on restricted forms of enumeration reducibility

Annals of Pure and Applied Logic 49 (1990) 75-96 North-Holland

75

ON RESTRICTED FORMS OF ENUMERATION REDUCIBILITY

Phil WATSON Royal Holloway and Bedford New College, University of London, Egham Hill, Egham, Surrey TW2OOEX, UK

Communicated by A. Nerode Received 5 June 1989

We consider enumeration reducibility and its various restricted froms, especially s- reducibility. We give a constructive proof in the case of & degrees of Zakharov’s result that there does not exist an s-contiguous e-degree, and strengthen this result in the A, and &-high cases to show that no A, or &,-high s-degree is minimal in its e-degree.

We briefly consider pc-degrees and show that there is a l7, pc-degree containing only one J7, s-degree.

Finally we introduce two new forms of restricted enumeration reducibility which occur frequently in proofs in this area but have not previously been recognised.

0. Introduction

De6nltlon. An enumeration of a set A is a sequence of finite sets {A,},,, such that

(i) Us,,& =A (4 W) A, c A,+*,

(iii) (Vs) IA,+1 -AsI G 1. (Note. (iii) is often weakened to require JA,,, - A,1 to be finite for every s 2 0.)

Definition. Let {De}_, be the set of all (canonically-indexed) finite sets of natural numbers.

Dehition (Friedberg and Rogers [4], Myhill [8]). For any e, B:

Y=(B) = {x : (3u)[(x, u) E Ye & 0, E B])

where Ye is an r.e. set called an enumeration operafor. (We call each (x, u) an axiom of Ye. 0, is the use of the axiom.)

Then for sets A, B we say A is enumeration reducible to B (or A is e-reducible to B or A 6, B) iff @)[A = ‘Pi(B)]. Then we say A se B via Yi.

Define A se B iff A ee B and B <,A. Since Ge is an equivalence relation, Ge partitions 2” into equivalence classes,

called enumeration degrees or just e-degrees. The ordering < on the e-degrees is the ordering induced by 6,.

The e-degree of a set A is the class of all sets B such that A =c B.

0168-0072/90/$03.50 @ 1990-EElsevier Science Publishers B.V. (North-Holland)

76 P. Watson

We may regard the e-degree of the r.e. sets as the lowest degree in our theory. We denote it by 0,.

Definition. A set A is in & iff it has an approximation {A,},,, such that for all x

x E A e @)(Vt 2 S)[X E A,].

Note that for a ,‘Zz set A it is possible that

x E A & (VS)@ 2 s)[x E A,].

Thus an approximation to a & set is already very difficult to handle, and there is no sensible approximation at all for sets which are not in &, so we shall confine ourselves to the e-degrees of & sets.

1. Restricted forms of e-reducibility

Definition. An e-operator Y is an s-operator if for all axioms {x, u) in Y,

lDul s 1. A is s-reducible to B (A ss B) if A is e-reducible to B via an s-operator.

Clearly A Ss B + A Se B. Also we can effectively list all s-operators using {Ye},,,, our standard list of all

e-operators, restricted to axioms (x, u) where D, s 1. Note that for every x an s-operator Y defines an r.e. set W, the union of all the

(singleton) sets D, such that (x, u) is an axiom, and then if A ss B via Y

Definition. An s-operator Y is an &operator if for every x there are finitely many u such that {x, u) E Y.

Clearly A ss B +A C, B. It is not possible to list all $-operators effectively because l{u : (x, u) E Y}l

does not have to be bounded by a recursive function. An alternative is bounded e-reducibility, defined as follows.

Definition. A is bounded e-reducible to B (A cbe B) if A 6, B via some Y and there exists n such that for all axioms (x, u) in Y, lDul d n.

Note that she is transitive. If A she B via Y with bound n and B S& C via @ with bound p, then A she C via some composite operator Y(G) with bound np.

Further it is possible to list all be-operators in the obvious way by listing all pairs (e, n) where Ye is an e-operator and n is a bound and taking the be-operator a(=,,,) to be Ye restricted to axioms (x, u ) where 1 D, I =S n.

Restricted forms of enumeration reducibility 77

Also

Instead of limiting the use of every axiom of an e-operator, we can limit the number of axioms which are allowed to enumerate any particular element x.

Deli&ion. A is pc-reducible to B(A spc B) if A se B via some Y and for every x:

((X,U)EY&(X,tJ)EW)+%=Z1.

We obviously list all pc-operators by listing all partial functions. A pc-operator Y such that (VX)(~U)[(X, u) E Y] is called a c-operator, and we

define c-reducibility in the obvious way. We cannot list all c-operators effectively, as this is equivalent to listing all total recursive functions.

We may also express the well-known reducibilities $,, and =$ as restricted versions of e-reducibility.

Definition. Say A is m-reducible to B (A srn B) if A se B via some Y and for every x there exists exactly one u such that

(x,r+Y

and furthermore ID”] = 1.

We cannot list all m-operators effectively, as this would correspond to a list of all total recursive functions.

Definition. Say Y is a pm-operator if for every x there exists at most one u such that (x, u) E Y, and furthermore l&l = 1. Define pm-reducibility in the obvious way.

We can list all pm-operators effectively from our standard list recursive functions.

Definition. Say A is l-reducible to B (A s1 B) if A G,,, B via some every axiom (x, u ) in Y:

(LL=D,& (y,v)EY)$y=x.

of partial

Y and for

The diagram which follows shows the known implications between restricted versions of e-reducibility and is a simplified version of a diagram in Cooper [2].

Known implications between forms of restricted e-reducibility

78 P. Watson

2. Contiguous degrees

It is easily shown that all of the above implications are strict, i.e. there exist ,Y2 sets A, B, C, and D such that

As,B but A&B,

CsbeD but C&D,

etc.

We might ask ‘how close’ two reducibilities are in the following way.

Definition. Given two reducibilities =s,, and + with A sY B +A cr B for all sets A and B we say a y-contiguous t-degree is a non-zero z-degree a, such that for all A, B E a,, A sY B.

It seems reasonable to think of two reducibilities as ‘close’ if one has a contiguous degree in the other. We might reason that they do not agree on all degrees a, but at least they agree on some.

For a long time the question of whether there was an s-contiguous e-degree was open, but this was answered negatively by Zakharov [lo]. We will give our own proof of Zakharov’s Theorem in the case of the & degrees which is constructive whereas Zakharov’s proof is not because we believe a constructive proof will give a better introduction to the results that follow.

Definition. Given a set A in & define A* to be the set comprising all n-tuples of elements in A, for every IZ.

Then it is not hard to see that A =,A * and for any B se A, B sc,A *. Hence A*

is of the highest s-degree of all sets eYeequivalent to A.

Definition. A thin approximation to a set A is an approximation {A,},,, such that for any I:

(3”s)[As [l =A [l].

It is not difficult to show that every set A has a thin approximation. In the proof of Zakharov’s Theorem which follows we make use of a thin

approximation to ensure that initial segments of a set are ‘well-behaved’ during an approximation.

Theorem 1. Given any non-r.e. set A in .X2 there exists a X2 set B such that B =,A

but B<,A*.

Restricted fom of enumeration reducibility 79

Proof.

Requirements We satisfy for every i

Ri: A*#@

and also overall requirements

P: A*=TB and Q: B=A*

where { @ii)isw is our standard list of all s-operators; r is an e-operator which we construct; A is an s-operator which we construct.

Approximation Assume we are given a 2; thin approximation of A, {A,},,, and assume also

that at all stages s, x, y E A, e (x, y ) E A,*, etc.

Strategy for Ri 1. Given a length of agreement I, wait for setup:

(Vn<f)(Vx, ,..., x,<l)[(x1,..., x,,)~A:j(x~ ,..., x,)E@;:]

at some stage S. 2. When setup occurs find the finite set D, G B, such that ( (x1, . . . , x,), u) is

an axiom of @,. Fix D, in B,, priority i, and increase I by a factor of 2. Go to 1.

Our idea is to build increasing finite segments of @B so that

(Vl)(3s)(3Du satisfying setup) + @f is r.e.

If

(3l)(Vs)(Vu: D, does not satisfy setup)

then clearly A* # @F. This is where we use the properties of our thin approximation.

Otherwise as A and therefore A* is not r.e., there must be some element x which enters and leaves the approximation of A* infinitely often, yet x E @, so A*#@?.

Strategy for P (on an element x) (i) At stage x, choose the first x + 2 elements y,, . . . , Y~+~, which have not yet

been chosen in this way, and set

(x) E r{Yl*....Yx+z)

For every (x1,. . . , x,)suchthatx~{x~,., . , x,} and (Vi)[(xi) E P] set

(x1, ~~~,Xn)E~~lU~'~U~"U~Yl....~Y,+z)~

80 P. Watson

(ii) If x enters the approximation for the first time at stage s, put

{Yip * . . 9 Y*+*] c & (iii) If x later leaves A,, c > s, we want to remove one of the Yj from B,. If: 1. None of the Yj is fixed in B,, remove them all from B,. 2. Say y,, . . . , yk are all fixed by the same Ri. Then we have

(pi) E @i(‘l), . . . , (Pk) E @jyk’

but we must also have

(PI,. f . ,pk) E @j=’

by the way the strategy for Ri fkes elements. Then we remove from B, all the Yj except

(a) 2, if present: (b) the first Yj from the list to be fixed by Ri, otherwise.

3. If neither 1. nor 2. holds, remove from B, all the yj fixed by a requirement of lower priority than Rx (or not fixed at all). N.B. There must be at least one such

Yj* (iv) If x later re-enters A,, all the elements removed under (iii) re-enter B,.

Go to (iii).

Strategy for Q If y enters B, because x enters A,, set y E AtI). Then if x E A, y E B so A is correct on y. If x is removed from A, and y remains in B, because some other element of the

r-use of x is removed, make y E A0 an axiom.

Construction

Stage 0

B,=&=r,=g, ii(O) = i + 1 for all i.

Stage s + 1. For each i = 0, 1, . . . , s in order of priority, perform the strategy for Ri as follows:

1. Check whether we have setup, i.e.

DE;B (V~<I,(~))(V~~,...,~,<~,(~))[(~~,...,X,)EA,*~((X~,...,X,)EQZ~ *]

2. If not, go to Ri+l (set li(S + 1) = ii(S)). If we have setup, set li(S + 1) = 2fi(S)). Fix D s B,+*, priority i. GO to Ri+l. Perform the strategies for P and Q to ensure that

A* s+l = rB*+l and B,,, = AA:+1

Note that in the strategy for P, if given the choice between options 2. and 3., we always choose 2.


Proof Construction Works

Lemma 1. Every Ri succeeds, so A* & B.

Sublemma 1A. Every Ri is injured finitely often, by P.

Proof. Fix i. Ri is only injured when some element fixed in B by Ri is removed to satisfy the overall requirement P. It appears that this can be due to either (ii)2. or (ii)3. of the strategy for P.

First we prove that there is no injury to Ri from (ii)2. Suppose we have

and at some stage s, y, is removed from B, due to the action of (ii)2. But we notice that for some p2, . . . , pk we have

where z is some element fixed in B, priority i, and z is not removed from B at stage s, by (ii)2. Now we note that in fact there is no injury at all. When we originally fixed y, in B we intended that either p1 E A or A(pJ # @F{pl), winning Ri, whereas now we have either (pl, . . . , pk) E A* (and hence p1 EA) or

A*((PI, . . * 9 Pk))+ @i”((pb - * . , pk)), again winning Ri- Note further that although (ii)2. can cause this ‘transfer’ from yr to z finitely

often for a given p,, some z’ must eventually be fixed in B because only finitely many qi were fixed in @f before p1 (clause (ii)2(b)).

Now consider injury due to (ii)3. Ri can only be injured here when some element x 6 i leaves A for the first time. There are only finitely many such x and each causes the removal from B of finitely many Yj fixed there by Ri, so Ri is finitely injured. Cl

Sublemma 1B. A* = *+A* is r.e.

Proof. Assume A* = G-$, and let the finite set of elements removed from B by P which were fixed there by Ri be E (the set of elements on which Ri was injured).

Now

A * = @F j for every f&) we achieve setup.

Thus if y E B, @{y} is eventually fixed in @F. Now we give a recursive enumeration {W,} of A* = 0: simply as follows. Stage 0. W, = 0. Stage s + 1. Find all the elements which are fixed in B,,, but are not in the

Iinite set E. Let this set be X,,,. Set

w,,, = w, u @+I.

Then clearly W = @r = A*, and A* is r.e. Cl

82 P. Watson

Proof of Lemma 1. By assumption A is not r.e., and so A* is not r.e. Then by Sublemma lB, A * f @i” for any i. So every Ri succeeds and A * & B. Cl

Lemma2. A*s.,Bviar

Proof, Consider an element x. x E A e (x) E A*. Assume (x ) E A*. Then for some elements y,, . . . , Y,+~ we have

(x) e pyt. ‘. . Yx+d*

Also since (x) EA*,

(3t)(Vu > t)[ (x) E A,*].

Then by the construction

(Vu > t)(Vj E { 1, . . . , x + 2})[y, E B,]

so {YIP - * * , Y,+~} e B and (n> E rB. Assume (x) E r{y1,...,yx+2) and (x) $A*, then

W’O[(x) $ Xl. So by the construction

WWMY~ $41.

Since {yI, . . . , yx+2} is finite

WWO[Y~ $41.

So as only one r axiom can enumerate (x) we have (x) $ rB. Note also that for an element (x1, . . . , x,, ) :

(Xl, * * .,x,)~A*~(x~)~A*di-&(x,&A*

W~Yl.1, - - . ) yl,ml)[(xl) E r(y-. . . ry~m)

cfc {Yl,l, - * - 9 YI,~,) E Bl & . ..&

@Y?bl9 * - . , y_)[ (xn ) E W-J, . . . ’ Y~~~J

& {Yn.b - - - 9 ~n.m.1 E Bl- But by the construction,

(x1, * * . , x,) E p U h)_

so

(Xl, * * .,x,&A*(S(xl ,..., X&Z-~. Cl

Lemma 3. Ba,A via A.

Proof. Consider an element y. If y never enters the approximation to B, then no A axiom ever enumerates y, so B(y) = A”(y).

Restricted form of enumeration reducibility 83

So assume y enters the approximation to B because some x enters the approximation to A. First we note that if an element in the approximation to A*

((z*, a. - , z,) say) has y in its r-use then one of the zi must be X, by the way we choose r axioms, so only the removal of x from A (removal of (x) from A*) can cause the removal of y from B. Also y E AtX1.

IfxEA then

(3t)(Vu L t)[x E A,].

At stage t, x re-enters A for the last time and y re-enters B under the strategy for P. Then

(VU 2 t)[y E B,].

So B(y) = AA(y). Assume x $ a. Then

(=)[x $ A,].

If for all such t, y $ B,, then no other A axiom enumerates y, so y $ AA and

B(Y) = AA(y). If for some t

then at this stage we set y E A’, so now certainly y E AA. But by the strategy for P we now know that y will never be removed from B again so B(y) = AA.

So B=AA, and Bs,A. Cl

End of Proof Theorem 1

Corollary. There does not exkt an s-contiguous & e-degree.

3. s-Degrees within e-degrees

However, Zakharov’s proof only uses the highest s-degree in a given e-degree to show that there is some other s-degree in the same e-degree and our constructive proof does much the same thing; possibly this is the only other s-degree for some e-degree?

We are able to show that there are in fact infinitely many s-degrees in any e-degree which is

(i) in A2 or (ii) &high.

Detinition. A set A is in A2 if there exists an approximation {A,},,, to A such that for any x

x E A a (3s)(Vf 2 s)[x E A,],

x E Al C=s (3s)(Vt 2 s)[x E Al].

84 P. Watson

Definition (McEvoy and Cooper [7]). A set A is &-high iff there exists an approximation {A,},EW to A for which

cA(x) = (P > x)[A, ]x E Al

is total and dominates any total recursive function.

Theorem 2. For any non-r.e. set A in A*, there exd a set B in A2 such that B<,A butAEeB.

Proof.

Requirements

Ri: A f @F, P: A =r’, Q: B=AA

where { @j}jem is our standard list of all s-operators; r is an e-operator which we construct; A is an s-operator which we construct; B is a A2 set which we construct to satisfy the Theorem.

In addition, we will define a function I(i, s) for use during the construction. To prove the Theorem we must satisfy every Ri in conjunction with overall

requirements P and Q.

Approximations

Let {A,I,eo be a A2 approximation to A. We shall construct {B,},,,, a AZ approximation to B.

Strategy for P (on an element x) (i) At stage x choose the first x + 2 elements y,, . . . , yX+2 which have not yet

been chosen in this way. Make x E r(ylP...SyX+z) an axiom. (ii) If s is the first stage after x for which x E A,, put y,, . . . , yX+2 in B,. (iii) If x is removed from A,, for some t > s, remove from B, all yi which are

fixed by requirements of lower priority than R,. N. B. There must be at least one such element, as each Ri fixes at most one

element, and there are x + 2 elements in the r-use of x. (iv) If x re-enters A,, for some u > t, put all of the elements removed under

(iii) back into B, and go to (iii).

Strategy for Q (on an element y) (i) Zf y E B, because x E A,, make y E AIX} an axiom, unless it already is. (ii) If x E A,_1 -A, at some stage t > s:

(a) if y $ B,,, to to (iii); (b) otherwise make y E A0 an axiom, because y is fixed in B,,, with highest

priority. (iii) IfxEA,-A,_iforsomeu>t, thenyEB,. Go to(ii).


Strategy for Ri (i) If s is the latest stage for which Z(i, s) is defined, wait until it appears that

A, [ f(i, s) = @1”, [ Z(i, s)

at some stage t >s. Set I(i, t) = l(i, s) + 1. (ii) If at some stage t > s we had

A, [ l(i, s) = @F [ I(i, s)

but at stage t + 1 some x < l(i, s) was removed from A, then fix in B,,, that y such that x G @jy) and y was in B,, priority i.

(iii) If x re-enters A at some u > t + 1 cancel the fixing of y in B and go to (i).

Note on Injury The only possible way for the requirements to injure each other is if the

strategy for P removes from B some element which was fixed there by some Ri.

Construction

Stage 0

Ao=Bo=&,=A,,=O, l(i, 0) = i + 1 for all i.

Stage s + 1. Work on strategies for Rot . . . , R, in order of priority, as follows. For each Ri, if it is possible’to act for some x under (ii) then do so. Otherwise

act under (iii) if some element is fixed in B, or (i) otherwise. At no time will more than one element be fixed in B by Rio

Do as instructed by the strategy for P on y for each y = 0, 1, . . . , s.

Do as instructed by the strategy for Q on each x = 0, 1, . . . , s.


Lemma 1. Every Ri is injured finitely often and eventually satisfied.

Proof. Note first that for every x which is ever in A, there is exactly one axiom x E rD for some D and D has at least two members, at most one of which is fixed in B at any stage by Ro, so P never injures Ro. R. is satisfied as follows.

Eventually the strategy for R. must be left waiting in (i) indefinitely or we will be able to fix a disagreement y E @B and y E A. Otherwise A [ I(0, 0), A [ Z(0, sl),

A [l(O, sz), . . . will give a recursive enumeration of A contradicting our hypothesis (where sl, s2, . . . is the sequence of stages at which I(i, s) is defined). Thus R0 is satisfied.

Now assume that every R-requirement up to Ri is satisfied by stage s, i.e. either the strategy has fixed a disagreement which is never injured after s or the strategy is always waiting under (i) after s.

NOW P injures Ri+l finitely often as follows. Only an x G i has i + 2 or less elements in its corresponding D where x E TD. (Other elements have more than

86 P. Watson

i + 2 elements in their corresponding D, so always avoid injuring R,,, . . . , R,+l.) There are only finitely many such x, so Ri+l is injured finitely often by P. Assume Ri+l is never injured after stage f. Let u = max{s, t}. Then the strategy for Ri+l is either left waiting indefinitely in (i) or is able to fix a disagreement under (ii). Otherwise A 1 Z(i + 1, ul), A [ l(i + 1, uz), . . . is a recursive enumeration of A, contradicting our hypothesis, where ul, u2, . . . are the stages after u at which Z(i + 1, s) is defined.

So every Ri is eventually satisfied. Cl

Lemma 2. P is satisfied.

Proof. Fix x. If x never enters A, then if x E r(Yl* .. . *fi+d is an axiom,

Yl, * * - 9 Yx+2 never enter B, so x $ AB. If x enters A but later leaves A and is ultimately in A, then x E r{J’l* . . * h+d is

the only axiom enumerating x in rB and one of the yi in {yl, . . . , Y~+~} was removed from B when x left A for the last time, so x $ TB.

If x is in A, then for some stage U, x is in A, for all u 2 V, and x E r(Yl, . . ’ yx+z}

for some {yl, . . . , Y~+~} and

(Vu > V)[{Yl, . * * 9 Yx+zl c&l

so x E P. 0

Lemma 3. Q is satisfied.

Proof. As Theorem 1, Lemma 3. Cl

Lemma 4. B is in A2.

Proof. Fii y. If y never enters B there is no problem. If y enters B at some stage it is because some x entered A at that stage. Now y

can leave B only when x leaves A, and y can re-enter B only when x re-enters A. There is some stage u after which the value of A(x) never changes, because A is A2, so after u the value of B(y) never changes and so B is AZ. 0


Corollary. Any non-r.e. AZ e-degree contains an infinite sequence of distinct A2 s-degrees sl, s2, . . . , s,, . . . such that

(Vi)[Si+l <sSi]*

Proof. Repeated application of the Theorem. Cl

We are also able to prove the more difficult result which follows.


Theorem 3. Given a &-high set A there exist& a & set B such that B <,A but B=,A.

haa Let WA,, be a &-high approximation to A. Then we can define

c”(x) = (P >x)[& [x GA]

and we know that C? dominates every total recursive function.

Requirements We shall give a Z; approximation of B to satisfy the following:

Ri: A# @BP P: A=Z+, Q: B=AA

where { Qji}ieo is our standard list of s-operators; r is an e-operator which we construct; A is an s-operator which we construct.

For the purposes of the construction we shall also define for every i a partial recursive function Si by:

Si(x) = (P ‘x)[A, [X E @;:I

Strategy for Ri We shall split the strategy for Ri into countable many sub-strategies Ri,,j,. The

sub-strategies will have a linear priority ordering as follows: if (i, j) = n, Rii,j, is of (n + 1)th highest priority.

At any time Rii,j, will have an associated ‘limit’ which we will call r(i,j). Our intention will be that Rii,j, will attempt to find a witness y to the sucuss of Rip such that Y < r(i,j).

Ri will succeed if we can find a pair (x, y ) such that

n E @/y)

andxeA andyEB. If we think we have found such a pair (x, y ) with y < ‘(id) we fix y in B priority

R{i,j,. Exactly how this is done will be explained in the construction.

Strategy for P (on an element x) Same as Strategy for P in Theorem 2.

Strategy for Q (on an element y) Same as Strategy for Q in Theorem 2.

Note on Injury The only possible injury between requirements is if the strategy for P removes

from B some y which was fixed by some Ri,j,. This can happen in one of two ways:

(a) An element x is removed from A, where x E rlyl*.. *yx+2) and y,,, is fixed in B by R{i,jj and y,, . . . , y,,, are all fixed in B by requirements of higher priority than R{i,j>.

88 P. Watson

(b) As for (a), except that {y,, . . . , y,,,}, for some m, are frxed in B because

{Yl, * * . 9Y&A0. We shall show later that these amount to finite injury.

construction Stage 0

Ao=Bo=G=A,,=O,

(Vi)[@i,O = 81, (vi9 _i)[r(i,j) =i + l]* Stage s + 1. In order of priority, and for i =GS, work on as many Rii,j, as

possible, i.e. until A,+1 1 r(i,j) 4 @$?I or ‘(i,j) 3 s + 1, as follows. If Rii,j, fixed y E B,, check that Y was not removed due to Q-injury. As long as

it was not, fix Y in B,+l, priority Rii,jk. If y was removed from B due to Q-injury, Rii,j> is injured and proceed as

follows. Freeze every Rii,k), k 2 j, until a stage t > s + 1 such that Si(z) = t, for some z. At this stage set r(i,j) = t and for all k, k’ such that k’ > k 2 j, set T(i,k’) > r(i,k), and unfreeze all the R i, i k). (If no such t exists we win Ri because A $ @i”.)

If Rii,j, fixed no element Y E B,, or such a fixing was injured, and r(i,j) = z with si(z) = s + 1 we must act as follows. For all y E B, [ z check whether

(3X)[X E @{*$]

and y is not already fixed in B,,, by a requirement of higher priority than Rii,j), and x is not enumerated by any lixed y. There may be several such y. Fix the least such y, if any, in B s+l, priority Rii,j,. GO tO Rii,j)+l.

Do as instructed by the strategy for P for each x 6 s. Do as instructed by the strategy for Q for.each y c s.

Proof Construction Work-s

Lemma 1. P succeeds, i.e. A = TB.

Proof. As in Theorem 2, with the following addition. If x leaves and re-enters A infinitely often (meaning x $ A), then part (iii) of

the Strategy for P must act infinitely often. Thus at least one of the Yj from the finite set {yi, . . . , yX+*} is removed from B infinitely often. Then D $ B and

X$P.

Thus A(x) = TB(x). 0

Lemma 2. Q succeeds, i.e. B = A*.

Proof. As for Theorem 2. Cl

Lemma 3. For all i, Ri succeeds, i.e. B <,A.

Restricted form of enumeration reducibility 89

Proof. Fix i. If s&x) is not total, A # !Df and Ri succeeds, so assume si(x) is total.

Sublemma 3A. There are co-finitely many z such that

(3x& < z

& (3, t)[s > t > Si(Z)

&x$A,UA

&X EA,

& (3y)[ y E B, & x E @!,:‘I

Proof. We have

S&X) = (P > x)[A, 1 x c @:; r xl

a total recursive function, by assumption. Then si(x) is dominated by

c”(x) = (/.Js >x)[A, [x c A].

SO

i.e.

(W(Vz > z’)[@;$& r z $4,

(3z’)(Vz > z’)[(3y)(3x < z)[x E @$x, & y E B,(,) & x $ A]

as required. q

Sublemma 3B. Each Ri,,j, is injured finitely often.

Proof. Assume that R{o,o), . . . , Rti,j)-1 have all fixed in B at most one witness each and are never injured again, and will never again fix any other element after stage t. Also, up to stage t finitely many Zj have been fixed in B by virtue of there being an axiom Zj E l-@. Say no element larger than n is ever fixed.

Then Rii,i, is never injured in connection with r-axioms with n + 3 or more elements in their use because even the worst such case has a r-use of

where y is the element fixed by Rii,j, and q is some other element fixed by a requirement of lower priority than Rii,j), or not fixed at all.

There are finitely many A axioms with less than n + 3 elements in their use, namely those which enumerate some x SIZ. R{,,jl is injured at most once in connection with each of these, so Rii,j) is injured finitely often. Cl

Sablemma 3C. For each i, some R{,,j, succeeds in satisfying Ri.

90 P. Watson

Proof. Fix the first R{i,j, such that r(i,j) > z’, i.e.

CA(r(i,j)) ‘si(r(i.j))

after Rti,jj is injured for the last time, where z’ is defined in the proof of Sublemma 3A.

Then at stage Si(r(i,j)) we choose a new candidate pair (n, y ) and fix

Y E R,(r<i.i,)

so y E B. There may be several possible pairs. Sublemma 3A tells us that at least one of

these pairs, say (x’, y’), will satisfy Ri. Suppose we do not choose (x’, y’), but instead choose some other pair (p, 4).

Then we show that cA(r(i,j)) = Si(r(i,j)), contrary to hypothesis. The only possible problem pair is (p, 4). However we know

p E @{4)

and 4 E B and yet (p, q ) will not serve as a witness pair, by assumption. So p E A and therefore

Asi(rcu,) ] r(i.1) G A so cA(r(i,j)) = ri(r(i.j))

contrary to assumption. So by contradiction we must choose (p, q) = (x’, y’), y’ is fixed so y’ E B and

n $ A. SO Ri is satisfied. Cl


Corollary. For any &-high e-degree a, there exists an infinite sequence

Sl, s2, . . . , s,, . . . of s-degrees such that for all n:

s,,=,a, and s,,+~~~s,,

Proof. Repeated application of the Theorem. Cl

Note. The proof of the Theorem above does not carry over to all Z2 e-degrees. Highness is required to prove that every Ri is satisfied, specifically in Lemmas 3A and 3C.

This leaves the question of whether there exists a (non-A2, non-high) X2 e-degree which contains only finitely many distinct s-degrees.

Copestake [3] has shown using the notion of terse set that all l-generic e-degrees contain infinitely many distinct s-degrees, and that there exist Z2 l-generic e-degrees which are not in A2.

Although we have not been able to answer this question, we conjecture that there is no such e-degree, and that every e-degree contains infinitely many distinct s-degrees.


4. pc-Degrees

By now, having seen the rich structure which it is possible to have within a single degree, we might speculate that for any distinct reducibilities sY and Q, such that A s,, B +A 6, B, there is no such thing as a y-contiguous z-degree. The following result is not a counter-example, but suggests that things may not be so simple.

Theorem 4. There exists a 17, PC-degree a: 0 < a < 0’, such that if A and B are Ill sets in a then A ss B.

Proof. We construct a set A of the required degree. We must satisfy for all i the requirements:

Ri: [A= yF&+=]+[A=@&@+Af]

where & and Ai are s-operators which we construct; { (M$, Yiy;:, @ii)}ioo is a list of all possible triples from r.e. sets, pc-operators and pc-operators, respectively.

Strategy for Ni Initially A0 = w. Choose x $ Wi and wait until x enters W;:, then extract x

from A.

Strategy for Ri Wait for a stage s at which

A [Z(i) = YF [ Z(i) & I& [ Z(i) = C$ [ Z(i).

Consider each axiom of Yi:

Xe y~YI#n~....Y"l

for some n, where yl, . . . , y,, E I8$. We must have for each j E {1,2, . . . , n}:

yi = @a.. . s &tlsA

for some m. Look at the axiom for each j. If every axiom is of one of the forms:

(a) yi E * then set (a) Yj E A@, (b) Yj E S/” then set (b) yj E Aczl, (c) Yj E S,cZ,, . . pxs . ’ Irnl then set (c) yj E Atxl,

then proceed as follows. If some axiom is of the form (a), set x E rtyj) for that j. If not, but some axiom is of the form (b), set x E rtyjl for that j. If not, but some axiom is of the form (c), set x E r{fil for that j. Otherwise set x E r{fi) for the smallest yj in the original axiom.

92 P. Watson

However, if one of the Yj has an axiom of the form

y, e @fZl.. .s %I1 J

withx${zr,..., z,}, then consider every zk in the use of every Yj. If one of the zk is not fixed in A by a requirement of higher priority than Ri,

then fix x in A, priority Ri, and extract that zk from A. If all of the zk are fixed in A by higher priority requirements we can do nothing.

We say Ri is obstructed by the higher priority requirements.

Construction

Stage 0

A,,=w, &,, = Ai,, = 0

and define l(i) = i + 1 for all i. Attempt to satisfy No. Fix 0 in A, priority No (highest).

Stage s + 1. For each i E (0, 1, . . . , s} such that

A, [ Z(i) = Y$‘.’ ” & 8’i.s [ Z(i) = e; Is [ l(i)

perform the strategy for Ri up to the highest x in A [ I(i). Increase l(i) by a factor Of 2. GO tO Ri+l.

See if there is any Ni which is not satisfied on s (i.e. for which A, [s = W,,, [s). If so choose the least such i. For the first x on which it is possible, create and/or preserve the disagreement

A(x) + I%(x)

priority Ni (respecting higher priorities).


Lemma 1. NO is satisfied.

Proof. NO is the requirement of highest priority and so is never injured. By the construction, No chooses 0 as its witness. Initially, 0 E A0 and 0 $ W,,,. If

0 $ W, then no action is ever needed to satisfy No. Otherwise 0 E W, so at some stage s we remove 0 from A, and No is satisfied. q

Lemma 2. Every Ri and Ni, i > 0, 13 injured finitely ofren.

Proof. Without loss of generality, consider Ni. Assume all N, and R, for e < i are injured finitely often and eventually satisfied.

Then Ni is injured finitely often, by the following argument. When N,, e <i, causes x to leave A, N, injures Nip but N, is forever satisfied

after that, so each N,, e < i, injures Ni at most once.


When R,, e < i, causes x to leave A, R, is satisfied at that stage, and by assumption R, is finitely injured, so each R,, e < i, is eventually satisfied forever. So each of the finitely many requirements of higher priority than Ni injures Ni finitely often. Cl

T.R~UM 3. Every Ni in eventually satisfied.

Proof. By Lemma 2 there is a stage after which Ni is never injured again. After .this stage Ni chooses a new candidate and this candidate satisfies Ni by a similar argument to that used in Lemma 1 (to show that N,, is satisfied). Cl

L~IIUIUI 4. Every Ri is eventually satisfied.

Proof. We have shown above that Ri is injured finitely often. Consider the attempt to satisfy Ri begun at this stage.

If we are able to extract from A some z such that

y E @~Z,*. . . IZ,. . . ,z,) and XE [email protected]

and x is fixed by Ri, then because Oi and !& are pc-operators no other axioms will ever enumerate x and Ri succeeds.

If we are not able to create a disagreement in this way, then we attempt to satisfy Ri by duplicating, as far as possible, Si axioms in Ai and Yi axioms in &. The strategy for Ri can succeed in this on every x except those (possibly infinitely many) on which the strategy is obstructed by higher priority requirements.

Note, however, that each requirement can only fix a single element at any one time, and because we are assuming that every R, for e <i has been satisfied this gives a fixed finite set D say. 07 is the set of elements on which R, is obstructed.

Now the set enumerated by SD will be r.e. So in the limit we will have

Wi=AtU Wk and A=TFU W,

for some k and 1. This is enough to satisfy Ri because

cu w<,c

for any r.e. set W and any set C. Thus we have shown that

Wi=AfU WkS-,Af and A = r,* U W, es r,*

which is enough to show

%.<,A and A SsWi

as required. Cl End of Proof Theorem 4

94 P. Watson

Note that we have not shown that every set in the PC-degree of A is in the same s-degree, because there may be sets in Ir;; - 17, which are not covered by the above proof.

5. nm- and npm-Reducibiity

We now define two new forms of restricted e-reducibility which slightly strengthen all the results in this paper for no extra work.

Definition. We say an e-operator Y is an nm-operator if for every element x there is a total recursive function f and at most two axioms which enumerate,x:

(i) exactly one axiom of the form x E Y(f(X)), (ii) at most one axiom of the form n E Y’@.

Definition. We say an e-operator Y is an npm-operator if for every element x there is a partial recursive function f and at most two axioms that enumerate x:

(i) exactly one axiom of the form x E !PytrcX)), iff f (x) is defined, (ii) at most one axiom of the form x E Y@.

Definition. A is nearly m-reducible to B (write A G,,,,, B) if A <e B via an nm-operator Y.

A is nearly partial m-reducible to B (A d,,,,,, B) if A 6, B via an npm-operator Y.

Our intuition (and justification for the names) is as follows.

As,,B iff A=@UW

where 8 is an m-operator and W is an r.e. set. In, for example, s-reducibility, A = Y’ U W (where Y is an s-operator) is good

enough to make A ss B, and we have used this fact in Theorem 4. However in the case of m-reduciblity we can have A = eB U W where 63 is an m-operator, but not A G,,, B. So we say A is nearly m-reducible to B.

npm-reducibility and pm-reducibility are related in a similar way. Note also that both dnpm and snrn are transitive and reflexive so it makes sense

to talk about nm-degrees and npm-degrees. We can list all npm-operators (pairs of partial recursive functions and r-e. sets)

but not all nm-operators, as this would require a list of all total recursive functions.

We justify inflicting yet more forms of restricted enumeration reducibility on the world, as follows.

nm- and npm-operators are in fact very common in constructive proofs in the


literature, but have never been recognised as classes of operators in their own right.

In particular, in each of Theorems 1,2 and 3 in this paper, A is an nm-operator (this is easily seen from the Strategy for Q in each case). Furthermore, in each of these Theorems r is a c-operator, so we have actually proved:

Theorem IA. For any & set A, there exists a & set B such that:

(9 B=%,A, (ii) A *S,B

(iii) A &B. ’

‘llreorem 2A (3A). For any set A which is in A2 (&-high) there em& a A2 (&high) set B such that:

(9 B%,A, (ii) A C, B,

(iii) A #S B.

These two new reducibilities appear follows:

in our diagram of known implications as

See [2] for definitions of i-, pm-, and s-reducibility.

6. Further note

If we examine further the Strategy for P in Theorems 1 to 3 we see that in each case we are using the same c-operator r, and that this r can be specified before the construction begins, i.e. if we define a total recursive function

f (0) = 1, f(s+l)=f(s)+s+2,

then 0 E rto* ‘) and for every x > 0,

X E IWx-l)+l.f(x-1)+2.. .f(x)l

and this is the only r-axiom enumerating X. This allows us to reason that requirement Ri is only injured when elements 0, 1, . . . , i are removed from A in Theorems 2 and 3 (Lemma 1 in each case).

Thus if we modify the Strategy for Rj in Theorems 2 and 3 so that Ri never fixes elements less than or equal to f(i) in B, Ri is never injured. Furthermore we still satisfy Ri this way because

A = ,@+A = @;-(o*. . >f(i)) u W 8

where W = @F rf(i)+l is r.e. (We can split the use of <pi like this because Gi is an s-operator.) Now the stages at which Z(i, s) increase still give an effective enumeration of A as follows:

Stage 0 Set A0 = 0.

Stage s + 1

Set A,+1 =A, U C, U W,,,. where C, = @;;-tov 1, .f(i)l and t is the (s + 1)th stage at which I(i, s) increases.

In this way we achieve a zero-injury proof Theorems 2 and 3.

Acknowledgements

The results in this paper form part of the author’s Ph.D. Thesis (University of Leeds) and the author would particularly like to thank his supervisor Barry Cooper for help and encouragement.

During this time the author received financial support from the Science and Engineering Research Council, which he gratefully acknowledges.

References

[l] S.B. Cooper, Partial degrees and the density problem, Part II: the enumeration degrees of the & sets are dense, J. Symbolic Logic 49 (1984) 503-511.

[2] S.B. Cooper, Enumeration reducibility using bounded information: counting minimal covers, Preprint, Department of Pure Mathematics, Univ. of Leeds, 1987.

(31 C.S. Copestake, Enumeration degrees of & sets, Ph.D. Thesis, Department of Pure Mathemat- ics, Univ. of Leeds, 1987.

[4] R. Friedberg and H. Rogers, Jr., Reducibility and completeness for sets of integers, Math. Logik Grundlag. Math. 5 (1959) 117-125.

[S] K. McEvoy, The structure of the enumeration degrees, Ph.D. Thesis, Department of Pure Mathematics, Univ. of Leeds, 1984.

(61 K. McEvoy, Jumps of quasi-minimal enumeration degrees, J. Symbolic Logic 50 (1985) 839-848. [7] K. McEvoy and S. B. Cooper, On minimal pairs of enumeration degrees, J. Symbolic Logic 50

(1985) 983-1001. [8] J. Myhill, Note on degrees of partial functions, Proc. Amer. Math. Sot. 12 (1961) 519-521. [9] P. Watson, On d-recursively enumerable sets and s-reducibility of X2 sets, Ph.D. Thesis,

Department of Pure Mathematics, Univ. of Leeds, 1988. [lo] S.D. Zakharov, e- and s-Degrees, Algebra and Logic 23(4) (1984) 273-281.

on restricted forms of enumeration reducibility

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