Bounding Sequences of Fast & Slow-Growing Order Functions

I. 3.1 (Partial) Recursive Functions and Sets

II.3 Fast & Slow-Growing Order Functions

II.3.1 Bounding Sequences of Fast & Slow-Growing Order Functions

As our interest lies inEw, we must show that deg_w(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, defineψ_π(3)(a,b,e)∶ ⊆N→Nby

ψ_π(3)(a,b,e)(x) ≃

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

ψ₀(y) ifay+b=x, ϕ_e(x) otherwise.

Then ψ is linearly universal if and only if there exist a, b, e∈ N such that ψ= ψ_π⁽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,deg_w(LUA(p)) ∈ E_w.

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 thatϕ_f(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 deg_w(LUA(p)) ∈ E_w.

a slow-growing upper bound.

Proposition II.3.2. Suppose⟨p_k⟩_k∈N is a recursive sequence of slow-growing order functions.

(a) There is a slow-growing order function q⁻ such thatq⁻≤dompk for allk∈N.

(b) Suppose, additionally, thatpk≤dompk+1 for allk∈N. Then there is a slow-growing order funcitonq⁺ such that pk≤domq⁺ for allk∈N.

Proof.

(a) We simultaneously define the valuesq⁻(n)and natural numbersMn by recursion. We start by setting q⁻(0) ∶=p0(0)andM0∶=0.

Givenq⁻(0), q⁻(1), . . . , q⁻(n)andM0, M1, . . . , Mnhave been defined, letMn+1equalMn+1 ifMn+1≤ min0≤k≤M_n+1pk(n+1)and otherwise equal to Mn. Then define

q⁻(n+1) ∶=min{p0(n+1), p1(n+1), . . . , pMn+1(n+1), Mn+1+1}.

We now claim thatq⁻is a slow-growing order function dominated by each p_k. Nondecreasing. Given n∈N,

q⁻(n) =min{p0(n), p1(n), . . . , pM_n(n), Mn+1}

≤min{p₀(n+1), p₁(n+1), . . . , p_Mn+1(n+1), M_n+1}

≤min{p0(n+1), p1(n+1), . . . , pM_n+1(n+1), Mn+1+1}

=q⁻(n+1) as pk is nondecreasing for eachk.

Unbounded. It suffices to show that limn→∞Mn = ∞. Suppose for the sake of a contradiction that limn→∞Mn < ∞, so that Mn is eventually constant, say to M. For Mn to be eventually con- stant, it must be the case that M >min0≤k≤Mpk(n+1)for all n∈ N. Butp0 is unbounded, so min_0≤k≤Mp_k(n+1)is unbounded as a function ofn, yielding a contradiction.

Dominated by p_k. Givenk, there existsn∈Nsuch thatM_n≥k. Thenq⁻(n) ≤p_k(n)for alln≥M_n. Slow-Growing. As∑^∞_n=0p0(n)⁻¹= ∞andq⁻is dominated byp0, it follows by Direct Comparison that

∑^∞_n=0q⁻(n)⁻¹= ∞.

Recursive. The uniform recursiveness of thepk’s implies that the simultaneous construction ofq⁻and the sequence⟨M ⟩ _∈Nis recursive.

(b) We start by recursively defining natural numbersNm∈N. LetN0=0, and givenNmhas been defined, define N_m+1 to be the least natural number greater thanN_m such that

N_m+1−1

∑

n=N_m

p_m(n)⁻¹≥1

and for whichpm(Nm+1−1) ≤pm+1(Nm+1). This is possible becausepmis slow-growing andpm≤dom

pm+1. We defineq⁺∶N→ (0,∞)as follows: givenn∈N, letmbe the unique natural number for which Nm≤n<Nm+1, and define

q⁺(n) ∶=pm(n).

We claim thatq⁺is a slow-growing order function dominating eachpk.

Nondecreasing. By definition, q⁺ is nondecreasing on the intervalNm≤n<Nm+1 since it agrees with the nondecreasing functionp_mon that interval. Thus, to show thatq⁺is nondecreasing, it suffices to show thatq⁺(N_m+1−1) ≤q⁺(N_m+1)for eachm∈N. But by the definition ofN_m+1andq⁺, we have q⁺(N_m+1−1) =p_m(N_m+1−1) ≤p_m+1(N_m+1) =q⁺(N_m+1).

Unbounded. By definition, p0(n) ≤q⁺(n)for all n∈ N. Since p0 is unbounded, it follows that q⁺ is unbounded.

Slow-Growing. For each m∈N, by the definition of⟨Nm⟩m∈Nandq⁺ we have

N_m−1

∑

n=0

q⁺(n)⁻¹=

N₁−1

∑

n=0

p0(n)⁻¹+

N₂−1

∑

n=N₁

p1(n)⁻¹+ ⋯ +

N_m−1

∑

n=N_m−1

pm(n)⁻¹≥m.

Thus,∑^∞_n=0q⁺(n)⁻¹=limm→∞∑^N_n=^m₀⁻¹q⁺(N)⁻¹=limm→∞m= ∞, soq⁺ is slow-growing.

Dominatespk. By the definition of ⟨Nm⟩m∈N and q⁺, for each k ∈ N we have q⁺(n) ≥pk(n) for all n≥Nk, so q⁺≥dompk.

Recursive. The uniform recursiveness of the p_k’s implies that the sequence⟨N_m⟩_m∈Nis recursive, and subsequently that the functionq⁺is recursive.

Without the additional hypotheses in Proposition II.3.2(b), an upper bound may not exist, however:

Example II.3.3. We simultaneously define two slow-growing order functionsp1, p2∶N→ (0,∞)and a strictly increasing sequence⟨Nm⟩m∈N. The role of⟨Nm⟩m∈Nwill be that the behaviors ofp1 orp2 will be consistent between Nm and Nm+1−1, with those behaviors switching upon incrementing m. We start by defining p1(0) =p2(0) ∶=1,N0∶=0, andN1∶=1. Suppose Nmhas been defined and that p1(n)and p2(n)have been defined for alln<N_m. We split into two cases, depending on whethermis even or not.

Case 1: meven. LetNm+1be the least natural number greater thanNmsuch that∑^N_n=^m₀⁻¹p1(n)⁻¹+ (Nm+1− N_m) ⋅p₁(N_m−1)⁻¹≥m+1, then define

p₁(n) ∶=p₁(N_m−1), p2(n) ∶=n²,

forNm≤n<Nm+1.

Case 2: modd. Identical to the case where mis even, but withp₁ andp₂ switched.

In other words, we continually switch between being constant and being equal to the square function, withp₁andp₂having the opposite behavior of the other. By construction, bothp₁andp₂are slow-growing order functions, but max{p1(n), p2(n)} =n² for every n∈ N>0, so there is no slow-growing order function q∶N→ (0,∞)which dominates bothp1andp2 since∑^∞_n=1n¹2 < ∞.

For the fast-growing case, an upper bound always exists, but the existence of a lower bound requires an additional hypothesis. Unlike in the slow-growing case, this additional hypothesis is not only on the form of the sequence⟨pk⟩k∈N, but on the constituentpk’s themselves.

Lemma II.3.4. Supposep∶N→ (0,∞)is a fast-growing order function. Then∑^∞n=0p(n)⁻¹is a left r.e. real.

Proof. ⟨∑^k_n=0p(n)⁻¹⟩

k∈Nis a sequence of uniformly recursive reals converging monotonically to∑^∞n=0p(n)⁻¹ from below.

Proposition II.3.5. Suppose⟨pk⟩k∈Nis a recursive sequence of fast-growing order functionspk∶N→ (0,∞).

(a) There is a fast-growing order funcitonq⁺such that pk≤domq⁺ for allk∈N.

(b) Suppose, additionally, that⟨∑^∞n=0pk(n)⁻¹⟩k∈N is a sequence of uniformly recursive reals. Then there is a fast-growing order function q⁻ such that q⁻ ≤_dom pk for all k∈ N and for which ∑^∞n=0q⁻(n)⁻¹ is a recursive real.

Proof.

(a) Forn∈Nwe define

q⁺(n) ∶=max

k≤n p_k(n).

We claim thatq⁺is a fast-growing order function dominating each pk. Nondecreasing. For alln∈N, that eachp_k is nondecreasing implies

q⁺(n) =maxp_k(n) ≤maxp_k(n+1) ≤ maxp_k(n+1) =q⁺(n+1).

Unbounded. By construction, q⁺(n) ≥p0(n)for alln∈N. Becausep0 is unbounded, q⁺is as well.

Fast-Growing. By construction,q⁺(n) ≥p0(n)for alln, so Direct Comparison shows∑^∞_n=0q⁺(n)⁻¹< ∞ sincep0 is fast-growing.

Dominatesp_k. Given k∈N, for alln≥kwe haveq⁺(n) =max_m≤np_m(n) ≥p_k(n). Thus,p_k≤_domq⁺. Recursive. The uniform recursiveness of thepk’s immediately shows thatq⁺is recursive.

(b) We start by recursively defining natural numbesrNm∈N. LetN0∶=0, and givenNmhas been defined, define Nm+1 to be the least natural number greater thanNm such that

m+1

∑

k=0

∞

∑

n=Nm+1

1 p_k(n)≤

1 2^m+1 which exists sincepk is fast-growing for eachk.

Now defineq⁻as follows. Givenn∈N, letmbe the unique natural number for whichN_m≤n<N_m+1. Then define

q⁻(n) ∶=min

k≤mpk(n).

We claim that q⁻ is a nondecreasing, unbounded, fast-growing function which is dominated by each pk.

Nondecreasing. Forn∈N, letmbe such thatNm≤n<Nm+1. Then

q⁻(n+1) =

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

min_k≤mp_k(n+1) ifN_m≤n+1<N_m+1, min_k≤m+1p_k(n+1) ifN_m+1≤n+1,

≥min

k≤mpk(n+1) ≥min

k≤mpk(n) =q⁻(n).

Unbounded. Observe that if Nm≤n, then pk(n) ≥m for all k≤m+1. Thus, ifNm≤n<Nm+1, we have

q⁻(n) =min

k≤mp_k(n) ≥m.

It follows thatq⁻ is unbounded.

Fast-Growing. By the definition ofNm form≥1,

m

∑

k=0 Nm+1−1

∑

n=N_m

pk(n)⁻¹≤

m

∑

k=0

∞

∑

n=N_m

pk(n)⁻¹≤2^−m. Thus,

∞

∑

n=0

q⁻(n)⁻¹=

N1−1

∑

n=0

q⁻(n)⁻¹+

∞

∑

m=1 N_m+1−1

∑

n=N_m

max

k≤mp_k(n)⁻¹

≤

N1−1

∑

n=0

q⁻(n)⁻¹+

∞

∑

m=1 m

∑

k=0 Nm+1−1

∑

n=N_m

p_k(n)⁻¹

≤

N₁−1

∑

n=0

q⁻(n)⁻¹+

∞

∑

m=1

2^−m

=

N₁−1

∑

n=0

q⁻(n)⁻¹+1

< ∞.

Dominated by pk. By construction, for eachk,q⁻(n) ≤pk(n)for alln>Nk.

If the both the functionsp_kand the realsα_k ∶= ∑^∞_n=0p_k(n)⁻¹are uniformly recursive, then the function m↦N_m is recursive sinceN_m+1 is the least natural number greater thanN_msuch that

m+1

∑

k=0

⎛

⎝ α_k−

N_m+1−1

∑

n=Nm

p_k(n)⁻¹⎞

⎠

≤2^−m. The recursiveness of the mapm↦Nmthen implies that q⁻is recursive.

Finally, we show that β∶= ∑^∞_n=0q⁻(n)⁻¹ is a recursive real. Define, fori≥1, βi∶=

N_i−1

∑

n=0

q⁻(n)⁻¹+2⁻⁽ⁱ⁻¹⁾.

We claim that ⟨βi⟩_i∈N≥1 is a recursive sequence of uniformly recursive reals converging monotonically to β from above. Since lim_i→∞∑^N_n=ⁱ⁻₀¹q⁻(n)⁻¹ =β and lim_i→∞2⁻⁽ⁱ⁻¹⁾ =0, an argument analogous to the proof that q⁻was fast-growing shows lim_i→∞β_i=β. Additionally,⟨β_i⟩_i∈N≥1 is nonincreasing:

β_i+1=

Ni−1

∑

n=0

q⁻(n)⁻¹+

N_i+1−1

∑

n=Ni

q⁻(n)⁻¹+2⁻⁽ⁱ⁺¹⁾≤

Ni−1

∑

n=0

q⁻(n)⁻¹+2⁻⁽ⁱ⁺¹⁾+2⁻⁽ⁱ⁺¹⁾=

Ni−1

∑

n=0

q⁻(n)⁻¹+2⁻ⁱ=β_i. Thus,β is right r.e. and hence recursive.

Corollary II.3.6. Supposep∶N→ (0,∞) is a fast-growing order function such that ∑^∞_n=0p(n)⁻¹ is recur- sive. Then there exists a fast-growing order function q such that p(n)/q(n) ↗ ∞ asn→ ∞ and for which

∑^∞n=0q(n)⁻¹ is recursive.

Proof. Letα∶= ∑^∞_n=0p(n)⁻¹ and let p_m denote the function defined byp_k(n) ∶=p(n)/2^k for k, n∈N. Note that∑^∞_n=0p_k(n)⁻¹=2^kαis recursive, and the sequences⟨p_k⟩_k∈Nand⟨2^kα⟩_k∈Nare uniformly recursive.

By Proposition II.3.5(b), there exists a fast-growing order function q such that q≤dom pk for all k∈N and where∑^∞_n=0q(n)⁻¹ is a recursive real. Moreover, the proof of Proposition II.3.5(b) shows that there is such aqfor whichp(n)/q(n) ↗ ∞asn→ ∞.