Stephen G. Simpson
Department of Mathematics
August 10, 2004
Pennsylvania State University
A principal object of study in recursion theory going back to the
seminal work of Turing [36] and Post [25] has been the
countable upper semilattice
of recursively enumerable
Turing degrees, i.e., Turing degrees of recursively enumerable sets
of positive integers. See the monographs of Sacks [27],
Rogers [26], Soare [35], and Odifreddi
[22,23].
A major difficulty or obstacle in the study of
has been the lack
of natural examples. Although it has long been known that
is
infinite and structurally rich, to this day no specific, natural
examples of recursively enumerable Turing degrees are known, beyond
the two examples originally noted by Turing:
the Turing
degree of the Halting Problem, and
the Turing degree of
solvable problems. Furthermore,
and
are respectively
the top and bottom elements of
. This paucity of examples in
is striking, because it is widely recognized that most other
branches of mathematics are motivated and nurtured by a rich stock of
examples. Clearly it ought to be of interest to somehow overcome this
deficiency in the study of
.
In 1999 [30,31] we defined a degree
structure, here denoted
, which is closely related to
, but
superior to
in at least two respects. First,
exhibits
better structural behavior than
, in the sense that
is a
countable distributive lattice, while
is not even a lattice.
Second and more importantly, there are plenty of specific, natural
degrees in
which are intermediate between
and
, the
top and bottom elements of
. Thus
does not suffer from the
above-mentioned lack of examples, which plagues
.
In more detail, let
be the lattice of weak degrees
(a.k.a., Muchnik degrees) of mass problems given by
nonempty
subsets of
. In 1999
[30] we showed that among the intermediate degrees in
is the specific degree
associated with the set of
-random reals. The concept of
-randomness was already well
known from algorithmic information theory [19]. After
1999, we and other authors [2,3,4,5,32,33,34] continued
the study of
, using priority arguments to prove structural
properties, just as for
. In addition, we [33]
discovered families of specific, natural, intermediate degrees in
related to foundationally interesting topics such as reverse
mathematics, Gentzen-style proof theory, and computational complexity.
Some additional degrees of this kind are presented in Sections
3 and 4 below.
The purpose of the present paper is to further clarify the
relationship between the semilattice
and the lattice
.
Namely, we exhibit a specific, natural embedding
which is one-to-one, preserves the semilattice structure of
, and
carries the top and bottom elements of
to the top and bottom
elements of
. See Theorem 5.5 below. By identifying
with its image in
, we place the recursively enumerable
Turing degrees into a wider context, where natural intermediate
degrees occur. We view this as a step toward overcoming the
above-mentioned difficulties concerning
.
At the same time, our embedding of
into
fails to solve the
long standing open problem of finding a specific, natural,
intermediate degree within
itself. Indeed, we shall see below
(Theorem 5.6) that, regrettably, all of the intermediate
degrees belonging to the image of
under our embedding are
incomparable with all of the known natural intermediate degrees in
.
In this section we review some basic information concerning the
semilattice
and the lattice
.
Throughout this paper we shall use standard recursion-theoretic
notation from Rogers [26] and Soare [35]. For
special aspects of mass problems and
sets, a convenient
reference is [33].
We write
for the set of natural numbers,
for the space of total functions from
into
, and
for the space of total functions from
into
. We sometimes identify a set
with its characteristic function
given by
if
,
if
. For
and
we write
to mean that the Turing machine with Gödel number
and oracle
and input
eventually halts with output
. In the
absence of an oracle
, we write
. For
we consider recursive functionals
given by
for some
and all
and
. A function
is said to be recursive if there exists
such that
for all
. A set
is said to be recursively enumerable if it
is the image of a recursive function, i.e.,
for some recursive
.
For
we write
to mean that
is
Turing reducible to
, i.e.,
. The Turing degree of
, denoted
, is the set of all
such that
, i.e.,
and
. The set
of all Turing degrees is
partially ordered by putting
if and only if
. Under this partial ordering, the bottom element of
is
is recursive
. Within
, the least upper bound of
and
is
where
and
for all
. A Turing degree
is said
to be recursively enumerable if
where
is recursively enumerable. The set of all
recursively enumerable Turing degrees is denoted
. It is easy to
see that
is closed under the least upper bound operation
inherited from
. The top and bottom elements of
are
and
respectively, where
is the Turing
degree of the Halting Problem,
.
Thus
is a countable upper semilattice with a top and bottom
element.
Proof.The least upper bound of
and
in
is
where
Proof.See [33, Theorems 4.7 and 4.10].![]()
Proof.This is immediate from Theorem 2.7.![]()
Proof.If
and
are
subsets of
, then so are
and
. Thus
is closed under the least
upper bound and greatest lower bound operations inherited from
. Clearly
is countable, because there are only countably
many
subsets of
. Clearly
is the bottom element of
. Let
be the set of completions of Peano Arithmetic. Identifying
sentences with their Gödel numbers, we may view
as a
subset of
. By Scott
[29],
is the
top element of
. See also [33, Section 6].![]()
We end this section by mentioning some technical notions and results concerning trees and almost recursiveness.
A finite sequence of natural numbers
is called a
string of length
. The set of all strings is denoted
. The set of strings of
's and
's is denoted
. If
are strings of length
respectively, then the concatenation
A tree is a set
such that, for all
,
. A path through
is an
such that
. The set of all paths through
is denoted
. We sometimes identify a string
with its Gödel
number
. A tree
is said to be
recursive if
is recursive.
Proof.See [33, Theorem 4.3].![]()
Proof.See [33, Theorem 4.18].![]()
The following result is known as the Almost Recursive Basis Theorem.
Proof.This is a restatement of the Hyperimmune-Free Basis Theorem of
[15, Theorem 2.4]. See also [33, Theorem 4.19].![]()
In this section we identify and characterize some specific, natural
degrees in
, and we investigate their degree-theoretic
properties.
Thus
is
-random if and only if
is random in the
sense of Martin-Löf [20], and
is
-random if
and only if
is random relative to the Halting Problem. For a
thorough treatment of randomness and
-randomness, see Kautz
[16] or Downey/Hirschfeldt [8]. We write
Proof.It is is well known (see for instance [33, Theorem 8.3])
that
is
. Relativizing to
we see that
is
-random
is
,
i.e.,
relative to
. Putting
, we see
that
is
relative to
. From this it
follows easily that
is
.![]()
Proof.First use a Skolem function technique to reduce to the case where
is
. Namely, fix a recursive predicate
such that
, and
replace
by the set of all
such that
holds.
Clearly the latter set is
and
. Assuming now
that
is a
subset of
, let
be a
recursive subtree of
such that
is the set of
paths through
. We may safely assume that, for all
and
the length of
,
. Let
be a
recursive subtree of
such that
is the set of paths
through
. Define
to be the set of strings
of the form
Proof.By Lemmas 3.2 and 3.3 we can find
sets
such that
and
. Since
is nonempty and
, there is a nonempty
set
.
Then
, hence
, hence
.![]()
Proof.By Lemma 3.4 there are
sets
such that
and
. Thus
and
. Clearly
, hence
. Clearly
has no recursive member, i.e.,
. By [33, Section 7] or [15, Corollary
5.4], the Turing upward closure of
is of measure
,
i.e.,
for all
of positive
measure. In particular
, i.e.,
. Summarizing, we have now shown that
.
It remains to show that
, i.e.,
. By Lemma 3.2, let
be a
nonempty
subset of
. By the Almost Recursive Basis
Theorem 2.15, let
be almost recursive. By Kautz
[16, Theorem IV.2.4(iv)] or Dobrinen/Simpson
[7, Remark 2.8], there is no almost recursive
.
In particular, there is no
such that
. Suppose
there were
such that
. By Theorem 2.14
let
be a total recursive functional such
that
. Put
. We have
, hence
, hence
is nonempty. Since
and
are
, it follows by [33, Theorem 4.4] that
is
. Hence, by
[33, Lemma 8.8],
is of positive measure.
Since
, it follows that the Turing upward
closure of
is of positive measure, but this is a
contradiction. We have now shown that, for a particular
,
there is no
such that
. Thus
, and this completes the proof.![]()
Proof.The first statement is [33, Theorem 8.10]. For the
second statement, assume that
is a
subset of
whose Turing upward closure
is of positive
measure. A computation in the style of Tarski and Kuratowski
(compare Rogers [26, Section 14.3]) shows that
is
. Since
is
of positive measure, let
be
of
positive measure. By Kautz [16, Lemma II.1.4(ii)] or
Dobrinen/Simpson [7, Theorem 3.3], we may assume that
is
relative to
. Relativizing [33, Lemma
8.7] to
, we have
. Since
, it follows that
,
hence
. Furthermore, since
is a nonempty
subset of
, we have
. We now see that
, i.e.,
. On the other
hand, by Theorem 3.6 let
be a
subset of
such that
. Note that
, hence
is of positive
measure. We have now shown that
is the maximum
such that
is of positive measure.
This completes the proof of our theorem.![]()
Proof.If
then trivially
, hence
is of measure
. Conversely, if
is of
positive measure, then by Theorem 3.8 we have
.![]()
Proof.By Theorem 3.6 let
be
such
that
. By Theorem 3.8,
is of positive measure. If there were a
set
of positive measure, then by Theorem
3.8 we would have
, contradicting Theorem
3.6.![]()
Proof.The first statement is [33, Theorem 8.12, part 3]. For
the second statement, let
be nonempty
subsets of
with
and
. Trivially
.
Moreover
, hence
is of positive measure if and only if at least
one of
and
is of positive measure. By
Corollary 3.9 this means that
if and only if
at
least one of
and
.![]()
Proof.This is a special case of [33, Theorem 7.5].![]()
Proof.Assume
. By the definition of
, we have
, hence
. Since
is separating and
, Theorem 3.13
tells us that
, hence
.![]()
Proof.This follows from Theorem 3.11 and Corollary 3.14.![]()
In this section we identify some additional specific, natural,
intermediate degrees in
related to diagonal nonrecursiveness.
Proof.Put
is diagonally nonrecursive
.
Obviously
is a nonempty
subset of
. By
Lemma 3.3 we can find a nonempty
set
such that
. But
, hence
. We now see that
.![]()
Proof.By Theorem 3.6 we have
.
Clearly
has no recursive member, i.e.,
. By
Giusto/Simpson [11, Lemma 6.18], for all
there
exists
such that
. Thus we have
. It remains to show that
.
By Kumabe [18] there is a diagonally nonrecursive function
which is of minimal Turing degree. But if
is
-random, then
is not of minimal Turing degree, because the
functions
and
defined by
are Turing
incomparable (see for instance van Lambalgen [37]). This
proves
. An alternative reference for the
conclusion
is Ambos-Spies et al [1, Theorems
1.4 and 2.1].![]()
Proof.A Tarski/Kuratowski computation shows that
is
. Let
be as in the proof of Theorem
4.2. By Lemma 3.3 we can find a nonempty
set
such that
. Thus
. By Ambos-Spies
et al [1, Theorems 1.4, 1.8, 1.9] we have
, and the rest is from Theorem
4.3.![]()
In this section we exhibit a specific, natural embedding of the
countable upper semilattice
into the countable distributive
lattice
.
Proof.The first statement is the special case of Lemma 3.3
with
and
. For the second statement, note that for
any
and
we have
, and
implies
. In particular this
holds when
and
are
singletons and
.![]()
Proof.The first statement is a special case of Lemma 5.2,
because
if and only if
is
(see
Kleene [17, Theorem XI, page 293]), which implies that
is a
singleton. It is easy to see that
. To show that
, by Kleene
[17, Theorem 38*, pages 401-402] let
be
. Then
and
, hence
.![]()
Proof.Let
be such that
and
. We must show that
if and only if
. The ``only if'' part is trivial.
For the ``if'' part, suppose
. In
particular,
. If
, then
, hence
, hence
by
the Arslanov Completeness Criterion [12], hence
, a contradiction. Thus
. This proves our lemma.![]()
We now obtain our main result.
Proof.This result is obtained by combining Lemmas 5.3 and
5.4.![]()
Proof.By the Arslanov Completeness Criterion [12] we have
. It remains to show that
. This is obvious, because for any
nonrecursive
we have
[27, §10, Theorem 1].![]()
We finish by noting some generalizations of Theorem 5.5.
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