THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER

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1 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER JASPER DEANTONIO Abstract. This paper is a study of the Turing Degrees, which are levels of incomputability naturally arising from sets of natural numbers. I explore the structure of the Turing Degrees, concluding with a proof that there exist Turing Degrees which are not comparable so as to illustrate the nonlinearity of the Turing Degrees. Contents 1. An Introduction to Computability 1 2. Computably Enumberable Sets and K Computing with Oracles and the Turing Equivalence 6 4. The Turing Degrees are not Linearly Ordered by the Relation 9 Acknowledgments 11 References An Introduction to Computability Computability Theory is the study of degrees of computability. For this paper, we shall specifically study the computability of sets of the natural numbers. Intuitively, a computable set is a set whose elements can be determined with finitely many precise directions, i.e., with a computer program. Thus, one can determine both what is and what is not in the set in finitely many steps. So, for a set to be computable, the complement of the set must also be computable. For instance, the prime numbers are computable: if we fix n N, we simply check every number less than n and greater than 1 to see if that number divides n. So, in n 2 steps, we know whether or not n is prime. Thus, the primes are computable; for any natural number, we can, in finitely many steps, tell whether or not the number is prime. In order to rigorously define this intuitive notion, we shall use a structure called the Turing Machine: Definition 1.1. Turing Machine A Turing machine uses five substructures: A Tape T : An infinite strip with an infinite line of boxes on it. Each box can contain a symbol. Date: September 24,

3 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER 3 Let T be a Turing machine with quadruples: q 0 10q 1 and q 1 0Rq 0. Now, consider what happens at input n. We input a tape with n + 1 ones into the Turing machine. At q 0, the starting state, the reading head encounters 1, so we use our first quadruple. So, we write 0 on the tape and move to state q 1. Next, we use our second quadruple because we encounter q 1 and 0. This causes us to move right, where we ll find another 1. Thus, one can see how this process can turn any tape of consecutive ones into a tape containing only zeros, making the output zero, which is precisely what we desire. How does the intuitive computability described at the beginning of this section relate to Turing computable? One might think that the intuitive idea is more broad. In fact, by the Church-Turing Thesis, these two notions are equivalent. Therefore, we shall justify computability from now on using our intuitive notion as opposed to Turing computability. Next, we consider how many computable functions there are. Lemma 1.4. There are countably many computable sets. Proof. Every computable set has a unique computable function. Every computable function is expressible as a Turing machine. Every Turing machine has a program which is composed of finitely many quadruples of the form q i SAq j where q i and q j are internal states, S is a symbol on the tape, and A is an action. There are countably many internal states, finitely many symbols, and finitely many actions (L,R, and one for each symbol). Thus, there are countably many quadruples available and each Turing machine can only choose finitely many of these. Therefore, there are only countably many Turing machines and thus countably many computable sets. Note that we could also prove this lemma using the intuitive computability discussed earlier with a very similar argument. Because there are countably many computable functions, we can put these functions into a one-to-one correspondence with the natural numbers as follows: Definition 1.5. φ i and W i We define φ i to be the i-th computable function. We define W i to be the domain of φ i. Using this defintion, we can enumerate all computable functions and domains of these functions, giving us much more control over these computable functions. 2. Computably Enumberable Sets and K 0 Having considered what is computable, we now explore what is not computable. In order to do so, we define a new notation: Definition 2.1. The pairing function We define a pairing function to be a computable bijection function <, >: N 2 N. Using these pairing functions, we can examine subsets of N 2 as subsets of N and thus study their computability. In fact, using multiple pairing functions, for any k, we can consider the computability of subsets of N k.

4 4 JASPER DEANTONIO Specifically, we study the set K 0 = { i, x φ i (x) halts }. This set is known as the Halting Problem. At first one may be confused about why we even consider programs which don t halt. Recall that one can easily write a computer program which does not halt; a simple example is any loop that has no end condition. These are still computer programs and can be encoded as Turing machines. Thus, there exists m N such that φ m is exactly a program which, given every input, loops infinitely. Therefore, in studying the computable functions, we must be aware of functions which do not halt, which is one reason why we shall study K 0. Theorem 2.2. K 0 is not computable Proof. Assume K 0 is computable. Then, χ K0 (i, n) = 0 or 1 for all i, n N, i.e., for all i, n, we can computably tell whether or not n W i. Therefore, we can define the function H : N N as follows: { 1 χk0 (i, i) = 0 H(i) = 0 χ K0 (i, i) = 1 H is computable since χ K0 is. Therefore, there exists e N such that H = φ e because we can enumerate all the computable functions. We then consider H(e). { 1 χk0 (e, e) = H(e) = 0 H(e) = 0 χ K0 (e, e) = H(e) = 1 Either way, H(e) = 1 0 = H(e), which is a contradiction. This illustrates that K 0 cannot be computable. Remark 2.3. Since K 0 is not computable, what is it? It seems like K 0 is almost computable, as we can enumerate the elements of K 0 using the following process: Step through the triples i, x, s, running φ i (x) for s steps. If φ i (x) halts in s steps, add φ i (x) to K 0. Unfortunately, this is not all we need for K 0 to be computable. The problem is that we cannot computably determine if any pair is not in K 0. On a pair i, x, if φ i (x) has not halted in s steps, we don t know if it will halt on the (s + 1)-th step or never halt. Thus, we can never state that a pair i, x is not in K 0 because we cannot look ahead an infinite number of steps to see if φ i (x) ever halts. Intuitively, this is why K 0 is not computable. Because K 0 is incomputable, it cannot be generated by any computer. No possible algorithm could give the exact set which is K 0. Yet, K 0 is a relatively standard set in the sense that it is easy for mathematicians to understand and simple for us to construct. This illustrates the hidden difficulty of actually creating many of the sets which mathematicians commonly use. Definition 2.4. Computably Enumerable We define a computably enumerable set (abbreviated c.e. set) to be a set A N such that A = range(f) for some computable function f. Intuitively, this means that the elements of A can be enumerated using f. So, for every a A, there are n, s N such that, after s steps, f(n) halts and f(n) = a. Example 2.5. K 0 and W i are c.e. K 0 is a c.e. set, since, using the process described in Remark 2.3, the elements of K 0 can be enumerated.

5 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER 5 Let g : N 2 N such that g(x, s) does the first s steps of φ i on input x, returning x if φ i halts. Let h : N N 2 such that h is bijective. Then, if we run (g h)(n) for each n N starting with n = 1, we will be able to enumerate the elements of W i. Therefore, W i is c.e. Next we provide a few theorems about computably enumerable sets to give the reader a better understanding of them. Theorem 2.6. Every computable set is c.e. Proof. Let A be a computable set. We want to show that A is c.e. so we will find a computable f such that f(n) = A. If A =, then let f be the Turing machine that just loops infinitely for any input. Then, f never returns any value. Thus, f(n) =. Now, assume A. Let a A. Then, since χ A is computable (because A is a computable set), consider the function g : N N: { n χa (n) = 1 g(n) = a χ A (n) = 0 Then g(n) = A. g is computable since χ A is computable. Thus, A is c.e. There is another way to describe sets which are c.e., as we illustrate below: Definition 2.7. Existential Sets We define an existential set to be a set A N such that for all x, x A if and only if there exists y such that R(x, y) holds for some computable relation R. Example 2.8. K 0 is an existential set. i, x K 0 if and only if there exists s N such that φ i (x) halts in s steps. Thus, our relation here is composed of three variables i, x, s. So, i, x K 0 exactly when (i, x, s) is in this relation R. Note that R is computable since we only run the algorithm s times on the given input, then see if it has halted or not. Theorem 2.9. A set A N is c.e. iff A is a existential set. Proof. First we show that if A N is c.e., then A is a existential set. If A is c.e., then A is the range of some computable function f. So, x A ( s)[f(s) = x] Since f is computable, this relation is computable as well. Therefore, A is existential due to this relation. Next, we show that if A is a existential set, then A N is c.e. If A is an existential set, then we have, for some computable relation R, x A ( s)[r(s, x)] Define the function ψ : N N as follows: { 0 ( s)[r(s, x)] ψ(x) = loop infinitely otherwise ψ is an algorithm a computer could reproduce given the relation R. But the relation R is computable. So, ψ is also computable. Therefore, the domain of ψ, which is the set A, is equal to W e for some e N. But, as stated in example 2.5, W e is c.e. Therefore, A is also c.e.

7 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER 7 Example 3.6. For all e N, W e T K 0 If x N, x W e if and only if e, x K 0. Therefore, W e is K 0 -computable, and W e T K 0 We can now generalize the notion of the Halting Problem, or K 0, as shown below. Definition 3.7. The Jump Operator We define the jump of a set A N to be A = { i, x : Wi A (x) halts} Specifically, note that K 0 = Also, given that K 0 =, we can then take the jump of K 0 =. We can continue this process indefinitely, yielding countably many sets, { (n) } n N. Similar to the proof of < as seen in Theorem 2.1, for every n N, (n) < T (n+1). Therefore, each (n) is less computable than every (m), where m < n. So, we have a sequence { (n) } n N which gets less and less computable without ever stopping. Lemma 3.8. (1) Relative computability is a transitive relation, as is Turing equivalence. (2) The Turing equivalence is well defined, i.e., if A T B and A T C and B T D, then C T D Proof. (1)Let A T B and B T C. Then, for any n N, we can computably determine whether or not n A using answers to finitely many questions Is m 0 B?, Is m 1 B?,... Is m k B?. Since each of these questions can be computably answered using answers to finitely many questions Is p 0 C?, Is p 1 C?,... Is p l C?, we can computably determine whether or not n A using answers to finitely many questions about elements of C. Thus, A is C-computable, and A T C. As Turing equivalence is based upon relative computability, it follows immediately that if A T B and B T C, then A T C. (2)If A T B, then there exists a computable function f using B as an oracle which computes A. So, for any n N, f(n) B-computably returns a 1 or a 0 if n A or n / A, respectively. Similarly, since A T C and B T D, C T A, A T B, and B T D. Therefore, by part 1 of this lemma, C T D. Example 3.9. Let L = {e : W e }. Then, K 0 T L. First we show that K 0 T L. Consider the pair i, n. We want to know L-computably if i, n K 0, i.e., if n is in the domain of φ i. In order to do so, we computably define a function φ i as follows: { φ Loop infinitely k n i(k) = φ i (n) k = n Because this program φ i is Turing computable (due to the fact that φ i(n) is), there exists an e N such that φ e = φ i. So, W e is the domain of φ i. For every natural number x that is not equal to n, φ i (x) does not halt, so x / W e. If φ i (n) halts, then {n} is the domain of φ. Therefore, W e, and so e L. Otherwise W e =, so e / L. So, to determine if i, n K 0, we need only ask if e is in L. Thus, K 0 T L. Next, we show that L T K 0 Note that x L if and only if there exists n, s such that φ x (n) halts in s steps. Therefore, L is an existential set. By Theorem 2.7, there exists e N such that W e = L. Since W e T K 0 for all e, L T K 0. Therefore, L T K 0 and K 0 T L, so K 0 T L.

8 8 JASPER DEANTONIO Remark Since L T K 0 and < T K 0, < T L. Therefore, L is not computable. Yet, we can compute L from K 0. This illustrates the power of relative computability. It allows us to use the otherwise-inaccessible information of K 0 to computably generate incomputable sets. So, we could say that K 0 has some computing ability, or information, stored in its set. In fact, any incomputable set A has some of this information, as there are other incomputable sets which can be A-computed. Since any two Turing equivalent sets can generate each other computably, any two Turing equivalent sets have the same amount of information. Thus, we shall shift our study to these sets of Turing equivalent sets so as to better understand these varying degrees of information. Definition Turing Degrees (1) We define the Turing Degree of a set A N, denoted deg(a), to be the set {X N : X T A}. (2) We define deg( ) to be the Turing degree of the computable sets. (3) We define D to be the set of all Turing degrees, i.e., D = {deg(a) : A N} (4) We define the operator for Turing Degrees A and B as follows: A B if there exists X A and Y B such that X T Y. Note that for all X A, Y B, X T Y by Lemma 3.8 so long as X Y. So, when A B, every set in A is B-computable. Example The Least Turing Degree We show that deg( ) is the least Turing Degree, i.e., that deg( ) deg(a) for all A N. Let X be a computable set and A N. Then, X T A by Example 3.4. Therefore, since X deg( ), deg( ) deg(a) forall A N. Theorem D has uncountably many degrees. Proof. For ease of notation in this proof, for any set A, let car(a) be the cardinality of A. We prove this theorem by contradiction. Assume that D has countably many degrees. Then, note that for every A N, A deg(b) for some B N. Therefore, car(p(n)) car(d) sup{car(deg(a)) : A N} As stated in Remark 3.2, for any set A N, the set of A-computable sets is countable. So, sup{car(deg(a)) : A N} is countable since each deg(a) is countable. Thus, D sup{deg(a) : A N} must be countable, as a countable set cross a countable set is still countable. Then, P(N) must also be countable. This is a contradiction, as P(N) is uncountable. Therefore, D must have uncountably many degrees. Corollary D has no greatest element. Proof. We prove this by contradiction. Assume that D has a greatest element. Then, there exists A D such that B A for all B D. Let X A. The set of X-computable sets is countable. Therefore, the set of Turing degrees less than A is also countable. However, the set of Turing degrees less than A is D. By the above theorem, D is uncountable. Thus, we arrive at a contradiction. So, D cannot have a greatest element.

9 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER 9 4. The Turing Degrees are not Linearly Ordered by the Relation In order to prove that the Turing Degrees are not linearly ordered by, we will first need some more definitions. Definition 4.1. Strings (1) We define a string, generally denoted σ or τ, to be a finite sequence of 0 s and 1 s. For example is a string. Note that a string can also be considered as the beginning of a characteristic function of a set. For example, for some set A, the string would designate that 0 / A, 1 A, 2 / A, 3 / A, etc. (2) We define the length of a string σ, denoted σ, to be the number of entries in σ. For example, = 6. (3) We define the concatenation of two strings σ and τ, denoted σ τ, to be the string σ followed by τ. For example, if σ = and τ = 11111, σ τ = (4) We say that τ is an extension of σ, denoted σ τ, if in every place in which σ is defined, τ has the same value, and σ τ. For example, is an extention of 010. Definition 4.2. The Use of a Computation Let A be an expression whose computation involves a finite set of queries n 0,..., n k to oracle A. We define the use of A to be the maximum of the set {n i : 0 i k}. Intuitively, the use is the largest element of A one needs to compute A. Since every computation is only finitely many steps long, there is always a use for an expression. Finally, we prove that the Turing Degrees are not linearly ordered by the operation : Theorem 4.3. The Turing Degrees are not linearly ordered - that is, there exist C, E D such that neither C E nor E C. Proof. In order to show that there exist C, E D such that neither C E nor E C, we will construct two sets, A, B N, such that A is not B-computable and B is not A-computable. Given these sets A and B, let C = deg(a) and E = deg(b). Next, assume, without loss of generality, that C E. Then, as illustrated in Definition , A T B, which would be a contradiction. Therefore, neither C E nor E C. So, if we can construct A and B, then the theorem is proven. We construct A, B N such that A is not B-computable and B is not A-computable. Recall that there are countably many A-computable sets and countably many B- computable sets. So, to ensure that A is not B-computable and B is not A computable, we will satisfy the following relations: R 2i : χ A φ B i R 2i+1 : χ B φ A i With these relations satisfied, A cannot be B-computable because A is not any of the B-computable sets. Similarly, B cannot be A-computable. To construct A and B, we shall use sequences of strings {σ i } i N and {τ i } i N, where σ 0 σ 1 σ 2... and τ 0 τ 1 τ 2... Then, we shall define the characteristic functions of A and B by: χ A = i N σ i

10 10 JASPER DEANTONIO χ B = i N τ i In order to satisfy the relations, at stage i + 1, we will define σ i+1 and τ i+1 in such a way as to ensure that the i-th requirement is satisfied. The Construction Stage 0: Define σ 0 = τ 0 =. Stage i + 1 = 2e + 1: We specifically handle R 2e and leave R 2e+1 to the reader as the process is exactly the same. Assume σ 0 σ 1 σ 2... σ i and τ 0 τ 1 τ 2... τ i are already defined. Let x i = σ i We look for a string τ τ i such that φ τ e(x i ) halts. CASE I: If τ exists Then computably choose a τ and let k = φ τ e(x i ). Then define: σ i+1 = σ i (1 k) τ i+1 = τ We show that φ B e (x i ) = φ τ e(x i ) by contradiction. Assume φ B e (x i ) φ τ e(x i ). Then, since τ is the beginning of the characteristic function of B, τ and B agree up to τ. Therefore, there must exist k N such that k > τ and the k-th value of B changes the output of φ B e (x i ) so that it is not equal to φ τ e(x i ). Then, φ τ e(x i ) must query the k-th value of τ as well. However, as τ is not defined at this value, φ τ e(x i ) could not halt. This would be a contradiction of our initial assumption, which states that φ τ e(x i ) halts. Therefore, φ B e (x i ) = φ τ e(x i ), and so φ B e (x i ) = φ τ e(x i ) = k (1 k) = χ A (x i ). The characteristic functions of φ B e and A disagree on input x i and thus A cannot be the i-th B-computable set. Therefore, the relation R 2e is satisfied. CASE II: If τ does not exist Then define σ i+1 = σ i 0 τ i+1 = τ i 0 I claim that, regardless of how τ n, n > i, are picked, φ B e (x i ) will not halt. Given that φ B e (x i ) does not halt, our relation is satisfied because φ A (x i ) = 0 but φ B e (x i ) is undefined. So A is not the i-th B-computable set, as the characteristic functions disagree at input x i. Thus, the relation R 2e is satisfied. Proof of Claim We prove this claim by contradiction. Assume that φ B e (x i ) halts. Let w be greater than τ i and the use of φ B e (x i ). Then, for τ equal to the first w elements of B, φ B e (x i ) = φ τ e(x i ) because for every element of B used to compute φ B e (x i ), τ has the exact same elements. Note also that τ τ i. We have found

11 THE TURING DEGREES AND THEIR LACK OF LINEAR ORDER 11 a τ such that φ τ e(x i ) halts, so we cannot be in Case II. Therefore, we arrive at a contradiction. By this contradiction, φ B e (x i ) cannot halt. Corollary 4.4. In fact, we can find A, B N such that A, B T K 0 and A is not B-computable and B is not A-computable. So, not even the computably enumerable sets are linearly ordered. Proof. We need to verify that the construction in the above theorem can be carried out computably using an oracle V which is K 0 -computable (because then A, B T V T K 0 ). Note that the only part of the above construction which is not computable is in determining whether or not there exists τ such that τ extends τ i and φ τ e(x i ) halts. Therefore, we need to be able to tell V -computably whether or not τ exists. Note that τ exists if and only if there exists a pair s, τ such that τ τ i and φ τ e(x i ) halts in s steps. Define V = { π, π : ( s, τ )[τ π and φ τ e( π ) halts in s steps]} Then, τ exists at stage e + 1 if and only if τ e, σ e V Note that V is an existential set, so V is equal to some W e, e N, as illustrated by Theorem 2.7. Then, by Example 3.4, W e T K 0, so V T K 0. Therefore, A, B T K 0. Conclusion The above theorem and associated corollary give us a very interesting result: there exist sets which are not Turing comparable. Note that neither of the two sets A and B generated are computable (if, say A, was computable, then by Example 3.5, A T B, which is a contradiction). This means that both A and B are sets which hold some computing information. Since each A and B hold computing information, but A is not B-computable and B is not A-computable, both A and B have some computing information which the other does not. Thus, there must be at least two different kinds of this computing information: a set does not have just an amount of computing information but also a kind. The above corollary is surprising because it shows that there are different kinds of information even in the computably enumerable sets. Thus, this division of information is immediate. Acknowledgments. It is my pleasure to thank my mentors, Damir Dzhafarov and Eric Astor, for their help in my studies. They guided me towards useful books, helped me write this paper, and weathered my questions which were both frequent and at times more obscure than the concept I had questions about. References [1] Barry Cooper. Computability Theory. Chapman and Hall/CRC [2] Michael Sipser. Introduction to the Theory of Computation. PWS Publishing Company

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