# Mass problems and intuitionism

Stephen G. Simpson1

Department of Mathematics

First draft: July 25, 2007
This draft: April 28, 2008

Notre Dame Journal of Formal Logic, 49, 2008, 127-136

Pennsylvania State University

### Abstract:

Let be the lattice of Muchnik degrees of nonempty subsets of . The lattice has been studied extensively in previous publications. In this note we prove that the lattice is not Brouwerian.

# Introduction

Definition 1   Let denote the set of natural numbers, . Let denote the Baire space, . Following Medvedev [27] and Rogers [32, §13.7] we define a mass problem to be an arbitrary subset of . For mass problems and we say that is Medvedev reducible or strongly reducible to , abbreviated , if there exists a partial recursive functional such that for all . We say that is Muchnik reducible or weakly reducible to , abbreviated , if for all there exists such that is Turing reducible to . Clearly Medvedev reducibility implies Muchnik reducibility, but the converse does not hold.

Definition 2   A Medvedev degree or degree of difficulty or strong degree is an equivalence class of mass problems under mutual Medvedev reducibility. A Muchnik degree or weak degree is an equivalence class of mass problems under mutual Muchnik reducibility. We write the Medvedev degree of . We write the Muchnik degree of . Let be the set of Medvedev degrees, partially ordered by Medvedev reducibility. There is a natural embedding of the Turing degrees into given by . Let be the set of Muchnik degrees, partially ordered by Muchnik reducibility. There is a natural embedding of the Turing degrees into given by . Here is the singleton set whose only element is .

Definition 3   Let be a lattice. For we define to be the unique minimum such that . Note that may or may not exist in . Following Birkhoff [8,9] (first two editions) and McKinsey/Tarski [25] we say that is Brouwerian if exists in for all and has a top element. It is known (see Birkhoff [9, §IX.12] [10, §II.11] or McKinsey/Tarski [25] or Rasiowa/Sikorski [31, §I.12]) that if is Brouwerian then is distributive and has a bottom element and for all in the sublattice

is again Brouwerian.

Remark 1   Given a Brouwerian lattice , we may view as a model of first-order intuitionistic propositional calculus. Namely, for we define , , as above, and where is the top element of . We may also define if and only if in . There is a completeness theorem (see Tarski [52] or McKinsey/Tarski [24,25,26] or Rasiowa/Sikorski [31, §IX.3] or Rasiowa [30, § XI.8]) saying that a first-order propositional formula is intuitionistically provable if and only if it evaluates identically to the bottom element in all Brouwerian lattices.

Remark 2   Brouwerian lattices have also been studied under other names and with other notation and terminology. A pseudo-Boolean algebra is a lattice such that the dual of is Brouwerian; see Rasiowa/Sikorski [31] and Rasiowa [30]. Pseudo-Boolean algebras are also known as Heyting algebras; see Balbes/Dwinger [2, Chapter IX], Fourman/Scott [18], and Grätzer [19]. Brouwerian lattices are also known as Brouwer algebras; see Sorbi [48,49], Sorbi/Terwijn [51], and Terwijn [53,54,55,56,57]. Remarkably, the so-called Brouwerian lattices of Birkhoff [10] (third edition) are dual to those of Birkhoff [8,9] (first two editions). We adhere to the terminology of Birkhoff [8,9].

Remark 3   It is known that and are Brouwerian lattices. There is a natural homomorphism of onto given by . This homomorphism preserves the binary lattice operations and and the top and bottom elements, but it does not preserve the binary if-then operation .

Remark 4   The relationship between mass problems and intuitionism has a considerable history. Indeed, it seems fair to say that the entire subject of mass problems originated from intuitionistic considerations. The impetus came from Kolmogorov 1932 [22,23] who informally proposed to view Heyting's intuitionistic propositional calculus [20] as a calculus of problems'' (Aufgabenrechnung''). This idea amounts to what is now known as the BHK or Brouwer/Heyting/Kolmogorov interpretation of the intuitionistic propositional connectives; see Troelstra/van Dalen [59, §§1.3.1 and 1.5.3]. Elaborating Kolmogorov's idea, Medvedev 1955 [27] introduced and noted that is a Brouwerian lattice. Later Muchnik 1963 [28] introduced and noted that is a Brouwerian lattice. Some further papers in this line are Skvortsova [47], Sorbi [48,49,50], Sorbi/Terwijn [51], and Terwijn [54,53,55,56,57].

Definition 4   Let denote the Cantor space, . Following Simpson [40] let be the sublattice of consisting of the Medvedev degrees of nonempty subsets of , and let be the sublattice of consisting of the Muchnik degrees of nonempty subsets of .

Remark 5   The lattices and are mathematically rich and have been studied extensively. See Alfeld [1], Binns [3,4,5,6], Binns/Simpson [7], Cenzer/Hinman [11], Cole/Simpson [13], Kjos-Hanssen/Simpson [21], Simpson [34,35,37,38,39,40,41,42,43,45,44], Simpson/Slaman [46], and Terwijn [54]. It is known that contains not only the recursively enumerable Turing degrees [42] but also many specific, natural Muchnik degrees which arise from foundationally interesting topics. Among these foundationally interesting topics are algorithmic randomness [40,42], reverse mathematics [36,40,41,43], almost everywhere domination [43], hyperarithmeticity [13], diagonal nonrecursiveness [40,42], subrecursive hierarchies [21,40], resource-bounded computational complexity [21,40], and Kolmogorov complexity [21]. Recently Simpson [44] has applied and to prove a new theorem in symbolic dynamics.

Remark 6   It is known that and are distributive lattices with top and bottom elements. Moreover, the natural lattice homomorphism of onto restricts to a natural lattice homomorphism of onto preserving top and bottom elements.

Remark 7   In view of Remarks 3, 4, 5 and 6, it is natural to ask whether and are Brouwerian lattices. The purpose of this note is to show that is not a Brouwerian lattice. Letting denote the top element of , we shall produce a family of Muchnik degrees such that does not exist in . In other words, does not exist in .

Remark 8   It remains open whether is a Brouwerian lattice. Terwijn [54] has shown that the dual of is not a Brouwerian lattice. It remains open whether the dual of is a Brouwerian lattice.

# Proof that is not Brouwerian

In this section we prove that the lattice is not Brouwerian.

Definition 5   For we write to mean that is Turing reducible to , i.e., is computable relative to the Turing oracle . We write the Turing jump of . In particular the halting problem the Turing jump of . We use standard recursion-theoretic notation from Rogers [32]. We say that is majorized by if for all .

We begin with four well known lemmas.

Lemma 1   Given we can find such that is .

Proof.Since , it follows by Post's Theorem (see for instance [32, §14.5, Theorem VIII]) that is . From this it follows that the singleton set is . Let be a recursive predicate such that our is the unique such that holds. Let where is defined by the least such that holds. It is easy to verify that and is .

Lemma 2   If is and is nonrecursive, then is not majorized by any recursive function.

Proof.This lemma is equivalent to, for instance, [40, Theorem 4.15].

Lemma 3   For all nonempty sets we have .

Proof.This lemma is a restatement of the well known Kleene Basis Theorem. Namely, every nonempty subset of contains an element which is . See for instance the proof of [42, Lemma 5.3].

Lemma 4   Let be nonempty such that no element of is recursive. Then we can find such that and .

Proof.By Lemma 3 it suffices to find such that and . To construct we may proceed as in the proof of Lemma 5 below. The construction is easier than in Lemma 5, because we can ignore .

Lemma 5   Let be nonempty . Let be such that and . Then we can find such that and and .

Proof.We adapt the technique of Posner/Robinson [29].

Let be a recursive tree such that paths through . By Lemmas 1 and 2 we may safely assume that is not majorized by any recursive function.

For integers and strings we write

where the least such that either or . Note that the mapping is recursive and monotonic, i.e., implies . Moreover, for all we have if and only if . Here we are writing

In order to prove Lemma 5, we shall inductively define an increasing sequence of strings , . We shall then let . In presenting the construction, we shall identify strings with their Gödel numbers.

Stage . Let the empty string.

Stage . Assume that has been defined. The definition of will be given in a finite number of substages.

Substage . Let .

Substage . Assume that has been defined. Let the least such that either

and
or
and .
Note that exists, because otherwise would be majorized by the recursive function least such that and . If (1) holds with let . If (2) holds with let . This completes our description of the construction.

We claim that, within each stage , (2) holds for some . Otherwise, we would have infinite increasing sequences of strings

and

with for all . Moreover, these sequences would be recursive relative to , namely where least such that (1) holds. Thus, letting , we would have and . Thus , a contradiction. This proves our claim.

From the previous claim it follows that is defined for all . By construction, the sequence , is recursive relative to . Moreover, is recursive relative to , because for all we have .

Finally let . Clearly .

We claim that the sequence is . Namely, given , we may use and as oracles to compute as follows. We begin with . Given we use the oracle to compute . Then, using the oracle , we ask whether there exists such that and . If so, we compute the least such . If not, we use the oracle to compute . This proves our claim.

From the previous claim it follows that . Hence .

We claim that . To see this, let be such that is a total function. Consider what happened at stage of the construction. Consider the least such that (2) holds, i.e., . Since (2) holds, there does not exist such that and . In particular, letting be an initial segment of such that and , we have . Hence . This proves our claim.

From the two previous claims, it follows that . The proof of Lemma 5 is now finished.

Remark 9   By a similar argument we can prove the following. Let be . Let be of hyperimmune Turing degree such that . Let be such that . Then we can find such that and and .

Lemma 6   Let be nonempty . Let be . Then

Proof.This is Simpson's Embedding Lemma. See [42, Lemma 3.3] or [45].

We are now ready to prove our main result.

Theorem 1   is not Brouwerian.

Proof.Let be the set of completions of Peano Arithmetic. Recall from Simpson [40] that the top element of . By Lemma 4 let be such that and . Let

and note that . By Lemmas 1 and 6 we have .

It is well known (see for instance [40, Remark 3.9]) that is a complete lattice. This means that for all the least upper bound and the greatest lower bound exist in . Therefore, within , let

and note that in . In other words, in .

We claim that . Otherwise, let where is nonempty . Since , we have for all . Since , it follows that . By Lemma 5 let be such that and and . Let

and note that . By Lemmas 1 and 6 we have . By Lemma 3 we have , hence contradicting the definition of . This proves our claim.

Because it follows that does not exist in . Thus is not Brouwerian.

Remark 10   The same proof shows that for all in we can find in such that does not exist in . On the other hand, we know at least a few nontrivial instances where exists in . For example, letting be the Muchnik degree of the set of 1-random reals, Theorem 8.12 of Simpson [40] tells us that in and exists in . In fact, in is equal to in , which is equal to . We do not know any instances of where exists in and both and are in .

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Mass problems and intuitionism

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#### Footnotes

... Simpson1
The author thanks Sebastiaan Terwijn for helpful correspondence. The author's research was partially supported by the United States National Science Foundation, grants DMS-0600823 and DMS-0652637, and by the Cada and Susan Grove Mathematics Enhancement Endowment at the Pennsylvania State University.

Stephen G Simpson 2008-04-28