Recursion Theory and Computational Complexity (C.I.M.E. Summer Schools, 79)

G. Lolli ( E d.)

Recursion Theory and Computational Complexity Lectures given at a Summer School of the Centro Internazionale Matematico Estivo (C.I.M.E.), held in Bressanone (Bolzano), Italy, June 14-23, 1979

C.I.M.E. Foundation c/o Dipartimento di Matematica “U. Dini” Viale Morgagni n. 67/a 50134 Firenze Italy [email protected]

ISBN 978-3-642-11071-9 e-ISBN: 978-3-642-11072-6 DOI:10.1007/978-3-642-11072-6 Springer Heidelberg Dordrecht London New York

©Springer-Verlag Berlin Heidelberg 2010 st Reprint of the 1 ed. C.I.M.E., Ed. Liguori, Napoli & Birkhäuser 1981 With kind permission of C.I.M.E.

Printed on acid-free paper

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C O N T E N T S

.

S HOMER

: Admissible Recursion 'Iheory

Pag.

7

B. E.

: Computational Complexity and recursion theory : A Survey of S e t Recursion

Pag.

Pag.

31 101

Pag.

109

: C o n s t r u c t i o n s i n t h e Recursively Enumerable Degrees Pag.

172

: Recursively I n v a r i a n t Recursion meory

229

D. G.E. R.I.

JACOBS

NORMANN

SACKS SOARE

W. MAASS

: P r i o r i t y Arguments in Higgler Recursion

Pag.

CEN TRO I N TERNAZIONALE MATEMATICO ESTIVO (c.I.M.E.1

ADMISSIBLE RECURSION THEORY

STEVE HOMER

ADMISSIBLE RECURSION THEORY Steve Homer DePaul University INTRODUCTION The purpose of these lectures is to develop some deeper results in a-recursion theory which will hold in somewhat more general setting than L(a) and in particular in many other admissible sets and structures.

In addition, I will briefly mention

some applications to structures which arise from other areas of recursion theory and which are inadmissible. An important underlying idea behind many of these theorems is the notion of a dynamic argument. by this is the following:

In general what is meant

In w-recursion theory, the starting

point for all of these generalizations, the exceedingly strong closure properties of w are used in almost every construction, often without a second glance. Almost any operation on finite sets yields a finite set and in particular the image of a finite set under any function is finite. Uhen we consider a-recursion theory, a C1-admissible, we of course loose a great deal of the closure present in w-recursion theory. have a certain weak closure property

However we still

- the image of an a-finite

set under a CI (La 1 function is a-finite. But as soon as we do a construction in which a C (La) (or C 3 (La) .) function arises,

...

and such does occur in almost every priority argument, then we immediately run into trouble. (We will see a concrete example of this shortly when we start talking about Post's problem in this setting. While we have lost the strong closure properties in

L(a)

we have gained the use of much work in set theory, mainly by ~ o d e land Jensen, in which many deep properties of L have been developed.

It is just these properties of L which often save us

when we run into trouble because of a lack of closure under certain functions. Now some properties of L work, under suitable conditions, in other settings, while others are really peculiar to L. icular many of the deeper results about L depend

In part-

on taking

Skolem hulls of certain sets in L(a), taking the tranitive collapse of that Skolem hull, and being able to determine exactly what that transitive collapse looks like, namely, an initial segment of L(a)

(an L (y) for some ~ ( a ) .

These collapsing arguments tend to be very specific to L and almost never work in other settings.

In particular, they fail

when the universe is changed by constructing L relative to a given predicate.

So, to get constructions to work in more gen-

eral settings, we need to eliminate, if at all possible, these collapsing arguments peculiar to L.

We want to give a "dynamicn

argument (basically more similar to the original one for w), which views r.e.

sets is being listed and increasing in an

effective manner, not as being defined by a El-formula which is really the crucial property making collapsing arguments work. In order to make this more explicit

I want now to turn to a

couple of concrete examples.

Both examples depend upon the same

method and so the second will be given in much less detail. After presenting these examples I will indicate some extensions and applications of these results to other areas and in particular to admissible sets and structures. We will also, and of course this is the reason these methods were first developed, gain some knowledge of the structure of a-r.e. degrees. In what follows I am assuming a familiarity with the basic facts and definitions of a-recursion theory.

These are

given in the first of Sacks' lectures in this volume.

For a

more detailed account see the papers by simpsonpg or ~hore(141. 1.

Post's Problem for a-Recursion Theory Let o be a C1-admissible ordinal.

We want to prove the

following theorem. Theorem: There exists two a-r.e. sets which are incomparable

.

with respect to < This theorem was originally proved by Sacks -a and Simpson 1111. Their argument depended heavily on using properties of Skolem hulls in L. I will present another way of proving this theorem based

largely on ideas of R. ~horeb41,(161

.

applied to this problem by Simpson t191

Shore's ideas were The proof is more

dynamic, more "constructive" if you will, than the original proof and, as we will see, is more adaptable to other settings. It is not so dependent on the special properities of initial segments of L. The argument will be presented by starting out with the basic ideas for solving Post's problem from w-recursion theory.

I assume some familiarity with that argument.

As we try to

carry out the argument in the setting of L(a) we will meet with various difficulties for which we will propose solutions. Finally, we will put all of this together to get the actual construction and proof. Now, we are going to construct two a-r.e. We require that A g a B and BfaA.

sets A and B.

In fact we will construct them

to satisfy the stronger incomparability, A dWaB and B & , A . That is, for each e e L(a), we want to ensure that

SAe : {elA

+B

and

A bit of notation is necessary here. form

Requirements of the B are called A-requirements, requirements of the form Se

~ft

are B-requirements. In these requirements we are identifying a set C with its characteristic function,

For any set CSa, we let

F

= a-C.

In the construction of A and

B we let 'A (Ba) = set of elements enumerated in A (B) by stage a.

=

Finally let A"

A ~ ,B<'

6
=

U

B6.

6
The method used to satisfy the above requirements is the A same as for U-recursion theory. Consider (el # B. At various stages of the construction it will appear that we can use A to to enumerate

E.

That is, at a stage a we will see that

-

(e]ti%O for some x which is in B".

We would like to put x into

~(x) = 1, and at the same time try to keep

Bar insuring that

A< a

the computation {el

(x) = 0 correct. This computations uses

-

a-finite subsets of A'' finite subset of''A

- and so we try to keep the

and''A

u-

from intersecting A.

Of course, we do the same for requirements {e'l

B

# A.

For

such requirements we try to keep elements out of B and to put them into A,so they tend to conflict with the A-requirements. Such conflicts ariseinclassical recursion theory as well and are resolved by giving priority to a requirement s over a B if, say, e requirement Set el. If we do this, an inductive argument then proves that each requirement is eventually acted upon successfully.

In addition, for any e, there is a stage of

the construction by which any action taken on behalf of requireor sB , where eO, el < e, has already been taken. In 0 1 other words all activity on behalf of such requirements has al-

ments

sA

ready settled down. This gives rise to the first major problem in doing this argument for a.

The function taking the eth requirement to the

first stage at which it has settled down is C2(L(a)) and so we can't use the admissibility to get a bound on the activity of the requirement. For example, lets look at the particular admissible ordinal a =

It could be that 5: does not stop acting until after

stage

x2,

I

etc.

S:

until after stage

I A1, the S 1 until after stage A

If this were the case then there might be no stage A

past which the requirements,

is never injured, there being no

bound on the activity of the first ,-many requirements.

Picture:

requirements

stages past which requirements have settled down

The problem is that our priority listing of requirements is too long.

If we just list our requirements in the usual, most B straightforward way, st, ,:s S$, e set --, e
6, 4,

..... 8,

an initial segment of this list may never settle down.* Shore devised the following method to avoid this problem. Note that requirements of the same kind do not conflict.

We

take advantage of this trivial fact by arranging our requirements in groups or blocks, each block containing requirements In fact it can be shown that this problem is avoided for this particular a. problem.

However for many admissible a this is the crucial

We essentially treat each block as a A B B ; , R1.. single requirement. So we will have blocks Ro, Ro, R B where :R is a block of A-requirements, Re a block of B-requireof the same type.

.

ments.

Each block consists of lots of requirements but as

they don't conflict we give each requirement the same priority. By doing this we can, for troublesome a, get away with less than a-many priorities and blocks and so will be able to bound the activity of an initial segment of blocks, even though the function giving this bound is

5 (L(a)1.

How short do we need to make the list of blocks? enough so that no

5

Short

function from an initial segment of the

list of blocks to a is unbouriddd. Definition: is an

Let A = Z, cf(a) = least

5 a such that the there

f : B + a , f % ( L ( a ) ) and range(f) unbounded in a.

We will make the list of blocks have length A, thus ensuring the above property. Now this takes care of part of the problem as there are now not too many blocks. don't conflict.

Within each block, requirements

We need to make sure that no block is too

large so that we can find a stage at which all activity in a given block has settled down. block.

Let's look at the very first

The requirements in this block, since they have the

highest priority, are all acted upon once at most. of requirements which are acted upon is a-r.e..

The set

If we could

ensure that this set is a-finite, we could then conclude there is a stage past which no requirement in the first block is acted upon.

How can we do this? Make the block bounded

below a*, for any a-r.e.

set bounded below a* is a-finite.

(For a proof see Devlin 131

or Sacks' lectures in this

To accomplish this we will use a projection

volume.) 1-1

p: a --+a* a*.

to enumerate our requirements in a list of length

Our blocks will then be subsets of a*. We have one more problem to surmount.

We would like to

We could divide a* into

divide a* into C 2cf(a) many blocks.

C 2cf (a*) many blocks, where C 2 ~ (a*) f = least B 5 a such that there is an f: (3

+

a*, f C (L(a)) and range (f) unbounded in

a*, but this could be too many blocks.

The following crucial

lemma says that this cannot happen.

C2cf (a) = C 2 ~ (a*). f

Lemma:

Proof (simpson 1191

):

1- 1

Let p: a +a* be a-rec.

Using the admissibility of

u ~t is straightforward to show that if X S a is unbounded then

Pm is unbounded in a*. NOW

let 6

<

unbounded in a.

a, g: 6-

be C

The peg: 6---*

(L(a)) with range (g) is

C2 (L(a)) and by the

above fact, range(pog) is unbounded in a*.

So Z2cf(a*) 5

.

C2cf (a)

Conversely let h: y - ~ * be C (L(a)) and range (h) unbounded in a*.

k is C SO

cp

Define k: y--4-a

by, for v < y,

(L(a)) and range (k) is unbounded in a. f (a)

5 C2cf (a*).

2.

The Construction and Proof Using the above lemma we can now put all of our consid-

erations together and give the blocking construct-ion. Let H: A = C 2cf (a) = C2cf (a*)

-4a*

be a C2 ( ~ ( a)) function.

Without loss of generality we can assume,

. . . -< H(y) 5 . . . < a*

1.

~(02 ) H(1) 2

2.

H ( 0 = y y 6 ~ ( y ) rfor any limit ordinal 6.

and

For details of this and other properties of the blocking function H and its approximation see Simpson rl91

.

Using the above H function we can now say precisely what

4, R . . . of requirements will be. Ro = SO, S1, S2, . . . , Se, . . . e < H(0) B B B B Ro = So, S1, . . . , Se, . . . e < H(1) our blocks,:R A

:R

A

A

A A * So, S,,

B = So, B S1, B R2

R ,

A

A

A . . . , Sen ... B . . . . . . , SeI

e < H(2) e < H(3)

The picture is:

We will now give the construction of the a-r.e.

sets

A and B.

In this construction three "details" are left out.

These require some care but tend to obscure the proof so I will briefly describe them here and leave them to you. 1.

It is convenient to keep the witnesses x for different

requirements disjoint and necessary to make sure we have an endless supply of them.

To do this, break a up into a-many

disjoint a-rec. sets, each unbounded in a. set to each requirement in each block.

Assign one such

The witnesses for

a requirement are constrained to come from the associated set. 2.

Since H is E *(L(a)) we cannot effectively calculate its

value.

Instead we approximate H with an a-rec. function G.

G has the property that when restricted to an initial segment

of A, G equals H on all but initial segment of stages.

We

will refer to H in the construction but keep in mind that what we really do is calculate values of G, a "nice" approximation to H. 1

3.

Using the a-rec. projection p: a

-

1

-+- a*

we can assume

that the requirements are enumerated via ordinals less than B Se a*. So from now on we assume we have requirements

<,

where e < a*.

Of course many of these requirements are

undefined but all of the original requirements are contained in this enumeration. The construction takes place in a-many stages. AO = B 0 =

Initially

8.

Stage a: Step 1:

For each

odd

v < A proceed as follows.

Find

the least e < H(v) and the least x < o in the witness set for :S

such that

A requirement for Se.

1.

There is no

2.

B<'(x)

3.

x is not being kept out of B for some requirement in an

= 0 = {ela (XI.

earlier block.

That is, a requirement in some block deter-

mined by H ( v ' ) where v ' < v. If no such e and x exist do nothing and proceed with step 2. Otherwise put x in 'B

and create a negative requirement

keeping out of A all elements of A" = 0.

of {elf:x)

used in the computation

If x is contained in any negative require-

ment on B that requirement is destroyed. Step 2:

Do the same as in step 1 only for even v < X

interchanging the roles of A and B.

The following claim is the main ingredient in showing that the construction works. Claim 1:

For each v < A there is a stage a.

activity on all blocks past a.

2

v has ceased.

such that all

i.e. At no stage

is any requirement in any block 5 v injured or acted

upon. Proof: Case 1:

By induction on v < A . v = 0

The proof here is the same as in the successor case 2. Case 2:

v = v'

+

1

By induction there is a stage a such that by stage a 1 1 all blocks up to and including v ' have settled down. Then past stage al

no requirement in block v can be injured.

Consider the set I = {ele < H ( v ) and for some a2

1

alr

A is acted upon a't stage up} (Without loss of generality Se we are assuming here that v is odd.)

I is a-r.e.

and

< a*, hence I is a-infinite.

I "(v)

Define f: I ---+ a by f (el = least a (st is acted upon at stage 0).

f

[a is bounded by the desired stage ao.

Case 3:

Then

v a limit ordinal

Define g: u

4

a by, for y < v, g (y) = least a (block

y has settled down by stage a), Then g is C

(L(a)) and SO

since v < A , g rv] is bounded by the desired stage oo.

__I

We can now prove the theorem. Claim 2: Proof:

A

f waB and B

1 waA.

We will show A f waB.

Assume not, then A = {elB for some e < a*. least even ordinal such that e < H(v). past which all blocks

Let v < A be the

Let uo be a stage

5 v have settled down.

an x < u0 in the witness set for

sB <5

Then there is

as a v-requirement and

a stage al > X, u0 such that {elB (XI = A<',(X) = 0. AS O1 x > ao, x is not in any negative requirement for A of higher priority and x jf A.

So at stage ul either x is put into

~'1, contradicting x # A or there is a requirement for B S present at stage al and hence never injured, insuring e that {elB f A.

+

3.

The Splitting Theorem The blocking method described in the previous sections was

first developed by Shore to prove the following generalization of Sacks' splitting theorem. Theorem (Shore C163 : D not a-recursive. that

~n

B =

$,

Let C and D be regular a-r.e. sets with

Then there exists a-r.e. sets A and B such

A U B = C, A ,

c

D{~A

and D&B.

All of the usual corollaries of the splittinq theorem concerning Turing degrees are true for a-degrees as a consequence of this theorem.

For instance below any non-zero

a-r.e. degree there are two incomparable a-r.e. degrees. We have already encountered in our proof of Post's problem all of the major obstacles to provinu Shore's splitting theorem. This being so I will constrain myself to briefly mentionins the main ideas of the proof. The requirements which we satisfy are of three types:

::s :s

{elA

f

D

{elB f D

R : x E C + X E A U B As

usual we construct A and B in a-many stages.

To

satisfy :S we follow Sacks' original strategy of trying to preserve the equality {elA = D on lonsek and longer initial segments of a.

The idea here is that if we succeed in pre-

serving eauality all the way up to a then we could a-recursively.

compute D

Since D is not a-recursive this can't happen,

but if {elA = D we could show that the preservation up to a A must be satisfied. : would succeed. So Se S is handled similarly.

To satisfy R we insist that when an ordinal x

in this volume or Kleene r9J.

Harrington

1 7 1 was

able to apply

the results of a-recursion theory to solve a particular version In order to do this

of Post's problem for higher types.

Harrington used the following strong solution to Post's-problem for a-recursion theory. There exists a pair of integers m,n

Theorem (Shore h7)):

such that for a11 B and a where (L~KB], m and n are indices for a-B-r.e,

r,

B)

is admissible,

hyperregular sets such that

neither is a-B-recursive in the other. The proof of this theorem is a refinement and extension of the blocking proof presented earlier.

In particular, extra

requirements are necessary to make sure the sets constructed are hyperregular.

To make the construction uniform the para-

meters used in the construction, a* and E2cf (a) , cannot be used outright but must be "guessed at", that is approximated, during the construction. Harrington's version of Post's problem considers reduction procedures which allow subconstructive parameters.

These para-

meters are essentially ordinals which arise from computations in an object of type n individual, parameters. two r.e.

+

2 with type n

Harrington

- 1, that is sub-

171 shows that there are

sets of subconstructive ordinals which are incomparable

with respect to the above reducibility. The proof proceeds by reducing PosC1s problem in the above setting to Post's problem for an admissible structure of the

187,

form < L ~

E,

8)

and then applying Shore' s result.

Recently, a strong version of Post's problem for higher types has been proved by Sacks.

The blocking method is also an

ingredient of his proof.

For a more detailed discussion of

this see Sacks' article in this volume. A second area of application for these methods has been

to B-recursion theory where B is a weakly admissible ordinal (that is Zlcf(B) 2 B*).

Post's problem was originally solved

for weakly admissible ordinals 6 by Sy Friedman

147 .

Friedman's proof was very close to the above argument for a-recursion theory.

The one important difference is that

requirements have to be added to insure that the two sets constructed are tamely-0-r.e..

The splitting theorem works

similarly in this case, (see Homer

MI.

that the admissible collapse of Maass

It should be noted

501 can be

used to get

these results directly from the corresponding results of a-recursion theory. Admissible Sets In this section I want to present the axioms for admissible sets due to Kripke and Platek.

Models for these axioms were

envisioned as structures within which to do recursion theory. As such, it is natural to try to determine the solution to Post's problem in various admissible sets.

In particular, it

would be interesting to see if there are certain necessary and sufficient conditions for Post's problem to have a positive solution in an admissible set.

Before addressing this

question we need to briefly investigate How much recursion theory can be developed in these structures.

For an in-depth

look at admissible sets see Barwise r21. Definition: 1.

union:

-

An admissible

x

E

A

UX

E

set A A

is a transitive set satisfying

2.

pairing:

3.

AO-comprehension:

VaJb)/x(x 4.

x,y

c

b

-

b0-bounding:

(\ds3tVx

E

~

E

A

x

If E

{x,yl

E A

4 (x) is any A0 formula,

a 4 #(XI).

if ((x,y) is any A0 formula, (Vx3y((xIy))-r 3

E

y~('$(xIY))).

There are many stronger forms of comprehension and bounding which are provable from these axioms.

We now state one of

these in order to relate these axioms to the replacement principle of a-recursion theory. Theorem (C1-replacement):

If

A

Zl(A) then ffa is an element of

is admissible, a

E

A

and f is

A.

Hence we see that the same replacement principle is present in both a-recursion theory and the theory of admissible sets. The hope is that by studying this theory we will gain some understanding of exactly what properties are most important to carry out various recursion theoretic arguments.

Studying

admissible sets allows us to try to apply our methods to structures other than L(a) without giving up the strong closure properties provided by admissibility. Typical examples of admissible sets are the collection of hereditarily finite sets, sets of hereditary cardinality less than

K,

for any cardinal

K,

and models of ZF.

The

following definition links up the notions of admissible set and admissible ordinal. Definition:

Let A be admissible.

A = least a(a is an ordinal and a

Then

/

I A ~=

A).

Claim: a is an admissible ordinal iff a = sible set A.

ordinal height of

IAl

for some admis-

Sketch of proof:

Assume a is an admissible ordinal.

Since

then L(a) satisfies C1-replacement it is easy to check that L (a) satisfies the axioms for an admissible set and Conversely assume a =

I A ~ for

some admissible A.

out the construction of L(a) within A. is the real L(a) C -replacement.

IL

(a)1 = a.

Now carry

By absoluteness, this

and by Z1-replacement in A, L(a) satisfies So a is an admissible ordinal.

1

--i

We now turn to exploring recursion theory on admissible sets. Definition:

5

Let A be admissible.

A is A-r.e.

if B is E1(A).

1.

B

2.

B S A is A-rec. if B is A1(A).

3.

B

A is A-finite if B

E

A.

Theorem (Enumeration Theorem):

There is a universal A-r.e.

relation W(x,y) such that, if Wa = { y l ~ W(a,y)), Wa ranges over all A-r;e. Corollary:

sets as a ranges over A.

There is a non-A-recursive, A-r.e.

set.

The above two facts are about as far as we can go without any additional assumptions on the admissible set.

In particular,

there is no hope of a positive solution to Post's problem for all admissible sets as the following theorem of Harrington shows. Theorem (Harington

171):

which every A-r.e.

set is A ~ ( A )or zl-complete.

There exists an admissible set A in

There are a couple of interesting open problems associated with this result. structs has height

The admissible set which Harrington con:

a2.

Tt is not known if Post's problem can fail in an admissible

set of height

or in one which is locally countable.

Some even simpler properties of r.e. sets in classical recursion theory are not provable in arbitrary admissible sets.

For example, consider the following property of r.e.

sets. Separations for r.e.

sets:

For any disjoint r.e. sets A and C and B 0 C =

B there is a recursive C with A

/.

Picture:

The separation property fails for r.e. sets. are r.e. sets A . C with A

That is there

and Bo such that there is no recursive set

5 C, B n C =

#.

Even this property of r.e. sets is

not true in every admissible set. Theorem (Harington

171) :

There exists an admissible set A

in which separation for A-r.e.

sets holds.

In fact Abramson Ll] has shown that such an admissible set can have height any countable admissible ordinal. One reason that many of the important arguments from w-recursion theory fail to work in an arbitrary admissible set is that an admissible set may lack a definable wellordering. It is just such a wellordering which enables the dynamic approach to all of these arguments to succeed.

Without it

there is no canonical way of carrying out an effective

inductive construction as is done in priority arguments.

Even

a simple selection theorem, as is necessary to prove elementary principles like non-separation for r.e. sets, cannot be proved for admissible sets without wellorderings. Something like the following notion is necessary. Definition:

An admissible set A with

IA~=

a has a recursive

wellordering if there is an A-recursive bijection from a onto A. Theorem:

If A has a recursive wellordering then Post's

problem has a positive solution for A. Proof:

For such an admissible set all of the machinery

developed in sections 1 and 2 can be brought to bear. Essentially the same proof that was given there can be carried out for an admissible set with a recursive wellordering. Actually a slightly weaker property than a recursive wellordering will suffice to give a positive solution to Post's problem for an admissible set A.

Simpson

a many-one A-recursive projection from A to are a-finite a

thin prewellorderinq.

argument in [ld

IAI

calls

whose levels

He notes that the

can be used to show that an admissible set

with a thin prewellordering yields a positive solution to Post's problem. The general problem of giving necessary and sufficient conditions for a positive solution to Post's problem is wide open.

It would be most interesting to be able to find such

conditions for an admissible set or an admissible structure.

Bibliography Abramson, F., C1-Separation, to appear in

JSL.

Barwise, J., Admissible Sets and Structures, Springer, 1975. Devlin, K.J., Aspects of Constructibility, Springer Lecture Notes 8354, 1973. Friedman, S.D., Recursion on Inadmissable Ordinals, Ph.D. Thesis, M.I.T., 1976. Friedman, S.D., AMS

.

@-Recursion Theory, to appear, Transactions

Harrington, L., Contributions to recursion theory in higher types, Ph.D. Thesis, M.I.T., 1973. Harrington, L., An Admissible Set with no Intermediate El-Degrees, to appear. Homer, S.E., Priority Arguments in 0-Recursion Theory, Ph.D. Thesis, M.I.T., 1978. Kleene, S.C., Recursive functionals and quantifiers of finite type I and 11, TAMS 91, 1959 and 1963. Maass, W., Contributaions to a- and 0-Recursion Theory, Habilitationschrift, University of Munich, 1977. Sacks,. G.E. and Simpson, S.G., The a-Finite Injury Method, Ann. Math. Logic 4, 1972. Sacks, G.E., Post's problem, absoluteness and recursion in finite types, 1978 Kleene Symposium volume, to appear. Sacks, G.E., Three theorems on recursive enumerability in a finite type object, to appear. Shore, R., a-recursion theory, in: Barwise, editor, Handbook of Mathematical Logic, North Holland, 1977. Shore, R., Priority Arguments in a-Recursion Theory, Ph.D. Thesis, M.I.T., 1972. Shore, R., Splitting an a-R.E.

Set, TAMS 204, 1975.

Shore, R., XnSets which are An Incomparable, JSL 39, 1974. Simpson, S.G., Short course on admissible recursion theory, in: Fenstod, Gandy and Sacks, editors, Generalized Recursion Theory 11, North Holland, 1977.

19.

Simpson, S.G., Degree Theory on Admissible Ordinals, in: Fenstad and Hinman editors, Generalized Recursion Theory I, North Holland, 1974.

20.

Simpson, S.G., Post's Problem for Admissible Sets, in: Fenstad and Hinman editors, Generalized Recursion Theory I, North Holland, 1974.

CENTRO INTERNAZIONALE MATEMATICO ESTIVO (c.I.M.E.)

COMPUTATIONAL COMPLEX1 TY AND RECURSION THEORY

BARRY E. JACOBS

COMPUTATIONAL COMPLEXITY AND RECURSION THEORY Barry E. Jacobs University of Maryland Department of Computer Science College Park, Maryland 20742 Contents

0.

Introduction

1.

Abstract Computational Complexity Measures

2.

Blum-Rabin, Compression and Speedup Theorems

3.

Complexity Classes

4.

Speedable and Nonspeedable Sets

5.

Generalizations References

0.

Introduction

A general axiomatic approach to computational complexity was formulated by Blum 111.

His work was greatly influenced by

the work of Rabin 1231 and served to generalize to all complexity measures, results derived by Hartmanis and Sterns 181 for Turing machine computation times.

The speed-up theorem, due to

Blum, was probably most responsible for attracting great interest into the area.

Since then abstract or axiomatic complexity

theory has seen a great deal of activity.

This paper represents

a survey of some of the major products of that research. An outline of this paper is as follows. In Section 1 we introduce the notion of an abstract

computational complexity measure and later by Blum.

Examples

first considered by Rabin

of complexity measures

are given

as well as examples of measures that fail to satisfy Blum's axioms.

Several easy results are shown.

Namely, that recursive

increasing functions applied to the step counters of a measure produce new measures.

Recursive relations exist between step

counting functions across different measures (for the same partial recursive functions). to compute

In any measure, arbitrarily hard

recursive functions exist.

recursive way

Although there is no

to bound the complexity of a function by its

output, the opposite does hold.

We conclude Section 1 with some

results that relate the complexity of a computation and its

.

subcomputations

In Section 2 we prove results essentialy due to M. Blum. The first, called the Blum Theorem, tells us that in any complexarbitrarily hard to compute partial recursive

ity measure, functions exist.

The second

is Blum's Compression Theorem

which tells us how'heasure~setsof functions serving as complexity bounds (on classes of recursive functions) can be effectively increased to produce larger bounds.

The third result of Section

2 is Blum's well-known Speed-up Theorem.

any measure

recursive functions exist

This tells us that in which have no best

algorithm. In Section 3 we consider classes of recursive functions whose step-counting functions are bounded by a tion.

recursive func-

Such collections are called comptexity c Z a s s e s . We first

prove results that tell us when we may effectively increase a complexity class to include more functions. Borodin's Gap Theorem

We then consider

which tells us that there are very few

step-counting functions compared to the recursive functions. The McCreight-Meyer Union Theorem tells us that for certain sequences of functions bounding complexity classes, a recursive function exists whose complexity class is precisely the union of these classes.

The case in which classes are bounded by primitive

recursive functions is given as an illustration.

Section 3 is

concluded with the Meyer-Moll-McCreight Honesty or Naming Theorem. This tells us that for any sequence of partial recursive functions a "measured" sequence can be generated so that none of the complexity classes defined by the former are lost. In Section 4 we explore some close interaction between computational complexity and pure recursion theory. We consider classes of sets called speedable

and nonspeedable

and their

relation to various classes of recursively enumerable sets. Soare's index set characterization is proven for nonspeedable sets.

That is, an r.e.-set A is nonspeedable precisely when the

set of r.e. indices of sets not contained in ible to 0'.

A

is Turing reduc-

This is used to prove Marque's result that every

r.e. Turing degree contains a nonspeedable set.

It is also used

to show that the Turing degrees containing speedable sets are precisely the high ones (i.e. its Turing jump is

On).

We

conclude Section 4 with a discussion of speedable sets and their relation to well-studied classes of r.e. sets.

In particular, we

see that every maximal r.e. set is speedable. In Section 5 we consider an infinitary ahlogue to Blum complexity theory called a-complexity theory.

Specifically, we

examine the generalization of the Blum axioms in Kripke-Platek admissible recursion theory.

Several analogues to important

results of ordinary complexity theory are formulated and proven;

namely, the a-Blum, a-Rabin, and a-Gap Theorems.

We conclude

Section 5 with a discussion of those results of the o-theory that have been shown to generalize to the admissible case. 1. Abstract Computational Complexity Measures In this section we introduce the notion of an abstract computational complexity measure first introduced by Rabin I231 and later greatly exploited by Blum Ill.

Several illustrations

of complexity measures are given as well as examples of measures that do not satisfy the defining axioms.

For the remainder of

the section we prove same relatively easy results about computational complexity measures demonstrating the power of the theory. We begin by showing that any recursive increasing function applied to all the step-counters yields another canplexity measure. relations exist

i n a complexity measure always

We then show that recursive

between complexities of corresponding algo-

rithms in two different measures.

That is, "easy to computew

functions in one measure are "easy to compute" in another meaeure We next show that in any complexity measure, arbitrarily hard to compute recursive functions always exist. ed

to the case of 0-1 valued

This result is extend-

functions or predicates

yielding

the fact that there is no recursive way of bounding the complexity of a function from its output.

However, the converse does

hold,namely, that one can recursively bound an output by the complexity of a computation.

We conclude the section with tech-

nical lemmas that relate the complexity of a computation with the complexities of subcomputations. Definition.

An abstract computationat comptexity measure @

(or simply a comptexity measure) is an enumeration of the partial

recursive f d c t i o n s

1

e < u3

t o which a r e associated t h e p a r t i a l r e c u r s i v e step-counting functions

1

e < wl

f o r which t h e following axioms

hold:

COe

(1) $, (x) i s defined on a given x i f f Oe (2)

the

(XI i s defined;

predicate

i s recursive. W e s h a l l a l s o assume t h a t t h e

s :

and Universal Function

Theorems (Davis 15 1 ) hold f o r t h e enumerations { +e

I

e < w 1.

W e a l s o assume t h a t given any index e, both t h e function $,

t h e form of an algorithm " e m , and its step-counter

,

in

Oe may be

effectively retrieved. W e i l l u s t r a t e t h e notion of complexity measure with s e v e r a l

examples. (1) Let $e be t h e p a r t i a l recursive function (of one argu-

ment) computed by t h e Turing machine with Godel number e. Let Oe(x) be t h e number of s t e p s taken t o compute $,(XI by machine "e".

Clearly, t h e sequences

{Oel

c o n s t i t u t e a complexity

measure. (2)

Let

cpe

be t h e same a s i n (1). Let

Oe (x)

be t h e

number of squares used by "en i n t h e computation on x with t h e proviso t h a t i f

+e(XI 4 then Oe (XI t

.

The sequences { $e 1, I Qe1

a l s o c o n s t i t u t e a complexity measure. (3)

L e t 4,

be t h e p r i m i t i v e r e c u r s i v e function computed by

t h e equations with Godel number e and Oe

t h e number of recur-

s i o n s i n "em. Such does n o t c o n s t i t u t e a complexity measure s i n c e t h e schema are n o t capable of capturing a l l p a r t i a l recurs i v e functions. (4)

L e t 9, be t h e same as i n (1)- Let @,(XI be defined by

'

Qe("1

$=(x)t

0

if

t

otherwise

This doesn't c o n s t i t u t e a complexity m e a s u r e s i n c e t h e predicate M(e,x,y) * @,(x) = y

(Its recursiveness would

is n o t recursive.

imply t h e recursive s o l v a b i l i t y of t h e h a l t i n g problem.) For t h e remainder of this s e c t i o n w e present s e v e r a l r e s u l t s r e l a t i v e l y e a s i l y obtainable from the d e f i n i t i o n of complexity measures.

(The author i s n o t c e r t a i n who are t h e individuals t o

be c r e d i t e d f o r t h e s e r e s u l t s .

However, most of them can be

found i n t h e survey a r t i c l e by Hartmanis and Hopcroft [71.) W e f i r s t see t h a t

a r e c u r s i v e i n c r e a s i n g function applied

t o step-counting functions 1.1 Theorem,

y i e l d s step-countinv functions.

L e t O be a complexity measure with enumeration L e t f ( y ) be a r e c u r s i v e

I+,) and step-counting functions {(Pel. function such t h a t f ( y )

2 y.

Then if oOe)

i s a set of step-

counting functions f o r {$,). Proof.

For (11, s i n c e f i s recursive fo9 (XI+

For (2) , l e t M ( e , x , ~ )=

+

iff

@,(XI

iff

$e(xl t

.

0 if

f o

@,(XI = y

1i f

f

@,(XI # y

0

Since f ( y ) 2 y, w e check t o see i f t h e r e i s a w 5 y such t h a t Qe(x) =

W.

This can be done recursively s i n c e t h e search i s

bounded and {ae) s a t i s f i e s ( 2 ) . M(e,x,y) = 1.

I f no such w is found,

I f such a w is found, then t h e value of M(e,x,y)

is 0 i f f(w) = y; otherwise, it is 1. W e next show t h a t w e can find a recursive r e l a t i o n between

t h e complexities of corresponding algorithms i n two complexity measures. This tells us t h a t functions t h a t a r e "easy t o compute" i n one measure a r e "easy t o compute" i n a l l measures. L e t @,

i

be two complexity measures such t h a t {(,I.

a r e t h e i r corresponding enumerations. W e c a l l Q and

{ie)

h

@

paratZel

A

i f f o r a l l e < w, $Ie Z $Ie. 1.2

Theorem.

Let

@

and

h

two p a r a l l e l complexity measures.

@ be

Then t h e r e e x i s t s a recursive function r such t h a t f o r a l l e < w and f o r a l l but a f i n i t e set of x. Proof.

Let

c ( x , y ) = {e

/

e 5 x

A

&

Pe(x) = Y v ae(x) = Y).

Then our desired function r i s defined a s

10,

otherwise

Clearly, r i s a r e c u r s i v e function. W e check to see t h a t r s a t i s f i e s t h e conditions. Let x

and suppose

$ (x)

i s defined. Then

e

~ ( x , (XI Q )~ = max{

max {ae, (XI1, max 1 5 ~(.X I 1 ) e 'EC(x, Qe (x) 1 e'EC(x,ae (x) 1

Since x 2 e , t h e index e was one of those i n d i c e s checked in the computation of r ( x , e e ( x f )

it follows t h a t

If Qe(x) i s undefined then both l e f t and r i g h t s i d e s a r e

undefined

and t h e r e l a t i o n holds*

It follows t h a t the d e s i r e d

i n e q u a l i t y holds everywhere, b u t i n t h e bounded s e t Ixlx < e1.O We next show t h a t f o r any complexity measure, given an a r b i t r a r y recursive function f , t h e r e e x i s t s

another recursive

4 such t h a t any way of computing $ takes more than f s t e p s on

an i n f i n i t e s e t of inputs. 1.3

Theorem (Rabin [23 1 )

a recursive function.

.

L e t 9 be a complexity measure and f

Then t h e r e e x i s t s a recursive function

such t h a t f o r any index e o f $ (in Q) Qe(XI > f

(x)

on an i n f i n i t e s e t of x. Proof.

Define t h e recursive function $ by

lo

otherwise

cnl(x) 's 2 ( X I > = x i s some standard p a i r i n g function. I t follows from t h e d e f i n i t i o n of complexity measure and t h e

where

recurs$veness of f t h a t $ is recursive. To s e e

4 s a t i s f i e s t h e theorem, assume t h e r e e x i s t s some

index e f o r 4 and a

y

suchthat f o r a l l

x,

x > Y

oe

5f

(x)

(x)

.

L e t z be

.

y such t h a t nl

(2)

= e. Hence, f o r

this z

since

(be ( 2 ) ( f

( z ) by t h e clefinition of $.

But nl

(2)

= e, hence

and s i n c e 4e is total, t h i s i u a contradiction. I n Section 2 we w i l l prove t h e following: 1.4

Theorem (Rabin E231).

a r e c u r s i v e function.

Let O be a complexity measure and f

Then t h e r e e x i s t s a recursive 0-1 valued

function such t h a t f o r any index e f o r $

f o r a l l but a f i n i t e set of x. W e w i l l prove Theorem 1 . 4 i n a more general form (i.e.,

f

p a r t i a l recursive i n s t e a d of f r e c u r s i v e ) . One may i n t u i t i v e l y f e e l t h a t t h e s i z e of an output i n d i c a t e s the complexity of t h e computation required t o o b t a i n t h a t output. Theorem 1.5 t e l l s us t h a t we c a n n o t recursively bound t h e complexity of a computation by its output. 1.5

Theorem.

Let

be a complexity measure.

Then t h e r e does

n o t e x i s t a recursive function k such t h a t f o r each e

k ( ~ , $ ,(XI 1 ) Qe(x) f o r a l l but a f i n i t e set of x. Proof.

Suppose such a recursive k e x i s t s . Then t h e

complexity of any 0-1 valued recursive Ge must s a t i s f y

f o r a l l b u t a f i n i t e set of x. recursive. $

Clearly, k(x,O) + k ( x , l ) is

Hence, by Theorem 1.4 t h e r e is a 0-1 valued r e c u r s i v e

such t h a t f o r any index e f o r

f o r a l l but a f i n i t e set o f x.

$I

Formulas (1) and (2) y i e l d an

obvious contradiction. Although Theorem 1.5 tells us that we cannot recursively bound t h e complexity of a computation by i t s output, our next r e s u l t shows t h a t the opposite does hold.

Namely, t h a t w e may

recursively bound t h e value of a computation by its complexity. 1.6

Theorem.

L e t 4 be a complexity measure.

Then there e x i s t s

a recursive function h such t h a t f o r each e

f o r a l l b u t a f i n i t e s e t of x. Proof. -

Define t h e a u x i l i a r y function H(erxry)

H is e v i d e n t l y recursive.

t

$,(XI i f Oe(x) = Y 1 otherwise

The desired function h is defined as

h(x,y) = max H(e',x.y) e 'LX Clearly, h i s recursive.

.

To see t h e conditions o f t h e theorem a r e s a t i s f i e d suppose x

2

e and t h a t 9, (x) is defined. Thus,

Since x

2 e,

it follows t h a t

max ~ l ( e ' . x , @ ~ ( x 2 ) ) @,(XI e 'zx

I f qe (XI i s undefined, then t h e r e l a t i o n h (x,4,(x) 1 holds.

Oe ( X I s t i l l

I t then follows t h a t t h e set f o r which t h e r e l a t i o n i n

t h e theorem f a i l s i s bounded by e.

0

W e conclude t h i s s e c t i o n v i t h s e v e r a l t e c h n i c a l lemmas t h a t

w i l l be used in a l a t e r chapter.

Intuitively, in a single

complexity measure, when w e 'combine1 computations t h e complexi t y of t h e new computation i s r e c u r s i v e l y bounded by t h e complexity of t h e component computatione. 1.7

-

Lemma

( F i r s t Combining L8ma).

Let O be a complexity

measure and l e t C(e,y) be a recursive function such t h a t i f Ge(x) and $el ( x ) a r e defined ( f o r a given x ) , then s o i s @ ~ ( e , e n(x) ) a l s o defined. such t h a t

Then t h e r e i s a recursive function h

@c(e,el) ( X I 2 h (x, ae (XI ,@e ( X I

)

f o r a l l but a f i n i t e set of x. Proof.

Define t h e intermediate function

(XI i f p ( e , e l ,x,u,v)

ee (XI=

u and

eel (XI

= v

= otherwise

p is evidently recursive. h(x,u,v) = h i s a l s o recursive. tions, let x

Define t h e function h by max p(;,;* {Z,5'~1

,x,u,v)

To s e e t h a t h s a t i s f i e s t h e desired condi-

2 e,e8. From t h e d e f i n i t i o n of h

I f both Qe(XI and Oe ,(XI a r e defined, by hypothesis, s o is

@c(eve' (XI defined.

Fmm the d e f i n i t i o n o f p and s i n c e x

2 e,e'

Should e i t h e r Oe (x) o r Oe, (XIbe undefined, then the i n e q u a l i t y

holds independently of whether o r not

(x) i s

,eS)

defined.

It follows t h a t t h e set of a l l x f o r which the i n e q u a l i t y fails

i s bounded by max Ce,e'l.

D

Another form involves a C w h i c h is a function o f a s i n g l e argument. 1.8

.

Lemma (Secand Combining Lemma) Let @ be a complexity

m a s u r e and l e t C(e) be a r e c u r s i v e function such t h a t i f $,(XI i s defined

then

(XI i s defined.

Then t h e r e i s a recur-

s i v e h such t h a t

f o r a l l but a f i n i t e set of x. Proof. -

Define t h e function p(e,x,y) =

p is evidently

@C(e) (XI i f 4,

(XI = y

otherwise

recursive, Define the function h a s

h(x,y) = max p(e',x,y) Eev(xl h i s r e c u r s i v e and f o r any x 2 e,

(whethel o r n o t < @C(e) (XI 1

0(,

(XI i s defined).

i s bounded by e.

Hence, the set {xi h (x,Be (x) )

D

I t should be clear t h a t Lemmas 1.7 and 1.8 can be general-

i z e d t o any f i n i t e mnber o f arguments.

W e prove a last v e r s i o n

which allows us t o e l i m i n a t e t h e f i r s t argument of t h e bounding function. 1.9

Lemma

(Third Combining Lemma).

Suppose t h e same hypothesis

a s Lemma 1.8 holds and i n a d d i t i o n (Pe(x) 2 x f o r a l l x. Then t h e r e i s a r e c u r s i v e f u n c t i o n h such t h a t

(XI 5 h (Oe (XI

BC

f o r a l l b u t a f i n i t e set of x. Proof.

Define th@ function p(e,x,y) =

p is e v i d e n t l y r e c u r s i v e . h(y) = h i s a l s o recursive.

2

x

(XI i f Qe(x) = y otherwise

Define t h e f u n c t i o n h by max

p(el,x*,y)

{e*' , x g ~ . )

For x 2 e

h(Qe(x)) = Since Oe(x)

@C(e)

max

( e ' ,x'(ee(x) 1

~(e',x*,@~(x)

e,

max p ( e ' ,x* ,Be (XI 1 { e *,xm5Qe(x) 1 It follows t h a t t h e set Ex lh (ae (XI )

< OC

(XI

(XI 1 i s bounded

by e. 2.

Blum-Rabin Compression and Speed-Up Theorems In

t h i s s e c t i o n we prove r e s u l t s on t h e complexity of compu-

t a t i o n s t h a t a r e b a s i c a l l y due t o M. Blum [ l l

.

The first r e s u l t ,

c a l l e d t h e Blum Theorem, t e l l s us t h a t i n any complexity measure t h e r e a r e a r b i t r a r i l y complex p a r t i a l r e c u r s i v e functions, i - e . ,

than a

functions such t h a t any way of computing them takes more predetermined p a r t i a l recursive g

almost everywhere- A s a

special case we prove Theorem 1.4 of Section 1. The second major r e s u l t is t h e Blum

Compression Theorem

This e s s e n t i a l l y tells how lneasured'sets

functions and

a r b i t r a r y step-counting functions a r e interweaved; i n p a r t i c u l a r , when t h e measured funotions.

collection is a sequence of step-counting

We conclude Section 2 with a proof of B l u m ' s well

known Speed-up Theorem. W e proved i n Theorem 1.3 t h a t f o r a r b i t r a r y complexity

measure 4 and recursive function

there always e x i s t s some

f

0-1 valued + s u c h t h a t any way of computing

f ( x ) s t e p s o r an inEinite set of x.

+

takes more

than

W e next prove t h a t such an

inequality holds on not j u s t an i n f i n i t e s e t but for a l l but a finite

set.

This r e s u l t is further generalized from recursive

t o p a r t i a l recursive functions. 2.1

Theorem ( B l u m i l l ) .

L e t 4 be a complexity measure and g a

p a r t i a l recursive function.

Then there corresponds t o g a 0 - 1

valued p a r t i a l recursive function f , with t h e same domain a s g, such t h a t i f e i s any index f o r f then cOe (XI 2 g ( x )

f o r a l l but a f i n i t e set of x. Assuming t h a t we've proven Theorem 2.1 we have 2.2 g

Corollary (Rabin 12311. be a recursive function.

Let 4 be a complexity measure and Then there corresponds t o g a 0-1

valued recursive function f such t h a t for any index e f o r f

f o r a l l but a f i n i t e set o f x. Proof.

L e t g be recursive.

By Theorem 2.1 t h e r e e x i s t s a

p a r t i a l recursive f such t h a t f o r any index e f o r f t h e set { x ~ @ ~ <( xg)( x ) ) is f i n i t e .

Since g is recursive and has t h e

same domain as f , f i s a l s o recursive. Corollary 2.2 w a s a c t u a l l y Theorem 1.4 of Section 1 whose proof was promised e a r l i e r . W e now prove Theorem 2.1.

Proof. -

The function f w i l l be defined by a construction

s p e c i f i e d below, will

be

The construction w i l l proceed i n s t e p s s

accumulating

,a t

The s e t A'

two sets

A and

s t a g e s, w i l l be a

F

and

defined a s follows.

f i n i t e c o l l e c t i o n of

i n d i c e s t h a t have been "cancelled" p r i o r t o o r during s t a g e s. The set A<' w i l l denote those cancelled i n d i c e s j u s t p r i o r t o s t a g e s.

The set FS w i l l represent t h e graph of our d e s i r e d

function f c o l l e c t e d

a t s t a g e s. The set F<" w i l l be FS

just

p r i o r t o s t a g e s. Construction:

For an input value k t h e following construction

t e l l s us how t o compute f (k).

Stage 0:

B e t A'

= F0 = 0.

Stage s:

Let x = sl(s) and y = r 2 ( s )where rl,r2 a r e p a i r i n g

functions. I f @,(XI # y, where g = 0, go t o s t a g e s + l . g(x)S, hence If

then set AS

-

A'",

FS = FCS and

In t h i s s i t u a t i o n , it is not c e r t a i n whether

w e add nothing t o A o r F.

@,(x) = y then set V = {ele z x 61 Qe(x) < g ( x ) ) .

t h i s s i t u a t i o n we know g(x1S.

In

What we do is to c o l l e c t i n V the

set of indices e which a r e " e l i g i b l e f o r cancellation".

W e next ask i f s e t V contains any new i n d i c e s not already

cancelled: If A < = ~ A<' and go t o CHECK.

U V

then set A = A<'

and F' = F < U~ I < x , l > )

I n t h e case where every e l i g i b l e index has

already been cancelled w e merely set As = A<'

and assign t o f (x)

t h e value I. ~f A<'

z

A<'

u

v

then set

where 8 is e i t h e r 0 o r 1 but n o t $=(x) and go t o CHECK. H e r e t h e r e is an e l i g i b l e index which can be cancelled.

We choose it

t o be t h a t index with t h e h i g h e s t p r i o r i t y ( s m a l l e s t ) of a l l t h e W e then set f ( x ) t o e i t h e r 0 o r 1 b u t unequal t o

eligibles.

t h e value of t h e function with cancelled index. CHECK:

I f x = k then output t h e value 8 and h a l t , otherwise go t o s t a g e s+l.

This completes t h e construction. FS and d e f i n e f ( x ) = y * <x,y> E F. W e next s show t h a t f s a t i s f i e s a l l requirements of t h e theorem. Let

F

3

U

w e see f i s w e l l defined,

By examining t h e construction

dom(f) = dom(g) and has range { O , I ) .

By t h e e f f e c t i v e nature of

t h e construction, f i s p a r t i a l recursive. Remark.

A value f ( k ) f o r t h e function defined above i s

dependent on input k and on t h e index R of g. Hence, it i s a p a r t i a l recursive function of two arguments, m(a,k). follows from t h e s 3 t h e o r e m

that

OS (El (k) = m(E,k)

It

f o r some

W e w i l l use t h i s result l a t e r on i n t h i s s e c t i o n .

r e c u r s i v e s.

W e complete t h e proof of t h e theorem by proving t h e follow-

ing* Claim. ix

I

L e t e be any index for f .

Then t h e set

Oe(x) < g ( x ) I is bounded. Proof. -

Assume the contrary.

That i e , f o r some index e f o r Then t h e r e w i l l be a

f t h e set {xlOe(x) :g ( x ) 3 is i n f i n i t e .

s t a g e so p a s t which no index s m a l l e r t h a n e w i l l e v e r be cancelled.

Since {xi Oe (XI< g (x) 1 is i n f i n i t e , it must have a

member x ' > so

. Since x'

is i n t h e s e t , @,(x1), hence Oa(x') i s

defined. Furthermore, by a b a s i c property of p a i r i n g f u n c t i o n s =

S1

<XI

,aQ (x') > 2

X'

>

So

F r m t h e d e t a i l s of t h e c o n s t r u c t i o n e i t h e r (a)

A t s t a g e sl

(b)

A t s t a g e 2,

,e

i s added t o A

or

s2 < sl, e is added t o A.

In e i t h e r case, t h e r e e x i s t s a stage added t o A. However,

By t h e c o n s t r u c t i o n , ,<;

6 was

chosen s o t h a t

assumption t h a t f Definition.

+ $,

k * <2,3> a t which e

B">

is added t o F a t

is

;.

(2) c o n t r a d i c t i n g t h e

@e.

0

A sequence of f u n c t i o n s {ye) i s s a i d t o be

r e o u r s i v e t y enumerabte i f t h e f u n c t i o n t ( e , x ) = ye(x) i s p a r t i a l

recursive.

A sequence o f f u n c t i o n s

{ye)

which is r e c u r s i v e l y

enumerable and f o r which we can r e c u r s i v e l y decide f o r a l l e,x,y whether

ye(x) = y i s c a l l e d a measured s e t of functions.

By our assumptions, t h e step-counting functions

{ae)

o f any

complexity measure forms a measured s e t of functions. 2.3

Lemma.

L e t {ye) be a measured set of f u n c t i o n s and l e t O

be a complexity measure,

Then there e x i s t s a r e c u r s i v e r such

that

Proof. M ' (e,x,y)

By t h e d e f i n i t i o n of a measured s e t t h e p r e d i c a t e

* ye (x) = y i s r e c u r s i v e .

Hence, t h e f u n c t i o n

f (e,xl = min iM' (e,x,yl I i s p a r t i a l r e c u r s i v e . By t h e .:-theorem

Y

t h e r e i s a r e c u r s i v e s such t h a t $r(e) (x) = f ( e , x ) = y e ( x ) . Our n e x t r e s u l t r e l a t e s measured s e t s and complexity measures.

It t e l l s us t h a t i f {ye] i s measured,

is a complexity measure and h

ye E {ye},

a recursive function,

@

that there

e x i s t s a p a r t i a l r e c u r s i v e f u n c t i o n f w i t h an i n d e x Z! such t h a t f o r a l l b u t a f i n i t e set

Hence, t h e r e i s some a l g o r i t h m f o r f whose complexity i s compressed between h (x, ye ( x ) ) and ye (x) 2.4

.

Theorem (Blum [ I ] ) .

complexity measure.

Let {y 1 be a measured s e t and @ a e Then t h e r e e x i s t s a r e c u r s i v e h and f o r

each p a r t i a l r e c u r s i v e y

e

t h e r e i s a 0-1 valued p a r t i a l recur-

s i v e f u n c t i o n f w i t h t h e same domain a s ye such t h a t (1) I f e ' i s any index f o r f t h e n ye(x) < Q e ,( x ) f o r a l l b u t a

f i n i t e set of x , (2)

t h e r e e x i s t s an index e * f o r f such t h a t

@,* (x) < h ( x I y e ( x ) ) f o r a l l b u t a f i n i t e s e t of x, (3)

and

t h e r e e x i s t s a r e c u r s i v e f u n c t i o n t such t h a t f o r each ye,

e* = t ( e ) s a t i s f i e s ( 2 ) . Proof.

By Lemma 2.3 t h e r e is a r e c u r s i v e f u n c t i o n r such

ye = $ r (e)' L e t s be t h e r e c u r s i v e f u n c t i o n whose e x i s t e n c e w a s i n t r o d u c e d i n t h e remark i n t h e proof o f Theorem 2.1. that

W e w i l l see t h a t t h e f u n c t i o n f corresponding t o ye i s p r e c i s e l y

@ s o r(el

:

By Theorem 2.1,

i s a 0-1 valued p a r t i a l r e c u r s i v e .

@sor

F u r t h e r , s i n c e @ r ( e )= ye

, by

t h e same theorem,

( e l ) = dom(ye) and f o r each index e ' Of $ s o r ( e ) (x) I= ye ( x ) ) f o r a l l b u t a f i n i t e s e t of x. (x) >

dom(@sor

ae,

This proves (1). To show (21, let otherwise

( 0

C l e a r l y , p i s r e c u r s i v e s i n c e Eye) i s measured and dom(ye

)=

( @ s o r ( e).Let ) h ( x , y ) = max p ( e l , x , y ) + l . h i s r e c u r s i v e . e ' x We show t h a t e * = s o r ( e l s a t i s f i e s ( 2 ) L e t x > e. Then by

.

definition, h(x,ye ( X I ) = max p ( e ' ,x,ye ( x ) )+1 e'Lx But s i n c e x > e

and (2) i s s a t i s f i e d . Since s and r a r e both r e c u r s i v e , t h e f u n c t i o n t = sor is recursive.

F u r t h e r , t h e map t sends index e f o r ye i n t o index

s o t ( e ) f o r f which s a t i s f i e s ( 2 ) . Hence,

( 3 ) follows.

0

Our next r e s u l t i s t h e w e l l known Blum speed-up phenomenon. A s seen i n Theorem 2.1,

there exists

p a r t i a l recursive functions.

a r b i t r a r i l y complex

We s e e h e r e t h a t t h e r e e x i s t s

r e c u r s i v e f u n c t i o n s w i t h no f a s t e s t a l g o r i t h m . any complexity measure

Specifically i n

9 , given a r e c u r s i v e f u n c t i o n r ( x , y )

t h e r e e x i s t s a r e c u r s i v e f u n c t i o n f ( x ) which can always be "r-sped-up".

A s a consequence, of t h e speedup theorem w e see

t h a t r e c u r s i v e f u n c t i o n s cannot be c h a r a c t e r i z e d by t h e i r f a s t e s t

algorithm. The proof t h a t follows is t h a t of Young 2.4

Theorem

(Blum

1301.

For any r e c u r s i v e f u n c t i o n r ( x , y )

[ll).

t h e r e e x i s t s a 0-1 valued r e c u r s i v e f such t h a t f o r any index e f o r f t h e r e e x i s t s another index e ' such t h a t

@,(XI > r (x,Qe, (XI 1 f o r a l l b u t a f i n i t e set of x. Proof.

A

r e c u r s i v e f u n c t i o n w i l l be defined i n terms of a

c o n s t r u c t i o n given below. parameters

u,v,ll and x

construction runs

The c o n s t r u c t i o n w i l l depend on f o u r and w i l l proceed in s t a g e s .

As

the

w e w i l l be accumulating two s e t s ARuv and FRuv

S The set AQuv a t stage s w i l l serve a s a f i n i t e collection

of i n d i c e s which have been c a n c e l l e d p r i o r t o o r during s t a g e s. The s e t

w i l l consist of those indices cancelled j u s t p r i o r

t o s t a g e a.

S e t ,F:

w i l l c o n s i s t of t h e graph of t h e d e s i r e d <s c o n t a i n s members function accumulated by s t a g e s. The s e t Ftuv

of F ; ~e n~t e r e d p r i o r t o s t a g e s. D e f i n i t i o n . A set B i s a f i n i t e f u n c t i o n i f B i s

the

graph of a f i n i t e function w i t h domain an* i n i t i a l segment of w . L e t { G ~ denote } some canonical enumeration of a l l f i n i t e func-

tions.

I n a d d i t i o n , l e t {Ke) denote some canonical enumeration

of t h e f i n i t e s u b s e t s o f w . The c o n s t r u c t i o n t h a t n W follows w i l l compute a function The r e s u l t a l s o ddpends on t h r e e o t h e r

value f o r argument x. parameters

u, v and a.

Construction Stage 0 :

0 S e t AQuv

Stage s:

If s

E

0

f

FQuv = 8 .

dom(G,

)

1

(where rl i s a p a i r i n g f u n c t i o n ) ,

t h e n set

Afuv =

<s A ~ U V

and go t o CHECK. W e check

t o see i f t h e nl(v)-th f i n i t e function w i l l give

us a value on s.

I f s o , we use it i n o b t a i n i n g t h e value f o r

t h e constructed function. Otherwise (2)

Compute t h e s e t

V i s t h e s e t of a l l i n d i c e s e eligible for cancetlation

a t s t a g e s. We w i l l s e e below t h a t f o r a l l s , v , x , u

and a p a r t i c u l a r R,

t h e c o n d i t i o n s i n t h e s e t may be r e c u r s i v e l y checked. We next check t o s e e i f t h e r e a r e any new i n d i c e s i n V n o t previously cancelled. (3)

If

W e then s e t

#

A;&

U

V

then we compute e' = min {EEV e

6

E$!A;:~}

and S

e u v = Fkuv u { < s , 9 > 1 where

Here we have found i n d i c e s which a r e e l i g i b l e f o r c a n c e l l a t i o n and which were never cancelled.

W e cancel

t h e smallest

( h i g h e s t p r i o r i t y ) by p l a c i n g ft i n t o

W e then assign t o

our c o n s t r u c t e d function a v a l u e making it d i f f e r e n t from t h e function whose index was cancelled.

W e then

go to CHECK. 4 1

I f A;&

=

U

t h e n r e set

V

,A :

Iluv

0

and go t o CHECK. Here we cannot cancel any index.

W e a r b i t r a r i l y a s s i g n t h e value

0 t o our function f o r argument s. CHECK:

If s = x

then o u t p u t 8; otherwise go t o s t a g e s + l .

This ends t h e c o n s t r u c t i o n . Define AQuv =

S

U AQuv

s

0

and FQuv =

function f a u v a s follows. QUV

(XI = Y *

<XIY>

U

s

s FQuv

.

W e d e f i n e our

E FRUV* <x,y> is placed i n t o FEU, a t Stage X.

I t Follows from t h e d e t a i l s of t h e c o n s t r u c t i o n t h a t f Q u v i s

a 0-1 valued, well defined, p a r t i a l r e c u r s i v e function. By t h e #-theorem fR

u v ( ~= ) L

f o r r e c u r s i v e s1,s2.

! ( L~) (u,v,x) ~

= (s2( R , ~ , v( X) I

By t h e r e c u r s i o n theorem t h e r e i s an Ro

such t h a t (u,v,x)

.

OS

(, ) (u,v,x) 1 0 The f u n o t i o n f r e q u i r e d b y t h e theorem i s p r e c i s e t y

$

=

Ro

Recall t h a t f R O o 0 ( x )=

0

(0,O.x).

of g e n e r a l i t y t h a t Kn ( o ) = GT2 (o) = 1

fROo0 (x)

.

W e assume without l o s s

8 . The main c o m p u t a t i o n

of 4

(u,v,x) refers to the main body computation and not to any &o recursive calls which may be produced. 2.4.1

Cam. Proof.

on d = s 2.4.2 (

-

u.

Claim. u,v,x

)

For each x and u,v < w,

fgou,v(XI is defined.

By an outer induction on s and an inner induction The details are left to Let u,v,x < w.

U

the reader.

Then in the main computation of

no index e ever gets cancelled more than once.

&o

2.4.3

proof.

Immediate.

Claim.

Let u < w.

Then in the main computation of

4,

(O,O,x) there is an su > u such that no e < u ever gets cancel0 cancelled after stage sU.

2.4.4

Proof. -

Immediate.

Claim.

Let u < w and sU be as in Claim 2.4.3.

0

Then there

exists a vU such that

and Kr2

= {e le is cancelled in computing 4

for some x<s -u

Proof.

Since 4

Claim.

For all x < w

Proof.

Immediate from the details of the construction and

(O,O,x)/sU is a finite function, let d be &o such that G (x) = (O,O,x)/su. Let Kn = Iele is cancelled in du 0 u computing 4 (O,O,x) for some x f sU). The result follows by &o taking vU = 2.4.5

t h e d e f i n i t i o n s o f sUand vu. Claim.

2.4.6

For any index e f o r f R 00 0

f o r a l l b u t a f i n i t e set of x. Proof. -

Suppose o t h e r w i s e t h a t t h e set

A f t e r a s t a g e so no index o f

is i n f i n i t e .

higher p r i o r i t y

than e g e t s c a n c e l l e d d u r i n g t h e main computation o f f

ve+l' By t h e d e t a i l s of t h e c o n s t r u c t i o n , e i t h e r a t - s t a g e x > so and x

€

B

index e i s c a n c e l l e d implying

4, f f

D

R0e+l ve+l

The remainder of t h e theorem follows immediately from Claim 2 . 4 . 6

and the i d e n t i t y

of Claim 2.4.5. 3.

Complexity C l a s s e s I n this s e c t i o n w e c o n c e n t r a t e on

sets of r e c u r s i v e

functions

a l l of whose step-counters a r e bounded by a r e c u r s i v e

function.

Such sets a r e c a l l e d "complexity c l a s s e s " .

We f i r s t

consider t h e problem o f , given an algorithm f o r t h e bounding function, e f f e c t i v e l y i n c r e a s i n g t h e complexity c l a s s t o i n c l u d e more f u n c t i o n s .

Several r e s u l t s a r e provided which s o l v e t h i s

problem. We n e x t prove t h e well-known Borodin Gap Theorem. This t e l l s us, e s s e n t i a l l y , t h a t t h e s t e p c o u n t i n g f u n c t i o n s of a complexity

measure a r e functions.

r e l a t h e l y s p a r s e compared t o t h e s e t o f r e c u r s i v e We t h e n prove McCreight-Meyer's

T h i s t e l l s us t h a t whenever an i n c r e a s i n g

Union Theorem. sequence o f r e c u r s i v e

f u n c t i o n s d e f i n e s a sequence of complexity c l a s s e s , t h e r e w i l l b e a r e c u r s i v e f u n c t i o n d e f i n i n g p r e c i s e l y t h e union of t h e s e classes.

A s an a p p l i c a t i o n we s e e t h a t t h e union of complexity

c l a s s e s d e f i n e d by t h e p r i m i t i v e r e c u r s i v e f u n c t i o n s i s a complex complexity c l a s s

d e f i n e d by a s i n g l e r e c u r s i v e f u n c t i o n .

W e conclude S e c t i o n 3 w i t h a proof o f t h e Meyer-Moll-

McCreight

Honesty o r Naming Theorem.

T h i s p r o v i d e s us w i t h a

means of g e n e r a t i n g from any sequence o f p a r t i a l r e c u r s i v e functions, a

measured

sequence of p a r t i a l r e c u r s i v e f u n c t i o n s .

Furthermore,: t h e c o n s t r u c t e d f u n c t i o n s a r e such t h a t none o f t h e complexity c l a s s e s named by t h e f i r s t set i s l o s t . Definition.

For any computational complexity

measure

a,

we

d e f i n e f o r a given recursive function t , t h e c o m p l e x i t y c l a s s C:

=

{(sl(s

is t o t a l

&

P ( x ) z t ( x ) f o r a l l but a f i n i t e s e t of x S

Thus C t0 o r simply Ct

, when

P is understood, i s t h e set o f

a l l r e c u r s i v e f u n c t i o n s whose complexity is bounded by t f o r a l l b u t a f i n i t e s e t o f x.

, does n o t imply t h a t Ps'$e $e almost everywhere. I t o n l y i m p l i e s t h a t t h e r e i s an i n d e x k f o r Remark.

The f a c t t h a t

$s, 6, = $k where

Pk

(s

E

C

5 $e almost everywhere.

Without l o s s of

g e n e r a l i t y , we assume throughout t h a t t h e c o r r e c t index i s chosen. We begin by c o n s i d e r i n g t h e problem o f extending a complexi t y class C

@e

e f f e c t i v e l y when given t h e index e f o r t h e bounding

function.

(Again, t h e author is not c e r t a i n of t h e names of

those responsible f o r t h e s e r e s u l t s and once again p o i n t s t h e reader t o t h e survey by Hartmanis and Hopcroft 171.) 3.1

Theorem.

I n any measure

t h e r e e x i s t s a recursive

function h such t h a t f o r each t o t a l 4, t h e r e i s a corresponding @ such t h a t $ 9 C $e

Proof. -

and $ E C A x h l x , a e ( x ) l .

Recall t h e p a i r i n g function wl has t h e property

t h a t f o r any s, nl (x) = s f o r an unbounded set of x. Let ( x ) + l if 4,

1 otherwise

$(XI =

(XI 2

@e(~)

Since +e i s t o t a l , $ is recursive. To see

.

, l e t $ = Os and suppose it w e r e i n C e @e Then f o r s u f f i c i e n t l y l a r g e x both 4s (x) ( ge (x) and n1 (x) = s. @

9

C

@

By t h e construction,$ f .$,

To continue t h e proof, l e t

$ =

$s f o r some s.

Clearly,

t h i s s i s dependent on e, and furthermore t h e r e e x i s t s a recurrecursive C (el such t h a t bC(,) = Lemma 1.8, s i n c e $e i s t o t a l , a f i n i t e set of x. If

Qe(x) > x

Hence,

Theorem.

$

Proof.

-

(x) < h (x,ee (x) ) f o r a l l b u t

E CAxh(x,ae(x))

Qe(x) (and n o t x)

In any measure

.

0

such t h a t

.

t h e r e i s a r e c u r s i v e function h where

such t h a t f o r a l l t o t a l 4, e x i s t s an

(DC

By t h e Second Combining

then t h e recursive function h need only

depend on t h e value 3.2

.

Qe(x)

) X

.

f o r a l l x, t h e r e

and $ E Ch 4e O Q e Exactly t h e same a s t h e proof of Theorem 3.1 except $

C

t h e Third Combining Lemma 1.9 is used.

W e modify the above proof t o o b t a i n a new complexity c l a s s

which properly c o n t a i n s t h e o r i g i n a l one. '3.3

Theorem.

I n any measure Q

there exists

f u n c t i o n R such t h a t f o r a l l t o t a l

.

me

a recursive

C

+e C C h X R [ X , f ( X ) ~ * A s i n t h e proof of Theorem 3.1 (and 3.2). we can

Proof.

find recursive

and h s o t h a t

+C (,)

a l l b u t a f i n i t e set o f x

(XI

QC

.

By Theorem 1.6, +e for a l l we can f i n d a r e c u r s i v e h' such t h a t h' ( x , Q e ( x ) ) 2 b u t a f i n i t e s e t o f x. W e next show C

for a l l x

where

( h(xIQe(x))for

9 C

Let R(XIy) = max{h(x,y),h'(x,y)).

+e c C R ( x I Q e ( X )

From t h e above. w e s e e t h a t

R[x,Qe(x) 1 = max {htx,Qe ( x ) ) , h ' (x,Qe(x) ) 1

-> h ' ( x , Q e ( x ) ) By t h e c o n s t r u c t i o n o f h '

-> 4,

(x)

f o r a 1 1 b u t a f i n i t e set of x. t h e containment i s proper

Hence, C

observe t h a t

me

5 CRIxIQe(x)l.

To s e e

R[x,Qe(x) I 2 h ( x I Q e(XI 1 f o r a l l x.

By o u r c o n s t r u c t i o n

f o r a l l b u t a f i n i t e set of x. recall that

mC

.

complexity c l a s s d e f i n e d by on Qe(x) 3.4

E

C R [ x I Q e ( x ) l *But

-

Hence. C c CR[x,Qe(X)I +e e then the recursive function increasing t h e

F'C

I f Qe(x) > x

I-Ience.

+,

can be made t o depend only on

.

Theorem.

I n any measure Q l e t 4,

be a r e c u r s i v e f u n c t i o n

where Qe(x)

c @

e

2 x f o r a l l x.

Then t h e r e is a r e c u r s i v e B such t h a t

= CR.Qe. Proof.

Following t h e proof o f Theorem 3.2, we can f i n d

r e c u r s i v e f u n c t i o n s h and

so that

a l l b u t a f i n i t e set o f x.

(x)

(PC

5 h(@, (x) ) f o r

F u r t h e r , we have t h a t $C(e) P C @

3.4.1

Claim.

Let

@

be a measure.

f u n c t i o n such t h a t @,(XI

Suppose $e i s a r e c u r s i v e

2 x f o r a l l x.

r e c u r s i v e h' such t h a t for e a c h e

e

Then t h e r e e x i s t s a

h ' ((De(x)1

2

(x) f o r a l l but

a f i n i t e set o f x. Proof.

Let H(~,x,Y) =

H is recursive.

[

x

if

ee ( x )

= Y

otherwise

max H(eg,x',y1. h ' i s fe' ,x'lyl r e c u r s i v e and we show t h a t t h e set I x l h ' (Qe( x ) ) < Oe ( x ) 1 i s finite.

Let x

Define h ' as h' (y) =

2 e. Then

h'(Oe(xI) = Since

max H ( e ' , ~ ' , @ ~ () x ) I e ' ,x'f" ,(XI 1

ae (XII. x 1 e,

-> $,(XI . Hence, t h e s e t i s bounded by e.

0

Returning t o t h e p r o o f , we d e f i n e t h e f u n c t i o n R(y) = max { h ( y ) , h ' ( y ) l .

A3

i n t h e proof o f

Theorem 3.3,

Our n e x t r e s u l t i s an a s s e r t i o n t h a t f o r any complexity measure Q

t h e step-counting f u n c t i o n s (9,)

t o t h e recursive functions.

a r e sparse r e l a t i v e

I n terms of t h e previous r e s u l t s ,

Borodin's Gap Theorem t e l l s u s t h a t t h e r e i s no r e c u r s i v e r, where r ( x ) > x , t h a t w i l l expand every p o s s i b l e complexity c l a s s .

The proof here is that of Young I301. Theorem (Borodin [31).

3.6

In any complexity measure Q for any

recursive function r with r(x)

2 x, for all x, there exists a

recursive monotonically increasing function t such that

Proof.

Let {Qel be the enumeration of the step counting

functions of measure Q.

We define function t by

t(0) = 0 t(x) = max[t(s)+m(s,x,t(s))

Is <

XI

where m(s,x,y) = min Ve,I m

-

(Qe(XI

5 r(y+m)

We first show the function m(s,x,y)

+

(Qe (XI f y+m)

is recursive.

Clearly,

the predicate

is recursive.

Thus, it can be tested for m = 0,1,...

. Further,

for fixed s, x and y, there will always be a corresponding m. For certainty, there exists an m such that Qe(x) ( y+m for all e < s, Qe (x) is

defined, (Qe(x) > r (y+m), otherwise .) Hence,

m(s,x,y) is recursive.

It follows from

its definition that t

is also recursive (and monotonically increasing). We neft show Ct = Crot 4,

E Crot

. Then we

when x > xo.

. Since r(x) 2 x, Ct 5 Grot. Suppose

know that for some xo > e, Qe (x) 5 rot (x)

But

= max[t(s)

+

min V {(Qe(x) Lr(t(s)+m)) m ezs + (oe(x) 5 t(s) + m) 1 I s < XI.

Thus, f o r x > x, > e , t ( x )

2 Qe(x) making

+e

E

Ct.

Hence,

We next show t h a t f o r any sequence of increasing recursive functions, t h e r e is a complexity c l a s s comprised of t h e union of t h e c l a s s e s bounded by functions from t h e sequence. 3.7

Theorem (McCreight-Meyer 1191

measure.

I

Suppose {fe

of recursive functions

e < w)

.

L e t Q be a complexity

i s a recursively enumerable s e t

such t h a t f o r each e , e l , x

f o r e > el. Then t h e r e is a

recursive function k ( x ) such t h a t

Observe t h a t t h e function k (x) = f x ( x ) i s g r e a t e r than f e ( x ) f o r each e klx)

on a l l but a f i n i t e set of x.

.

For when x

2 e,

2 f e (x) Hence, f o r each e t h e s e t {XIk (x) < f e (x)

f i n i t e and we concLude t h a t

U

is

C

Ck. e fe However, t h e r e may e x i s t some +I f o r which (1) Q I ( x ) 5 k ( x )

f o r a l l b u t a f i n i t e set of x, and yet ( 2 ) GQ(x) > f e ( x ) f o r each e on an unbounded s e t of x.

This would cause + I t o be i n Ck but

, thus, implying only U C C Ck. fe e 'e The way we avoid t h i s d i f f i c u l t y i s t o "guess" f o r each

not i n t h e union of t h e C

and some e t h a t f e ( x )

1 Q E ( x )f o r a l l but

a f i n i t e s e t of x. I f

w e discover t h a t f o r some x, f e ( x ) < Q I ( x ) , w e t r y t o assign a

value t o k ( x ) which i s l e s s than Q Q ( x ) . We a l s o "guessn again t h a t f o r a l a r g e r e ' ( e ' > e l t h a t f e , (x) ) Q Q ( x )f o r a l l but a f i n i t e set.

If

, then t h e r e w i l l be an e fe (x) f o r a l l but a f i n i t e s e t . However,

QI i s a c t u a l l y i n

eo such t h a t @ & ( X I

f

0

it w i l l a l s o be t h e case t h a t f

eo

u C

(x) 5 k ( x ) f o r a l l but a f i n i t e

set.

Hence

union o f t h e C set.

@a B

Ck.

E

i s not i n t h e

On t h e o t h e r hand, i f

t h e n f o r some e , f e (x) < Q R (x) f o r an i n f i n i t e

e

( x ) f o r an i n f i n i t e set and hence,

This w i l l f o r c e k (x) < Ck-

A r e c u r s i v e f u n c t i o n k w i l l be d e f i n e d i n terms o f

Proof.

The c o n s t r u c t i o n w i l l compute k(m)

a c o n s t r u c t i o n g i v e n below.

Throughout t h e c o n s t r u c t i o n we w i l l be accumulating

f o r any m.

t h r e e s e t s K , I, and TO The s e t KS constructed.

A

,at

d e f i n e d a s follows.

s t a g e s,

represents t h e function k(x) being

p a i r <x,y> i s placed i n KX a t some s t a g e x i f and

o n l y i f k ( x ) = y.

The set IS, a t s t a g e s , r e p r e s e n t s encodings

o f p a i r s such t h a t we have made t h e guess f k ( x ) f o r a l l but a f i n i t e s e t .

A t stage s

no more t h a n 21s guesses.

The s e t

QR(x)

t h e r e w i l l be an e x i s t e n c e

TO^

a t s t a g e s, c o n s i s t s o f

encodings of p a i r s f o r i n c o r r e c t -"guesses". S e t s K < ~ , I < ~ and

TO<^

represent

respectively, prior to

K ~ ,I~ and TO",

S t a g e s. We n e x t d e f i n e t h e c o n s t r u c t i o n which computes k(m) f o r any

m < u. The C o n s t r u c t i o n S t a g e 0.

S e t I' = TO'

S t a g e s.

Compute t h e s e t V = ( r l r

= KO =

9. E

1"

&

r 9 ~ 0 &' ~

W e s e a r c h through o u r l i s t o f guesses and choose one o f t h o s e

f o r which we guessed i n c o r r e c t l y . If V = I'

= 1'

U

9 t h e set

{<s,s>)

6 = f s ( s ) , K~ =

, TO"

=

TO<^

and go t o

U

{<s,B>j,

CHECK;

Here it

seems t h a t s o far we have guessed c o r r e c t l y . We then

make a new guess t h a t f s ( x ) ) Q S ( x ) f o r a11 b u t a f i n i t e set and set k ( s ) = f S ( s ) . I f V # $3 then s e t k = min { r 2 ( r )/ r v

5

min { n l ( r )

1

r

E

€

V) and

V & r 2 ( r ) = k).

The value o f k i s t h e number w i t h h i g h e s t p r i o r i t y such t h a t t h e I f more t h a n one member

value r e p r e s e n t s a wrong guess.

has a second component k, we choose t h e one w i t h s m a l l e s t

of 1'

f i r s t component. set 1' = 1"

ms U

= TO"

u {
u {<s,B>I,

We a r e a l s o guessing t h a t

2 e V ( x ) and f s ( x ) 2 Q s ( x ) f o r

CHECK:

K"

{ < v , s > , < s , s > ) and go t o CHECK.

Here we a r e s e t t i n g k (s) = f k ( s ) . fs(x)

, K'=

a l l b u t a f i n i t e s e t of

X.

I f s = m then o u t p u t 8 and h a l t , otherwise, go t o S t e p s t e p s+l.

This ends t h e c o n s t r u c t i o n .

0

We d e f i n e t h e following sets. K = U K S ,

I s U I

s

s

8

and

TO = u TO' S

.

Our r e q u i r e d f u n c t i o n k i s d e f i n e d a s

3.7.1

k ( x ) is a well-defined r e c u r s i v e f u n c t i o n .

Claim.

Proof.

I n order t o compute k (x) , f o r any x

compute through t h e x-th s t a g e .

,

we merely

C l e a r l y , a t any s t a g e s e x a c t l y

one value i s assigned t o k ( s ) and every s t a g e x can be e f f e c t i v e l y pel'f ormed

.

The remainder of t h e proof i s t h e demonstration t h a t Ck=UC e fe

.

d a n m-pair i f s 2 ( d ) = m.

W e c a l l the value

W e say t h a t

an m-pair d i s popped a t s t a g e s, i f d i s placed i n TO'

a t stage

8.

3.7.2

Claim.

stages s > s

e

Proof.

For each e t h e r e i s a s t a g e se such t h a t f o r a l l i f an m-pair i s popped t h e n m > e. Immediate.

0

Claim. Proof.

E Ck. Then w e have O (x) ( k ( x ) f o r a l l 9 9 We have t o show t h a t t h e r e m u s t b e an f k where

Suppose @

but a f i n i t e s e t .

where Q ( x ) 5 f k ( x ) 4 @gE C and @ E U *k g set D~ = { x l f k ( x ) < 9

f o r a l l but a f i n i t e s e t . C

fe

.

That i s , f o r any k

Suppose n o t .

O (x)1 9

This w i l l imply the

is infinite.

= { s l a t s t a g e s a p a i r of t h e form 9 < g , s > i s p u t i n t o 1' w h i l e another p a i r i s p u t i n t o TO') 3.7.4

Claim.

The s e t A

is infinite. The proof o f C l a i m 3.4.3

f o l l o w s from Claim 3.7.4.

For by

t h e d e t a i l s of t h e c o n s t r u c t i o n , i f t h i s s e t i s unbounded t h e n s o i s t h e s e t { s 1 k ( s ) < Q ( s )1 which c o n t r a d i c t s t h e f a c t t h a t 9

Proof (of Claim 3 . 7 . 4 ) .

is f i n i t e . 9 t h e r e must be a s t a g e ko

Suppose t h a t t h e s e t A

From t h e d e t a i l s of t h e c o n s t r u c t i o n

s u ~ h t h a t ~ g , k ~ ~ E ~a ~n - dT ~O ~ k ~ * ~ g , k ~ ~ E i ~ That i s , a f t e r s t a g e kg

, the

p a i r endures. By Claim3.7.2

t h e r e is a stage

such t h a t no m-pair w i t h m having Ska h i g h e r o r e q u a l p r i o r i t y t h a n ko , i s e v e r popped a f t e r s t a g e

s

. ko

D 05 { x l f k ( x ) < (Pg(x)l 0 E D where

By o u r o r i g i n a l assumption t h e s e t

i s i n f i n i t e . Hence, t h e r e is

an

s'

9

By t h e d e t a i l s of t h e c o n s t r u c t i o n a t s t a g e s t

s ' > maxisk ,kg). 0

t h e p a i r w i l l be p u t i n t o TO".

But t h i s c o n t r a d i c t s t h e

f a c t t h a t ko bounds a l l such s t a g e s .

0

Before proving t h e second i n c l u s i o n w e v e r i f y t h e following. 3.7.5

For each e, f e (x) 5 k (x)

Claim.

f o r a l l but a f i n i t e

s e t of x. Proof. -

t h e r e e x i s t s a s t a g e se such t h a t

By C l a i p 3.7.2,

,if

f o r a l l s t a g e s s > se

an m-pair is popped t h e n m > e. From

the d e t a i l s of t h e construction a t

al& stages

either

k(x) =

fX(x) > fe(x)

or

k(x) =

fm(x) > fe(x)

( s i n c e m > e)

.

x > maxfse,e)

Hence, t h e set {x.

f e (XI> k ( x ) 1

is finite. 3.7.6

0

Claim. Proof.

u

a

5 Ck.

C

fe Suppose $

U

€

Q

and, hence,

e

C

fe

.

Then t h e r e e x i s t s an eo such

0

(*I

QI

9

f o r a l l b u t a f i n i t e set. t h a t O (x) 3 '

It f o l l o w s from Claim 3.7.5

5 k ( x ) f o r a l l b u t a f i n i t e set. Hence, 4g

The e q u a l i t y and 3.7.6.

(XI 2 f eo (XI

u C

e<w

fe

and (*I E Ck.

0

= Ck f o l l o w s from Claims 3.7.3

A s a n a p p l i c a t i o n o f t h e Theorem 3.7

0

consider t h e c l a s s

of p r i m i t i v e r e c u r s i v e f u n c t i o n s and t h e complexity measure @ which c o u n t s t h e s t e p s o f a s i n g l e - t a p e Turing machine.

3.8.1

Claim.

If g is primitive recursive then there is a primi-

tive recursive t such that 9 Proof. sive g.

E

0 Ct.

By induction on the definition of primitive recur-

Details are left to the reader.

3.8.2. Claim.

0

If t is primitive recursive then

C:

is made up of

only primitive recursive functions. Proof.

It is easy to show that primitive recursive

functions are sufficient to simulate Turing machines for a primitive recursive number of steps.

Again, details are left to

the reader. 3.8.3

Claim.

There exists a recursively enumerable strictly

increasing sequence af primitive recursive functions fe 1 such that every primitive recursive function is majorized by some fe. Proof.

Let {g.) be a recursively enumerable sequence of 1

all the primitive recursive functions. fi (x) = max {gl (XI ,g2 (XI,.

Let fi(x) be defined by

..,gy(XI

1 + 1.

Then the sequence if.) is a recursively enumerable sequence of 1 increasing functions majorizing every primitive recursive U

function. 3.8

Theorem.

There exists a recursive function t such that

the set of functions computable on a one-tapeaTuring machine in time bound t is exactly the set of primitive recursive functions. Proof.

We first show that the set of primitive recursive

where {fe) is the sequence of Claim 3.8.3. C functions is u : e e By Claim 3,8,1 Let 4 be primitive recursive to see $I E U C@ e fe

.

t h e r e e x i s t s a p r i m i t i v e r e c u r s i v e g such t h a t $ 3.8.3,

.

e t h e popposite i n c l u s i o n , by Claim 3.8.3,

c:~

By Claim 3.8.2,

r e c u r s i v e functions; hence, r e c u r s i v e functions.

u

e

By Claim

and ( E U C i For e fe f e i s p r i m i t i v e recur-

g < f e f o r some e ; thus, ( E C:

s i v e f o r a l l e.

0

E Cg.

c o n t a i n s only p r i m i t i v e

i a made up of only p r i m i t i v e

C'

fe

0

Let { f e ) be t h e r e c u r s i v e l y enumerable s t r i c t l y i n c r e a s i n g sequence of Claim 3.8.3.

By Theorem 3.7

i s a r e c u r s i v e t such t h a t

C:

recursive functions.

-

u e

C'

*e

and t h e above

there

= t h e c l a s s of p r i m i t i v e

W e conclude t h i s s e c t i o n with a proof of t h e Xeyer-

McCreight-Moll Honesty

o r Naming Theorem.

This r e s u l t t e l l s us

t h a t given any sequence of p a r t i a l r e c u r s i v e f u n c t i o n s a measured sequence can be constructed so t h a t none of

the

complexity c l a s s e s of r e c u r s i v e functions a r e l o s t . 3.9

Theorem (Meyer-3lcCreight-Moll

[19,21]).Let @ be a complex-

i t y measure with p a r t i a l r e c u r s i v e functions {4e) ; then t h e r e e x i s t s a measured s e t {gg(,)} naming every complexity c l a s s . Proof.

W e w i l l g i v e a procedure which f o r any p a r t i a l

r e c u r s i v e Ot produces a p a r t i a l r e c u r s i v e $ g ( t ) such t h a t M(trx,y)

* $g(t) (XI

Y

i s a r e c u r s i v e p r e d i c a t e . That is, t h e s e t measured set of p a r t i a l r e c u r s i v e functions. be

uniform i n t

{ $ g ( t l ) w i l l be a Our procedure w i l l

implying t h a t g i s a r e c u r s i v e function. We

remark t h a t $g ( t ) w i l l n o t be t o t a l even i f $ t . i s t o t a l . We then show how t o o b t a i n a r e c u r s i v e implies

$g, ( t )

t o t a l . Further,

g' from g such t h a t $t t o t a l

w i l l be r e c u r s i v e and

.

= C f o r a l l et t o t a l 4t $g'(t) The c o n s t r u c t i o n d e s c r i b e d below f o r computing + g ( t ) (x)'

C

makes use of s e v e r a l a u x i l i a r y s e t s .

,at

The set '0

s t a g e s,

o f t h e c o n s t r u c t i o n , r e p r e s e n t s a queue i n which elements a r e "popped o u t " and moved f u r t h e r dawn t h e queue. Q'

are triples. (actually

.

The v

The elements o f

. having t h e

encodings of t r i p l e s )

r e p r e s e n t s t h e p r i o r i t y o f index e ; t h a t i s ,

e ' s r e l a t i v e p o s i t i o n on t h e queue.

The v a l u e of r i s e i t h e r

depending on whether o r n o t e i s "poppable".

1or 0

form

An i n d e x

e i s made poppable when i t is d i s c o v e r e d t h a t f o r some x , Q e ( x ) > 4t ( X I .

The index e i s l a t e r popped when an assignment

t o 4 g ( t ) i s made making Q e ( x ) > $ point

( x ) f o r some x. A t t h i s g (t) index e once a g a i n becomes u n p o p p a b t e . The s e t TO', at

s t a g e s , a c t s os a c o l l e c t i o n of t r i p l e s coming from Q" which have been d i s c a r d e d . < v , e , r > is e j e c t e d from QS The s e t

TP',

,

s'c s ,

Whenever a t r i p l e o f t h e form it i s done s o by p l a c i n g it i n t o TO".

a t s t a g e s , r e p r e s e n t s t h e f u n c t i o n $g ( t ) b e i n g

constructed.

A p a i r <x,y> i s placed i n t o TpS

only i f

4

( x ) = y.

and TP',

r e s p e c t i v e l y , j u s t p r i o r t o s t a g e s.

g (t)

Sets

QCS

, TO"

a t s t a g e s i f and

and TP<'

a r e QS, TO'

W e n e x t g i v e t h e procedure f o r computing $g ( t ) ( k ) f o r a

given k

.

Construction Stage 0.

S e t 'Q

Stage s.

L e t b = r l ( s ) and m = a 2 ( s ) where w1,r2

functions.

= TO'

= TP'

= $8.

(We assume t h a t *2 is i n c r e a s i n g i.e.

are pairing

* r Z (XI

x < Y

< n2 ( ~ 1 . 1 .

Next s e t QS = Q"

U

{<s,s,O>}.

p u t t i n g index s i n t o t h e queue g i v i n g it p r i o r i t y s.

We are

We a r e a l s o

making index s unpoppabZe ( z e r o ) . I f Q t ( b ) = m t o go t o ( 2 ) o t h e r w i s e , go t o ( 3 ) . Here we

at

check t o s e e i f (2)

(b) converges.

S e t TOs = TO" &

U

Qe( b ) > $ t ( b ) I

Q'

= Q'

-

{ < g , e , r > l < g , e , r >E

TOS

and set

u {l E

Q'

- TO'

&

a e ( b ) > (t (b) I .

That i s , w e make poppabte a l l t h o s e i n d i c e s e which t a k e more than $ t ( b ) s t e p s t o compute an i n p u t b.

Notice t h a t i n making p u t t i n g it

an index poppable w e have t o e j e c t t h e o l d t r i p l e i n t o TO'. (3)

I f 9 TP"

Otherwise set TP'

then go t o ( 4 ) .

= TP"

and go t o s t a g e s + l . H e r e it i s

discovered t h a t $ g ( t ) ( b ) has a l r e a d y been d e f i n e d . (4)

- TO'

Let v = min { f o r some e , < v , e , l > E QS v<s & Vv,,vI ( E TO' + pel ( b )

-

-

ml

&

Qe(b) > m

I.

That is, e i s t h e poppable index of h i g h e s t p r i o r i t y , l e s s t h a n o r e q u a l t o s, such t h a t at2 unpoppable i n d i c e s o f h i g h e r p r i o r i t y t a k e no more t h a n m s t e p s on i n p u t b.

F u r t h e r , index e ,

i t s e l f , t a k e s more t h a n m s t e p s on i n p u t b. If v = 0

t h e n set TpS = T P < ~and go t o s t a g e s + l . Here, no

such index e x i s t s . Otherwise, s e t

-

TO' & a e ( b ) > m e = rnin { ~ v , e , l >E e & v v ' < v[ ( < v 1 , e v , ~E> as TO') * Qel(bl 5 m11

-

-

e i s t h a t index w i t h p r i o r i t y v

causing a change t o be made.

I f t h e r e i s more t h a n one such e l i g i b l e index, e i s t h e s m a l l e s t

one. Set TpS = TP<' =

Set TO'

U {). That is, $g(t) (b) = m.

TO<^

U

{l and Q~ = 'Q

U

{cs,e,O>). That is,

pop index e from the queue to the s-th position and make it unpoppabZe. CHECK:

If k g b

then halt the procedure and output m,

otherwise go to stage s+l. This ends the construction. The function created by the construction was denoted symbolically as

@g(t).

represented.

We at no point made reference to what g

Bat it is clear that the function created is a

partial recursive function 8(t,k) of two arguments. Applying that sg-theorem 3.9.1

justifies our notation.

Claim. Proof.

{$g(t)) is a measured set. We have to show that M(t,x,y) * $g(t) (XI = Y

is a recursive predicate. it is clear that if

From the details of the construction

(x) is defined at stage s, then g(t) s = < x , $ ~ ( (x)>. ~) Hence, to decide if $ (x) = y we just run g (t) the construction until stage <x,y>. I#

We next modify g to g'

so that if $t is total then

$9' (t) is total, and (2) C @t = C @g'(t)

.

We define

(XI = min {$g(t) (x)I$t(x) + Og' (t) Again, the s:-theorem 3.9.2

Claim.

at (XI 1

justifies our use of notation.

{+g, (t) is a measured set.

Proof.

Since {$g(tl 1 i s measured

decided whether o r n o t Og ( t l (x) = y . since

qt (x)S * at (x)S,

it can be r e c u r s i v e l y Since {ae) i s measured and

it f o l l o w s t h a t we can r e c u r s i v e l y

decide i f + t ( ~ ) + Q t ( x ) = y.

I f e i t h e r one e q u a l s y , we can

t e s t t o see i f t h e o t h e r is ( y. The r e s u l t follows.

0

I t i s c l e a r from t h e d e f i n i t i o n o r $

t h a t +t t o t a l g1(t) Thus what remains t o be shown, i s t h a t f o r

total. implies $ g' (t) f o r +t t o t a l , C = C $t @g'(t) t h i s reduces t o showing

.

Claim.

3.9.3

S i n c e both +t

, +gl (t)

For $t t o t a l and any e , { x i ae ( x )

i f and o n l y i f

5

are total

$t (x) 1 f i n i t e

{ x l Q e(XI > + g l ( t ) (XI 1 f i n i t e .

Before proceeding we d e f i n e some u s e f u l terminology regarding

t h e c o n s t r u c t i o n of t h e f u n c t i o n $ Definition.

g ( t )'

An index e i s s a i d t o be s t a b l e i f t h e r e e x i s t s

e x i s t s some s t a g e so i n t h e c o n s t r u c t i o n some g and r E {0,11

(1) < g , e , r > E Q'

-

such t h a t f o r +g ( t ) TO" f o r a l l s > so.

Of

Index e i s s t a b l e a t O ( 1 ) i f t h e r above i s t h a t e becomes s t a b i l i z e d a f t e r so

o(1).

We a l s o s a y

.

An index e i s s a i d t o be u n s t a b l e i f it is n o t s t a b l e . That

is, e i s unstable i f f o r a l l stages s there e x i s t s a stage

s' > s

such t h a t f o r g , g l and r , r l E {0,11,

and where

g # g l o r r # r t . For a f i x e d e an e - t r i p l e i s a t r i p l e o f t h e form < g , e , r >

f o r some g and r E I 0 , l ) .

I f index e i s

Claim.

3.9.4

s t a b l e ( a t 0 o r 1) t h e n t h e r e

e x i s t s a s t a g e so such t h a t a f t e r s t a g e so

,

(1) e has a l r e a d y become s t a b i l i z e d ( a t 0 o r 1 1 , and

a l l i n d i c e s w i t h h i g h e r p r i o r i t y than e , and which

(2)

s t a b i l i z e , have done s o a l r e a d y . Proof.

Immediate.

Lemma.

3.9.5

0

Then f o r any index e i f e is

Let $t be t o t a l .

u n s t a b l e t h e n t h e s e t s A = {XIa e ( x ) > $t (x) ) and are infinite.

A' = { ~ l @ ~ (> x$)g l ( t ) (x) 1

Suppose e were u n s t a b l e . Then by t h e d e t a i l s of

Proof.

t h e construction t h e s e t s B = {S

I

an e - t u p l e goes from an unpoppable t o a poppable s t a t e

a t s t a g e sl, and B1=

is

I

an e - t u p l e g e t s popped a t s t a g e s ) ,

a r e both i n f i n i t e . tion.

Observe

then a t stage

Suppose

A

were f i n i t e t o o b t a i n a c o n t r a d i c -

5 A.

s e t C = {XIS = < x , Q t ( x ) > E B}

For i f x E C ,

s = < x , a t ( x ) > an e - t u p l e goes from an unpoppable

t o a poppable s t a t e . By t h e d e t a i l s o f t h e c o n s t r u c t i o n , t h i s o c c u r s o n l y i f a e ( x ) > $ t ( ~ ) . Hence, x E A. Thus, s i n c e C

5

A,

and A i s f i n i t e , C is f i n i t e . Define t h e map f ( x ) = < x , a t ( x ) > t o s e e B

s E B, t h e n a t s t a g e

C

f [C].

For i f

s = < x , a t ( x ) > , an e - t u p l e goes from an

unp ppable t o a poppable s t a t e . C l e a r l y , x

E

C and s = f ( x ) E f [ C ] .

However, s i n c e C is f i n i t e , f [ C ] i s f i n i t e , hence B must a l s o be f i n i t e

c o n t r a d i c t i n g our i n i t i a l observation.

For t h e second h a l f suppose A ' were bounded t o d e r i v e a contradiction.

Define t h e mapping k ( x ) = < x , $ J ~ (( X ~ I)> t o s e e

B 1 C k[A1]. For suppose

s

E

B1.

Then a t s t a g e s an e - t u p l e

g e t s popped.

By t h e c o n s t r u c t i o n s = < ~ , c p ~ (( x~) > ) .

t h a t is, Oe ( x ) > (g, ( t ) (x)

i f we show x E A ' .

.

We a r e done

By t h e construc-

tion Oe

(*

( X I > 9g(t) (xt

.

By d e f i n i t i o n +g, ( t ) (x) = min { @ g ( t )(x) , Q t ( x )+Ot (x) 1, and t h u s

(**I

+g(t)

(XI

@g'( t )( X I

'

Combining ( * ) and (**I w e have v e r i f i e d t h a t B' C k [A']. A s i n t h e f i r s t h a l f t h e assumption t h a t A '

i s bounded

t o g e t h e r w i t h t h e f a c t B1 C k [ A 8 ] l e a d s t o t h e conclusion B' i s finite. This ends t h e proof of Lemma 3.9.5. Lemma.

3.9.6

Let

0

Then f o r a l l i n d i c e s e , i f e is

be t o t a l .

s t a b l e a t 0 t h e n t h e s e t s A = { x l o e ( x ) > I $ ~ ( x ) and ) A'

= {xloe(x) >

Proof. C = {s

2

(g8

(t)

(x) 1 a r e f i n i t e .

Consider t h e set

e l d u r i n g s t a g e s , e i s poppable}

U

{sls ( e l .

Since e i s s t a b l e t h e s e t C i s f i n i t e , hence, s o i s nl[Cl. To s e e t h a t A f nl[Cl must be a s t a g e

x E A. Then $ t ( ~ )i s d e f i n e d and t h e r e

so = < x , O t ( x ) > . I f so

x = a l ( s O ) E al[Cl. construction

, let

2e

I f so < e , so E C, and hence, then by t h e d e t a i l s of t h e

s i n c e it is d i s c o v e r e d a t s t a g e so t h a t Q , ( x ) > @ ~ ( x ) ,

t h e index e must be poppable a t s t a g e so.

.

5

Since A x = a 1 ( S 0) E n1 [C] proved t h e f i r s t h a l f . Consider, n e x t

Hence,

nl [C] and al [Cl i s f i n i t e , we

t h e set

F = Ex

I

<x,O> > s o l

where so i s t h e s t a g e o f C l a i m 3.9.4.

Since <

,>

is increasing

i n t h e second argument a l l t h e x ' s i n F appear i n t h e c o n s t r u c t i o n of $

f o r t h e f i r s t time a f t e r s t a g e so.

g (tl

Claim.

3.9.7

Proof.

Tf x

€

F

then Q e ( x )

L $g '

(t)

'

is g' (t) t h e r e a r e two subcases.

S i n c e t h e value of $J

min {$Jg(t)( x ) '41.t ( x ) , Q t ( x ) I

.

Subcase (1)

$ g l ( t ) (XI i s $Jg( t ) (XI

.

From t h e d e t a i l s of

(x) i s made a f t e r g(t) it is done by popping an index of tower p r i o r i t y than

t h e c o n s t r u c t i o n , i f an assignment t o $ s t a g e so

,

e . F u r t h e r , from t h e d e t a i l s

of t h e c o n s t r u c t i o n , t h e assignment

i s made t o $g ( t ) ( x ) such t h a t $g ( t ) ( x ) ) Oe (x) result

.

Hence, t h e

follows.

Subcase ( 2 ) . t h a t $t ( x ) < Oe ( x )

.

+

+ g , ( t ) ( x ) i s +t ( x )

Qe(x).

Ot ( x )

+ OT (x)

Suppose

.

We have t o show

i n s t e a d t h a t $ t ( x ) + Q t( x )

Then +t ( x ) < Oe ( x ) and from t h e d e t a i l s of t h e constru-

construction

e would be poppabZe a t s t a g e

< x , Q t ( x ) > . This,

however, c o n t r a d i c t s t h e assumption t h a t e i s s t a b l e a t 0. From Claim 3.9.7

- (xl<x,O> 2 s o ) . c t h e former, A'

it follows t h a t

XI$^,(^) ( X I

< Oe(x) 1

But t h e l a t e r s e t i s f i n i t e , hence, s o i s

.

This completes t h e proof of Lemma 3.9.6. Lemma.

3.9.8

L e t +t be t o t a l .

0

Then i f e i s s t a b l e a t 1 t h e n

t h e sets A = {XIee ( X I > $t ( x ) 1 and A ' =

{xi Ot ( x ) > $g , ( t ) (XI 1 a r e

finite. Proof.

Consider any x having t h e p r o p e r t y t h a t <x,O> > so,

where so i s t h e s t a g e of Claim 3.9.4.

Since <

,>

is i n c r e a s i n g

i n t h e second argurrlent, it f o l l o w s t h a t p r i o r t o s t a g e "<x,O>" no mention of i n p u t x h a s e v e r been made i n t h e c o n s t r u c t i o n .

3.9.9

Claim.

I f x i s such t h a t <x,O> > so t h e n Q e ( x )

5 $t(x)

and Q e ( x )

5

$g 1 ( t )

'

Once we've proven Claim 3.9.9 A , A 1 C_ {xl<x,O> f s o ) ,

hence A,A1 a r e f i n i t e

Before showing Claim 3.9.9 3.9.10

it f o l l o w s t h a t

we i n t r o d u c e

For any x s u c h t h a t <x,O> > s

Claim.

,,

Qe

1

(XI

I$t ( x ) ,4g1 ( t ) ( X I ) f o r a l l e ' having higher p r i o r i t y t h a n

-< min

e and such t h a t e ' i s unpoppable. Proof.

S i n c e e n h a s h i g h e r p r i o r i t y t h a n e , e ' must be

s t a b l e a t 0.

From t h e d e t a i l s o f t h e proof o f Claim 3.9.7,

<x,O> > so then Oe, (x) ( Q g 1( t ) ( X I

.

since

It must a l s o be t h a t

Q e , ( x ) f $It(x) f o r o t h e r w i s e , e ' would be poppable c o n t r a r y t o o u r assumption.

Hence, 6,'

( x ) f min {$t ( x ) ,$g,

(x)) for a l l x

such t h a t <x,O> > so. We r e t u r n t o t h e proof o f Claim 3.9.9. where <x,O> > so.

Let m = max { Q e , ( x )l e '

Consider x fixed has h i g h e r p r i o r i t y

than e and e ' i s unpoppable a f t e r s t a g e sol.

m < min {$It(x) ,$Ig,

(*I

From Claim 3.9.10

(XI)

I f we can show

(**I

(Pe

we a r e done.

For by ( * ) and Qe(x)

5

(XI ( m

(**I,

min C$t ( x ) ,$g, ( t ) (XI 1

which v e r i f i e s Claim 3.9.9. 3.9.11

Claim.

A t s t a g e <x,m> is t h e earliest s t a g e a t which

$g (t) (x) c o u l d be d e f i n e d . Proof.

It i s c l e a r t h a t $ g ( t ) ( x ) cannot b e d e f i n e d b e f o r e

s t a g e <x,O>, s i n c e x never a p p e a r s f o r c o n s i d e r a t i o n before. Suppose + g ( t ) ( x ) was d e f i n e d a t s t a g e < x , y > , y < m. Then it must have been t h a t s i n c e m = Q e , ( x ) (e' having h i g h e r

p r i o r i t y than

e)

that m =

Qe,(k) > y = $Si(t) ( X I .

But t h i s c a n n o t o c c u r by

t h e d e t a i l s o f t h e c o n s t r u c t i o n . Hence, Claim 3.9.11 A l l t h a t remains is t o v e r i f y ( * * ) .

i s proved.D

C e r t a i n l y , Qe,( x ) 5 m

f o r a l l e ' having h i g h e r p r i o r i t y t h a n e and unpoppable. By Claim 3.9.11

a t s t a g e <x,m> i n d e x e would b e popped.

But t h i s

c o n t r a d i c t s t h e s t a b i l i t y o f e a f t e r s t a g e so. T h i s completes t h e proof o f Lemma 3.9.8. The proof o f Claim 3.9.3

D

f o l l o w s from Lemmas 3.9.5,

3.9.6

F o r a l l t h r e e imply t h a t f o r $t t o t a l and e any

and 3.9.8. index,

{x

I

@ e ( ~> ) $t

(XI 1

i s f i n i t e i f and o n l y i f (X

1

Qe( x ) > $g, ( t ) (XI 1

is f i n i t e .

4.

Speedable and

Nonspeedable S e t s

I n t h i s s e c t i o n w e examine some c l o s e r e l a t i o n s h i p s between a b s t r a c t complexity a n d p u r e

recursion theory.

In particular,

we c o n s i d e r c o n n e c t i o n s between s p e c i a l t y p e s o f sets c a l l e d speedable and nonspeedable and classes o f r e c u r s i v e l y enumerable

(r.e.) sets.

The f i r s t theorem i s S o a r e ' s 1281 p u r e r e c u r s i o n

t h e o r e t i c c h a r a c t e r i z a t i o n o f nonspeedable sets. Namely, s e t A is nonspeedable p r e c i s e l y when t h e c o l 1 e c t i o n o f r . e .

sets h a v i n g nonempty i n t e r s e c t i o n w i t h t o 0'.

IIJe c a l l any

i s Turing r e d u c i b l e

s a t i s f y i n g t h e l a t t e r c o n d i t i o n semilow.

A second theorem t e l l s

i n e v e r y r.e. T u r i n g d e g r e e . :larques,

indices of

u s t h a t t h e r e are nonspeedable sets Althouqh t h i s was f i r s t shown by

t h e proof g i v e n h e r e

is t h a t o f Soare.

h i s s e m i - l o w n e s s c r i t e r i o n makes t h e proof s i m p l e r . )

( H i s u s e of

We nex*

c h a r a c t e r i z e t h e s e r.e. Turing degrees t h a t c o n t a i n speedable sets.

Namely, an r.e. d e g r e e h a s a speedable s e t i f and o n l y i f

its Turing jump is above

9'.

We conclude t h e s e c t i o n w i t h a

d i s c u s s i o n o f t h e s t r u c t u r e o f t h e speedable s e t s and t h e i r r e l a t i o n t o w e l l - s t u d i e d c l a s s e s of r . e .

sets.

W e assume t h e r e a d e r is f a m i l i a r with t h e b a s i c n o t i o n s o f

o r d i n a r y and r e l a t i v i z e d r e c u r s i o n theory.

In particular, the

concepts o f r e c u r s i v e l y enumerable i n a set, Turing degree, jump of a s e t and a degree, O w , Turing r e d u c i b i l i t y

(zT),m-reducibil-

i t y (2nd and m-completeness a r e assumed. For d e t a i l s we r e f e r t h e r e a d e r t o Roqers [ 2 4 1

.

From now on {Re) and (

B ~ denote ~ 3 some c a n o n i c a l enumeration

of t h e r e c u r s i v e l y enumerable

sets and t h e s e t s r e c u r s i v e l y

enumerable i n B, r e s p e c t i v e l y . 4.1

Definition.(Blum and Marques 1 2 1 ) .

speedable i f f o r a l l e such t h a t Re =

A

An r.e.

s e t A is

and f o r a l l r e c u r s i v e

f u n c t i o n s h, t h e r e e x i s t s e' such t h a t Re,= A and

{XI

Qe(x) > h (x,Qe, ( x ) ) 1

is infinite.

I n t u i t i v e l y , i f A is s p e e d a b l e then f o r e v e r y a l g o r i t h m " e n f o r A and every r e c u r s i v e h t h e r e i s a n o t h e r a l g o r i t h m " e n " f o r A which i s a n h speed-up o f t h e f i r s t ( i . e . , i n f i n i t e l y often) 4.2

Definition.

n o t speedable.

Qe(x) > h (x, G e l (x)

. An r . e .

set A i s c a l l e d nonspeedabte i f it i s

More p r e c i s e l y , r . e .

s e t A i s nonspeedabte i f and

only i f t h e r e e x i s t s e such t h a t Re = A and a r e c u r s i v e f u n c t i o n h such t h a t f o r a l l e 1 where Re,

= A

,

x E A * (De(x) 5 h ( x , Q e , ( x ) ) f o r a l l b u t a f i n i t e set.

E s s e n t i a l l y a n r.e.

set A

i s nonspeedable i f t h e r e e x i s t s a

r e c u r s i v e h and an i n d e x e f o r A s u c h t h a t e v e r y a l g o r i t h m e' f o r A

c a n n o t be a n h speedup o f e. Given a set A t h e weak j u m p of A i s t h e set

Definition. HA =

el^^ " A

A set

# g).

A i s c a l l e d semi-Zou i f i t s weak

jump i s Turing r e d u c i b l e t o 0' (HA

L~ 0 '

.

S e v e r a l o f o u r r e s u l t s r e a u i r e t h e w e l l known l i m i t theorem of Schoenfield [261. 4.1

L i m i t Lemma.

Given S r w , S

L~ 0' i f and o n l y i f t h e r e i s

a r e c u r s i v e p r e d i c a t e S (e,s) such t h a t S ( e ) = l i m S ( e , s ) e x i s t s S

f o r a l l e. Proof (*)

. Suppose

and e q u a l s S ( e ) . L e t

S ( e , s ) i s r e c u r s i v e and l i m S ( e n s )e x i s t s

A,Z

S

b e d e f i n e d by

-

is r . e .

Since

t h e r e is a r e c u r s i v e f s u c h

and 0' i s m-complete

t h a t < e , t >E A * f ( < e , t > ) 9 0'. Hence,

g(e) *

and

Thus, S

. Then

3 t < e , t >E A

&

S(e,t)

2T 0 ' . Assume S f T 0' and t h a t S ( e ) = Cell

S ( e ) * 3 n 3m < e , l , n , m >

E

Re

&

1

Kn

5

0'

( e l f o r some el.

- 5'

0' & Km C

& K~ c - O' & K~ 5 5' el (By t h e d e f i n i t i o n o f r e c u r s i v e i n 0 ' where {Ki) i s a n enumera-

5(e) * 3 n 3m <e,O,n,m>

E R

t i o n of the f i n i t e subsets of w ) . 1

r e c u r s i v e p r e d i c a t e Os ( e ) where

S i n c e 0' i s r.e. 0' =

l i m 0;.

there is a

Define

i

1 if 3h.m < s <e,l,n,n> RS & KnC_ 0; S(e,s) I el 0 otherwise Clearly S(e,s) is recursive and from S = tell0',

&

,K

-

5 0;

lim S(e,s) S

0

exists and equals S(e).

We next prove Soare's recursion theoreticcharacterization of nonspeedability. Namely, nonspeedable sets are those r.e. sets whose complements are semi-low. Theorem (Soare [ 2 8 1 ) .

4.2

A recursively enumerable set A is

nonspeedable if and only if Proof

el^, "

# 81 $ 0'.

Let A be nonspeedable and let el and h satisfy

(*):

= Re

Define recursive function f by Rf

Definition 4.2. and Hz(e,s) by 1 HA(e,s) = 0

i

if (3X) [X E R:-R~

A

(XI > h (x,@f (el(x)

and 0 1

U

1

otherwise

We show that Hx(e) = lim HX(e,s) s is the characteristic function of Hz. By the Limit Lemma 4.1 HX(e,s) is clearly recursive.

HJi LT 0'. Suppose x

€

Re n

z (i.e.

e E Hz). Then x E :R

- 'A

= R:-RS

el Thus, since Re = A, 0 (x) diverges and since 1 el Rf(e) , h(@f(e) (x),x) is defined. Hence, HJi(e,s) = 1, s >so. Suppose Re n = (i.e., e 9 HA) then Re C_ A and hence,

for all s > so. Re

5

Rf (el

= Re U A = A. By the nonspeedability of A, since A = Rf(e),

is bounded by some xo. Hence, there will be some stage so such s S (= A ) for all s > so. Hence, (A Q x0)n(R:-$ ) 1 that A x0 Rel = $3 and thus HE(e,s) = 0 for all s > so.

(-1

Assume

is semilow,

By the Limit Lemma 4.1

that is, Hz= (elRen

# 91

there are Hji(e,s) and HE(e) where

zT 0'.

.

Hx (e) = lim Hx (e,s) s G(x,y,e) as follows: (1) If Qe(x) # y

Let el be such that Re = A and define 1

set

G(x,y,e) = 0,

t = min [Qe (x) = s or Hji(e,s) = 11 s2x 1 (a) If Qe (XI = t set G(x,y,e) = t let

(2) If Qe(x) = y 1

A

(b) Otherwise,

set

G(x,y,e)=O.

Observe that t exists in (2) when Qe (x) = y. Re = A or else x 1 for all but a finite set of s.

hence, either x

E

E

5 =

Let x

jB, HJi(e,s) = 0 for s > so.

2

,

x) to see that el and h

E

A where

h(Qe(x) ,XI.

x > maxis0,e,Qe (x)1, to see that Q (x) 1 el h (x,Qe(x) = max {G(x,Q~(x),e I e XI. Since x > e,

Re

For suppose Re = A. Then

witness the nonspeedability of A. since Re n

5

E

and then Hx(e,s) = 1

Re n

Define h(x,y) = max {G(x,y,el) le'

For then x

For then

G (x,Qe(x),el.

But since x > so, by the definition of G,

(XI. 0 el An easy method of demonstrating nonspeedability is given = @

in the following: 4.3

Corollary (Soare [281).

able it suffices for all e

to provide enumeration

set A to be nonspeed-

{ ~ ~ lof ~A such < ~ that

there are infinite sets of s such that

implies Re n Proof.

For an r.e.

RE-A~#

jB

x # 9. Define

RE-A~# 9

1

if

0

otherwise

Hx(e,s) = We will see that Hx(e) = lim Hx(e,s) exists for all e. Hence, s by Theorem 4.2,and Limit Lemma 4.1, A is nonspeedable. Suppose e

E

Hz

then there is an x E Re n

x.

Hence, for all

s > so. x E R:

n

6'

and t h u s Hi(e,s)

HJi(e,s)=L Then by h y p o t h e s i s

= 1.

Suppose f o r a l l s > so,

# $3 and t h u s e E Hz.

Re n

0

The n e x t r e s u l t was f i r s t shown by Marques i n [17]. I n 1281

Soare e x p l o i t s t h i s semi-lowness c r i t e r i o n t o o b t a i n a s h o r t e r proof. 4.4

Theorem (Soare I281, Marques 1171).

s e t s i n e v e r y r.e. Proof.

There a r e nonspeedable

Turing degree.

r.e. set i n degree a and l e t BS be some

L e t B be an

r e c u r s i v e enumeration o f B. Let IFn) be some c a n o n i c a l sequence of d i s j o i n t f i n i t e sets such t h a t (1)

IF,^

= n + 1 n # m, and

(2)

Fn n Fm = $3,

(3)

t h e f i n i t e set Fn i s r e c u r s i v e l y o b t a i n a b l e from

n.

We d e f i n e A by a c o n s t r u c t i o n . Stage 0.

S e t A'

Stage s + l :

=

0.

Define

min[*R:-As] e

I f n E B

set

A'+'

l e t xn

B',

-

u {xn).

= A'

i f such x e x i s t s otherwise

min Ix

E

Pn-

U

{dzle < n l l .

X

End o f c o n s t r u c t i o n . 4.41

Claim.

Proof.

s > so

,if

de = l i m d:

s For any e

n E B'+'-B~

construction

,s

e x i s t s f o r a l l e.

t h e r e i s a s t a g e so such t h a t f o r a l l t h e n n > e.

By t h e d e t a i l s of t h e

> so cannot be added i n t o AS.

Hence t h e

o n l y way i t can be changed is by being r e p l a c e d by

a smaller

d:

member of Re.

I t f o l l o w s t h a t a f t e r some s t a g e sl, dg is s t a b l e .

4.4.2

A i s nonspeedable.

Claim. Proof.

Suppose t h a t f o r an i n f i n i t e s e t o f s R:-As

# 0.

By the details of the construction it must be that Ren Hence, A's 4.4.3

Claim.

&

A

$3,

nonspeedability follows from Corollary 4.3. n E B * Fn n

A

# $3.

By this claim and the fact that B

g

and

A

IT B.

IF

fl

n

AI

( 1 we have

Hence, A and B are in the same Turing degree.

Proof (of Claim 4.4.3).

(*)

If n

E

B, let n E B'+~-B

for

some s~+l.Since IFn[ = n+l it follows that some xnE Fn is added to A at stage s+l. Hence, Fn n A # 8. ('1

then n

Since the {I?,) E

are mutually disjoint, if x

E

Fn n A

B.

This ends the proof of Theorem 4.3. Definition.

An r.e. set A is h i g h

if A' ET 0' and Zow if

-

A' ET 0'. An r.e. degree a is h i g h (Zoo) if it contains a high (low) set.

An r.e. set which is

4.5 Lemma. Proof.

low is semi-low. Let B be any set and let HB= el^,B # $3). We will

show that (1)

H~

H~

where

&,

and

H~

(2)

&"

B1

denotes m-reducible. From these it follows that HB

and that if B' ( T 0' then HB For (1) e

E

HB * Re n B # $3

For (2) e

E

nB * :R

&"

B'

L~ 0'. * :R (,) #

for some recursive f, *£(el H ~ . # $3 * 3m,n,x <x8nIm>ERe& Kn C - B & Km 5 !ij

& K~ c B & K~ 5 i where * 3m,n E R r (el R~ { 13x <x,m,n> E Re * B' 9

f(e) E B'

Marques proved

where

f (el = -

U

that if an r.e. degree contained a speedable

set then it must be high. Soare [ 2 8 1 completed the characterization by showing that

each such degree contains a speedable set

4.6

, Soare

Theorem (Marques 117I

[28])

.

An r. e. degree a

-

-

contains a speedable set if and only if a' > 0'. Proof.

-

Suppose a is low. Then for some A

(*)

E

-

a, A is low

and by Lemma 4.5 A is semilow. By Theorem 4.2, A is nonspeedable.

*

2' > 0-'. Then choose 9). (Here ( ~ is~ some 1

Suppose

A = 1 x 1 n ~ B~ f

-

B E a and let enumeration of the finite

sets where "xn recursively encodes set Kx.) Clearly, A ET B and 0' <

B'. T 4.6.1 Claim.

B'

LT

Hz.

From this and the above it follows that 0'
Hz fT 0'

. By Theorem 4.2,

A is speedable. B Recall that B'= {ele€~~). Define

Proof (of Claim 4.6.1).

as in Rogers [24]. (That is, p(e) S e (x) = y whenever Km <x,y,m,n> E Rp(e) By the s:-theorem, Ra (el Then e

5S

&

%5

8.)

there is a recursive u(e) such that

= {m13n [<e,e,n,m> E

, hence

B' * (317)[v

By definition of A,

E

R p(e)

E

R,(,)

&

a

R,

n

By definition of Hz *

- B).

C

& n'

Kv

5 El.

f pl.

~ ( eE ) Hz.

(Remark. B' is not only Turing reducible to HE

but also

1-reducible.) This completes the proof of Theorem 4.4.

0

We conclude the section by relating speedable sets to well structured classes of recursively enumerable sets. 4.5

Definition.

hypersimple

A confinite r.e. set A is f i n i t e l y s t r o n g l y

'if there is no recursive function f such that for

all x and y (4.5.1) (4.5.3)

x # y * Rf(x) n Rf(y) = Rf(x) isfinite, and

0 (4.5.2) Rf(x) n z f 0 (4.5.4)

U R f ( x ) -3 X

X.

The diagram below summarizes relationships between simple sets where "P

+

Qn indicates that property P implies

property Q and the lack of an arrow indicates that the implication fails.

Deleting condition (4.5.4) we obtain the definition

of A being h y p e r h y p e r s i m p l e and deleting (4.5.3) we have A The definitions of the others can be

s t r o n g l y hypersimple.

found in Rogers [24]

.

strongly r-maximal * hypersimple

maximal ir L quasi-maximal 4.7

,finite strongly hypersimple

T +

hyperhypersimple

Theorem (Jockusch [14], Soare [28])

+

\3

dense simple

. If A

+

hypersimple

J.

simple is finitely

strongly hypersimple then A is speedable. Proof.

Jockusch [14, Theorem 6.11

proves that if

is

infinite and low there is a recursive function satisfying 4.5.1-4.5.4.

His proof uses only the hypothesis HE I?, @ ' .

The theorem follows from Theorem 4.2. 4.8

Corollary (Marques [17]). Proof.

5.

0

Every maximal set is speedable.

Immediate from Theorem 4.7 and the diagram.

0

Generalizations In this section we examine the infinitary analogue of

abstract complexity theory in Kriple-Platek theory (cf. [I51 , [22])

.

admissible recursion

We introduce a generalization to the

definition of computational complexity measure.

Using this we

formulate generalizations to several important results from the w

theory and show these generalizations hold in the new theory.

In particular we prove analogues

to the

Blum, Rabin, and Gap

t h e r e e x i s t s some s t a n d a r d enumeration {K

0

In

< a ) of a l l t h e

a - f i n i t e subsets of a. 5.1

F S .

Let f b e a - p a r t i a l r e c u r s i v e and K an a - f i n i t e s u b s e t

of t h e domain of f . Then f[K] i s a - f i n i t e . o f a , denoted a*, i s t h e l e a s t o r d i n a l 8

The projectum such t h a t t h e r e 8.

i s a one-to-one

a - r e c u r s i v e mapping from a i n t o

Unlike w , s e t s which a r e bounded and a-r.e.

i l y a-finite.

5a

a r e not necessar-

However, a - f i n i t e n e s s w i l l occur i f t h e bound i s

s m a l l enough. 5.2

Fact.

I f 0 < a* and A i s an a - r e c u r s i v e l y enumerable s u b s e t

of 0 , then A is a - f i n i t e . We employ t h e a - r e c u r s i v e f u n c t i o n s , i n g a one-to-one

'rrl,s2

correspondence between a and a

and < x

,>

obtain-

a (cf. [16]).

These enjoy t h e u s u a l p r o p e r t i e s of p a i r i n g f u n c t i o n s . Namely, f o r B1,B2,y

( i ) < * 1 ( ~ )= rY ~ ; (ii) 2(~ al() )> = B1

< a,

a2() = B 2 ; and n l ( y ) = B1

and

f o r an unbounded s e t of y.

Sacks [15] proves t h e e x i s t e n c e of an a - r e c u r s i v e r ( a , & )

2 r implies

such t h a t ( i ) a

Kr ( u , E )

, and

5

a - r e c u r s i v e l y enumerable i m p l i e s B=U{K Definition. represents the

Let R:

= Kr

.

(ii)B

l a < a ) f o r some € < a .

r(u,~)

Then t h e s e t R ~ = ~ { R : I U C a]

a-recursiveZy enumerable s e t . We c a l l

E~~

E

the

t

index of R E . This formulation provides us w i t h an enumeration of t h e a - p a r t i a l r e c u r s i v e f u n c t i o n s . Namely, t h e a - p a r t i a l r e c u r s i v e f u n c t i o n w i t h index

E

is

.

-

Note t h i s c a r r i e s a l o n g w i t h it an a-analogue t o t h e Normal Form Theorem ( c f . A

(51

.

g e n e r a l i z a t i o n t o Blumts n o t i o n of a b s t r a c t c o m ~ l e x i t y

measure i s introduced. Definition.

i s an enumeration

An a-complexity measure, @,

( i n a ) of t h e a - p a r t i a l r e c u r s i v e f u n c t i o n s {@,I&< a } t o which a r e a s s o c i a t e d t h e a - p a r t i a l r e c u r s i v e a-step counting f u n c t i o n s I@,le
is d e f i n e d i f f @,(B)

(1) $,(B) (2)

f o l l o w i n g axioms hold:

The p r e d i c a t e M(c,B,y) *

i s defined QE

(0) = y i s a-recursive.

It is a l s o assumed t h a t a - r e c u r s i v e ve'rsions of t h e

Universal Function and Recursion Theorems hold tion

,

1E

s:,

f o r t h e enumera-

Implicit i n t h i s definition is the capability

< a .

t o r e t r i e v e , given any index form of an ' a l g o r i t h m ' ,

E,

,

b o t h t h e f u n c t i o n 4,

and t h e a - s t e p c o u n t e r , Q,

in the

.

iJe conclude t h e s e c t i o n w i t h s e v e r a l examples of a-complex-

i t y measures.

(1) For t h e c a s e a = w ,

it i s c l e a r t h e n o t i o n of

a-complexity measure c o i n c i d e s w i t h Blum's n o t i o n . ( 2 ) L e t {0,lc
be t h e enumeration of t h e a - p a r t i a l r e c u r -

s i v e f u n c t i o n s o b t a i n e d v i a t h e Kripke Equation C a l c u l u s (EC) D5]. Namely, 8,

i s t h e f u n c t i o n computed from t h e system of e q u a t i o n s

with Giidel number Godel number

E

&

< a. L e t G N - ~ ( E ) be t h e s e t of e q u a t i o n s w i t h

and SE t h e a - f i n i t e s e t of e q u a t i o n s d e r i v e d from

Y

s e t E by s t a g e y according t o t h e r u l e s of EC. A set o f a - s t e p c o u n t i n g f u n c t i o n s { O , l ~ < a }i s d e f i n e d by

OE(B) = y * BE(@) Clearly,

is f i r s t definea a t

0, ($1f i f f 0, ( B ) f

.

G N ' ~(E S

Y

Further, the predicate

PI(c,B,y) * t h e e q u a t i o n d e f i n i n g t h e v a l u e of 0, ( 6 ) f i r s t G N - ~( E

a p p e a r s i n S ' N - ~ ( ~ )i n some s t a n d a r d enumeration of Sa

Y

is a-recursive.

( 3 ) L e t ( 6 e l ~ < a l be t h e enumeration o f t h e a - p a r t i a l r e c u r s i v e f u n c t i o n s d e f i n e d i n ( * ) . L e t { A , ]& < a }be a corresponding s e t o f a - s t e p c o u n t i n g f u n c t i o n s d e f i n e d by a2-(v A,(B) = min { < B , n l ( ~ ) > R E I. YRE 1 y
prove a g e n e r a l i z a t i o n t o Blum's theorem, c a l l e d ,

a p p r o p r i a t e l y enough, t h e a-Blum theorem.

An analogue t o t h e

Rabin theorem f o l l o w s a s an immediate c o r o l l a r y . 5.2 a-Blum Theorem (Jacobs [ l o ] .

L e t 4' be an a-computational

complexity measure and g an a - p a r t i a l r e c u r s i v e f u n c t i o n . Then there

corresponds t o g a 0-1 valued a - p a r t i a l r e c u r s i v e f u n c t i o n

f , w i t h t h e same domain a s g , such t h a t i f

E

i s any i n d e x f o r f

t h e n Q E ( B ) ) g(B) f o r a l l b u t a bounded s e t of 6. An immediate consequence i s t h e 5.3 a-Rabin Theorem (Jacobs [ l o ] ) . L e t

Q be an a-computational

complexity measure and g an a - r e c u r s i v e f u n c t i o n . Then t h e r e corresponds t o g a 0-1 valued a - r e c u r s i v e f u n c t i o n f such t h a t if

E

is any index f o r f t h e n O E ( B ) 2 g(B) f o r a l l b u t an a - f i n i t e

s e t of 6. Proof.

L e t f be t h e 0-1 valued a - p a r t i a l

corresponding t o g by t h e a-Blum Theorem. and dom(g) = d o m ( f ) , f is a - r e c u r s i v e .

recursive function

Since g i s a - r e c u r s i v e

For any index

E

f o r f , the

s e t B = (BIOE(B) < g ( 6 ) ) is bounded. S i n c e g i s a - r e c u r s i v e , a-recursive

, hence,

a-f i n i t e

.

B is

0

Rabin and B l u m ' s o r i g i n a l c o n s t r u c t i o n s h i n g e upon a fundamental p r o p e r t y of w; namely, t h a t unbounded sequences c a n n o t be one-one mapped i n t o p r o p e r i n i t i a l segments.

I n t h e c a s e of

admissibles greater than w , such can occur due to the presence of limits.

Consequently, in a-constructions in which objects are

"cancelledn based upon natural order priority ( < a), one cannot automatically assume that objects with unbounded opportunities for cancellation will ultimately do so.

For it may occur that

every time an object with priority 0
be cancelled.

The way in which we solve this problem is by formulating constructions with priorities founded, not upon natural order, but rather upon ordering induced by an a-recursive projection. By making use of the properties that a-r.e. a-f inite (Fact 5.2

)

subsets below a* are

and that a satisfies El-replacement (Fact 5.1)

we argue that objects trying hard enough will ultimately be cancelled. We now prove the a-Blum Theorem. Proof.

By definition of projectum, there exists an a-recur-

sive one-to-one map t: a

+

a*. The function t serves to generate

our "prioritiesn. For any index E
pseudo index for

is called the

E.

Since t is one-to-one and a-recursive, it has an a-partial recursive inverse t-l: a * . Fix c0 in our enumeration (4,) = t-l. Let ',t 0 ,
that (E

so

be the a p p r o r i m a t i o n t o t-l t h r o u g h s t a g e

($1

if

(0) 5 a

@, 0

otherwise Clearly, for ,,$
s p e c i f i e d belbw.

T h i s c o n s t r u c t i o n proceeds i n s t a g e s a < a and i s

d e f i n e d by t r a n s f i n i t e i n d u c t i o n on a . Throughout t h e c o n s t r u c t i o n we w i l l be accumulating two s e t s A and F. The s e t A',

a t s t a g e a , w i l l s e r v e a s t h e c o l l e c t i o n of

pseudo i n d i c e s which have been " c a n c e l l e d n p r i o r t o o r d u r i n g s t a g e a . The s e t A<'

w i l l denote t h o s e c a n c e l l e d pseudo i n d i c e s

j u s t p r i o r t o s t a g e a ; i . e . A<'

=

A'.

U T
The s e t F',

a t s t a g e a s e r v e s a s t h a t p a r t of t h e graph of

o u r d e s i r e d f u n c t i o n c o l l e c t e d a t s t a g e a . The s e t F<' w i l l be a F j u s t p r i o r t o s t a g e a ; i . e . F<' = u F'. '
Stage 0.

= F'

=

9.

Stage a. (1) S e t B = (2)

I T ~ ( U I)

= n 2 ( a ) where

IJ

I I ~ , I Ia~r e

I f @ I ~ ( B )# 1 1 , where g = @ A t h e n s e t A'

d e f i n e d above.

= A<';

FU = F";

go t o s t a g e a + l . (3)

set

v

(4)

If

A<'

= { ~ l t ( l ' ( ~ ) L+ =A<'

Fa = F"

U

u V

E

then

5 t (8) set

A'

{1; go t o s t e p

(5) S e t p = min ( T

E

V

& T

T

-1 (6) < g ( 8 ) ) . t a (E) = A < U ; 0 = 1;

L

9 A<'};

(6). A'=

where 0 i s e i t h e r 0 o r 1 b u t n o t 4 (6)

If 8 =

K

Set F =

A<'U

t - h

t h e n o u t p u t 0 and h a l t ; else U

a
{PI; F ~ = F < ~ U { < B , ~ > ~ (8). t o stage a+l.

Fa, A =

U A' and l e t f be d e f i n e d by a E F

.

We show f s a t i s f i e s a l l requirements v i a a sequence of lemmas. 5.2.1 and

Lemma.

( i ) f i s a 0-1-valued

(ii) dorn(g) = dom(f1.

a-partial recursive function

Proof.

(i) The 0-1 valuedness is immediate.

definedness we observe that a value

For well

is entered into F at

stage a only if a = and aX(B) = p (g = only one a =

Since there is

and at each stage at most one entry is put

into F, the result follows.

The construction is a-effective,

hence f is a-partial recursive. (ii) B E dom(g1 iff @h(B) = p (g = $A) for p < a; iff at stage a = an entry is placed into F; iff At this

B

E dom(f).

point we introduce some useful notation. For v
E v = { p < a* 1 p E range (t) & p < v and p gets

cancelled during the construction}. Ev (=A (1 v)

is the set of all pseudo indices less than v which

are ultimately cancelled. Define the mapping k from a*

+

a

as

k ( ~ = ) the stage at which pseudo index

E

gets cancelled during the construction.

Since a pseudo index is cancelled at most once, k is well defined.

It is understood that for ~ < a *not in t [a], k (E) is

undefined. By the a-effectiveness of the construction, k is a-partial recursive. 5.2.2

Lemma.

For all v
(i) Ev is a-finite, Proof.

and

(ii) k[EVl is a-finite.

(i) Clearly, Ev is a-recursively enumerable and The result follows from Fact 5.2.

bounded below a*.

(ii) By (i), EV is a-f inite. a-partial recursive.

From the definitions EV

result follows from Fact 5.1. 5.2.3

Corollary.

Let v
following stage a

0

By the above discussion, k is

5 dom(k). The 0

Then there is a stage oo such that

no pseudo index smaller than v will ever be

91

c a n c e l l e d by t h e c o n s t r u c t i o n . The s e t k[Ev] i s comprised o f a l l o f t h e s t a g e s a t

Proof.

which pseudo i n d i c e s below v a r e c a n c e l l e d .

set i s a - f i n i t e , 5.2.4

Lemma.

a > a

2

By Lemma 5-2.2

hence bounded by some o r d i n a l a.

L e t v
this

< a.

Then t h e r e e x i s t s a 02
i m p l i e s t ( a ) > v. The s e t CV=

Proof.

IE

< a*[€ E t[a] b

E

is

(v}

a - r e c u r s i v e l y enumerable and bounded below a*; by F a c t 5.2 it i s a-finite.

S i n c e C,z

domft-l)

by F a c t 5.1 ti' [ c V ]i s a - f i n i t e ,

and t-'

is a-partial recursive.

hence bounded by u2ca. The lemma

f o l l o w s from t h e d e f i n i t i o n s . We a r e now r e a d y t o prove 5.2.5 {B

Lemma.

1 aE ( B )

Let

<$(B) 1

Proof.

E

b e any i n d e x f o r

f.

Then t h e set

i s bounded.

Assume f o r some i n d e x

E

for f , the s e t

{ B [ @ ~ (
L e t t ( ~= )E ' and l e t al b e t h e

-1 l e a s t s t a g e where to ( E ' )

By C o r o l l a r y 5.2.3

1

=

E.

s u c h t h a t no pseudo i n d e x s m a l l e r t h a n s t a g e ao. t(a) >

E'.

B ' 2 maxi0 defined.

By Lemma 5.2.4

there is a

E'

t h e r e i s a oo

is cancelled p a s t

s o t h a t a > a2 i m p l i e s

S i n c e i B I @ E ( B ) < g ( B ) 1 i s unbounded, it h a s a member al ,a 2 1.

C e r t a i n l y , Q E ( 8 ' 1 S and hence < B

Let a ' = < B o 8 O E ( B ' ) > .

' ,eE ( B '

> is

By p r o p e r t i e s o f p a i r i n g

f u n c t i o n s ( c f . G6del [ 6 ] ) , 0 ' < a ' < a , and hence a ' > max {a0,a1,a2).

From t h e d e t a i l s o f t h e c o n s t r u c t i o n

must be c a n c e l l e d no l a t e r t h a n s t a g e a ' . f #

@E,

E'

However, t h i s i m p l i e s

c o n t r a d i c t i n g t h e hypothesis.

N e remark t h a t i n b o t h B l u m ' s and R a b i n ' s r e s u l t s (a = w )

t h e complexity o f t h e c o n s t r u c t e d f u n c t i o n exceeded t h e g i v e n

0

f u n c t i o n on a l l b u t a ' f i n i t e '

set of w.

In t h e generalization,

however, one analogue t o f i n i t e was a - f i n i t e

The reason f o r t h i s is t h a t t h e

t h e o t h e r was bounded (a-Blum). excluded set { $ I QC ( 8 ) < g ( $ )

(a-Rabin), w h i l e

a l t h o u g h bounded does n o t appear

a-recursive f o r g nontotal.

I n 1291, Stoltenberg-Hansen

proves

an analogue t o t h e B l u m Theorem where f i n i t e i s , i n f a c t , Furthermore, it i s a l s o s e e n i n [291 t h a t

r e p l a c e d by a - f i n i t e .

even when a i s n o t admissable.

an analogue h o l d s

For any a-complexity measure 0 and a - r e c u r s i v e

Definition.

f u n c t i o n s t h e a-comptexity c l a s s bounded by s i s : C

= {(,I(,

Thus C 0

is t o t a l

, or

&

@,($)

simply Cs

5 s ( $ ) f o r a l l b u t a n a - f i n i t e set

,

o f $1. when 0 i s understood, i s t h e s e t

of a l l a - r e c u r s i v e f u n c t i o n s whose complexity i s bounded by s on a l l b u t an a - f i n i t e s u b s e t o f a. We n e x t prove a g e n e r a l i z a t i o n of t h e Borodin Gap Theorem Our r e s u l t , t h e a-Gap Theorem, t e l l s us t h a t t h e r e e x i s t s no uniform way t o i n c r e a s e bounds on a-complexity c l a s s e s t h a t w i l l always y i e l d l a r g e r

classes.

r e s u l t is t h e a - s t e p

Another way of i n t e r p r e t i n g t h e

counting functions a r e sparse r e l a t i v e t o

the a-recursive functions. 5.4 a-Gap Theorem (Jacobs [ l o ] ) .

a ( $ ) and r ( B , y )

(with r ( $ , y )

L

For a l l a - r e c u r s i v e f u n c t i o n s

y)

we can f i n d an a - r e c u r s i v e

f u n c t i o n s such t h a t

(i) s ( B ) 2 a ( $ ) f o r a l l (ii)For each

E

$ c a

< a , OE($)

, and

2 ~ ( $ 1i m p l i e s

>- r(B,s(B)

f o r a l l b u t an a - f i n i t e set o f 0 . The a-Gap Theorem may also b e viewed i n terms of a-complexit classes.

5.5

Corollary.

Assume t h e h y p o t h e s i s o f t h e a-Gap Theorem.

Then w e can f i n d an a - r e c u r s i v e f u n c t i o n s s u c h t h a t (i)

~ ( $ 1> a(B) f o r a l l 8 < a , and

(ii)

'XBS(B)

Proof.

0

C ~ ~ r ( ~1 , s ( ~ )

-1 L e t t and to b e a s i n t h e p r o o f o f Theorem 5.2.

An a - r e c u r s i v e

f u n c t i o n s ( B ) w i l l be d e f i n e d i n terms o f a

c o n s t r u c t i o n below.

For each s t a g e o < a , t h e construction a s p l i t s i n t o s u b s t a g e s 6 < a , where t h e v a l u e s sB a r e computed. A f t e r enough s u b s t a g e s a r e performed, some sa i s chosen as t h e 80 value of s (a)

.

Construction. S t a g e 0.

Set s ( 0 ) = a ( 0 ) .

Stage a. S u b s t a g e 0.

(3 S e t so = a(o)

S u b s t a g e B. a = s u p t ( u , s:) + 1 set sg 6<8 I f f o r nc E ' ( t ( 0 ) -1 1. t + and a 2s B( 0 - 1 (a) 5r(u8 t, ( € ' I t h e n s e t s ( a ) = s:; go t o s t a g e

si) a+l;

else go t o s u b s t a g e @+1. end o f s u b s t a g e B. end o f s t a g e a . 5.4.1

Lemma.

For e a c h a < a t h e r e e x i s t s a B,

s ( a ) i s d e f i n e d a t s u b s t a g e B,

< a such t h a t

o f s t a g e a.

Certainly, t h e a-recursjveness o f s ( a ) follows. For t h e moment we w i l l assume Lemma 5.4.1 5.4.2

Lemma.

s(al

2

i s proven.

a ( a ) f o r a l l a < a.

By a simple induction on 8 , w e s e e t h a t a t s t a g e a ,

Proof.

si 2 a ( u ) f o r a l l 5.4.3

Lemma.

The r e s u l t follows by Lemma 5.4.1.

8.

For each

&

< a

t h e r e e x i s t s a a, < a such t h a t

whenever a > a&:

2

rP,(a)

i s a-r.e.

@&

(01 > r ( a , s ( o ) 1 .

Cv z ~ ' l v € r n g ( t ) l .

and bounded below a*; hence, applying Fact 5.2

E& is a - f i n i t e .

is

implies

Let t ( &=) E ' and l e t E,=

Proof. E,

s(a)

Define F& = t - l [ + l

~ ( E ) I .

t

-1

-1

is a-finite.

bounded above by al. Since -1( E ' ) o > a2 i m p l i e s ta

5

i s a - f i n i t e and EE 5 dom(t

a - p a r t i a l r e c u r s i v e , E,

hence by F a c t 5.1 F,

= {a < a l t ( u )

&'

1;

s o t h a t F, i s

Choose a1 < a

r n g ( t ) , l e t a 2 be such t h a t

E

+.

5.4.4

Claim. Suppose

By d e f i n i t i o n of u1 , a 2

of a. I f OE ( a )

E'.

2 s (a)

C = {al@,(a) 2 S ( U )

hold).

6

@,(a)

(i.e.

.

2

E

< a, t h e s e t

r(a, s(a))I

where t h e i m p l i c a t i o n f a i l s t o

Since s and r a r e a-recursive,

a-recursive;

and

2 @,(a) ( r ( o , s ( a ) ) ;

we conclude t h a t f o r

i s bounded above by some a,

, ti1( E ' ) (.

by t h e d e t a i l s of t h e c o n s t r u c t i o n

( s i n c e sa = s ( u ) ) it c a n l t be t h a t s ( u ) a hence it must be t h a t r ( o , s ( a ) ) < @ , ( a ) From Lemma 5.4.3,

s ( a ) i s defined

By Lemma 5.4.1,

o > max {al,a2).

a t substage t(a) >

i s max {al,a2}.

The required a,

it follows t h a t C i s

hence, C i s a - f i n i t e .

Before v e r i f y i n g t h e l a s t p a r t we introduce preliminary 5.4.5

For a < a , t h e set H = {@,(a)It(,) ( t ( a )

Lemma.

til(t + (E))

Proof.

& @& ( a ) + )

is bounded by some

6' < a

&

.

-

F i r a < a and l e t Du = { c l < t ( u ) ~ t ( y l ( ~ ' &) +

a)+}. @ t Zll ((&

D,

i s a-r.e.

and bounded below a*; hence,

by Fact 5.2, Do is a-finite.

Define g,:

a*

+

a

by

( ~ ) i f t ~ ~ ( c ~ )( 0t) )~ ~ - ~ t, (E') otherwise g, is a-partial recursive and Do g, [D,]

5

dom(ga); thus by Fact 5.1,

bounded by some 13 '

is a-finite, and

from the observation that H = g, [Do]

.

The lemma follows

.

Finally, Proof.(of Lemma 5.4.1): defined for all o' 1s; 18 < a)

< a.

Fix a < a and assume s(oj is

By a simple induction, the sequence

is strictly increasing.

This regarded as a one-to-

one order preserving map of a iinto a, implies that rng(si) is cofinal with a.

Let 8' < a be as in Lemma 5.4.5,

8' > (OE(o) It(€) 5 t(o) of 1s;)

there is a

BO

&

til(t(€))+

< a, where s

0

6

OE(ol+). > 8'

that is, By cofinality

. Thus,

at substage

-< r(u, s" ) . Hence, by 0 0 u at most substage B0 of a, s(a) is defined; concluding s(o) = s , @a for some 8, < a B0 , for all investigated E, @€(a)

< si

.

Most of the major results of abstract computational complexity theory (as found in Hartmanis and Hopcroft [7]) have been shown in [ 9 1 to generalize (in one form or another) to a-recursion theory. Besides those presented here, these include Blum's Speed-up, Lewis-Landweber-Robertson's Nonrecursively Enumerable Complexity Class and McCreight-Meyer's Honesty and Union Theorems.

The generalization of the last result, the Union

Theorem, required a rather extensive overhaul of the w-Proof. Namely, instead of a typical cancellation argument, the generalization necessitated a cancellation atop a finite injury priority

construction.

Further, the listing provided by a* did not

suffice here.

Consequently, a shorter listing together with a

blocking strategy (cf. [27]) had to be implemented.

References [I] Blum, M.,

"A machine-independent theory of the complexity of

recursive functions," J. ACM [21

Blum, M.,

and Marques, I.,

14 (1967) pp.

"On complexity

322-336. properties of

recursively enumerable sets," J. Symbolic Logic 38 (1973) 579-593. [3] Borodin, A. ,"Computational complexity and the existence of complexity gaps," J. ACM [41

Constable, R.,

[5] Davis, M.,

2

(1972) 185-194.

"The operator gap," J. ACM

19

(1972) 175-183.

Computability and UnsoZvabiZity, ~ ~ G r a w - H i l l

(1958), New York, N. Y. [61 Godel, K.,

The Consistency of the Continuum HypothesCs,

Princeton Univ. Press, Princeton, New Jersey, 1940. [7] Hartmanis, J.,

and Hopcroft, J., "An overview of the theory

-

of computational complexity," J. ACM 18 (19711, 444-475. [81

Hartmanis, J.,

and Stearns, R. E., "Computational complexity

of recursive sequences," IEEE Proc. Fifth Annual Symp. on Switching Circuit Theory and Logical Design, 1964, 82-90. 191

Jacobs, B. E., a-ComputationaZ CompZexity, Ph.D. Thesis, New York Univ., Tech. Rep. IMM-408, Courant Inst., NYU, 1975.

[lo] Jacobs, B. E.,"On J. Symb. Logic, [ll] Jacobs, B. E.,

generalized computational complexity,"

42

(1) (1977) 47-58.

"The a-union theorem and generalized primi-

tive recursion," Trans. AMS [12] Jacobs, B. E.,

237

(1978) 63-81.

"a-Naming and aSpeedup Theorems," Notre Dame

J. Symb. Logic, XX (2) (April 1979) 241-261. [13] Jacobs, B. E., " a-Speedable and non a-speedable sets," Can. J. Math. ,. to appear. [14] Jockusch, C. G., "The degrees

of hyperhyperimmune sets,"

J. Symb. Logic 34 (1969) 489-493. [15] Kripke, S., I, 11,

Transfinite recursion on admissible ordinals,

(abstracts),

J. Symb. Logic

29

(1964) 161-162.

[16] Kripke, S., The Theory of Transfinite Recursion, unpublished unpublished lecture notes by A. Thomas Tymoczko. [17.] Marques, I., "On degrees of unsolvability and complexity properties," J. Symb. Logic (1975) 529-540. [18] McCreight, E. M.,

CZasses of Computabte

Functions Defined

by Bounds on Computation, Ph.D. Thesis, Computer Sci. Dept., Carnegie-Mellon Univ., Pittsburgh, Pa., 1969. [19] McCreight, E., and Meyer, A,, "Classes of computable functions

Defined by bounds on computation," Preliminary

report, Proc. ACM Symp. on Theory of Computing (1969) 79-88. [20] Meyer, A., and Fischer, P., ~8Computationalspeedup by effective

operators:

J. Symb. Logic

[21] Moll, R., and Meyer,

A.

37

(1972) 55-68.

R., "Honest bounds for complexity

classes of recursive functions: J. Syrnb. Logic

39

(1974)

127-138. [22] Platek, R., Foundations of Recursion Theory, Ph.D. Thesis, Stanford, 1966. [23] Rabin, M. O., "Degrees of difficulty of computing a function and a partial ordering of recursive sets," Tech. Rep. 2, Hebrew Univ., Jerusalem, Israel (1960). [24] Rogers, H.,

Theory of Recursive Functions and Effective

Computability, McGraw-Hill, 1967, New York, N. Y. [25] Sacks, G. E., Righer Recursion Theory, Springer, Berlin, to appear. 1261 Shoenfield, J. R., Amsterdam, 19 [271 Shore, R. A.,

Degrees of UneoZvabitity, North-Holland,

. "Splitting an a-recursively enumerable set:

Trans. Am. Math. Soc.

204

(1975) 65-78.

1281 Soare, R., "Computational complexity, levelable sets,* J. Symb. Logic [291 Stoltenberg-Hansen, V.,

42

speedable and

(1977) 545-563.

"On computational complexity in

weakly admissible structures," Tech. Rept. ISBN 82-553-0342-1 Inst. of Mathem., University of Oslo. [30] Young, P.,

gap and

"Easy constructions in complexity theory: speedup theorems: Proc. AMS

37

(1973) 555-563.

CEN TRO I N TERN AZIONALE MATEMATICO ESTIVO (c.I.M.E.

A SURVEY OF S E T RECURSION DAG N ORMANN

A SURYEX

OF SET HXCURSION.

Dag Normann, University of Oslo, Norway.

a-recursion-theory, p -recursion-theory and recursion in normal type-3 objects has one thing in common, one investigates certain notions of computability on certsin sets, L(*), L(p) ~ n dthe type-1 objects I respectively. But while in d - and p-recursion theory o m mqy carry out the resp. L(p) , we nust e;o outside I in order to capture the finer points concerning . are several ways to do this. recursion in, say 3 ~ There One way is to follow Sacks ( this volume ) and construct a hierarchy for the subsets of I recursive in 'E, or in analysis within the structure L(4)

, This involves bringing in a certain' general, in 3 ~ F. superstructure of sets, e.g. ordinals and subsets of I, and history has shown that any finer analysis of 3~ involves the construction on one superstructure or another. In this paper we will take the full and most general consequence of this. We will start with an arbitrary set x and an arbitrary relation R and we will ask : About which other sets do x , R contain perfect inforaation? An alternative way of putting the question is : iVhich sets are *recursivet in q , H ?

It turns out that when we have answered this question we have not only constructed a sensible notion of set-recursion but we have also fully captured and generalized recursion in norvnal functionals. So we have given an alternative to the defirdtion of recursion in normal functiollals given by Sacks ( this volutne )

.

We will now do a bit of History-forging and develop Set-recursion as it ought to have beendeveloped. * We have given a relation lf and sets x = (xl, and we will just find out which,operations on 2 relative

... 3 2

to R

we may within reason call set-recursive. We must

forget absolute notions of finite computations and agree upon the following crucial point : Once we accept a set x as an argument of ? set-recursive function then this function may think that x In

a-

is finite. and p-recursion theory we investigate r.e.

sets

nnd recursion sets but the notion of a computation is rather

innlicit. In Sackst definition of recursion in 3~ we feel that from a notation for an Hbset Ifa and an index for a lst order formula over Hi

, there i a

p we

may compute the set defined by y

some sort of transfintte procedure

leading from (8,e) to the set defined. Here we will take the full computatio~vrlpoint of view. We will define a class of pernittable computations by induction over the ordinals. This scheaatic approach is the same aa Kleene used when he first defined recursion in Mgher types. At the botto~nwe need a few combinatorial operations : f( 2 ) = xi ( 1 4 ) is set-recursive with indcr
= xi\ x

f);(

j

= [xi,xJ)

f);(

(i,jgk)

is rot-rccuraivt:rith

index ( 2 ,i,j,k > (I,j4k) is sat-recursive with

index ( 3 ,i, j,k> X'e will describe two methods of constructing new set-recursive functions from old ones.

In the first we really think that

x is finite : If f is set-recursive with index e,, g(;)

=

ygl (f(y,x2, ... ,x 1 ) n

we let

be aet-recursive

with indt~x (4,e,,k>

.

Less controversial is to close the set-recursive functions under composition : are set-recursive with indices If fl, vfn el, 'en respectively and they all accept the same nuqber

...

.*.

of arguments, and h

is set-recursive for n arguments and

with index d then

.!?(?I

= h(fl(?),

. .f n( 1 )

is set-recursive with

index ( 5 ,d ,el, We have not brought in the R

... ,en,k>

yet but the sirnplest

way to do so is just by : f ( 2 ) = xin R is set-recursive relative to R

with

index ( 6 ,i,n> The class of functions we have defined so far is well-

.

known and called the functions rudimentary in R But we have taken the liberty to enumerate the llalgorithms"for the rudimentary functions by the indices and as every recursion 3 theorist knows, given a n .argument x and an index e for an algorithm we may uniformly in ( {elR(;))

2

and

e compute f e) (?)

. So we add the last part of the definition :

f(e,?)

=

R 4 Eel (x) is set-recursive relative to R with index <7,k>

Theoretically we have n o w given seven clauses in an R inductive definition of the relation {el (34) = y , and such notions as denght of computation and subcomputations are easily defined. We may also prove the recursion-theorem and other standard helpful1 results in recursion theory. O u r first lemma is trivial, but it is the most important result about set-recursion Lemma

t

-

There is a set-recursive function f such that given -b

e,x,x

is defined if and only if [el (y,?)

then f (e,x,;)

are defined for all y E x

, and

then

Proof Let g(e,x,E'I = yk/xx[e3(~,z) Then g(e,x,'jZ3 = P C? Vy x Ee3 (y,3 = 16 f is constructed from g using a rudimentary function. Tho following results are simple but tedious and we leave them for the reader : 1.

The function f(x) = ordinal rank of x recursive.

2.

If x

3.

We 3ay set-recursively decide if a set x nuaber or not.

4.

g(x) = TC(x) is set-recursive. If x is infinite ( < = 7 rank x . ) w )

5.

=

is set-

is infinite then w is set-recursive in x

the closure of x

.

is a natural

then Cl(x)

under finite subsets, is set-

recursive uniforaly in x 6.

If

x = I end

a

.

is sn F-notsti on for H~ then R

.

8a

is

uniformly set-recursive in I, a relative to Y 6. shows that if A 6 3E , F, a then A is set-recursive in I, a relative to F , The opposite is also true, but the argument involves a cod in^ machinery which we won't give here. Theorem Let A S I, a s 1

. Let

Y

bc

R

normal type-3 functional.

is recursive in F, a if and only if A

Then A

set-recursive in I, a relative to F A

set or class A

relative to H

.

is

+

is called semi-set-recursive in x

if there is nn index e such that

re3R (x,;) 2 0 We then also have the following result : A G Tp(2)

xEA

r.e, in 3 ~ F, ,

<=)

R

if and only if A

in I, a relative to F

is

is semi-set-recurstvc

.

This shows that the r.e. deeree-theory for 3~

may be

represeneed as a theory for set-recursive d~gcees. Our final comparison with recursion in 3~ comes in showing that we have natural alternatives to the W-sets. Let x

be given.and assume that x

is transitive and

closed under finite subsets. Each: computation [el 4

YeX

has a certain length 11 <e,?,x)l\ M,(y) = [ z : 3 e [e\(y,x) = z

u

. Let A

(F,x)

for

Jl<e,y,x>l\I Q ]

.

Let M, = Mc(y) Then each I , is recursive in x YGX some y c x , and they will fully play the role of H-sets.

and

We will not go deeply into degree-problems here, that is dealt with by Sacks. For each infinite set x with suitable closure-properties there will be the following

natural degree-sfructure to investieate say on X

gl

= ( a: 38, J G

x

I e x > =.c

3

using

R

-b

set-recursive reducibility set-recursive reducibility modulo elements of a given z + x

-C

set-recursive reducibility modulo elements of X: ( corresponding to Sacks* absolute version of Post's

problem). Thls ought .t;obe a rich area for future research.

CENTRO I N TERSAZIOh ALE MATEMATIC0 ESTIVO

(c.I.M.E.

PRIORITY ARGUMENTS I N HIGHER RECURSION THEORY

GERALD

E. SACKS

Centro Internazionale Matematico Estivo

Priority Arguments in Higher Recursion Theory

Gerald E. Sacks

1.

Regularity of a-R.E.

Degrees

This is the first of six lectures on priority arguments in higher recursion theory.

The term "priority argument" refers to

a mode of combinatorial reasoning introduced independently by Friedberg [l] and Muchnik (21 to obtain a positive solution to Post's problem, the existence of two recursively enumerable sets such that neither is recursive in the other.

I will try to show

how their idea lifts to three generalizations of ordinary recursion theory.

The three are:

admissible ordinal;

a-recursion, where

8-recursion, where

0

a limit ordinal; and Kleene recursion in type 3.

a

is a

C1

is little more than

F,

a normal object of

Some of the results to be described are new and, in the

case of Kleene recursion, not hitherto announced.

Others are old

but viewed, it is hoped, in the light of a new day. One obstacle encountered in all three generalizations is the failure of typical recursively enumerable objects to be regular. Today's lecture focuses on that problem in the case of

a-recur-

sion theory. But first a whirlwind review of the fundamentals of recursion.

For any set

x,

order definable subsets of

let X.

L,

Fod(X)

a-

be the set of all first

G6del's universe of construc-

tible sets, is definable by iterating

Fod

through the ordinals.

L(i) = V{L(~) 1

1

L=V{L(6)

6 < A}

6

(A

is an ordinal}.

The Levy hierarchy of formulas of is .A bounded.

G

no

(also

is

G

a L(a)

Co)

A formula is

Bn-l.

where

or

Cn

An

is said to be satisfies

function from

Bn

is

into

by means of some

Then

f[a]

{f (b)

I

A formula

(ExIG, where

if it is of the form

(x)G,

".

Cn n

admissible (or simply admissible) if

C1

L(a)

is as follows.

if it is of the form

replacement.

C1

L(a)

(=

ZF

if all of its quantifiers are

A formula is

is

is a limit).

To elaborate, let

f

be a

L(a) whose graph is definable over formula with parameters in

C1

b E a))

is a member of

L(a)

L(a)

.

for all

a E L(a). A function g : a --> graph of be:

g

is

C1

a

over

is said to be

L(a).

a-recursive if the

Suppose A C a.

a-recursive if the characteristic function of

recursive, and

a-recursively enumerable if

a-finite if

a-finite, then

A E L(a). g[Hl

(=

If

{g(x)

g

I

is

is

A

Z1

A n H

L(a). is

x E HI)

is

If A

is

a-recursive and

H

a-finite.

If

A # 0,

is

enumerable iff

A

then A

is the range of an

A

a-finite.

C1

is

is

over

a-recursive and

last is nothing more than a restatement of the ity of

is said to

A

aL(a),

H

is

This

admissibila-finite, then

a-recursively

a-recursive function.

There is a natural, one-to-one correspondence between a and

L(a)

that is

A1

over

this correspondence yield:

L(a).

Standard arguments based on

(a) an effective coding of the

finite sets by ordinals less than

a-

ar and (b) an enumeration

theorem for the are

a-recursively enumerable sets.

a-recursive functions t

ties:

for each

I

It(%)

6, let Kg

x < ~(6)); then

out repetitions of all index of the versal,

v

be the

{K6

1

withthe following propera-finite set

6 < a}

is an enumeration with-

a-finite sets.

a-finite set

Kg.

6

is the canonical

(b) means there exists a uni-

a-recursively enumerable set W.

sively enumerable set A,

I

A = {x such a

and

6

(a) means there

For each

there exists a

6

a-recur-

such that

<x,6> E W};

is termed an index for A,

and the situation is

summed up by saying A = W6. From now on let f : a ->

function set B c a such that (1)

H, J, K,

a

=

y

f Gwa B.

6, y < a.

&

H c B

&

J c a-B]

The matter is rendered in symbols by

The enumeration theorem for

a-recursively enumerable

a-recursive rednction procedures.

{ ~ } ~ ( 6 )is said to be defined and equal to replaced by such that f = { E } B

R

.

in

B

if

CA

a-recursive in C1

,

a-recursive in a

iff

sets yields one for weak

with

WE.

A

Thus

f gwa B

y

admissible a

if (1) holds

iff there is an

is (by definition) weakly

the characteristic function of CB.

A

a-recursively enumerable R

(EH)(EJ)[ E R

for all

denote a-finite sets.

is said to be weakly

if there exists an f(6)

...

A,

E

a-recursive is weakly

Driscoll [ 3 ] showed that there exists a such that Gwa , restricted to the

recursively enumerable sets, is not transitive.

a-

Shore [ 4 1

characterizes all such

a's.

The following notion of reducibility is transitive. .a-recursive in

B;

in symbols A
recursively enumerable

R

and

if there exist

such that for all

S

A

is

a-

a-finite

K: K C A <->

(2)

(EH)(EJ) [ E R

H C B

&

K C a-B]

&

and K c a-A c->

(3)

Clearly

(EH) (EJ) [ E

B ->

A

A <wa are said to have the same

A < B a

and

B,

and


sc

H c B

L

is transitive.

a-degree (in symbols A E

a

Assume

A

is

Since

A

R = {K

is

I

for all Proof.

{6

if

1

S

Then satis-

KCA)

a-recursively enumerable.

Proposition 1.2. be

B)

B

a-recursively enumerable, the set

Then

R,

will satisfy clause (2) of the definition of

A*

and

a-recursively enumerable.

A < B iff there exists an a-recursively enumerable a fying clause (3) of the definition of Go.

is also

A

B
Proposition 1.1.

Proof.

K c a-B].

Assume

K6 n A # 0 ) .

A

is Then

slightly modified,
0

a-recursively enumerable. A* ra A,

and

B. There exists an

a-recursive function

t

such that

Let

for all

Then

y.

And so by 1.1,

for all

B.

The substitution of

in (1) yields A

A
An

A

degree as some regular,

Let

without repetitions.

is

a-finite for every a-

An

A

must have the same

a-recursively enumerable set

f

be an

a-

B.

Such

enumeration of

B

a-recursive enumeration of

(Assume B

In other words, for each

a-finite

a-finite set Let

A c a,

Thus

subsets of

is unbounded.)

K,

B n K.

B

Then

there is a stage in the

after which no element of

If not, there would be a one-one

L(A).

B

has a dynamic property that is extremely useful in prior-

ity arguments.

the

for

need not be regular, but the main

result of today's lecture is that

B

A n 6

a-recursive set is of necessity regular.

recursively enumerable set

a

(respectively A)

(respectively A* Ca A).

is said to be regular if

6 < a.

A*

K

is added to

a-recursive map of

a

B.

onto

Cf. 1.5 ( 2 ) .

and relativize Godel's construction of

L

to

L(~,6+1) is the set of all first order definable .

(The predicate

x E A

is treated as

atomic along with L(A,a) over

=

L(a).

x E y.)

In addition,

C1

the least

B

projecturn of

iff

is

B

is arbitrary and

a,

denoted by

A1

A

alpa

definable

is regular. or

a

*

,

is

such that

y

(Ef) [f is

*

is regular iff

A

B Gwa A

, where The

a

Then

a-recursive

f :a

&

1-1 into> y]. w

makes it possible to lift some combinatoric tricks from

to

a.

Note that there exists a nonregplar,

enumerable set iff Proposition 1.3.

a

a

*

*

< a. is the least 8

sively enumerable subset of Proof.

Suppose 6 < a

some

*

.

enumeration of map of

a

Let range of

A If A

such that some

is not

a-finite.

a-recursively enumerable and

were not

a-recur-

a-finite, then any

A C 6

for

a-recursive

6.

f : a -> is

f3

without repetitions would define a one-one

into

f

is A

a-recursively

a

*

be a one-one,

a-recursive map.

a-recursively enumerable but not

Then the

a-finite.

R. B. Jensen has shown by means of a collapsing argument that 1.3 remains true when

a .is replaced by an arbitrary limit

ordinal. Theorem 1.4

(Sacks [S]).

Let

Then there exists a regular, the same Proof. A*

a-degree as

Let

be

a-recursively enumerable.

a-recursively enumerable

B

of

A.

Following Maass [6].

be as in 1.2.

A

It is safe to assume

f : a ->

A*

0 E A.

be a one-one, onto,

Let

a - r e c u r s i v e map, a n d . l e t r e c u r s i v e map.

The d e s i r e d

(Ey) Ix To s e e t h a t

p : a ->

<

y

<

pf ( x )

is regular, f i x

B

be a one-one,

i s d e f i n e d by:

B

pf ( y )

&

a

T

h

f (y)

and d e f i n e

<w,x>

a€

iff

B

< wl. <w,x> E B~

by

Y

Note t h a t

Suppose

is not

B n T~

i n f i n i t e sequence

y

a-finite.

<

y1

< y2 <

Then t h e r e e x i s t s an

...

such t h a t f o r a l l

i

< j,

But t h e n (1) y i e l d s an i n f i n i . t e descending sequence of o r d i n a l s . To show Fix

is

z.


it s u f f i c e s by 1.2 t o show

a - f i n i t e by 1.3, and s o

xo

A

*

Gwa B.

The set

t h e r e is a

Let

A

yo

p'l

[K] i s

such t h a t f o r a l l

be t h e unique

Note t h a t f o r a l l

xr

y

>

yo

a-f i n i t e .

I t follows

y,

t h a t minimizes

pf(y).

Then

Thus

TO prove

B
let

K

be an

a-finite subset of

a2 ,

and define =

K

u

({vIv<w

t

pv
<w,x>EK

K

P

is

a-finite, and

Exercise 1.5 (1) A predicate Q(x,y)

P (x,y) is uniformizable by a predicate

if P(x,y) I,

(x) (y)fQ(x,y) -> (XI [ (Ey)P (xry) -> Assume

A

is regular.

formized by some (2) Let repetitions.

f

be an

B

P(x,y) Gwa A

>

is regular.

can be uni-

A. a-recursive enumeration of

Show

(6) (Eu)(T) [T

holds iff

(Ely)Q (x,y)I

Show each

Q(x,y)

and

u ->

f (T) > 6 1

B

without

2.

Existence of Hyperregular Sets

A c a.

Suppose f <wa A

and

6

Note that if

A

hyperregular.

A

< a,

is said to be hyperregular if for every

it is the case that

B
is hyperregular and

then

B

is

Hyperregularity does not imply regularity, but

note well the next lemma. Lemma 2.1. then

A

Proof.

If

A

is

a-recursively enumerable and hyperregular,

is regular. Fix

6

range of some

to see

A n 6

is

a-finite.

a-recursive function

g.

uo(g(a) = X)

if

x

if

x $ A.

Assume

A

is the

Let E A

f (x) =

Since

f <wa A,

A n 6

is

Lemma 2.2.

there is a

g[yl n 6. A

and

Suppose

f <wa A.

f (61 c y.

such that

Then

13

is regular and hyperregular iff

(f) (6)16 Proob.

y

A


&

f

f

r6

is

is regular and hyperregular.

There exists an

such that for all

A -->

x, y < a:

a-finite]

~ i x6 < a

a-recursively enumerable f (XI = y

iff

.

R

(EH) (EK) [ E R

(1)

&

H C A

K C a-A].

&

According to exercise 1.5 (I), the regularity of A that any

P(x,y) , <

Q(x,y)

A.

A

n y

can be uniformized by some

It follows there are functions ElX,

such that for all

Since A

A

x,

there is a y

is hyperregular, there is a

is a-finite, since A

substituted for A tion of

f

r

implies

A

such that

such that

y

is regular.

If A n y

in (l), the result is an

is

a-finite defini-

6.

Today's priority argument requires some familiar combinaSuppose B < a.

toric considerations in an effective setting.

B

is an

a-cardinal if there is no one-to-one, a-finite 8

correspondence between L (a)

is a cardinal]

[$

Note that if olpa < a, If an

a-finite set A

then every

and some 6 < 6 .

.

then alpa has

In short

is the greatest a-cardinal.

a-cardinality less than

a-recursively enumerable subset of

H

ulpa,

is a-finite

by Proposition 1.3.

An

a-cardinal

is said to be regular if there is no

a-finite function with domain less than bounded subset of

0.

Otherwise 6

f3

and range an un-

is singular.

The next

lemma is useful in many priority arguments, but is not needed

today, so its proof is left for the reader. is said to be :Il formula.

An

if it is definable over

a-cardinal

6

is

.function with domain less than set of

801 6

A function L(a)

by a

f

C

a*

111

regular if there is no :Il

and range an unbounded sub-

6.

Lemma 2.3.

If

f3

is a regular

a-cardinal, then

0

is :TI

regular.

An

a-recursively enumerable set

C

is called complete if

B < C for every 6-recursively enumerable B. The existence a of such a C, and the fact that it cannot be a-recursive, are consequences of the enumeration theorem.

An

a-recursively

enumerable set is intermediate if it is neither complete nor a-recursive. Theorem 2.4.

There exists an intermediate, hyperregular,

a-

recursively enumerable set. Proof.

Let

p : a ->

a*

be one-one, into, and

a-recursive.

Define

t fl for each

E

< a*.

otherwise Thus

{R€

recursive enumeration of all

I

E

*

< a. ) is a simultaneous a-

a-recursively enumerable sets.

[clA is defined similarlyf thus

is defined lclA(x)

is

I

undefined otherwise.

for all

E

< a

*

x < a.

and

It is helpful to think of an union of a noncontracting, sets. each

:R

XEU

Let

I

RE

a-recursively set

R

a-recursive sequence of

be an

as the

a-finite

a-recursive function such for

E,

is simply that part of

RE

enumerated prior to stage

the simultaneous enumeration of recursively enumerable set, then approximated as follows. enumerated in

B

Let

[E] ( Y )

B<'

prior to stage

(E < a

RE

).

If

can be

B

a

of

is an

a-

a-recursively

be the set of all ordinals a

of the enumeration of

B.

Then

is defined and equal to

J

such that

(In order to make (1) single valued, take the "least"



that satisfies (2). E, u fied.)

and

y,

6

if there exist

H

and

In order to make (1) a total function of

give (1) the value

Thus (1) is an

[E]~(~). To be precise

a

if (2) cannot be satis-

a-recursive function that converges to [ E ] ~ ( ~ is ) defined and equal to

only if for all sufficiently large

o r the value of (1) is

6

6.

To prove the theorem a set

B

will be

a-recursively

enumerated with three sets of requirements in mind. (3)

a-B

is an unbounded subset of

(4)

If

(5)

For certain

WE

is unbounded, then

a. JE

yls, the value of

B

+ Z.

[ c ] r (y) will be

preserved. ( 3 ) and ( 4 ) combine to make

B

A re-

non-a-recursive.

quirement of type (4) is said to be positive because it is met by adding an element to

B.

Requirements of type (3) or (5)

are said to be negative because they are met by keeping elements out of

B.

The negative character of (5) arises as follows.

The computation of (1) is based on some H C Ba and

J

bers of

from

J

C

a-Ba.

(1)

H

and

J

such that

is preserved by excluding all mem-

3.

Whenever the potential conflict between negative and positive requirements becomes actual, a decision is made in favor of Priorities are indexed by

the requirement of higher priority. ordinals less than

a

*

,

and "higher" means less than with res-

pect to the usual wellordering of the ordinals. The construction of sive definitions of Stage a.

Xa

B

I

Ba,

consists of simultaneous Xac

r (arc) is the least

I

r ( a , ~ ) and B

such that

Xac

a-recur-

I

m(a,c).

6

m(6,c)

6

,

6 < a, and

if

is designed to protect the computation of

if (6) has a value less than

a.

Thus m(a,e)

least upper bound of all ordinals in some J y = r(u,c).

( 2 ) when

Stage a Let h

be an

less than

~z~

' n'8

a

*

= 8,

is the strict needed to satisfy

Otherwise m(u,c) = 0.

ends with an attempt on a positive requirement. a-recursive function that enumerates each ordinal unboundedly often.

Suppose h(a) = E,.

and if there exists a

6

then put the least such n B <'

The clause R

=

such that

Otherwise 'B

in B.

If

= . ' < B

8 is important. It implies:

€0

(8)

For each

c

there is at most one stage a

an ordinal is added to B

at which

for the sake of positive requirement

(4).

If

6

is added to

positive requirement IE = {u

I

E

B

at stage a

and

6

< m(o,c),

is said to be injured at stage

positive reg.

E

then a.

Let

is injured at stage a).

IE

is

a-recursively: enumerable.

The assignment of priorities

implies (9)

IE

To prove (9) fix a e IE.

h[IEl

h f IE

1.3.

is

E

.


(7) yields

is one-one by ( 8 ) .

The function

h(u) < c

a-recursively enumerable, hence

a-cardinality at most that of

creases,

a-card (IE) < a-card ( e l .

is a-f inite and

Xu

I

So

I&

is

for all a-finite by

a-finite, and has

E.

is nondecreasing.

r(o,&)

r(u,~) changes its value only

by the same reasoning used to prove (9).

As

u

in-

a-finitely many times Thus

r(E) = lim r ( u t ~ ) u

< a*. TO prove (3) observe that r ( ~ )9 B and that r(py) > y for all Y < a exists for each

E

To prove (4) it suffices to show { m ( a , ~ ~ I) E~

are bounded sets. proof of (9). stage

uo

<

E

u
&

The bound on

The bound on

R

m

is established as in the

is more delicate.

<

such that for all

E~

-

p 1E~ < uo

p 1c0

defined

and no element of

R

-->

and

E:

-

is added to

There is a

B

, at or after stage

uo

0

for the sake of requirement (2) as in clause (7) above. r

, = ruo,c0

for all

E~

<

E

and

o > . , a

Then

For each

E

0

<

at any stage

E,

,

a > .a

rn(a,~~)

protects the computation (if there is one) of

otherwise

rn(u,co) = 0.

If there is a computation of (lo), then

it can never be subsequently injured, because all injuries to preservation (or positive) requirement stage

oo.

For each

if there is such a

<

eO a,

u(c0)

enumerable subset of

c,

a

>

Then

u(E,)

occurred prior to be the least

corresponds to an and so is

rn(c,,~)

> .a ,

a-recursively

a-finite by 1.3.

= rn(~~,a) for all

E~

<

Let E

T

be

and

T.

The hyperregularity of (11)

If

is

A

B

will follow from (11).

a-recursively enumerable but not hyperregu-

lar, then there exists a regular f GWa A

such that

f

r8

a-cardinal

is not

joined with (12) to show (12)

Let

f.?

B

y

< 6.

y

< 8.

such that

is defined for all

and a function

(11) will be

is hyperregular.

be a regular

is defined for all

f.?

a-finite.

The proof of (11) is left as an exercise.

a's

a

such that a computation of (10) exists.

The set of all such

its supremum.

let

E,

c0

8-cardinal, and assume

[~]~(y)

Then there exist unboundedly many

The proof of (12) is a closed unbounded sets argument that has been applied in the past by S. Friedman, D. Normann and other recursion theorists. function such that for each subset of recursion. sup {t(p) to

An

8.

Let

I

Q

h(6).

(6

be an

< 0, h'l(6)

a-finite

is an unbounded

t : 8 ->

a

be the least

a

is defined by

greater than

exists since (13) equals

= a*.

1 6<8

Thus there is a

&

a

I

6

< 8).

Then (13) is defined for all

To check this last assertion, fix h(6) = y

6-sequence

has the same T

of

limit of regular

a

*

< a.

Since

whose limit is

a

in the sequence.

a

8.

a B and

For

is either regular or the

a-cardinals, it is safe to assume

enumeration theorem for the

y < B.

a-cardinality as

a's

such that (13) is defined for every the moment assume

[clB(y) for all

a.

a, = SUP (t(6)

< 0 when a

The set

8

< 6) and such that (13) is defined with y equal

sufficiently large

y

6

a-finite function t (6)

t(6)

Let

h : 8 ->

Let

c < 8.

The

a-recursively enumerable sets has

all of the elementary properties associated with the enumeration theorem for ordinary recursion theory. an

a-recursive function q

for all

y

It follows there exists

such that

< fi and all x < a.

Since

E

< 8 and

f3 is an

cardinal, it is not difficult to arrange (15)

q(c,y)

< 8 for all y < 8.

By (14) a decision at stage

a

to protect the computation of

a-

is also a decision to protect the computation of (13).

a

follows from (9) that the set of stages

a

defined at stage finite and has

such that (13) is

and undefined at some stage

a-cardinality less than

It

$.

>a

T

0

Since

is

a-

is regu-

lar, it must be that (13) is defined by a fixed computation for all sufficiently large

a

in the

f3-sequence

same computation will define (13) when of

o =

Of3

T.

'

But then that the supremum

T. To finish the proof of (12). it is,necessary to remove the

assumption that yo

<

a

*

<

a

such that

a

Suppose

a.

q(e,y)

< yo

*

= a.

for all

that for all sufficiently large

or

Then there is a y < 6.

It follows

(16) will not be injured

for any

y

< 8, and so (13) will have attained its final value

for all

y

< 8. B

To see that

is hyperregular, fix

[elB(y) is defined for all

y

< B.

*

< a and assume

E

There exists an

s(e,y)

such that

for all

x.

defined for

The left side of (17) is defined iff y

< 8 by a single a-finite computation.

(13) there are unboundedly many

is defined.

a

a-finite, and so

B

By

such that

It follows from (9) that some

of (18) is never injured. is

[ c ] ~ ( ~ )is

a-finite computation

Consequently the right side of (17) is hyperregular by (11).

To show

B

enumerable C of

C,

add

is not complete, define an

as follows.

r ( a , ~ ) to

wise ' 6 = .'
B

Fix

E

via

E.

a > 1

At stage

C

a-recursively of the enumeration

if the value of (6) is

< a* to see

C

1;

is not weakly

othera-

Clearly

It suffices to show

Let

o

be the first stage such that (6) has the value

r(u,~) is

XUE

I

r(a,e)

r(~). Since implies that

of (6) at stage

a

r(a,s) = r(e), r(a,c) 9 C

0

and

the definition of and that the computation

will not be injured at any later stage.

Thus (19) holds. The above is only a firsk step towards priority arguments in

a-recursion theory, but it is a first step in the right

direction. and :C

Deeper results are obtained with the help of

projecta and cofinal sequences, and their

approximations. Exercise 2.4. (i) Prove Lemma 2.3. (ii) Prove (11).

O:

a-recursive

Recursion on Inadmissible ordinals

3.

8-recursion theory is the study of recursion theory on arbitrary initial segments of

L.

recursion theory is the Jensen J

The proper setting for

8-

hierarchy, but the Gbdel

hierarchy of the previous two lectures will suffice for present purposes.

B is a limit ordinal. The universe of

From now on

recursion theory on

is L(8).

fj

Since

8 need not be

admissible, there need not be a natural wellordering of

B.

of ordertype

L(f3)

L (8), origi-

The standard wellordering of

nated by Gbdel, has ordertype w8

C1

(ordinal exponentiation).

It

possesses effectiveness and uniformity properties that are essential to recursion-theoretic arguments.

In order to exploit

them, it will be necessary in the course of a construction to do more than one thing at each stage. Let if A A

c I,(@).

A

is

A

is said to be

definable over

C1

is p-recursive if both

enumerable.

A

definitions of

is

A

L(B)

p-recursively enumerable (with parameters in

and

L(8)-A

@-finite if A

L(f3).

are

L(8) )

8-recursively

Thus the basic

8-recursion theory differ from those of

a-

recursion theory only in that the universe is taken to be an initial segment of nals.

A

graph is

function

C1

over

recursive set and a

L

rather than an initial segment of ordi-

f : L(B) --> L(8).

L(B)

is

8-recursive if its

Note that the intersection of a

8-finite set need not be

f3-finite.

8-

This

.

last is perhaps the most bitter pill that must be swallowed by the would be

0-recursion theorist.

Suppose A of A

is a

A = U

{f (6)

ing

is 0-recursively enumerable. An enumeration

0-recursive function f : 0 ->

1

6 < 0).

L(0)

such that

There is in general no hope of enumerat-

one element at a time as in a-recursion theory.

A

The notions of < (weakly 0-recursive in) and < w0 B (0recursive in) are defined as in Lecture 1 with L(a) replaced by

L(B).

If 0 is

iff B <0 0, where

C1 0

admissible, then B

is the empty set.

admissible, then it can happen that B

is said to be tamely

B

if

{H

denote

I

H C B}

is

If B

is

is

0-recursive, but not

0-r.e., but not t.r.e.,

for any

C1

0-recursively enumerable (t.r.e.) (H, J, K,

...

then it is diffi-

in the style of Lecture 2, equation

since some a-finite H contained in

tained in B<"

0 is not

In the admissible case every r.e. set

cult to approximate {clB (2),

If

@-recursively enumerable.

0-finite sets.)

is t.r.e.

is 0-recursive

a < 0.

B may not be con-

It turns out that failures of

tameness are numerous and unavoidable when

0 is inadmissible,

and so some new tricks are needed to do priority arguments In the inadmissible case. Two significant parameters are cofinality of alcf(0)

0)

and

a

(or simply

in the definition of

(the sigma one

(the sigma one projectum of

is the least y < B

sive function f, f[yJ alp(B)

alp(B)

alcf(0)

such that for some

is an unbounded subset of

0-recur-

B.

B*) is defined by substituting a

in Lecture 1.

0).

0

for

Maass [6J suggested the following division of inadmissible $Is, and it has proved useful. 8

is said to be weakly inadmissible if

B > olcf(6) > f3".

Many, but far from all, ideas carry over from the admissible case to the weakly inadmissible case.

For example, the solution

to Post's problem carries over, but not the regular sets theorem. $

is said to be strongly inadmissible if

$

*

> ulcf(8).

S. Friedman [ 7 ] has shown that Post's problem has a positive solution, as in the admissible case, for some strongly inadmissible Bus, and a negative solution for others. Some of Jensen's fine structure results for L play a necessary part in the sigma n

the definition of

$*

by

Define onp ( $ 1

$-recursion theory.

projectum of

$,

by replacing

"P:".

appear to

,

"6-recursive" in

Jensen has shown by a series

of collapsing arguments that (1)

If A s t:

and A c 6 < onp(B), then A

E

L($).

Clearly (1) is a powerful generalization of proposition 1.3, which was proved by purely dynamic (or combinatoric) methods. The trouble with collapsing arguments is that they do not as a rule relativize from L($) cate A,

to

that is the structure

Dynamic arguments do relativize. means if it is assumed that fined by substituting Z: alcf (13).

L($)

with an additional predi-


,€,A>, where A c L(8).

(1) can be proved by dynamic

oncf ( $ 1 > unp ( $ 1

.

oncf ( B )

is de-

for a-recursive in the definition of

L(B)

[y

is a cardinal].

B

Let one,

be the least

y

a.

(S. Friedman [ 7 1 ) .

8

y

If

B is admissible, then B

weakly inadmissible.

gc(B)

>

B*.

= max(8*,alcf (B)1 .

.

A

Proof.

>

gc(B),

< 8 such that there exists a oneonto

B-recursive map of

Proposition 3.1

B-cardinal, denoted by

It canshappen that. B

is inadmissible.

such that

An important consequence of (1) is

the fact there must be a greatest if

y < B

0-cardinal, w before, is an ordinal

A

= 6 = alcf (B)

B

Let

be

(The strongly inadmissible case is left to

the reader with a warning that the following does not work.) g

be a strictly increasing

(=

alcf (8))

8-recursive function with domain

and range unbounded in

B

B-recursive map of

into

B*.

that portion of the graph of

p

of the natural enumeration of

H6 = domain p6

-

U

ity of map of

H6 H6

is onto

* 6 . B

*

Let

L(B).

p6

be g(6)

Define

{domain py

I

t6

be the least

y

< 6).

0-cardinal-

8-finite, one-one

("Leastn means least with respect to the

.)

Then

B-recursive map of

Define

The one-to-one correspondence between provided by 3.1 when

let

enumerated prior to stage

L (8)

is a one-one,

be a one-one,

6 < K,

For each

natural enumeration of

h

p

$-recursiveness so that the

Let

.

8.

K

B makes it possible to modify g

The weak inadmissibility of without destroying its

Let

6

onto

K x

alcf(B)

and

. B,

B is weakly inadmissible, is extremely

0

useful.

A construction having the form of a

normally done in

6

C1

steps, can now be done in

recursion,

alcf($)

steps.

It will then follow that the entire construction up to any intermediate stage takes place in some bounded part of the universe of the construction.

L(B),

Thus some vestige of

1

admissibility is retained in the weekly inadmissible case.

This

trick goes back to Driscoll's density theorem for metarecursion theory [ 3 ] , and is central in Shore's density theorem for arecursion theory 181 Assume A denoted by

.

is a regular subset of

A',

E

A,

is treated as atomic.)

several helpful consequences: over

in A;



If

6 = w,

set for the

then A'

The regularity of L(A,f3) = L(B);

A

is simply has

a predicate is

if and only if it is weakly

6-recursive

and predicates of the latter sort can be uniformized by

predicates of that sort. Further assume A~'*,

C1

(Remember that the additional predicate,

the familiar Turing jump of A.

A1

The jump of A,

is the complete (or universal)

structure .

x

L(B).

6

Cf. 1.5 (1). is inadmissible over

is inadmissible.

A,

that is

The half-jump of A,

denoted by

was discovered by S. Friedman in the course of studying

Post's problem in the inadmissible case. complete

A1

It is in one view the

set for the structure ,

has no counterpart in ordinary recursion theory.

and so viewed Let

alcfA(B),

A

alpA(B)

and

BA

be the relativizations of

alcf (B),

alp(6)

A

and

6 obtained by substituting "weakly

for

"6-recursive" in the absolute definitions.

be a function with domain

alcfA(B)

B-recursive in Let

A"

f Gw6 A

and range unbounded in

B.

is defined by <E,X,U> E A 1/2 <->

E, x

where

[{elA (x) is defined],



< 0 and 6 < olcfA(@).

({rlA(x)

is defined as in

Lecture 1.) Theorem 3.2

Friedman).

(S.

inadmissible over

A.

Assume

A

is regular and

B

is

Then

(1) All2 CW0 A. (2)

B CwB A ->

(3)

A' GwB A1/2

(4)

A <8 All2

Proof.

<0

B

A

<@A'.

To prove (2), choose

so

and

el

so that

x E B <->

{co} A (x) is defined, and

x 9 B <->

{el}A (x) is defined.

K c B <->

{ E ~ x} K ~ x K~

Then

J C 8-B <->

where

K~

is

J x

2 C f3

x K~

cA C 02

x

K

-

A

A

-

1/2

,

and

ulcfA(B).

( 3 ) follows from

x 9 A'

-

{

-

{(x),}

x

A

( (x)1 ) is undefined A

x{(x)~} x K

~ @2C

x

A

1/2

.

TO prove (4) suppose

Then by ( 3 ) .

All2 4@ A.

A' Cwg A,

an impossibility by the usual diagonal argument that shows the Turing jump of

A

A1

is not

in

A.

Finally suppose

A' < All2. Let C: be the complete Z2 set for . 6 It will be shown in a moment that C$ GWB A'. Then cA 4 All2, 2 wB and so for some B-recursively enumerable R,

~ h u s:C

is

A2

( 6 )A

over

,

an impossibility by the

usual diagonal argument that shows the complete

C2

set is not

A2. TO verify over

Gwg A',

C;

< L (6),€,A>

let

R " ( X , ~ , ~ ) be a predicate .A

such that

Then

According to 3.1 (relativized to that

g

is a one-one map of

There exists x,y

,

xo

a

kA

A) onto

there is a 8.

such

Then

@-recursive function of

E A' <->

g GwB A

{x0) A (y) is defined.

x,

such that

A A {xl} ( 2 ) = {x0}. (g(z)).

Let

<-->

{xl}

Then

6

BA

x

C A'.

0

There is a paradoxical element in the above proof, namely the observation that weakly with

<,

$-recursive in

A",

the complete

A'.

Z2

It is tempting to identify

and then conclude (falsely) that

course admissibility of equivalent to

0

set (over A

over

A

A" GwB A'.

is needed to show

well defined on

0-degrees (i.e. the

B-degree of

the complete

C :

But of

<

is

A".

It follows from (1) and (2) of Theorem 3.2 that

the

is

Al

A~"),

and that

0-degree of

A 'I2

A

A 1/2

is

determines

is rightfully called

set relative to A.

( 4 ) yields a weak positive solution to Post's problem for

.

L (0)

Let

A = 0.

Then

is a

0-recursive set such that

*w: , then 1 0-recursively enumerable degree.

S. Friedman 191 has shown:

only intermediate

0112

if

0 =

uL

o1I2

is the

His remark-

able negative result demonstrates the impossibility of priority arguments in highly inadmissible situations.

On the other hand

he has obtained a strong positive solution to Post's problem 171 via a genuine priority argument that succeeds for a large class of strongly inadmissible that

B's,

in particular those

f3* is regular with respect to all

0's

such

0-recursive functions.

The next lecture will include his solution to Post's problem when

a

is weakly admissible.

The proof of 3.2

(3), slightly modified and with

shows that every tamely 0lY2.

recursive in

A = 0,

0-recursively enumerable set is

8-

W. Maass has unearthed a great wealth of

useful information concerning tame enumeration and regular sets in the weakly inadmissible case.

Some of his results are as

follows. Theorem 3.3 b

is a

(ii) b (iii) b Theorem 3.4 is a

(ii) b either

contains a tamely contains a

0112

(iii) b

then

tamely

0

is weakly inadmissible and

0-recursively enumerable set.

contains a

0-recursively enumerable set, and

4:

is weakly admissible over

b

or

0

contains a

B c L(o1cf (0)) such that

0-immune L(olcf(B))

A,

where

0112.

and is tamely

A c L(ulcf(B)),

B

0-recursively A GB B

and

0-recursively enumerable. 0-immune means:

K E L (olcf (0))

comprehend at first. being

Assume

contains a regular

is tamely is

0-recursively enumerable set.

Then (i), (ii) and (iii) are equivalent.

enumerable over some

B

is weakly inadmissible and

0-recursive set.

(Maass [6]).

is a regular subset of

A

0

contains a regular, recursive, t. r. e. set.

0-degree.

(i) b

Assume

Then (i), (ii) and (iii) are equivalent.

$-degree.

(i) b

b

(Maass [6]).

0-r.e. 0-r.e.

.

if

K c B

or

K c L(alcf(0))

B,

Clause (iii) of 3.4 is difficult to In essence it says (modulo degrees) that

is the same as being tamely

0-r.e.

over something

It follows that the construction of a

set can be approached somewhat indirectly in the weakly inadmissible case.

-

0-r.e.

Corollary 3.5

(Maass's Regular Sets Theorem)

weakly admissible over

0

ol/*.

and

enumerable set has the same

.

Assume

Then each

8

is

$-recursively

$-degree as some regular

$-

recursively enumerable set. Corollary 3.6

(Homer

admissible over

0

&

Sacks 1101

and

o1I2.

.

Assume

Then the

is weakly

$

$-recursively enumer-

able degrees are dense, that is between any two that are comparable there is a third. 3.6 is almost immediate from 3.4 and the proof of Shore's density theorem for the admissible case [8].

One aspect of the

matter will be discussed in the next lecture. hypothesis of 3.6 holds when

Note that the

is admissible.

$

Today's final

result applies some of the thinking behind 3.3 to characterize the half-jump. Theorem 3.7

(Homer

sible over

0,

$-degree.

&

Sacks [lo]).

and weakly admissible over

is

o ~ / ~Let .

8-recursively enumerable, and

(ii) There exists a tamely such that

al"

= b

and

The proof that 3.7 priority argument. details.

is weakly inadmis-

$

b

be a

Then (i) and (ii) are equivalent.

(i) b

a

Assume

a

o ~
$-recursively enumerable degree

< 01/2 .

(i) implies 3.7

(ii) is of course a

Time does not allow much discussion of its

One point is worth singling out.

weak admissibility of lar representative for

over

$

b,

recursive approximations of

o1I2

The assumption of

not only provides a regu-

but also help to control the

{€la

8-

needed in that part of the

argument that insures

0112 $ a.

Corollary 4.2 further

clarifies the significance of weak admissibility over Maass [6] has shown: degrees of the tamely

if

0

0112.

is weakly inadmissible, then the

B-recursively enumerable sets are not an

initial segment of degrees of the

0-recursively enumerable sets.

Exercise 3.8. (1) Suppose

B is inadmissible and B is a tamely B-

recursively enumerable set. (2) Suppose

a tamely

,3

Show B <

B ol/*.

is weakly inadmissible.

0-recursively enumerable C

Show there exists

such that C E

B

o1l2.

4.

Weak Admissibility and Priority Arguments

Today's lecture includes a positive solution to Post's problem in the weakly inadmissible case.

The proof is similar

to that given in lecture 2 for the admissible case, but there are some conceptual differences occasioned by the loss of admissibility.

If time allows, some remarks on the strongly

inadmissibility case will be made at the end of the lecture. It is a curious fact that the study of weak admissibility began (somewhat unconsciously) with the density theorems of Driscoll (31 and Shore [8].

Thenexttheorem, proved rather

sketchily in [8], is central to the proof of density. Theorem 4.1

(Shore).

Suppose

regular, incomplete,

Let

and let

B

B.

relative to

for some

D y

<

is a

Then

a

ulpB(a),

a-recursively enumerable set, the sigma one projectum of

a

First it is necessary to verify the

relativization of Proposition 1.3 to mind, let

B

B.

be a regular,

denote

P

is admissible and

a-recursively enumerable set.

is weakly admissible over Proof.

a

over

B.

With that end in

be

El

, and assume

p.

It is necessary to show that

D

D c y

is

a-

finite.

The first half of the proof of 1.3 is inapplicable,

because



siqma over

n L(a)

need not be admissible.

projectum of for

a,

anpa,

the

is defined by substituting

a-recursive in the definition of

ulpa.

En

Jensen has shown

E

and

is

Cn

over

L(a),

E

then

is

a-finite.

Jensen's proof involves Skolem hulls and Mostowski collapses, and does not assume relativize to

B

a

is admissible.

His proof does not

because the collapse of a subset of

little chance of being an initial segment of D D

is

is

C2

over

a-finite.

partial

So assume

(over L(a))

C2

.

L(a)

If

y

a2pa

is admissible, there exists an

for every

x

y.

Let

u2pa

t(x)

within

t.

g

L(a).

to show that all false values of

onto g

be a one-one, a.

Since

a

such that

is obtained by guessing Admissibility is needed

t(x)

less than the true

value can be recognized as false inside ciently large Let

H

L(y) u {p},

be the where

C1 p

Skolem hull in

hence

a-finite.

is

If

H

such that

sufficiently large H

such that

H = L(a).

H

of

is bounded below So assume

hope of a contradiction. x < a2pa

for all suf fi-

is the parameter occurring in the

g.

D,

L(u)

a.

definition of

Thus

t

a-recursive

in the domain of

at the value of

E

B.

lim g(o,x) = t(x) u+a

(2)

a,

has

< o2pa, then (1) implies C

map of

B

Fix

y

t(x) = y. a.

Since

g(ao,x) = y.

H

< a.

a,

then

H,

is unbounded in the There exists a unique

By (21, g(a,x) = y H

C1

for all

is unbounded, there is a Since

x, p E H,

y E H.

can be construed 9s the range of a one-one,

partial

C1

z;'~

~ u tthen there is a one-one, partial than

alpa

onto

B

Now assume

a, a

intent of showing h : $ ->

and let

C

that

be an $

<

.

a

be a one-one, strictly increasing map

domain

over

C

C

is regular.

is a subset of

$2


alcfB(a),

.

The assumption By Theorem 1.4,

Define

El

definable over

It was shown above that 1.3 remains true when

Then $,

Z1

denote

with the

is complete.

a-recursively enumerable set.

j

relativized to finite.

0

Let

B

B

it is safe to assume

The graph of

map of something less

is not weakly admissible over

will be used to prove

p

{PI.

a contradiction.

with unbounded range and a definition Let

L(y) u

(over ) map applied to

B.

Since

<

p,

it follows that

is

j

C n h ( ~ ) , regarded as function (of x)

is weakly

Corollary 4.2.

$

a-recursive in

Suppose

regular, incomplete, weakly admissible over

$

B,

and so

C

is weakly admissible and

$-recursively enumerable set.

a-

with


O

is a

Then

B

is

B.

The proof of 4.2 is virtually a replay of that of 4.1; however, Maass's notion of admissible collapse 161 is essential to any understanding of why this is so. S. Friedman [71 was the first to solve Post's problem in the weakly admissible case.

His argument was an adaptation of

Shore's "blocking' method [ll].

The proof of Theorem 4.3 below

is a straightforward modification of the closed unbounded sets

argument of Theorem 2.4.

It is intended to suggest how

admissibility proofs may be extended to the weakly inadmissible case. Theorem 4.3.

Suppose

exists a tamely

0

is weakly inadmissible.

Then there

B c alcf(6)

@-recursively enumerable set

such that alcfg(B) = alcf (0) and

0

Proof.

<0 B <@ 01/2 . Let

K

0-recursive map

denote h

of

alcf(0). K

onto

By 3.1 there is a one-one, 0.

h

is the remedy for all

problems that arise in the course of compressing the proof of 2.4 to the weakly admissible case.

The construction as before

takes place in stages, but now the stages are indexed by ordinals less than a

K

Thus prior to stage

0-finite set of events has occurred.

a

(a

< K), only

At stage

a

the

structure L(sup h(y) Y
K

steps, each one

along the way the entire universe of this understanding,

:R

and

are defined as in 2.4.

Thanks to

L(0)

0-recursive functions B
h

the

0-finite, yet

is unraveled.

With

B

The set

is enumerated with four sets of

C K

requirements in mind. (1)

K-B

(2)

If

Let

is an unbounded subset of K

n WE

K.

is an unbounded subset of

K,

then

WE n B # 0 For certain

(3)

yes, the value of

[ EB
served.

If

(4

K

n Kg

K6 n (K-B) # 0 . of the

is an unbounded subset of

{K*

(Let

1

6

< 8.1

8-r.e.

as its objectives

olcfB(a) =

larity is analogous to

K

K C B

has

and

B <8 01i2.

K's

less than

ly

(3) has

(Hyperregu-

(4) insures that

K. )

B

is

If (4) succeeds, then every

is

K,

and conse-

0-recursively enumerable.

[ E I ~ ( ~is ) defined, then it is defined by a

8-finite computation whose positive part has

ends.

enumeration

non-@-recursive.

0-cardinality less than

quently the set of all such In addition if

B

alcfB(a) =

0-recursively enumerable.

8-finite

B

then

@-finite sets.)

(1) and (2) combine to make

tamely

be a

K,

8-cardinality

and so is developed before the construction of

K,

Thus the true purpose of ( 4 ) is not to make

B

tame-

0-recursively enumerable, but rather to make sure that [c]r(y)

converges to

as in the admissible case.

[EI~(~)

Note that (1) and (4) together

imply that each neighborhood condition, hence each computation, satisfied by

B

is an element of

L(K).

Stage

a

is very much the same as it was in 2.4

save for

some additional negative requirements for the sake of (4). 6-recursive function r ( u , ~ ) in 2.4.

t(o,6)

K6 n (K-B<'). subset of

since

K,

(2) for some

B,

<

E

of

as

is, if possible, a member of

B<'

a

-

6.

6,

K,

a ->

it

I,

is

a,

in 2.4.

I

The limit of

The assumption of weak The set of

a's

such that

6-recursively enumer-

(The proof of 1.3 works perfectly well

is less than B*.

Since

In addition the B*

<

K,

In other words weak admissibility implies the 6-finite set of

@-cardinality

the enumeration of

will be completed before the construction of

enumeration of a

6-

< 0 , the set in question, call

E

in the weakly admissible case.) of

6.

is

exists for the same reasons as the limit

Since

6-finite.

t(a,6)

the set of such increases is

increases is correlated with a E.

t (a,6)

K.

If that happens, the value of

admissibility is used as follows.

able subset of

is an unbounded

is a bounded subset of

0-cardinality at most that of

r(a,~), as

t(a,6)

K6

unless it has to be put in for the sake of

For each

finite and of t(o,B),

is devised that behaves as did

Such a choice is possible if

is kept out of

increased.

t(o,6)

A

B

I

is finished. 6-recursive

0-cardinality less than

B*

goes to completion at some intermediate stage of a construction taking place in

K

stages.

In this manner the proof of 2.4 stretches to a proof of 4.2 without too much fuss.

Weak admissibility is strong enough

to push through the priority argument developed for the admissible case.

In order to enable the priority argument to

achieve its ends, the set

B

is a subset of

ulcf(6)

rather

B,

than

and requirements of type (4) are added to force

computations from

B

to converge quickly.

The proof that

olcfB($) = alcf(f3) follows very closely the argument given in 2.4 to establish hyperregularity. B
ness. B

And the proof that

follows the argument given in 2.4 for incompleteOne detail worth mentioning is that

is tamely recursively enumerable.

B <@ 0

because

This last point is based

on reasoning similar to that behind Theorem 3.2 (3). Corollary 4.3.

Suppose

6

is weakly inadmissible.

exists a regular,

$-recursive

inadmissible over

B

Proof.

and

Apply 3.3 to the

B

such that

8

0

Then there

is weakly

0 C B B.

B

provided by 4.2.

0

There are still many open questions concerning the recursively enumerable degrees when (A f3-degree is said to be tives is

0-r.e.1

B

0-

is weakly inadmissible.

0-r.e. if one of its representa-

For example, are the

able degrees densely ordered?

8-recursively enumer-

At the moment this question has

plausible lines of attack in both positive and negative directions. The principal results in the strongly inadmissible case are due to S. Friedman [ 7 , 9 ] . incomparable

He has shown that there exist

B-recursively enumerable degrees if

recursively regular.

(This last means there is no

sive function with domain less that subset of

6.)

*

$

B

*

is

B-

B-recur-

and range an unbounded

The trick associated with negative requirements

of type (4) in the proof of Theorem 4.2 is of some use in the strongly inadmissible case, but it fails to solve difficulties

associated with a lack of tamely recursive enumerations. Friedman invokes an effective version of Jensen's to help him guess the

8-finite subsets of the

enumerable sets he constructs. has shown that there exists a

0-principle

0-recursively

In the opposite direction he

B

B-

such that the only

recursively enumerable degrees are

O1

o1l2

and

0'.

One

B is Nu * u of L.

such

The proof of this remarkable result 1 combines ideas from Silver's work on the Generalized Continuum Hypothesis with recursion theoretic ideas.

Homer [12] has

pointed out that the strongly inadmissible case has implications for the weakly inadmissible case:

if

6

inadmissible but strongly inadmissible over

o

then / ~ the ~

~

o ~ /and~

8-recursively enumerable degrees between in some cases be identified with the

is weakly

0'

can

y-recursively enumerable

degrees for some strongly inadmissible

y.

Exercise 4.4. (1) Supply the details of the proof of Theorem 4.3. (2) Suppose is that

2:.

B

is weakly admissible and

Show there exists a lim t(u,x) u

is

g(x)

8-recursive for all

x < 6.

t :

g : 6 ->

B2

->

B

B such

5.

Kleene Recursion in

3~

Today's lecture and the next are devoted to Kleene's theory of recursion in normal objects of type 3.

All that

follows applies equally well to type 4 and higher, but it seemed discrete to avoid higher type complexities in order to achieve a more valiant exposition at the type 3 level.

Today

is largely review, and tomorrow is regular sets and Post's problem. 2w

is the set of all reals.

cate for sets of reals. false otherwise, where domain is

w 2w

Thus

is the equality predi-

3~

s

E(x,y)

x, y C 2W.

Let

is true if F

F

and

be a function whose

and whose range is contained in

[13] defines recursion in

x = y

w.

Kleene

by means of schemes not unlike

those in common use at the first type level. schemes amounts to generating computations.

Unraveling those If

F

is normal,

then the computations lend themselves to a natural enumeration in the form of a hierarchy that resembles L. means

3~

is recursive in

F.

F

is normal

Since "recursive inw has not

yet been defined, it may be best to think of a normal object as consisting of a pair

< 3E,F>,

where

F

is arbitrary.

The use of hierarchies to characterize Kleene recursion in the normal case originates with Shoenfield [14] and Grilliot [151. The idea is to construe

F

as a jump operator acting on sets

of real's, and then to generate the sets recursive in

F

and a

arbitrary real in a fashion analogous to the generation of the

hyperarlthmetic sets by iteration of the Turing jump through the recursive ordinals.

Nothing could be more reasonable once

Kleene had shown that the hyperarithmetic sets (of integers) are just those sets recursive in Let

X C 2W.

The

F

2 ~ .

jump of

X,

denoted by

jX,

is the

effective disjoint union of certain definable subsets of

1

Let

be the

e-th

function

(e-th

according to some

standard GGdel numbering of formulas) from graph is definable over formula whose

X. let jx

all

<

be

{a

I

a

E

is the set of all

<e,a,y+l>

where

<X,a>

encodes the

whose

z

2W

y

E

z

and

z

E

X,

x

is a real variable and

&

<e,a,O>

X {ell(a) = 01. such that

a E

where de-

<,

and

such that

is a set of reals that encodes the pair

The first part of finable from

w

parameters are nonnegative integers, and

is an integer variable,

notes

2W into

<2W,c,~> by means of a first order

whose atomic parts are of the form y

2W.

X

jX

encodes all subsets of

X, a.

2W

de-

via integer parameters, and the second part

F-images of all subsets of

wW

definable via

real parameters. The sets '6 Ordinals, bers of If

a

E

oF oF,

and

$,

and the function

1 lF

are defined by a simultaneous recursion.

:

oF

->

The mem-

are reals that serve as notations for ordinals. F then lalF is the ordinal denoted by a. icl

is the least ordinal without a notation in

oF.

For each

a

< KF , H

jump

a

is in essence the result of iterating the

times.

F

The only complexity in the following defini-

A notation for a limit ordinal

tion occurs at limit stages.

is introduced in a predicative fashion from below that corresponds to an effective closure property:

if

H

is recursive

and each member of H is a notation for an ordinal less than K~F , then the supremum of the ordinals denoted by members of H is an ordinal less than K F~ . (For notational in

3 ~ , F

simplicity, the superscript

(1)



(2

Suppose

<2e,a>

E

0,

<e,a> E 0

1<2e,a>l = o+l

Suppose Then

<3m*5e,a>

<m,a> 0

E

is often omitted below.)

Il = 0

0,

E

F

E

and

and and

0,

HA

The hierarchy

1 <3m*se,a> 1

a <

F

kl)

Then

Ha+l = jH,.

= A,

and

We C 0.

A

where

J<m,a>l and all

is the set of all

I FH ~I

(a E 2W).

I<e,a>l = a.

I<m,a>l = o

least limit ordinal greater than that b E We In addition

Ho = 0

and



is the Ibl

such

such that

of sets of reals is equiva-

lent to a hierarchy of all the computations that arise in the computation of all functions (from 2W

F, 3 ~ ,a

into

w)

recursive in

for all real a. Eel( 3E,F,a) is said to be defined and equal to x if Ha e = <eo,el>, <eo,a> E 0, I<eo,a>l = a and {elln (a) = x.

X

{el( 3E,F,a)

To compute eo

is to iterate the F

is seen to be a notation for some ordinal a ,

extract the value

I {el ( 3 ~ , ~ , 1a )=

the matter is expressed by

there is an

f : 2@ -> e

such that

T.he symbol

3

f < E, F, a

means

a.

f

3 ~ F ,

A

if its characteristic

"<" means "recursive in". Thus

is constructive in F, a if a is <e,a> E 0F a is recursive in F, a

.

for some

is the height of a prewellordering of

3 ~ F, , a.

{e 1 ( 3 ~ ,a)~,, and

f

An ordinal a

a

is

is recursive in 3 ~ , , that is 3 f (b) = {el( E,F,a,b) for all b.

for some e,

I<e,a>l

u

is recursive in 3 ~ F, if 3 f(a) = {el( E,F,a) for all a.

2W.

set R C 2@ is recursive in function is.

and then to

in an elementary manner from Ha.

x

regarded as the length of the computation of

Suppose

jump until

2@

if

recursive in

Each constructive ordinal is recursive, but not

conversely. Clause (3) above implies a powerful bounding principle: if

oF

f : 2' ->

exists a

a

constructive in F, a

all real b. sive in in

F, a

follows.

is recursive in

3 ~ F, , a, then there such that

If(b) lF < a

for

It is a consequence that a set R c 2@ is recur-

3 ~ F, , a

iff there exists an ordinal a

such that

R

is one-one reducible to H

constructive as

For some e,

for all real b. A

an

e

set R

is recursively enumerable in F, a

such that for all b,

if there is

The downward arrow

means "convergesn or "is defined".

4

upward arrow t

means "diverges" or "is undefined". {el (3~,~,b)+,then lie) ( 3E.F.~) 1 <.:K Of course R

3 ~ F, , a

iff

R

and

An

Thus if

2 W - ~ are recursively enumerable in

is recursive in

3 ~ F, , a.

This last fact is

proved with the help of Gandy's selection principle [16], the first important insight into Kleene recursion in the normal case.

It provides a uniform method for selecting an integer

from a nonempty recursively enumerable (in 3 ~ F, , a) of integers.

a

set

R

Gandy's method amounts to arguing that the least

in the hierarchy by which some member of

an ordinal recursive in F. Gandy's method fails if

R

R

stage a

F.

recursive in

is developed is

Moschovakis [171 shows that is a set of reals; he exhibits a

nonernpty recursively enumerable (in 3 ~ F) , such that no member of

R

set R

of reals

is developed (or enumerated) at any

In order to formulate Post's problem for recursion in a normal type 3 object F

it is necessary to consider once

again what is meant by "finite" and "recursively enumerable" F 1 a < KF when the universe is {Ha ~ } . Let A be a recursively enumerable subclass of the universe.

Experience with ordinary

recursion theory suggests three guiding principles: member of

A

(a) each

is a finite object; (b) e'ach member of A

to be a member of A

gets

by virtue of a computation which is a

finite object; (c) each proper initial segment of the universe, and every rudimentary modification of it, is a finite object.

Suppose ( a ) , (b) and (c) are accepted as guides. "finite" sets are simply the sets recursive in some real a.

Then the

3 ~ F, , a

And the bounding principle becomes:

for

if

{el ( 3 ~ , ~ , - )is defined on some "finite" set, then the collection of all relevant computations is contained in, and hence equal to, some "finite" set.

(It now becomes very tempting to

think that the bounding principle is equivalent to some global form of

C1

Let A sive in

admissibility, but this is far from true.) be a class of "finite" sets, that is sets recur-

3 ~ F, , b

enumerable in

for all pair

3 ~ F, , a

R C 2W.

F, R.)

for some real b.

is recursively

A

if there is an

e

such that

( is a type 3 object that encodes the

Definition (4) is not as transparent as it seems.

Note that the right side of (4) must not converge for any that is not "finite", since all such from

A

in accord with (a) above.

R

R

have been excluded

The requirement that every

member of a recursively enumerable class be "finite" is more than a mere formality in the light of the next theorem and its corollary.

Let

-<

be a wellordering of

2W.

Normann [la]

calls 4 F-recursively regular if there is no function f < 3E, F, a for some a such that the domain of f is a proper initial segment of in

-(

and the range of

f

is unbounded

4.

Theorem 5.1

(Sacks 1191 )

.

Suppose -( is

regular, and is recursive in class

3 ~ a ,

3~-recursively

for some a.

Then the

{R

I

(Eb)(R < 3 ~ , b I)

is not recursively enumerable in Corollary 5.2. regular.

3 ~ ,a

for any

a.

Suppose the cardinality of the continuum is

Then there exists a normal type 3

F

such that the

class

is not recursively enumerable in

F, a

for any

a.

The proof of 5.1 is a forcing argument based on the first half of the next result. Theorem 5.3

(Normann [18])

.

regular, and is recursive in (i) Call a set R

R

is bounded in<.

-<

Suppose J

E, a.

is

3~-recursively

Then (i) and (ii) hold.

"small" if

R

< 3E, b for some b and

Then there is a predicate

P(brc) < 3E, a

such that

is an enumeration of all "small* sets.

Q

(ii) Suppose 3 ~ )subset of such that

{c

c < 'E,

is a nonempty recursively enumerable (in

I

c 4 b}

a, b

.

Then there exists some

c

E

Q

.

(uniformly)

The proof of (i) is a Gijdel style collapsing argument. (i) makes it possible to treat the collection of all "small" sets as a "finite" set. sion theory:

if

finite subsets of

a

*

a

(i) is similar to a truth of

< a,

*

then the collection of all

bounded below

a

*

is an

a-recura-

a-finite set.

(ii) follows from (i) by means of an argument not unlike that used in the proof of Grilliot's selection theorem.

Grilliot

[15] showed (with gaps filled in by Harrington and MacQueen [20]): 'E)

suppose

Q

is a nonempty recursively enumerable (in

set of reals; then there exists a nonempty

that

QO

is recursive in

wellordering of

*E.

Qo

C

Q

such

(If in addition there is a

zY recursive in 'E,

down to a single real recursive in

then

Qo

can be cut

'E.)

It is not difficult to live with the anomaly of 5.1. ObF serve that the class {Hu ( u < rl) is recursive in 3 ~ ,F, since to be an iterating the

Ha F

set is to be a hierarchy generated by jump short of the closure point K F And

every set recursive in

3 ~ , F, b

definition relative to some

Hu.

for some

b

Thus the

Ha

has a primitive sets are

cofinal in the "finite" sets. and so it seems safe to restrict attention to recursively enumerable classes of

HZ

sets are a bit more flexible than the

H,F

set if

set.

R

has the same degree

To be precise there exist R = {eo)( 3E,F,H,)

The collection of all 3 ~ ,F,

H,

and

eo

HE

(mod 3E,F) and

el

Ho

sets.

sets.

R

The is an

as some such that

Ha = {el)( 3E,F,R).

sets is recursively enumerable in

and is closed under rudimentary operations such as

pairing [19].

Another definition of

H,

set, perhaps more

intrinsic, is: R is an H, set if R has the same degree 3 (mod E,F) as some prewellordering W of 2W such that W < 3E, F, b for some b. The height of W is of course F some ordinal less than K ~ . Thus the H, sets can be loosely

identified with the ordinals in the universe generated by

F.

That universe has a useful description in terms of a relativized version of Godel's

X

E

L.

L(a+l,F)

X

if

definable with parameters over

L(X,F) =

U

L(a,F)

C

I

(L(a,F)

and

X

is first order

M(a,F).

a

< A)

if

is a limit.

F L(tcl,F)

It is not difficult to verify that

and

F { H ~I a

< tcl)F

are equivalent structures. This means that a transitive set X F belongs to L(tcl,F) iff the structure < X I € > is isomorphic to some binary relation

R

< 3E, F, a for some a.

Let

L(F) be the language of ZF augmented by ranked variables x a y a constants that name elements of L(F),

.

and a function constant An ordinal sentence

G

of

F.

is said to be

T

L(F)

a-reflecting if for each

with parameters in

w

U

{a},

C1

it is the

case that M(T,F) (K:

G

implies

M(K:,F)

G

.

is the least ordinal not constructive in

F, a. )

The use

of reflecting ordinals in the study of Kleene recursion goes back to the proof of the plus-one theorem [21]. A typical C~ F1 sentence is (Eo) [<e.a> E Ha since each encodes all

.

first order facts about To say

T

is

F

(Ha

I

Y

a < y)

a-reflecting is to say:

in a positive fashion. an

8a

satisfying a

,A

stage

T

<e,a> E

F, a.

HZ

is recursive in F over Ha. Let

is developed before

only if one is developed before stage

recursive in and

o u {a))

formula (with parameters in

:r

Note that if

for some

is recursive in

u < y,

holds for some F, a

y

y

y

F, a

then the least such

because it is first order definable

be the greatest

a-ref lecting ordinal.

Some re-

sults of Moschovakis 1171 and Harrington [221 show

A powerful characterization of

K,

a

due to Harrington, and its

application to Post's problem for recursion in discussed in the next lecture.

Very briefly:

3 ~ F, ,

will be

certain proper-

rr are needed to show that each requirement of negative type is injured only "finitely" many times. ties of

Post's Problem and Regularity for 3~

6.

It seems wise, before time runs out, to say explicitly what is meant by a positive solution to Post's problem for a normal type 3 object assume F O IE and A

B

F

,

F.

For the remainder of this lecture,

is normal in the most direct sense, that is where

Po

is an arbitrary type 3 object.

HZ

be classes of

is said to be recursive

symbols A < B, F, H) every H,F set K,

B, F, H

H

on

if there exists an

Note that the behavior of

HZ

sets, and let

is

be an

Let A F H, set.

HZ

sets (in

the e

F

such that for

{~)(B,F,H,z) when

Z

is not an

set is left open; it may not even converge.

Dictum 6.0. Post's problem for there exist an

HZ

sets such that

A

set and

H B

F

has a positive solution if: and classes A and B of H,F

are recursively enumerable in

F, H I

and such that A for any

6

B, K set

and

B +A, K

K.

Some of the reasoning that led to the above formulation of Post's problem was discussed in [191. Time permits only one

aspect of the matter to be reviewed. that if

A

then every

Spector [23] observed

is a nonhyperarithmetic, I[; set of integers, 1 IT1 set B of integers is hyperarithmetic in

His argument with inessential changes shows: of reals such that some real

is a set

A

is recursively enumerable in

A

a, but not recursive in

F, b

F, a

for any real

then every recursively enumerable (in F, c) is recursive in

if

set

A.

B

for

b,

of reals

It follows that Post's problem for F

F, A.

has to be formulated in terms of classes of sets of reals rather than sets of reals.

This sort of conclusion was first

reached at the type 2 level by Kreisel [ 2 4 ] in the early days of metarecursion theory. Normann [I83 has shown that Post's problem has a positive solution when

F = 3E

regular wellordering real

a.

and there exists a

-<

of

2°

3~-recursively

recursive in

3 ~ , a

for some

His proof relies heavily on a powerful reflection

principle (Theorem 5.3 (ii) above) to show each negative requirement is injured at most countably many times. able bound is not unexpected when

1(1,

A

count-

the height of the

wellordering 4 , is aleph-one, but is certainly remarkable when

> aleph-one. Normann derives reflection from a

G6del style collapse that does not relativize from arbitrary normal type 3 F.

3~

to an

The solution to Post's problem

described below avoids collapsing arguments in favor of dynamic and combinatoric principles that make sense for all normal F.

The countable bound on injury sets will be dropped.

The dynamic approach relies on several parameters associated with

F.

.

The greater projectum of F is denoted by PF It is the F such that there exists an HF-set Iil with the least o C rl following property: as

cH1,HT>

for some

exists a one-one map

{HE I

<

0

every T

$-set

< o.

f

such that both

pF}

F

recursive in

and some

a*.

is analogous to { F H ~I YI < K F~ ) into

pF

from

(mod F)

has the same degree

f

and

#-set.

is the most distinctive feature of

i1

There

(where defined) are

The computability of f-I F p , and is essential for

priority arguments. The lesser projectum of least

o

<

I C

F

such that there exists an

.

It is the

Iil

with the

nF

is denoted by IiF-set

following property:

there exists a recursively enumerable (in F F, H1) class of H -sets contained in {H, 1 T < el, but not 2 recursive in F, Ii2 for any HF-set H

.

Proposition 6.1 (i)

in

nF <

pF.

(iil

Suppose there exists a wellordering of

F, a

for some real

2W

recursive

a.

Then the least height attained by any such wellordering equals pF

.

(iii

Suppose there exists a

wellordering of Then Proof.

n 3~

= p

3~

2W

3~-recursivelyregular

recursive in

3 ~ , a

for some real

.

To prove (iii) it suffices to fix

supremum of

{K:

a.

I

c < b)

is less than

shows that this supremum is at most that -4is recursive in

K

b 3~

rl

,

and show the

.

Normann I181

where

a

is such

As promised in the previous 3~ C < I C The bound lecture, it will be shown below that I 3 ~ ,a.

.

K

~ 'can ~ also be derived directly from 5.3 (ii) via a result r of Harrington [ 2 2 ] trr the effect that xzpb # :K o r all b

and

c. F

Let y

K ~ .

by

recfF(y).

an

IiF-set

The r.e. in

F

It is the least a

H1

an unbounded subclass of {T

If y < K~F ordinal 6 < K~F tion

(

y

I ,

HT

E

A}

then

is an

f that maps

{HT

I

T

y

class A

< y},

there exists a such that A

and such that the

is o.

recfp(y)

is an

F-cardinal.

I

r

An

< 6) into a proper initial segment

of itself and is recursive in F and some HF-set. F F p < K then pF and nF are F-cardinals, and greatest F-cardinal.

is

F-cardinal if there is no one-one func-

{H:

.

is denoted

such that there exists

with the following property:

recursively enumerable (in F, HI)

ordertype of

cofinality of

If

is the An example due to Ted Slaman shows pF

can be a successor cardinal without being

pF

F-recursively

regular. In a moment it will be shown that same r.e. cofinality. 1111 that says a*

n and

K~

have the

This lemma is analogous to one of Shore

and

a

have the same tame :L

cofinality.

however the proof of the finite types result draws on reflection phenomena that have no counterpart in

a-recursion theory.

Those phenomena originate in a theorem of Moschovakis that is central to the theory of Kleene recursion in objects of finite

Theorem 6.2 (Moschovakis [17]). enumerable (in F)

set of reals.

Let A

be a recursively

Then there exists a

recursively enumerable (in F)

set

B

of reals such that

for all real a. Proof.

a 9 A

To say

tation indexed by

is the same aa saying that some compu-

a

fails to converge.

tation has the form of a wellfounded tree. defined eo that indexed by

a

" E B"

means:

A convergent canpuThus

B

must be

the computation tree

has an infinite descending path encpded by

For the sake of simplicity suppose A

is

oF. oF

b. is

both a typical, and a universal, recursively enumerable (in F) set.

The inductive definition of

oF

readily yields a

recursively enumerable (in F) predicate (1)

if

lclF

> !dlF; and

(2)

R(c,d)

c $3 oF,

if

R(c,d)

holds and

R(c,d)

holds and c

d 9

oF,

then there exists a

R (c,d) such that: then d E

d

oF

and

such that

oF.

is an artful variation of the predicate:

d

indexes an immediate subcomputation of the computation indexed by

d.

If

a



P

A,

E

B

then

lows from (1).

0

is defined by

(Eb)[

E

Bl

by ( 2 ) .

The converse fol-

Corollary 6.3. (i) (in F)

(Moschovakis)

The class of recursively enumerable

predicates of reals is not closed under existential

auantification.

6.3 (ii) follows from 6.2 and the existence of a nonrecursive, recursively enumerable set of reals. K~a was defined above for an arbitrary real type 3 F. the form degree

a

relative to a normal

The definition extends easily from reals to sets of

8. If 1 a 1 (mod F),

and

Lemma 6.4 ( 1 2 5 1 ) .

If

= u, K

Hu

qF

then :H

and

a

have the same

is (by definition) .:K

> y, then rlF > sup {rrH,

I

IJ

< y}.

(In addition the supremum can be computed effectively from The proof of 6.4 is technical, but not difficult.

y.)

It

begins with the observation that

and makes use of the fact that

qF

is an

F-cardinal.

The

parenthetical portion of 6.4 follows from 6.2. Theorem 6.5

.

([251)

recfp(n) = recfF(K1).

The proof of 6.5 is based on 6.4

(including the parentheti-

cal part) and 6.2. Let

H

be a set of the form

characterization of

:K

8. The

Harrington

begins with a slight variant of 6.2:

there exists a W e 1 number

n

such that

for all

e

If

A.

and

In) (F,e,~,blZ, then the ordinal

1 in1 (~,e,~,b) 1

is

said to be a Moschovakis witness to the divergence of {el (F,H). (Harrington 1221 1.

Theorem 6.6 for all

er

if

{el(F,H)

is the least

:K

such that

diverges, then there is a Moschovakis

witness to its divergence at or below

a.

The proof of 6.6 combines the details of the proof of 6.2 with the following basis theorem of Harrington Suppose A reals; if that

&

Kechris 1221:

is a recursively enumerable (in F, H)

set of

2 W - ~ is nonempty, then it has a member c such <..K H The basis theorem is applied countably many

":K

times in succession to obtain a.suitable b

that encodes an

infinite, descending path through the computation tree of {el (FIB). Theorem 6.7 for

F

.

(Sacks [251)

If

nF =

pF,

then Post's problem

has a positive solution (in the sense of 6.0).

The principal features of the proof of 6.7 are as follows. All requirements, positive and negative, are indexed by ordinals less than

p.

Among them are the usual Friedberg-

Muchnik requirements designed to make the recursively enumerable tin

F)

classes A

and

B

computations of height less than ments have as their object the

incomparable with respect to F K ~ . The remaining require-

F K~-subgenericity of

A

and

B,

that is

All of the subgenericity requirements are negative.

A typical

one proceeds by committing to preservation carefully selected parts of the computation tree of

{e)(F,A,H),

where

H

is

4-set. The parts are selected with the aim of locating below K~F a Moschovakis witness to the divergence of some

{e)(F,A,H).

The selection is careful to help make sure that

all activity associated with the computation tree of Eel(F,A,H)

F

eventually comes to end at some stage prior to

K ~ .

It turns out that if no Moschovakis witness is developed below F K~ , then {ef(F,A,II) converges via a computation of length F less than K ~ . Thus convergent computations are forced to conF

verge rapidly, that is below tations to diverge rapidly.

K~

,

by forcing divergent compu-

This rather indirect approach

seems necessary in order to bound the amount of activity associated with each requirement. The requirements are divided into blocks as in Shore [ll], and the number of such blocks is

A,

the

recfF(p).

Since

TI = p ,

6.4 and 6.6 can be used to bound the activity within one

block.

If

6 < A,

then the activity within the first

blocks can be bounded below 6.6.

6.5 and the equality of

K~

6

many

with the help of 6.5, 6.4 and and

p

imply that

parenthetical portion of 6.4 combines with r e ~ f ~ = ( A. ~ ~ The ) 6.6 to yield a recursively enumerable class

{H:(~)

I

Y < 6)

such that each

H, (y) bounds the activity within block

But then

,

where

6

= sup a

)

1

y < 6 ,

y.

bounds

6

because

u(6)

6 many blocks.

the activity within the first

<

K~

.

< recfF (kl)

The proof of 6.7 can be stretched to cover some of the situations that occur when 0 < p. One such is: nF < p F , nF and

are the only uncountable F-cardinals, end F Such an F has been devised by T. Slaman in a recfp (p) = 0 pF

.

generic extension of

L

of his own making.

Splitting and density theorems [ 2 5 ] can be proved for the recursively enumerable (in F recfF(p) = p = n, the best possible.

etc.)

classes of

4-sets if

but it is unlikely that these results are The above hypothesis has also been used by

E, Griffor to obtain a minimal pair.

He builds on minimal

pair results of Normann [18] and others. The proof of the density theorem of course needs a regular sets theorem.

Let

be a class of $-sets. A is said to be regular if for each a < klF , the intersection of A and H 1 T < I is recursive in F and some HF-set. A

Theorem 6.8 (Sacks [ 2 5 ] ) . (in F

etc. ) class of

Let

$-sets.

recursively enumerable (in F such that

A G F , B, H

A

and

be a recursively enumerable Then there exists a regular,

etc.) class B < F , A, H

B

of

for some

$-sets $-set

H.

Note that the above regular sets theorem does not require any assumptions about

or

p.

Grazie dell'ascolto, e buon giorno.

References [l]

R. Friedberg, Two recursively enumerable sets of incomparable degrees of unsolvability, Proc. Nat. Acad. Sci. 43 (19571, 236-238.

[2] A. A. Muchnik, On the unsolvability of the problem of reducibility in the theory of algorithms, Doklady Akad. Nauk SSSR 108 (19561, 194-197. [33

G. Driscoll, Metarecursively enumerable sets and their metadegrees, Jour. Symb. Log. 33 (19681, 389-411.

141

R. Shore, The irregular and non-hyperregular a-r.e. degrees, Israel Jour. Math. 22 (19751, 28-41.

[5] G. E. Sacks, Post's problem, admissible ordinals and regularity, Trans. Amer. Math. Soc. 124 (19661, 1-23. [6] W. Maass, Contributions to a- and 6-Recursion Theory, Habilitationschrift, University of Munich 1977. [71

S. Friedman, Recursion on Inadmissible Ordinals, Ph.D. Thesis, Massachusetts Institute of Technology 1976.

[8] R. Shore, The recursively enumerable a-degrees are dense, Ann. of Math. Log. 9 (19761, 123-155.

I93

S. Friedman, Negative solutions to Post's problem 11, to appear.

[lOl

S. Homer and G. E. Sacks, Inverting the half-jump, to appear.

I

R. Shore, Splitting an a-recursively enumerable set, Trans. Amer. Math. Soc. 204 (19751, 65-78.

1121

S. Homer, Priority Arguments in 6-Recursion Theory, Ph.D. Thesis, Massachusetts Institute of Technology, 1978

1131

S. C. Kleene, Recursive functionals and quantifiers of finite type, Trans. Amer. Math. Soc. 91 (19591, 1-52; and 10.8 (19631, 106-142.

[14] J. Shoenfield, A hierarchy based on a type 2 object, Trans. Amer. Math. Soc. 134 (19681, 103-108.

[15] T. Grilliot, Selection functions for recursive functionals, Notre Dame Jour. Form. Log. 10 (19691, 229-234. 1161

R. Gandy, General recursive functions of finite type and hierarchies of functions, University of Clermont-Ferrand 1962.

1171

Y. N. Moschovakis, Hyperanalytic predicates, .Trans. Amer. Math. SOC. 138 (19671, 249-282.

[l81 D. Normann, Degrees of functionals, Ann. of Math. Log., to appear. [191 G. E. Sacks, Post's problem, absoluteness and recursion in finite types, 1978 Kleene Symposium volume, to appear. 1203

L. Harrington and D. MacQueen, Grilliot's selection principle, Jour. Symb. Log. 41 (1976), 153-158.

[211 G. E. Sacks, The k-section of a type n object, Amer. Jour. Math. 99 (1977), 901-917. 1221

L. Harrington, Contributions to recursion theory in higher types, Ph.D. Thesis, Massachusetts Institute of Technology 1973.

[23] C. Spector, Recursive well-orderings, Jour. Symb. Log. 20 (19551, 151-163. [24] G. Kreisel, Set theoretic problems suggested by the notion of potential totality, in Infinitistic Methods (Pergamon Press, Oxford) 1961, 103-140. (253 G. E. Sacks, Three theorem on recursive enumerability in a finite type object, forthcoming.

Harvard University Massachusetts Institute of Technology

Acknowledgement:

The author is grateful to the Centro Inter-

nazionale Matematico Estivo for inviting him to lecture, and to the National Science Foundation for their support of the venture.

CEN TRO I N TERN AZIONALE MATEMATICO ESTIVO

(c.I.M.E.)

CONSTRUCTIONS I N THE RECURSIVELY ENUMERABLE DEGREES

ROBERT I. SOARE

CONSTRUCTIONS IN THE RECURSIVELY ENUMERABLE DEGREES Robert I. Soare1

CONTENTS Introduction Lecture 1. Simple sets, fixed points, and a completeness criterion for r. e. sets Lecture 2.

The finite injury priority method

Lecture 3.

The infinite injury priority method

Lecture 4.

The minimal pair method

Lecture 5.

Embedding distributive lattices in the r. e. degrees

Lecture 6.

The non-diamond theorem

Lecture 7.

The Renaissance in classical recursion theory, the Sacks density theorem. and beyond

'This work was partially supported by NSF Grant MCS-76-07033.

CONSTRUCTIONS IN

THE RECURSIVELY ENUMERABLE DEGREES Robert I. Soare University of Chicago

INTRODUCTION In these lectures we survey some of the most important results and the fundamental methods concerning degrees of recursively enumerable sets.

(r. e. )

We begin $ 1 with Post's simple sets and a recent elegant generaliza-

tion of the recursion theorem.

In $ 2 we give the finite injury priority method,

the solution of Post's problem, and the Sacks splitting theorem,

In

5 3 the

infinite injury method i s introduced and applied to prove the thickness lemma and the Sacks density theorem.

In $4 and $5 we develop the minimal pair

method for embedding distributive lattices in the r. e. degrees by maps preserving infimum6 a s well a s supremums.

In $6 we present the non-diamond

theorem which a s s e r t s that such embeddings cannot always preserve greatest and least elements.

F o r background reading we suggest Rogers [17],

Shoenfield [23], and Soare [25]. Our notation i s standard as in Rogers [17], with a few additions. sets A, B E o we say B i s recursive in A reduction @ such that @(A)= B. B STA.

(B < A) i f there i s a Turing -T

Let A z T B denote that A

The degree of A, dg(A) = {B :B

=

For

A),

STB

and

Lower case bold-face

letters

2 ,&,

... denote degrees.

of all Turing reductions. so {wJ i s A'

ee o

Let

{%Ie

A We )

.

and W = We e

T h e m of A

The jump is well-defined on degrees. F o r example, K

We write d g ( ~ 1 ) dg(B) if A IT B.

d g ( ~ ) Udg(B)= d g ( A @ ~ where )

An r. e. set A

= $' = {e :e c We)

is

Let

A @ B ={Zx:xcA) U ( 2 x t l : x e B ) .

A degree is r. e, if i t contains an r. e. sea.

The r. e. degrees

form an upper semi-lattice with least element

= dg(p).

B,

weA = {x:@~(A;x)converges),

i s complete i f W < A for a l l e. e -T

ment

be an acceptable numbering

i s a n acceptable numbering of all t. e. sets.

= {e :e l

complete.

Let

Let @ (A; x) = y if evs

ee

2

(5,S ,u)

= dg(@ and greatest ele-

with oracle A and input x

As A yields output y in s steps. We also use {e)s (x) = y and {e) (x) = y for A Q~,;(X) = y and QefA;x)=y,respectively. Pix a l:l recursive function from w X o onto o and let <x, y> denote the image of the ordered pair (x, y) under

this map.

We identify a set A with its characteristic function, and let

~ [ x denote ) the restriction of this function to arguments (_ x

.

1. SIMPLE SETS, FMED POINT THEOREMS, AND A COMPLETENESS CRITERION FOR R. E. SETS The study of degrees of r. e. eets began with the famous paper of Post [14].

and

Post's problem was to find an r. e. set A of degree different from

2.

2

Post attempted to construct such an incomplete r. e. set A by

considering structural properties on A such a s simplicity which he hoped would guarantee

incompleteness.

Although unsuccessful, this approach led

ultimately to a very pleasing fixed-point theorem which generalizes Kleene's

recursion theorem and Which yields a necessary and sufficient condition f o r an r.e. s e t to be complete. An r. e. set C is creative if there i s a recurrive fwction f(x) such that if Wx

identity function. B = {bo, bt,

- Wx.

then f(x) e

F o r example, X i s creative via the

F o r every creative set C,

...)

a coinfinite r.e.

where Wb

g,

-C contains an infinite r. e. set = Wb (f(bn)) . Post defined n

and W U bn+1 s e t A to be simple if contains no infinite r.e.

o

=

set.

He

were sufficiently @thinnwith respect to containment of

hoped that if

infinite r. e. sets then A would be incomplete.

Post constructed simple sets

and proved them incomplete with respect to a stronger reducibility called m-reducibility.

A set A is m-reducible to B (A

-C

8 ) if there i s a

recursive function f such that, for all1 x, x r A iff f(x) e B. see that if C <

-m A

simple).

It i s easy to

and C i s creative, then A i s creative (and thus not

Hence no simple set A can be m-complete.

Theorem 1.1 (Post):

There exists a simple set S.

Proof. Fix a recursive enumeration {a ) of the r. e. set s sro A = { ( e , x ) : x r We & x

a

s

> 2e).

Define an r. e. set B

A by enumerating

= ( e , x ) into 3 if there i s no at = < e , y) for t < s and y #x. Since

B i s single-valued i t represents the graph of some partial recursive (p. r. ) function

JI;

i. e.

,

+(x) = y iff (x, y> r B.

Let S = range

JI.

(Intuitively,

enumerate W until the first element +(e) > 2e appears in W and then e e put JI(e) into S.) (1) S i s r. e.

The following facts give the simplicity of S. (S i s the range of a p. r. function. )

(2) % i s infinite.

To see this, note that S contains at most e elements

out of {0,1,2,.

..,2e),

15n [0,2e] 1 >

e,

so

'S

namely

+(O),+(l),

...,+(e-1).

Hence

is infinite.

(3) If We i s infinite, then W

e

n S # $,

because < e , x > e A for some

x > 2e and so $(e) i s defined and +(e) e S A W

.

A simple set A i s effectively simple if t h e r e i s a recursive function f

such that ( V e ) [We Cwhere

lwei

=+

i s the cardinality of We.

Pel-< f(e)l* Note that Post's simple s e t i s

effectively simple via f(e) = 2e. P o s t realized that simple s e t s could be complete (indeed S i s complete by Corollary 1.5), and s o he defined coinfinite r. e. s e t s with still thinner complements called hypersimple and hyperhypersimple.

A coinfinite r . e.

s e t A i s hypersimple (h-simple) if there i s no recursive function f which majorizes

A

in the sense that f o r a l l n,

principal function of

h his

x, is defined by

f(n) 2 p-(n),

where p-

A

p-(n) = an, where

x = a.

, the

A

A

< a1 <

... .

i s not Post's original definition but can easily be shown equivalent to i t

[17,p. 1391.)

Clearly, h-simple s e t s a r e simple.

The converse i s false

-

since f o r Post's simple s e t S the function f(x) = 2x majorizes S. m o r e , h-simple sets a r e not necessarily incomplete.

Further-

Indeed, Dekker [2]

proved that every nonzero r. e . degree contains an h-simple (and therefore simple) set. Theorem 1.2 (Dekker):

is a n h-simple s e t B Proof.

t

F o r every nonrecursive r. e. set A t h e r e

A.

Let A = rng(f),

f a 1:i recursive function, and let a

= f(s),

and As = {f(0). s e t of -

...,f(s)}.

Define B

A f o r the enumeration f. Next, B is h-simple

i s infinite.

I I:

( 3 t > s ) [at < a#]}, the deficiency

Clearly B is C i and hence r. e., and f o r if g(x) 2 p-(x)

=

f o r some recursive

B

...,ag(x)) , which would imply A because x c A iff { a O , a i , . .. , a P, ( 41 .

function g, then x c A iff x 4 {ao, a * , Similarly, A < B -T

recursive.

xc

B B < A since to test whether s c B, we A-recursively compute -T

Finally,

t such that ~ ~ [ = aA[aS]. ~ ]

Now s c B iff a U< a e f o r some u such that

n

s
The crucial point about the deficiency s e t B i s that any nondeficiency stage s e

5

i s a "true" stage in the enumeration { A ~ ) ~ of A in that

~ ~ [ f ( s=) ~] [ f ( s ) ] . These stages will prove v e r y useful in o u r study of the infinite injury priority method in 53. Martin [lZ] found a f a i r l y general sufficient condition for a n r. e, s e t to be complete.

His condition applied t o many "effectivelyggnonrecursive r. e.

sets including t r e a t i v e s e t s and effectively simple sets. ~ a c h l a n [ S modified ] Martinge condition s o that i t became both n e c e s s a r y and sufficient.

Arslanov

[1] then converted i t t o the following f o r m which can be viewed a s a generalization of Kleene's recursion theorem.

It a s s e r t s that not only recursive

functions b ut a l l those functions of r. e. degree l e s s than point.

have a fixed

( ~ e c a l that l the recursion theorem [17, p. 1801 a s s e r t s that every

recursive function f has a "fixed pointn n such that W = W n f(n). ) Theorem 1.3 ( ~ r s l a n o v ) : An r.e. function f < A such that W -T f(x)

Proof.

( d )T r i v i a l

# Wx

s e t A i s complete iff t h e r e is a

f o r all x.

since {x: Wx =

fl

8'.

Assume (g)

( V x ) [Wf6)

# Wx]

A where f(x) = {e) (x) and g(x) i s

the greatest element used in the latter computation. {Ks)s

'

O

be recursive enumerations of A and K.

x c K, and m(x) = 0 otherwise.

Let { A ~ ), ~ ~ Let m(x) = ps[xc Ks] if

Define the recursive function

At At f (s,x) = {elt (x) where t L s i s minimal such that {e) (x) i s defined. t A

By

the recursion theorem with parameters define the recursive function h by if x e K

W otherwise Since g

sT A

we can define the function r < A by -T

Now if x e K and r(x) s m(x) then A

[gh(x)] = A[gh(x)], so m(x)

?(m(x), h(x)) = f (h(x)) and Wf(h(x)) = Wh(x) contrary to the hypothesis on f. Hence, for all xc K, m(x) < r(x).

Corollary 1.4. f c

2

Thus for all x,

Given an r. e. degree

5 , 2 '< 2 iff every function

has a fixed point. Corollary 1.5.

An r.e.

set A i s complete if A i s either

(a) creative, o r (b) effectively simple. Proof. -

(a) Let A be creative via g(x).

Choose a

t

and define

Let A be effectively simple via g(x).

(b)

W

f(x)

-

where A = a

0

< ai C

.., .

= {ao, ai,

Define f < A by -T

..., a s(x9 *

In each case, it i s clear that W f Wx for f(x)

n

all x so A i s complete by Theorem 1.3.

After Post's problem was solved by an entirely different method, some structural properties guaranteeing incompleteness were discovered [ l i ] , [26, p. 5501.

2.

THE FINITE INJURY PRIORITY METHOD

A positive solution to Post's problem was finally achieved by

Friedberg [3] and independently by Muchnik [13].

In their method, known a s

the priority method, the desired r. e. set i s constructed by stages to meet a certain sequence of conditions {R ) n new

called requirements.

If n

< m,

requirement R i s given priority over Rm and action taken for R at n m some stage s may a t a later stage t thereby injuring

R m at stage t.

> s be undone for the sake of Rn*

The original priority method of Friedberg

and Muchnik has the property that each requirement i s injured at most finitely often. We illustrate the finite injury method by proving the FriedbergMuchnik theorem using a variation of Sacks [I81 which i s more powerful than the standard method and which will be used in $3.

In our constructions, the

requirements {R ) will be divided into the negative requirements e ec w

N

e

=R

2e

which attempt to keep elements

structed, and positive requirements

e'

out of

the r. e. set A being conwhich attempt to put

= R~e+i

elements

KO A.

The negative requirements will be of the form C

# Q!e(A),

where C is a fixed nonrecursive r. e. set, so that the negative requirements together a s s e r t that C d

T ~ .Sacks observed that the

requirement Ne can

be met by attempting to preserve agreement between Cs(x) and Oe,.(AS;x)

rather than disagreement as one might suppose.

(The point i s

that i f we preserve this agreement sufficiently often and i f C = Q (A) then e C will be recursive contrary to hypothesis. )

The positive requirements

will ensure a s in Theorem l.l that A i s siriiple and hence nonrecursive.

F o r every nonrecursive r. e. set

Theorem 2.1 (Friedberg-Muchnik):

C there i s a simple set A such that C $ A (and hence j# <

T

Proof. -

A<

$').

It clearly suffices to construct A to be coinfinite and to

satisfy, for all e, the requirements:

Ne :C

# Q ~ ( A,)

Pe: We infinite Let {Cs)s

-7 We

,be a recursive enumeration of

C.

n

A

#$.

Define A. = $.

Given As

define the following three recursive functions whose roles a r e obvious f r o m their names: (use function)

min{z: Q (A*[%];x) defined) i f z exists e*5 u(e, x, s) = otherwise

0

(length function)

1( e , )

(restraint function)

=

m={r

.

w y < x)[Cs(y) = Qe,s(~s:y)])

r(e,s) =.max{u(e,x, s): x 5 ~ ( es)) ,

.

F o r each e 5 s , if W n As = e,s

$

and

then enumerate the least such x in A sti'

Define A =

usAs .

(~ntuitively,u(e,x, s ) i s the maximum element used in the above com-

< r(e, s )

putation, and the elements x

a r e restrained from A

by require4

sti

N in o r d e r to preserve the length of agreement measured by I (e, s). ) e

The

negative requirement N i s injured at stage s t i by element x i f e x I. r(e, s) and xc A (injury set)

st1

,iK

These elements form an r. e. set:

- As & x 5 r(e, s)]).

Ie = { x : ( 3 s ) [ x c Asti

Note that each I

Pi

- As.

e

i s finite because N

e

i s injured at most once for each

e, whereupon P. i s satisfied thereafter. 1

(Positive requirements,

of course, a r e never injured.) Lemma 2.2. Proof. -

( \de)[c

#q(~)].

Assume for a contradiction that C = Choose s t such that N

l i m I (e, s ) = oo. s

e

Q,

e

(A).

i s never injured after stage st.

We shall recursively compute C(x) contrary to hypothesis. C(p) f o r p

E

Then

o find some s > s1 such that l(e, s) > p.

To compute

It follows by induc-

tion on t 2 s that

and hbnce that

Q,

e,s

( A ~p); = Q e ( ~ sp) ; = ( P ~ ( AP) ; = C(P). Since s

clearly holds unless C (x) t

# Cs(x)

> s f , (2. 2)

for some t 2 s and x 5 p ; but if x and t

a r e minimal then our use of "5I (e, t)" r a t h e r than nition of r(e, t ) i n s u r e s that the disagreement C (x) t

< l ( e , t)" i n the defi-

# @e ,t ( A ~ ; x ) is

Note that

preserved forever, contrary to the hypothesis that C = ee(A).

even though the Sacks strategy is always described a s one which preserves agreements, it is crucial that we preserve at least one disagreement a s well whenever possible. Lemma 2.3. Proof. -

(\de)[lim r(e.s) exists and i s finite]. s

By Lemma 1.1 choose p = Fx[C(x)

sufficiently l a r g e such that, f o r all s (Vx

# Se(~;x)].

Choose s *

2 sB,

< ~ ) r @ ~ , ~ ( A= ~%(A: ; x ) 41,

( V x 5 p)[Cs(x) = C(x)l, and

Ne is not injured a t stage s. Case 1. s

,

( V s 2 s*)[\,,(A,;

p) undefined].

Then r(e,s) = r ( e , s l ) for all

sB.

C a s e 2.

4e,t(At; p) i s defined f o r some t 2 s'.

Then 4

e,s

(As; P) = q , t ( A t i ~ )

f o r all s 2 t because l(e,s) 2 p, and so, by the definition of r(e,s), the computation @ (At; p) is preserved and N i s not injured after stage sf. e ,t e Thus ae(A; p) = S

Hence

(As; p).

, r(e, t) = limsr(e, Lemma 2.4,

Proof. max(r(i):

we

e,s

But C(p)

# ae(A; p).

Thus

s).

( V e)[We

infinite d We

nA

# $1.

By Lemma 1.2, let r(e) = limsr(e,s) and R(e)

i (e l .

Now if

(3x)[xt

We & x

> RCe)

& x

=

> Ze] then

* A # $. Note that

i s simple.

i s infinite by the clause IBx> 2e

in (2.1). and hence A

Sacks invented-the above preservation method (which plays a crucial role in the l a t e r infinite injury argument) to prove the following theorem. Theorem 2.5 (Sacks Splitting Theorem [18]): such that C i s nonrecursive.

Let B and C be r. e. s e t s

Then t h e r e exist r.e.

sets A0 and A1 such

that (a)

AOU A 1 = B and A O n A l =

(b)

G

$,

and

I T ~fori ,i = 0 , i .

Proof. -

st

Let

and C such that Bo =

$

and

( c ~ )be ~recursive ~

enumerations of B

-

and ) B ~ +Bs~ I = 1 f o r all s. It suffices to give

recursive enumerations {A

}

i,s s c

0'

i

= O,l,

satisfying the single positive

requirement P : x c B stl -Bs

@ [ X ~ A ~ o, r~x + c A ~i , s t l l *

and the negative requirements for i = 0 , 1 and a l l e,

: N

c # \(Ai).

Define A = $. Given A.1,s define the recursive functions ll(e.s) i,O i Let x c Bsti Bs and r (e,s) a s above but with A in place of A i,s so i Choose <el, i*) t o be t h e least <e, i ) such that x ( r (e, s ) and enumerate

-

' Ai-i*,s+i.

If ( e l , i * )

.

fails to exist, enumerate x c A0 , s t l ' This

defines Ai, i = 0,1. i To s e e that the construction succeeds, define the injury s e t Ie a s above but with Ai i n place of A. i

= 0, i and a l l e ,

It follows by induction on (e, i ) that, for

(1) C

# @e(Ai)8

i (2) lim r (e, s ) exists and i s finite, and s (3) 1:

i s finite.

D

It can be shown [26, p. 5251 that the r. e. s e t s Ai a r e automatically low, namely -

A: s T$I, where the jump of A was defined e a r l i e r to be

A At = {e: e r We ). By setting C = B in Theorem 2.5, it follows that any nonrecursive r. e. set B can be split a s the disjoint union of low r. e. s e t s A.

and A which a r e Turing incomparable and such that i

d g ( ~ =) dg(AO) U dg(Ai).

Thus, there i s no minimal r.e. degree.

Finite injury arguments a r e characterized by the fact that the injury s e t Ie i s finite for each e.

In $3 we will consider c a s e s where I

infinite although usually recursive.

e

is

Note that Lemma 2.2 holds by virtually

the s a m e proof as above if we a s s u m e "Ie recursiven in place of "Ie finitet1. This i s what allows the infinite injury method to succeed.

3. THE INFINITE INJURY PRIORITY METHOD

Shoenfield [22] and

, independently,

Sacks [19], [ZO], [21] discovered a

technique for handling a r e q u i r e m e n t which may be injured infinitely often (namely, the injury s e t Ie m a y be infinite).

Sacks considerably developed

this technique into what he called the "infinite injury priority method" and he used i t to prove many important results on r. e. degrees, the most striking of which i s the density theorem which a s s e r t s that for any r.e. d

CI

< 5 t h e r e i s an r. e. degree 2 such that

sketch of the method.

cJ,

<2 <2

.

degrees

We now give a brief

More details and applications can be found i n [ 2 6 ] .

We wish to coostruct an r.e. Ne a r e a s in

(where X

set A where the negative requirements

9 2 and the positive requirements a r e of the form

C*

Y denotes that X

-Y

i s finite) so that a single positive

requirement may contribute infinitely many elements to A. c a s e s , the r.e.

s e t s {W 1 will be recursive. p(e) e c w

In the simplest

F o r each Ne we

would like a restraint function ?(e, s ) so that exactly a s in f 2 we can enumerate x in A x

s +l

and

f o r the sake of Pe just if x c W p(e), s + l

> qi, s), f o r all i <_ e. The negative requirement Ne can now be injured

P ~ I *i < e ,

infinitely often by those P., i < e, but the recursiveness of W 1

enable us to meet Ne a s in Lemma 2.2.

will

The main difficulty will be that

some P remains unsatisfied because of the restraint functions :(is s), e i 5 e , which may now be unbounded in s (i. e., lim sup G(i, s) = w).

To

satisfy P i t clearly suffices to define "re, s ) such that A

lim infs ~ ( es ,)

(3.1)

< oo ,

C1

where R(e, s ) = max{$(i, s): i 5 e), because then Pe has a ltwindown through the negative r e s t r a i n t s a t least infinitely often. The first

obstacle to achieving (3.1) i s that i f we let ?(e, s ) be r(e,s)

a s defined in f 2, then we may have lim

S

r(e, s) = co for some e.

example, suppose @ (X; 0) = C(0) just i f n 1 ss

/x

(For

for some even n < s , but

P0 eventually forces every even number into A so that Ql(A; 0) i s undefined.

Then N1 i s satisfied by divergence but

C(0) = @ (As;O) for 1,s

almost every s, s o limsu(l,O, 6) = a, lim r(1.s) = oo and PI is not S

satisfied.)

This difficulty a r i s e s only if there a r e infinitely many stages s

# As+l [u]

such that As[u]

Thus, we can easily remove

where u = ~ ( 1 . 0 ,s).

A

the f i r s t obstacle by replacing Q e v e r w h e r e by @ defined below, and e ,s e, s letting Z(e, s ) be the resulting r e s t r a i n t function.

If C

# Qe(A)

we then

haee lim inf ?(e, s ) < oo. S

A

The second obstacle to (3.1) i s that lim R(e, s) = ao even though lid inf ;(i, 5 ) S

< oo f o r each i <,e.

(For example, N

1

and N may together 2

permanently r e s t r a i n a l l elements because their r e s t r a i n t functions do not drop back simultaneously.)

4

Surprisingly, the Q solution to the f i r s t e,s

obstacle automatically removes t h e second, a s Lachlan f i r s t observed [7]. Suppose we wish to give a recursive enumeration { A }

S BE W

s e t A.

Given {A :t t

s ) , define

1rnax(ASv { s ) ) $

of an r.e.

otherwise ; if defined and u(e, x, s ) < as,

e, s

(As;x) = undefined

othe w i s e ;

if

f e,s (AS:x)

i s defined,

othe rwise; and

If

,i s any recursive enumeration of an r. e.

to T a s the s e t of

true ( n o n d e f i ~ i e n c stages ~) of

T i s infinite and T r

A uniformly i n A.

lims { , . s ( ~ s ; x ) = y a s before.

s e t A we r e f e r

this enumeration.

If Qe(A; X) = y then clearly h

The crucial point about Q e, h

Note that

6

i s that for any

true stage t any apparent computation Q (At;x) = y i s a t r u e computation e,t

Qe(A;x) = y.

Namely, using the fact that u(e,x.t) ( t

u(e,x, t)

because if qSt(At;x) i s defined then

, we have

.

< at and At[at] = ~ [ a ~ ]

The simplest application of this method i s the thickness lemma. any s e t A and x and

=

AtX) =*

w, define the ncolumnn A(X) = {
6

U{A"):

-

A subset A S B i s a thick scbset of B i f

z < x).

(We write X

(X - Y ) u ( Y - X )

=* Y

i f the symmetric difference

is finite.)

Theorem 3 . 1

(Thickness Lemma

- Shoenfield

thick subset A of B such that C Proof. -

$.

$

set B there i s a

A.

Fix recursive enumerations { Bs )s c ; {=s}s. Given { A :t t

maining functions a s i n (length function)

< s)

5 2 with

A

define O (As) a s above. e,s A

cb

e,s

in place of cb e,s

A

f(e. 6) = max{x:

(restraint function) :(e,s) h

(injury set)

1,

of B and C.

Define the r e -

.

(V Y < x)[CS(Y) = Af ,s(As;~)l)

= max{t(e, x, s): x

A

f (e, s))

,

=us$,swhere #

Ie,s = {x: ( 3 v S s ) [ x I ; ( e , v )

&

To meet the requirements e

0

, namely,

A

P :B(e) =*

Given a

[ 2 2 ] ):

nonrecursive r. e. set C and a piecewise recursive r.e.

Let A = 0

= x) ,

for a l l x, and B i s piecewise recursive if B ( ~ )i s recursive

for a l l x.

,

For

and

Ne :C

#

%(A)

,

- Av]}.

(e) and x > $(i. s) for a l l i we enumerate x in A(') just if x r Bs+l s+l Let A =

us

Te C A

because N

5

e

i s injured by P. only if i < e. 1

Thus

A

s T ~ ( c e )because if x e A ( < ~ ' , say x t A(<e), s then x e Ie just

Fix e and assume by induction that C i

.

As '

Note that we have

5a

=*

# ei(A)

and A ( ~ ) B(i) for all h

< e. Then A ( < ~=* ) B(<e) i s recursive and hence I i s recursive. e Lemma 3.2 ( ~ n j u r yLemma):

C

# %(A).

Proof. Assume for a contradiction that A

lims ~ ( es) , = m.

Fixing

contrary to hypothesis. i ( e , s)

ie

C = .Qle(A). Then

a s an oracle we compute C, so C i s recursive

To compute C(p) f o r p

t u find

some s such that

> p and (Vx
Iu(e.x.s)

3

[ r 4?e o r z

c

As]].

Such s exists since C = a e ( ~ ) . By the same remarks as in Lemma 2.2, it follows by induction on t

2 s that

and hence that @

e,s

Lemma 3.3

(As~p)='P,(As;p)=Qe(A:~)=c(~)-

(Window Lemma):

in the enumeration {A } s st ( ~ e n c e .since C

# mi(A),

A

R(e, s) = max{:(i,

Proof.

5):

W

of A.

Let T be the set of true stages If C

for a l l i 5 e , lim

# Qi(A)

then limt

, :(is t ) < m .

A

ttT

R(e, t) < a, where

i S e) thereby satisfying (3.1). )

We know C j/ IPi(A) for all i 5 e.

Fix i 5 e. Define

Choose s f sufficiently large such that, for all s 2 s t , P = ~ [ C ( X#) Q).(A;x)]. 1

C a s e 1. such that Case 2.

.

1, s

(v t 2 sl)[t t

E

T

A

t

A

, we have

l ( i , t) = p

and

iiSt(At: p) is defined for some t

(As; p) for a l l s 2 t by (3.2). A

(V s 2 t ) [f(i,s)

t

But C(p)

TI t

:(i ,t) = max{u(i,x,sl): x < p}.

2

s t . Then

# @&A;p).

ai(A; p) =

Hence, we have

= p & :(i, s) = :(i, t)].

(Thickness Lemma

Theorem 3 . 4

Then for any t 2 s f ,

T =;> QiSt(At:p) undefined].

d

- Strong F o r m ):

Given a non-

recursive r. e. s e t C and an r. e. set B there i s an r. e. s e t A that A < 'T

B such

B and

(a) ( V e ) [ c $ , B ( < ~ ) ] d [C $,A (b)

=

(V e ) [ PT ~ B(<

3

& A i s a thick subset of B]

( V i 5 e)[C

.

8 B(~)]]

# Qi(A) &

F u r t h e r m o r e , an index for A can be computed uniformly in indices for B and C. This strong form [26, p. 5201 follows by carefully examining the above proof and by replacing ;(e, s ) in the definition of ;(el s) by A

A t e , s ) = max(x :(3 v <_ s)[x ~f (e, v)

(modified length function)

.

& ( V Y~ x ) [ ~ ~ [ ~ ( e ,=yA~~v[ :)( ~ l .Y.V)~II

C o r o l l a r y 3.5 (Sacks [21]):

Let

i0 < i1 < d < ... "2

sequence of simultaneously r. e. degrees. bound

.

2 such that d < d < .. < 2 < 2' NO

-1

upper bound f o r the sequence. )

be an infinite

Then there exists an r. e. upper (Hence,

2' i s not a

minimal

Proof.

Fix a recursive function h such that dg(W ) = d for a l l x. h(x) 3

Define the r. e. set B by B(X) = { ( x . y ) -:y t w ~ ( ~ $ . Let C = K and apply Theorem 3.4 to obtain a thick r. e. subset A G B such that K By. thickness,

Wh(x)

=

-T

.(XI

=+ A(x] so that d < dg(A) f o r a l l i. "i

ITA. 0

The Sacks density theorem can be derived from the thickness lemma using the following results of Yates on index s e t s [31, pp. 312,3141 which can be proved without priority methods. that there i s a predicate R~

If V i s a n r. e. set, let S

v.

{x: Wx s T V) t Z3

Lemma B (Yates):

F o r any set S

.

(3e ) [ ~ ( ~ ) v h(x)

/ S 3 (V e

T

T

C

v 3

there i s a recursive function

ir ( V i < e)[w(i) h(x1

i s recursive]],

(el ) [ ~ ( ~ i)s recursive 1.

Theorem 3.6 (lndex Set Theorem

D such that D <

t

and

h(x) such that, for a11 x. Wh(x) 5T V

(b) x

ZV denote 3

recursive in V such that x e S iff

Lemma A (Yates):

(a) x t S 5

t

C and S e Z

C 3

- Yates [30]):

Given r. e. s e t s C and

there i s a recursive function g(x) such

that, for a l l x,

(b) x t S W W g(x) = T C . Corollary 3.7 and D

ensit it^ Theorem

< C then t h e r e e x i s t s an T

-

Sacks [20]): If D and C a r e r. e.

r. e. s e t A such that D

< A< C T

.

Proof (Cor6llary 3.7): -

s

6

D

z3

and hence S a

zC 3'

L e t S = { x :Wx

0

such that Wx

0

=

Proof (Theorem 3.6): -

W g(x0)

~

=

(

o

C for a l l Bx 4T '

,

)

D:

C.

C .

F o r each x define the r. e. s e t Bx

and ~ Bx(e+i)= { ( e t ~ , :~y )t W$;?)) h(x)

<

Theorem 3.4(b) with e = 0.

A

x

T

because A?)

.

(Note that

F o r each x apply

5. C and D S T C . )

Theorem 3.4(b) to Bx and C to find A x s Bx, s o that Ax

T'

0

Fix V = C . S a 'Z and h(x) the recursive 3

x because W

Moreover, f o r each x, D

A,

By the recursion theorem

. T h e n D X T WI3(xo)
function for S according to Lemma B. by

Dl. Then, by Lemma

Apply Theorem 3.6 to find g(x) such that

D s T W g ( x ) IT C and Wx s T D just if W &!(XI choose x

T

STBx C T C

.

=* B?) s T D by

By the uniformity of Theorem 3.4, there i s a

recursive function g(x) such that W = Ax. g(x)

Now i f x

/S

then ) !B

is

recursive for a l l e > 0 by Lemma ~ ( b ) whence , C $T Ax by Theorem 3.4(b).

If x

E

S then, by Lemma ~ ( a and ) the definition of B, choose e such

that

~

Hence, Bf

=*

A(~) x

x

( s~ c1 & T

zT D
3 )B !

i s recursive].

Therefore, by Theorem 3.4(b).

B(e) s C , s o that C x T

Ax STBx IT C*

(V i)[O < i c e

ITAx.

Thus, Ax sTC b e u u a e

D

The density theorem can also be proved directly without index s e t s [ 2 6 , p. 5251 by combining the infinite injury method with a c l e v e r coding

method of Sacks.

4.

THE MINIMAL PAIR METHOD

If P is any countable partially ordered set (poset), then the finite injury priority method enables us to embed P (by an order preserving map)into the r.e.

degrees f? [Zl], and the infinite injury priority method

allows P to be even embedded in any interval a

N

[15].

<&

[z,&I,

for

2 .k c

and

If P happens to be also a lattice then these embeddings

naturally preserve supremums (sups) but not necessarily infimums (infs).

We now introduce a new method for embedding certain lattices into

5

preserving both infs and sups.

algebra

Consider the four element Boolean

0 which we call the Diamond.

In 54 we prove that the Diamond

can be embedded into

5

element a s well.

5 we replace the Diamond by an arbitrary countable

In

distributive lattice.

preserving not only sups and infs but the least

On the other hand, in 5 6 we prove the surprising

fact that not even the Diamond can be embedded as a lattice in f? by a map preserving both least and greatest elements.

A corollary i s that the

r. e. degrees a r e not closed under infs and thus fail to form a lattice. The embedding technique of

4 4 (called

the minimal

pair method

because of Definition 4.1) i s also one where some requirements a r e infinitary, but i t i s quite different from the infinite injury method of The version we present here (derived from

3.

[ 7 ] ) uses special stages

analogous to the nondeficiency stages of 5 3 so that the negative restraints drop back simultaneously, and the proof closely resembles a finite injury argument.

As F o r notational convenience we let {els (x) denote

@

from now on.

e,s

(AB:

X)

Nonzero r , e. degrees

Definition 4.1.

and

2

&

form a minimal

pair if -

- Yates [30]):

Theorem 4.2 (Lachlan [4] of r.e.

degrees

and

2

Corollary 4.3.

There exists a minimal pair

h.

The Diamond lattice can be embedded in

pre-

serving sups, infs, and least element.

Proof.

Let

5

=

Diamond lattice.

5 V& .

Then {,$

,2,k ,z)

embeds the

l3

Proof of Theorem 4.2.

It suffices to construct r.e.

s e t s A and B

satisfying, fbr a l l e , i, j, the requirements

Pte: A

# Ge

'2e+l : B N

(i,j)

:{i)

A

.

# We ,

and

= {j}B = f total =.> f i s recursive.

The following r e m a r k allows us to simplify the f o r m of the negative requirements. R e m a r k 4.4 (Posner):

To satisfy a l l N

(i,j)

,

isj e w, it suffices to satisfy

f o r a l l e the requirement N: :{ e l A = {elB = f total Proof. that A

# B,

+ f ,is

recursive.

We may a s s u m e without loss of generality that we can a r r a n g e s a y no e A

- B.

F o r each i and j t h e r e i s an index e such that

The remark follows immediately.

0

F r o m now on we will replace all occurrences of negative requirements similar to N latter as Ne

0.j)

and we will write the by equivalent requirements N' e '

.

Given {A ;t 5 s) and { B :t 5 s) we define a s usual the functions t t (length function)

1(e.s) = max{x:

(maximum length function)

r ( 0 , s) =

,

m(e,s) = max{ f (e,t) : t 5 s )

A stage s is called 0-maximal

function

As B (y) = {els s(y)]}

( V < x) [{els

i"

i f l(0, s) > m(0, sai).

Define the restraint

if s i s 0-maximal

,

otherwise

.

the greatest 0-maximal stage t < s

(Notice that we can define the restraint function in terms of a stage s rather than an element z used in a computation at stage s since we may assume z 5 u ( A ~e , x , s)

s s,

where u(As, e , x, s) i s u(e, x, s ) where A

s

i s the

oracle used in the computation. The strategy uo for meeting a single negative requirement N i s to 0 allow x to enter A U B at stage s t 1 only if s i s 0-maximal, and at most one of the sets -

A,B receives an element x at such a stage.

Thus, if x

As s destroys one of the computations {0Is (p) = q o r { o ) ~(p) = q for some As Bs p < l(0, s), say { o ) ~(p). then the other computation { o } ~ (p) = q will be pre-

served until the A-computation i s restored, and outputs q again

In this way

if {0}* = ( 0 )

B

= f i s a total function then f is recursive.

f ( ~ ) we find the least s such that p

F u r t h e r m o r e , lim inf

s

(To compute

< 1(0, s ) and we set f(p) = (0)

r(0, s ) < a, since lim inf r(0, s ) s

=

As S

(p). )

0 unless t h e r e i s

a l a r g e s t 0-maximal stage t , in which c a s e r(0, s) = t for a l l s

2 t.

This fundamental strategy of having one side o r the other hold the computation a t a l l times is applied to the other negative requirements Ne,

e

>

0, but with some c r u c i a l modifications to force the negative r e s t r a i n t s

t o drop back simultaneously, thus creating mwindowsn through the restraints a s in

5 3. F o r example, t o drop back simultaneously with N N must guess the 0' 1

value of k = l i m infs r(0, s). many strategies

"1

'

Thus, N

1

must simultaneously play infinitely

kc o, one for each possible value of k.

Each strategy

k u k i s played like a but with S = {s :r(0, s ) = k) in place of o a s the set 1 0 of stages during which it i s active, and on which i t s length functions I and m a r e defined.

This allows uk to open i t s window more often since its length 1 k Strategy u k still succeeds if functions ignore the stages in w - S 1 k any restraint i t imposes is maintained during intermediate stages s / S

.

while cr:

i s dormant.

k Thus, a t stage s i f k = r(0, s ) , we play ul

,

i maintain the r e s t r a i n t s previously imposed by the dormant a 1' i < k, and

discard r e s t r a i n t s imposed by u: (1)

.

j > k.

Thus i f k = l i m i d s r(0. s), then: i the strategies ul, i < k, strategy u k succeeds in meeting Nl; ( 2 ) 1

impose finitely much r e s t r a i n t over the whole construction; and (3) strategies crj 1

, j > k,

entire restraint r(1, s ) imposed by No and Ni together has l i m ids r(1, s ) < w

.

.

k

drop a l l restraint a t each stage s r S

the

Thus, the

Construction of A and B. Stage s = 0. Stage s t 1 r(e, s ) f o r N

e

.

Do nothing. Given As and Bs, define the restraint function

by induction on e a s follows.

Define r(0, s ) a s above.

A stage s i s let1)-maximal i f

Let . r(et1, s ) be the maximum of

0)

r(e,s),

(ii) those t (iii) those t

< s such that r(e, t ) < r(e, s ) , and < s such that r(e,t) = r(e, s) and t i s (et1)-maximal,

if s i s not (et1)-maximal. Requirement P requires attention if 2e

we,s n nS = ft , and

(4-1,

(3x)[x t

(4.2)

W

e,s

& 2e

< x & r(e, s ) < x]

and likewise for P with B in place of A. 2et1

Choose the highest priority

requirement P which requires attention and the least x corresponding to e that e .

Enumerate x i n A id e i s even (in B i f e is odd).

Lemma 1. Proof. -

( y e ) [lim ids r(e, s ) < a].

We f i r s t prove the case e = 0 . If there a r e infinitely many

Q ~ m a x i m astages l then lim infs r(0,s) = 0. largest 0-maximal stage. k = lim inf

that

s

r(e, s)

r(e, s).

< k.

Otherwise lims r ( O , s ) i s the

F o r the inductive step, fix e and assume

Then there a r e only finitely manv stages s such Let t be the largest such.

Let

Either there a r e infinitely many (et1)-maximal stages in

S = {s: r(e, s ) = k).

S, in which case E m inf l ( e t 1 . s ) = max{t, k), o r else there i s a largest S

(et1)-maximal stage v c S , in which c a s e lim inf Lemma 2.

s

r ( e t 1 , s ) = max{t, k,v).

Every positive requirement i s satisfied and acts at most

once. Proof. -

Consider requirement P (since P is similar). 2e 2eti

C

A i s infinite a s usual by the second clause of (4.2).

then W contains some x e

> lim

inf

s

First

Now if We i s infinite

r(e, s), and some such x i s eventually

enumerated in A satisfying P 2e ' Lemma 3. Proof.

( v e ) [requirement N i s met]. e

Fix e and let k = lim inf r(e-1, s), and S = {s: r(e-1, s ) = k). s

I

(1f e = 0 let S = o and k = 0.)

Choose s t such that no P i < e , a c t s iv

after stage s t and r(e-1, s ) 2 k for a l l s 2 s t . { e l A = { e l B = f i s a total function. :find an e-maximal stage s" e S, s n

Now assume that

To recursively compute f(p). p c o s t , such that ~ ( e s, t * )> p.

,

Let

AStt Bsn q = {e)stt (p) = {els (p). We will prove by induction on t that for all t either

A

and hence that f(p) = q. (i) o r (ii).

2 sn

Suppose that x destroys the last of the computations

Now i f x enters A V B a t any stage s t 1 such that s

s must have been e-maximal, s o both (i) and (ii) hold for t = s.

t

S then

But x can

destroy a t most one of the computations, s o the other holds at t = s t l .

F u r t h e r m o r e , x cannot e n t e r A u B a t stage s + l f o r s r(e, s) 2 x by clause (ii) in the definition of ~ ( es). ,

/ S,

s > s " , since

Q

This construction can be modified in a number of ways.

F i r s t , one

can construct an r. e. sequence of r. e. degrees {2i: i e w} such that a a i s a minimal pair for each i wiP-j requirements a s in

5 3,

Next, by allowing infinitary positive

one c a n construct a minimal pair

degrees which a r e high (i. e.

5.

# j.

.

5 ,i of r.

e.

= 2")(see Lachlan [4]).

=

EMBEDDING DISTRIBUTIVE LATTICES IN THE R. E. DEGREES

Using a fairly e a s y modification of the preceding method we will now replace the Diamond lattice of Theorem 4.2 by lattice.

any countable

distributive

Since any countable distributive lattice can be embedded in the

countable atomless Boolean algebra i t suffices to prove the following. Theorem 5.1 (Lachlan-Thomason [29]): countable atomless Boolean algebra

There i s an embedding of a

8 into the r. e. degrees

f? which p r e -

s e r v e s sups, infe, and least element. Proof. -

Let {a.: i c w} 1

be any uniformly recursive sequence of

recursive s e t s (i. e., ( ( x, i ) :x c ai} i s a recursive relation) which f o r m s an atomless Boolean algebra 73 under U tains o and has jd

a s i t s only finite member.

A i , i c y and define A

= { < i,x (I

we immediately have

, n , and complementation, con-

> :x r A .

& i c a)

We will construct r.e.

sets

f o r a c 73. Notice that

"g)

(5.1)

dr(Aa

(5.2)

a Z B => de(A,)

= ds(AJ

We will further meet f o r a l l i.a,

N

Aa

8

= {jfp

a ! , , j : {j

I

v dg(Ap) dg(Ag)

.

, and

the requirements

= f total

< A -T a n e '

5 f

These requirements i n s u r e

Note that (5.1)-(5.5) guarantee that the m a p a

-,

dg(A ) i s the desired Q

embedding, and (5.5) guarantees that the map i s 1: 1. negative requirements insure (5.5) suppose: (2) a

$

f3, s a y i r a

- B.

(To s e e that the

(1) dg(Aa) 5 dg(AB): but

Then dg(Ai) (dg(Aa) I dg(A ) by (5.2) and (l),

but d g ( ~ , )5 dg(A- ) by (2).

B

B

Hence dg(Ai)

dg(A

(5.4)' contradicting A nonrecursive. ) i

-)

B"B

= dg(A4 = 2 by

The strategy f o r meeting the negative requirements N a s before.

Denoting N

e = ( i l . i2.j a s in

5 4.

)

(aoB.j)

by Ne

, where

a!

=a

ii'

(a9B.j

= a.

I2

,

begins

i

and

we define the restraint function by induction on e exactly

However, new difficulties in proving Lemma 3 (that N

e

i s satis-

fied) require greater c a r e i n enumerating elements for the positive requirements.

To meet requirement P
> we will appoint followers

x r w so-called because the eventual enumeration of x in A. will satisfy

'
> (although x may be cancelled before this happens).

If x i s a follower

of P. and y a follower of P. then we say x has higher priority than y (x< y) 1 1 if i < j o r i = j and x was appointed before y. x4 y

followers x and y existing a t stage s that Construction of A

i'

Stage s t 1 Jv 8

< y.

Do nothing.

.

Requirement

P i s satisfied i f

<,,j>

# 8.

A W.

1,s

iff x

i r o:

Stage s =, 0.

A.

We will arrange for all

Requirement P requires attention if P

i s not

satisfied and either x

(5.6)

and x

W.

t

> r ( < i , j ) , s ) f o r some uncancelled follower

x of

$ 9 6

P or ' (5,7)

xew. JI

Let P

s



attention.

ioreveryuncancelledfollowerxof P Ci, j)

.

be the highest priority requirement which requires

If (5.6) holds for some x enumerate the least such x in A..

If

(5.6) fails, and (5.7) holds, then appoint x = s t 1 a s a follower of P



In either c a s e cancel a l l followers y of lower priority than x (i. e. (If no P

Ci, j

>

( v e ) [lim infs r(e, s ) < a].

Exactly a s in $ 4 , Lemma 1.

Lemma 2. i s met

x d y).

requires attention, then do nothing.)

Lemma 1. Proof.

.

(t/ e)[Pe

receives attention a t most finitely often and

1.

Proof.

Fix e and chooge s

receive attention after stage s

0'

0

such that for no el < e does Pel

Let k = lim idsr(e, s) by Lemma 1.

Let

e =
Nowif

P



receives attention infinitely often, then some

follower x > k i s appointed to follow P

C i . j)

and x i s never cancelled.

such that F u r t h e r m o r e , x o W. by (5.7). Hence, there i s a stage t t l > s J 0 Now x o r some s m a l l e r follower of P i s enumer r(e,t) < x and x E W. ~rt' e

-

ated i n A at stage t + l , P i s met, and P never again requires attention. e e Therefore, P receives attention at most finitely often. e met because otherwise P(i,j)and'(is j)

( v u ) ( v B) ( V j) [requirement N

Fix N = N e

A, Assume that {j)

of (4.

A {e)sv* '(x) To A

= W. , x t W. for every uncancelled follower x of J J

receives attention infinitely often under (5.7).

Lemma 3. Proof. -

i

Finally, P is (iIj)

(u,B,j)*

Choose k,S and stage s1 a s in Lemma 3

= { j t S = f is a total function.

is A ,-correct if A

6s

[u] = A,[u]

A computation

where u = U ( A , , ~ ,e. x , s).

-recursively compute f(p), find an e-maximal stage s A

U ~ B

t

S, s

> s',

such that l ( e , s) > p and both computations {e) = q and S A {els S's(p) = q a r e A -correct. We will show by induction on t that f o r unB a l l t z s either

via an A

an S

-correct computation.

(5.8) o r (5.9) by entering A u u A e-maximal and t o S

B

Now i f x destroys e i t h e r computation at stage

t t i then t must have been

( a s in Lemma 3 of 14) s o & t J computations existed at

the end of stage t.

By inductive hypothesis a t least one computation, say

(5.8). i s A

-

n B

~

Suppose x i s enumerated in A

destroying this computation.

~

~

~

~

~

~

.u

a t stage t t l ,

Then x cancels at stage t t l a l l followers y

such that x

< y (since these a r e exactly those followers y such that

Furthermore, z > t t l 2 u appointed.

But x

troyed by x. was A

f f nB

A

B-

p =dfn U(Ap,t' , ,p) for any follower z later

uu =an~(Au,t,e, t, p) since the A-computation is des

Also A

-correct.

Hence A ff

-

-

u ] since the Au-computation (5.8)

u n ~ . t [ ~ u= '

Brt

p[ u [ttl] = A

computation (5.9) now becomes Au Corollary 5.2.

x < Y).

fffi

[ t t l ] , and u 5 t t l s o the

B

B

-correct.

Any countable distributive lattice can be embedded into

the r.e. degrees f? by a map which preserves sups, i d s , and least element. Embedding nondistributive lattices into f? i s much more difficult. Lachlan [6] showed that the following two 5-element nondistributive lattices

M 5 (a modular lattice) and N 5 (a nonmodular lattice) can be embedded in

5

by a map preserving sups, infs, and least element.

This partial success led many to believe in the Embedding Conjecture which a s s e r t e d that every finite lattice can be embedded in lattice.

5 as a

finite

This conjecture was recently refuted by Lachlan and Soare [lo] who

showed that the following lattice S8 cannot be embedded in f? a s a lattice.

The obstacle to embedding S8 i s that for

&

to be the sup of the lower

M 5 lattice, an elaborate system of t r a c e s i s required for enumerating e l e ments into the s e t of degree

2.

This i n t e r f e r e s with the delicate minimal

pair machinery above which insures that

2

The most important open question on elementary theory.

i s the inf of a -0

5 i s that of

and a

-1

'

the decidability of i t s

Considerably more structural results (such a s

embedding and nonembedding theorems) will be required to meet this goal.

6.

THE NON-DIAMOND THEOREM

One might expect to extend Theorem 5.1 by constructing lattice embeddings which p r e s e r v e both g r e a t e s t a s well a s least elements.

The

following surprising theorem shows this i s impossible even for the Diamond lattice.

6.1

Non-Diamond Theorem (Lachlan [4]):

2u

r. e. degree & such that

and $ ,

Proof. and

<5

a n d 2 a r e non-

.

Let A and B be r. e. s e t s in degrees

,,. {bs) ,

2

= 2' then there i s a nonrecursive

recursive r. e. degrees such that

2 <5

If

5,i

respectively,

be recursive enumerations of A and B.

As = {at: t 5 s) and Bs = {bt: t <_ s ) .

We will construct coinfinite r. e.

s e t s E and F such that one of these s e t s has the desired degree attempt to meet for each i and j, i

< j,

Let

2.

We

the requirement

0 As a n aid in the construction we will a l s o construct a A 2 s e t D. Since dg(A 8 B) = 21 we may a s s u m e (using the Recursion

he or em) that

we can fix a t the beginning bn index e such that D = {elA'

Let u(k, s)

(For X = D, E o r F , let Xs

denote the use function u(AsQ Bs,e,k, s).

denote the elements in X at the end of stage s.) In order to insure that E < A and E < B we will enumerate a -T -T number x in E at stage s only i f numbers c x a r e enumerated in F o r example, to satisfy the requirement R.: W. 1

a

s

and u < x

1 x,

u = u(k, s).

for some

k

1

n E j $d

both A and B.

suppose x c WiSs

ASQBS {e} (k) = Ds(k) s

such that

,

,

We can now attack this requirement by inserting o r extracting

the attacker k from D according as D (k) = 0 Q r 1. S

Since D = {e}A'B

this forces a number z I u to be enumerated in either A o r B.

If z i s

enumerated in B then the attack is successful and we enumerate x in E satisfying R. forever. 1

If there a r e infinitely many unsuccessful attacks on R

i'

we must insure

that F n W. # jif for all infinite W.. J J

Thus, we do not attack as above unless

there i s t < s such that y c W. b 3,t' t

< y,

and u < y.

Now after the attack,

if z i s enumerated in A, then we can enumerate y in F and argue that

F I T A and F STB. ment P.

1. j

Thus we do not attack R. directly but rather require1

via the attacker k = j.

satisfying P.

1. j

forever so P. 1.

j

An attack on P. succeeds in 1.j i s never attacked again.

Thus, D i s A

0 2

because each j i s inserted in (or extracted from) D finitely often, namely at

. , i < j.

most once for each P.

1. J

Construption. Stage s = 0 . Stage s + 1

.

Do nothing. Requirement P. 1.

j

requires attention if i < j,

P.

1, j

isnotyetsatisfied(i.e.,

E s n W i s s = $ and F s n W .

J*S

=fi,a n d t h e r e

i s a pair (x, y) satisfying: AsCBBs (1) {els (j) = Ds(j)

(3) x

e

Wi,

, and

(4) t h e r e i s some t

(b)

<s

such that:

and

A8B '(j) converges. (c) (As@BS)[u(j.t)l = (At@Bt)[u(j.t)l 8 {el (Note that these conditions imply u(j, s) = u(j, t). ) Choose the least (i, j

>

least corresponding p a i r < x , y Ds+i(j)

such that P.

>.

1, j

requires attention and the

Insert o r extract j from D t o insure

Enumerate A and B until the first number z

# Ds(j)

appears i n AU B.

(If no such z appears, the construction stops.

< u(j, t) How-

ever, i f e i s the index obtained by the recursion theorem satisfying D = {e}A8B,

then z must appear.)

If z appears in A, enumerate y in F.

If z appears i n B, enumerate x i n E. Lemma 1. Proof. -

0

The s e t D is A2

and hence D <

-T

f#.

An integer j i s inserted o r removed from D only when some P.1sJ.

with i < j receives attention, but each P.

1. j

receives attention a t most once,

s o j i s inserted at most finitely often. Lemma 2.

F ITA.

The s e t s E and F a r e coinfinite, E IT A, E

< B, and -T

Proof. and (4).

The

The s e t s a r e coinfinite by the conditions x > 2 i , y > 2j of (2)


reductions follow by the usual permitting method.

(For

example, to A-recursively decide whether x c E , find s such that a

t

> x for

Now x e E iff x c Es .)

t > s.

Lemma 3.

If E i s recursive, then

(i) F i s nonrecursive, and (it) F

sTB. -

Proof of (i).

F i x i such that W. = E. 0 lo

If F i s recursive, choose

the least j > i such that W. = F. Choose s such that for a l l s 0 J 0

2s

0 '

u(js s ) = ~ ( jso) , and (A 8 B)[u] = (AS O BS )[u], where u = u(j, so). Now 0 0 since B i s nonrecursive, t h e r e exist y > u and t > s such that y c W. 0 Jet0 and y i s permitted by B at stage t , i. e . , such that (4)(a), (b) and ( c ) hold f o r y and any s > t.

But since A i s nonrecursive t h e r e exist infinitely many

satisfying (1),(2) and (3) f o r some s > t. Hence requirement P . 0 iosJ receives attention, and e i t h e r E 0 W. # $ o r F n W. # $ contrary to '0 J hypothesis. x c

Wi

Proof of (ii).

B we m a y f i r s t assume that we know

To prove F < 'T

those finitely many y contributed t o F by some P. 1.

j

with j <_ i 0'

F o r each

remaining y, to decide whether y c F we B-recursively find the largest t such that b <_ y. t

Consider a l l j

> i0 such that (4)(a), (4)(b) and the second

clause of (4)(c) hold f o r y and t. F o r e a c h such j find the least s I t such that either: (5)

(As@ BSICu(j.t)l

f

(At@ BtICuO.t)J.

(in which c a s e clause (4)(c) prevents P. for anv i from putting y into F a f t e r 1.j stage s ) , o r

(6)

w.J *s

Fs

fa.

in which case if'y r

E

f o r P.

. then y r

Fs. Note that if (5) fails then (6) must

1.3

hold because a s in part (i) there a r e infinitely many x r W.

1

'and (3) and thus eligible to form a pair with y for P. I,,*

.,

J

satisfying (I() 2 ), .

0

.

s o P.

will receive

lo*J

'attention. Theorem 6.2 ( ~ a c h l a n[4]). degree

(2

52 then there is an

and

,$
{ilB= D

sets A r

f o r some s e t

2 ,i a r e

If

r.e. degrees and such that

r.e. degree

.

5 Br&

$

2

is a

3 22 ,

and indices e, i such that

D r 2 . Let { A ~ )

be re-

and (8,)

cursive enumerations of A and B. As usual define the recursive length function ,

As

l ( s ) = max{x:

(vY < x)[{eIs

Bs

(y) =

At (x) where t = max{v: v

# {e)t

<s

(Y)]1.

At stage s , if l ( s ) 7 k and

F o r each x define an r.e. s e t Cx a s follows. As {e), (x)

{iIs

& l(v) > x), then enumerate in Cx

a l l s t < s.

(Hence, Cx contains a l l those stages t such that the common A B. values y = {el+'(x) = {i) '(x) i s l a t e r replaced by a new common value &

As s y1 = {els (x) = {iIs (x) a t some stage s

> t. Thus. for any s

As A B then {els (x) = { i ) P ( x ) = {e) (x) = {i) (x).) C1earlyCisr.e..

infimum.

0

-T

There a r e r. e. degrees

5 ,&

with no

Hence, the r. e. degrees do not form a lattice.

Proof.

Let 2 ,& be incomparable low r. e. degrees with fiuk = 0'

P'

(obtain these by the Sacks splitting theorem.) b o t h 2 and

if I(s) 7 x

Now set C = O {Cx: x r u).

D < C, and C + A , B .

Corollary 6.3 ( ~ a c h l a n[4]).

/ Cx,

2.

Now

$ =$

Theorem 6.1 produces

since

2' = 2'.

Let

s o the relativization to

5 7 2 , such that 2 < fi

we may assume that & i s r. e.

Thus

2

and

be any degree below

2

.5

<

2

of

and by Lemma 6.2

have no infimum in the

upper semi-lattice of r. e. degrees o r even in the upper semi-lattice of= degrees.

0

Recently, Jockusch has given an easy direct proof of Corollary 6.3. Jockusch has a l s o noted that the method of Theorem 6.1 easily yields the following generalization.

g

Let

= ( 2 :2 i s r. e. and one half of a minimal pair).

N o finite supremum of degrees in

is

2'.

Hence, by applying the Sacks

splitting theorem to K t h e r e is a low degree which is not below any finite supremum of elements of

g.

Lachlan [ 4 ] raised the question of whether incomparable r. e. degrees

2.-b

satisfying

2

u

2 = 2'

-

can e v e r have a n infimum.

Shoenfield and

Soare [ 2 4 ] and independently Lachlan [ 9 ] showed that they could.

Indeed,

Lachlan combined a new method of preserving infimums with the. Sacks preservation method fj 2 to obtain the following pleasing generalization of the Sacks splitting theorem. Theorem 6.4 (Lachlan Splitting Theorem [ 9 ] ) : Let A be a nonrecursive

r. e. set.

T h e r e exist r. e. s e t s

B ,B and C such that 0 1'

(a) C S T A (b) B ~ U =BA ~ and ( c ) Bi,i

CT Bi

8 C

B0nB1=$

, i = 0.1

(d) dg(C) = dg(Bo B) C) A dg(B1 @ C ) .

Thus, f o r any r. e . degree d e g r e e s b = d g ( ~ o BC) -0

a and infimum

U

2

= dg(C).

and

b

-1

2>

there a r e incomparable r. e .

= d g (8 ~ C~ ) which have supremum

7.

THE RENAISSANCE IN CLASSICAL RECURSION THEORY,

THE SACKS DENSITY THEOREM AND BEYOND

This lecture should be numbered Lecture 3.5 since it was delivered a s a special lecture between Lectures 3 and 4 and relies on the material and methods developed in Lecture 3.

The other six lectures were written before

the actual meeting, while this lecture was written after the meeting.

The

main mathematical content of this lecture i s the density theorem a s well a s some recent generalizations and related work.

We begin, however, with

some informal remarks which were inspired during the meeting and were not included in the earlier written text.

The extra references given here a r e

numbered to follow consecutively those of the previous lectures. The term ordinary recursion theory ( ORT)

has occasionally been

used to describe recursion theory on o to distinguish i t from various forms of generalized recursion theory ( G R T )

such a s &-recursion theory,

p-recursion theory, o r recursion on higher types.

The former term i s of

course a misnomer and has now been replaced by the term classical recursion theory (CRT). An analogy between classical recursion theory and classical a r t seems appropriate for these lectures delivered in Italy, the birthplace of the Renaissance.

The a r t and architecture of the Italian Renaissance is

characterized by balance, harmony, simplicity, and a self -contained world on a human scale. In a Renaissance courtyard the human figure feels a t home and in scale with his surroundings, unlike the human figure standing before amonumental a r c h o r a modern skyscraper where it i s dwarfed by the

by the immense size of the architecture.

Likewise, in CRT the universe i s

merely the natural numbers which the human mind can readily grasp, and not some large ordinal o r higher type object.

A Renaissance painting is

characterized by simplicity of line and composition. in the great simplicity of its fundamental notions.

The beauty of CRT lies

F o r example, the notion

of an r. e. s e t a s one which can be effectively listed i s one of the fewnotions i n higher mathematics which can easily be explained to the common man. In a Renaissance painting one i s immediately etruct by the symmetry and balance of the composition.

Not only a r e the background buildings in

s yrnmetry, but the composition of the figures themselves reveals a harmony and balance.

F o r example, in a typical Renaissance painting, the gesture of

each figure c a r r i e s the eye to another figure and on to the next in carefully designed circular patterns, emphasizing a self-contained and harmonious world.

In CRT the most beautiful work i s characterized by a delicate balance

and resolution of conflict between opposing requirements.

The action taken

f o r the negative require Ne must be sufficiently strong to satisfy Ne but not so strong as to prevent the next positive requirement Pe from being satisfied.

(In other words, Ne must not f i l l the whole painting but must lie in

balance on the canvas with Pe. )

This became dramatically apparent in our

development of the infinite injury priority method in

5 3.

Our naive first

attempt to satisfy requirement Ne using the old restraint function r(e, s) would have thrown all the weight in favor of Ne, leaving P , forever unsatisfied.

A proper balance was restored with the "hat trick" of keplacing

r(e, s) everywhere by

?(e, s) which enabled us still to satisfy Ne

(Lemma 3.2 Injury Lemma) but yet to keep Ne in proper proportion

(Lemma 3.3 Window Lemma) s o that P saw a %vindow' through the wall e R(e,s) of restraint functions a t least infinitely often (lim inf R(e,s) s

< a)and

s o P was eventually satisfied. e In contrast, recursion theory on admissible ordinals (&-recursion theory) resembles a Gothic cathedral in that the human figure stands before lofty pointed s p i r e s c a r r y i n g the eye ever higher toward the sky.

This eye

movement is not continuous, however, but proceeds step by step along the regular niches and protrueions of the Gothic lines just a s the a-recursion theorist climbs ordinal by ordinal e v e r toward a.

The Gothic architect

thickened the walls of his cathedrals to the absolute limits in o r d e r t o support still higher spires.

His t h i r s t f o r height remained unquenched, however,

until he l a t e r developed flying b u t t r e s s e s t o support the walls and achieve the dramatically soaring Gothic s p i r e s of the cathedrals a t C h a r t r e s o r Notre Dame.

In developing p-recursion theory the ordinal recursion theorist

abandoned the thick walls of Z -admissibility f o r the flying buttresses of the i fine s t r u c t u r e of L in his passion to c a r r y recursion theory e v e r on into the heavens. Baroque painting of the sixteenth and seventheenth centuries i s analogous to recursion on higher types.

In a typical Baroque painting the human

figures a r e i n a room with a n open window in say the upper right corner. The figures a r e arranged in a strong diagonal pattern from lower left to upper right, pointing toward the open window.

The eye i s c a r r i e d

dramatically by the composition and gestures of the figures toward the open window and a vast unknown space beyond. contained world of the high Renaissance.)

(Gone i s the symmetry and selfAnalogously, i n recursion on

higher types the human mind i s c a r r i e d suddenly to a vast and unknown space whose geography i s only hinted a t by analogy with the human figures (CRT) in the foreground. Returning t o the analogy between Renaissance a r t and CRT, the period of Graeco-Roman antiquity corresponds to the period f o r CRT of 1931-1944 which included the formal definitions of recursive functions, the existence of nonrecursive r. e. s e t s and t h e i r use in G5de11s Incompleteness Theorem.

This period culminated in the remarkable 1944 paper of Post [14]

i n which he discarded the formalism associated with the development of recursive functions i n the 1930's and revealed in a c l e a r informal style the fundamental properties of r. e. sets.

The period of 1945-1955 constitutes the

Dark Ages when relatively little progress was made on Post's problem o r CRT with a few exceptions such a s Dekker [2] and Kleene-Post [35]. The e a r l y Renaissance for CRT is the introduction of the priority method by F r i e d b s r g [3] and Muchnik [13].

The analogous a r t i s t of the

Italian Renaissance i s Giotto, the f i r s t of the great Florentine painters. paintings of Giotto's contemporaries a r e flat and two-dimensional.

The

What a

contrast to Giotto who gives his figures with light and shade and functional line such a massiveness and three-dimensional quality that we believe we could reach out and touch them.

This dramatic beginning of the conversion

of the flat canvas into a three-dimensional world i s equivalent i n CRT t o the priority method of

52

which has been of central importance ever since.

The Florentine school of painting developed over the next two centuries and into the high Renaissance a remarkably accurate mathematical perspective and drawing which not only created a full three-dimensional

space f o r the figures on canvas but endowed them with a vibrant energy a s well.

The Florentines said disparagingly of their rivals in Venice (who were

renowned for their use of color and subtle hues like those of Venice itself) that the Venetians would be good painters if they could only learn t o draw. This supreme emphasis on discipline and drawing corresponds in CRT to the emphasis on h a r d combinatorial arguments and ever more difficult devices to resolve the conflict between opposing requirements.

The

high Renaiesance for CRT which continues to the present day began in the e a r l y 1960's with the introduction of the infinite injury priority method, The most important and

whose development was described in Lecture 3.

beautiful result during this period is the Sacks density theorem which r e mains the Mona Lisa f o r r. e. degrees. However, one difference between Renaissance a r t and CRT i s that the beauty of a painting i s instantly apparent while a difficult proof may stand f o r years before the theorem is sufficiently well understood to admit a simple proof revealing its inherent elegance and beauty.

The following proof [25] of

the density theory combines the simplified infinite injury method of Lecture 3 with the original coding strategy of Sacks [ 2 0 ] . Before giving the proof of the density theorem we must develop a few more properties of the simplified infinite injury method of § 3. can be found in [25]. ) r.e.

Let C be a fixed nonrecursive r. e. set and A an

s e t which we a r e constructing.

Let {A,\

cursive enumerations of AandC' respectively. the enumeration of A as defined in $ 3 . and the injury set

A

(More details

and

{C,} sc

be r e -

Let T be the true stages of

Let the restraint function 3(e, s )

Ie be defined a s in $ 3 but with the length function

A

l ( e , s) replaced everywhere by &(e, s ) defined in 9 3 following Theorem 3.4. The point of

% I

2e,v (AV) established at

is to record a length of agreement between Cv and

6

some stage v for a s long a s the

(A) side. re-

mains unchanged even though a change in the "C sideu at some stage s > v may cause ;(e,s)

The use of

< ;(e,v).

?I

immediately yields by (3. 2) c e r -

tain convenient properties such a s Remark 7.4 below and also

Now virtually the same proofs a s in Lemmas 3.2 and 3.3 establish the following lemmas. Lemma 7.1

(Injury Lemma):

Lemma 7.2

(Window Lemma):

( ~ e n c e ,i f C

#

A

If

C L T I e then C

R(e, s) = max{f.(i, s) : i

Qe(A).

If C f Q ~ ( A then ) limt

Qi(A), f o r a l l i S e, then lim

A

#

A

ttT

~ ( et),

< m,

;(e, t )

< m.

where

.

e) )

In Theorem 3.1 (Thickness Lemma), if we assume

bT

("1'

instead of B piecewise recursive, then the e a r l i e r

=*

A

proof still suffices, because Ie C T A(<el

e em ma 7.1) requires only that

.)

B ( < ~ ) and the Injury Lemma

.

C & T ~ e Indeed, locally we have proved

that, for each e,

Remark 7.3.

Define As

=U{A!)

: i < e) and let Te be the s e t of

true stages in the enumeration { A ~ B(e, s ) = mu{:(i,s)

: i 5 e)

.

Then T

e

)sew

of the r.e.

STA(<

set A(<e).

and if *Ie C_ A

Let then

(V t t T ~ ) ( V 5 X~

(7.2)

( et))[;est(~t;x) ,

= y

* @,(A; x) = y]

and

( v t ~ T ~ ) ( v2st)[&(e,t) 5 &(e,s) & g ( e , t) s f i ( e , s)].

(7.3) Proof. -

A

Since 1 5 A(< e

(7.2) follows a s (3.2) and (7.3) follows

a s (7.1). F o r A and B a s in Lemma 3.1 we have automatically achieved

Remark 7.4. ASTB. Proof. -

-

Fix e and assume that for all i < e we have B effectively

computed g(i) such that and

= 45

g(i)

NOWif x. B ( ~ ) , say x

TO.

x t A ( ~ )just if x t A$)

t

(B).

Then B-effectively compute A(< e )

B r ) , let tt = pt[t 2 s & t t T ~ ] . Hence,

by (7.3).

All these remarks combined yield a proof of Theorem 3.4 (Thickness Lemma

- Strong Form).

Theorem 7.5 ( ~ e n s i t yTheorem

- Sacks [20]):

Given r. e. Bets D and

C such that D < C there exists a n r. e. s e t A such that D < A T T Proof. Define A)!'


Fix recursive enumerations {C a)s t d { D s ) s t o of C and D.

= { ( 0, y ) :y r D,)

s o that D

< A. We shall arrange A -T < C

T '

a s i n Remark 7.4 by finding a C-recursive function g such that A(.)

=@

(C) for a l l e. s(e)

meet, for a l l e

> 0,

To make both inequalities strict, i t suffices to

the requirements Ne: C

f ee(A) and Pe: A # Qe(D).

To meet Pe we attempt to code C into A(e) so that if A = q ( D ) then C contrary to hypothesis.

D

Let A.

=

#.

s) define :(e, s ) a s in $ 2, and define

Given {A,: t

r* D (e, s ) = max{x: ( V y < x)[As(y)

< e,x, s )

Let

denote

<e,x, t) in A *D r < 1 (e,v)

.

<e, <x, s>>

f o r a l l v. t-< v

UsA,

> :x, y e w )

to <e,x, t

a t some stage s t

a r e arranged in a plane.

> at stage t if >t

enumerate L

Cs+i and

.

(We visualize this coding strategy a s follows. {<e, x, y

>i

s ) for a l l 0 < i <_ e, x

Let A =

s.

e .s (Ds, Y)]).

A Q,

F o r t <_ s, and e

> ;(i,

just if <e,x, t)

s ti

=

Fixing e, the elements

A "coding marker" i s assigned

D (e, t) > x.

The m a r k e r i s later removed (forever)

AD i f I (e, s ) 5 x.

If x r Cs+l then a l l elements (e, x, t > ,

I

s, still possessing m a r k e r s a t stage s and not restrained with higher

priority a r e enumerated in A

s+i

'1

To s e e that A succeeds we shall verify the requirements by induction on e ahd simultaneously C-recursively define g(e) such that A(e) = Q g(e)(c)* Clearly, g(0) exists because A(O)

D < C. T

IT

Fix e

> 0 and assume, for

a l l i, 0 < i < e, that

(7.6)

Ag) is recursive, and

(7.7)

A") = @ (C). where g(i) has been C-recursively computed. g(i) Lemma i. Proof.

7

C

#

%(A)

.

Now A(<e) < -T D because A@) i s recursive for O
ie be the injury s e t f o r Ne defined in $3. Thus,

< D < C. Hence, C -T

T

Then

ieEA(<.)

and hence

I

Let

e(<el .~

# Q ~ ( Aby ) the injury Lemma (Lemma7.1).

~

A # Qe,D).

Lemma 2.

By Lemma 1 and the Window Lemma (Lemma 7.2),

Proof. A

$(el = lim inf ~ ( es), where B(e,s) = rnax{:(i,s):o < i e). If A = f (D) then s AD limsI (e,s) = co. Since D i s r.e., we can D-recursively compute the modulus function:

M(X)= (ps)(\d t

F o r x>;((e), x

t

2 s)[:

C just if < e . x , ~ ( x ) ) t A.

t r a r y to hypothesis.

D(e, t) > XI. Hence. C S T A L D con-

T

(Note that M is a D-recursive function because D r. e.

rD implies that M(x) = ( ~ s ) [ l (e, s) > x & D ~ [ u=] D[u]], where

u = m a x ( u ( ~ ~ , e , ys):, y l x ) . ) Lemma 3. Proof.

A ' ~ ) i s recursive.

By Lemn.a 2, let pe = (px)[A (x) #

9;x)].

Then

AD lim inf I (e, s) = p F o r x 2 p given x and t find s 2 t such t t a t s e e AD F o r x < pe fix I ( e , s ) = p e . Then < e , x . t > t A justif < e , x , t ) c A s .

.

A

s1 such that C [ >,= C[pe] and R(e,s) = B(e) for a11 s st

Then < e , x , t > t A justif Lemma 4.

Proof. -

(e,x, t

Then

(e,x,t

< e , x , t > a Av,.

We can C-recursively compute g(e) such that

Define T~ a s in Remark 7.3.

C-recursively compute A then

> / A.

>

t

-

> st. Given t define

A

If x

andhence T t

just i f

C, say x

(e,x,t

t

>

Cs

t

e

F r o m {g(i):i < e) , we

. Fix ( e , x , t ) .

, define

Avt by (7. 3)

1f X / C

(Notice ihat we do not claim here that (7.6) and (7.7) can be combined i s the characteristic to produce a C-recursive function g such that cp g(e) function of A(e) f o r a l l e

> 0 , but merely that A ( ~=) QI

i s that even though A ( ~ )i s recursive for a l l e

> 0,

g(e!

(C).

The point

the proof of Lemma 3

A

above depends upon parameters p and R(e) which cannot be C-recursively e computed uniformly i n e , and the proof of Lemma 4 clearly uses a C-oracle f o r each x.) The abbve coding procedure has many other applications such a s the following.

(The nonuniform version of the theorem follows immediately The uniform version requires an infinite

f r o m t h e Sacks Splitting Theorem.

injury argument and was proved by Yates [ 3 i ] using index sets.) Theorem 7.6 p a c k s - ~ a t e s ) : Given any r. e. s e t C such that

$<

C

<

there exists an r. e. set A such that A i s Turing incom-

$I

Furthermore, an index f o r A can be found uniformly from

parable with C. one for C.

It suffices to meet for all e 2 0 the requirements

Proof. Ne: G

#

Pe: A # .Q),(C).

QIe(A), and

the hypothesis D

Let K = {e: e c

we) 3T 8'

< G of Theorem 7.5 we use the hypothesis C < K so that T

T

C and K play the f o r m e r roles of D and C respectively. {Cs) S E b, *

sc w A

1

For t ( s

C

(e, s ) = m a 4 x: ( V y < x)[As(y) =

sti'

Let A. =

8.

Given

and { A ~ :t C. s) a s usual define

enumerate

all i c e , x c K

. In place of

< e,x,t >

in A

s ti

AC and x < l ( e , v ) . f o r a l l

QI

e ,s

just if v,

(Cs; Y)D.

<e,x,t

t i v i s .

> > 3(i, s )

for

Let A z U S A s .

F.ix e and assume by induction that, for all i < e, C A

{ ei(C), and A ( ~ )i s recursive.

#

@.(A), 1

The proofs above establish (with C a i d X

in place of D and C respectively in Lemmas 2 and 3): C f 4 (A); A # e

Q)

e

(C);

and A(e) i s recursive.

0

An interesting generalization of the density theorem has recently been discovered.

An r. e. degree

Definition 7.7. parable r.e.

2 and

degrees

2

i s branching i f there a r e incom-

2 such that 2 i s the infimum of

and

2

g), and a i s nonbranching otherwise.

(2 = &

2

F o r example, Theor em 4.2 shows that

i s branching, and i t follows

a t once from Theorem 5. i that there a r e nonzero branching degrees. Lachlan [4] showed the existence of nonbranching degrees and recently F e j e r [33] has proved them dense in the r. e. degrees. Theorem 7.8

(Fejer Density Theorem): If

then there i s a nonbranching r. e. degree Corollary 7.9

2

such that

(Fejer Splitting he or em):

degree then there a r e nonbranching degrees

a r e r. e. degrees

B

$< 2 < s

.

If a i s a nonzero r. e.

and 2 such that 2 =

u 2.

It follows that the nonbranching degrees when closed under 1) generate all the r.e. degrees

E

.

nontrivial definable subset of generate

&.

The nonbranching degrees a r e the first

$ to be shown to be dense and hence to

If, as suggested by Jockusch and Lerman, the r. e. degrees

a r e given the order topology where a typical subasic open set has the form

[Q,s)=

{Q : b_< g) o r

.

(a,%I]= {h :k> a ) ,

then the branching degrees,

together with

a r e precisely the isolated points.

Theorem 7.8 that the Cantor-Bendixson rank of

It follows f r o m with this topology i s i.

The Sacks coding strategy of the density theorem has recently been applied to give a new result on index s e t s which immediately yields several index s e t results of Yates [25] and [26] a s pointed out in [34].

The following

result was f i r s t stated by Kallibekov [34] with an entirely different method of Unfortunately, the verification that his ingenious

proof f o r infinite injury.

method succeeds s e e m s t o contain an e r r o r .

Stob [36] gave a completely

different proof -by combining the Sacks coding strategy with the thickness lemma method of $3. Let C and D be r.e.

Theorem 7.10.

9 CTC

CT

9'.

D

Let S be any set in Cg.

s e t s such that C I D and

T

Then there i s a uniformly r. e.

sequence of r. e. s e t s { A ~ ) such ~ ~ that f o r a l l k,

k c S *Ak==D k

/S

d A

k

and C a r e Turing incomparable.

In Stob's proof t h e r e i s a new element beyond what we have seen h e r e and in [ 2 5 ] , since if k e S we may have f o r certain e that lim infs r ( e , s )

= m.

Stob needed a new insight in this c a s e to show that Ak l T D . In a different direction another conjecture on r. e. degrees has

recently been refuted.

After seeing the density theorem, Shoenfield

formulated a conjecture that the r.e.

degrees

form a dense structure a s

a partially o r d e r e d s e t analogously a s the rationals f o r m a dense structure a s a linearly o r d e r e d set. q(xi,.

.., n ,y) x

Shoenfieldts conjecture implies that if

i s a quantifier f r e e formula of the language L(&

U,0 , i )

and

.

q,. .,

e

then there exists

2t

R mch that

(p

(gi,. ..,sn,b)

holds

unless the existence of ]Z would lead to an winconsistency.m. The existence of a minimal pair (Theorem 4.2) refutes Shoenfieldfs conjecture.

Jockusch

then suggested that since minimal pairs a r e the source of much difficulty, perhaps Shoenfield's conjecture holds for the simple degrees, those nonzero degrees which a r e not half of a minmal pair.

Recently, Klaus Ambos has

refuted this by combining the Yates construction [30] of a simple degree with the Lachlan construction of a nonzero branching degree simple branching degree.

( 3 5) to produce a

(Since simple degrees a r e closed upwardly, the

branches must also be .simple. )

REFERENCES 1.

M. R. Arslanov, R.F. Nadirov and V. D. Solovev, Completeness c r i t e r i a for recursively enumerable s e t s and some general theorems on fixed points, Matematica University News, 1977 (179) No.4.

2.

J. Dekker, A theorem on hypersimple s e t s , Proc. Amer.Math.Soc.,

-5 (1954). 3.

791-796.

R. M. Friedberg,

Two recursively enumerable s e t s of incomparable

degrees of unsolvability, P r o c . Natl. Acad. Sciences, U. S. A. 43 (1957), 4.

A.H.

236-238.

Lachlan,

MR 18, 867.

Lower bounds for p a i r s of r . e . degrees, P r o c . London

-

Math. Soc. (3). 16 (1966). 537-569.

5.

,

MR 34 no.4126.

, Complete recursively enumerable s e t s , Proc. Amer. Math SOC. 2 (1968). 99-102.

6.

, Embedding nondistributive lattices in the recursively enumerable d e g r e e s , Conf. Mathematical Logic, London 1970, Lecture Notes in Math., no. 255, Springer-Verlag, Berlin and New York, 1972, pp. 149-177.

,

7.

The priority method f o r the construction of recursively

enumerable s e t s , P r o c . Cambridge Summer School in Logic, 1971, Springer-Verlag, Berlin & New York, 1973. 8.

,

A recursively enumerable degree which wid1 not split over

a l l l e s s e r ones, Ann. Math. Logic, 9.

,

9 (1975), 307-365.

Decomposition of recursively enumerable degrees, Proc.

Amer. Math. Soc., t o appear. 10.

, and R.I. Soare, Not e v e r y finite lattice i s embeddable in the recursively enumerable degrees,

to appear.

11. S. S. Marchenkov,

A c l a s s of partial sets. Mathematicheskie Zametki

Vol. 20, No. 4 (1976), 473-478. 12.

D. A. Martin.

Completeness, the recursion theorem, and effectively

simple s e t s , 13.

A. A. Muchnik,

Proc. Amer. Math. Soc..

(1966). 838-842.

On the unsolvability of the problem of reducibility in

the theory of algorithms (Russian), Doklady Akademii Nauk,

n. s., 14.

108 (1956).

194-197.

E.L. Poet, Recursively enumerable s e t s of positive integers and their decision problems,

Bull. Amer. Math. Soc.,

0 (1944),

284- 3 16. 15.

R. W. Robinson, Interpolation and embedding in the recursively enumerable degrees,

16.

,

, 2 (1971), 285-314.

Jump r e s t r i c t e d interpolation in the r. e. degrees, Annals

2 (1971).

Math., 17.

Annals Math.

586-596.

H. Rogers, Jr. , Theory of recursive functions and effective computability,

18. G. E. Sacks,

McGraw-Hill, N. Y.,

1967.

On the d e g r e e s l e s s than 08,

Annals Math.,

71 (1963),

211-231. 19.

,

Recursive enumerability and the jump operator, T rans.

Amer. Math. Soc., 20.

, The

108

(1963). 223-239.

recursively enumerable degrees a r e dense, Annals of

-

Math. (2). 80 (1964). 300-312. 2 1.

,

Degrees of unsolvability, rev. ed., Annals of Math.

Studies, No. 55, Princeton Univ. P r e s s , Princeton, N. J. 1966. 22.

J. R. Shoenfield, Undecidable and creative theories, Fundamenta

-

Mathematicae, 49 (1961), 171-179. 23.

, 1971.

Degrees of unsolvability,

North-Holland, Amsterdam

J. R. Shoenfield and R. I. Soare,

The generalized diamond theorem,

( ~ b s t r a c t )Recursive function theory newsletter, 19 C1978), # 2 1 9 The infinite injury priority method, J.Symbolic Logic,

R.I. Soare,

41 (1976). -

513-530.

Computational complexity, speedable and levelable s e t s ,

,

J. Symbolic Logic,

,

42 (1 977),

545-563.

Recursively enumerable s e t s and degrees, Bull. A.M. S.

Vol. 84, No.6 (1978), 1149-1181.

,

Recursively enumerable s e t s and degrees, (Omega Series),

Springer-Verlag, Berlin and New York (to appear). S. K. Thomason,

Sublattices of the recursively enumerable degrees,

2. Math. Logik und Grundlagen d. Math., C.E. M. Yates,

A minimal pair of r.e.

31 (1966), -

,

17 (1971).

273-280.

degrees, J. Symbolic Logic,

159-168.

On the degrees of index s e t s , Trans. Amer. Math. Soc.,

-

121, (1966), 309-328.

,

On the degrees of index s e t s , 11. Trans. Amer. Math. Soc.

135, (1969). -

P. A. F e j e r ,

Ph.D.

249-266. Dissertation, University of Chicago, 1980.

See

Density of nonbranching degrees (abstract 7 9 ~ - E 3 2 )Notices A. M.S.

S. Kallibekov,

June 1979, p. A-390. Index s e t s of degrees of unsolvability, Algebra i Logika

10 (1971), 316-326 (Russian). S. C. Kleene and E. L. Post,

The upper semi-lattice of degrees of

recursive unsolvability, Ann. of Math. (2) 59 (1954), 379-407. 36.

M. Stob,

Index s e t s and degrees of unsolvability,

t o appear.

CEK TRO I h TERNAZIOh ALE MATEMATICO E S T I V O (c.I.M.E.)

RECURSIVELY I N V A R I A N T

-RECURSION

WOLFGANG MAAS

THEORY

Recursively Invariant (3-Recursion Theory (Preliminary Survey)

Wolfgang Maass Massachusetts Institute of Technology, Cambridge, USA

I

In my lecture I want to sketch a new branch of generalized recursion theory: invariant (I-recursion theory. (3 is any limit ordinal in the following.

A set A E Lp is called (3-recursively enumerable ( (3-r.e. ) if it is definable over LC) by some TIformula q

(see

Friedman and Sacks [I]). Observe that this is really a very intuitive definition. Generate successively the levels L0,LI,..,L8 to (3

,..

(rd(3)

of the constructible hierarchy up

. Enumerate at every step

which satisfy Ll k q(z)

y

those elements z into A

and which have not already been

enumerated before. The example shows that the general concept of a recursively enumerable set -as described by Post 121 in 1944- does not require any strong closure conditions of the underlying domain like admissiblity. *The author is supported by the Heisenberg-program of the Deutsche Forschungsgemeinschaft.

A function f : Lp* Llj is called (3-recursive if its graph is (3-r.e.. Consider the group of all p-recursive functions which map Lp one-one onto Lp

together with composition of maps. A

property of subsets of Lp is called G-invariant sively invariant if for every f e G some set B this property if and only if f[B]

or recur-

ti Lp

has

has it.

Felix Klein suggested in his Erlanger Programm (1872) to define branches of mathematics in terms of a space X and a group G of transformations acting on that space., The branch of mathematics determined by

X and G is the study of G-invar-

iant properties. Lp

and the previously defined group G determine for

(3= w classical recursion theory and for

(3 =

(a admissible)

OL

a-recursion theory. Let us now look whether there is an appropriate notion of finiteness in invariant @-recursion theory. Any recursively invariant class of ()-recursive bounded (i .e. y e (3 )

S

P for some

L

subsets of Lo is a candidate for such a notion. It is

obvious that there exiets a largest such class which we call I. We will see in the following that there are several good reasons to take I as the notion of finiteness in invariant (3-recursion theory. The elements of I are called i-finite sets. If (3 is an admissible ordinal u IY

then i-finite is equivalent to

-finite. Define wlcf() :=

(3-recursive f : 6 -B

the least 6 + (3 (there exists some

0

with range unbounded in

0

).

. In fact

Lemma 1 : I is a (3-recursive subset of L,,

I = f x E Lp 1 Lo b [cardinality(x) < vlcfp ] f

.

The proof is not difficult but relies heavily on the fine structure of L

(collapsing of Skolem hulls).

Every (3-r.e. set A

can be enumerated in

clcfp many

steps, i.e. there exists a (l-recursive function f : ulcfp L

y x ~ A y , such that A = U{Ay

every single x e A

I x *

elcfp)

. Thus

is enumerated after an i-finite number of

steps. It is easy to see that I is the only recursively invariant class of @-recursive bounded subsets of

Lo which

is in this sense coherent with the notion of a 0-r.e. set. Further for any i-finite subset K for some i-finite

of A

we have K 5 Ar

. This property is important for priority

constructions. It implies that every true i-finite neighborhood condition about A settles down at some point of the-construction. Another useful property is the following : Every (3-recursive subset of an i-finite set is again i-finite.

Consider for any limit ordinal p the structure ePP :=


- I;I,F,T>

where

Z

e := c P Lo

*

the canonical (3-recursive truth predicate for

.

I

and T is

AO Lo

formulas

dlo is construed as a set with urelements as in L o is the underlying collection in Barwise [3], where LP I

-

of urelements.

Theorem 2 :

&e

is an admissible structure with urelements.

Foreveryset M c L p

M

Z1(4) zR

is

:

=

M

is

Further the sets in the structure

Z l ( ~ ,Q ) P

.

P are exactly the i-finite

sets. Corollary 3 : Let

c,

p-r.e.

Assume that (3

is a countable limit ordinal.

Lo be some 9-recursive language and let T be a

set of sentences in the language

*

with i-finite dis-

junctions and conjunctions.

If every i-finite set To G T has a model, then T has a model. Proof of the Corollary : Apply the Barwise Compactness Theorem

[?I

to C L p .

Remark : The compactness theorem does not hold for any larger notion of Itfinitenin L p

.

The preceding compactness theorem (Corollary 3) can be used to show that for every countable (3

invariant p-recursion

theory can be characterized in terms of absoluteness or model theoretic invariance as this effect was called by Kreisel [ 4 ] . The concept of model theoretic invariance is useful in order to understand the mathematical meaning of computations in recursion theory. The situation is analogous as in first order logic where the completeness theorem gives a mathematical meaning to formal proofs. The connection between model theoretic invariance and recursive invariance is the following : The notion of a "finiten

set is recursively invariant in every recursion theory which can be characterized in terms of model theoretic invariance.

In order to get an intrinsic notion of a computation relative to an oracle B

q

Lp one can extend the Kripke equation

calculus in a canonical way. The essential rule allows to survey i-finite many subcomputations in a computation. Every computation has the structure of an i-finitely branching tree :

axioms of the form 2

L & B , F ( E , ~ )=

C

B

,

where F

is some rudimentary function We say that A tic function of A

is computable from B

if the characteris-

can be computed from B

in this equation

calculus. We say that A

is i-finitely computable from B

this can be done by using i-finite computations only.

B

if is

called semigeneric if every equation which can be computed from

B can be computed from B with an i-finite computation.For a semigeneric set B

the preceding two notions of reducibility

coincide for every set A

.

Lemma 4 : a)

For countable (3

A

is computable from

implicitly invariantly definable from B b)

A

0-r.e.

B iff A is

(see [ 6 ] ) .

is i-finitely computable from B iff there exists a set W

such that for every x e Lo

:

cA(x) = i

3 i-finite,K,H(< X , ~ , K , B I E W

O

K

A

B

4

H

A

5

Lp

- B)

(cA is the characteristic function of A ). c) B is semigeneric iff for every relation R s L p x L P the form R(X,Y)

3 i-finite K,H(

H

with W (3-r.e.

x,g,K,H, e W

and dom R

i-finite function f

G

Ks B

A

i-finite

A

of

H o L B- B)

there exists an

.

R with &om f = dom R

The relation in b) is not transitive and therefore we codsider instead the following reducibility relation : A Sf B

:

<=>

there exists a (3-r.e. set W

such that for

all i-finite H,,H2

3i-finite K,H(
HI S A

i

W

A

K

5

B

A

H c Lp

- B)

and HZ

5 La

-B

U

3 i-finite K,H(< H2.2,K,H>rW

r

KGB

A

H sLp

- B).

The associated equivalence classes are called i-degrees. For admissible

Q

they coincide with the d-degrees.

Every i-degree i a recursively invariant. The i-degree 0 (i.e. the equivalence class of the empty set) contains exactly the fl-recursive eets. As usual one gets immediately that there exists a maximal (3-r.e. i-degree 0' which is strictly greater than 0

. There is no trivial way to show the existence of an

intermediate

(3

-r. e. i-degree.

Except for a few

(where it is still open) one can

understand the structure of the (9-r.e. (+degrees (see[l]) substructure of the I-degrees.

as a

Theorem 5 : sets A

,B

The proof

For every limit ordinal such that A #i B

there exist p-r.e.

and B $i A

.

(3 is

is given in the most interesting case where

strongly inadmissible (i.e. u fcfp c (3*) by a priority construction following Friedman [ 5 ] . The combinatorial principle 0 can here be eleminated (this may

be helpful for applications to

other inadmissible sets). f3 the i-degrees coincide with the

Oberelcve that for every

degrees in the admissible collapse

a O Thus . Theorem

5

con-

tains as a special case the solution of Post's Problem for some enormously fat admissible sets.

Theorem 6 : For every limit ordinal (\ there exist sets A

,B

such that A

is not computable from A

is not computable from B

.

p-r.e. and B

The proof is slightly more difficult than the proof of Theorem

5

. We make

neede

A

and B in addition semigeneric. For this one

0.

Theorem 7 : For many strongly inadmissible (3 there are (3-r.e. sets A

such that 0 ci A but S

pi A

for every simple set

S (see [6] for the definition of simple). The proof

is a first example of an infinite preservation stra-

tegy in the strongly inadmissible case. Besides new combinatorial argument

0 it uses a

. We expect that refinements of the

applied strategy will lead to a splitting theorem for i-degrees.

We had mentioned the definition of a semigeneric set because at this point an important new effect arises in the step from al-

to (3-recursion theory. Several equivalent definitions

of nhyperregularninot-recurslon theopy lead to different classes in (I-recursion theory. For some strongly Inadmissible (3 there are C)-r.e. sets B such that every computation from B has an i-finite length but B is not eemigeneric. All details can be found in the forthcoming paper [6].

Literature

[I]

S.D.Friedman and G.E.Sacks, Inadmissible recursion theory, Bul1.Am.Math.Soc.

83 (1977) 255-256

[2] E.L.Post, Recursively enumerable sets of positive integers and their decision problems, Bu1l.Am.Math.Soc.

50 (1944)

284-3 1 6 (31 J.Barwise, Admissible Sets and Structures, (Springer,Berlin, 1975)

[4] G.Kreir~e1, Model theoretic invariants: applications to reoursive and hyperarithmetic operations, in: J.W.Addison,

.

L.Henkin,, A.Tarsk1, eds ,The Theory of Models (~orth Holland ,Amsterdam,1965)

f 51 S.D.Friedman, Post's problem without admissibility, to appear (63 W .Maass, Recureively invariant (3-recursion theory, to appear

.