I. 3.1 (Partial) Recursive Functions and Sets
II.4 Depth
II.4.4 Depth for non-r.b. Π 0 1 Sets
Nothing in our definition of depth necessitates thatP be a r.b. Π01 class in order for the definition to make sense. However, there are two reasons for our restriction to only r.b. Π01 classes. The first is that our interest in depth is ultimately relegated to r.b. Π01 subsets of NN. The second is that it is unclear whether ‘depth’
is a useful notion for arbitrary subsets ofNN, and if so, whether the verbatim extension of the definition of depth to any subset ofNN provides the ‘right’ definition.
In fact, we can show that extending our definition of depth even to only Π01 subsets ofNNor Π02subsets of{0,1}Ncauses us to lose the guarantee of Theorem II.4.11 that no difference random computes a member of a ‘deep set’.
Lemma II.4.23. SupposeX∈ {0,1}NandX≤T0′. Then X is aΠ02 singleton, i.e.,{X} isΠ02. Proof. X≤T0′impliesX (as a subset of N) is ∆02, so there are recursive predicatesR andS such that
x∈X ⇐⇒ ∀n∃m R(x, n, m) ⇐⇒ ∃n∀m S(x, n, m).
Then
{X} = {Y ∈ {0,1}N∣ ∀x((x∈Y → ∀n∃m R(x, n, m)) ∧ (∃n∀m S(x, n, m) →x∈Y))}
shows thatX is a Π02singleton.
Proposition II.4.24. There exists a subsetP ⊆NNwhich is deep in the sense that there is an order function rsuch thatM(P↾r(n)) ≤2−n for alln∈N, but for which there are difference random sequences that compute members of P. Moreover, P may be taken to either be a Π02 subset of{0,1}N or aΠ01 subset ofNN.
Proof. LetQ= {X∈ {0,1}N∣ ∀nKA(X↾n) ≥n−c}, wherecis sufficiently large so that Q≠ ∅. By the Low Basis Theorem [14] there is anA∈Q such thatA<T0′. Lemma II.4.23 then impliesA is a Π02 singleton, but being an incomplete Martin-L¨of random sequence means that it is also difference random. SinceA∈Q, we also know thatM(A↾ (n+c)) ≤2c⋅2−(n+c)=2−n for alln∈N. Thus,{A}is deep in the extended sense, but the difference randomAcomputes a member of{A}.
Being a Π02 singleton, there is a recursive predicate R such that A is the only sequence X satisfying
∀n∃m R(X, n, m). Definef∶N→Nby
f(n) ∶=leastmsuch that⟨A, n, m⟩ ∈R.
Then define B ∈ NN by B(n) ∶= π(2)(A(n), f(n)). Given X ∈ NN and i ∈ {0,1}, let (X)i be defined by (X)i(n) = (X(n))i forn∈N, where (π(2)(n0, n1))i =ni. Then {B} = {X ∣ ∀n R((X)0, n,(X)1(n))} is Π01 and recursively homeomorphic to{A}.
Let Ψ∶NN → NN be the total recursive functional defined by Ψ(X) ∶= (X)0 for X ∈ NN. By Proposi- tion II.4.3, there is a partial recursive functional Ψ0∶ ⊆ {0,1}N→ NN such that M(σ) = λ(Ψ−01(σ)). Then defineν to be the left r.e. continuous semimeasure corresponding to the partial recursive functional Ψ○Ψ0, i.e.,
ν(σ) =λ({Z∈ {0,1}N∣ (Ψ○Ψ0)Z⊇σ}).
By the universality ofM, there is ac∈Nsuch that ν(σ) ≤c′⋅M(σ)for all σ∈ {0,1}∗. Letm∈N be such that 2c′≤m. Then
M({B}↾ (n+c+m)) =λ({Z∈ {0,1}N∣ΨZ0 ⊇B↾ (n+c+m)})
≤λ({Z∈ {0,1}N∣ (Ψ○Ψ0)Z⊇A↾ (n+c+m)})
=ν(A↾ (n+c+m))
≤c′⋅M(A↾ (n+c+m))
≤2−n.
Thus,{B} is a Π01subset ofNNwhich is deep in the extended sense, but the difference randomAcomputes a member of{B}.
CHAPTER III
COMPLEXITY AND FAST-GROWING AVOIDANCE
Looking downward, the COMPLEX and LUA hierarchies are closely coupled based on the following reseult of Kjos-Hanssen, Merkle, and Stephan:
Theorem. [17, Theorem 2.3.2] SupposeX∈ {0,1}N. The following are equivalent.
(i) X ∈COMPLEX.
(ii) There is a total recursive functionalΨ∶ {0,1}N→NN such thatΨ(X) ∈DNR.
In terms of the mass problems COMPLEX and DNRrec∶= ⋃{DNR(p) ∣precursive}, [17, Theorem 2.3.2]
implies:
Corollary III.0.1. COMPLEX≡wDNRrec.
Proof. SupposeX∈DNRrec, so that there is an order functionpsuch that X∈DNRp. Let Ψ∶ {0,1}N→pN be a recursive homeomorphism. Then Ψ(Ψ−1(X)) =X shows that Ψ−1(X) ∈COMPLEX by [17, Theorem 2.3.2]. This shows COMPLEX≤wDNRrec.
Now supposeX ∈COMPLEX. By [17, Theorem 2.3.2], there is a total recursive functional Ψ∶ {0,1}N→NN such that Y ∶= Ψ(X) ∈ DNR. Lemma II.4.13 shows that there exist nondecreasing recursive functions ψ∶ {0,1}∗→N∗andj∶N→Nsuch thatY↾n=ψ(X↾j(n)). Letp∶N→Nbe defined byp(n) ∶=max{ψ(σ)(n) ∣ σ∈ {0,1}j(n+1)} +1, so thatY(n) <p(n)for alln∈N. pis recursive, showing Y is recursively bounded, i.e., Y ∈DNRrec. This shows DNRrec≤wCOMPLEX.
Let LUArec∶= ⋃{LUA(p) ∣precursive}.
Corollary III.0.2. COMPLEX≡wLUArec.
Proof. By Corollary III.0.1, it suffices to show that LUArec≡w DNRrec. Because [17, Theorem 2.3.2] (and by extension Corollary III.0.1) holds with DNR defined with respect to any admissible enumeration ϕ●, we may assume without loss of generality that DNR is interpreted with respect to an admissible enumerationϕ● whose diagonalψis linearly universal, so that DNRrec⊆LUArec and hence LUArec≤wDNRrec. Conversely, givenX∈LUArec, there is a linearly universal partial recursive function ˜ψand an order functionpsuch that X∈Avoidψ˜(p). Because ˜ψis linearly universal, there exista, b∈Nsuch that ˜ψ(an+b) ≃ψ(n)for alln∈N. LetY ∈NNbe defined by Y(n) ∶=X(an+b)forn∈N. ThenY ∈Avoidψ(λn.p(an+b)) ⊆DNRrec, showing DNRrec≤wLUArec.
Knowing COMPLEX≡wLUArec alone does not reveal how the complexity and fast-growing LUA hier- archies intertwine (if at all) when going downward, prompting the following questions.
Question III.0.3. Given a sub-identical order function f∶N→ [0,∞), is there a fast-growing order function q∶N→ (1,∞)such that LUA(q) ≤wCOMPLEX(f)?
Question III.0.4. Given a fast-growing order functionp∶N→ (1,∞), is there a sub-identical order function g∶N→ [0,∞)such that COMPLEX(g) ≤wLUA(p)?
We will answer Questions III.0.3 and III.0.4 in the affirmative, proving:
Theorem III.0.5. To each sub-identical order functionf∶N→ [0,∞)there is a fast-growing order function q∶N→ (1,∞) such that LUA(q) ≤s COMPLEX(f), and to each fast-growing order function p∶N→ (1,∞) there is a sub-identical order function g∶N→ [0,∞)such that COMPLEX(g) ≤sLUA(p).
Theorem III.0.5 will follow as a direct consequence of the following theorems, which additionally provide explicit bounds onq andgin terms off andp, respectively.
Theorem III.1.1. Supposef∶N→ [0,∞) is a sub-identical order function,k is a nonzero natural number, andε>0is a rational number. Then
LUA(λn.exp2(finv(log2n+log22n+ ⋯ +logk2−1n+ (1+ε)logk2n) +1)) ≤sCOMPLEX(f).
Theorem III.2.1. Supposep∶N→ (1,∞)is an order function, and letr∶N→Nbe any order function such that limn→∞r(n)/2n= ∞. Then
COMPLEX((λn.∑i<r(n)⌊log2p(i)⌋)inv) ≤sLUA(p).
Less is known in the opposite direction.
Question III.0.6. Given a sub-identical order function f∶N→ [0,∞), is there a fast-growing order function q∶N→ (1,∞)such that COMPLEX(f) ≤wLUA(q)?
Question III.0.7. Given a fast-growing order functionp∶N→ (1,∞), is there a sub-identical order function g∶N→ [0,∞)such that LUA(p) ≤wCOMPLEX(g)?
While we have no general answer to Question III.0.6, we will give a partial affirmative answer to Ques- tion III.0.7.
Theorem III.3.3. If p∶N→ (1,∞) is a fast-growing order function such that ∑∞n=0p(n)−1 is a recursive real, then there exists a convex sub-identical order functiong such that LUA(p) ≤sCOMPLEX(g) ≠MLR.
Theorem III.3.3 follows from a more general result:
Theorem III.3.4. Suppose p∶N→ (1,∞) is a fast-growing order function such that ∑∞n=0p(n)−1 is a re- cursive real. Then for any order functionp∶˜N→ (1,∞)such thatp(n)/p(n) ↗ ∞˜ asn→ ∞ and for which
∑∞n=0p(n)˜ −1 is a recursive real,
LUA(p) ≤sCOMPLEX(λn.log2p(p˜ inv(2n+1) −1)).
Moreover, such ap˜exists and for any suchp˜the functionλn.log2p(p˜ inv(2n+1) −1)is dominated by a convex sub-identical recursive function g∶N→ [0,∞).