In the first section of this chapter, we already noted thatGK(b) =O(2b). This
can be made even stronger.
Fact 9.1([DH10, Theorem 10.1.11]). P b∈NGK(b)/2 b<∞. It follows that lim b GK(b) 2b = 0. This fact is exploited in the proof thatGK∈/∆02.
Proposition 9.2 ([DH10, Theorem 10.1.13]). The functionGK is not∆02. Proof. We assume that GK does have an effective approximation (Gs)s∈N, and derive a contradiction from that. This we do by employing this approximation in the construction of a prefix-free machine that renders more setsK-trivial via a certain constant thanGK actually asserts.
As mentioned before, by the Recursion Theorem we may assume we know in advance the indexdof the machine we construct. Now takerto be a constant via which the computable set ∅ isK-trivial. Letbsbe the least number above rsuch that
Gs(bs) 2bs <2
−d.
Then bs might not be defined for all s – but for our purposes it is enough that it is for sufficiently large s, and this is guaranteed by the result that GK(b)/2b
approaches zero asbincreases.
Construction. At stages+ 1, see if the approximation tobhas changed, so bs6=bs−1.
• If it does, we start a new attempt to build too manyK-trivial sets. Define the setsAσ as 0s1σ0ω for allσof lengthbs−d.
• If b does not change at this stage, sobs =bs−1, lets0 be the last stage
where b did change. We try keeping the original sets Aσ K-trivial via bs0. Note that by the choice of bs above the constant r via which ∅
is K-trivial, we do not have to worry about the initial segments 0n for n ≤ s0. Therefore we only enumerate for the n with s0 < n ≤ s new
descriptions of Aσ nof length K(n)[s] +bs−dinto M (if indeed still KM(Aσn)[s]> K(n)[s] +bs0−d).
Verification. The weight ofM is bounded by that of the universal machine,
because for everyn∈N, only the initial segmentsAσnfor theσof lengthbs−d with s = n get new descriptions. For if this stage has not been reached yet, no initial segments of this length will be considered; on reaching this stage, as long as the approximationbso =bs stays the same new descriptions of the currentAσn may be given; and if later on the approximationbt ofb has changed frombs, only new descriptions for initial segments of length greater thant > s are enumerated. The initial segments up to nof the 2bs−d different Aσ all receive a description of
lengthK(n)[s] +bs−d, so in the end there is no more weight added toM than the descriptions ofnadd to the universal machine.
Now at some pointbs settles (so it is the least such thatGK(bs)/2bs <2−d),
and we will no longer have to redefine theAσ. We can be sure thatK(Aσ n)≤ K(n) +bs for n≤ s because Aσ n = 0n and K(0n)≤ K(n) +r ≤K(n) +bs.
Forn > s, the construction ascertains thatKM(Aσ n)[s] =K(n)[s] +bs−d, so K(Aσ n)[s] = K(n)[s] +bs. Thus ∀n (K(Aσ n)≤K(n) +bs) for all our Aσ, and we have 2bs−d sets that are K-trivial viabs. But that is in conflict with the
valueGK(bs) gave previously, since 2bs−d/2bs = 2−d.
We conclude that there can be no computable approximation to functionGK,
so it is not a ∆0
2function.
We have arrived at the answer to our question.
Theorem 9.3. The function GK is∆03.
Proof. Our function GK behaves exactly the same as the G-function cor-
responding to the uniformly c.e. family of K-trivial trees of Theorem 7.5. This function is computable in ∅00, as follows. Given b ∈
N, we can determine a Φ∅ 0 e that computes the treeTb and by Corollary8.6we can then uniformly compute its low2-ness index. Finally, by Proposition5.5, we uniformly compute the number of
paths.
ThusGKis computable in∅00as well. That makes it ∆03.
Now that we have established the arithmetical complexity ofGK, we may still
wonder how strong it is as an oracle. Is the halting set, or even the double jump, encoded in the information about the number ofK-trivials via each constant? This is a question that could still depend on the particular universal machine we choose.
Question 9.4. Is∅0or even∅00computable inG
K? Does this depend on the choice
of the underlying universal prefix-free machine?
10. The number of low for K sets
SetA is low for K via constant b ifA as an oracle will not help to compress any string more thanb bits. So the class Mb consists of the setsA such that for all stringsτ we have K(τ)≤KA(τ) +b. Then the paths of the ∆02 tree
TbM={σ| ∀s∈N∀l >|σ| ∀τ∈2<|σ|∃ρ∈2l(ρσ&K(τ)≤Kρ(τ)[s] +b)} are precisely the sets inMb. Admittedly, this is not as easy to see as it was for the K-triviality trees.
Proposition 10.1. Set A is low for K via b if and only if A is a path of TbM. Moreover,TbM is a∆02 tree.
Proof. Take any initial segmentAnof givenAthat is low forKvia constant
b. Then for every stagesand level l > n, for extensionρ=Alwe haveK(τ)≤
Kρ(τ)[s]+bfrom the facts thatK(τ)≤KA(τ)+bandKA(τ)≤Kρ(τ)≤Kρ(τ)[s]. So all initial segments ofA are in the tree, givingA∈[TbM].
Conversely, ifAisnot low forK viab, soK(τ)> KA(τ) +bfor at least oneτ, there is some part σ≺A that is used in giving such a short description of τ. So, for all extensionsρσ, we haveK(τ)> Kρ(τ)[s] +bfor large enoughs. But then initial segmentσcannot be in the treeTbM, andAis certainly not a path.
To show that the trees are ∆02, we explain how ∅0 can decide the membership
question. Given stringσ, it first computes the values ofK(τ) for all shorter strings τ. Then it is a Π01 question whether for all stagessand higher levels lthere is an
extensionρσsuch thatKρ(τ)[s] +bis above all the previously computed values,