X. Creative Setyuxi/teaching/computability2013/slides/1… · Quotation from Post The terminology...

X. Creative Set

Yuxi Fu

BASICS, Shanghai Jiao Tong University

Quotation from Post

The terminology ‘creative set’ was introduced by E. Post in

Recursively Enumerable Sets of Positive Integers and their DecisionProblems. Bulletin of American Mathematical Society, 1944.

“. . . every symbolic logic is incomplete and extensible relative tothe class of propositions”.

“The conclusion is inescapable that even for such fixed,well-defined body of mathematical propositions, mathematicalthinking is, and must remain, essentially creative.”

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What are the Most Difficult Semi-Decidable Problems?

We know that K is the most difficult semi-decidable problem.

What is then the m-degree dm(K )?

What is an r.e. set C s.t. A ≤m C for every r.e. set A?

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What are the Most Difficult Semi-Decidable Problems?

An r.e. set is very difficult if it is very non-recursive.

An r.e. set is very non-recursive if its complement is very non-r.e..

A set is very non-r.e. if it is easy to distinguish it from any r.e. set.

These sets are creative respectively productive.

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Synopsis

1. Productive Set

2. Creative Set

3. The Lattice of m-Degrees

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1. Productive Set

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Suppose Wx ⊆ K . Then x ∈ K \Wx .

So x witnesses the strict inclusion Wx ( K .

In other words the identity function is an effective proof that Kdiffers from every r.e. set.

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Productive Set

A set A is productive if there is a total computable function p suchthat whenever Wx ⊆ A, then p(x) ∈ A \Wx .

The function p is called a productive function for A.

A productive set is not r.e. by definition.

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Example

1. K is productive.

2. {x | c /∈Wx} is productive.

3. {x | c /∈ Ex} is productive.

4. {x | φx(x) 6= 0} is productive.

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Example

Suppose A = {x | φx(x) 6= 0}.

By S-m-n Theorem one gets a primitive recursive function p(x)such that φp(x)(y) = 0 if and only if φx(y) is defined. Then

p(x) ∈Wx ⇔ p(x) /∈ A.

So if Wx ⊆ A we must have p(x) ∈ A \Wx .

Thus p is a productive function for A.

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Productive Set

Lemma. If A ≤m B and A is productive, then B is productive.

Proof.Suppose r : A ≤m B and p is a production function for A.

By applying S-m-n Theorem to φx(r(y)), one gets a primitiverecursive function k(x) such that Wk(x) = r−1(Wx).

Then rpk is a production function for B.

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Productive Set

Theorem. Suppose that B is a set of unary computable functionswith f∅ ∈ B and B 6= C1. Then B = {x | φx ∈ B} is productive.

Proof.Suppose g /∈ B. Consider the function f defined by

f (x , y) '{

g(y), if x ∈Wx ,↑, if x /∈Wx .

By S-m-n Theorem there is a primitive recursive function k(x)such that φk(x)(y) ' f (x , y).

Clearly x /∈Wx iff φk(x) = f∅ iff φk(x) ∈ B iff k(x) ∈ B.

Hence k : K ≤m B.

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Property of Productive Set

Lemma. Suppose that g is a total computable function. Thenthere is a primitive recursive function p such that for all x ,Wp(x) = Wx ∪ {g(x)}.

Proof.Using S-m-n Theorem, take p(x) to be a primitive recursivefunction such that

φp(x)(y) '{

1, if y ∈Wx ∨ y = g(x),↑, otherwise.

We are done.

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Property of Productive Set

Theorem. A productive set contains an infinite r.e. subset.

Proof.Suppose p is a production function for A.

Take e0 to be some index for ∅. Then p(e0) ∈ A by definition.

By the Lemma there is a primitive recursive function k such thatfor all x , Wk(x) = Wx ∪ {p(x)}.

Apparently {e0, . . . , kn(e0), . . .} is r.e.

Consequently {p(e0), . . . , p(kn(e0)), . . .} is a r.e. subset of A,which must be infinite by the definition of k .

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Productive Function via a Partial Function

Proposition. A set A is productive iff there is a partial recursivefunction p such that

∀x .(Wx ⊆ A⇒ (p(x) ↓ ∧p(x) ∈ A \Wx)). (1)

Proof.Suppose p is a partial recursive function satisfying (1). Let s be aprimitive recursive function such that

φs(x)(y) '{

y , p(x)↓ ∧ y ∈Wx ,↑, otherwise.

A productive function q can be defined by running p(x) andp(s(x)) in parallel and stops when either terminates.

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Productive Function Made Injective

Proposition. A productive set has an injective productive function.

Proof.Suppose p is a productive function of A. Let

Wh(x) = Wx ∪ {p(x)}.

ClearlyWx ⊆ A⇒Wh(x) ⊆ A. (2)

Define q(0) = p(0).

I If p(x+1), ph(x+1), . . . , phx+1(x+1) are pairwise distinct, letq(x+1) be the smallest one not in {q(0), . . . , q(x)}.

I Otherwise we can let q(x+1) be µy .y /∈ {q(0), . . . , q(x)}.This is fine since Wx 6⊆ A due to (2).

It is easily seen that q is an injective production function for A.

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Myhill’s Characterization of Productive Set

Theorem. (Myhill, 1955) A is productive iff K ≤1 A iff K ≤m A.

K ≤1 A implies K ≤m A, which in turn implies “A is productive”.

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Proof

Suppose p is a productive function for A. Define

f (x , y , z) '{

0, if z = p(x) and y ∈ K ,↑, otherwise.

By S-m-n Theorem there is an injective primitive recursive functions(x , y) such that

φs(x ,y)(z) ' f (x , y , z).

By definition,

Ws(x ,y) =

{{p(x)}, if y ∈ K ,∅, otherwise.

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Proof

By Recursion Theorem there is an injective primitive recursivefunction n(y) such that Ws(n(y),y) = Wn(y) for all y . So

Wn(y) =

{{p(n(y))}, if y ∈ K ,∅, otherwise.

We claim that K ≤m A.

y ∈ K ⇒ Wn(y) = {p(n(y))} ⇒ p(n(y)) /∈ A.

y /∈ K ⇒ Wn(y) = ∅ ⇒ p(n(y)) ∈ A.

By the previous theorem we may assume that p is injective. So thereduction function p(n( )) is injective. Conclude K ≤1 A.

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2. Creative Set

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Creative Set

A set A is creative if it is r.e. and its complement A is productive.

Intuitively a creative set A is effectively non-recursive in the sensethat the non-recursiveness of A, hence the non-recursiveness of A,can be effectively demonstrated.

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Creative Set

1. K is creative.

2. {x | c ∈Wx} is creative.

3. {x | c ∈ Ex} is creative.

4. {x | φx(x) = 0} is creative.

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Creative Set

Theorem. Suppose that A ⊆ C1 and let A = {x | φx ∈ A}. If A isr.e. and A 6= ∅,N, then A is creative.

Proof.Suppose A is r.e. and A 6= ∅,N. If f∅ ∈ A, then A is productive bya previous theorem. This is a contradiction.

So A is productive by the same theorem. Hence A is creative.

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Creative Set

The set K0 = {x |Wx 6= ∅} is creative. It corresponds to the setA = {f ∈ C1 | f 6= f∅}.

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Creative Sets are m-Complete

Theorem. (Myhill, 1955)

C is creative iff C is m-complete iff C is 1-complete iff C ≡ K .

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3. The Lattice of m-Degrees

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What Else?

Q: In the world of recursively enumerable sets, is there anythingbetween the recursive sets and the creative sets?

A: There is plenty.

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What Else?

Q: In the world of recursively enumerable sets, is there anythingbetween the recursive sets and the creative sets?

A: There is plenty.

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Trivial m-Degrees

1. o = {∅}.

2. n = {N}.

3. o ≤m a provided a 6= n.

4. n ≤m a provided a 6= o.

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Nontrivial m-Degrees

5. The recursive m-degree 0m consists of all the nontrivialrecursive sets.

6. An r.e. m-degree contains only r.e. sets.

7. The maximum r.e. m-degree dm(K ) is denoted by 0′m.

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The Distributive Lattice of m-Degrees

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0m

0′m

ab

c

...

The m-degrees ordered by ≤m form a distributive lattice.

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Problem with m-Degree

The m-reducibility has two unsatisfactory features:

(i) The exceptional behavior of ∅ and N.

(ii) The invalidity of A 6≡m A in general.

The problem is due to the restricted use of oracles.

We shall remove this restriction in Turing reducibility.

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