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Annals of Mathematics Studies Number 5 5
ANNALS OF MATHEMATICS STUDIES Edited by Robert C. Gunning, John C. Moore, and Marston Morse 1.
Algebraic Theory of Numbers, by He r m a n n W eyl
3. Consistency of the Continuum Hypothesis, by K urt G odel 11.
Introduction to Nonlinear Mechanics, by N. K r y l o ff and N. B o g o liu bo ff
20. Contributions to the Theory of Nonlinear Oscillations, Vol. I, edited by S . L efsc h etz 2 1.
Functional Operators, Vol. 1 , by J ohn
24.
Contributions to the Theory of Games, Vol. I, edited by H. W. K uh n and A. W. T u c k e r
25.
Contributions to Fourier Analysis, edited by A. Zy g m u n d , W. T r a n su e , M. M o r se , A. P. C a ld e ro n , and S. B ochner
27.
Isoperimetric Inequalities in Mathematical Physics, by G. P o l y a and G. S zego
vo n
Neum ann
28. Contributions to the Theory of Games, Vol. II, edited by H. W. K uh n and A. W.T u c k e r 30. Contributions to the Theory of Riemann Surfaces, edited by L. A h l f o r s et al. 33. Contributions to the Theory of Partial Differential Equations, edited by L. B e r s ,S. B o ch ner , and F. J ohn 34.
Automata Studies, edited by C. E. S h an n o n and J. M c C a r t h y
38.
Linear Inequalities and Related Systems, edited by H. W. K uhn and A. W. T u c k e r
39. Contributions to the Theory of Games, Vol. Ill, edited by M. D re sh er , A. W. T u c k e r and P. W o l f e 40.
Contributions to the Theory of Games, Vol. IV, edited by R. D u n c a n L u c e and A. W .
41.
Contributions to the Theory of Nonlinear Oscillations, Vol. IV, edited by S. L e f sc h e t z
42.
Lectures on Fourier Integrals, by S. B o chn er
43.
Ramification Theoretic Methods in Algebraic Geometry, by S. A b h y a n k a r
T u c k er
44.
Stationary Processes and Prediction Theory, by H. F u rste n be rg
45.
Contributions to the Theory of Nonlinear Oscillations, Vol. V, edited by L. C e s a r i , J. L a S a l l e , and S. L e fsc h etz
46.
Seminar on Transformation Groups, by A. B orel et al.
47.
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48.
Lectures on Modular Forms, by R. C. G u n n in g
49.
Composition Methods in Homotopy Groups o f Spheres, by H. T o d a
50.
Cohomology Operations, lectures by N. E. S t e e n r o d , written and revised by D. B. A.
51.
Morse Theory, by J. W. M il n o r
52.
Advances in Game Theory, edited by M. D resh er , L. S h a p l e y , and A. W . T u c ker
E pstein
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54.
Elementary Differential Topology, by J. R. M unkres
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Degrees of Unsolvability, by G. E. S a c k s
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Knot Groups, by L. P. N e u w ir t h
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Topology Seminar, Wisconsin, 1 9 6 5 , edited by R. H. B in g and R. J. B e a n
DEGREES OF UNSOLVABILITY BY
Gerald E. Sacks
SECOND E D I T I O N
PRINCETON, NEW JERSEY PRINCETON UNIVERSITY PRESS 1966
Copyright © 1963,1966 by Princeton University Press All Rights Reserved L. C. Card 63-9996 1966 1993
SECOND EDITION REPRINTED
Printed in the United States o f America Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, Chichester, West Sussex Printed in the United States of America
Princeton University Press books are printed on acid-free paper and meet the guidelines for permanence and durability o f the Committee on Production Guidelines for Book Longevity of the Council on Library Resources
This Book Is For CLIFFORD SPECTOR, 1930-1961 'HS£> 7] i\ov fivr/fir] TtdvrjKoros EPICURUS
PREFACE TO REVISED EDITION It is intended that Section 1 be read before any section that follows it; the same applies to Sections b and 8 . There is a continuous commentary on the priority method which starts in Section b and ends in Section 10. We wish to thank the many persons who helped up prepare and re vise this monograph at one stage or another. Among them were R. Friedberg, S. C. KLeene, A. Nerode, A. Robinson, H. Rogers, Jr., J. Rosenstein, J. R. Shoenfield, R. Smullyan, and C. Spector. Special thanks are owed to J. Barkley Rosser. We also thank the National Science Foundation and the Army Research Office (Durham) for financial support. Many results on degrees were obtained after the appearance of the first printing of this monograph. We discuss some of them in Section 1 2 . Nonetheless, we still feel that the subject of degrees is far from fin ished.
Many of the weapons developed to attack degrees are now making
inspiring appearances on the other battlefields of mathematical logic, particularly set theory. We hope that what follows will hasten the final victory.
Ithaca, New York April 1, 1 9 6 6
CONTENTS
§1. §2. §3. §4. §5. §6. §7. §8. §9. §10.
P r e lim in a rie s ...................................................................................................... ... a Continuum of M utually Incomparable D egrees....................................... Uncountable Suborderings of D e g re e s ...................................................... The P r io rity Method of Friedberg and Muchnik........................................ , An E xistence Theorem fo r R ecursively Enumerable Degrees . . . The Jump O p e ra to r.............................................................................................. An In te rp o la tio n Theorem fo r R ecursively Enumerable Degrees . Minimal Upper Bounds fo r Sequences of D e g re e s ................................ Minimal D e g re e s................................................................................................... M easure-the ore t i c , Category and D escrip tiv e S e t-th e o re tic A rgum ents...................................................................................................... §11. I n i t i a l segnents of D e g re e s......................................................................... § 1 2 . F u rth er R esults and C o n je c tu r e s ............................................................
55 77 117 123 135
BIBLIOGRAPHY........................................................................................................................
173
ix
1
7 21
*3
153 163 1 69
DEGREES OF IMSOLVABILITY
DEGREES OF UN SOLVABILITY
§1.
Preliminaries
The natural numbers are 0 ,1 ,2 ,...;
let f
and
g
from the natural numbers into the natural numbers. We say have the same degree of recursive unsolvability if f and
g
is recursive in
solvability. f
f
by
f;
g
f
and
Let
and
g
is recursive in
g
is the set of all functions is recursive in
g
U
£ 0.
There are only three cases at stage CASE 1.
such that
t
such that
n > o and i < 2;
let
di(p|+n)
t( j)
It
has not het been
be the least such
t.
We
define di(Pj(J)+n) for all
n > o
and
CASE 2.
-
V
n)
i < 2.
(s)Q = 0
and
(s)3 = 1.
Let
(s)1 = e
and
(s)2 = f.
The purpose of this case it to insure that any function recursive in with Godel number
e
and recursive in
recursive in one of the
bi,s.
d1
with Godel number
i < 2,
For each
partial function whose domain consists of all defined prior to For each
i < 2,the notion
extensions of of and
b
stages
d?.
If
b
and whose
is a finite extension of
of
d® and
is defined and
g
of
that
d^(n)
d^,
either
proceed directly to stage
s + 1.
n
d^(n)
has been
d^(n) = di(n) .
d^,
n such that
dQ
will be
the unique
of weight well-orders the set of all
CASE 2.1 . There exists an b
such
is not defined and also of the values of
for which b(n)
be
values are given by
is an effective encoding of those
d^(n)
n
let d|
f
finite
then the weight
b(n) b(n)
is defined for those
n
is not. such that for any finite extensions
(e}t>(n)
or
(f)g(n)
isundefined.
We
18
DEGREES OP UNSOLVABILITT CASE 2.2.
Case 2.1 does not hold, and there exist an
finite extensions,
b
and
c,
of
(e)b (n) are defined but are not equal.
n
and
d3 such that and
(e}c(n)
Among all such triads
(n, b, c)
there is
a unique one with the property that 2n . 3weight of b . ^weight of c has
the least possible value; let it be denoted by
Case
2.1 does nothold, there exists afinite extension
that
(e}s (ng) isdefined; let
be that member of
{bs, c3}
gs
We define
dQ(n),
for each d3(n)
n
d.,(n)
for which
ly to stage
d3(n)
hs(n),
gs(n)
.
hs(n),
gs(n)
respectively,
respectively, is defined and
Neither Case 2.1 nor Case 2.2 holds.
s is otherwise.
d3(n),
We proceed direct
We define
Let
r a be the least n such that s vg be the least n such that d3(n) is un
= d^ vs^ 55 °*
That completes the construction. fined everywhere.
By Case 1, each
We show that the degrees of
dQ
By Case 3,
dQ and
is recursive in both
and
d1
d1 dQ
and
f
bi
Case 2
d 1.
have no greatest lower bound. dQ
whose degree is greater than the degree of
Let
and w.
Let
be Godel numbers such that do di » = (e) 0 = {f} 1
s
are de
and
be a function whose degree is a lower bound on the degrees of
d1. We will find a
n
hs
of least weight such that
respectively,to be
is undefined, and let
defined.
Let
such
s + 1.
CASE 3.
an
ofd3
respectively, is not defined. CASE 2.3.
e
g
be the one ofleast weight.Let
(e}hS(ns) * {f)gS(ng)
w
(ng, bs, cs) . Since
be such that
(s)0 * o,
holds at stage
s.
(s)3 = i,
.
(s)., = e
and (s)2 - f.
Thus
If Case 2.1 holds at stage s, then there exists
such that either (e)
is undefined.
do
(n)
or
{f}
di (n)
It follows that Case 2.1 does not hold at stage
s.
If
§2.
19
A CONTINUUM OF MUTUALLY INCOMPARABLE DEGREES
Case 2.2 holds, then there exists an w(n) = {e}
dn
n
(namely,
such that
di (n) ^ Cf) (n) = w(n)
It follows that Case 2.3 holds at stage (i)
ng)
s.
there is a finite extension
Thus for each b
of
n, we have:
d®such that
(e)b (n)
is defined; (ii)
if
b
and
and
c
(e}c(n)
We will use (i) in
bs. Let
Dg
are finite extensions of are both defined, then
and
(ii)
be the domain of
to show that
Dg
|2J
0j-
is a finite extension of
&
where
(e}b (n)
It also follows from (L2) that the values of d® are
effectively computable from j < s.
such that
d®. Then
Dg = F U where
d^
d^
be a function such that for each
Xn|B(i, n) = bi . It is easily checked that xin|A(i, n) . We saw above that the degree of
Xin|B(i, n) Xin|A(i, n)
are at i,
is recursive in was at most
£ T.
20
DEGREES OF UNSOLVABILITY
If we can show that both xin|B(i, n)
dQ
and
d1
are recursive in a composition of
and a predicate of degree
the degrees of
dQ
and
d1
< o",
are at most
then it will be clear that
o".
stage
s
of the simultaneous definition of
then
t(j)
is obtained effectively from
Consider what happens at
dQ
and
d^ and
d1.
d®,
If Case i holds,
and
are finitely extended with the help ofxin|B(i, n).
d^
and
d^
Suppose Case 2 holds.
We can tell which of the three cases, 2.1, 2.2, or 2.3, holds by composing xinlB(i, n)
and a predicate of degree
£ n.
(We do not exhibit the predi
cate since it is similar in nature to the one-quantifier form occurring in our argument that
xin|A(i, n)
has degree at most
£ ’.)
With the help of
this predicate we can single out the desired extensions of If Case 3 holds, we extend explicitly define
d^*1
d^
and
and d^ d^+1 in
of a predicate of degree at most
o"
d®
in a trivial fashion. terms of
d®
and
and the function
d^
d^.
Thus we can with the aid
xin|B(i, n) .
follows with the help of the closing remarks of Section 1 that are of degree at most
and
dQ
It
and
d1
o".
Each of the theorems of the present section was proved by means of the diagonal method.
Each of the above constructions amounted to a defini
tion of a function by induction; the function had to meet countably many requirements, and stage requirement.
s
of the induction was devoted to meeting the
3th
Godel !s construction of an undecidable, arithmetical predi
cate and Kleene!s construction of a non-recursive, one-quantifier form made similar use of the diagonal method.
In Sections ^-7, 9 and 11 we will
see that the diagonal method lacks the power needed to obtain results about degrees deeper than those of Sections 2 and 3.
§ 3 . UNCOUNTABLE SUBORDERINGS OF DEGREES
in urns sec Lion we sLuay conai Lions wmcn are suiricienL ana, in some cases, necessary for a p a rtia lly ordered set to be imbeddable in the upper sem i-lattice of degrees. Two p a rtia lly ordered sets, M and M1, are called order-isomorphic i f there is a map m-^mT of M onto M’ such th at m < n i f and only i f m' < n*. A p a rtia lly ordered set P is said to be imbeddable in a p a rtia lly ordered set Q if P is order-iscmorphic to some subset of Q. Let T be a p a rtia lly ordered set of cardinality at most th at of the continuum such that each member of T has a t most alephone successors; we show T is imbeddable in the degrees i f and only i f each member of T has a t most countably many predecessors. Let A and B be sets of degrees such that A is countable, B has cardinality less than th at of the continuum, and no member of B is less than or equal to any fin ite union of members of A; we show there exists a degree d such that d is greater than every member of A and incomparable with every member of B. Finally, we show that i f T is a p a rtia lly ordered set of cardinality at most that of the continuum with the property th at each member of T has a t most fin ite ly many predecessors, then T is imbeddable in the degrees. We assume complete fam iliarity with the arguments of Section 2. THEOREM 1 . Let T be ap a rtia lly ordered set, and le t M and Nbe d isjo in t subsets of T such th at M has cardinality less than th at of the continuum, N is countable, and no memberof N is less than any member of M. For each n € N, the set Mn = {m|m e M& m 0,
will denote the unique member of
(B^)
recursive in
v =
if and only if for each
2v(k) - 1 < v(k+i) < 2v(k).
whose index is
(B^)
will have cardinality of the con
be the set of all sequences of natural numbers such that
(v(k)|k = o, 1, 2, ...}
if
a^.
< n^j
of functions will be extracted from a sequence
of sets of functions.
B^
will constitute
must meet five sets of requirements: if n^ < n^j
the members of
be such
be a function of degree
b^
1 < v(k) < 2k
M
if and only if
2, •••)
at
bj recursive in
The sequence
S
let
and
{b^li = 0, 1, 2 , ...}
fbjji = o>
t c V,
N = {i^li = 0,
A
at < a^
(R1)
(R5)
Bi
V,
We will obtain a sequence of functions
Let
and
and let the given order-isomorphism between
that for each m^
o
and
Bi^m ^
B^(m)
B^
if
r € R^,
*
During the course of the construc
is defined if and only if for sane
Note that if
' then
will be a member of
such that for each
f^ = % n v ( n ) (r)
tion we will say
23
n,
Qj_nv(n)
is defined by and and hence equal to
ls deflned for a11
v c S
with the property that
w(n) = v(n). Since for each
c N,
is countable, we write
■ 0, ^ For each of
A
and
i
and
j,
let
corresponding to
a^
m^j
2, ...} 55 {tit < ^
.
be a function whose degree is that member under the given order-iscmorphism between
M. The construction of the functions
definition by induction.
& t € m|
At stage
s
{Q^ ^ }
A
takes the form of a
of the construction, either or both
of two kinds of activity may take place: (1) for
some
i
and
their common set of arguments (2)
n the functions
R^
and
irrevocable commitments are made concerning the values to be
taken by some of the functions them.
CfynV l1 < k < 2n}
are defined;
when it becomes time to define
All commitments will be capable of expression by one of two utter
ances: (a)
for all
v,
as soon as one of
Bj(m)
and
B^(pjI+I^ 1^ )
is assigned the value it must take when it is defined, the other must be assigned the same value and must take that value when it is defined; (b)
if
then for all
6 • p“+t(iJ) iS put in k
such that
assigned the value
a ^ (m)
R^
1 < k < 2n ,
when
is defined,
Qink(6*p“+t^i^ )must be
and must take that value when it is
defined. All commitments will be honored at the earliest possible stage. describe the status of
p“+r(13),
6 • p1^ ^ )
We will
respectively in utterance
DEGREES OP UNSOLVABILITY
2k
(a), (b) respectively, by saying that
p^l+r^i^ ,
6•
respectively,
has been committedto a reserved, closed respectively,classification in the ith partition.
We will
say that B^(m)
has received a
value if it has been
defined or if it has been assigned a value it must take when it is defined. If a commitment of type (a) has been made at stage the first one of
Bj(m)
value of the other.
and
B^(p“+r^i^ )
s,
then we will say
to receive a value induces the
Before we proceed with stage
s
of the construction,
we must state an induction hypothesis concerning what has happened prior to stage
s: (Hi)
Only a finite number of the sets
been defined. has been
For each
defined for all
such that
and k o
and
and
j : if
j
p1^,
positive integer ber of the form
p^,
6 • p1^
respectively,
is restricted to a finite range of values. 6 • p®
r(ij), p“
For each
respectively, has been committed to a reserved,
closed respectively, classification in the t(ij)
i^*1 partition, then there is a
respectively, such that every natural num
(m > r(ij)),
6 • p1^
(m > t(ij))
respectively, has
been committed similarly and such that no natural number of the form (m < r(ij)),
6 • p: “
furthermore, for each
(m < t(ij)) v € S
and natural number
m,
Bj(m)
of type (a) and if
been put in
then
R ^ prior to stage
s,
= for all
k
such that
(H3)
For each
p“
respectively, has been committed similarly:
are joined together by a commitment
6 •
g
was committed to a reserved classification in
classification in some partition are of the form where
Rik
(i, j, m)
has been committed to a reserved or closed classification in the tition.
have
has been defined, then
1 < g < 2n . There are only finitely many triples
such that for some the
i
{R^^li > 0 , n > o)
and
B^(pjl+r^i^
6 • p^+b(ij)
a1;)(m)
1 < k < 2n . i
and
j,
a natural number of the form
p®,
respectively, has been committed to a reserved, closed respectively,
§3.
classification in the 22 •3J • 51
< s
25
UNCOUNTABLE SUBORDERINGS OP DEGREES ith partition only if
2 • 3^ •
< s
and
j ^ i,
respectively.
At stage
s
there are five possible cases.
Cases 1-3 correspond
to (R1 ) - (R3). CASE 1.
s= 2 • 3J* •
only finitely many m prior to stage finitely many stage
s.
s. m
and
such that
such that for some
Suppose
(p?)
there must be a
B^(p*?)
has received a value
k
v,
B^(p^)
was defined prior to
had its value induced prior to stage
closed classification in the
s
for sane v,
The first part of (Hi) tells us that there are only
follows from (H2) and (H3) that
stage
n^ < ni . We claim that there are
p^
It
was not committed to a reserved or
ittl partition prior to stage
such that
s.
B^(p^)
(t = p^1)
and thereby induced the value of
s.
But then
received a value prior to
B^(p^). Thus
p^
was committed
to a reserved classification in the
ktJl partition and
received a value prior to stage
The last part of (H1) says that there
are only finitely many triples
s.
(k, i, t)
duced prior to stage
s.
have it induced or defined. least positive
r
m,
p^
tion in the
Bj(m)
can receive a value is to
m > 0
v,
s.
B^(p^)
had its value in
Let
such that for all
was
and all
r(ij)
be the
B^(p^+r)
has not
It follows from (H2) and (H3) that for
has not yet been committed to a reserved or closed classifica ith partition.
fication in the of
B^(p^)
B^(p^)
p£
Consequently, there are
That proves our claim.
received a value prior to stage all
s.
for some v,
The only way
v,
kth partition and
subsequently received a value prior to stage such that
subsequently
such that for some
committed to a reserved classification in the
only finitely many m
B^(p^)
and
We now commit
i^*1 partition for all B^(p^+r^ ^ )
to a reserved classi
m > 0:
for all
v,
as soon as one
receives a value, the other must receive the
same value; if the first has received a value prior to stage
s,
the other
must receive that value now. CASE 2. many stage
m
s= 22 • 3^ • 51 .We claim that there are only finitely
such that for some s.
The only way
duced or defined. in Case 1 . Let
v,
B^(6 • p3^)
B^(6 • pj1)
has received a value prior to
can receive a value is to have it in
Our claim is easily proved by means of the argument given t(ij)
be the least, positive
t
such that for
m > 0
and
26
DEGREES OF UNSOLVABILITY
all
v,
B^(6 • p^+ 0:
such that
Before we proceed with Case 3, we fix what is meant by a finite extension of
B^
if
6*
arguments of
B^
B^(m^)
m
Let
B^(m0)
some value of
B^
v s.
B^
at stage
For example, the assignment Bj
which might induce
B^fm^.
B^n^)
and
make matters still worse, the assignment of a value to for infinitely many
Suppose we assign a value to values in two different ways.
may be induced.
has not be the
(j < z). Unfortunately, the situation is com
which might induce a value for
m
B^(m)
s might consist simply of
be possible to assign values independently to
induced; second,
It follows from
such that
might induce some value of
duce the value of B^(m)
and consider
< m 1 < ... < m z-1
plicated by the possibility of induced values. of a value to
and
which have not received a value prior to stage
Then a finite extension of
assigning values to
s.
i
1 < k < 2n .
at stage
(H1) and (H2) that there are infinitely many received a value prior to stage
s.
6•
at some future stage, then at that stage,
must be set equal to
z
Of
has not yet been committed to a reserved or closed classifi
closed classification in the
first
s.
Thus it may not B^n^).
B^Cn^)
To
might in
m.
B^(m). This assignment may induce
First, the value of
may be of the form
Bj(p“+
may be
p£+ r (ik) and the value of
B^(n)
It follows from (H2) that the sets ■jj lvalue of
Bj(p“+r^ i^)
induced by
B^(m)j-
and j^(k, n) lvalue of B^(n) induced by B^(m) & m = p£+ r ^iic^ jare finite.
However, each value thus induced may itself induce a value in
two different ways. such that value of ..., at
ak+1
and
aQ, a1, ..., a^
be a finite sequence of values
such that for each
ak+1
k < t,
in one of the two ways described above.
a chain of induced values of length
Suppose that
aQ = B^(m)
Let
t.
a^
induces the
We call
We claim that
aQ, a1, t < 2s.
is an induced value of the first kind; then it is easily seen cannot be an induced value of the second kind.
Thus to prove
§3.
UNCOUNTABLE SUBORDERINGS OF DEGREES
27
our claim, we need only show that if all the induced values in a chain are of the same kind, then the chain has length at most lows the fact that there are only finitely many n
s.
k
But this last fol
such that for some
has been committed to a reserved classification in the
(H2).
Since each chain has length at most
2s
n,
k*'*1 partition
and since each induced value
can induce only finitely more values in each of the two ways, it follows that the set {(k, n) |B^(n)
receives a value determined by the value assigned to
B^(m) is finite.
and by commitments made prior to stage
Thus the assignment of a value to
B^(m)
sj-
has only finitely
many repercussions despite the commitments we are forced to honor immedi ately.
Note that it is perfectly possible for the assignment of a value to
B^(mQ)
to determine a value for
B^Cm^.
By a finite extension of
B^
We cannot rule out such an event.
of length
not only the receiving of values by the first have not received values prior to stage
s,
z
z
(j < z)
CASE 3.
which
but also the assignment of all tB^I;) =
1> 2> •••)
course the values assigned
B^(nij)
defined prior to stage
n s.
0;
Let
to
B^(mj)
let
f
Bl[(m)
B^
for which
and if the act of putting
u2,
tension of
B^
B^(m)
f(m)
induces the value of
Let
q(2), q(l)
respectively, be the greatest mem
ug, u 1
q(2), q(l)
respectively,
respectively,
be if
-1.At stage f(m) »
has received a value prior to stage equal to s
f(m)
for all
m
for which
B^(n),
We require
then
s bJ
would constitute a finite ex f
to be complete in the
is defined and if assigning the value
equal to the induced value of
re
respectively, has not been
is a finite extension of B^
as described above. if
nk |
R ^ , Rin
has not received a value prior to stage
following sense:
independently.
respectively, if
otherwise,
s we say a partial function m
and
such that
Rk#q(2)_i> Ri,q(i)-i
is greater than
for all
(j < z)
s = 23 • 3k • 5e • 71
spectively, be the least
B^(m)
we mean
must be consistent with our commitments, since it may not be pos
sible to assign values to
ber of
Of
s
argument of
values thereby determined in any of the functions as a result of commitments.
at stage
f(n)
f(m)
to
is also defined and is
B^(n). Thus the partial function
f
assigns
28
DEGREES OF UNSOLVABILITY
consistent values to the first ceived values prior to stage thereby induced in
B^.
z
arguments of
s,
which have not re
and in addition, assigns all values
We define the weight of
of Theorem 1 of Section 2.
f
The partial function
as we did in the proof
(e) , where
f
is a
partial function, was defined in Section 1. CASE 3a.
q(l) = q(2) = q.
we will define
^
^iqg
the property that
v(q) = g,
Godel number
e
1 < g < 2q,
let w(g)
g
such that
such a way
we will have
B^
be the unique member of
(k)(k > q r bethe least
n
for a11
For each
S
with
B^
g
with
such that
defined by
w(g)(k+l) = 2w(g)(k)).
such that
f
v € S
;
received a value prior to stage is a partial function
1 < g < 2q ,
not recursive in
when the construction is complete.
v(g)(q) = q
Let
For each
p£ > u2
s;
r
and such that
B^(p£)
is well-defined by (H3).
which is a finite extension of B ^ g^
has not If there
and which
has the property that (1)
fe)f(p£)
is defined, we say
g
is not null.
weight) such finite extension of let
If
g
B ^ g^
is not null, let the least (by
be the needed one at stage
Zg be the resulting value of (1), and let
ment of B ^ g^
If
g
is
null,
let t
be
t = max |t 11 < g < 2qj- + p£
R iq " {n,U1 < n
(2)
g
be the greatest argu
which receives a value as a result of making the needed
finiteextension.
For each
tg
s,
and h
< 4 J V
such that
=
1 < g < 2q
u 1 + 1.
We define:
;
{n,U2 < n < and
4
•
h € Riq, we define:
value received at some stage prior to stage s = i value received above at stage s (g not null) o otherwise.
It follows from (H3) that for any
v € S,
B^(p£)
could not have had its
value induced as a consequence of the finite extensions described by (2). Thus we are at liberty to define:
§3.
1 - z if g is notnull and h = p£ = -j value received at stage sor prior to ^ 0 otherwise and g such that h € R ^ and 1 < g < 2q
(h) for
each h
CASE 3b.
q(l) ^ q(2).
Rk, q(2)+h = ^u 2 +
1 q(2).
.
We first define
otherwise whenever
q(2) < q(2) +
h 0,
be
-1 . Define
(s)1 = k
such that
let
u
R^
(s)2 =e1
let
and
(s)3 = e2 . Let
has not been defined prior to stage Rk ;otherwise,
be the greatest member of
w(g) as in Case 3a.
eachd e D,
For
Suppose
stage s
q(2)+h g^u 2 + h) = value received at some
s~7 0
stage prior to
29
UNCOUNTABLE SUBORDERINGS OP DEGREES
d(o) = (d)Q andd(i) = (dj^. For each
will finitely extend
)
and
s.
let
D = {2a •3^11 < a < b< 2%
Let
B ^ ^ 1))
q
u .
d € D
we
j_n SUch a way that when the
construction is complete, either (3)
u(uyr](§£(d(0))(y), e,, m, y))
will be false for some some
m
m
=
u(nyT] (§£(d(1 ^ (y), e2, m, y))
or at least one side of (3) will be undefined for
or the left side of (3) will be a function recursive in
a^, a ^ ,
..., akg. The order in which these pairs of finite extensions will occur is the natural order of since of
1
2 . 3
2
B a n d
suffer
2^-1
D.
Thus we begin by extending
is the least member of extend
,
Thus for each
b£^
n € {i|i < d) n D,
g,
at stage
s.
Let
will
d € D 3^3
we have extended
We now dispose of
we may have already extended
n < d;
,
successive, finite extensions, and the union of these will be
according to the above plan. and
and
Then we make a further extension
and so on.
the needed finite extension of that for each
D.
1^
d.
and suppose gj*(n(l))
Thus we extend for the sake of some
in that event, we will further extend it.
We regard
B ^ ^ 1^
similarly. CASE ^a. (U)
There is an
m
such that the left side of
u(nyr](^(y), e,, m, y))
=
u(nyT] (?2 (y), e2, m, y))
is not defined for any partial function of
B ^ d^0^ ,
or there is an
m
defined for any partial function
f1
which is a finite extension
such that the right side of (U) is not f2
which is a finite extension of
30
DEGREES OP UNSOLVABILITY
3^(d(l))^
The needed finite extensions are both trivial (empty). CASE hb.
gW(d (°))
There is an
and an
m,
an
f1 which is a finite extension of
which is a finite extension of
both sides of (4) are defined but are not equal. (m, fv f1
and
f2) v
we assign the weight is the weight of
such that
2m •3M • 5V ,
To each such triple where
n
is the weight of
fg . Then there is a unique triple of least
weight from which we extract the needed finite extensions of gwCdO)) ^ an
m
and
follows that when the construction is complete, there will be
(namely, the
m
associated with the triple of least weight) such
that both sides of (3) are defined but are not equal. (ha.) and (^b) are false.
CASE ^c.
side of (U) is defined for some and for each
m
Then for each
m
the left
f1 which is a finite extension of
all finite extensions of
side of (h) give it the same value.
b^
^ 0^ which define the left
This means that when the construction
is complete, the left side of (3) will have the property that if it is a function defined for all that portion of Fix
m
m,
B ^ d ^0^
then its values are completely determined by
which has been determined prior to stage
s.
and suppose the left side of (3) is defined when the construction
is complete.
To determine the value of the left side of (3), we merely
take any finite extension of
b £ ^ 0^
at stage
s
which defines the left
side of (fc) and insert that finite extension in the left side of ( h ) ; result will be the value of the left side of (3).
the
The hypothesis of Case
(he) tells us that the finite extension we need does exist and that it does not matter what finite extension we take.
Suppose that when the construc
tion is complete, the left side of (3) is
a function defined for all
let
g
be that function.
number
Then
g
isrecursive in
e1 . But as we have just seen
portion of
b J ^ ^ 0^
g
b£
^°^
m;
with GSdel
is actually recursive in that
determined prior to stage
s.
We now find ourselves
in the same situation we encountered in Case 2.3 of the second half of the proof of Theorem 3 of Section 2. whose domain consists of all value prior to stage Then
g
s
m
Let
B s be the unique partial function
such that
B ^ d ^°^(m)
and whose values are given by
is computable from B g, D g, where
follows from (Hi) and (H2) that
Bg
and
Dg
Dg
has received a B s(m) =
is the domain of
(m). Bg.
are each recursive in
It
§3.
UNCOUNTABLE SUBORDERINGS OP DEGREES
ako> ®ki * •••* aks' sive in
fact> D s
ako, akl, ..., a ^ .
is recursive.)
31
But then
g
is recur
This last follows by an argument not essenti
ally different from that given in the analysis of Case 2.3 of the second half of Theorem 3 of Section 2.
We note only that (H1)-(H3) provide us
with a picture of the stage of consists of the functions
at stage
akQ, a ^ , ..., a ^ ,
s,
a picture which
and finitely many remarks
about commitments. Thus we dispose of
d € D
as described in (^a), (^b), and (^c).
In Case (ha) and (^c), we make trivial (null) extensions of Bk ^ 0^ ^(dO)).
Case
ig Wi2ere tlie non-trivial extensions take place.
and After
all the members
of D
be the greatest
integer which has received a value during theabove series
of finite extensions;
have been exhausted in this manner, we define
that is,
z
those arguments of the functions
z
to
is the greatest argument among all
*{b ^|v = w (i ), w (2), ..., w(2q);
i > oj*
which received values during the above series of finite extensions. Remember that we include induced values. we take
z
to be
If all the extensions were trivial
u + 1; u was defined at the beginning of Case 4.
We de
fine Rkq = {h^u < h ^ and rvalue received above at stage ^ k q g ^ = “jo otherwise for all of
D
h
and
g
such that
we assumed that
and (^c),
h € If
and q = 0,
and proceed directly to stage CASE 5.
such that
R^
s
is otherwise.
1 < g < 2m ^
,
i < s,
R^
that
B^(p^)
In our definition
we simply ignore (ha.), (^b), s + 1.
For each
if
i,
let
*n(i) > 0
m(i) be the least s,
and
u.±
be
otherwise.
If
and let
-1
• ! » . . « • ! ’ . ... . 4 ( i ) ■ S"'1’
n
•
we define "«1,
v(i)
1 < g < 2^.
s
then we write 6-
where
or prior to
has not been defined prior to stage
the greatest member of
For each
q>0.
s
is the least
■ {»'»!< » < P f l>*V o}
is recursive in
if
< nk
B^
by Case 2 ;
by Case 3.
we define
Rk = *jv|v € S & (Et) (t € V & at and
and
has cardinality equal to
of functions which satisfy (R1), (R2), and (R3) : B^ if
b£
recursive
in
&mt
z,
and k > n. where
has not been defined prior to stage
By the definition of
z,
s.
there must be a
V
§3. d € D
UNCOUNTABLE SUBORDERINGS OF DEGREES
33
such that {d(0), d(i)} = (u(q), v(q)}
.
The measures we took in Case it guarantee that either (6)
(S£(y), e2, a , y))
tj(njTj(B^(y), e,, m, y)) =
will be false for some
m
or at least one side of (6) will be undefined
for some
m
aks*
both sides of (6) are equal to
sive in
or the left side of (6) will be recursive in
a ^ , akl,
a^.
that each finite subset of member of m^. J; nk if
Mk
and
at;
a ^ , akl, ...,
consequently,
at
is recur
It follows from the hypothesis of Theorem 1 Mk
has an upper bound in Mk.Let
with the property that for all m^ < nk, we must have
i < s,
be a
m ^ < m^.
Since
m^. J: mg. But this is absurd, because
a
is the element of A corresponding to m under the given order© © isomorphism between A and M, thena ^ is recursive in a^ for all i < s,
at
is recursive in
For each Since Let
V
k,
ag, and m^ < mg. Sk = jv|v € S & (Et)(t € V 8s B^ recursive in
let
has cardinality less than that of the continuum, so must
H = U{Rk U Sk lk > 0}.
continuum, it follows a sequence
(B^li > o)
Since
S -
H
H
is not empty.
For each v € S - H,we have
of functions which satisfy (Rl)-(R5).
to define a unique member
w
of
S - H
The main difficulty in our
argument arose from the fact that we were given If
M
M
M
as having cardinality
were countable, we could have pro
ceeded quite simply in the vein of Section 2. bility of
It is easy
by induction.
That finishes the proof of Theorem 1.
less than that of the continuum.
S^.
has cardinality less than that of the
Since we allowed the possi
being uncountable, we had to make use of a counting argument
based on Case ^ in order to satisfy (&b) and (R5). If
M,
and hence
V,
were countable, we could have satisfied (R^) and (R5) by a direct construc tion similar to those occurring in Section 2. We say a partially ordered set exists an ordinal that
cc and a collection
P = UCB^lr < a)
P
is completely normal if there
(B^lr < a)
of subsets of
and such that for each ordinal
y < a:
P
such
DEGREES OP UNSOLVABILITY
3b
(1)
U {B& |s < 7 ) has cardinality less than that of the continuum;
(2)
B
(3) (b)
is at most countable and is disjoint from U (Be l8 < 7 ) > 7 o no member of B^ is less than any member of U {Bc |8 < 7 ); 7 o for each n € B^, the set = {a|a < n & a e U CB& 16 < 7 }}
is at most countable and any two members of We say a partially ordered set
Q
completely normal, partially ordered set Note that if
P
P
l£
in which
is normal, then each member of
number of predecessors, and
P
have an upper bound in
L^
is normal if there exists a
P
Q
can be imbedded.
has at most a countable
has cardinality at most that of the con
tinuum. LEMMA 2. Let P be a partially ordered set of cardi nality at most aleph-one. Then P is normal if and only if each member of P has at most a countable number of predecessors. PROOF. aleph-one
Let
P
be a partially ordered set of cardinality at most
such that each member of
predecessors.
For each
p € P,
P
has at most a countable number of
let
p* = (u|u c P & u < p} let
P* = (p'lp € P)
bers of
P*.
P*
and let
;
P ! be the set of all finite unions of mem
is an upper semi-lattice partially ordered by set-lnclu-
sion and with set-theoretic union as its join operation; furthermore, is imbeddable in
P f,
of predecessors, and
each member of
all ordinals less than a one-one 7 < a,
P 1 has at most a countable number
P ! has cardinality at most aleph-one.
is completely normal.
Let a
a
P
We claim
be the least ordinal such that the set of
has the same cardinality as
map of the ordinals less than
a onto
P*.
P 1,
and let
f
For each ordinal
let Cr = (f(8 ) 16 < 7 }
,
= set of all finite unions of members of
Ar = CalaeP' & (Eb)(b € b 7 = a 7 - U (A 6 l 8 < 7 ) It is readily seen that
p*
{B^ 17 < or)
& a < b))
Cy
,
,
. is a collection of subsets of
that possesses the properties needed to show
P*
Pf
is completely normal.
be
§5.
35
UNCOUNTABLE SUBORDERINGS OF DEGREES
THEOREM 3. Let A and B be sets of degrees such that A Is countable and B has cardinality less than that of the continuum; let D be the set of all degrees greater than every member of A and in comparable with every member of B. Let T be a non-empty, normal, partially ordered set. Then T is imbeddable in D if and only if no member of B is less than or equal to any finite union of members of A. PROOF.
If
T
is non-empty and imbeddable in
sity of the condition on normal.
Let
T
A
partially ordered set.
Let
collection of subsets of be normal.
B
a
T*
is clear.
T',
Let
A
T
T 1 is a completely normal,
be an ordinal, and let
be a
that has the properties necessary for
is a member of
A
and that
T T to
(A U B) n T*
T" = A U B U T 1. We define a partial ordering for
that each member of
ordinals less than
and of
B.
by
A
and incompar
By means of a transfinite induction on the
cr, we construct a one-one function
the degrees which imbed
T”
is
T 1 and by specifying
T ! is greater than every member of
able with every member of
g
We show sufficiency for
where
retaining the partial orderings of A U B
and
then the neces
Without loss of generality we assume that the least upper bound
of any two members of empty.
and
be imbeddable in
D,
T 1 in
D.
Suppose
y
g
from
T”
into
is an ordinal less than
a
is already defined on A U B U (U {Bo 16 < r))
in such A U B
is
a way that g
imbeds
U CB5 1 s < y)
the identity function.
in
D
and
g
restricted to
Inorder to extendg to
A U B U (U {Bo 15 — < y)) in such
U
< y)
a way that g
Imbeds
in
D,
weneed only verify:
(1) A U B U
(U {B^l5 < 7 )) has cardinalityless than that of
continuum; (2)
is at most countable and is disjoint from A U B U (U CB& la < y ) );
(3)
no member of
is less than any member of
A U B U (U {B^16 < r));
the
36
DEGREES OF UNSOLVABILITY (^)
for each
n e B^,
the set
-ja|a < n & a e A U B U and any two members of
=
(u (Bgle < L^
is countable
have an upper bound in
L^.
A direct application of Theorem 1 of the present section completes the proof. COROLLARY 1 . Let T be a partially ordered set of cardinality at most aleph-one. Then T is imbeddable in the upper semi-lattice of degrees if and only if each member of T has at most a countable number of predecessors. COROLLARY 2 . Let T be a partially ordered set of cardinality at most that of the continuum with the property that each member of T has at most alephone successors. Then T is imbeddable in the upper semi-lattice of degrees if and only if each member of T has at most a countable number of predecessors. PROOF.
The first corollary follows from the second.
Let
T
be
a partially ordered set of cardinality at most that of the continuum with the property that each member of at most
countably many predecessors.
tinuumhypothesis holds, then tinuum hypothesis is false. every member of T1
and
T
T2
T1
T If
has at most aleph-one successors and We show T
is normal.
is normal by Lemma T1
and
T2
2.
Suppose the con
are subsets of
is incomparable with every member of
are incomparable subsets of
If the con
T.
T
T2,
We claim that
T
such that then we say is the
union of a collection of disjoint, mutually incomparable subsets of
T
that each member of the collection has cardinality at most aleph-one. prove this last by transfinite induction. (T^lr < a) T
a
We
be an ordinal, and let
be a collection of disjoint, mutually incomparable subsets of
such that each
T^
has cardinality at most aleph-one; furthermore,
suppose every member of% T is a member of that let
Let
such
T^;
t € T - U (T^17 < a).
which is comparable with some member of some
finally, suppose Let
be the set of all members of
T
H(o) = (t);
T - U {T^lr < a) for each
n > o,
is non-empty, let
H(n+1 )
which are comparable with some member of
H(n). Let Ta = U (H(n) in > 0 )
.
T^
§3.
UNCOUNTABLE SUBORDERINGS OP DEGREES
Since each member of
T
has at most aleph-one successors and at most
countably many predecessors, it follows one.
Since each member of
Tr(r < °0
T
that each member of member of
Ta
has cardinality at most aleph-
which is comparable with some member of some
is a member of that
and incomparable with
37
T ,
it follows that
Ta
is disjoint from
U {T^lr < or} . It is clear from the definition of
T
which is
comparable
with some member of
Ta
H
is a
Ta .
Thus there exists
an ordinal
6
disjoint, mutually incomparable subsets of property that each
T^
CT^ Ir
and a collection T
whose union is
has cardinality at most aleph-one.
T
< 6)of
with the
By repeating
the argument of Lemma 2 of the present section, we can obtain for each
7 < 8 , a completely normal partially ordered set T^ imbeddable in
T^
and such that
T^
We can assume that the members of U {T^lr < 5).
T^
are disjoint. T1
Let
5,
every member of
T is imbeddable in
Then
is
every member of
T^ T !,
T* =
by retaining the par
and by specifying that for each pair
distinct ordinals less than T^.
T^
has cardinality at most aleph-one.
(T^lr < or}
We define a partial ordering for
tial ordering of each
such that
(7 , a)
is incomparable with and T f is completely
normal. COROLLARY 3. Let A and B be sets of degrees such that A iscountable, B has cardinality less than that of thecontinuum, and no member of B is less than or equal to any finite union of members of A. Then there exists a degree d such that d is greater than every member of A and incomparable with every member of B. The notion of independent degrees was defined in Section 1 . We call the functions in a set independent if the degrees of the members of the set are independent. degrees are.
of
We say two functions are incomparable if their
Since Theorem k is a consequence solely of the methods used
in the proof of Theorem 1 , we are content to sketch its proof. THEOREM 4. Let A and B be sets of functions such that A iscountable and B has cardinality less than that of thecontinuum. Then (1 ) and (2 ) are equivalent: (1 ) there is a set C of independent functions such
DEGREES OP UNSOLVABILITY
38
that C has cardinality of the continuum, each member of C is incomparable with every member of B, and each member of A is recursive in every member of C; (2) no member of B is recursive in the members of any finite subset of A. PROOF. and let
S
We show (2) implies (1).
be defined as in Theorem 1.
Let
A = {aQ, a1, a2,
A set of integer-valued functions
{Q^ln > o & 1 < k < 2n} will be defined by stages. a non-empty, finite set least member ofRn+1 ber of
will
D
will have
arguments. For each
D
of functions such that
if and only if there is a member
o, the restriction of
f to
Rn
is
v
of
S
f
0 ^ ^ ) • After straightforward
commitments and finite extensions apply here. s
s.
If
stage
Rn
and
then for each n < k,
prior to stage (H2)
s
for each
g
s
such that
are of the form
to a finite range of values.
p^+t^ ^
(m > o)
furthermore, for each then
has been defined prior to have also been defined
1 < g < 2n .
where j,
m > o
and
j
is restricted
if some natural number of the
has been committed to a closed classification prior to stage
natural number of the form
s,
p^,
For each
then there is a positive integer the form
R^
{Rn ln =* o, 1, 2, ...}
All natural numbers committed to a closed classification
prior to stage
p^
The induction hypothesis at
Only a finite number of the sets
have been defined prior to stage
form
concerning
of the construction is: (Hi)
s,
will be a
such that for all
modifications, the remarksmade in the proof of Theorem 1 above
stage
n, the
be the Immediate successor of the greatest mem
Rn . We will define a set
member of n >
Rn as its domain of
Each
Qng(Pj+t(J) )
m,
t(j)
s,
such that every natural number of
has been similarly committed and such that no p^
(m < t(j))
if
p^+t^ ^
has been similarly committed;
has been put in
has been set equal to
a^(m)
Rn
prior to stage
for all
g
such that
1 < g < 2n . (H3)
For all
classification only if CASE 1. each member of
A
m
and
j,
p^
has been committed to a closed
2 • 3^ < s.
s = 2 • 3J .
The object of this case is to insure that
is recursive in every member of
D.
Similar to Case 2
§3.
UNCOUNTABLE SUBORDERINGS OP DEGREES
39
of Theorem 1. CASE 2.
(s)Q = 3,
(s)1 = e1 and
(s)2 = e2. The purpose of
this case is to guarantee that for each pair (u, v) of S,
if a function
f
is recursive in Du
in Dv
with Godel number
members of D
(s)Q =
independent. Let
not been defined prior to stage 1 < g < 2q = p,
define
finitely extend Dw ^
with Godel number
e1 and
f is recursive in the members of
Similar to Case k of Theorem 1. k and (s)1 = e. This case is needed to make the
some finite subset of A. CASE 3.
e2, then
of distinct members
w( g)
, Dw ^
q
be the least
s.
n
such that Rn
Let p = 2q . For each
as in Case
g
has
such that
k of Theorem 1. For each g we
, ..., Dw ^
in such a way that
Dw (8>(b ) = u(nyT^1'"-»1(Dw(l)(y), 0)
S » ^ ,)(y),
will be raise ror some some m
S^y),
or tne rignt siae or (i) will be undefined. ior
m
when the construction is complete.
finitely extended
In the process each Dw ^
2q times and the union of these consecutive
sions Is the needed finite extension at stage after Cases 3 and CASE
e, m, y))
s.
2q
Is
exten
This case is modeled
k of Theorem 1.
k.
s is otherwise.
be the greatest member of r
Define
q
if q > o and
as in Case 3. -1
otherwise.
Let
u
We define:
Rq = {u + 1} 0
for all
g
(u+1)
such that
By Case members of D continuum.
f value assigned prior to stage s t o otherwise, 1 < g < 2q.
k, the members of D are well-defined. By Case 3, the
are independent and
Let F
D
has cardinality equal to that of the
be the set of all members of D
with some member of B. continuum and A
s
Since
B
which are comparable
has cardinality less than that of the
is countable, itfollows from Case 2 that
nality less thanthat of the continuum.
D
- F
F
has cardi
is the desired set
COROLLARY 1. There exists a set of independent de grees whose cardinality is that of the continuum.
C.
DEGREES OF UNSOLVABILITY
IfO
COROLLARY 2. Let T be a partially ordered set of cardinality less than or equal to that of the con tinuum. If each member of T has only finitely many predecessors, then T is imbeddable in the upper semi-lattice of degrees. In Theorem k take
PROOF. G,
A
and
B
to be empty and thus obtain
a set of independent degrees whose cardinality is that of the continuum.
Let
T
be a partially ordered set as described in the hypothesis of Corol
lary 2.
Let f
be a one-one map of
T
into
G. For each
t € T,
let
t* = {w|w € T & w < t } Since for each For each
t,
t e T,
{t”|t € T).
let
Then
means of the map
T
t 1isfinite, we write
t* = (uQ ,
,..., •
t" = f(uQ) U f(Uj) U ... U
Let
T" =
is order-iscmorphic to the set of degrees T"
by
t -► t".
Note that these last two corollaries are proved without any use of the axiom of choice. needed
In the proof of Theorem3, the axiom
of choice is
only to show F has cardinality less than that of the continuum.
In Corollaries 1 and
2, F is empty because
B
is empty.
Theorem 3 tells us that normality isa sufficient partially ordered set to be imbeddable in thedegrees.
In
condition for a Corollary 2 to
Theorem 3, we saw that it is possible to show certain partially ordered sets of cardinality of the continuum are normal without any use of the continuum hypothesis.
Unfortunately, the continuum hypothesis is closely tied to the
question of which partially ordered sets are normal. LEMMA 5. Let T be a partially ordered set of cardinality of the continuum such that any two members of Thave an upper bound in T and such that any member of T has at most countably many predecessors. Then T is normal if and only if the continuum hypothesis holds. PROOF. T
is normal.
mal, partially
If the continuum hypothesis holds, then by Lemma 2 above, Suppose T
is normal.
Then there exists
ordered set P and a one-one map
f
of
a completely nor T into
Pwhich
§3. imbeds
T
in P.
UNCOUNTABLE SCJBORDERINGS OF DEGREES
Let
a
be an ordinal and
of subsets
of P
normality.
We assume that each
CB^Ir < 7 *-
definition of complete normality tells us that in B^
for some
With the help
because
such that
B^
be such an upper t < n,
where
t e B^*,
This last is ruled out by clauses (2 ) and (3) of the
definition of complete normality.
bound
7
be the least Let
P,
and because any two elements
7 < s.
t
Then clause (k) of the and
d
have an upper
This last contradicts the definition of
8.
of Lemma 5, it is immediate that normality is a
necessary condition for a par-daily ordered set to be imbeddable in the upper semi-lattice of degrees only if the continuum hypothesis holds. After all, the upper semi-lattice of degrees itself is normal only if the continuum hypothesis holds.
Nonetheless, the notion of normality is useful,
since it made possible the proof of Corollary 2 to Theorem 3 without the
1*2 continuum hypothesis.
DEGREES OP TJNSOLVABILITY Corollary 2 of Theorem h provides us with very simple
partially ordered sets which are imbeddable in the degrees but which are not provably normal without the continuum hypothesis.
One such partially order
ed set is the set of all finite subsets of real numbers ordered by setinclusion.
§1*.
THE PRIORITY METHOD OP FRIEDEERG AND MUCHNIK
In [1 7 ] Post raised but did not answer the following problem: do all non-recursive, recursively enumerable sets have the same degree of re cursive unsolvability?
The solution to Post!s problem was found almost
simultaneously by Friedberg [1 ] and Muchnik [1 3 ]; these authors independ ently discovered a new technique which we will call the priority method. We will use the term ’’priority method” somewhat ambiguously to designate any method of proof which owes a large portion of its inspiration to [1 ] and [1 3 ].
Thus we say that Theorem k of [2 3 ] and Theorem 2 of C1 83 were
proved with the help of the priority method. Our purpose in formulating Theorem 1 of this section is to sepa rate (insofar as is possible) the combinatorial aspects of the priority method as manifested in [1 ] from the recursion-theoretic aspects.
We do
not claim that Theorem 1 stands as a fundamental principle from which all results so far obtained by the priority method readily follow, but we do believe that Theorem 1 and its proof will be useful to anyone who wishes to develop an intuitive understanding of the workings of the priority method in all of its manifestations.
We will put Theorem 1 to practical use in
this section by deriving from it as corollaries the solution to Post's problem and the fact that every countable, partially ordered set can be imbedded in the upper semi-lattice of degrees of recursively enumerable sets. A requirement
R = {(Fi, H ^ l i cl)
is a collection (possibly
empty) of ordered pairs of disjoint, finite sets of natural numbers. T
of natural numbers meets requirement
Fi C T
and If
R
if there is an
i € I
A set
such that
H^ n T = 0 . L = (hQ, ht, ..., h^}
is a finite set of natural numbers, we h3
U
DEGREES OP ^SOLVABILITY
h0 define j (L) =2 + 2 map of
h-
h +...+2
. The function j
is a one-to-one
the set of all finite sets of natural numbers onto the set of all J(0 ) = 0 . Both
natural numbers, if it is understood that verse
j-1
and its in
are effective.
Let
t be a function defined on the natural numbers with natural
numbers as values. s > 0,
j
We say
j”1((t(s))0)
and
t
enumerates requirements if for each
j”1((t(s)) ^
requirements enumerated by
t
are disjoint, finite sets.
are denoted by
RQ, R1, R 2, ...;
The
for each
Ek = {(j-’ Otf-Ho). J“1((t(3))i))|(t(s))2 = k} • If
t
enumerates requirements we denote
(t(s))2
by
F s, H s
and g(s)
j’1((t(s))0) , j”1 ((t(s)) 1)
respectively.
Thus for each
Rk = {(Fs, H s)|g(s) = k} A if
A
set
A
.
T
t
F* C Tr
s = 0. TQ = 0
Stage
s > 0 . Tg = Tg-1if (a) or there is an r < s
(1 Ts_, = 0
and
such that
there Is an r < s
and
if n Ts_, - 0;
(c)
Hs n T , ft o. s-i
and
k,
Fs £
T
s
we say Tg. If
such that
= T
Rk Rk
, U Fs s-i
is recursively
(b) or (c) is true: g(r) < g(s), r > 0 , F1* £ Tr_«|>
g(r) = g ( s ) , r > 0 , F r £ T
otherwise.
is met at stage
is met at stage
since clause (c) must be false at stage all
T
if n F s ? 0;
(b)
For each F s £ Tg-1
called the priority
t.
Stage
(a)
t
and meets every member of a certain subclass of the class
of requirements enumerated by
Fr C T r , f
T
is defined by stages, and as we shall see,
enumerable in
f
f . With each function
which enumerates requirements, we associate a set t.
k,
is said to be recursively enumerable in a function
is the range of a function recursive in
set of
and
s,
s
s,
T =
U T0 . s=0 s
if s > o, then
and since
,
k = g(s),
H3 n Tg = 0,
Hs n Fs = o
for
s. For each
r < s
k,
we say
such that Rk was met
Rk
is injured at stage
at stage
r,
s if there is an
i f n Tg-1 = o and
H1* n Tg ^ o.
THE PRIORITY METHOD OF FRIEDBERG AND MCJCHNIK
$k.
If
Rk
is met at stage
r
^
and is not injured at any stage after stage
r,
then
Fr C T and if fi T = o for all s > r. But then Fr C T and — r s — — H17 n T = 0 . Thus if Rk is met at stage r and is not injured at any succeeding stage, then For each there is an
k,
s > o
T
meets
we say
such that
R^.
Rk
is
t-dense if for each finite set Hs n T g_ 1 = o
F s $ Tg__1,
g(s) = k,
L,
and
LflF3 = 0 .
setting pair
The definition of
T
TQ - o.
s > 0,
(Fs, Hs)
At stage
may be described as follows. the function
of disjoint, finite sets.
t
We begin by
presents us with a
This pair is of interest to us
because it may represent a chance to make
T
meet
^
Rg(s) •
v e set
T S = T S—I - U Fs,7 and if we can manage to have T„ u — > s,7 U n Hs = 0 for all then T will meet Rg(s)clause (c) is true at stage s, then Hs n T g - 1 ^ o, Rg(s)-
and stage
s
clearly does not represent a chance to meet
If clause (b) is true, then there was a stage
such that
Rg(s)
was met
jured at any stage after
stage
r
and such that
rand before
with the hope of not injuring
Rg (s)
s;
Rg(r)
s
Rg(r )
Againwe set
T g- 1
U F s,
then
Rg(s)
s,
was met, and
g(r) < g(s);
and before s,
Rg(r)will be
Hg(s) at stage
have assigned a higher priority to
* T g_ 1
Tg
then there was a stage
Rg(r)
s
r
in addition,
but if
Tg
injured at stage
Tg = Tg__1, because we do not wish to injure
forthe sake of meeting
s
vas not in
consequently we set
was not injured at any stage after r
is set equal to
s
at which
prior to stage
at any future stage.
If clause (a) is true at stage prior to stage
r
when
than to
Rg(r)
g(r)
s. atstage
< g(s). Thus
Rg(s)
when
we
g(r) < g(s).
LEMMA 1 . If r < s and R^ is met at stage r and at stage s, then there is a u such that r < u < s and Rk is injured at stage u. PROOF. follows that stage
s
Tp -1 = 0 ,
and
We have
g(r) = g(s) = k,
H 17 n T g __1 7^ 0 , R g(s)
Fr
Tr - 1
and
Fr £ Tr .
since otherwise clause (b) would be true at
would not be met at stage
s.
We know that
since otherwise clause (c) would be true at stage
would not be met at stage
r.
It
Then
H37 n T
= 0,
since
r
H 17 n and
H37 n Fr = 0
Rg(r) for
DEGREES OF ^SOLVABILITY
k6
all
r.Thus there is
and
H1* n Tu ^ o.
But then
LEMMA 2. at stage PROOF. i < k, 2*.
a unique u
such
Rk is injured atstage
Hr n T u l = o
u.
For each k, the set CslRv,is injured lr s) has cardinality less than 2 . By induction on
the set
that r < u < s,
{s|Ri
k. Let
k > o,
is injured at stage
and suppose that for each
s)has cardinality less than
Then the set R = (sI(Ei)(i < k & R1 ^
has cardinality less than
- k.
is injured at stage Let
S = (s|(Ei)(i < k & R.^ With the help of Lemma 1, dinality of stage
s).
is met at stage
s)}
we proceed to obtain an upper bound on the car
S. Let m. be the cardinality of the set {s|R^ is met at ^ v k-i 1 Then the cardinality of S is at most / i_Q . By Lemma 1^ = (slR^^
1, the cardinality of the set least
s)}
- 1.
Since for each
i < k,
is injured at stage
s)
is at
Ii has cardinality less than
2^
is must be that
Thus the cardinality of
k-i £ (n^ - 1) < 2k - k i=0 If S is less than % .
Wenow show that if i < k set
such that ts|Rk
Ri
s
is injured at stage
and suppose
such that
R^
Thus wehave Fs / o.
Rk
S
0
0}
s1
be such that
s and
n such that
be the greatest member of
s" > s 1 and
j < yi#(s,
n, e*).
It follows
e*)|s > 0} is infinite.
Suppose that the set
m ! be its greatestmember.
all
’
is finite;
s' > s* s>
{m^fs, e*)|s > s').
s' Let
let
andc(s, n) and n < m*. s"
be such
m^# (s", e*) = m” < m f.
We now show by induction on
s > s”
that
=
m^Cs, e*) = m"
and
DEGREES OP UNSOLVABILITY
6k
Kj*(s>
e*)*=&j*(3"> e*) e*) = m"
and
for a11
3 > 3"* Suppose then that
Kj*(s, e*) » Kj*(sl!,e*). By Lemma 3, 7 j*(s, n, e*) = y±*(s+i, n, e*)
s",
n < m^fs, n, e*),
since
s
s > s" > s ! > s*.
This means that
P±*(s, n, e*) = P±*(s+i, n, e*) = c(s, n) for all
n < m^fs, e*). But
n < m" < m f,
since
s > s*.
c(s+i, n) = c(s, n) = c(n)
for all
Hence
pA.( s+i, n, e*) = c(s+i, n) for all
n < mn * m^fs, e*),
and consequently,
follows from the definition of
m”
that
m" < m^# (s+i, e*). It
m^# (s+i, e*) = m ” . But then by
remark (R2), y±*(s, n, e*) = y1#(B+i, n, e*) < Kj*(s, e*) for all
n < rn^Cs+i,e*) * m 11 = m^fs, e*). Prom this last and
nition of
Kj*,
it is clear that
Thus we have shown
Kj*(s+1, e*) = Kj*(s, e*).
Kj*(s, e*) * K^Cs", e*)
but this is absurd, since by definition of { K ^ s , e*) Is > 0}
e*
and
for all i*,
the set
is infinite.
LEMMA 5. If s > s* and c(n) * P±*(s, n, e*). PROOF. Let
s > a*
and
n < m^Cs, e*),
then
n < m^Cs, e*). By Lemma 3,
y±*(s, n, e*) = y^fs', n, e*) < s for all
By Lemma k ,
s* > s.
there is an
s" > s such that
(s", e*) > ^*(3, e*)
and
c(s", n) = lim3 c(s, n) = c(n) But then
c(s", n) - Pi#(s", n, e*) * u(yi#(sM, n, e*)),
n < m^# (sfl,e*)
and
y^*(sfl,
n, e*) < s < s".
c(n) = u(yi#(s", n, since
t(n)
since
It follows that
e*)) = P±*(s, n, e*)
y1#(s", n, e*) = y±*(s, n, e*) < s. Let
the defi
denote the partial function nsfs > s* & n < m^Cs, e*) ]
;
,
s > sMj
AN EXISTENCE THEOREM FOR R.E. DEGREES
§5. by Lemma k 9
t(n)
is defined for all
n.
65
By Lemma 5,
c(n) = P±*(t(n), n, e*) for all
n.
Since each of the functions
recursive in
B,
it must be that
our argument that for each is finite. D±
This
for any
odd.
i < 2
i #di(n)
Let
Cy(n) In < m}.
Let
n < m.
n< m
and all
j
B U Di
is
only
since
B U
Di
for any
b • d^. e)|s > o}.
c(n)
for some
Then e, n, j ) '
3
for all
n < m.
s be such that
< y.
B,
,
is either undefined or unequal to
c(s, n) * c(n) for all
B UD^
m be thelargest member of (K^Cs,
y(n) “ nyrj( II J< 7
member of
{K^(s, e)|s > 0}
C is not recursive in
C is notrecursive in
TJ(y(n)) = c(n)
is
That completes
& n € B U D^)
Suppose this last is not the case.
is defined and
the set
are each recursive in
The representing function of
We now show
e,
Pi.Cs, n, e)
B.
contains only even numbers and C
1 ► (n is odd
Thus it is sufficient to show
i
and
«— ► (n is even & n € B U D i)
n €^
n < m.
and each
and
n € B
Fix
m^fs, e)
is recursive in
is all we need to know to show
This means that
i < 2.
C
Let
y
be the largest
y < s,
and d± (s, j) * d± (j) Then y ^ s , n, e) * y(n) < s for all
It follows that P± (s, n, e) = u(yi (s, n, e)) = u(y(n)) = c(n) * c(s, n)
for all
n < m.This means
y(m) < K^Cs, e) < m.
m < m^fs,e) . But then by remark (R2),
Thislast is absurd, since
It remains only to see that It is clear that
Di ^ 0,
For each
D1
would equal
i < 2,
if
^((n^,
(n),) = 0
gj/n) - { nt(t € Di)
otherwise
D
and
B. C
we define a function
and which enumerates (n),
GND.
is recursively enumerable in
since otherwise
would be recursive in B, which is recursive in B
Di
m < y(m) by
.
g^^
66
DEGREES OF UNSOLVABILITY We need the following elementary lemma to obtain some corollaries
to Theorem 1. LEMMA 6. If AQ for each i < 2, Aq U A 1, then the least upper bound PROOF. A0 U A 1
Let
and
A1;
in both,
and
Clearly
n € AQ .
no, then n t AQ . Aq
aQ, a1
respectively.
wish toknow if
and A 1 are disjoint sets and if A1 is recursively enumerable in degree of AQ U A 1 equals the of the degrees of AQ and A 1
If the
eventually
£g
be the degrees of
Aq, A 1
U a^. We show
< a.
a
£0 U £1
b < £, < £.
then
= £•
We saw in Section h that there exists a recursively enumerable degree
d
such that
Corollary 3, than
d.
d
£ o)
such that
i,
.
Our construction is quite similar to that of Theorem 1.
We define six
§5. functions, and
t(s),
K(i, s, e),
f (s)
AN EXISTENCE THEOREM FOR R.E. DEGREES
d(i, s, n),
y(i, s, n, e), P(i, s, n, e), m(i, s, e)
simultaneously by induction on
into just one of the D1 ,s
D.
will be the representing function of D^.
member of
B
B.
s
d(o, o, f(o)) = o. We set
Stage
D^s
will
limg d(i, s, n)
Each of our six functions will
1 for
P(i, o, n, e)
= 2, d(i+i,
and m(i, o, e) = K(i, o,
s > o. We define
z(s) =
e) « o
^ °J
We set
all
for
n ^ f(o), o, n) « all i,
and we t(o) *
e and
n.
t(s) and z(s):
< t(s) &
z(s) < f(s).
is odd.
d(o, o, n) =
& f(s) < K(i, s-1, e))J
t(s) = iixx < f (s^(El)(Ee)(x = p^+e
Note that
i,
s we put
We assume without any loss of generalitythat every
= o. We set
y(i, o, n, e) = i
At stage
For each
is even and thatevery member of D
Stage set
s.
thereby guaranteeing that the
be disjoint and that their union will be
be recursive in
71
;
•
d(z(s), s, f(s)) » o
and
d(i, s, n) = d(l, s-1, n) for all f(s)
i
into
and
n such that
1 ^ z(s)
or
n ^ f(s). Thus we have
put
D Z(S). Let d1 ^ , J) =
,11
d(k, s, j)
k^i & k < s Wedefiney(i, s, n, e)
and
P(l, s, n, e)
y(i, s, n, e) * uyy < s[T](
forall
i,
n
Pj(J) * d
u(y(i, s, n, e))
if
and
e:
e, n,y)] y(i, s, n, e) < s
P(i, s, n, e) = | s+2 Before we define
m(i, s, e),
otherwise.
we observe that
(i)(e)(Et)Ct < s & d(s, t) ? P(i, s, t, e)] This last is clear, since = s+ 2
for all
i
and
d(s, s) = 1 ,
and since by
P(i, s, s, e)
e:
m(i, s, e) = nt[d(s, t) ^ P(i, s, t, e)] " K(i, s-i, e) K(i, s, e) =
0}
is not recursive in
e
such that the set
B, D 1
is finite, and then use this for any
i*
and
Suppose there is
{K(i, s, e) Is > 0} is infinite.
t* = ut(Ei) (Ee)£t = p^+e &{K(i,s, e)|s > Let
i.
0}
is infiniteJ
Let .
e*be such that t* = p±*
All that is needed now is to repeat the arguments of Lemmas 1-5 in order to show
D
is recursive in
B,
which is impossible, since
b < d.
There
would be little profit in actually repeating these arguments, since the only real difference between the constructions of Theorems 1 and 2 resides in the assignment of priorities; all other differences are merely notational. We content ourselves with proving the counterpart of Lemma 1 and stating the counterparts of Lemmas b and 5. LEMMA 7. z(s) = i* PROOF. and
of
Let
and
repetitions.) S and a
S
Since t**
S
exist
for all i** and
e**
ssuch that
(Recall that
f
z(s)
t(s) « t**
for all
f(s) > t* > t(s) = t**
enumerates
D
without
for all
s € R. t(s)
such that
and f(s)
s 1,
be an infinite set such that for all
f(s) > t* > t(s).
Thus we have > t(s)
such that for all
Suppose there are infinitely many
t* > t(s)*.
z(s) ^ i*
There is an s' or t* < t(s).
3 -1 , e**)
Since
f(s)
that there
§5. for all since
s c R.
R
AN EXISTENCE THEOREM FOR R.E. DEGREES
But then the set
{K(i**, s, e**)ls > 0}
is infinite, and since
This means that t(s) = t*
t* < t**,
for all
s € R.
Let
consequently
But
then
= i*
s1
n < m(i*,
s, e*),then
Theorem 1, that m(i*, s, e*) that
D
D
and
is recursive in
B,
P(i*, s, n, e*)
becomes: if
e,
the set
D
is not recursive in
B, D1
It remains only to see that the
and D
recursive in B
For each
and enumerates the if*
| (nt(d(i, D± *
of Lemma 6.
then
taneously enumerate the D^s,
Di ,s
since otherwise
D1
The function
would
g(i, n)
is
simultaneously:
d(l, (n)Q,(n),) - o
n i D .. J
otherwise
;
. D,
n € By
we repeat the argument
Ve first ask if
n € D.
If the answer is yes, then we simul-
and eventually
n
will turn up in exactly
thereby answering our initial question.
have just shown that
the D ^ s are uniformly recursive in
there is a recursive
function h
Godel number
is
are simultaneously recur
B, B^.
is recursive in
D^'s,
Since we are given
easily repeated in order
(t)Q , (t)n). o))1
Supposewe wish to know if
If the answer is no,
one of the
D^'s
Cg(i, n) In > 0 }
To see that each
B.
for any i.
i, Di ^ o,
would be recursive in
| (n))
as in
{K(i, s, e)|s > o}
to show
D,
s > s* and
that completes our argument by reductio
The concluding argument of Theorem l is
equal
plays the
since each of the functions
finite.
B.
t* < f(s)
s*
There is now no difficulty in
is recursive in
for each i and
sively enumerable in
s > s !,
P(i*, s, n, e*). It then follows,
is not recursive in B,
ad absurdum that
s € R,
Lemma k becomes: the set
Lemma 5
d(n) =
for all
The natural number
mimicing the arguments of Lemmas 2-5. s, e*)|s > 0)is infinite.
It follows that
s e S.
such that for all
ort* < t(s).
is infinite,
without repetitions.
t* = t**.
for all
same role it did in the proof of Theorem l.
(m(i*,
D
z(s) = i** = i*
z(s) ^ i*
s*be the least z(s)
enumerates
and
which is impossible, since
and either
f
73
such that D 1
Actually, we D;
that is,
is recursive in
D
h(i). By a similar argument, it follows that for each
with i,
DEGREES OF ^SOLVABILITY
Ik
the members of the sequence D0> D i> •••' Di-i* Di+1’ ••• are uniformly recursive in For each
i,let
cause every member of dition the F^>s
F^BUD^
all
m, vn|F(m, n) s > o,
B.
Let
D is
let
F1
xmn|F(m, n)
In ad
be a function such that for
n € U (D1 1i < n))
recursive in xmnlF(m, n),
F^'s
is odd.
and are simultaneously re
Fm - Recall that for
This means
(n) (n € D «— ► n
i,
D
f(s) e D z(s)> 5111(1 since
s > o,
We show that the
is recursive in F b e
is the representing function of
(n) (n € D
that
B
B is even and every member of
z(s) < f(s).
since for all
Then
are uniformly recursive in
cursively enumerable in each
D1 .
, € D o*
follow,s
since
is odd &(Ei) (i < n & F(i, n) = 0))
are recursively independent (in sequence).
For each
denote the function xmn|F(m + sg((m+i) — i), n) .
First we observe that (m)(n)jF(m + sg((m+l) - i), r.) = 0— It follows that
F*
n e B v [n e Dm+gg( (m+,^ is recursive in B, D1, D0>
D
F^
D*.
is not recursive in
cursive in F1 ,
since
is not recursive in
D
since the sets
,##> Di-i* Di+l'
are uniformly recursive in since
n > m ♦ Sg((m+1) - l)] } .
But then D B, D*.
is not recursive in
This means
is recursive in
xmn|F(m, n)
xmn|F(m, n).
F1 ,
is not re
It follows that
F*. recursive, universal, partial ordering relation
described in the proof of Corollary 3 of Section k . For each
i,
let
Bi = {pl+ni» « F±} Then the
B^'s
are uniformly recursive in
D,
simultaneously recursively
§5 .
enumerable in
B
AN EXISTENCE THEOREM FOR R . E . DEGREES
and recursively independent (in sequence); also
recursive in each
B^.
For each
u,
Cu
u, B
is recursive in
is recursively enumerable in
B
is
let
Cu - u CB± |i < R u) Then for each
75
Cu ,
Cu
. is recursive in D
and
B.
We conclude the proof of our theorem in exactly the same manner Section k.
that we concluded theproof of Corollary 3 of
It is necessary
to show (u)(v)(u < R v «— ► Cu If
u { R v,
then
independence of the
Cu
is recursive in
Is not recursive in
Bi ,s.
If
because of the recursiveness of
u < R v, 0 [h ^ 1 < 4h(u)& P? = It is clear
that the predicate
less than or equal to
o ’.
i ■ t] }
(Ey)K((y)Q, (y).,, e, d,
h, c)has degree
It follows the predicate
(Ey)K((y)0, (y),, e, d, h, c(e)) is recursive
in the function c,
since
c > 0/.
Our plan is to define three functions, simultaneously by induction on sive in
c.
e;
y(e), h(e),
and
d(e, i),
each of these functions will be recur
In addition, the function
d(e, i)
will have the property
that d(i, i) = d(e, i) whenever
e> i.
The desired function
d(i)
will be definedby
d(i) = d(i, i) Stage all
e * o.
We set
.
y(o) = h(o) * i.
We define
for
i: c(0) f c(
d(0,
L l Stage
e > 0.
if
(Em)(pQ = i & m > o)
otherwise
We set:
nyK((y)0 , (y),,e, (2T(e, h(e - i)))e_,, y(e) *
if such a 1
otherwise ;
Note that d(e, i) for
- 1),c(e))
-
mind, we define
i:
c(e)
lf
h(e - 1) < i 0 (i = p“ )
d(e - 1, i)
otherwise
That concludes the construction. Since > e
h(e
exists,
£h((y(e))Q) < h(e) . With this lastin
all
d(e, 1) =
y
h(e) a h(e-i) + y(e).
' (y(e))o,l
h(e)
d(o, i)
for all e. Since
.
y(e) > o
h(e - 1) < h(e)
for all
whenever
e,
e > o,
wehave it
follows
§6 . THE JUMP OPERATOR d(e,
i)=
d(e - 1, i)
whenever
79
i < e -1 . But then
d(e, i) = d(i, i) whenever
e > i.
We define
d(i) = d(i, i). Let
d be the degree of
d(i). LEMMA 1 . PROOF.
c < d*.
It is clear from the construction that d(t, p“ ) = c(t)
whenever
h( t)
0.
We show
d(e, p*) = c(t) whenever and
m
h(t) < p!j?,
m > o
be such that
and
h(t) < p“ ,
t < e
by an induction on
m > o and
e > t,
e.
Let t, e
and suppose
d(e - 1, p“) = c(t) Let
p® = i.
since
Now
i ^ p^.
d(e, i)
Suppose
equals either
d(e, i)
d(e - 1, i)
or
h(e - 1) < i < *h((y(e))Q) & p “ = i & t < e & since the first caseof the definition of follows from the definition of
K
and
d(e, i)
y(e)
y(e) ^ 1
must hold.
,
But
then it
that
(r(®>)o,i ■ c(t) since
(y(e))0
^ d(e - l, i). Then we have
*
y(e) ^ 1. Thus
d(e, p^) = c(t)
whenever
h(t)
0
and
t < e.
It follows d(p^) = d(p“ , p“) = c(t) whenever
h(t) < p^
and
m > o,
since
P t > t for all
t.
Then we have
(t)(En)[(m)m > n (d(p“ ) = o) V (m)m ^ n (d(p“) = i)] sincec(t) < 1 n.
It
for all t;
isclear that
for each
t,
let
k(t)
be the least
k has degree less than or equal to c(t) = d(pt(t))
, such
d! , and that
80
DEGREES OF ^SOLVABILITY LEMMA 2.
d 1 < c.
PROOF.
Since
y
is recursive in
c,
it is sufficient to show
(Ey)T](cf(y), e, e, y) «-► y(e) ^ 1 for
all
e > o. Fix
e > o
and suppose
y(e) ^ 1.
all
i
eand
i < £h((y(e))0) < y(e) < h(e) < h(t - 1) since
t> e
nition of
t
If
is non-decreasing. that
that
d(t, i) = d(e, i)
Since
then
y(e) ^ 1 ,
i < h(e - 1) < h(e),
,
But then it follows from the defi
d(t, i) = d(t - 1, i).
h(e - 1) < i,
i < h(e - 1). Since
h
d(t, i)
duction on d(e, i).
and
for
for all
Thus we have shown by in t > e.
It follows
(y(e))0 ^ . have (y(e))0 ±
d(e, i) =
we must
we must have
d(i) =
Suppose * ± . Since
y(e) ^ 1,
t ]((y(e))0,
it follows
e, e, (yfe)),) & (y(e)), < £h((y(e))Q)
We knew d(i) = (y(e))0 ±
for all
.
i < £h((y(e))0). It follows
(Ey)T’(d(y), e, e, y)
.
That completes the first half of the proof of Lemma 2. Now suppose y
has the property that
show
y(e) ^ 1.
Since
T^(d(y), e, e, y). 0
Again let
e > 0.
We wish to
is not the Godel number of a deduction, it will
be sufficient to show K ^ y ) , y, e, (cfte, h(e - i)))e_,, h(e - i),c(e))
.
But then it will be enough to prove: (1)
(l)[i < h(e - 1) - d(l) = d(e - i, 1)]
(2)
(D(t)t < e (m)m > 0 [h(e - 1) < 1 < y & p“ = 1
To prove (1),
suppose
andconsequently, d(e - 1, i).
i < h(e - 1).
d(t, i) = d(t - 1,i)
To prove (2), suppose
Then
;
i < h(t - 1)
for all
t
- d(i) =c(t)]
.
for all
t > e,
> e.It follows
d(i)
h(e - 1) < i < y,
t < e,
p* = i
and
§6 . THE JUMP OPERATOR m >
0 . Then
h( t) < p*£,
Lemma 1 that
d(p“ ) = c(t)
By Theorem 1 , £ ! > £ !.
since
t < e.
whenever
It was shown in the proof of
h(t) < p“
the equation
x’ = £
has a solution if and only if have a unique solution? c;
showed that the answer is no regardless of the value of
his result states: for each degree b < b ^ b < b1
such that
m > 0»
and
x* = c
It is now natural to ask: does
Spector [2 5 ]
81
b,
there are degrees
and
b«j
bg U b 1 = b 1 = b^ = b • We combine the
and
idea of his proof with the system of priorities of Theorem 1 of Section 5 in order to prove Theorem 2 of the present section.
We could, if we wished,
obtain Theorem 2 as a ccmplicated corollary of Theorem 1 of Section k ; how ever, the price of such elegance would be a total lack of clarity. THEOREM 2 . Let A and B be sets such that B is recursively enumerable in A. Then there exist dis joint sets Bq and B 1 such that BQ U B 1 = B and such that for each i < 2 , (B.^ U A) 1 is recursive in A f, and Bi is recursively enumerable in A. PROOF. is
B -A.
b^s,
Let f
be a one-one function recursive in
We will define six functions simultaneously by induction on
n), y^s, n) and
be recursive in
A.
(i = 0 , 1 ).
t^s)
sand
n.
sively enumerable in we put
f(s)(the
We assume
all
n.We set
We set all
A,
namely,
sth member of
Ci . At stage B
s = 0 . We set bi (0 , n) =
bQ (o, f(0 )) = 1
i < 2
and all
Stage
+ 1 ,n) > 0
is the representing function of a set recur
B - A is infinite, since Stage
Each of these functions will
It follows from this last (cf. Section 5) that for
i < 2 , lims bi (s, n)
each
s:
i < 2 , we will have
For each
1 > b± (s, n) > b ^ s
for all
Awhose range
and
1
- A)
s
in CQ
or
of the construction, C1
but not in both.
otherwise there is nothing to prove.
y^Co, n) = t^o) = 0 for all
i < 2
for all
and all n
i < 2
t A U (f(o)}.
b ^ o , f(o)) = 0 . We set bi (o,n) = 0
n c A.
s > 0 . For each
i < 2 , we define
t± (s) = nn^ < g[f (s) < yi (s-i, n)]
.
and
for
8a
DEGREES OP nNSOLVABILITY
Let
z(s)
. 1 if tQ(s)
b^(s, n)
< t^s), and let
forall i < 2
z(s) = o
otherwise.
We define
and all n:
b,(s, n) =
r o if i = z(s) & n = f(s) -j *• b^(s - 1, n) otherwise.
We conclude the construction by defining _1 /
yi (s, n) for all
TT
Pi
o
n:
. n » n> y)
v J o
and
A 1.
i
and
n
be k
does not exist and 2m + k
n + 1 & yi(s, n) ^ y ^ s - 1, n) > o since
i < 2.
lims yi (s, n)
does not exist.
i. There must be an infinite set
forall
have BQ n B 1 = o and i (B^ U A) 1 = ^ is recursive in
and each
limg yi(s, n)
b1 -
we
We use the method of infinite descent.
such that
2n +
be the set whose representing function is
TT
yi (s - 1, n) > o.
bj (S-1 ,j) V Pj , n, n, j )
There must be a
,
j < yi(s - i, n)
§ 6 . THE JUMP OPERATOR such that
hi(s - 1, J) ^ ^ ( s , j),
But then
j * f(s),
because
f(s) < y ^ s - 1, n)
83
y^fs, n) ^ yi (s - 1, n).
and
b± (s, f(s)) t b ^ s - i, f(s)) This last can happen only if Since
f(s) < y ^ s - 1, n) Since
i = z(s).
and
o < t1-i(s) < n
infinite subset
T
of
S
.
It follows
n < s,
^ ^ ( s ) < t^Cs).
we must have
for all
t^s) < n.
s € S, there is an
m
and an
such that Vi(s) = m < n
for all then
s
2m +
t^s) = n
€ T.
Letk = l - i.
k < 2m + 1< 2n < 2n + i. for all
We saw above that and
We show
2m + k < 2n +
Suppose m = n.
s€ T,and consequently,
z(s) = i
for all
i. If
Then
z(s) » i
m < n,
t1_l (s) =
for all
s € T.
i = 1,
k = 0,
s e S. It follows
2m + k < 2n + i. It remains onlyto see that
suffices to show the set
limg y^Cs, m)
(yk (s, m)|s > o}
does not exist.
is infinite.
For each
It s € T,
we have tk (s) = m < n < s But
then f(s) < yk (s, m)
for all
infinite, it follows the set Now we fix
i < 2
s e T.
.
Since
(yk (s, m)|s > o) and show
f
isone-one and
T
is
is infinite.
is recursive in
A f.
Lemma 3
tells us that (n)(Et)(s)a> t(y1 (s, n) = y± (s + 1, n))
.
We define: t(n) - nt(s)a> t(y± (s, n) - y ^ s + i, n)) y^n) - y^tCn), n) The function
y^(n)
is recursive in
A.
.
is recursive in A' Clearly,
j
because the function
yi (n) = limg yi (s, n)
for all
y^(s, n) n.
defined in Section 1; it is the representing set of the predicate (Ey)T](bi (y), n, n, y)
.
C£ was
DEGREES OP UNSOLVABILITY
8b
We show
is recursive in
A 1 by showing
(EyiT^b^y), n, n, y) «— y^n) > 0 for all
n. Fix
n.
First suppose
y^n) > o.
Then
y±(s + i, n) « y(s, n) > o for all
s > t(n).
Since
o
is not the Godel number of a deduction, it
follows v i~
y^n) = y ^ s , n) » yi(t(n), n) = for all
s > t(n). Let
s f be so large that
’ ’ n ’ n ' y)
s* > t(n)
and
b± (s', J) = b± Q) for all
J < y^(n). Then we have Ti(t>l(7i (n))> n, n, y1(n))
Now suppose
.
(Ey)T] (b^Cy), n, n, y). Let y ’ = ►‘y T i ^ C y ) , n, n, y) > 0
Let
t
be so large that
t > y*
and
b± (t, i ) for all
.
= b± (3)
i o
for all
s > t.
This last means
y^(n)
= y 1 > o.
COROLLARY 1. Let a and b be degrees such that a < b and b is recursively enumerable in a. Then there exist degrees bg and b1 such that ^ U b 1 = b , a ’ = b^ = bj, and such that for each i < 2, a < b^ and b^ is recursively enumerable in a. PROOF. has degree
b,
bers and B B1
B
A
and
BQ U B 1 = B A 1,
B
be sets such that A
is recursively enumerable in
has only odd.
such that
recursive in
Let
and
Bi
A, A
has degree
a, B
has only even mem
By Theorem 2, there are disjoint sets and such that for each i < 2, is recursively enumerable in
BQ and
(Bi U A) 1 is A.
For each
i