I. 3.1 (Partial) Recursive Functions and Sets
II.2 DNR and Avoidance
II.2.2 Properties of Linearly Universal Partial Recursive Functions
Every linearly universal partial recursive function ψ produces an effective enumeration ϕ● of the partial recursive functions by setting
ϕe(x) ≃ψ((e)0x+ (e)1)
fore, x∈N, where (−)i∶=πi○ (π(2))−1fori∈ {0,1} (soπ(2)((n)0,(n)1) =nfor alln∈N). This enumeration ϕ●is admissible.
Proposition II.2.10(Parametrization Theorem for Linearly Universal Partial Recursive Functions). Sup- pose ψ is a linearly universal partial recursive function and θ∶ ⊆Nk+1 → N is a partial recursive function.
There exist elementary recursive functionsA, B∶Nk→Nsuch that ψ(A(x)y+B(x)) ≃θ(x, y)
for allx∈Nk andy∈N.
Proof. Because ψ is linearly universal, there exists a, b∈N such that ψ(aπ(k+1)(x, y) +b) ≃θ(x, y)for all x∈Nk andy∈N. Thus, definingA, B∶Nk→Nby
A(x) ∶=a2π(k)(x)+1 and B(x) ∶=a(2π(k)(x)−1) +b gives
ψ(A(x)y+B(x)) ≃ψ(a2π(k)(x)+1y+a(2π(k)(x)−1) +b)
≃ψ(a(2π(k)(x)(2y+1) −1) +b)
≃ψ(aπ(k+1)(x, y) +b)
≃θ(x, y) for allx∈Nk andy∈N.
Corollary II.2.11. Supposeψis a linearly universal partial recursive function. Then the effective enumer- ation ϕ● defined byϕe(n) ∶=ψ((e)0n+ (e)1)is admissible.
Corollary II.2.12 (Recursion Theorem for Linearly Universal Partial Recursive Functions). Supposeψ is a linearly universal partial recursive function and θ∶ ⊆Nk+2→N is a partial recursive function. Then there exist a, b∈Nsuch that
ψ(aπ(k)(x) +b) ≃θ(a, b,x) for allx∈Nk.
Proof. Let ˜θ∶ ⊆Nk+1→Nbe defined by
θ(c,˜ x) ≃θ((c)0,(c)1,x)
for allc∈Nandx∈Nk. Corollary II.2.11 and Proposition I.3.12 show that that there exists ane∈Nsuch that
ψ((e)0π(k)(x) + (e)1) ≃ϕ(ek)(x) ≃θ(e,˜ x) ≃θ((e)0,(e)1,x) for allx∈N. Thus, we may leta= (e)0andb= (e)1.
Moreover, the diagonal of ϕ●is linearly universal.
Proposition II.2.13. If ψ0 is a linearly universal partial recursive function, then the partial function ψ defined byψ(e) ≃ψ0((e)0e+ (e)1)fore∈Nis also linearly universal partial recursive.
Proof. Supposeθ∶ ⊆N→Nis a partial recursive function. There areaandbsuch that∀x(ψ0(ax+b) ≃θ(x)).
For anyx∈N,π(2)(0, x) =20(2x+1) −1=2x, so
ψ(2x) ≃ψ(π(2)(0, x)) ≃ψ0(0⋅x+x) =ψ0(x).
It follows thatψis linearly universal.
Together, Corollary II.2.11 and Proposition II.2.13 allow us to enjoy the benefits of linearly universal partial recursive functions and of admissible enumerations simultaneously, e.g., having access to the par- ticularly nice versions of the Parametrization and Recursion Theorems in the forms of Proposition II.2.10 and Corollary II.2.12, respectively.
Another convenient property of linearly universal partial recursive functions is that we may we edit any finite number of values without affecting the linear universality.
Proposition II.2.14. Suppose ψ, χ are partial recursive functions such that ψ(n) ≃χ(n) for almost all n∈N. Thenψ is linearly universal if and only ifχ is linearly universal.
Proof. Supposeψ is linearly universal. Let N ∈ Nbe such that ψ(x) ≃ χ(x) for allx≥N. Consider the partial recursive functionθ∶ ⊆N→Ndefined by
θ(x) ≃
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
ψ(x−N) ifx≥N,
↑ otherwise.
Becauseψ is linearly universal, there area, b∈Nsuch thatψ(ax+b) ≃θ(x)for allx∈N. θis nonconstant,
soa≠0 and henceax+b≥xfor allx∈N. Thus, for all x∈N,
χ(ax+ (aN+b)) ≃ψ(ax+ (aN+b)) ≃ψ(a(x+N) +b) ≃θ(x+N) ≃ψ(x), showingχis linearly universal.
Convention II.2.15. Given p∶ ⊆N→R, supposep↾N≥a is a total, computable, nondecreasing, unbounded function with image in(1,∞)for somea∈N. Define ˜p∶N→ (1,∞)by ˜p(x) ∶=p(x)forx≥aand ˜p(x) ∶=p(a) otherwise. Then ˜pis an order function, and we let LUA(p)denote the class LUA(˜p). This allows us to make sense of something like, e.g., LUA(log2).
II.2.2.1 Basic Properties of the LUA Hierarchy
The regularity of the manner in which linearly universal partial recursive functions express their universality allows us to prove some simple but useful strong and weak reductions.
Proposition II.2.16. Let p, q∶N→ (1,∞) be recursive functions.
(a) Ifq(x) =p(ax+b)for some a∈N>0 andb∈Nfor allx∈N, thenLUA(p) ≡wLUA(q).
(b) IfX∈LUA(p),Y ∈pN, andX(x) =Y(x)for almost allx∈N, thenY ∈LUA(p).
(c) Ifp≤domq, then LUA(q) ≤sLUA(p).
(d) If for alla, b∈Nwe have q(ax+b) ≤p(x) for almost all x∈Nand ψ0 is a partial recursive function, then Avoidψ0(p) ≤wLUA(q).
Proof.
(a) Letψbe a linearly universal partial recursive function. IfX ∈Avoidψ(p), thenX∈Avoidψ(q) ⊆LUA(q) sincep(x) ≤q(x)for allx∈N. Thus, the recursive functionalX↦X shows LUA(q) ≤sLUA(p).
Conversely, if X∈Avoidψ(q), define ˜ψ by
ψ(y) ≃˜
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
ψ(x) ify=ax+b,
↑ otherwise.
Because ˜ψ(ax+b) ≃ ψ(x), it follows that ˜ψ is linearly universal partial recursive. Similarly define X˜ ∈NNby
X˜(y) ≃
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
X(x) ify=ax+b, 0 otherwise.
IfX∩ψ= ∅ andX isq-bounded, then ˜X∩ψ˜= ∅and ˜X isp-bounded. Thus, the recursive functional X ↦X˜ shows LUA(p) ≤sLUA(q).
(b) Let ψ be a linearly universal partial recursive function and suppose X ∈Avoidψp. LetN ∈Nbe such that X(x) =Y(x)for allx≥N and define ˜ψby
ψ(x) ≃˜
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
ψ(x) ifx≥N,
↑ otherwise,
so Y ∈Avoidψ˜(p). By Proposition II.2.14, ˜ψis linearly universal, soY ∈LUA(p).
(c) Supposep(x) ≤q(x)for allx≥N. GivenX∈NN, let ˜X be defined by
X˜(x) ∶=
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
X(x) ifx≥N c otherwise,
wherecis a rational number such that 1<c<q(0). IfX isp-bounded, then ˜Xisq-bounded. (b) above then shows that ˜X ∈LUA(q). This process defines a total recursive functional Ψ, so LUA(q) ≤sLUA(p).
(d) Let ψ be a linearly universal partial recursive function and suppose X ∈Avoidψ(q). Let a, b∈ N be such thatψ(ax+b) ≃ψ0(x)for all x∈Nand let ˜X be defined by ˜X(x) ∶=X(ax+b). Then ˜X∩ψ0= ∅ and ˜X(x) =X(ax+b) <q(ax+b) ≤p(x)shows that ˜X∈Avoidψ0(p).
Some other general basic reductions we make use of are given below.
Proposition II.2.17. Let ψ∶ ⊆N→Nandp∶N→ (1,∞)be given.
(a) Supposeq∶N→ (1,∞)is an order function dominatingp. ThenAvoidψ(q) ≤sAvoidψ(p).
(b) Supposeu∶N→Nis recursive. Then Avoidψ○u(p○u) ≤sAvoidψ(p).
Proof.
(a) Suppose p(n) ≤ q(n) for all n ≥ N. Let τ ∈ {0,1}N be any string such that τ(n) ≄ ψ(n) for all n<N. Then the recursive functionalX ↦τ⌢(X↾ [N,∞)) gives a strong reduction from Avoidψ(q)to Avoidψ(p).
(b) The recursive functionalX↦X○ugives a strong reduction from Avoidψ(p)to Avoidψ○u(p○u).
As our interest lies inEw, we must show that degw(LUA(p)) ∈ Ewfor any recursivep.
Lemma II.2.18. There is an effective enumeration of the linearly universal partial recursive functions.
Proof. Let ϕ● be an admissible enumeration of the partial recursive functions, let ψ0 be a fixed linearly universal partial recursive function, and lete0 be such thatψ0=ϕe0.
The central observation we make is that for any partial recursive ψ, ψ is linearly universal if and only if there are a, b ∈ N such that ψ(ax+b) ≃ ψ0(x) for all x ∈ N. With this in mind, we modify ϕ● to produce an enumeration of the linearly universal partial recursive functions as follows: given a, b, e ∈ N, defineψπ(3)(a,b,e)∶ ⊆N→Nby
ψπ(3)(a,b,e)(x) ≃
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
ψ0(y) ifay+b=x, ϕe(x) otherwise.
Then ψ is linearly universal if and only if there exist a, b, e∈ N such that ψ= ψπ(3)(a,b,e), so ψ● gives an effective enumeration of the linearly universal partial recursive functions.
Proposition II.2.19. Supposepis a recursive function. ThenLUA(p)isΣ02. Consequently,degw(LUA(p)) ∈ Ew.
Proof. By Lemma II.2.18, there exists an effective enumerationψ●of the linearly universal partial recursive functions. Let ϕ● be any admissible enumeration; the Parametrization Theorem implies there is a total recursive functionf∶N→Nsuch thatϕf(e)=ψefor alle∈N. Then
X∈LUA(p) ≡ ∃e∀n∀s∀m(ϕf(e),s(n)↓ =m→m≠X(n)) ∧ ∀n(X(n) <p(n)) shows that LUA(p)is Σ02. The Embedding Lemma then implies degw(LUA(p)) ∈ Ew.