Cupping Classes of Σ 0 2. Enumeration Degrees. Mariya I. Soskova Department of Pure Mathematics University of Leeds

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1 Cuppng Classes of Σ 0 2 Enumeraton Degrees Marya I. Soskova Department of Pure Mathematcs Unversty of Leeds marya@maths.leeds.ac.uk

2 The local structure of the enumeraton degrees Σ 2 e-degrees 0 e 2 e-degrees Π 1 e-degrees 0 e

3 Transferrng results from the Turng degrees There s a natural embeddng of the Turng degrees n the enumeraton degrees. The mages of Turng degrees under ths embeddng are the total e-degrees. D T 0 0 e D e ι Σ Tot- 2 C.E Π e

4 Cuppng We say that a degree a s cuppable f there exsts a degree b < 0 e such that a b = 0 e. Negatve Results: (Cooper, Sorb, Y): There exsts a nonzero Σ 2 enumeraton degree that s not cuppable. Postve Results: (Cooper, Sorb and Y): Every nonzero 2 e-degree s cuppable by a total ncomplete 2 e-degree. (S, Wu): Every nonzero 2 e-degree s cuppable by a partal and low 2 e-degree.

5 Cuppng partners Queston How much further can we lmt the the search for cuppng partners. 0 e a c Low Partal- 2 Tot Π 1 b 0 e

6 Reachng the frst lmt Theorem For every unform sequence of ncomplete 2 enumeraton degrees {a n } n<ω there s a non-zero 2 enumeraton degree b such that a n b 0 e for every n. Proof: The Constructon of a non-cuppable Σ 2 enumeraton degree carred out aganst a unform sequence of ncomplete 2 enumeraton degrees.

7 Proof sketch Let {A n } n<ω be a lst of representatves of the gven enumeraton degrees. Let {A n,s } s<ω be a good 2 approxmaton to A n. ( s)(a s A). Lm s A s (x).

8 Proof sketch We shall construct a 2 set B satsfyng the followng requrements: For every natural number e we have a requrement: N e : W e B. For every j and every n we wll have a requrement : P j,n : Θ A n,b j K.

9 The N-strategy N e : W e B d w Select a wtness x as a fresh number. If x / W e - do nothng (outcome w) If x W e then extract x from B (outcome d)

10 The P-strategy P j,n : Θ An,B j K w Construct an e-operator Γ threatenng to prove that Γ An = K. Perform cycles k of ncreasng length, montorng each number n < k.

11 The P-strategy P j,n : Θ A n,b j K w n K : Search for an axom n Θ j that s vald on almost all stages. Ax(n) = n, D A, D B. Vald Ax(n) Enumerate n, D A n Γ, go on to n + 1. Invald Ax(n) Then outcome. Redefne Ax(n), move on to n + 1. Infntely many tmes outcome n / Θ A n,b j.

12 The P-strategy P j,n : Θ An,B j K w n / K Rectfy Γ A (n). Incorrect For each axom n, D A Γ, enumerate D B back n B, outcome s w. Do not move on to next element. On all but fntely many stages: outcome w n Θ An,B j. Correct Looks lke n / Γ A, restore B and go on to n + 1.

13 The P-strategy P j,n : Θ A n,b j K w A n s ncomplete. Hence Γ An K. Let n be the least dfference. If n K \Γ An then Θ j has faled to provde us wth a vald axom. Infntely often - outcome. If n Γ A n \K then we have restored an axom n Θ j and t s vald forever. Cofntely often outcome w.

14 The set B s 2 P w Θ... Γ... n, D A, D B n, D A N x d w

15 Lookng at the local structure more closely Defnton 1. A set A s n-c.e. f there s a computable functon f such that for each x, f (x, 0) = 0, {s + 1 f (x, s) f (x, s + 1)} n and A(x) = lm s f (x, s). 2. A s ω-c.e. f there are two computable functons f (x, s), g(x) such that for all x, f (x, 0) = 0, {s + 1 f (x, s) f (x, s + 1)} g(x) and lm s f (x, s) = A(x). 3. A degree a s n-c.e.(ω-c.e.) f t contans a n-c.e.(ω-c.e.) set.

16 Lookng at the local structure more closely 0 e Σ 2 2 a ω-c.e. 3-c.e b Π 1 = 2-c.e Wu, S: For every non-zero ω-c.e. enumeraton degree a there exsts an ncomplete 3-c.e. enumeraton degree b that cups a. 0 e

17 Another approach to the lmt 0 e Σ 2 2 ω-c.e. 3-c.e Π 1 = 2-c.e (Cooper, Seetapun and L): In the Turng degrees there exsts a sngle ncomplete 2 Turng degree d that cups every non-zero c.e. Turng degree. Can we fnd a smlar result for bgger classes? 0 e

18 The second lmt Theorem For every ncomplete Σ 2 enumeraton degree a there exsts a non-zero 3-c.e. enumeraton degree b such that a does not cup b. Proof: Let A be a representatve of the gven Σ 2 e-degree wth good approxmaton {A s }. We shall construct two 3-c.e. sets X and Y so that one of them wll have the requred propertes.

19 Requrements For every natural number e we have a requrement: N e : W e X W e Y. For every we wll have a par of requrements: P 0 P 1 : Θ A,X K. : Ψ A,Y K. We wll ensure that: ( )(P 0 ) ( )(P 1 ).

20 The P-strategy P,j : Θ A,X K Ψ A,Y j K X, 0 Y, 0... X, n Y, n... X, w Y, w Construct an e-operator Γ threatenng to prove that A s complete. Run cycles k scannng each element n < k. For every element n act as n the prevous constructon.

21 The P-strategy P,j : Θ A,X K Ψ A,Y j K X, 0 Y, 0... X, n Y, n... X, w Y, w n K : Search for a vald Ax θ (n) = n, D A,θ, D X and Ax ψ = n, D A,ψ, D Y. Invald Ax θ (n) Then outcome X, n. Redefne Ax θ (n), move on to n + 1. Invald Ax ψ (n) Then outcome Y, n. Redefne Ax ψ (n), move on to n + 1. Vald Enumerate n, D A,θ D A,ψ n Γ, go on to n + 1.

22 The P-strategy P,j : Θ A,X K Ψ A,Y j K X, 0 Y, 0... X, n Y, n... X, w Y, w n / K Rectfy Γ A (n). Incorrect For each axom n, D A,θ D A,ψ Γ, enumerate D X back n X or D Y back n Y. Choose the axom for n vald the longest n Γ. If Θ was restored: outcome X, w. If Ψ was restored: outcome Y, w. Correct Looks lke n / Γ A then go on to n + 1.

23 The P-strategy P,j : Θ A,X K Ψ A,Y j K X, 0 Y, 0... X, n Y, n... X, w Y, w A s ncomplete. Hence Γ A K. Let n be the least dfference. After a certan stage s outcomes X, m and Y, m are not accessble. If n K \Γ A then Θ has faled to provde us wth a vald axom. Ψj has faled to provde us wth a vald axom. If n Γ A \K then We have restored an axom n Θ and t s vald forever. We have restored an axom n Ψ j and t s vald forever.

24 The tree P 0 P 1 j... X, n Y, n... X, w Y, w P 0 +1 P1 j P 0 P 1 j+1 P 0 +1 P1 j P 0 P 1 j+1

25 The N-strategy N e : W e X W e Y d w Select a wtness x as a fresh number. If x / W e - do nothng (outcome w) If x W e then extract x from both X and Y (outcome d)

26 The N-strategy N e : W e X W e Y d w Permanently restran x out of X but allow t to be enumerated back n Y. Select a second wtness y - one that does not appear n any axom seen sofar n the constructon. If y / W e then - do nothng (outcome w) If y W e then extract and permanently restran y from Y (outcome d)

27 Conflcts resolved P Θ Γ Ψ n, D A,θ, D X n, D A,ψ, D Y N X, w Y, w n, D A,θ D A,ψ d w x y

28 Bblography S. B. Cooper, A. Sorb, X. Y, Cuppng and noncuppng n the enumeraton degrees of Σ 0 2 sets, Ann. Pure Appl. Logc 82 (1996), A. H. Lachlan and R. A. Shore, The n-rea enumeraton degrees are dense, Arch. Math. Logc 31 (1992), M. Soskova, G. Wu, Cuppng 0 2 enumeraton degrees to 0, Computablty n Europe 2007, Lecture Notes n Computer Scence 4497 (2007),