment on why there is no difficulty in dovetailing the arguments in order to handle the countably many Ii. To simplify the notation, we will drop the subscript so that for the fixed i, I = 1; and F = Fi. We will also let B = A’ and A = @ jzi A’.
During the construction
we will have finite approximations,
I” and E”, to I and
F. Associated to each pair (I’, E”) is a string q in the tree 2’” defined by
0
q(e)
if W, EP, = 1 if W, EFF, i T otherwise.
For a, 7~~2~“’ U 2” we will always use (TE t to indicate that t is an extension of a, and o i t to indicate that (Tis Turing reducible to t (in the case that a, r E 2”‘). We will use the symbol =Cfor the Kleene-Brouwer ordering on 2’” U 2”. Definition 6.1 (Kleene-Brouwer).
a=Ct iff
(i.e.,
oEror a(e) < r(e) where e is least such that o(e) # r(e)
I
o < r iff o is to the ‘left’ of t).
To make F a strong filter, we will build, for each level 4 stage s, a set P such that F” c W, for all W, in E” U {B}. In addition we define 1’ = @ 1’ C3A. We need to ensure that I” 3 F” for all s. Otherwise there is an ‘obstruction’ (see Fig. 4). Definition 6.2. There is an obstruction such that C@= F.
at level 4 stage s if there is a reduction
CJ~
When there is an obstruction, we will try to ‘go around’ until our path is barred. More precisely, at level 4 stage s we ask “is there an obstruction at stage s - l?” If not, let e be least such that W, 4 F”-’ U I’-‘. Let 1’ = I’-’ U {W,} and F = I’-‘. On the other hand, if there is an obstruction at s - 1 and I’-’ # 0, then let e be largest such that W,EI,_,. Let P=P-‘\{W,} and F”=F”-‘U{W,}\ {y: j > e}. (We are moving to the ‘right’ of the obstruction in 2’“.)
50
W. C. Calhoun
Fig. 4. If I” 2 F” there is an obstruction
at s.
In either case we then play (a strategy on level 3 which plays) a strategy on level 2 which builds (using a strategy on level 1) an r.e. set F” and builds and protects a reduction YS such that (ly,)” = F” for all X E E” U {B}. (To simplify the notation, we are using only one reduction, Y, for all the X’s, by the usual trick: for any (a, p) E YS, cr is long enough to distinguish between the different X’s in F” U {B}.) In this case, since we are not building the X’s, we will ensure the totality of Yx by putting a sequence of strategies along level 1, such that the nth strategy restrains the use of Yx(n) from growing longer once the stronger priority strategies have settled down. Since at each point on level 1 we will only have to respect finitely many such strategies, they will amount to a finite restraint, and will not be noticed in our construction. To complete the description of the construction, we must tell what to do at level 4 stage s if there is an obstruction at s - 1 and IS-’ = 0. We will describe below a level 1 strategy, S, to be employed by level 4 in this case. The strategy S produces a contradiction to the assumption that this case has arisen, thus our intention of employing S if necessary guarantees that this case will not arise on level 4. It follows that lim P and lim F” exists, and if we let I = lim P and F = lim F, then I is an ideal with complement F, which is a strong filter. That is, I is a prime ideal. Before we describe S, note that we need only be concerned that S respect level 2 strategies T such that T was played by level 4 at some stage t
Incomparable
prime ideals of r.e. degrees
51
F* and finite collection of r.e. sets F*, Yx = F* for all X E F* U {B}. Since F* is not one of F’, F2, . . . , FSwl, S will have no need to put numbers in F*, and S can respect T simply by restraining F*. Thus, although we are using 0C4)to build I, the r.e. sets A, B, and F are essentially built by the level 1 construction. So, most of the ‘action’ in the proof takes place in the finite injury argument on level 1. (Although, for the proof to succeed, we need that the level 1 and level 2 strategies have the ‘knowledge’ given them by level 4.) We now describe the strategy S to be used in the case that there is an obstruction at s - 1 and I’-’ = 0. In this case our path through 2<” is ‘barred’. To make this precise, we make some definitions. Definition 6.3. A bar, 2, is an antichain of strings such that for every
f E2”
(3crE.z)f 30. Note that every bar is finite, since 2”’ is compact and a bar corresponds minimal basic open cover of 2”.
to a
Definition 6.4. There is a bar at s if the set
_Y= {at: t
at t}
is a bar. (See Fig. 5.) Lemma s.
6.5.
If there is an obstruction
at s - 1, and P-’ = 0, then there is a bar at
Proof. Let 2 be as given in the definition of a bar at s. We must show that 2 is a bar. Let f E2”. Let t be the largest stage at which a,O). Also, there must be an obstruction at t, since there is an obstruction at s - 1, and
w4
w3
10
W2
w
wo Fig. 5. Obstructions
at 1,5,7,8,9
and 10 form a bar.
W. C. Calhoun
52
f
:.
Fig. 6. The situation
in Lemma
6.5.
if t
We will call x, x,, , xt2, . . . , x,, an s-set up at level Ostuge v if (i)-(v) above hold when F’, I’, cPl, Yr, B, and X are all replaced by their finite approximations at v. When we mean an s-set up as written, we will emphasize this by calling it a permanent s-set up. We need to know that we will have s-set ups to use.
Incomparable prime ideals of r.e. degrees
53
Lemma 6.7 (Window Property). For any r, s E w, if there is a bar at level 4 stage s, then there is a permanent s-set up x, xl,, x,*, . . . , x,,, such that x > r and x,, > r for all j. Proof. We need only require that the sets F’ we are building are coinfinite, and that the reductions Y, we are building satisfy the usual convention that if length(use( Y:(y))) 2 y for all y and (a; /3) E ‘y,, then length(a) 2 length(P), i.e., . Y. We ensure the latter by simply making it a condition on the construction of Y. To ensure the former, we put a sequence of strategies along level 1 (at every other level 1 stage, for instance) such that the nth strategy ensures that there are n numbers not in F’. Since the effect of the stronger level 1 strategies settles down to a finite restraint, each such strategy can succeed simply by restraining one more number from F. It is now easy to see that the window property holds. Let x be any number such that x > r and x 4 B. Then choose x,,, for i = 1, 2, . . . , k, so that xg > r, x,, C$F5, and use(@c(x,)) /X E use( Yf(x,)) for all i
Now we describe the level 1 strategy, S, which level 4 plays to produce contradiction in the case that there is a bar, 2, at s.
a
Step 1. S waits for a level 0 stage at which there is an s-set up x, xrl, xfz, . . . , xq such that none of the numbers x, xI,, x,*, . . . , xg is restrained, by the finite restraint imposed by level 1 strategies stronger than S, from entering its target set. Step 2. S restrains use( @(x,)) IA, and puts x into B. Subsequent steps. S puts in x,, , xu2, . . . , x,,, for some ul < u2 < . * . < I.+, where uj E T for all i. After putting in xq, S waits until @am before going on to xy+,. During this process, if Yf(x,) becomes undefined for some t E T and X E F’ U {B}, then S keeps the computation undefined until S gets x, into F” for some l4 z=t. S can do this, because if J’< k then Ytk is built with the knowledge that @,‘= F5. Thus, ‘yk can see that S will wait only finitely long for @$‘(xj)T once x5 E F?. From this it will be clear, once we finish describing S, that S eventually gets x, into F” for some u 2 tk. Thus, ‘yk computations can be kept undefined with the excuse that we are waiting for S to get x, into F”, for some u 2 tk. Now we describe inductively what S does in steps 3,4, . . . . Assume that S is to the point of putting x, into F’. Assume that
Yf(xJT
for all u 2 t and all X E F’ U {B}.
(1)
W. C. Calhoun
54
Notice that (1) is true at the end of step 2 above, for t = t,, since F’I = 0, and we have that Y$U)t for all u E T (i.e., we have a base for induction). Since (1) holds, S is free to put x, into F’. If t = tk, we will see below that we have reached a contradiction, so assume t < tk. Then, S waits until use( #(x,)) has changed from the value it had in step 1. (This must happen since otherwise @%t) f F’(G) Since use( @(x,)) rA is restrained, there must be some e such that W, E I’ and use(@:‘(x,)) rW, has changed. Fix such an e. Then WF(x,)T for all u E T such that u > t and We E F”. Consider f E 2” defined by 1
if o,(i) = 1 or i = e,
f(i) = { 0 otherwise. There must Furthermore, inductively.
be some
u E T such that a,, cf. Clearly, o, < a,, , so t < ~1. Hence, (1) holds for u, and we can continue
F” G E” U {W,}.
Thus eventually S puts x,, into F’*. We then reach a contradiction, since we have 1’~= 0, and so Z’k= A. Since the relevant use of A was restrained, we now have @~T(x~,)# F’k(xJ (and this computation will never be corrected since we maintain our restraint on A). This is the desired contradiction. We close the proof with a comment on why we can use the construction described above, for building one prime ideal, to build countably many prime ideals Ii, Iz, 13, . . . . On level 4 we simply dovetail strategies like those described above for each of the I;. Since these level 4 strategies do nothing but play level 2 strategies and level 1 strategies, a conflict could only arise on level 2 or level 1. But if i #j, there is no significant interaction between a strategy S associated with Ii and a strategy T associated with 11. If S and T are level 1 strategies, then they settle down to finite restraints. If S is a level 2 strategy building a reduction Y to compute F;, S has no effect on T, which is concerned with sets F;, F;, . . . , Fj. Thus, if S and T are level 1 or level 2 strategies associated with different ideals, the only connection between S and T is that all strategies control the construction of A’, A*, . . . . However, any effect of S (or T) on the Ak is included in the finite restraint on level 1. In fact, at most, the effect of S is to put one number into A’ and restrain BjziAi. Thus no conflicts will arise. (It is no worse than building an infinite antichain of r.e. degrees.)
7. Reducing the complexity,
a 0”’ construction
A recursion theorist would be remiss not to comment on the complexity of the objects constructed. As mentioned in the previous section, the complexity of the prime ideals built by this construction is not the lowest possible. We will discuss a modification of the construction that builds prime ideals of lower complexity, and we pose two interesting questions concerning the complexity of prime ideals. We will use the following definition of the complexity of an ideal.
Incomparable prime ideals of r.e. degrees
55
For any ideal I, let G(1) = {e: deg(W,) E I}. We say that I is _Z,,(QJ if G(1) is ,Yi (L?$. We say I is A, if I is Z;, and fl,. Definition
7.1.
Examining the proof, we see that I = lim P, where G(P) is recursive in OC4). Thus I is &. But F is also X6, so I is A+ However, we could have used a slightly more complicated construction on level 3, which we now describe. Our oracle is now only o”‘, so we cannot ask, at stage s, if there is an obstruction at stage s - 1. Instead of getting the answer to this question immediately at stage s, we will have go on with the construction. It will only be at a later stage that we see that there is an obstruction at s - 1. We break up the _Z”,question of whether there is an obstruction at s - 1 into the infinite sequence of questions “Does @ witness an obstruction at s - l?“, for each reduction @. These questions are rr(,), so 0”’ can answer them. Thus, if there is an obstruction at s - 1, we will eventually discover this fact at some level 3 stage t 2 s. Then, we set I’ and F’ as we would have at stage s in the level 4 construction. At any level 3 stage u 2 t, the strategies at stages s, s + 1, . . . , t - 1 are known to be ‘mistakes’, and can be disregarded since the sets they build will not be used in the rest of the construction. With these changes, the proof works as before. By analyzing this level 3 construction in the same way we analyzed the level 4 construction above, we see that the prime ideals built are A5. There does not seem to be any easy modification of our construction that would further lower the complexity of the prime ideals we build. However, there is a 17, prime ideal. The prime ideal M is L14, since M = R\PS, and, by inspecting the definition, we see that to be of promptly simple degree is a Z4 property. This raises the question: is there an infinite antichain of IL, prime ideals? On the other hand, M is not Z4, in fact M is Lr4-complete, as was shown by Schwarz [18]. This raises another question: is there a _Z4prime ideal?
References [l] K. Ambos-Spies, C.G. Jockusch Jr., R.A. Shore and R.I. Soare, An algebraic decomposition of the recursively enumerable degrees, Trans. Amer. Math. Sot. 281 (1984) 109-128. [2] K. Ambos-Spies and R.A. Shore, l-Types and undecidability in the r.e. degrees, to appear. [3] C.J. Ash, Recursive labelling systems and stability of recursive structures in hyperarithmetical degrees, Trans. Amer. Math. Sot. 298 (1986) 497-514; corrections, ibid., 310 (1988) 851. [4] C.J. Ash, Labelling systems and r.e. structures, Ann. Pure Appl. Logic 47 (1990), 99-119. [5] W.C. Calhoun, The Lattice of Ideals of Recursively Enumerable Degrees, Ph.D. Dissertation, University of California at Berkeley, 1990. [6] R.M. Friedberg, Two recursively enumerable sets of incomparable degrees of unsolvability, Proc. Natl. Acad. Sci. USA 43 (1957) 236-238. [7] L. Harrington and S. Shelah, The undecidability of the recursively enumerable degrees (research announcement), Bull. Amer. Math. Sot. (N.S.) 6 (1) (1982) 79-80. [8] J.F. Knight, A metatheorem for constructions by finitely many workers, J. Symbolic Logic 55 (1990) 787-804. (91 J.F. Knight, Constructions by transfinitely many workers, Preprint.
W. C. Calhoun
56 [lo]
A.H.
Lachlan,
Lower
bounds
for pairs of recursively
enumerable
degrees,
Proc.
Logic 63 (1993) 39-56
39
Incomparable prime ideals of recursively enumerable degrees William C. Calhoun* Department
of Mathematics,
Kalamazoo
College,
Kalamazoo,
MI 49006, USA
Communicated by A. Nerode Received 15 June 1991 Revised 18 November 1992
Abstract Calhoun, W.C., Incomparable prime and Applied Logic 63 (1993) 39-56. We show that there is a countably degrees. This solves a generalized
ideals of recursively
enumerable
infinite antichain of prime form of Post’s problem.
degrees,
Annals
ideals of recursively
of Pure
enumerable
1. Background Recursion theory may be broadly divided into two parts: degree theory and applied recursion theory. Applied recursion theory includes both effective mathematics and undecidability results such as the unsolvability of Hilbert’s 10th problem. This paper concerns degree theory. Simply put, the aim of degree theory is to understand the degree structures that arise naturally from various notions of relative computability. In particular, our focus here will be (R, c), the Turng degrees of recursively enumerable sets (r.e. degrees) ordered by Turing reducibility. The r.e. degrees are a natural object to study since they are ‘almost computable’ in the sense that they can be effectively generated. Turing reducibility is the most general kind of relative computability. In addition to being a natural object to study, the structure of (R, S) has turned out to be fascinatingly complicated and notoriously difficult to uncover. The main tool for investigating the r.e. degrees is the priority argument. Priority arguments were invented by Friedberg [6] and Muchnik [14] to solve Post’s problem [15]. Friedberg and Muchnik showed that there are incomparable Correspondence to: W.C. Calhoun, Department of Mathematics, Kalamazoo College, Kalamazoo, MI 49006, USA. *This paper is based on a section of the author’s Ph.D. thesis, written at the University of California at Berkeley. The author would like to thank his thesis advisor, Leo Harrington, for his invaluable help. 0168-0072/93/$06.00
@ 1993-
Elsevier
Science
Publishers
B.V. All rights reserved
40
W. C. Calhoun
r.e. degrees, thus affirmatively answering Post’s question of whether there are r.e. degrees strictly between 0, the degree of the recursive sets, and 0’, the degree of the halting problem. The difficulty in a priority argument is to design a construction in which r.e. sets are effectively generated while ensuring that the sets satisfy highly noneffective requirements. For instance, in the FriedbergMuchnik theorem, sets A and B must be constructed so that A and B are recursively enumerable and such that A is incomparable with B. The latter requirement is highly noneffective. In fact, the set {(i, i): w is incomparable with y} has degree 04. Friedberg and Muchnik solved the problem by breaking up the requirement “A is incomparable with B" into a countably infinite list of subrequirements of the form @ # B and @f #A. Since there are countably many subrequirements, they may be prioritized by choosing some effective w-ordering of them. At each stage of the construction an attempt is made to satisfy the strongest priority requirement that is not yet satisfied. Attempts to satisfy a stronger priority requirement may ‘injure’ weaker priority requirements. In Friedberg and Muchnik’s construction, no requirement is injured more than finitely many times, so each requirement is eventually satisfied. This type of construction is called a finite injury priority argument, or a 0’ priority argument, since a 0’ oracle is needed to determine how each requirement is finally satisfied. A more delicate kind of construction, the infinite injury, or 0” priority argument was developed by Sacks [16] and Shoenfield [19]. Sacks [17] used a 0” priority argument to show that (R, S) is dense as a partial ordering. This led Shoenfield [20] to conjecture that (R, =s) is dense as an upper semi-lattice. However this was refuted by Lachlan [lo] and Yates [25] who showed that some pairs of incomparable degrees in (R, S) have infima. In particular they showed that there is a pair a, b such that a A b = 0. Such a pair is called a minimal pair. The existence of minimal pairs, which contradicts Shoenfield’s conjecture, reveals that the structure of (R, S) is more complicated than Shoenfield had predicted. An even more difficult type of construction, the 0”’ priority argument, was invented by Lachlan [12] and used by him to show additional complexity of (R, s). Harrington found ways to simplify Lachlan’s method, which had been Harrington’s methods also have the something of a technical tour-de-force. advantage of giving a unified approach to 0” priority arguments for all IZE w (and even to 0” priority arguments for ordinals a 3 0). We will present a version of Harrington’s approach in Section 5. Harrington and Shelah [7] used a 0”’ priority argument to show that the first-order theory of (R, S) is undecidable. This was later extended by Harrington and Slaman to show that deg(theory of (R, s)) is O”, the degree of the set of true sentences in the language of first-order arithmetic. (Due to the complexity of the proofs, full proofs have not been published. Recently, simplified proofs have been found by Slaman and Woodin [22] and by Ambos-Spies and Shore [2].) Thus the interaction between degree theory and applied recursion theory has gone in both directions. Degree theory has given us a better understanding of the
Incomparable
prime ideals of r.e. degrees
41
relationships between the nonrecursive sets which arise in applied recursion theory. In the other direction, we have undecidability results about the degrees themselves.
2. Algebraic structure Although the results of Harrington and Shelah and of Harrington and Slaman show that the first order theory of (R, =s) is undecidable, and is in fact as complicated as the set of true sentences of first-order arithmetic, recursion theorists continue to try to understand the structure as best we can. Of course, most mathematicians strive to obtain deep, although necessarily incomplete, understanding of undecidable theories. Number theorists, for instance, are undeterred by the fact that the degree of true arithmetic is 0”. So our position is in no way unusual. One approach is to consider the algebraic structure of (R, G). It is easy to see that (R, S) is an upper semi-lattice since if a = deg(A) and b = deg(B) then aU b = deg(A Cl3B) where A @B is the disjoint union of A and B. However, (R, C) is not a lattice. As mentioned above, Lachlan and Yates showed that some pairs of r.e. degrees have an infimum, but others do not. Thus we consider the algebraic structure (R, C, v , A, 0, l), where v is the join operation and A is the meet partial operation. We may define an ideal as in lattice theory. Definition 2.1. An ideal I, is a nonempty subset which is closed downward and under join, that is, if a, b E I then a v b E I and if c c a then c E I. Since the infimum of two degrees may not exist, the appropriate generalization for an upper semi-lattice of the notion of a filter in lattice theory is the strong filter. Definition 2.2. A strong jilter, F, is a nonempty subset which is closed upward and contains a lower bound for each pair from F, that is, if a, b E F then (3cEF)(cSa,b) and if d?=a then dEF. There have been some interesting recent theorems concerning ideals of r.e. degrees. Shore [21] proved a noninversion theorem for jump, which implies that jump inversion fails for ideals, if we take the jump of an ideal to be its image under the jump operator. An investigation of the structure of the set of ideals of r.e. degrees was begun in the author’s Ph.D. thesis [5]. The main result of this paper, and another interesting recent result described below, concern prime ideals of (R, s).
W. C. Calhoun
42
Definition 2.3. A prime ideal is an ideal whose complement
is a strong filter.
Since the structure of (R, S) has proven to be so complicated, it was somewhat surprising when Ambos-Spies, Jockusch, Shore and Soare [l] showed that the well studied subset M of R is a prime ideal. A degree a is said to be half of a minimal pair if there is a degree b such that a, b is a minimal pair. The set M consists of all halves of minimal pairs.
3. Algebraic strengthening
of the Friedberg-Muchnik
theorem
We are now ready to state the main theorem of this paper, a solution to Post’s problem generalized to the set of prime ideals of (R, s). Theorem 3.1. There are incomparable prime ideals of R. (In fact, countably infinite antichain of prime ideals of R.)
there is a
As a corollary we get a strengthening of Friedberg and Muchnik’s theorem showing that there are r.e. degrees which are incomparable in a strong algebraic sense. Definition
3.2. The radical of an r.e. degree ideals containing a.
a is the intersection
of all prime
Corollary 3.3. There are r.e. degrees a and b whose radicals are incomparable. (In fact, there is a countable sequence of r.e. degrees whose radicals form an injinite antichain.) Proof. Let 1 and J be incomparable
prime ideals. Let a E I\J, and b E J\I. The second statement follows in a similar way from Theorem 3.1. 0 To see how this corollary algebraically strengthens the Friedberg-Muchnik theorem, we describe one consequence of the corollary involving the 1-3-1 or MS lattice (see Fig. 1). 1
@ 0 Fig. 1. The 1-3-1 or M, lattice.
Incomparable prime ideals of r.e. degrees
Fig. 2. Two r.e. degrees
with incomparable
radicals.
The 1-3-1 lattice was shown to be embeddable in R by Lachlan [ll]. Since then embeddings of the 1-3-1 have been studied extensively. For our purposes, the important observation is that if f : M5 + R is an embedding and f(0) E I, where I is a prime ideal, then f(l) E I. Thus if a, b are as in the corollary and f(0) c a then f(0) is in the radical of a, and hence f(1) is also in the radical of a. Therefore f(1) j b, since otherwise b is in the radical of a which contradicts the corollary. The same reasoning shows that if fi, f2, . . . , fn are embeddings of MS into R,f,(O) 4 a, and J(O) CA-r(1) for i > 1, then &(l) 3 b. So no ‘stack’ of 1-3-1’s starting below a can lead to a degree greater than or equal to b. (See Fig. 2.) Of course the same statement is true if we reverse the roles of a and b.
4. Notation Our notation will be fairly standard, similar to that in Soare [23]. Light face lower case Roman letters usually represent integers. Light face upper case Roman letters represent sets of integers, usually r.e. sets. Bold face lower case Roman letters represent degrees, usually r.e. degrees. Bold face upper case Roman letters represent sets of r.e. degrees, usually ideals of r.e. degrees. A string is a member of o<“‘. Lower case Greek letters are used for strings and finite unions of strings. We identify sets of integers with their characteristic functions so that it makes sense to say CYc A, for instance. We hope our typographical conventions will prevent a possible confusion here. When we say
W. C. Calhoun
44
a E B we mean the characteristic
function
of B extends
a, but when we say
A c B we mean A is a subset of B. 4.1. For A = U A,, to make the restraint LYon A at stage s means LYc_A, and we require that a c A, for all t > s such that the restraint is in force at t. If we wish the restraint to cease to be in force at some stage t we will say that we drop the restraint at t. Definition
functions) A, B, A 63 B is the disjoint union of A and B pairing function). To make the restraint a: CBp on A CI3B is the restraints a on A and p on B. In general, when we we mean a restraint of the form p = a, Cl3a2 Cl3- . - CBa;, on where the Ai are finitely many of the sets being A,@A,$--.@A,, constructed, and oi is a restraint on Ai. Upper case Greek letters are used for reductions, which we now define. For sets (or partial (coded by an effective the same as to make speak of a restraint p,
Qi is an r.e. set of pairs ((u, p) such that if en LY, # c+, and if (Y,c LX~then /3, c &. For a reduction (au,, P,), (a*, PJ E @ th @ and a set of integers A, 0“ = IJ {/?: (3a)(a, j3) E bi and (YCA}. Notice that ti may be total or partial. That is, its domain may be o or merely an integer. For a possibly partial function Y, we say Y(x) converges, and write Y(x)J, if x is in the domain of W, otherwise we say that Y(X) diverges, and write Y(X)?.
Definition
4.2. A reduction
Definition 4.3. If @(x)l,
the use of this computation
is the least a c A such that
@%)L For a string /3, we write @(/3)1 if /I c @. The use of this computation is the union of the uses for all x in the domain of /3. All r.e. sets (including reductions) are constructed in stages. The finite approximation at stage s is indicated by the subscript S. Thus we have A,, as,, etc. When an expression in parentheses is subscripted, all sets within the parentheses are taken to be subscripted. For example, (@3x)), means @$(x). We assume standard enumerations @Cand W, of the partial recursive functions and r.e. sets. To simplify the notation we will usually write @ and W, taking it as understood that we, in fact, mean not just the partial recursive function or r.e. set, but also a particular index e for @ or W.
5. Priority arguments
We will use a formulation of priority arguments modeled on that of Harrington. We give our own presentation of this method, but the ideas are his. For another presentation see Weinstein [24]. A related framework for priority
Incomparable
prime ideals of r.e. degrees
45
arguments has been proposed by Lempp and Lerman [13]. Also, Knight [8,9] has given a framework for constructions in recursive model theory that is inspired by Harrington’s method as well as work of Ash [3,4]. We assume that the reader is familiar with priority arguments as described in Soare [23], for instance. In a priority argument
Harrington’s
formulation,
many
for some ordinal
levels,
is a construction
CY.We will only treat
that
the case that
involves
(Y is finite.
Q A
level n argument is one which uses levels 0, 1,2, . . . , II. We begin ordering.
by describing
level 1 arguments.
Level 0 is a recursive
of members
of P,. In this paper,
Let (P,,, co) be a recursive
partial
sequence
P;, will consist
of the potential
finite approxima-
tions to the r.e. set (or sets) we wish to construct (which we will represent by strings), ordered by A 2 B iff A c B (as sets, not as strings). We will often call the subscript t of A, a level 0 stage. Level 1 is a recursive in 0’ sequence level 0. We will often call the A, = S,, S,, &, . . . of level 1 strategies controlling subscript s of S, a level 1 stage. We think of stages representing a progression of time and use terminology like ‘wait for * - . ’ to mean ‘at the least stage such that . . . ‘. A Level 1 strategy is a recursive function S : P;“-+ 8,) where 8, is the set of allowed level 1 environments for the construction. A level 1 environment is a nonempty recursive subset of PO, i.e. potential entries for level 0. Particular sets Y,, Z, of allowed level 1 strategies and allowed level 1 environments are chosen for a particular construction in such a way that Yi, %‘, and the inclusion relation on %‘,are recursive. The intention is that if S(& It) = E then S ‘wants’ A, to be chosen so that A, E E. For any A E p0, the level 1 environment EA = {B: B GA} is always allowed. So the environment E*,_, which allows all possible extensions of A,_, to appear is allowed. In fact, since we must have that A,
46
W. C. Calhoun
Proposition 5.1. For all sets A and B, and functions f recursive in A and g recursive in B, there are integers a and b such that c&’ = C$Ca,bjand @,S = cD$~,~). Proof. By the relativized Recursion Theorem
function function
with parameters there is a recursive G(y) such that for all y, @&Bo,j,Yj= CD&,,. Let h be the B-recursive
h(y) = g@(y), Y). Then there is a b such that @&,, = @f. Let a = B(b). Then e @6”= @&b).
= 4&,b,
and
0
The proposition tells us that in constructing level 0 and level 1 we may assume that we have indices for & and A,. Thus on level 1 we can know what happens at any level 0 stage and even use 0’ to see if Ai properties hold on level 0. In building level 1, we will informally say that we ‘ask 0” or simply that we ‘ask’ if some Ai property holds on level 0. This means that the use of 0’ in computing that level 1 stage is sufficiently large to answer this question. We need to use some caution in this regard, however. We cannot assume that & and A, are total. Thus, on level 1 we cannot ask questions about a level 1 stage s for which we have not yet specified S,. (We might get no answer because we wait forever for an answer, and therefore never specify S,.) However, we will construct level 0 so that & is total even if A, is not. Thus, we can safely ask A’: questions about level 0 on level 1 as described above. This type of consideration implies that for any level n argument, questions asked in order to decide what strategy should appear at stage s on level m must be about level m stages
Incomparable prime ideals of r.e. degrees
47
level 1 strategy for every i, and Al respects priority, then for every level 1 stage s there is a level 0 stage t, such that for all level 0 stages t > t,, S, is respected at t. Proof. By the limit lemma (see Soare [23, p. 571) there is a recursive double sequence S,,, such that lim, S,,, = S, for all s. At level 0 stage t let s, 6 t be largest so that S,,, E 9’i for all s G s, and &J/i, It) I> .!$+,,,(A, It) for all s
The fact that this construction always produces a total &, even if A, is partial, means that the & given by the Recursion Theorem will be total, so level 1 can assume level 0 is total, as mentioned above. In practice we will not explicitly describe how we are building level 0. We will describe how we are building level 1, and leave it as understood that level 0 is built as in the proof of the theorem. Occasionally, it is convenient to take some action on a lower level that is not for the purpose of respecting strategies on the level above. It will be clear that we can do this, and still build the lower level so as to eventually respect the strategies on the level above, as in the theorem. We now turn to the general case of level n arguments. Level m + 1, for m < n, is a recursive in Om+’ sequence A,,, of level m + 1 strategies controlling the construction of the sequence A,,, on level m. The definitions of level m + 1 strategy, level m + 1 environment, A,+I respects priority, etc. are obtained by relativizing the definitions for level 1. The relationship between level m + 1 and level m is simply the relativization of the relationship between level 1 and level 0. The partial ordering (P,, 6, ) on level m, corresponding to (P,,, +,), is given by P,,, = Ym, the allowed level m strategies, and s,,, is the ordering of the allowed level m strategies defined by Sam T * (Va E Pz’?,) S(a) 2 T(o). Notice that since the relation E, 1 E, on level m environments is recursive in Om-‘, the relation S, am S, on level m strategies is recursive in 0”. Thus (P,, G,) is a recursive in 0” partial ordering and the relativization to higher levels can be carried out as we have described. Furthermore, the relativization of Theorem 5.2 shows that for any recursive in 0” sequence of level n strategies, A,, which respects priority, there is a recursive in O”-’ sequence A,_, such that each level IZ strategy is eventually respected on level 12- 1. Then, by the definition of the ordering on level n - 1, A,-1 respects priority, so there is a recursive in O”-’ sequence A,,_, which eventually respects each level it - 1 strategy. Repeating this argument we finally get a recursive sequence & whose construction is controlled by level 1, whose construction is
W. C. Calhoun
0”
s, \
0’
Tt \
Fig. 3. A level 2 argument
controlled by level 2, and so on. (In fact we build levels 0, 1,2, . . . , n simultaneously.) Thus, in any level n argument, the construction of level 0 is ultimately controlled by level n, but as it gets larger this control becomes increasingly indirect (and the arguments, therefore, become increasingly delicate). We mention one particular example of a use of this framework which comes up in many priority arguments. We often wish to have a level 2 strategy S (see Fig. 3) build a reduction @ so that cP’ = B, for some sets A and B. To do so, S first requires that a strategy T appear on level 1 (we will say that S ‘plays’ T on level 1). T ensures that if ((u, 6) E CDand (Yc A then p c B. To ensure this, T does not allow ((u, p) to enter @ at level 0 stage s unless (YG A, and /3 G B,. Then, if at some later stage t, /I $ B,, T requires that a number enter A so that cy+ A. Thus, T insures that the reduction @ is ‘correct’. In addition, S must ensure that CC@ is total. We will say that S ‘protects’ @, meaning that S plays a level 2 environment to ensure that no strategy T’ may appear after T on level 1 which restrains pairs ((u, 0) from entering @ at any level 0 stage s without an ‘excuse’. An excuse is a II’: fact C$about level 0 which is ‘known’ to be true by S. (That is, the use of 0” in constructing level 2 up to S is enough to determine that # is true.) With the excuse $, T’ may restrain all new pairs from entering @. We will say in this case that @ has been ‘made finite’. However, at the first level 0 stage t at which a new witness to the truth of $ appears, the excuse ceases to be valid and T’ must drop its restraint on @. Thus, Qi can only be made finite temporarily. If Qi is protected, @ can be made total by placing strategies T,, T2, . . . on level 1 such that T ensures that @(x)i for all x < i.
6. Proof of the main theorem In this section we prove Theorem 3.1. The proof is a level 4 priority argument in which we build the ideals Ir, 12, . . . with complementary, strong filters, such that A’ E Ii t, i #j. For each i, and each and r.e. sets A’, A’, . . . F,, F,, . . . > r.e. set W,, we must decide if We E Ii or We E F;. ‘Injury’ occurs when an attempt to put an r.e. set We into Ii fails, and switches to an attempt to put We into Fi. In
Incomparable
prime ideals of r.e. degrees
49
the next section, we will discuss a slight modification of the construction which allows the argument to be done on level 3. The prime ideals built by the level 3 construction are A,. This is a better bound on the complexity of the prime ideals we build than we have for the level 4 construction, which we now describe. The argument dovetails arguments for building 1; and Fj for each i. Since the same construction is used for each i, and the interaction between the constructions of the different ideals is slight, we will simplify the notation by describing the argument as if we were only concerned with one fixed i. At the end of the proof we will comment on why there is no difficulty in dovetailing the arguments in order to handle the countably many Ii. To simplify the notation, we will drop the subscript so that for the fixed i, I = 1; and F = Fi. We will also let B = A’ and A = @ jzi A’.
During the construction
we will have finite approximations,
I” and E”, to I and
F. Associated to each pair (I’, E”) is a string q in the tree 2’” defined by
0
q(e)
if W, EP, = 1 if W, EFF, i T otherwise.
For a, 7~~2~“’ U 2” we will always use (TE t to indicate that t is an extension of a, and o i t to indicate that (Tis Turing reducible to t (in the case that a, r E 2”‘). We will use the symbol =Cfor the Kleene-Brouwer ordering on 2’” U 2”. Definition 6.1 (Kleene-Brouwer).
a=Ct iff
(i.e.,
oEror a(e) < r(e) where e is least such that o(e) # r(e)
I
o < r iff o is to the ‘left’ of t).
To make F a strong filter, we will build, for each level 4 stage s, a set P such that F” c W, for all W, in E” U {B}. In addition we define 1’ = @ 1’ C3A. We need to ensure that I” 3 F” for all s. Otherwise there is an ‘obstruction’ (see Fig. 4). Definition 6.2. There is an obstruction such that C@= F.
at level 4 stage s if there is a reduction
CJ~
When there is an obstruction, we will try to ‘go around’ until our path is barred. More precisely, at level 4 stage s we ask “is there an obstruction at stage s - l?” If not, let e be least such that W, 4 F”-’ U I’-‘. Let 1’ = I’-’ U {W,} and F = I’-‘. On the other hand, if there is an obstruction at s - 1 and I’-’ # 0, then let e be largest such that W,EI,_,. Let P=P-‘\{W,} and F”=F”-‘U{W,}\ {y: j > e}. (We are moving to the ‘right’ of the obstruction in 2’“.)
50
W. C. Calhoun
Fig. 4. If I” 2 F” there is an obstruction
at s.
In either case we then play (a strategy on level 3 which plays) a strategy on level 2 which builds (using a strategy on level 1) an r.e. set F” and builds and protects a reduction YS such that (ly,)” = F” for all X E E” U {B}. (To simplify the notation, we are using only one reduction, Y, for all the X’s, by the usual trick: for any (a, p) E YS, cr is long enough to distinguish between the different X’s in F” U {B}.) In this case, since we are not building the X’s, we will ensure the totality of Yx by putting a sequence of strategies along level 1, such that the nth strategy restrains the use of Yx(n) from growing longer once the stronger priority strategies have settled down. Since at each point on level 1 we will only have to respect finitely many such strategies, they will amount to a finite restraint, and will not be noticed in our construction. To complete the description of the construction, we must tell what to do at level 4 stage s if there is an obstruction at s - 1 and IS-’ = 0. We will describe below a level 1 strategy, S, to be employed by level 4 in this case. The strategy S produces a contradiction to the assumption that this case has arisen, thus our intention of employing S if necessary guarantees that this case will not arise on level 4. It follows that lim P and lim F” exists, and if we let I = lim P and F = lim F, then I is an ideal with complement F, which is a strong filter. That is, I is a prime ideal. Before we describe S, note that we need only be concerned that S respect level 2 strategies T such that T was played by level 4 at some stage t
Incomparable
prime ideals of r.e. degrees
51
F* and finite collection of r.e. sets F*, Yx = F* for all X E F* U {B}. Since F* is not one of F’, F2, . . . , FSwl, S will have no need to put numbers in F*, and S can respect T simply by restraining F*. Thus, although we are using 0C4)to build I, the r.e. sets A, B, and F are essentially built by the level 1 construction. So, most of the ‘action’ in the proof takes place in the finite injury argument on level 1. (Although, for the proof to succeed, we need that the level 1 and level 2 strategies have the ‘knowledge’ given them by level 4.) We now describe the strategy S to be used in the case that there is an obstruction at s - 1 and I’-’ = 0. In this case our path through 2<” is ‘barred’. To make this precise, we make some definitions. Definition 6.3. A bar, 2, is an antichain of strings such that for every
f E2”
(3crE.z)f 30. Note that every bar is finite, since 2”’ is compact and a bar corresponds minimal basic open cover of 2”.
to a
Definition 6.4. There is a bar at s if the set
_Y= {at: t
at t}
is a bar. (See Fig. 5.) Lemma s.
6.5.
If there is an obstruction
at s - 1, and P-’ = 0, then there is a bar at
Proof. Let 2 be as given in the definition of a bar at s. We must show that 2 is a bar. Let f E2”. Let t be the largest stage at which a,
w4
w3
10
W2
w
wo Fig. 5. Obstructions
at 1,5,7,8,9
and 10 form a bar.
W. C. Calhoun
52
f
:.
Fig. 6. The situation
in Lemma
6.5.
if t
We will call x, x,, , xt2, . . . , x,, an s-set up at level Ostuge v if (i)-(v) above hold when F’, I’, cPl, Yr, B, and X are all replaced by their finite approximations at v. When we mean an s-set up as written, we will emphasize this by calling it a permanent s-set up. We need to know that we will have s-set ups to use.
Incomparable prime ideals of r.e. degrees
53
Lemma 6.7 (Window Property). For any r, s E w, if there is a bar at level 4 stage s, then there is a permanent s-set up x, xl,, x,*, . . . , x,,, such that x > r and x,, > r for all j. Proof. We need only require that the sets F’ we are building are coinfinite, and that the reductions Y, we are building satisfy the usual convention that if length(use( Y:(y))) 2 y for all y and (a; /3) E ‘y,, then length(a) 2 length(P), i.e., . Y. We ensure the latter by simply making it a condition on the construction of Y. To ensure the former, we put a sequence of strategies along level 1 (at every other level 1 stage, for instance) such that the nth strategy ensures that there are n numbers not in F’. Since the effect of the stronger level 1 strategies settles down to a finite restraint, each such strategy can succeed simply by restraining one more number from F. It is now easy to see that the window property holds. Let x be any number such that x > r and x 4 B. Then choose x,,, for i = 1, 2, . . . , k, so that xg > r, x,, C$F5, and use(@c(x,)) /X E use( Yf(x,)) for all i
Now we describe the level 1 strategy, S, which level 4 plays to produce contradiction in the case that there is a bar, 2, at s.
a
Step 1. S waits for a level 0 stage at which there is an s-set up x, xrl, xfz, . . . , xq such that none of the numbers x, xI,, x,*, . . . , xg is restrained, by the finite restraint imposed by level 1 strategies stronger than S, from entering its target set. Step 2. S restrains use( @(x,)) IA, and puts x into B. Subsequent steps. S puts in x,, , xu2, . . . , x,,, for some ul < u2 < . * . < I.+, where uj E T for all i. After putting in xq, S waits until @am before going on to xy+,. During this process, if Yf(x,) becomes undefined for some t E T and X E F’ U {B}, then S keeps the computation undefined until S gets x, into F” for some l4 z=t. S can do this, because if J’< k then Ytk is built with the knowledge that @,‘= F5. Thus, ‘yk can see that S will wait only finitely long for @$‘(xj)T once x5 E F?. From this it will be clear, once we finish describing S, that S eventually gets x, into F” for some u 2 tk. Thus, ‘yk computations can be kept undefined with the excuse that we are waiting for S to get x, into F”, for some u 2 tk. Now we describe inductively what S does in steps 3,4, . . . . Assume that S is to the point of putting x, into F’. Assume that
Yf(xJT
for all u 2 t and all X E F’ U {B}.
(1)
W. C. Calhoun
54
Notice that (1) is true at the end of step 2 above, for t = t,, since F’I = 0, and we have that Y$U)t for all u E T (i.e., we have a base for induction). Since (1) holds, S is free to put x, into F’. If t = tk, we will see below that we have reached a contradiction, so assume t < tk. Then, S waits until use( #(x,)) has changed from the value it had in step 1. (This must happen since otherwise @%t) f F’(G) Since use( @(x,)) rA is restrained, there must be some e such that W, E I’ and use(@:‘(x,)) rW, has changed. Fix such an e. Then WF(x,)T for all u E T such that u > t and We E F”. Consider f E 2” defined by 1
if o,(i) = 1 or i = e,
f(i) = { 0 otherwise. There must Furthermore, inductively.
be some
u E T such that a,, cf. Clearly, o, < a,, , so t < ~1. Hence, (1) holds for u, and we can continue
F” G E” U {W,}.
Thus eventually S puts x,, into F’*. We then reach a contradiction, since we have 1’~= 0, and so Z’k= A. Since the relevant use of A was restrained, we now have @~T(x~,)# F’k(xJ (and this computation will never be corrected since we maintain our restraint on A). This is the desired contradiction. We close the proof with a comment on why we can use the construction described above, for building one prime ideal, to build countably many prime ideals Ii, Iz, 13, . . . . On level 4 we simply dovetail strategies like those described above for each of the I;. Since these level 4 strategies do nothing but play level 2 strategies and level 1 strategies, a conflict could only arise on level 2 or level 1. But if i #j, there is no significant interaction between a strategy S associated with Ii and a strategy T associated with 11. If S and T are level 1 strategies, then they settle down to finite restraints. If S is a level 2 strategy building a reduction Y to compute F;, S has no effect on T, which is concerned with sets F;, F;, . . . , Fj. Thus, if S and T are level 1 or level 2 strategies associated with different ideals, the only connection between S and T is that all strategies control the construction of A’, A*, . . . . However, any effect of S (or T) on the Ak is included in the finite restraint on level 1. In fact, at most, the effect of S is to put one number into A’ and restrain BjziAi. Thus no conflicts will arise. (It is no worse than building an infinite antichain of r.e. degrees.)
7. Reducing the complexity,
a 0”’ construction
A recursion theorist would be remiss not to comment on the complexity of the objects constructed. As mentioned in the previous section, the complexity of the prime ideals built by this construction is not the lowest possible. We will discuss a modification of the construction that builds prime ideals of lower complexity, and we pose two interesting questions concerning the complexity of prime ideals. We will use the following definition of the complexity of an ideal.
Incomparable prime ideals of r.e. degrees
55
For any ideal I, let G(1) = {e: deg(W,) E I}. We say that I is _Z,,(QJ if G(1) is ,Yi (L?$. We say I is A, if I is Z;, and fl,. Definition
7.1.
Examining the proof, we see that I = lim P, where G(P) is recursive in OC4). Thus I is &. But F is also X6, so I is A+ However, we could have used a slightly more complicated construction on level 3, which we now describe. Our oracle is now only o”‘, so we cannot ask, at stage s, if there is an obstruction at stage s - 1. Instead of getting the answer to this question immediately at stage s, we will have go on with the construction. It will only be at a later stage that we see that there is an obstruction at s - 1. We break up the _Z”,question of whether there is an obstruction at s - 1 into the infinite sequence of questions “Does @ witness an obstruction at s - l?“, for each reduction @. These questions are rr(,), so 0”’ can answer them. Thus, if there is an obstruction at s - 1, we will eventually discover this fact at some level 3 stage t 2 s. Then, we set I’ and F’ as we would have at stage s in the level 4 construction. At any level 3 stage u 2 t, the strategies at stages s, s + 1, . . . , t - 1 are known to be ‘mistakes’, and can be disregarded since the sets they build will not be used in the rest of the construction. With these changes, the proof works as before. By analyzing this level 3 construction in the same way we analyzed the level 4 construction above, we see that the prime ideals built are A5. There does not seem to be any easy modification of our construction that would further lower the complexity of the prime ideals we build. However, there is a 17, prime ideal. The prime ideal M is L14, since M = R\PS, and, by inspecting the definition, we see that to be of promptly simple degree is a Z4 property. This raises the question: is there an infinite antichain of IL, prime ideals? On the other hand, M is not Z4, in fact M is Lr4-complete, as was shown by Schwarz [18]. This raises another question: is there a _Z4prime ideal?
References [l] K. Ambos-Spies, C.G. Jockusch Jr., R.A. Shore and R.I. Soare, An algebraic decomposition of the recursively enumerable degrees, Trans. Amer. Math. Sot. 281 (1984) 109-128. [2] K. Ambos-Spies and R.A. Shore, l-Types and undecidability in the r.e. degrees, to appear. [3] C.J. Ash, Recursive labelling systems and stability of recursive structures in hyperarithmetical degrees, Trans. Amer. Math. Sot. 298 (1986) 497-514; corrections, ibid., 310 (1988) 851. [4] C.J. Ash, Labelling systems and r.e. structures, Ann. Pure Appl. Logic 47 (1990), 99-119. [5] W.C. Calhoun, The Lattice of Ideals of Recursively Enumerable Degrees, Ph.D. Dissertation, University of California at Berkeley, 1990. [6] R.M. Friedberg, Two recursively enumerable sets of incomparable degrees of unsolvability, Proc. Natl. Acad. Sci. USA 43 (1957) 236-238. [7] L. Harrington and S. Shelah, The undecidability of the recursively enumerable degrees (research announcement), Bull. Amer. Math. Sot. (N.S.) 6 (1) (1982) 79-80. [8] J.F. Knight, A metatheorem for constructions by finitely many workers, J. Symbolic Logic 55 (1990) 787-804. (91 J.F. Knight, Constructions by transfinitely many workers, Preprint.
W. C. Calhoun
56 [lo]
A.H.
Lachlan,
Lower
bounds
for pairs of recursively
enumerable
degrees,
Proc.