x. creative setyuxi/teaching/computability2013/slides/1… · quotation from post the terminology...
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X. Creative Set
Yuxi Fu
BASICS, Shanghai Jiao Tong University
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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.”
Computability Theory, by Y. Fu X. Creative Set 1 / 30
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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?
Computability Theory, by Y. Fu X. Creative Set 2 / 30
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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.
Computability Theory, by Y. Fu X. Creative Set 3 / 30
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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.
Computability Theory, by Y. Fu X. Creative Set 10 / 30
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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
Computability Theory, by Y. Fu X. Creative Set 19 / 30
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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.
Computability Theory, by Y. Fu X. Creative Set 20 / 30
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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.
Computability Theory, by Y. Fu X. Creative Set 26 / 30
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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
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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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