Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and...

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Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore Joint with F. Stephan and Y. Yang 18th May 2006

Transcript of Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and...

Page 1: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Ershov Hierarchies and Degree Theory

Liang YuDepartment of mathematics

National University of Singapore

Joint with F. Stephan and Y. Yang

18th May 2006

Page 2: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Analytical hierarchy

DefinitionA predicate R(x ,n) on ωω × ω is recursive if there is a partialrecursive function Φ(σ,n) on ω<ω × ω so that

(i) ∀σ∀τ∀n(Φ(σ,n) ↓ ∧σ � τ =⇒ Φ(σ,n) = Φ(τ,n)).(ii) ∀x∀n∃m(R(x ,n) ⇔ Φ(x � m,n) = 0).

(iii) ∀x∀n∃m(Φ(x � m,n) ↓).

A predicate P(x ,n) is Π11 if there is a recursive predicate

R(y , x ,n) so that

P(x ,n) ⇐⇒ ∀yR(y , x ,n).

The predicate ¬P(x ,n) is called Σ11.

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Recursive ordinals and Kleene’s O

Definition

The Π11-well-ordering <o on ω is defined by transfinite induction

as follows:0 <o 1;(∀n)n <o 2n;(∀n)Φe(n) <o Φe(n + 1) =⇒ (∀n)n <o 3 · 5e.

O is the field of <o.

An ordinal α is recursive if it is isomorphic toO � n = {m|m <o n} for some n ∈ O. ωCK

1 is the leastnon-recursive ordinal.

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Some properties of O

TheoremFix an enumeration {Wn}n of r.e. sets.

1 There is a recursive function p so that for all n ∈ O,Wp(n) = O � n.

2 There is a recursive function q so that for all n ∈ O,Wq(n) = {(n0,n1)|n0 <o n1 <o n}.

The recursive function +o has the following properties for allm,n.

(i) m,n ∈ O ⇐⇒ m +o n ∈ O.(ii) m,n ∈ O =⇒ |m +o n| = |m|+ |n|.

(iii) m,n ∈ O ∧ n 6= 1 =⇒ m <o m +o n.(iv) m ∈ O ∧ k <o n ⇐⇒ m +o k <o m +o n.(v) m ∈ O ∧ n = k ∈ O ⇐⇒ m +o n = m +o k .

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Facts in higher recursion theory

Theorem (Sacks)

There is a recursive function g so that for all e, if We ⊆ O and<o on We is linear, then g(e) ∈ O and∀n(n ∈ We =⇒ n <o g(e)).

Corollary

For all m ∈ O, there is a recursive function g so that for all e, ifWe ⊆ O and <o on We is linear, then g(e) ∈ O and∀n(n ∈ We =⇒ m +o n <o g(e)). Moreover, the function g canbe found uniformly.

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continued

Theorem (Feferman and Spector)

For each n ∈ O, there is a Π11 path T ⊂ O with |T | = ωCK

1 forwhich n ∈ T .

Theorem (Spector)

Each Σ11 well ordering is strictly below ωCK

1 .

Theorem (Kleene)

Given a Π11 set R ⊆ ω × ω, there is a Π1

1 function f so that1 ∀n∃m0∃m1((n,m) ∈ R =⇒ f (n) = m1 ∧ (n,m1) ∈ R).2 ∀n∀m(f (n) = m =⇒ (n,m) ∈ R).

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Ershov Hierarchies

Definition (Ershov)

For each n ∈ O, a subset A of ω is Σ−1n if there is a recursive

function f so that1 For all i <o j <o n, Wf (i) ⊆ Wf (j).2 For all k , k ∈ A if and only if there is a notation i <o n so

thatk ∈ Wf (i).|i | 6≡ |n|( mod 2).For all j <o i , k 6∈ Wf (j).

The set ω − A is said to be Π−1n .

A set B is said to be ∆−1n if B ∈ Σ−1

n ∩ Π−1n .

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Russian school’s results

Theorem (Ershov)

1 For all n <o m, Σ−1n ∪ Π−1

n ⊂ Σ−1m ∩ Π−1

m .2 ∆0

2 =⋃

n∈O∧|n|=ω2 Σ−1n .

3 For all m,n ∈ O with |m|, |n| < ω2, Σ−1m = Σ−1

n if and only if|n| = |m|.

4 For each path T ⊂ O, if |T | < ω3, then⋃

n∈T Σ−1n 6= ∆0

2.5 There is a path T ⊂ O with |T | = ω3 for which⋃

n∈T Σ−1n = ∆0

2.

Theorem (Selivanov )

For all n, there is a set A ∈ Σ−1n so that B 6∈

⋃k<on Σ−1

k for allB ≡T A.

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A basic fact

Theorem (Forklore)

For every n ∈ O and set A ⊆ ω, the following statements areequivalent:

1 A ∈ Σ−1n .

2 There is a recursive function f : ω × ω → Wp(2n) so that forall k,

1 For all i ≥ j , f (k , i) ≤o f (k , j).2 For all i , f (k , i + 1) 6= f (k , i) =⇒ |f (k , i + 1)| 6≡ |f (k , i)|(

mod 2).3 k ∈ A if and only if | lims f (k , s)| 6≡ |f (k ,0)|( mod 2).

Page 10: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Paths through O

Theorem (Stephan, Yang and Yu)

1 There is a path T ⊂ O with |T | = ωCK1 so that⋃

n∈T Σ−1n = ∆0

2.2 There is a path T ⊂ O with |T | = ωCK

1 so that⋃n∈T Σ−1

n 6= ∆02.

Proof.

For (1), highly non-uniformly putting ∆02 sets into Ershov’s

hierarchy. For (2), by Feferman and Spector’s results.

Proposition (Stephan, Yang and Yu)

If T is a Π11 path so that

⋃n∈T Σ−1

n = ∆02, then T is ∆1

1.

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Lachlan’s result and it’s generalization

Theorem (Stephan, Yang and Yu)

For each notation n = 2m ∈ O with 0 < |m| < ω and setA ∈ Σ−1

n − Σ−1m , there is a non-recursive set B ∈ Σ−1

m so thatB ≤T A.

Proposition (Stephan, Yang and Yu)

If n = n0 +o n1 ∈ O, then for every set A ∈ Σ−1n − (Σ−1

n0∪ Σ−1

22n1 ),

there is a non-recursive set B ∈ Σ−1n0

so that B ≤T A.

Corollary

For each n ∈ O with n >o 2, if A ∈ Σ−12n −Σ−1

n , then A computesa non-recursive Σ−1

n set.

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A counter example to generalize Lachlan’s result

Theorem (Ding, Jin and Wang)

If n ∈ O, then there is a minimal degree a < 0′ so that A 6∈ Σ−1n

for all A ∈ a.

Proposition (Stephan, Yang and Yu)There is a notation n = n0 +o n1 ∈ O with |n1| < |n| so thatthere is a minimal degree a for which a ∩

⋃m<on Σ−1

m = ∅ buta ∩ Σ−1

n 6= ∅.

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Finite levels of Ershov hierarchy

Definition (Putnam)A set A is n-r.e. if there is a recursive functionf : ω × ω → ω so that for each m,

f (0,m) = 0.A(m) = lims f (s,m).|{s|f (s + 1,m) 6= f (s,m)}| ≤ n.

A Turing degree is n-r.e. if it contains an n-r.e. set.

A set is n-r.e. iff it is Σ−1m where m ∈ O and |m| = n + 1.

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Model theory I

The partially ordered language, L(≤), L(≤) includes variablesa,b, c, x , y , z, ... and a binary relation ≤ intended to denote apartial order. Atomic formulas are x = y , x ≤ y . Σ0 formulasare built by the following induction definition.

Each atomic formula is Σ0.¬ψ for some Σ0 formula ψ.ψ1 ∨ ψ2 for two Σ0 formula ψ1, ψ2.ψ1 ∧ ψ2 for two Σ0 formula ψ1, ψ2ψ1 =⇒ ψ2 for two Σ0 formula ψ1, ψ2.

A formula ϕ is Σ1 if it is of the form ∃x1∃x2...∃xnψ(x1, x2, ..., xn)for some Σ0 formula ψ.A formula ϕ is Πn if it is the form ¬ψ for some Σn formula ψ anda formula ϕ is Σn+1 if it is the form ∃x1∃x2...∃xmψ(x1, x2, ..., xm)for some Πn formula ψ. A sentence is a formula without freevariables.

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Model theory II

Given two structures A(A,≤A) and B(B,≤B) for L(≤), we saythat A(A,≤A) is a substructure of B(B,≤B), writeA(A,≤A) ⊆ B(B,≤B), if A ⊆ B and the interpretation ≤A is arestriction to A of ≤B.Examples:D(≤ 0′) = (D(≤ 0′),≤). The structures of n-r.e. degreesDn = (Dn,≤).

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Definition(i) We say that A(A,≤A) is a Σn substructure of B(B,≤B),

write A(A,≤A) �Σn B(B,≤B), if A(A,≤A) ⊆ B(B,≤B) andfor all Σn formulas ϕ(

−→x ) and any −→a ⊆ A,

A(A,≤A) |= ϕ(−→a ) if and only if B(B,≤B) |= ϕ(

−→a ).

(ii) We say that A(A,≤A) is Σn-elementary-equivalent toB(B,≤B), write A(A,≤A) ≡Σn B(B,≤B), if for all Σnsentences ϕ,

A(A,≤A) |= ϕ if and only if B(B,≤B) |= ϕ.

Page 17: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Elementary difference among Ershov hierarchies

Theorem (Folklore)

For all n ∈ ω, Dn ≡Σ1 D(≤ 0′).

Theorem (Arslanov)For each natural number n > 1, D1 6≡Σ3 Dn.

Theorem (An accumulation of lots of results)

Dn 6≡Σ2 D(≤ 0′).For each natural number n > 1, D1 6≡Σ2 Dn.

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Downey’s conjecture and its solution

Conjecture (Downey )

For each n > 1 and k ≥ 0, Dn ≡Σk Dn+m.

Theorem (Arslanov, Kalimullin, Lempp )

D2 6≡Σ2 D3.

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Σ1-substructures of D(≤ 0′)

Theorem (Slaman)

(i) There are r.e. sets A,B and C and a ∆02 set E such that

∅ <T E ≤T A;C 6≤T B ⊕ E;For all r.e. set W (∅ <T W ≤T A ⇒ C ≤T W ⊕ B).

(ii) For each natural number n ≥ 1, Dn 6�Σ1 D(≤ 0′).

Proof.Take a Σ1 formula

ϕ(x1, x2, x3) ≡ ∃e∃y∃z(e ≤ x1 ∧ e ≥ y ∧ e 6= y ∧ z ≥ x2 ∧ z ≥ e ∧ z 6≥ x3).

Page 20: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Slaman’s conjecture and its generalization

Conjecture (Slaman)

For each n > 1, D1 �Σ1 Dn?

Conjecture (Arslanov and Lempp)For all n > m, Dm �Σ1 Dn?

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The solution

Theorem (Yang and Yu )

There are r.e. sets A,B,C and E and a d.r.e. set D such that1 D ≤T A and D 6≤T E;2 C 6≤T B ⊕ D;3 For all r.e. sets W (W ≤T A ⇒ either C ≤T W ⊕ B or

W ≤T E).

Theorem (Yang and Yu)For all n > 1, D1 6�Σ1 Dn.

Proof.

ϕ(x1, x2, x3, x4) ≡ ∃d∃g(d ≤ x1∧d 6≤ x4∧g ≥ x2∧g ≥ d∧x3 6≤ g).

Page 22: Ershov Hierarchies and Degree Theory - Nanjing Universityyuliang/tamc.pdf · Ershov Hierarchies and Degree Theory Liang Yu Department of mathematics National University of Singapore

Some questions

Question (Khoussainov)For n > 1, is there a function f : D1 → Dn so that for anyΣ1-formula ϕ(x1, ..., xm),

D1 |= ϕ(x1,x2, ...,xm) iff Dn |= ϕ(f (x1), f (x2), ..., f (xm)),

where x1,x2, ...,xm range over D1?

QuestionFinding out a proper Σ1-substructure of D(≤ 0′)

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Thank you