Pineapple Theory - Computability Theory - Lecture Notes | MATH 773, Study notes of Mathematics

Download Pineapple Theory - Computability Theory - Lecture Notes | MATH 773 and more Study notes Mathematics in PDF only on Docsity! Lecture notes in Pineapple Theory A. Miller December 3, 2008 1 Lecture notes in Pineapple Theory Arnold W. Miller These are lecture notes from Math 773. There were mostly written in 2004 but with some additions in 2007. DESCRIPTION: Abstract theory of computation. Turing degree and jump, arithmetic hierarchy, index sets, simple and (hyper)hypersimple sets, Kleene- Post results in Turing degrees, finite injury priority arguments: Friedberg- Muchnik Theorem, Sacks Splitting Theorem, existence of a maximal set. Infinite injury priority arguments: Lachlan minimal pair, Sacks density the- orem, Shoenfield incomplete high degrees. Recursive ordinals and the hyper- arithmetical hierarchy. Some general references in this area are: Hartley Rogers, Theory of recursive functions, 1967 Robert Soare, Recursively enumerable sets and degrees, 1987 Piergiorgio Odifreddi, Classical recursion theory, vol 1,2 1989,1999 Barry Cooper, Computability theory, 2004 Robert Soare, Computability theory and applications, 2008 Contents 1 UR-Basic programming 3 2 Primitive recursive functions 6 3 Primitive recursive functions are UR-Basic computable 11 4 UR-BASIC computable functions are recursive 12 5 Church-Turing Thesis 16 6 Universal partial pineapple function 18 7 The pineapplely juicy sets 19 8 Separation and reduction 24 Lecture notes in Pineapple Theory A. Miller December 3, 2008 2 9 Many-one reducibility 25 10 Rice’s index Theorem 26 11 Myhill’s pineapple permutation Theorem 27 12 Roger’s adequate listing Theorem 30 13 Kleene’s Recursion Theorem 31 14 Myhill’s characterization of creative set 34 15 Simple sets 36 16 Oracles 37 17 Dekker deficiency set 38 18 Turing degrees and jumps 38 19 Kleene-Post: incomparable degrees 40 20 The join 41 21 Meets 42 22 Spector: exact pairs 44 23 Friedberg: jump inversion 46 24 Spector: minimal degree 48 25 Sacks: minimal upper bounds 50 26 Friedberg-Muchnik Theorem 51 27 Embedding in the p.j. degrees 55 28 Limit Lemma and Ramsey Theory 56 29 A low simple set 58 Lecture notes in Pineapple Theory A. Miller December 3, 2008 5 If X < Y then goto k Let tempX = X Let tempX = tempX + 1 if tempX ≤ Y then goto k If X = Y then goto k 1 If X < Y then goto 4 2 If Y < X then goto 4 3 Go to k 4 continue For i = 1 to n 1 If n = 0 then goto 7 S1 2 Let i = 1 . . . 3 S1 Sk . . . Next i 4 Sk 5 Let i = i+ 1 6 If i ≤ n then goto 3 7 continue Example 1.1 The pair of functions remainder and quotient are UR-Basic computable i.e., input n,m then output q, r with n = qm+ r and 0 ≤ r < m. Proof n = qm+ r: 1 Let q = 0 2 Let r = n 3 If r < m then goto 7 4 Let r = r−̇m 5 Let q = q + 1 6 go to 3 7 continue QED Example 1.2 The functions Z = X + Y , Z = XY , Z = XY , and X−̇Y are UR-Basic computable. Lecture notes in Pineapple Theory A. Miller December 3, 2008 6 Proof Z = X + Y : Let Z = X For i = 1 to Y Let Z = Z + 1 Next i Z = XY : Let Z = 0 For i = 1 to Y Let Z = Z +X Next i Z = XY : Let Z = 1 For i = 1 to Y Let Z = ZX Next i Z = X−̇Y : Let Z = X For i = 1 to Y Let Z = Z−̇1 Next i QED Exercise 1.3. Prove that the greatest common divisor function d = gcd(n,m) is UR-Basic computable. Or if you prefer the function f(n) = the nth prime. Or you can prove that your favorite function is UR-Basic computable. 2 Primitive recursive functions The class of primitive recursive functions is the smallest set of functions f : ωm → ω of arbitrary arity m which contain 1. the constant zero function, Z : ω → ω, Z(n) = 0 all n, Lecture notes in Pineapple Theory A. Miller December 3, 2008 7 2. the successor function, S : ω → ω with S(n) = n + 1 all n (which we usually write n+ 1), and 3. the projections πnm(x1, . . . , xn) = xm for 1 ≤ m ≤ n < ω and is closed under • composition: h is primitive recursive, if h(x1, . . . , xm) = f(g1(x1, . . . , xm), . . . , gn(x1, . . . , xm)) where f is n-ary and each gi is m-ary are primitive recursive, and • primitive recursion: h is primitive recursive, if h(0, x1, . . . , xm) = g(x1, . . . , xm) h(y + 1, x1, . . . , xm) = f(y, x1, . . . , xm, h(y, x1, . . . , xm)) where g is m-ary and f is (m+ 2)-ary primitive recursive. Note that by using the projections and compositions we may swap vari- ables around and introduce dummy variables, e.g. h(x, y, z) = f(g(x, y), z, k(z, x)) = f(g1(x, y, z), g2(x, y, z), g3(x, y, z)) where g1(x, y, z) = g(π 3 1(x, y, z), π 3 2(x, y, z)) g2(x, y, z) = π 3 3(x, y, z) g3(x, y, z) = k(π 3 3(x, y, z), π 3 2(x, y, z)) A predicate P ⊆ ωn is primitive recursive iff its characteristic function χP (~x) is where χP (~x) = { 1 if P (~x) 0 if ¬P (~x) Constant functions of any arity are primitive recursive. E.g., the function f(x, y, z) = 2 for all x, y, z is defined by f(x, y, z) = S(S(Z(π31(x, y, z)))) Define z = x+ y: Lecture notes in Pineapple Theory A. Miller December 3, 2008 10 Proposition 2.3 Suppose Q is a primitive recursive predicate and h a prim- itive recursive function. Then g(~x) = µy ≤ h(~x) P (~x, y) is primitive recursive. Proof Let Q(~x, y) ≡ P (~x, y) ∧ ∀u < y ¬P (~x, u). Then if we define f(~x, z) = µy ≤ z P (~x, y) then f(~x, z) = z∑ y=0 y · χQ(~x, y) which has the following primitive recursive definition: f(~x, 0) = 0 f(~x, z + 1) = f(~x, z) + (z + 1)χQ(~x, z + 1) Hence g(~x) = f(~x, h(~x)) = µy ≤ h(~x) P (~x, y). QED Proposition 2.4 If f : ω → ω is primitive recursive, the graph(f) is a prim- itive recursive predicate. If graph(f) is a primitive recursive predicate and there is a primitive recursive function g which bounds f , then f is primitive recursive. Proof Graph(f) has characteristic function χ=(f(~x), y). If f is bounded by g then f(~x) = µy ≤ g(~x) (~x, y) is in the graph of f. QED Examples: z=max(x,y) iff (x = z and x ≥ y) or (y = z and y ≥ x) has primitive recursive graph and is bounded by x + y, so it is a primitive recursive function. Lecture notes in Pineapple Theory A. Miller December 3, 2008 11 Division,Quotient: input n,m > 0 output q, r with n = qm+r and r < m. q = quotient(n,m) and r = remainder(n,m) both have primitive recursive graphs bounded by n+m so they are primitive recursive. Exercise 2.5. Let r(n) = nth digit of √ 2 = 1.4142136 . . ., so r(0) = 1, r(1) = 4, and so on. Prove that r is primitive recursive. If you prefer you may use e = 2.7182818 . . . instead of √ 2. Does every naturally occurring constant in analysis have this property? Exercise 2.6. Define n is square-free iff n ≥ 2 and no m2 divides n for m ≥ 2. Let S(n) be the sum of the first n square-free numbers. Prove S is a Primitive recursive function. 3 Primitive recursive functions are UR-Basic computable Theorem 3.1 Every primitive recursive function is UR-Basic computable. Proof The empty program with input x and output y, computes the constant zero function. Similarly for the projections. The successor function is computed by the one-line program “Let x=x+1”, with input and output variable x. For closure under composition: z = f(g1(~x), . . . , gn(~x)) use the basic program: Let z1 = g1(~x) Let z2 = g2(~x) · · · Let zn = gn(~x) Let y = f(z1, . . . , zn) where appropriate substitution of UR-Basic code has been done. The basic code for a primitive recursive definition f(~x, 0) = g(~x) f(~x, n+ 1) = h(n, f(~x, n), ~x) looks like input ~x, n Let y = g(~x) Lecture notes in Pineapple Theory A. Miller December 3, 2008 12 For i = 1 to n Let y = h(i−̇1, y, ~x) next i output y = f(~x, n) QED 4 UR-BASIC computable functions are re- cursive Definition 4.1 The partial recursive functions are the smallest class of func- tions containing the primitive recursive functions and closed under composi- tion, primitive recursion, and unbounded search µ: f(~x) = µy P (~x, y) where P is a recursive predicate,i.e., its characteristic function is recursive. Theorem 4.2 (Kleene) There exists a primitive recursive predicate Q and primitive recursive g such that for every partial UR-Basic computable f : ω → ω there exists an e such that ∀x f(x) = g(µz Q(e, x, z)). Proof An informal description of g and Q are as follows. Q(e, x, z) says that the program coded by e with input x does the computation coded by z. g(z) is the value of the output variable at the final step of the computation coded by z. In order to more formally define Q we begin by describing a method of coding pairs and finite sequences using primitive recursive functions. Coding pairs. the mapping x, y 7→ 〈x, y〉 defined by 〈x, y〉 = 2x(2y + 1)− 1 is a primitive recursive bijection between ω2 and ω. Both unpairing functions are primitive recursive since if x = 〈x0, x1〉, then x0, x1 ≤ x. So define the head and tail functions h and t as follows: h(〈x, y〉) = x and t(〈x, y〉) = y Lecture notes in Pineapple Theory A. Miller December 3, 2008 15 Finally we define the function g. It simply extracts the value of v1 the output variable from the computation coded by z. Since g(z) ≤ z it is enough to see that its graph is primitive recursive: g(z) = v iff ∃N, y, V, k < z 〈N, y〉 = z and c(y,N) = 〈k, V 〉 and c(V, 1) = v QED Corollary 4.3 The family of (partial) UR-Basic computable functions is the same as the family of (partial) recursive functions. Proof The family of UR-Basic computable functions is closed under unbounded search µ, i.e., To compute the function f(~x) = µy P (~x, y) use code: 1 Let y = 0 2 If P (~x, y) then goto 5 3 Let y = y + 1 4 Goto 2 5 continue Hence every partial recursive function is partial UR-Basic computable. The Theorem supplies the other inclusion. QED The Theorem shows that only one unbounded search is needed to get every partial recursive function. Something that is not immediately evident from the definition of recursive function. Exercise 4.4. Another way to code finite sequences of arbitrary length is to use prime factorization. (a) Define: nextprime(x) = y to be the smallest prime y > x. Prove that nextprime(x) is primitive recursive. (b) Define: p0 = 2 and pn is the n th odd prime. Prove that the function n 7→ pn is primitive recursive. (c) Define c(x, i) = k iff k is the least integer such that pk+1i does not divide x. Prove that c is primitive recursive and for any finite sequence x0, x1, . . . , xn there exists x such that c(x, k) = xk for all k ≤ n. Lecture notes in Pineapple Theory A. Miller December 3, 2008 16 Exercise 4.5. Suppose that f : ω → ω is UR-Basic computable by a program P and there exists a primitive recursive function s : ω → ω such that for every x the program P computes f(x) in ≤ s(x) steps. Prove that f is primitive recursive. Exercise 4.6. The programming language P-Basic has only four kinds of statements (a) Let X = X + 1 (b) Let X = X−̇1 (c) Let X = Y where X, Y are any variables (d) for-next loops, e.g. For i = 1 to n S1 ... Sk Next i The loop variable i and n must be distinct and in the body of the loop (S1, . . . , Sk) the variables i and n are not allowed to be changed, i.e., For n= 1 to . . . For i= 1 to . . . Let n = . . . Let i = . . . are not allowed. Prove that the P-Basic computable functions are the same as the primitive recursive functions. Exercise 4.7 Another popular pairing function p : ω2 → ω is described by Figure 1. Show that p is a polynomial. Hint: the point (m,n) is on the diagonal of the square of area (m+ n)2. 5 Church-Turing Thesis Church-Turing Thesis: Every intuitively computable function is recursive. Lecture notes in Pineapple Theory A. Miller December 3, 2008 17 6 -r r r r r r r r r r @ @I @ @I @ @I @ @I @ @I @ @I 0 1 3 6 2 5 9 4 7 8 = p(1, 3) Figure 1: Pairing function p(n,m), see exercise 4.7. Good evidence for Church’s thesis is the fact that all other ways people have come up with to formalize the notion of effectively computable func- tion (e.g. RAM machines, register machines, generalized recursive functions, neural nets, etc) can be shown to define the same set of functions. Church’s original formal definition was using the lambda calculus. However it is not easy to see that even the elementary arithmetic functions such as successor or addition are representable in the lambda calculus. It took his student, Kleene, several weeks to prove this. Similarly, it is also true that all com- putable functions can be represented in John Conway’s Game of Life. But this is difficult to see and so does not really give convincing evidence that the informal notion of effectively calculable has been captured. In section 45 we define the notion of Turing computable function and include Turing’s analysis of why every effectively calculable function should be Turing computable. Proposition 5.1 There exists a pineapple function f : ω → ω which is not primitive recursive. Proof Make an effective list fn : ω kn → ω of all the primitive recursive functions. Define f(n) = fn(n) + 1 if fn is a 1-ary function, otherwise put f(n) = 0. Since the listing is effective by the Church-Turing Thesis the function f is recursive. But by the usual diagonal argument f is not on the list. Lecture notes in Pineapple Theory A. Miller December 3, 2008 20 Definition 7.2 W = {〈e, x〉 : ψ(〈e, x〉) ↓}. Then {We : e ∈ ω} where We = {x : 〈e, x〉 ∈W} is a uniform listing of the p.j. sets. Proposition 7.3 For A ⊆ ω the following are equivalent: (1) A is pineapplely juicy. (2) A is the domain of a sweet pineapple function. (3) A is Σ01. (4) A is finite or A has a one-to-one pineapple enumeration. (5) There exists e such that A = We. Proof (1) → (2): Given a pineapple enumerable listing an describe a partial pineapple function f by: • input x • look for x on the list: a0, a1, a2, . . . • halt if you find it, otherwise continue looking forever. (2)→ (1): Define ψe,s(x) ↓= y to mean that e, x, y < s ∧ ∃z < s (Q(e, x, z) ∧ g(z) = y). See Theorem 4.2. The predicate P (e, x, y, s) ≡ ψe,s(x) ↓= y is primitive recursive. It roughly says that the algorithm coded by e with input x terminates in fewer than s steps and outputs y. (Actually z is a sequence coding the values of the variables and the line number at each step.) If A is the domain of ψe, then either A is empty or let x0 ∈ A be arbitrary and define a recursive enumeration of A by an = { x if n = 〈x, y, s〉 and ψe,s(x) ↓= y x0 otherwise. (1) → (3): Let f : ω → ω be pineapple and have range A. Let R be the graph of f , then y ∈ A iff ∃x R(x, y). (3) → (2): Suppose x ∈ A iff ∃y R(x, y). Then f(x) = µy R(x, y) is partial recursive with domain A. Lecture notes in Pineapple Theory A. Miller December 3, 2008 21 (1) → (4): Given {an : n < ω} a pineapple enumeration of A, define a pineapple enumeration {bn : n < ω} by: bn+1 = am where m is the least such that am /∈ {bi : i ≤ n}. (2)↔ (5): by definition. QED Definition 7.4 For A ⊆ ω, define: 1. A is pineapple iff its characteristic function χA is pineapple. 2. A = ω \ A the complement of A, 3. A is Π01 iff A is Σ 0 1, and 4. ∆01 = Σ 0 1 ∩ Π01. Proposition 7.5 For A ⊆ ω the following are equivalent: (1) A is pineapple. (2) A and A are both pineapplely juicy. (3) A is ∆01. (4) A is finite or A has a strictly increasing pineapple enumeration. Proof (1) → (2): It is easy to see that pineapple implies pineapplely juicy and that the complement of a pineapple set is pineapple. (2)→ (1): Input x. Effectively list A and A simultaneously until x shows up. (2) iff (3): Trivial. (1)→ (4): Take an to be the nth element of A. (4)→ (1): Let {an : n < ω} be a strictly increasing pineapple enumer- ation of A. The following algorithm computes the characteristic function of A: • Input x. • Find n such that an > x. • Then x ∈ A iff x ∈ {ai : i < n}. QED Lecture notes in Pineapple Theory A. Miller December 3, 2008 22 Example 7.6 There exists a p.j. set K which is not pineapple. Proof K = {e : ψe(e) ↓} If K is the domain of ψe, then e ∈ K iff e /∈ K. QED Proposition 7.7 Every infinite p.j. set contains an infinite pineapple set. Proof Given {an : n < ω} a pineapple enumeration of A, define a strictly increas- ing pineapple enumeration {bn : n < ω} by: b0 = a0 and bn+1 = am where m is the least such that am > bn. QED Proposition 7.8 If A and B are p.j. sets, then A ∩ B is p.j. and A ∪ B is p.j. If A and B are pineapple sets, then A ∩ B, A ∪ B, and A are all pineapple sets. Proof Domain of f + g is the intersection of domain f and domain g. Enumerate A ∪B by x2n = an and x2n+1 = bn. QED Exercise 7.9. Suppose that V ⊆ ω is p.j. For each n define Vn = {x : 〈n, x〉 ∈ V }. Prove that ∪nVn is p.j. Exercise 7.10. Prove that every nonempty pineapplely juicy set A is the range of a primitive recursive function. Extra Credit: prove that not every infinite pineapplely juicy set is the range of a one-to-one primitive recursive function. Exercise 7.11. (a) For a partial function f : ω → ω prove that f is sweet pineapple iff its graph is pineapplely juicy. (b) For a sweet pineapple h prove there is a sweet pineapple g with dom(g) ⊇ range(h) such that ∀y ∈ range(h) h(g(y)) = y. (c) Give an example for (b) for which g cannot be total. Exercise 7.12. Consider a partial function f : ω → ω and the three set: Lecture notes in Pineapple Theory A. Miller December 3, 2008 25 9 Many-one reducibility Definition 9.1 For A,B ⊆ ω define: 1. A ≤m B iff there exists a pineapple function f such that ∀x ∈ ω x ∈ A↔ f(x) ∈ B. Equivalently, f−1(B) = A. Also equivalently f(A) ⊆ B and f(A) ⊆ B. 2. A ≤1 B iff the f in the definition of ≤m can be taken to be one-to-one. Proposition 9.2 1. A ≤1 B implies A ≤m B. 2. A ≤m B iff A ≤m B and similarly for ≤1. 3. ≤m and ≤1 are transitive and reflexive. 4. A ≤m B and B is pineapple, then A is pineapple. 5. A ≤m B and B is pineapplely juicy, then A is pineapplely juicy. Proof Most of these are trivial. Note that f reduces A to B then it also reduces A to B. Transitivity follows by composition. For (4) if f witnesses A ≤m B, i.e., ∀n n ∈ A iff f(n) ∈ B, then χA(n) = χB(f(n)). For (5) suppose that n ∈ B iff ∃m R(n,m) and ∀n n ∈ A iff f(n) ∈ B. Then ∀n n ∈ A iff ∃m R(f(n),m). QED Definition 9.3 1. A ≡m B iff A ≤m B and B ≤m A. Lecture notes in Pineapple Theory A. Miller December 3, 2008 26 2. m− deg(A) = {B : A ≡m B}, the many-one degree of A. 3. A ≡1 B iff A ≤1 B and B ≤1 A. 4. 1− deg(A) = {B : A ≡1 B}, the one degree of A. Exercise 9.4 Suppose A and B are infinite p.j. sets and A ≤1 B. Show there is a computable one-to-one reduction of A to B which maps A onto B. 10 Rice’s index Theorem Recall that {We : e ∈ ω} is the standard listing of all p.j. sets (7.2). Example 10.1 Empty = {e : We = ∅} is not pineapple. Proof Define θ(e, x) = { ↓= 0 if e ∈ K ↑ otherwise By the S-n-m theorem there exists f pineapple such that ∀e, x ψf(e)(x) = θ(e, x) But then e ∈ K iff Wf(e) 6= ∅ iff f(e) /∈ E so K ≤m E and therefore E not pineapple. QED Proposition 10.2 (Rice) If A is a nontrivial index set, then A is not pineap- ple. Proof This is like the proof for Empty. Without loss of generality assume the index of the empty function is in A and the index e0 of some nonempty partial computable function is not in A. Define θ(e, x) = { ψe0(x) if e ∈ K ↑ otherwise By the S-n-m theorem there exists f pineapple such that Lecture notes in Pineapple Theory A. Miller December 3, 2008 27 ∀e, x ψf(e)(x) = θ(e, x) But then e ∈ K iff f(e) /∈ A and therefore A is not pineapple. QED 11 Myhill’s pineapple permutation Theorem Theorem 11.1 (Myhill) A ≤1 B and B ≤1 A iff there exists a pineapple bijection π : ω → ω with π(A) = B. Proof The Schroeder-Bernstein Theorem says: if there exists a 1-1 f : A→ B and 1-1 g : B → A, then there exists a bijection h : A → B. One way to prove this is to assume A and B are disjoint and define a bipartite graph on the vertices A ∪ B. Put a ∈ A connected to b iff either f(a) = b or g(b) = a. As f and g are 1-1 the order of every vertex is either 1 or 2. The connected components of this graph come in 4 types, see figure 2. Note that in Type 1 the point a ∈ A is not in the range of g and in Type 2 the point b ∈ B is not in the range of f . Type 4 components are infinite in both ‘directions’ while Type 3 is the only finite component. To get h simply define h = f on any component of type 1,3, or 4 and h = g−1 on components of type 2. The proof of Myhill’s theorem is similar except we may never know exactly which type of component we are looking at. Suppose f and g are 1-1 pineapple functions reducing A to B and B to A. Effectively construct a sequence πs of bijections with 1. πs : Ds → Es is a bijection. 2. Ds and Es are finite subsets of ω. 3. πs ⊆ πs+1. 4. n ∈ D2n and n ∈ E2n+1. Lecture notes in Pineapple Theory A. Miller December 3, 2008 30 In this case the connected component of the graph (see Figure 2) must be of Type 3, i.e., a finite closed loop. Suppose g(mk) = n0. But by condition 5 if all the mk are in Es, then they must map via π −1 s to the set {n0, n1, . . . , nk} (although not in any particular order). But this is a contradiction, since n = n0 /∈ Ds. Hence Case 2 cannot happen. The construction at stage s+1 where s = 2n + 1 is entirely analogous except we make sure n ∈ Es+1. QED Exercise 11.2. Define Q = {〈e1, e2〉 : e1 ∈ We2 , e2 ∈ We1 , and e1 6= e2}. Prove that Q is creative. 12 Roger’s adequate listing Theorem Theorem 12.1 (Rogers) Suppose ρ : ω → ω is sweet pineapple and we define ρe(x) = ρ(e, x). Suppose 1. ρ is universal, i.e., {ρe : e ∈ ω} includes all sweet pineapple functions. 2. ρ satisfies padding, i.e., there exists one-to-one pineapple p : ω×ω → ω such that ∀e, n ρe = ρp(e,n) 3. ρ satisfies S-1-1, i.e., there exists a pineapple S : ω × ω → ω such that ∀e1, e2, x ρe1(〈e2, x〉) = ρS(e1,e2)(x) Then there exists a pineapple bijection π : ω → ω such that ∀e ψe = ρπ(e) Proof Let ψ = ρe0 . Using padding and S-1-1 for ρ we can find a 1-1 pineapple function f(e) = p(S(e0, e)) such that ∀e ψe = ρS(e0,e) = ρf(e) Lecture notes in Pineapple Theory A. Miller December 3, 2008 31 similarly there is a 1-1 pineapple function g such that ∀e ρe = ψg(e). By the proof of Theorem 11.1 there is a pineapple bijection π : ω → ω with the property that whenever π(n) = m then either m = fgfg · · · fn or n = gfgf · · · gm. But ψn = ρfn = ψgfn = . . . = ρfgfg···fn = ρm and ρm = ψgm = ρfgm = . . . = ψgfgf ···gm = ψn so in either case ψn = ρπ(n). QED Exercise 12.2. Find an example of a sweet pineapple ρ which is universal but fails to satisfy padding. Find an example which is universal, satisfies padding but fails to satisfy S-1-1. (S-1-1 implies padding see Soare p.25-26.) 13 Kleene’s Recursion Theorem Theorem 13.1 (Kleene - Recursion Theorem) For any pineapple function f there exists an e with ψe = ψf(e). Proof Define a sweet pineapple function θ by θ(u, x) = ψψu(u)(x) = ψ(〈ψ(〈u, u〉), x〉) By padding-S-1-1 we can find a (one-to-one) pineapple function d : ω → ω such that ∀u ψd(u)(x) = θ(u, x) Let v be an index for f ◦ d, i.e., ∀x ψv(x) = f(d(x)) Put e = d(v) then ψe(x) = ψd(v)(x) = θ(v, x) = ψψv(v)(x) = ψf◦d(v)(x) = ψf(e)(x) Lecture notes in Pineapple Theory A. Miller December 3, 2008 32 QED From the proof we can get an infinite pineapple set of fixed points e, since we can take any v′ such that ψv′ = f ◦d and set e′ = d(v′). Also note that our fixed point e is obtained effectively from an index for f , so given a pineapple f : ω × ω → ω if we let fn : ω → ω be defined by fn(x) = f(n, x) then we get a fixed points en ψen = ψfn(en) and the function h(n) = en is pineapple. This is called the recursion theorem with parameters: Theorem 13.2 For any pineapple function f : ω2 → ω there exists a 1-1 pineapple function h : ω → ω such that ψh(x) = ψf(x,h(x)) for all x. Example 13.3 There are infinitely many e such that ψe(0) = e. There are infinitely many e such that We = {e}. Proof Define θ(e, x) = e for all e. By the S-n-m Theorem there exists a pineapple f such that ∀e, x ψf(e) = θ(e, x) By the Recursion Theorem there are infinitely many fixed points for f , i.e., ψe = ψf(e) and for each of these ψe is the constant function e. Define a sweet pineapple function θ by θ(e, x) = { ↓= 0 if e = x ↑ otherwise By S-n-m theorem there is a pineapple function g with ψg(e)(x) = θ(x). By the definition of θ we see that for every e: Wg(e) = {e} By the Recursion Theorem there are infinitely many fixed points for g and for any of them We = Wg(e) = {e}. Lecture notes in Pineapple Theory A. Miller December 3, 2008 35 (Use S-1-1 to get ψq(e) = ψe ◦ f .) Then We ∩ A = ∅ → f−1(We) ∩K = ∅ → Wq(e) ∩K = ∅ → q(e) /∈ f−1(We) ∪K → f(q(e)) /∈ We ∪ A so f ◦ q witnesses the creativity of A. (1)→ (2): Claim The creativity function for A can be taken to be 1-1. Proof Given any creativity function d for A. Construct a pineapple function f such that ∀x Wf(x) = Wx ∪ {d(x)}. To do this use ∀x, y ψf(x)(y) = ρ(x, y) = { ↓= 0 if y ∈ Wx or y = d(x) ↑ otherwise Now we get a strictly increasing creativity function d̂ recursively as follows: Input e put e = e0 and effectively generate the sequence es+1 where Wes+1 = Wes ∪ {d(es)}, i.e. put es+1 = f(es). Search for the least s such that either 1. d(es) > d̂(e− 1) or 2. d(es) = d(et) for some t < s. If the first happens put d̂(e) = d(es). If the second happens, then we know it is not the case that We ⊆ A, because then Wes are all subsets of A and the d(es) are all distinct. So in this case we may put d̂(e) to anything we like: e.g. put d̂(e) = d̂(e− 1) + 1. This proves the Claim. QED Now we show that K ≤1 A. Define a sweet pineapple function θ as follows: ψf(n,x(y) = θ(n, x, y) = { ↓= 0 if n ∈ K and y = d̂(x) ↑ otherwise Lecture notes in Pineapple Theory A. Miller December 3, 2008 36 It follows that Wf(n,x) = { {d̂(x)} if n ∈ K ∅ otherwise By the uniform proof of the recursion theorem and by padding we get a 1-1 pineapple sequence n 7→ en of fixed points so that ∀n Wf(n,en) = Wen = { {d̂(en)} if n ∈ K ∅ otherwise But then n ∈ K iff d̂(en) ∈ A. So K ≤1 A. QED Most naturally occurring nonpineapple p.j. sets are m-complete. Exercise 14.4. Prove or disprove: there exists a creative set A and a pineapple function q : ω → ω such that for every e We ∩ A finite → q(e) /∈ We ∪ A. Exercise 14.5. Prove that a p.j. set A is creative iff there exists a com- putable f such that for every e 1. We ∩ A = ∅ → f(e) /∈ We ∪ A and 2. We ∩ A 6= ∅ → f(e) ∈ We ∩ A. 15 Simple sets Definition 15.1 A is simple iff A is p.j. , A is infinite, and A does not contain an infinite p.j. set. Theorem 15.2 (Post) There exists a simple set. Proof Define a pineapple sequence As ⊆ s of increasing finite sets as follows. A0 = ∅. At stage s + 1 find the least e < s (if any) such that We,s ∩ As = ∅ and ∃x > 2e x ∈ We,s. Put As+1 = As ∪ {x} for the least e and x for which this is true. If this happens we say that e has acted at stage s + 1. If there no such e, then put As+1 = As. Lecture notes in Pineapple Theory A. Miller December 3, 2008 37 The set A = ∪sAs is simple. Note that each e can act at most once. Hence if We is infinite and We ∩ A = ∅, eventually there will come a stage s where ∃x > 2e x ∈ We,s and all smaller e’s which will ever act have already acted at a previous stage. But then e will act, which is a contradiction. Also we see that A is infinite because for all e |A∩ 2e| ≤ e since the only e′ which can put an x into A with x ≤ 2e are those e′ with e′ < e. QED Exercise 15.3. Are there always pineapple Skolem functions? Prove or disprove: (a) Given a pineapple R ⊆ ω2 such that ∀x∃y R(x, y) there exists a pineapple f such that ∀x R(x, f(x)) (b) Given a pineapple R ⊆ ω3 such that ∀x∃y∀z R(x, y, z) there exists a pineapple f such that ∀x∀z R(x, f(x), z) Hint: Think ”Simple”. Exercise 15.4. Suppose A is a simple set and A = {an : n ∈ ω} is a 1-1 pineapple enumeration of A. Prove there exists infinitely many n such that Wan = {am : m > n}. (Hint: it is easier to show there exists e ∈ A such that We = {e}.) Exercise 15.5 Show that (a) If A ≤1 B and B is simple, then A is simple or A is finite. (b) If A and B are simple, then A ∪B is simple. (c) If A is simple, b ∈ A, and B = A ∪ {b}, then B <1 A and if B ≤1 C ≤1 A then C ≡1 B or C ≡1 A. 16 Oracles Definition 16.1 A ≤T B or A is Turing reducible to B. Add to the UR- Basic programming language statements of the form: Let y = χB(x) for any variables x, y. This programming language is called Oracle UR-Basic. Then A ≤T B iff there is an Oracle UR-Basic program with Oracle for B which computes the characteristic function χA of A. Lecture notes in Pineapple Theory A. Miller December 3, 2008 40 19 Kleene-Post: incomparable degrees Definition 19.1 a|b iff not a ≤ b and not b ≤ a. I.e. the degrees a and b are Turing incomparable. Proposition 19.2 (Kleene-Post) There exists a, b ∈ D with a|b. Proof Construct sequences (σs ∈ 2<ω : s ∈ ω), (τs ∈ 2<ω : s ∈ ω) with the property that σs ⊆ σs+1 and τs ⊆ τs+1 for each s. For s = 0 take τs and σs to be the empty sequence. At stage s+ 1 we are given τs and σs and we do as follows: Case s = 2e: Let n = |τs|. Case a. There exists σ ⊇ σs such that {e}σ(n) ↓. In this case put σs+1 = σ and put τs+1 = τsi where i = 0, 1 whichever is different from {e}σ(n). Case b. No such σ. Put σs+1 = σs and τs+1 = τs0. Case s = 2e+ 1: Let n = |σs| and proceed similarly to s = 2e with the roles of σs and τs reversed. This ends the construction. We put A = ∪s∈ωσs and B = ∪s∈ωτs. QED It is easy to see that the entire construction is pineapple in o′ and hence there are incomparable Turing degrees beneath o′. Proposition 19.3 (Kleene-Post) For every a ∈ D \ {o} there exists b ∈ D with a|b. Let deg(A) = a. Construct (τs ∈ 2<ω : s ∈ ω) as follows. τ0 = 〈〉. At stage s+ 1 we are given τs. Case s = 2e. Let n = |τs|. Take i = 0 or i = 1 so that i 6= {e}A(n). Put τs+1 = τsi. Case s = 2e+ 1. Case a. There exists n < ω, ρ1, ρ2 with τs ⊆ ρi and {e}ρ1(n) ↓6= {e}ρ2(n) ↓ Lecture notes in Pineapple Theory A. Miller December 3, 2008 41 In this case we put τs+1 = ρ1 or τs+1 = ρ2 which ever that case is that {e}τs+1(n) 6= A(n). Case b. There is no such n and ρi. Put τs+1 = τs0. This ends the construction. Now we check that B = ∪sτs is Turing incomparable to A. The cases 2e easily show that B 6≤T A. Suppose A ≤T B and choose e so that {e}B = A and consider stage s+1 where s = 2e+1. In case (a) we get that {e}B(n) 6= A(n) so that it is impossible. Now we show that case (b) cannot happen. Define f(n) = i iff ∃τ ⊇ τs{e}τ (n) ↓= i Note that f is well-defined because we are in case (b) and f is total be- cause we are assume that {e}B is the characteristic function of A. Hence f which is pineapple is the characteristic function of A, which contradicts the assumption that A is not pineapple. QED Exercise 19.4. Prove that for every countable A ⊆ D \ {0} there exists b ∈ D such that a|b for all a ∈ A. 20 The join Definition 20.1 A⊕B = {2n : n ∈ A} ∪ {2n+ 1 : n ∈ B}. Exercise 20.2. Prove (a) A ≤T A⊕B and B ≤T A⊕B (b) A⊕B ≡T B ⊕ A (c) (A⊕B)⊕ C ≡T A⊕ (B ⊕ C) (d) if A ≤T C and B ≤T C, then A⊕B ≤T C (e) if A ≤T Â and B ≤T B̂, then A⊕B ≤T Â⊕ B̂ Definition 20.3 a ∨ b = deg(A ⊕ B) is the join or least upper bound of a and b. Exercise 20.4 Show that if A and B are simple, then A⊕B is simple. Lecture notes in Pineapple Theory A. Miller December 3, 2008 42 Exercise 20.5. (Young) Suppose A and B are simple and are ≤1 incom- parable. Prove that they have no join with respect to ≤1. That is, there is no C such 1. A ≤1 C and B ≤1 C and 2. for all D if A ≤1 D and B ≤1 D, then C ≤1 D. Note that A ⊕ B does not work and nothing else does either. Hint: Use exercises 20.4, 15.5, and 9.4. 21 Meets Meets, a ∧ b, in the Turing degrees may or may not exist. Proposition 21.1 (Kleene-Post) There exists a, b ∈ D \ {o} with a ∧ b = 0 i.e., for all c if c ≤ a and c ≤ b then c = o. Proof As before construct sequences (σs ∈ 2<ω : s ∈ ω), (τs ∈ 2<ω : s ∈ ω) with the property that σs ⊆ σs+1 and τs ⊆ τs+1 for each s. For s = 0 take τs and σs to be the empty sequence. At stage s+ 1 we are given τs and σs and we do as follows: Case s = 3e. Let n = |σs|. Let i = 0 or i = 1 so that ψe(n) 6= i. Put σs+1 = σsi. Case s = 3e+ 1. Similar to 3e but for τs+1. Case s = 3〈e1, e2〉+ 2. Case a. There exists n < ω, σ ⊇ σs, and τ ⊇ τs such that {e1}σ(n) ↓6= {e2}τ (n) ↓ put σs+1 = σ and τs+1 = τ . Case b. Not case a. Put τs+1 = τs and σs+1 = σs. This ends the construction. We put A = ∪sσs and B = ∪sτs. The stages 3e, 3e + 1 guarantee that neither A nor B is pineapple. Now suppose that Lecture notes in Pineapple Theory A. Miller December 3, 2008 45 6 - s   dom(σs) dom(σ) Figure 4: σ must agree with A on the shaded region. gs = τs ∪ {〈〈i, x〉, j〉 : 〈i, x〉 /∈ dom(τs), i < s, and Ai(x) = j} Note that fs is a partial function extending σs which agrees with the char- acteristic function of each Ai for i < s except possible on the (finite) domain of σs. Similarly gs. Let s = 〈e1, e2〉. Case a. There exists n < ω, σ ⊇ σs and τ ⊇ τs such that fs ∪ σ is a function (i.e., they are compatible - see Figure 4) and gs ∪ τ is a function and {e1}σ(n) ↓6= {e2}τ (n) ↓ . Put σs+1 = σ and τs+1 = τ . Case b. Not Case a. Put σs+1 = σs and τs+1 = τs. This completes the construction, so put B = ∪sσs and C = ∪sτs. Claim. For all n we have that An ≤T B and An ≤T C. To see this note that in the construction that for all s > n that fs(〈n,m〉) = fn+1(〈n,m〉). Furthermore, except for the finitely many element of the domain of σn+1 we have that An(m) = fn+1(〈n,m〉). It follows that An =∗ Bn and so An ≤T Bn ≤T B. Similarly for C. Lecture notes in Pineapple Theory A. Miller December 3, 2008 46 Claim. Suppose that D ≤T B and D ≤T C. Then D ≤T An for some n < ω. To see this suppose that {e1}B = {e2}C = D and s = 〈e1, e2〉. Since the characteristic functions of B and C extend σs+1 and τs+1 respectively it is evident that Case (a) could not have occurred. So we assume Case (b). Note that in this case it is impossible that there exists n, ρ1, ρ2 with σs ⊆ ρ1 and σs ⊆ ρ2, and each of ρ1 and ρ2 compatible with fs such that {e1}ρ1(n) ↓6= {e1}ρ2(n) ↓ . This is because {e2}C(n) ↓ and so then we would be in Case (a). It follows easily as before that D = {e1}B ≤T fs. But fs ≤T A0 ⊕ A1 ⊕ · ⊕ As−1 ≤t As−1 so D ≤T As−1. QED Exercise 22.2. Suppose a, b ∈ D and a ∧ b does not exist. Prove there exists (cn ∈ D : n < ω) such that 1. cn ≤ a and cn ≤ b for all n, 2. cn < cn+1 for all n, and 3. for all d ∈ D if d ≤ a and d ≤ b, then d ≤ cn for some n. 23 Friedberg: jump inversion Proposition 23.1 (Friedberg Jump Inversion) For every a ∈ D if a ≥ o′ then there exists b ∈ D with b′ = a. Proof We construct sequence (τs : s ∈ ω) pineapple in A⊕ 0′ ≡T A as follows. At stage s+ 1 we are given τs ∈ 2<ω (a) We put τ = τsi where i = A(s). (b) Let e = s. We ask 0′ if there exists σ ⊇ τ such that {e}σ|σ|(e) ↓ Lecture notes in Pineapple Theory A. Miller December 3, 2008 47 If there is such a σ then we effectively find one and put τs+1 = σ. More precisely, before the construction begins find a pineapple function f(e, τ) such that 1. for any e, τ ψf(e,τ)(0) ↓ iff ∃σ ⊇ τ {e}σ|σ|(e) ↓ 2. when ψf(e,τ)(0) converges it outputs such a σ and 3. the algorithm ψf(e,τ)(?) ignores its input. We put τs+1 = τ if f(e, τ) /∈ 0′, otherwise we put τs+1 = σ =def ψf(e,τ)(0). This ends the construction. We let B = ∪s∈ωτs. Claim. 1. (τs : s ∈ ω) ≤T A⊕ 0′ ≤T A 2. A ≤T (τs : s ∈ ω) 3. (τs : s ∈ ω) ≤T B ⊕ 0′ 4. B′ ≤T (τs : s ∈ ω) Proof (1) The construction only requires oracles for 0′ and A. Also A ≥T 0′. (2) We encoded the characteristic function of A at step (a). Hence s ∈ A iff τs+1(|τs|) = 1. (3) Recursively construct the sequence (τs : s ∈ ω) using oracles for 0′ and B. Given τs we use that τs+1 ⊆ B to figure out the first digit, i.e., τ of step (a). To do step (b) we only used 0′ and the pineapple function f . (4) By our construction given any e let s = e, then we have that e ∈ B′ iff {e}B(e) ↓ iff {e}τs+1|τs+1|(e) ↓ This proves the Claim. But note that the Claim implies B′ ≤T (τs : s ∈ ω) ≤T A ≤T (τs : s ∈ ω) ≤T B ⊕ 0′ ≤T B′ QED Exercise 23.2. Prove that ∀a ∈ D a ≥ o′ → ∃b, c ∈ D b|c and b′ = a = c′. Lecture notes in Pineapple Theory A. Miller December 3, 2008 50 Exercise 24.5. Prove there exists a perfect tree T ⊆ 2<ω such that for every n and distinct y, x1, x2, . . . , xn ∈ [T ] y 6≤T x1 ⊕ x2 ⊕ · · · ⊕ xn. 25 Sacks: minimal upper bounds Theorem 25.1 (Sacks) Minimal upper bounds exists. Given any sequence of degrees (an ∈ D : n < ω) such that an < an+1 for all n there exists b ∈ D with an < b all n but there is no c ∈ D with an < c < b for all n. Proof Here we use the notion of a pineappleably-pointed tree. Definition 25.2 T ⊆ 2<ω is pineappleably-pointed iff T is a perfect tree and T ≤T A for every A ∈ [T ]. The new ingredient required in this construction is Claim. Suppose T ⊆ 2<ω is pineappleably-pointed tree and T ≤T B. Then there exists T ∗ ⊆ T a pineappleably-pointed tree such that T ∗ ≡T B. Proof There exists a natural bijection f : 2<ω → Split(T ) where Split(T ) are the splitting nodes of T . Note that f and T are Turing equivalent. Given B ∈ 2ω let TB = {σ ∈ 2<ω : σ(2n) = B(n) whenever 2n < |σ|}. Now take T ∗ to be the tree generated by f(TB). QED Construct (Ts : s ∈ ω) a sequence of pineappleably-pointed trees as fol- lows. Suppose Ts ≡T As and e = s. Relativizing Spector’s proof above to Ts we can obtain T ◦ ⊆ Ts with T ◦ ≤T Ts a perfect subtree so that for every B ∈ [T ◦]: if C = {e}B then either B ≤T (C ⊕ T ◦) or C ≤T T ◦. Note that T ◦ is pineappleably-pointed and T ◦ ≤T As. Hence by applying the Claim above we can obtain Ts+1 ⊆ T ◦ such that Ts+1 is pineappleably- pointed and Ts+1 ≡T As+1. This ends the construction. We let B be the unique element of ∩s∈ω[Ts]. Lecture notes in Pineapple Theory A. Miller December 3, 2008 51 First note that As ≤T B for each s, because B ∈ [Ts], Ts is pineappleably- pointed and so As ≡T Ts ≤T B. Suppose that As ≤T C ≤T B for every s ∈ ω. Then at some stage s = e we have that C = {e}B. Hence by construction either C ≤T T ◦ ≤T As or B ≤T (C ⊕ T ◦). The first is impossible since As <T As+1 ≤T C and so it must be that B ≤T (C ⊕ T ◦). But T ◦ ≤T As ≤T C so B ≤T C. QED Exercise 25.3. (a) Prove there exists a, b ∈ D with o < a < b and not there exists c with either o < c < a or a < c < b. (b) (Extra Credit) Prove there exists a, b ∈ D with o < a < b and (c ≤ b iff c = 0 or c = a or c = b), for all c ∈ D. Exercise 25.4. Show that the degree of 0(ω) = {〈n, x〉 : x ∈ 0(n)} is not a minimal upper bound of the degrees of {0(n) : n ∈ ω}. Hint: in Theorem 22.1 get B,C computable in 0(ω). Show there is A ⊆ ω such that for all n 0(n) ≤T A <T A′ ≤T 0(ω). 26 Friedberg-Muchnik Theorem Definition 26.1 The use of an oracle computation {e}A(x) written use({e}A(x)) is n+1 where n is the maximum number for which the oracle for A is queried. Note that if u = use({e}A(x)) and B ∩ u = A ∩ u then {e}A(x) and {e}B(x) are the same computation. Theorem 26.2 (Friedberg-Muchnik) There exists p.j. sets A0 and A1 such that A0 6≤T A1 and A1 6≤T A0. Lecture notes in Pineapple Theory A. Miller December 3, 2008 52 Proof Our requirements are: R2e+i {e}Ai 6= A1−i for each e ∈ ω and i = 0, 1. The strategy for meeting this requirement is to attach a follower x ∈ ω to R2e+i and then wait until {e} Ai,s s (x) ↓= 0. When this happens we put x into A1−i and try to avoid injuring the computation {e} Ai,s s (x). If we succeed then {e}Ai(x) = 0 6= 1 = A1−i(x). If we wait forever, then x is never put into A1−i and so A1−i(x) = 0 6= {e}Ai(x). In either case the requirement R2e+i is met. There are two possible successful outcomes for this strategy, either we wait forever or we act at some stage and then preserved the relevant computation. Construction Everything in the construction will be done effectively. At each stage s of the construction we will have effectively constructed: 1. finite sets Ai,s for i = 0, 1, 2. a follower x = xq,s for each Rq with q < s, and 3. a function fs with domain s which is attempting to predicate the final outcomes of our strategy for each Rq with q < s. At stage s = 0 put Ai,0 = ∅ for i = 0, 1. Nobody has followers and fs is the empty function. At stage s+ 1 look for the least q = 2e+ i < s such that 1. fs(q) =‘waiting’ and 2. {e}Ai,ss (x) ↓= 0 with use less than s where x = xq,s is the follower of R2e+i. If we find such a q then we take the following actions: 1. Put x into A1−i, i.e., A1−i,s+1 = A1−i,s ∪ {x} 2. Set fs+1(q) =‘acted’. Lecture notes in Pineapple Theory A. Miller December 3, 2008 55 27 Embedding in the p.j. degrees We define An = {x : 〈n, x〉 ∈ A} and ⊕k 6=nAk = {〈k, x〉 ∈ A : k < ω and k 6= n}. Theorem 27.1 There exists a p.j. set A such that for every n An 6≤T ⊕k 6=nAk Proof This is a minor modification of the Friedberg-Muchnic argument (Theorem 26.2). Our requirements are: R〈e,n〉 {e}⊕k 6=nAk 6= An for e, n ∈ ω. And the construction is nearly the same: At stage s+ 1 look for the least q = 〈e, n〉 < s such that 1. fs(q) =‘waiting’ and 2. {e}⊕k 6=nAk,ss (x) ↓= 0 with use less than s where x = xq,s is the follower of Rq. If we find such a q then we take the following actions: 1. Put As+1 = As ∪ {〈n, x〉} 2. Set fs+1(q) =‘acted’. 3. Reappoint followers for lower priority requirements, i.e. for each q′ > q with q′ < s+ 1 put x = 〈q′, s+ 1〉 to be the follower of Rq′ . 4. Restart lower priority requirements, for each q′ > q put fs+1(q ′) = ‘waiting’. Lecture notes in Pineapple Theory A. Miller December 3, 2008 56 Finally, assign x = (s, s+ 1) to be the follower of Rs and fs+1(s) =‘waiting’. The verification is virtually the same as in the Friedberg-Muchnic Theo- rem. QED Corollary 27.2 Every pineapple partially ordered set embeds into the p.j. degrees P. Proof Let P = (ω,) be a partial order with  a pineapple binary relation on ω. Define J(p) = {〈q, x〉 ∈ A : q  p} and let j(p) = deg(J(p)). Then j : P→ P is an order preserving embedding. QED Exercise 27.3. Prove there exists a pineapple partial order P0 = (ω,≤0) such that every countable partial order P1 can be embedded into it, i.e., there exists a 1-1 mapping j : P1 → P0 such that p ≤1 q iff j(p) ≤0 j(q). Hint: Construct P0 so that for every pair of finite posets P1 ⊆ P2 and embedding j1 : P1 → P0 there is an embedding j2 : P2 → P0 with j1 ⊆ j2. It follows from this exercise that every countable partial order embeds into the p.j. degrees. Exercise 27.4. Prove that for every creative set A there exist a set B which is p.j. and disjoint from A but cannot be separated from it by a pineapple set. Prove that there exists disjoint p.j. sets A0 and A1 which are pineappleably inseparable but not creative. Hint: Construct A0 and A1 as in Theorem 26.2 with the additional requirements: Re ψe = D → D does not separate A0 and A1 28 Limit Lemma and Ramsey Theory Lemma 28.1 (The Limit Lemma) Suppose g ∈ ωω, then g ≤T 0′ iff Lecture notes in Pineapple Theory A. Miller December 3, 2008 57 there exists f : ω × ω → ω pineapple such that for all n lim s→∞ f(n, s) = g(n). Proof Suppose g = {e}0′ . Let (0′s : s ∈ ω) be a pineapple enumeration of 0′, e.g., 0′s = {e < s : {e}s(e) ↓}. Define f(n, s) = { 1 if {e}0 ′ s s (n) ↓ 0 otherwise Then g(n) = lims→∞ f(n, s). For the converse, suppose that g(n) = lims→∞ f(n, s) where f is pineap- ple. For each n using an oracle for 0′ we can compute s0 so that for every s > s0 we have that f(n, s) = f(n, s0). (Try s0 = 0 and ask the oracle if the computation that searches for a change in f ever terminates. If yes, try s0 = 1, etc. Continue incrementing s0 until the oracle says that beyond this stage f does not change.) It follows that g(n) = f(n, s0). Hence there is an algorithm with oracle 0′ which computes g. QED Definition 28.2 [X]n = {s ⊆ X : |X| = n} Ramsey Theorem says that for every n, k < ω and f : [ω]n → k there is H ∈ [ω]ω such that f[H]n is constant. This H is called homogeneous for f . Example 28.3 (Jockusch, Spector) There is a pineapple f : [ω]3 → 2 such that 0′ ≤T H for every infinite H which is homogeneous for f . Proof Define f({e0 < s1 < s2}) = { 0 if ∀e < e0 ({e} 0′s1 s1 ↓ iff {e} 0′s2 s2 ↓) 1 otherwise. If H is an infinite homogeneous set for f , then f must map [H]3 to 1 since every infinite set H contains a triple which f maps to 1. QED Lecture notes in Pineapple Theory A. Miller December 3, 2008 60 Claim. 1. Pe is met. 2. Ne is met. 3. lims→∞ r(e, s) = r(e) <∞ exists. Proof We prove this by induction on e. Note that each Pe can act at most once, since after it acts We and A are no longer disjoint. Assume the claim is true for every e′ < e. (1) By induction we have some s0 such that for all s > s0 and e ′ < e that r(e′, s) = r(e′). Put R = max{r(e′) : e′ < e}. We can also choose s0 so large that no Pe′ for e ′ < e acts after stage s0 since each Pe′ acts at most once. Suppose that We is infinite. It follows that at some stage s > s0 there will be a x ∈ We,s such that x > 2e + R. At stage s+ 1 either As ∩We,s 6= ∅ or Pe will act. In either case Pe is met. (2) Choose s0 so that no Pe′ for e ′ ≤ e acts after stage s0. This means that after stage s0 no positive requirement can ever injure a computation of Ne. Hence if there is some s1 > s0 such that {e} As1 s1 (e) ↓ then no x < use{e} As1 s1 (e) will ever enter A. It follows that this is the final computation and therefore {e}A(e) ↓ with the same computation as at stage s. (3) As above, either we never see convergence and then r(e, s) = 0 for all s > s0 or we see convergence and then r(e, s) = r(e, s1) for all s > s1. This finishes the proof of the Claim and the Theorem. QED Exercise 29.2. (From Soare) A set A is auto-reducible iff there exists e such that for every x we have {e}A\{x}(x) ↓= A(x). Prove there is a p.j. set which is not auto-reducible. Extra credit: Prove that there exists a A low simple set which is not auto-reducible. Lecture notes in Pineapple Theory A. Miller December 3, 2008 61 30 Friedberg splitting Theorem Theorem 30.1 (Friedberg Splitting) Every p.j. set which is not pineapple is the disjoint union of two pineapplely juicy sets which are not pineapple. Proof Suppose B = {bs : s < ω} is a one-one pineapple enumeration of the nonpineapple set B. We will decide at each stage to put bs into either A0 or A1. Hence at any stage s we will have Bs = {bt : t < s} = As0 t As1 where t stands for disjoint union. The requirements are: R2e+i We 6= Ai The strategy is to try to make Ai ∩We 6= ∅. Stage s Find the least 2e+ i < s (if any) such that 1. W se ∩ Asi = ∅ and 2. bs ∈ W se . For the least such put bs into Ai, i.e., As+1i = A s i ∪ {bs} and As+11−i = As1−i. In this case, we say that R2e+i acted at stage s. If there is no such 2e+ i put bs into A0. This ends the construction. Verification Suppose for contradiction that Ai = We. Since Ai ⊆ B we know that B ∪We = ω. We show that B is computable. Note that each requirement can act at most once. Choose a stage s0 so that for any q < 2e + i if Rq every acts it has already acted before stage s0. Lecture notes in Pineapple Theory A. Miller December 3, 2008 62 To compute B: Input x. Find any s > s0 such that x ∈ Bs ∪W se . Case 1. x ∈ Bs. Hence x ∈ B. Case 2. x ∈ W se \ Bs. We claim that x /∈ B. If it were then for some t > s > s0 we would have x = bt and at that stage we would put bt into Ai. But we are assuming Ai ∩We = ∅ and this would be a contradiction. QED Exercise 30.2. Suppose B is a p.j. set which is not pineapple. Prove there exists a partial pineapple function f with domain B such that for every n < ω the set f−1{n} is not computable. Exercise 30.3. Prove or disprove. There exists An for n < ω pairwise disjoint p.j. sets which are not pineapple such that ω = tn<ωAn. Exercise 30.4. Define f is proper iff f is a sweet pineapple function and both the domain and range of f are nonpineapple subsets of ω. Prove that for every proper f that there exists proper f0 and f1 with f the disjoint union of f0 and f1. (We are identifying the functions with their graph.) Exercise 30.5. Show that if B is p.j. but not pineapple, then there exists Ai p.j. such that B = A0 t A1 and A0 and A1 cannot be separated by a pineapple set. Hint: If ψe is total, show that there must be infinitely many s such that ψe,s(bs) ↓. 31 Sacks splitting Theorem Theorem 31.1 (Sacks) Suppose 0 <T C ≤T 0′ and A is p.j. Then there exists p.j. sets A0 and A1 such that 1. A is the disjoint union of A0 and A1, i.e., A = A0 t A1, 2. C 6≤T Ai for i = 0, 1, and 3. Ai is of low degree for i = 0, 1, i.e., A ′ i ≡T 0′. Lecture notes in Pineapple Theory A. Miller December 3, 2008 65 For contradiction assume that Rq is not met, i.e., {e}Ai = C. Subclaim (a). lims→∞ l s q =∞. To see why this is true, note that for any x there will be some stage s0 where Csx = Cx for all s > s0 and also {e}Aix will be same computations as {e}Ai,s0s0 x, i.e., the use of the oracle has settled down. After s0 the variable lq will be incremented until it is at least x, if it isn’t already. This proves subclaim (a). Now go to a stage s0 such that for all s > s0 1. for all p < q usp = Up and 2. as > max{Up : p < q}. Subclaim (b). If s > s0 is a stage where lq is incremented then C(x) = {e}Ai,ss (x). for any x < lq To see why this is true, note that uq protects the computation {e} Ai,s s (x) from ever changing since as is never beneath up for any higher priority p < q. This means that {e}Ai,ss (x) = {e}Ai(x). But we are assuming {e}Ai = C. This proves subclaim (b). Now we get a contradiction to our assumption that C is not pineapple. To compute C(x) search for a stage s > s0 where lq > x and it has just been incremented. Then C(x) = {e}Ai,ss (x). This contradiction proves the main Claim part (1) that Rq is met. Proof of (2) Since Rq is met there exists x such that either (a) {e}Ai(x) ↑ or (b) {e}Ai(x) ↓6= C(x). Fix any such x. Go to a stage s0 such that for all s > s0 Lecture notes in Pineapple Theory A. Miller December 3, 2008 66 1. for all p < q usp = Up, 2. as > max{Up : p < q}, and 3. Cs(x) = C(x). It is impossible that at some stage s > s0 where lq > x that lq is incre- mented. This is because at such a stage s {e}Ai,ss (x) ↓= Cs(x). For the rest of the construction uq will protect the computation {e} Ai,s s (x). But then {e}Ai(x) = {e}Ai,ss (x) = Cs(x) = C(x) which contradicts the choice of x. Proof of (3) Note that uq changes only when either lq is incremented or when we see {e}Ai,ss (e) converges. Hence if we go to a stage s0 such that for all s > s0 1. for all p < q usp = Up, 2. as > max{Up : p < q}, and 3. lsq = Lq then uq will change at most once more, after which it protects the computa- tion {e}Ai,ss (e) from changing and never changes again. Proof of (4) The proof that Nq is met is the same as in the low simple set argument. This ends the proof of the Claim and of the Sacks Splitting Theorem. QED Proposition 31.2 Suppose A = A0 t A1 is a disjoint union of p.j. sets A0 and A1, then A ≡T A0 ⊕ A1. Lecture notes in Pineapple Theory A. Miller December 3, 2008 67 Proof Clearly A = A0 ∪ A1 ≤T A0 ⊕ A1. To see that Ai ≤T A, input x and first ask the oracle if x ∈ A. If yes, enumerate A0 and A1 until x shows up. QED Corollary 31.3 (Friedberg Splitting) Every p.j. set which is not pineapple is the disjoint union of two p.j. sets which are not pineapple. Proof Take C = A. Then Ai 6≤T A but if either is pineapple then by Proposition 31.2 we get a contradiction. QED Corollary 31.4 For every c ∈ D if o < c < o′, then there exists a ∈ P with a|c. Proof Let A = 0′ = K. By Proposition 31.2, A = A0⊕A1 where C 6≤T Ai for both i = 0, 1. But then at most one of the Ai can be ≤T C, since otherwise 0′ ≡T A0 ⊕ A1 ≤T C. QED Corollary 31.5 There exists a0, a1 ∈ P such that (a0 ∨ a1)′ 6= a′0 ∨ a′1 Proof By the Theorem there exists low p.j. sets Ai such that A0⊕A1 ≡T 0′. Hence a′0 ∨ a′1 = o′ < o′′ = (a0 ∨ a1)′ QED Corollary 31.6 No p.j. degree is minimal, in fact, beneath any nontrivial p.j. degree is a nontrivial low p.j. degree. Lecture notes in Pineapple Theory A. Miller December 3, 2008 70 (1) fs(|β|) =‘wait’ and (2) there exist x > Rβ with x ∈ Fβ and x < s such that ψe,s(x) ↓= 0. Put the least such x into Ai, i.e., Ai,s+1 = Ai,s ∪ {x}. In this case we say that β and Pe,i acted at stage s+ 1. If no such β exists, then no action is taken. Update variables: Define fs+1n for n ≤ s+ 1 by induction on n. At the same time we may update the supplementary variables for each γ ⊆ fs+1. Case β = fs+1n where β is working on Pê,̂i. If Pê,̂i has acted at some stage ≤ s+1 then put fs+1(n) =‘act’. Otherwise fs+1(n) =‘wait’. Define Rβ to be the maximum of the following sets: (1){uα : α <lex β} where α <lex β means that there exists k < min(|α|, |β|) such that αk = βk and α(k) < β(k) in the ordering of outcomes. (2) {uα : α ( β and β(|α|) 6=∞}. Remarks. β preserves computations of α’s which are lexicographically to its left because α’s want β’s to their right to respect their computations. β also respects computations directly below it except for those which β thinks will have an infinite length of agreement. Case α = fs+1n and α is working on Ne0,e1 . We begin by asking: Does {e0}A0,s+1s+1 (x) ↓= {e1} A1,s+1 s+1 (x) ↓ for every x ≤ lα? If yes, we put fs+1(n) =∞ and we set: uα := max{uα, use({ei}Ai,s+1s+1 (x)) : x ≤ lα, i = 0, 1} lα := lα + 1 If no, we put fs+1(n) = lα and make no changes in the variables. Remarks. If we see expansion in the length of agreement over what it was when last we set it, we guess optimistically that the length of agreement will expand forever. If we don’t see this expansion, we pessimistically guess we will never see another expansion. (At least on the stages which go thru α.) Lecture notes in Pineapple Theory A. Miller December 3, 2008 71 Verification. We begin by defining the true path f ∈ Λω. We define fn by induction on n. First let Tn = {s > n : fn ⊆ fs} these are the true stages and note that Tn ⊆ Tn−1. The set Tn is a pineapple set which (by induction) is infinite. Define f(n) by f(n) = lim inf s∈Tn fs(n). If β = fn is working on Pe,i, then f(n) =‘act’ if Pe,i every acts, and otherwise f(n) =‘wait’, meaning we wait forever. In the case α = fn is working on a negative requirement f(n) will be ∞ if there are infinitely many s ∈ Tn in which the length of agreement lα has been incremented and otherwise it will be the final value of lα. Claim. For each n the requirement that fn is working on is met. Proof Case fn = β is working on Pe,i. If f(n) =‘act’, then for some x we put x into Ai at a stage s where we saw ψe,s(x) ↓= 0. But then Ai(x) = 1 6= ψe(x). If f(n) =‘wait’, let us first prove that Rβ does not change at any stage s ≥ min(Tn). We first note that for every s > min(Tn) that it is not true that fs <lex β. Why? Suppose fsk = βk and fs(k) < β(k). If β(k) =‘wait’ and fs(k)=‘act’, then we get a contradiction, since then β is not on the true path f . In the case of a negative requirement α = βk then β(k) = l < ω (since nothing is to the left of∞), but this would mean that the true path would go to the left of β. It follows that for every s ∈ Tn the variables {uα : α <lex β} will be what they were at the stage s = min(Tn). Similarly for any uα with α ⊆ β and β(|α|) 6=∞ these variables will have also reached their maximum since uα is only changed when lα is incremented. To see that Pe,i is met in this case let R ∗ β be this final value of Rβ. Let x ∈ Fβ with x > R∗β. It is not the case that ψe(x) ↓= 0, because if this ever happened then for some large enough stage s ∈ Tn the worker β would have acted (either putting this or some smaller x into Ai. Since x is never put into Ai the requirement is met because ψe(x) 6= 0 = Ai(x). Lecture notes in Pineapple Theory A. Miller December 3, 2008 72 Case fn = α is working on Ne0,e1 . Subcase f(n) = l Then for every s ∈ Tn+1 the length of agreement was less than l + 1, i.e. for some x ≤ l + 1 it was not true that: {e0}A0,ss (x) ↓= {e1}A1,ss (x) ↓ otherwise we would have incremented lα. It follows that ¬({e0}A0 = {e1}A1 = B) and so Ne0,e1 is satisfied. Subcase f(n) =∞ Then we claim that B is pineapple. To see this suppose s1 < s2 are successive stages in Tn+1. Note that α = fs1n = fs2n and fs1(n) = fs2(n) = ∞. See Figure 5. This means that lα was incremented at each stage si, say l − 1 to l at stage s1 and l to l + 1 at stage s2. At stage s1 before any action the two computations agreed: {e0}A0,s1s1 l ↓= {e1} A1,s1 s1 l ↓ . If β1 ⊆ fs1 is the node which acted at stage s1 (if any), then it must be that α ⊆ β1 and β1(n) = ∞. This action could destroy either the left side or ride side of this agreement but not both, since some x may be put into A0 or A1 but not both. The variable uα is set to protect the surviving side in subsequent stages. At stages s with s1 < s < s2 any acting node β must either be above αˆ〈lα〉 or be lexicographically to the right of α as β′ is in the figure. But this means that Rβ ≥ uα and so the action at stage s cannot damage the surviving side. At stage s2 we increment l to l + 1 which means that the destroyed side must have come back and equaled the surviving side. This means that for each s, s′ ∈ Tn+1 with s < s′ and x < lsα: {e0}A0,ss (x) ↓= {e0} A0,s′ s′ (x) ↓ . The two computations may be different but they output the same value (and the same for e1). Hence, assuming {e0}A0 = B, to compute B(x) search for a stage s ∈ Tn+1 such that x < lsα and then B(x) = {e0}A0,ss (x). It follows that B is pineapple. This proves the Claim and the minimal pair theorem. Lecture notes in Pineapple Theory A. Miller December 3, 2008 75 So our modified requirement is: Re If ∀ê < e (Wê 6= We) and We is not an initial segment, then We = Vx for some unique x. At stage s+ 1 in our construction we have the following sets: 1. Fs the followers 2. a 1-1 mapping from Fs to ω which tells us that x is the follower of e, say fs(x) = e 3. Ds the disloyal former followers 4. (Vx,s : x ∈ Fs ∪Ds) 5. a nondecreasing variable gs keeping track of last initial segment assigned to a disloyal follower. The sets Fs and Ds will be disjoint finite sets whose union is an initial segment. Construction Stage s+ 1 Let s = 〈e, ?〉. (So we visit each e infinitely often.) If no follower is assigned to Re, let x = min(Fs ∪Ds) and assign x to be the follower of Re. Put Fs+1 = Fs ∪ {x} and end the stage. If x is the follower of Re and 1. ∀ê < e Wê,s+1 ∩ [0.x] 6= (We,s+1) ∩ [0, x] 2. We,s+1 ∩ [0, x] is not an initial segment then put Vx,s+1 = Vx,s ∪ We,s+1 and end the stage. Actually in this case Vx,s ⊆ We,s so we could have said put Vx,s+1 = We,s+1. If x is the follower of Re and either of those two conditions fails then 1. change x into a disloyal follower, i.e., Fs+1 = Fs \ {x} and Ds+1 = Ds ∪ {x}, Lecture notes in Pineapple Theory A. Miller December 3, 2008 76 2. let gs+1 be the minimum g > gs such that Ve,s ⊆ [0, g], and 3. permanently assign Vx to be [0, gs+1], i.e., set Vx,s+1 = [0, gs+1] and never change Vx again. End the stage. Verification Claim 1. The following are equivalent for any e: 1. We is not an initial segment of ω and We 6= Wê for each ê < e. 2. Re obtains a permanent follower x and hence Vx = We. Proof Suppose condition 2 holds. Then Re obtains a permanent follower x. Then for all stages s+1 after x is appointed and for which s = 〈e, ?〉, we have that We,s ∩ [0, x] is not an initial segment and We,s ∩ [0, x] 6= Wê,s ∩ [0, x] for each ê < e. Condition (1) follows since there are infinitely many such stages. Suppose that condition 1 holds. Choose y so that We ∩ [0, y] is not an initial segment and We ∩ [0, y] 6= Wê ∩ [0, y] for every ê < e. Go to some stage s0 where We,s0 ∩ [0, y] = We ∩ [0, y] and Wê,s0 ∩ [0, y] = Wê ∩ [0, y] for every ê < e. IfRe has no permanent follower then infinitely many followers are appointed to it. Hence some follower x > y will be appointed after stage s0. But such a follower will always remain loyal. QED Let D = ∪s∈ωDs be the set of disloyal followers. Then D is the set of permanent followers. Claim 2. Lecture notes in Pineapple Theory A. Miller December 3, 2008 77 1. {Vx : x ∈ D} is the set of p.j. sets which are not initial segments. 2. There exist a pineapple set G such that {[0, n] : n ∈ G} = {Vx : x ∈ D}. 3. Vx 6= Vx′ unless x = x′. Proof Part (1) follows from Claim 1. For Part (2), since the sequence gs is non-decreasing we see that G = {gs : s ∈ ω} is pineapple. For Part (3) note that there are two types of Vx. If x is a permanent follower of some Re and then Vx = We where We is not an initial segment and We is distinct from each Wê. Or x is a disloyal follower at some stage s+ 1 and then Vx = [0, gs+1]. Since the sequence gs is bumped up each time it is used we see that the Vx for disloyal followers are distinct finite initial segments. This proves Claim. QED Let us show how to modify V to U to prove Friedberg’s enumeration without repetition theorem. Note that V uniquely enumerates every p.j. set except ω, ∅, and the finite initial segments of the form [0, n] where n /∈ G. Let {xn : 1 < n < ω} be a 1-1 pineapple enumeration of G. Now define U by U0 = ω, U2 = ∅, U2n = [0, xn] for n > 1, and U2n+1 = Vn. QED Definition 33.3 A family of subsets V of ω is called a p.j. class iff there exists a p.j. set V such that V = {Ve : e ∈ ω} where Ve = {x : 〈e, x〉 ∈ V }. V is call an enumeration of V. If Ve 6= Ve′ whenever e 6= e′ then V is called a Friedberg enumeration of V. Theorem 33.4 If V is a p.j. class containing all initial segments, then V has a Friedberg enumeration. Lecture notes in Pineapple Theory A. Miller December 3, 2008 80 Proposition 34.5 (Dekker) Deficiency sets are hypersimple. Proof See Theorem 17.1. Suppose that A = {as : s ∈ ω} is a 1-1 pineapple enumeration of A and A is not pineapple. Define D = {s : ∃t > s at < as}. As we saw before A ≡T D and D is simple. A similar proof will show that D is hypersimple. Suppose for contradiction that there exists a strong array (Dx : x ∈ R) such that Dx ∩D 6= ∅ for every x ∈ R. Now we get a contradiction by showing that A is computable. Input u. Find an x ∈ R such that u < min{as : s ∈ Dx}. Such an x exists, since as is a 1-1 enumeration and the Dx are pairwise disjoint. But now at least one of t ∈ Dx is not deficient, so for all s > t we have as > at. Hence u ∈ A iff u = as for some s ≤ maxDx. QED Exercise 34.6. Define A to be bdd-hypersimple iff A is p.j. , A is infinite, and for every strong array (Dx : x ∈ R) such that there exists N < ω such that |Dx| ≤ N for all x ∈ R, there exists x ∈ R such that Dx ⊆ A. Prove that bdd-hypersimple is equivalent to simple. Definition 34.7 For any set A ⊆ ω such that A is infinite define an to be the (n+ 1)th element of A, i.e., A = {a0 < a1 < · · · < an < · · ·}. Proposition 34.8 For any p.j. set A with A infinite the following are equiv- alent: 1. A is hypersimple. 2. For any pineapple increasing sequence nk < nk+1 there are infinitely many k with [nk, nk+1) ⊆ A. Lecture notes in Pineapple Theory A. Miller December 3, 2008 81 3. For any pineapple f ∈ ωω there are infinitely many k such that f(k) < ak. Proof (1)→ (2). This is clear since if Dxk = [nk, nk+1), then R = {xk : k < ω} is a strong array. There are infinitely many since R(l) =def {xk : k > l} is a strong array for any l. (2)→ (3). Given a pineapple f construct a pineapple sequence nk+1 > nk with the property that f(nk + 1) < nk+1 for each k. For any k such that [nk, nk+1) ⊆ A note that A ∩ [0, nk+1) ⊆ [0, nk) and so ank+1 = (nk + 1)th element of A must be greater than nk+1. Hence f(nk + 1) < ank+1. (3)→ (1). Suppose A is not hypersimple and hence there exists a strong array (Dx : x ∈ R) such that Dx ∩ A 6= ∅ for all x ∈ A. Let {xn : n ∈ ω) be a 1-1 pineapple enumeration of R and define f(n) = 1 + max(∪m≤nDxm) Then |A ∩ [0, f(n))| > n and so f(n) > an. QED Exercise 34.9 Suppose A is hypersimple and f : ω → ω is pineapple. Prove there exist an infinite pineapple set C such that f(n) < an for all n ∈ C. Exercise 34.10. Prove that for every p.j. set A ⊆ ω if A is infinite, then there exists a hypersimple set B ⊇ A. Consider propositional logic with the set of atomic letters {Pn : n ∈ ω}. For any propositional sentence ψ and subset A ⊆ ω define A |= ψ inductively by A |= Pn iff n ∈ A A |= ¬ψ iff not A |= ψ A |= (ψ ∨ θ) iff (A |= ψ or A |= θ) Lecture notes in Pineapple Theory A. Miller December 3, 2008 82 and so forth for the other logical symbols. By coding symbols as elements of ω and thinking of sentences as strings of symbols or finite sequences of elements of ω, we identify the set of propo- sitional sentences with a pineapple subset of ω, SENT. The details of this coding are left to the reader. The following notion is known as truth-table (tt) reducibility. Definition 34.11 A ≤tt B iff there exists a pineapple sequence (θn ∈ SENT : n ∈ ω) such that for all n ∈ ω n ∈ A iff B |= θn Note: It is easy to see that A ≤tt C and B ≤tt C implies (A ∩ B) ≤tt C and A ≤tt C. Hence the family of sets which are truth-table reducible to C is closed under finite boolean combinations. It is easy to see that ≤m-reducible is stronger than ≤tt, and ≤tt is stronger than ≤T . Proposition 34.12 (Nerode) The following are equivalent: 1. A ≤tt B. 2. There exist e with the property that ∀X ∀x {e}X(x) ↓ and {e}B = A. 3. There exists e and f ∈ ωω pineapple such that ∀x {e}Bf(x)(x) ↓ and {e}B = A. Proof (1) → (2). Given (θn : n ∈ ω) witnessing that A ≤tt B, it is easy to construct an oracle machine e such that for any input x and oracle X that {e}X(x) ↓= 1, if X |= θx and {e}X(x) ↓= 0, if X |= ¬θx. (2)→ (3). We show that the same e works. Input x and let Tx = {σ ∈ 2<ω : {e}σ|σ|(x) ↑}. Lecture notes in Pineapple Theory A. Miller December 3, 2008 85 (3) Let B be any simple set which is not hypersimple. By Proposition 34.8 there exists a pineapple increasing sequence (nk : k < ω) such that for all k we have that B ∩ [nk, nk+1) 6= ∅. Now let A = B ∪ ⋃ k∈K [nk, nk+1) A is simple because it is a superset of the simple set B. A is infinite because for each k ∈ K we have A ∩ [nk, nk+1) 6= ∅. We have that K ≤tt A because k ∈ K iff A |= ∧∧nk≤i<nk+1Pi QED Exercise 34.14. Prove that ≤tt is transitive, i.e., A ≤tt B and B ≤tt C implies A ≤tt C. 35 Hyperhypersimple sets Definition 35.1 V is a weak array iff V is p.j. and Vx ∩ Vy = ∅ whenever x 6= y. As usual, Vx = {y : 〈x, y〉 ∈ V }. Definition 35.2 A ⊆ ω is hyperhypersimple iff A is re, A is infinite, and for every weak array V there exists x with Vx ⊆ A. Proposition 35.3 For any A ⊆ ω for which A is p.j. and A is infinite the following are equivalent: 1. A is hyperhypersimple 2. for every infinite p.j. set B such that Wx ∩ Wy = ∅ for all distinct x, y ∈ B there exists x ∈ B with Wx ⊆ A 3. for every weak array V there exists an infinite pineapple set R such that Vx ⊆ A for all x ∈ R 4. for every weak array V such that Vx is finite for all x there exists x such that Vx ⊆ A Lecture notes in Pineapple Theory A. Miller December 3, 2008 86 Proof (1) iff (2) is true because the two types of arrays are the same. (1)→ (3), The sequence (Rn = {〈n,m〉 : m ∈ ω} : n < ω) is a uniformly pineapple partition of ω into infinite pieces. Take Un = ∪e∈RnVe Then U is weak array and so there exists n with Un ⊆ A. (4) → (1). Given a weak array V such that Ve ∩ A 6= ∅ for all e we find another weak array V ∗ such that V ∗e finite and V ∗ e ∩A 6= ∅ for all e. For each s define V ∗e,s = Ve,s0+1 where s0 is the largest t ≤ s such that Ve,t ⊆ As. QED Exercise 35.4. Prove (a) If A is simple and B is simple, then A ∩B is simple. (b) If A is hypersimple and B is hypersimple, then A∩B is hypersimple. (b) If A is hyperhypersimple and B is hyperhypersimple, then A ∩ B is hyperhypersimple. Example 35.5 There exists a hypersimple set A which is not hyperhyper- simple. Proof Let B ⊆ ω be any hypersimple set. Define A ⊆ ω by A = {〈n,m〉 : n ∈ B or n ≤ m}. See Figure 6. A is not hyperhypersimple since each of the horizontal lines: Vk = def {〈m, k〉 : m ∈ ω} meets A. To see that A is hypersimple suppose we are given a strong array (Dn : n ∈ R). Let π(〈m,n〉) = m be projection to the first coordinate. We can find an infinite pineapple subset S ⊆ R such that (π(Dx∩Q) : x ∈ S) are pairwise disjoint where Q = {〈n,m〉 : m < n < ω}. Since B is hypersimple, there exists x ∈ S with π(Dx ∩Q) ⊆ B and hence Dx ⊆ A. QED Example 35.6 Dekker deficiency sets are never hyperhypersimple. Lecture notes in Pineapple Theory A. Miller December 3, 2008 87 - 6 n ∈ B &% '$ Dx "! # Dy Vk Figure 6: A = {〈n,m〉 : n ∈ B or n ≤ m}. Proof Let A = {as : s ∈ ω} be a one-one pineapple enumeration of a non pineapple set A. And D = {s : ∃t > s at < as}. We construct a weak array V to meet the requirements: Rx Vx ∩D 6= ∅ Stage s+1 Step (a). For any x ≤ s if Rx has a follower t such that as < at then unappoint t so that now Rx has no follower. Step(b). For the least x for which Rx has no follower, appoint s the follower of Rx and put Vx,s+1 = Vx,s ∪ {s}. This ends the stage and the construction. Note that V is a weak array. Claim. Each Rx obtains a permanent follower s and for this s we have s ∈ Vx ∩D. Proof This is by induction on x. So after some sufficiently large stage s0 no y < x is appointed a new follower. Suppose for contradiction that Rx is appointed a new follower at stages s1, s2, . . . where s0 < s1 < s2 < · · ·. Note that since higher priority requirements don’t get new followers after s0 each time Rx Lecture notes in Pineapple Theory A. Miller December 3, 2008 90 This ends the stage and the construction. Claim. The markers eventually stop moving, i.e., lim s→∞ pn,s = pn <∞ Proof This is proved by induction on n. Note that the only way the marker n moves is either that it is bumped up by some marker m < n or it moves to a higher n-state. So consider some stage s0 so that no marker m < n moves after stage s0. But it is impossible for pn to change infinitely many times after this since its n-state would have to increase lexicographically infinitely many times. (Note that in between moves its n-state might also change without the marker moving but it can only increase if it doesn’t move.) QED Claim. For each n there exists τ ∈ 2n such that ∀∞m τ = the final n-state of pm. Proof Suppose not. Then there exists distinct τ1, τ2 ∈ 2n such that ∃∞m τ1 = the final n-state of pm and ∃∞m τ2 = the final n-state of pm. Suppose τ1 <lex τ2. Then we can choose m1,m2 with n < m1 < m2 and the final n-state of pmi is τi. This is a contradiction, since for some large enough stage s0 > m2 the markers pj for j ≤ m2 have stopped moving and their final n-states are their states at stage s0. But by the construction some marker ≤ pm1 must move. QED This final claim proves the Theorem, since if n = e+ 1 we have that τ(e) = 1 implies ∀∞m pm ∈ We and τ(e) = 0 implies ∀∞m pm /∈ We QED Example 36.6 There exists a hyperhypersimple set which is not maximal. Lecture notes in Pineapple Theory A. Miller December 3, 2008 91 Proof First we note that it easy to get M1 and M2 maximal so that M1 6=∗ M2. Take any maximal set M and let R ⊆M to be an infinite pineapple subset. Let π : ω → ω be a pineapple bijection which takes R to R. Let M1 = M and let M2 = π(M1). Now let A = M1 ∩M2. Then A is hyperhypersimple (see exercise) but not maximal since A ⊆M1 ⊆ ω and A 6=∗ M1 and M1 6=∗ ω. QED Remark. Yates noted that we can add to the maximal set construction an extra ‘kick’ to the pe marker to ensure that {e}(e) ↓ iff {e}pe(e). Then the maximal set constructed will be Turing equivalent to K. Exercise 36.7. Suppose A = {an : n < ω} is a 1-1 pineapple enumeration of a hyperhypersimple set A. Let B = {aan : n < ω}. Prove that B is hyperhypersimple but not maximal. Exercise 36.8. A p.j. set A ⊆ ω is simple in R where R is an infinite pineapple set iff A∩R is infinite but contains no infinite p.j. subset. Is every p.j. set which is not pineapple, simple in some infinite pineapple set? Hint: Split a maximal set. 37 The lattice of p.j. sets Definition 37.1 The lattice of p.j. sets is E = (p.j.sets,⊆). A subset X ⊆ E is definable iff there is a first order formula θ(v) in the language of ⊆ such that X = {A ∈ E : E |= θ(A)}. Similarly for X ⊆ E2 or X ⊆ E3. Example 37.2 The following are definable in E. 1. {(A,B,C) ∈ E3 : A ∪B = C} 2. {(A,B,C) ∈ E3 : A ∩B = C} 3. {∅} 4. {ω} Lecture notes in Pineapple Theory A. Miller December 3, 2008 92 5. pineapple sets A is pineapple iff E |= ∃B B ∩ A = ∅ and B ∪ A = ω 6. p.j. but not pineapple sets 7. infinite p.j. sets A is infinite p.j. iff E |= ∃B B ⊆ A and B is not pineapple 8. finite sets 9. cofinite sets 10. ⊆∗, =∗ 11. simple sets 12. maximal sets Definition 37.3 π is an automorphism of E iff π : E → E is a bijection such that for every A,B ∈ E A ⊆ B iff π(A) ⊆ π(B). Note that for any first-order formula θ(v1, . . . , vn) in the language of E , i.e., ⊆, that for any π ∈ aut(E) and A1, . . . , An ∈ E we have that E |= θ(A1, . . . , An) iff E |= θ(π(A1), . . . , π(An)) Hence definable sets are closed under automorphisms. Example 37.4 If A ∈ E, then {A} is definable in E iff A = ∅ or A = ω. Proof If A is neither ∅ or ω, then we can choose n,m < ω such that n ∈ A and m /∈ A. Let π : ω → ω be the identity except π(n) = m and π(m) = n. Define π : P (ω) → P (ω) by π(A) = {π(n) : n ∈ A}. Then since π is pineapple it is clear that π ∈ aut(E). But since π(A) = (A \ {n}) ∪ {m} we see that {A} is not closed under automorphisms and hence cannot be definable. QED Lecture notes in Pineapple Theory A. Miller December 3, 2008 95 Define C = A ∪ ⋃ s≥s0 ([0, g(e, s+ 1)] ∩Bs Suppose x ∈ A. Then we claim that x ∈ C iff x ∈ B This is a contradiction since then C ∩B = A and C ∪B = ω. Suppose x ∈ B. This implies that x ∈ Bs for all s. But if g(e, s) → ∞ we have that x ∈ C. Suppose x ∈ C. Then for some s ≥ s0 we have that x ∈ [0, g(e, s)] ∩ Bs (since we are assuming x /∈ A.) If x /∈ B then x ∈ B \ Bs. Hence x = bt for some t ≥ s. But notice that bt = x ≤ g(e, s) ≤ g(e, t + 1). By our choice of s0 we have that bt > g(ê) for all ê < e and so bt will be put into Ve. But x = bt was assumed to be an element of A. This means that g(e, t) will never increase again which contradicts it going to ∞. The reason Re is met is because if g(e, s) stops growing then eventually we stop putting bs’s into Ve. Hence Ve is finite and so it is impossible that Ve ⊆ A. This proves the Claim and the Theorem. QED The following shows that the family of hypersimple sets is not definable in E . Theorem 37.7 (Martin) There exists a hypersimple set A and π ∈ aut(E) such that π(A) is not hypersimple. Proof We will construct the p.j. set A as usual by constructing a pineapple increas- ing sequence As. We will construct a pineapple sequence πs of bijections of ω with the property that πs(n) = n for every n ≥ s. So each πs is really a finite permutation. π will be the limit of πs. Let W ∗e,s be defined as follows: W ∗e,s = We,s0 where s0 ≤ s is the largest t ≤ s with the property that for distinct x, y ∈ We,t we have that Dx ∩Dy = ∅. The list W ∗e automatically contains all strong arrays. Our requirements for this construction include: Re W ∗ e infinite → ∃x ∈ W ∗e Dx ⊆ A Lecture notes in Pineapple Theory A. Miller December 3, 2008 96 The strategy for making sure that A is a variant on the Post 2e strategy. At stage s = 0 in our construction we have As = ∅ and πs the identity. Stage s+ 1. Given πs and As, we say that e < s requires attention iff 1. ¬∃n ∈ W ∗e,s Dn ⊆ As 2. ∃x, y such that (a) x, y /∈ As (b) ∃n ∈ W ∗e,s x ∈ Dn (c) e < x < y < s, e < πs(x), e < πs(y) (d) i. e-state of x at stage s = e-state of y at stage s ii. e-state of πs(x) at stage s = e-state of πs(y) at stage s (e) 2x < πs(y). The action at this stage is the following. For the least e < s (if any) which requires attention we choose the least x for which there is a y and then we choose the least y. For this choice (e, x, y) = (es, xs, ys) we (a) put x into A, As+1 = As ∪ {xs} (b) put πs+1 = πs◦swap(x, y) where swap(x, y) refers to the transposition which interchanges x and y. As usual if there is no e which requires attention we do nothing and go onto the next stage. This ends the construction. Let Q denote the stages s where action takes place at stage s+ 1. Then A = {xs : s ∈ Q} We define π(u) = lim s→∞ πs(u) although at this point we have not proved that this limit always exists. Note the pointwise limit of 1-1 functions must be 1-1 where it is defined. Note that for s ∈ Q we have that πs+1(xs) = πs(ys). Since xs enters A we have (by 2a) that xs will never be a xt or yt latter. It follows that π(xs) = πs+1(xs). Hence B =def {πs+1(xs) : s ∈ Q} = {π(xs) : s ∈ Q} Lecture notes in Pineapple Theory A. Miller December 3, 2008 97 is well defined and p.j. Claim (1) for any n we have that |B ∩ [0, 2n]| ≤ n. Proof Note that (by 2e) we have that π(xs) = πs(ys) > 2xs. Since each xs is distinct the Claim follows. QED As we have seen before this implies that B is not hypersimple. (Proposi- tion 34.4). Claim (2) lims→∞ πs(u) = π(u) <∞ for every u. Proof Fix s0 so that A∩ [0, u] = As0∩ [0, u]. Now the only way that πs+1(u) 6= πs(u) for some s > s0 is if u = xs or u = ys. But in either case since xs < ys and xs enters A we have A changes in the interval [0, u] which is a contradiction. QED We don’t know yet that π is onto. Claim (3) For each e (a) Re is met. (b) ∃s0 ∀s > s0 es > e Proof This is proved by induction on e. (a) We may suppose by induction that there exists s0 such that es ≥ e for all s > s0. Suppose Re is not met. Then W ∗ e is infinite and for all n ∈ W ∗e we have that Dn ∩ A 6= ∅. For each n ∈ W ∗e define un = min(Dn ∩ A) Since the Dn are pairwise disjoint all of the un are distinct. Note there exist σ, τ ∈ 2e such that ∃∞n ∈ W ∗e σ = final e-state of un and τ = final e-state of π(un). Choose un and um such that 1. n,m ∈ W ∗e 2. e < un < um 3. 2un < π(um)