Submitted on Febuary 23, 2013 to the Notre Dame Journal of Formal Logic

Volume ??, Number ??,

Controlling Eective Packing Dimension of ∆02 Degrees

Jonathan Stephenson

Abstract This paper presents a renement of a result by Conidis[3], who proved that there is a real X of eective packing dimension0 < α < 1 which cannot compute any real of eective packing dimen-sion 1. The original construction was carried out below ∅′′, and thispaper's result is an improvement in the eectiveness of the argument,constructing such an X by a limit-computable approximation to getX ≤T ∅′.

1 Introduction

Eective packing dimension is one of several common ways to study the wayin which information is encoded in a real number. It assigns to each X ∈ 2ω

a real number dimP (X) ∈ [0, 1]. The concept has been considered in a widercontext in a number of other publications see for instance Downey andHirschfeldt's book [4], Downey and Greenberg [5], and Downey and Ng [6].An eective packing dimension equal to 1 corresponds to the notion that in-nitely many of the initial segments of the real are unable to be signicantlycompressed by any algorithmic process, whereas an eective packing dimen-sion of 0 indicates that initial segments of the real number are easily deducedfrom a relatively small amount of information. Martin-Löf random reals haveeective packing dimension one, and in fact are characterised as the class ofreals for which all initial segments are largely incompressible; they are a well-studied class of reals, and have several other natural characterisations. At theother end of the spectrum of eective packing dimension are reals with eec-tive packing dimension 0. Included in this class are both computable realsand noncomputable K-trivial reals which encode information, but in a verysparse manner.

The strong links between algorithmic randomness, information content, andeective packing dimension lead to reals whose eective packing dimensions

2010 Mathematics Subject Classication: 03D32, 68Q30Keywords: Eective Packing Dimension

1

2 J. Stephenson

lie strictly between 0 and 1 being often regarded as being partially random.This notion of partial randomness does not respect the Turing degree structureparticularly well, because one can easily exhibit reals in every Turing degreewith eective packing dimension 0. Indeed, given a real Z of eective packingdimension α > 0 and some rational number q with 0 ≤ q < α, it is easy toproduce a real in the same Turing degree as Z with eective packing dimensionis less than q, by thinning the information coded by Z in a computable way.

The goal of this paper is to improve on a result of Conidis, who constructeda real X which has intrinsically intermediate eective packing dimension inthe following sense:

Theorem 1.1 (Conidis[3]) There is a real X ≤T ∅′′ with nonzero eective

packing dimension such that X cannot compute any reals of eective packing

dimension 1.

The main theorem of this paper increases the eectiveness of this resultby constructing a real with the same property, but which is below ∅′ in theTuring degree structure.

Theorem 1.2 There is a real X ≤T ∅′ with nonzero eective packing di-

mension such that X cannot compute any reals of eective packing dimension

1.

The reals constructed by this theorem and that of Conidis could be de-scribed as having intrinsically intermediate eective packing dimension: theyare not able to be obtained by applying the thinning process described aboveto a real with eective packing dimension 1 and which is in the same Turingdegree.

We will obtain a corollary by applying one of the theorems of Bienvenu,Doty, and Stephan [2], who repackage of a result of Fortnow, Hitchcock, Pavan,Vinodchandran, and Wang [7], stating that any Turing degree which containsa real of nonzero eective packing dimension contains reals of eective packingdimension arbitrarily close to 1. Recall that the eective packing dimensionof a set of reals is equal to the supremum of the eective packing dimensionsof its members, and consider the Turing degree of the real X of the previoustheorem, to deduce that:

Corollary 1.3 There is a ∆02 Turing degree with eective packing dimension

1 which contains no real of eective packing dimension 1.

It is worth noting that eective packing dimension is only one of ways tomeasure the complexity of a real number. A better known measure is theeective Hausdor dimension, introduced by Lutz in [10], and which has beenwell-studied; in particular, some questions analogous to those studied here areconsidered in papers by Greenberg and Miller [8], and Miller [11], and (onceagain) a good overview of the area is given by Downey and Hirschfeldt [4].

Both the means of construction and the improvements in eectiveness seenin this paper and that of Conidis [3] correspond very closely to that which oc-curred in the development of minimal Turing degrees in classical computabilitytheory. In that case the original construction of a minimal degree was carriedout by Spector in 1956 [15], with the construction occurring below ∅′′. In

Controlling Eective Packing Dimension of ∆02 Degrees 3

1961 Sacks [12] noted that the construction could also be carried out below∅′. In eect, each of the ∅′′oracle questions was replaced by sequences of∅′computable questions.

This bound was later improved by Yates [16] in a 1970 paper which showedthat a minimal degree could be found below any c.e. degree. In this case themethod used was a limit computable (full approximation) construction.

In each case the construction is given by nding a sequence of computabletrees Ti with the property that for each i, Ti+1 ⊆ Ti, and so that for each Tiand each path X through Ti such that ΦXi is a total reduction, either ΦXi iscomputable, or it is a real which computes X. Thus the unique real X whichis a path through each of the Ti is in a degree which is minimal with respectto Turing reductions.

The construction given for reals with nonzero eective packing dimensionand which cannot compute any reals of eective packing dimension 1 in Coni-dis [3] and that presented here follow a similar developmental history. As inthe case of the minimal degree arguments, the constructions are carried outby building a nested sequence of trees, chosen so that the unique path whichlies in all of the trees satises the requirements of the problem.

The original proof by Conidis, which constructed a degree below ∅′′, isherein replaced by an approximation by ∅′-computable procedures, yieldinga ∆0

2 real with the desired properties. It should be noted that the furtherimprovement given by Yates in the case of minimal degrees which allowed thatconstruction to be carried out below any c.e. degree will not have an analogyin our case, since there are noncomputable degrees which cannot computeany real with nonzero eective packing dimension (for instance the K-trivialdegrees). The question of exactly which Turing degrees can compute a realof nonzero eective packing dimension that cannot in turn compute a realof eective packing dimension 1 remains an interesting and relevant one, forwhich some thoughts and a conjectured partial solution are given in Section7.

Some of the technical machinery and lemmas used by Conidis [3] will beuseful here. The notation of that paper is followed wherever it is reasonable todo so: For σ, τ ∈ 2<ω write στ to indicate the string formed by concatenatingσ and τ . Let A,B ⊆ 2<ω. Write AB to mean στ : σ ∈ A, τ ∈ B, andsimilarly Aσ and σA for Aσ and σA, respectively. I denote by 2≤n theset of strings in 2<ω of length at most n, and by 2=n those of length exactlyequal to n. Write |σ| for the length of a string σ ∈ 2<ω.

By σ τ I will mean that σ is an initial segment of τ , and by σ ≺ τ thatσ τ and σ 6= τ .

I will use K to denote prex-free Kolmogorov complexity of strings. Thiswill be the only notion of complexity considered throughout the paper; forbrevity I shall simply refer to K as Kolmogorov complexity. It should benoted that the usage of prex-free complexity here is not important to theconstruction: plain Kolmogorov complexity could be used instead, yieldingthe same results.

As pointed out by Conidis [3], the following denition of eective packingdimension is not the standard one, but the proof that they are equivalent is

4 J. Stephenson

indicated by Athreya, Hitchcock, Lutz, and Mayordomo [1], noting that itfollows from a similar result by Lutz [10].

Denition 1.4 Let X ∈ 2ω. The eective packing dimension of X is

dimP (X) = lim supn→∞

K(X n)

n.

If d ⊆ 2ω then the eective packing dimension of d is

dimP (d) = supX∈d

lim supn→∞

K(X n)

n.

Denition 1.5 Let Φii∈ω be a computable listing of all oracle Turing ma-chines. By an oracle Turing machine Φe we will mean one which outputs 0 or1 as a result of any halting computation.

Notation shall otherwise be as in Soare [14, 13]. In particular: byΦρe,s(x) = y I mean that the oracle computation Φρe(x) halts with output ywithin s steps of computation, and with use no greater than |ρ|. I assumethat computations of this kind either halt at some stage s < |ρ|, or not atall, and that they may halt only if x < |ρ|, so that nite oracle strings onlycompute nitely many outputs.

Adopting the conventions used by Conidis [3], assume that the oracle com-putations are monotonic in their use, in the sense that if Φρe,s(x) ↓ and y < x,then Φρe,s(y) ↓ too. This convention will ensure that if ρ ∈ 2<ω, then there is

a string τ ∈ 2<ω with Φρ(x) ↓ precisely when x < |τ | and that Φρi (x) = τ(x)in this situation. We will denote this by Φρi = τ (notice that τ is the longeststring computed by ρ in this way). We will also assume a second monotonicitycondition that if ρ is obtained from a string τ by extending it by a single 0 or1, then there is at most one x such that Φρi (x) ↓ but Φτi (x) ↑. This will ensurethat if T is a tree then dening ΦTi = τ ∈ 2<ω : (∃ρ ∈ T )Φρi = τ, ΦTi willalso be a tree; it is worth noting that given a c.e. tree T and an enumerationfor T , we can eectively enumerate ΦTi .

Denition 1.6 A clump is a subset of 2<ω of form σ2≤|σ|, where σ ∈ 2<ω.We refer to σ as the root of the clump.

Denition 1.7 A pruned clump is a subset of 2<ω which is a downward closedsubset of a clump σ2≤|σ|, and which contains at least two of the leaves ofσ2≤|σ|. Once again, refer to σ as the root of the pruned clump.

Pruned clumps will be the main tool used to make sure that the real wewill construct has nonzero packing dimension, and will consequently play amajor role in the construction which forms Section 4 of this paper.

Denition 1.8 An extendible string on a tree T is an element τ ∈ T with theproperty that there is a path through T which has τ as an initial segment.

Denition 1.9 A pruned clumpy tree is a tree T ⊆ 2<ω such that wheneverτ ∈ T is an extendible string on T there is a string σ ∈ T such that τ ≺ σand a pruned clump with root σ is on T .

Controlling Eective Packing Dimension of ∆02 Degrees 5

2 Strategy

The construction which is laid out in this paper will build a sequence of c.e.pruned clumpy trees T0 ⊃ T1 ⊃ T2 · · · , choosing a string ξi ∈ Ti for eachi. We will construct the ξi so that for each i, j either ξi ≺ ξj , or vice versa.Therefore the ξi will specify a unique path X which lies in the intersectionof the Ti. The ξi will also be chosen so that for each i, ξi has some initialsegment ρ of length at least i, and which has high Kolmogorov complexity, andthis will ensure that the path X will have nonzero eective packing dimension(indeed the dimension will be at least 1

4 ). The trees Ti are chosen to haveenough leaves on each clump that such strings ρ exist, but the leaves whichare added to the clumps are chosen carefully, so that X does not compute anyreal of eective packing dimension 1. We will see that this is done by eitherforcing that ΦYi is nontotal for each path Y through Ti, or that there is someα < 1 which is an upper bound on the eective packing dimension of the realΦXi . This latter case is achieved by a process of majority vote: by pruningTi carefully, we can ensure that a large number of the reals computed withpaths through Ti as oracles agree on long initial segments, and thus guaranteethat ΦXi is a path through a c.e. tree which branches suciently slowly thatthe initial segments of ΦXi must have quite low Kolmogorov complexity wewill see that identifying such an initial segment may be achieved by givingits length, together with information about its position in the lexicographicalordering of strings of that length on ΦTii , which will require a number of bitsproportional to the length of the segment, with a constant of proportionalityless than 1.

The construction will proceed by constructing for each i a sequenceT 0i , T

1i , T

2i , · · · of nite trees, with the property that for each i and any

k < k′, we have T ki ⊂ T k′

i . We shall ensure that

Ti =⋃k

T ki

is a computably enumerable pruned clumpy tree. In addition we will ∅′-computably construct sequences ξki of strings which will converge to limits ξi.The string ξi will act as a root to the tree Ti in the sense that all extendiblestrings on Ti will either extend ξi or be initial segments of it. To ensurethat there is a real X ∈ 2ω which has each ξi as an initial segment, we willonly ever let ξki be an extension of ξk−1i , and ensure that every other ξkj we

construct is -comparable with ξki . Thus the ξi will be the limits of uniformly∅′-limit-computable sequences. However we will see that they nonetheless willdene a real X which is ∅′-computable.

In the construction given by Conidis in section 5 of [3], the ∅′′ oracle wasable to identify whether a given tree Ti−1 contained a string τ with the prop-erty that for some x ∈ ω and every extension ρ ∈ Ti−1 of τ , one had Φρi (x) ↑.This allowed detection of strings forcing each oracle computation ΦYi with τas an initial segment of a path Y in Ti−1 to be nontotal. This will not bepossible to achieve using only a ∅′ oracle.

At each stage k of the construction carried out in this paper we will con-struct a nite tree T ki for each i ≤ k. We will also build a string σk with

6 J. Stephenson

the property that σk(i) tells us whether we currently believe that a string τfullling a similar role to that in Conidis' construction exists on T ki−1. Wewill be hampered by two factors in this matter. Firstly, we do not know whatthe nal tree Ti−1 will be while we are building Ti. For this reason we willwork with the assumption that it will not contain such a string τ , until we

are able to identify one. Thus we will build trees T k,si for each k ≥ s, whichwill correspond to our current guess at T si under this assumption. When k is

large enough we will have T k,si = T si for all s. Secondly, we will not be ableto consult ∅′′ to ask whether any string on Ti−1 forces divergence of an oraclecomputation. Identifying such a string will be achieved by approximating thisquestion: we will ask only if a specic string forces divergence on some speciccomputation.

At each stage k of the construction we will follow one of two strategies:either construct extensions which output halting computations under the ith

oracle machine, or identify a string on our tree whose extensions all divergeunder this oracle computation.

Denition 2.1 Let T be a computable tree, τ ∈ T and consider the statement:

(∃x < n)(∀ρ ∈ T )(∀s ∈ ω)[τ ρ→ Φτi,s(x) ↑

]which asserts that τ forces divergence of the oracle machine Φi on all of its ex-tensions on some (bounded) input x. This statement is uniformly computablein ∅′ given T, τ, n.

The portion of the strategy which seeks to force our real X to have nonzeroeective packing dimension will make heavy use of the following lemma; it isput to the same use by Conidis [3].

Lemma 2.2 (Conidis [3]) Let q ∈ Q, 0 ≤ q ≤ 1, and τ ∈ 2<ω, and let qτ be the

least natural number that is greater than or equal to q|τ |. Then, for any given

pruned clump of the form A ⊆ τ2≤|τ | such that A contains at least 2qτ -many

leaves of τ2≤|τ | , there is a leaf σ ∈ τ2≤|τ | in A such that

K(σ) >q

2|σ| − 1.

If X is the real we construct, and we consider some i is such that ΦXigives a total reduction, then the proof that dimP (ΦXi ) < 1 will rely on esti-mates on the initial segment prex-free complexity of the real ΦXi obtainedby considering a pair of prex-free machines.

Denition 2.3 For the purposes of computing Kolmogorov complexity, a pre-x free machine M is a partial computable map from 2<ω to 2<ω with theproperty that if σ τ , then M(τ) ↓ implies M(τ) ↑.

We will use the fact that if M is a prex free machine, then there is someconstant C (depending on M) such that for each pair σ, τ of binary stringswith M(τ) = σ, we have K(σ) < |τ |+ C.

The idea will be to construct two machines M1 and M2 which, betweenthem, exhibit short descriptions of the initial segments of ΦXi . They will doso by making ecient use of the structure we build into the trees we constructto extract information about strings computed from those trees.

Controlling Eective Packing Dimension of ∆02 Degrees 7

The remainder of the paper is organised as follows: In Section 3 we willsee two procedures which will be used to carry out the construction of thetrees Ti which we seek. This construction is laid out in detail in Section 4.In Section 5 we will check that the construction yields c.e. trees, and will seethat there is a single path X common to all of the trees, with eective packingdimension at least 1

4 . This will be the path we seek. In Section 6 we will verifythat the construction carried out will guarantee that any real computed byX must have eective packing dimension strictly less than 1, by giving anexplicit bound via a combinatorial estimate.

3 Approximations

This section outlines the algorithms which will be used to compute niteapproximations T ki to a tree Ti based on the trees T kj for j < i at some stagek ≥ i.

Suppose that we have a computable map from ω onto the set of all nitetrees, i.e. an indexing of those trees.

Now dene two computable maps T : 2<ω×ω2 → ω and S : ω4 → ω, wherewe think of each map as outputting a nite tree. The maps T and S eachprovide a means to guess at what some of the trees Ti we are building in theoverall construction are.

The algorithm S The algorithm S will be used at stages of the construction atwhich we are searching for strings to add to the nite approximations to ourtrees Ti in order to obtain longer halting computations. The computationswhich we seek will be required to come in families which all agree on someinitial segments of xed length; this will help satisfy the requirement thatdimP (ΦXi ) < 1, where X is the real which we are constructing.

Let S : ω4 × 2<ω → ω be the following algorithm. Let S(m,n, e, t, ξ) = p,where p is an index for the nite tree R constructed as follows: search forthe leaves of the tree T with index m and which are extensions of ξ; noticethat this search terminates, since T is nite. Then for each leaf λ0 ξ of T ,suppose λ0 is of length l. Attempt (by searching within the tree S with indexn) to extend λ0 to a string λ which is the root of a pruned clump on the treeS, of length 4N > 4l · 22e+4 for some N ≥ e + 2. For each λ0, choose N tobe the minimal possible, choosing the lex-least λ if a tie-break is needed. Foreach string λ obtained in this way, let Lλ be the leaves of the pruned clumpon S with root λ. Now for each λ and each ρ ∈ Lλ we search for a string ρ ρon S such that for each x < 2−2e−4|λ|, we have Φρe(x) ↓. Accept only strings

ρ for which |ρ| < t and Φρe,t(x) ↓ for each relevant x. Call such extensionssuitable with bound t.

If there is no λ for which we nd suitable extensions ρ for each ρ ∈ Lλ, letR = T ; we are not making any changes to the input tree. In this case saythat S fails; otherwise it succeeds.

If S succeeds, then for each λ such that we found suitable extensions forevery ρ ∈ Lλ, nd the least i ≤ t such that each ρ ∈ Lλ admits a suitableextension with bound i, and for each such ρ let ρ be the least suitable extensionwith bound i, when strings are ordered by length, and strings of the same

8 J. Stephenson

length are ordered lexicographically. Let the set of all such ρ λ obtained inthis way be Dλ. Now choose the string τλ of length x for which we have

ρ ∈ Dλ : (∀y < x)[τλ(y) = Φρe(y)]

of maximal size, choosing the lex-least such τλ in the case of a tie. Then add toR the downward closure of the nodes ρ ∈ Dλ for which we have τλ(y) = Φρe(y)for each y < x.

The algorithm T The algorithm T will be used at stages of the construction ofTi at which we have already found a string which we believe is not an initialsegment of any path Y through Ti−1 for which ΦYi is a total map. In this casewe are simply keeping all extensions of that string.

Let T (ξ,m, n) = p, where m, n are indices for trees T , S respectively: let pbe an index for the computable tree R given by letting R consist of preciselythose strings τ such that either τ ∈ T , or τ ∈ S and τ extends both ξ andsome leaf of T .

4 Construction

In this section I construct a nested sequence Tii∈ω of computable trees andstrings ξi ∈ Ti which dene a path X ∈ 2ω. This path will be computablein ∅′ and will not compute any reals of packing dimension 1, but will havenonzero packing dimension. The entire construction will be carried out below∅′.

We will assume a default strategy for each i that in building the tree Tiwe will always nd halting computations when we want them; we will moveto our secondary strategy of forcing divergence if this ceases to be a viablestrategy.

At stage k of the construction we will choose a string σk ∈ −1, 0, 1<ω oflength k+1. For each 0 ≤ i ≤ k+1 will build a tree T ki = T k which will be ournite approximation to Ti, and a string ξki which will be our approximationto ξi. The string σk will tell us what strategies we are following at stage kof the construction the construction. If σk(i) = −1, we have not yet met therequirement that ξki has an initial segment of length at least i, and which isof high complexity. If σk(i) = 0 then we currently do not believe that we willbe able to choose ξi in a way that the paths Y on Ti which pass through ξiall have ΦYi nontotal. Otherwise σk(i) = 1, in which case we believe we haveselected ξki to force nontotality of such computations.

We will ensure that ξki only changes at nitely many stages k, and thereforethat it comes to a limit ξi. We will see that the ξi are all distinct strings whichform a total order under inclusion as initial segments, and therefore that theydene a unique X ∈ 2ω, which is the unique path through the Ti.

For each i, we will dene

Ti =⋃k

T ki .

The resulting tree will be c.e. and have the property that for every k, eachleaf of T ki extends some leaf of T k−1i .

Controlling Eective Packing Dimension of ∆02 Degrees 9

For the base case, begin by setting T 0−1 = ∅, 0, i.e. a tree with a single

leaf of length 1, and recursively letting T k−1 be the downward closure of⋃λ∈Tk−1

−1

λ a leaf

λ2=|λ|02|λ|

so that each T k−1 has leaves of length 4k; this tree establishes the clumpystructure which underlies all of the Ti which we will construct. The specicchoice of the root lengths for the pruned clumps here will ensure that somelater computations will give integer bounds in cases where it is convenientthat this should be so.

Now for the construction proper. At stage k ≥ 0, for each i < k we are givennite trees T k−1i and strings ξk−1i ∈ 2<ω as well as a string σk−1 of length

k. The trees given satisfy T k−1i−1 ⊆ T k−1i and ξk−1i−1 ξk−1i or ξk−1i−1 ξk−1i for0 ≤ i < k.

Proceed by a series of substages, one for each 0 ≤ i ≤ k. At substage i weare given T kj , ξ

kj , and σ

k(j) for each j < i. We will build auxiliary computable

trees T k,si for each i < k − 1 as uniformly computable sets of strings. Thesetrees will tell us which strings we currently believe are to be added to Tiat some stage. We will see that for each i and for large enough k we have

T si = T k,si for each s ≥ k, so that T k,si is best thought of as a prediction ofwhat T si will be, according to the information which is available at stage k.

That the T si are eventually equal to the T k,si will guarantee that the tree Tiwhich is constructed will be a c.e. set of strings.

Constructing T k−1. This step in the computation serves to initialise the con-

struction for stage k.At each stage k, let T k−1 be as described above, and let ξk−1 be the empty

string. Dene σk(−1) = 1, for convenience.

In addition, for each s ≥ k, let T k,s−1 = T s−1.

Constructing T ki for 0 ≤ i < k. In this section, suppose that we have built T kj

and dened ξkj and σk(j) for each j < i, as well as having dened T k−1j , ξk−1j ,

and σk−1(j) for each j < k. We also assume that for j < i− 1 we have built

T k,sj for each s ≥ k.Before building T ki , we will construct trees T

k,si−1 which give a uniformly c.e.

guess at Ti−1. In the case that i = 0, we have already built the T k,si−1, and canomit this step.

So suppose that i > 0, and that we are given T k,si−2 for each s. Dene

T k,ki−1 = T ki−1. Given Tk,s−1i−1 , let T k,s−1i−1 have index m, and T k,si−2 have index n,

and set T k,si−1 to be the tree given by S(m,n, i, s, ξki ) if σk(i− 1) = 0, and set

T k,si−1 to be the tree given by T (m,n, ξki ) if σk(i− 1) = 1.

First check whether any j < i has σk(j) = 1 but σk−1(j) = 0, or any suchj has σk(j) = −1. In this case our strategy for Ti has been interrupted, or is

not yet active. So we must start over; to do so, set ξki = ξk−1i , let T ki = T k−1i ,and set σk(i) = −1. Then proceed to the construction of T ki+1.

10 J. Stephenson

Otherwise, if σk−1(i) = 1, then we know that ξk−1i forces divergence as

an oracle for the machine Φi on a tree T k−1i−1 whose construction was given

earlier. In this case, let ξki = ξk−1i , let σk(i) = 1, and let T ki be the tree

given by T (m,n, ξki ), where m and n are indices for the trees T k−1i and T ki−1,

respectively. Then proceed to the construction of T ki+1.

If we have σk−1(i) = −1, we have not conrmed that K(ρ) > 14 |ρ| − 1 for

some ρ ξki with |ρ| ≥ i.If i is the least number with σk(i) = −1, use ∅′ to check whether there is a

string ρ on T ki−1 which is comparable to each ξkj for each j such that j < i and

to ξk−1j for each j such that i < j < k, and for which we have K(ρ) > 14 |ρ|−1

and |ρ| ≥ i. If such a string ρ exists, set ξki to be a leaf on T ki−1 which extends

ρ and set T ki to be equal to T ki−1, and set σk(i) = 0. If no such string exists

set T ki = T k−1i and ξki = ξk−1i and σk(i) = −1. Then move to construction ofT ki+1.

Otherwise σk(i) = 0 and we have met the complexity condition for index

i, and will use the approximation T k,si−1 to Tj for each j < i to check whether

we can force divergence by moving to a new string ξki . Take m to be an index

for T k−1i , n to be an index for T ki−1 and ns to be an index for T k,si−1, for eachs > k. Check whether for some s, S(m,ns, i, s, τ) succeeds, where τ is the

longest string in the set ξk−1, ξk0 , ξk1 , · · · ξki−1, ξk−1i , · · · , ξk−1k−1.

If it does succeed, let ξki = ξk−1i and T ki be S(m,n, i, k, ξki ), and setσk(i) = 0, then proceed to construction of T ki+1. Notice that we are ap-plying the algorithm S to a dierent string to the one which we checked forsuccess on; this is because we cannot in general computably nd the string τof the previous paragraph, and will want to construct our tree by a methodwhich is able to be computably approximated once our strategy has settled.

Otherwise, check whether there is some leaf λ0 on T k−1i which extends

every ξkj for j < i and every ξk−1j for i < j < k, and which has an extension

λ ∈ T ki−1 and such that for some x < 2−2i−4|λ| every extension ρ of λ in T k,si

for any s has Φρi (x) ↑. In this case set T ki to be the tree obtained by adding to

T k−1i the downward closure of the string λ. Set ξki to be λ, and set σk(i) = 1,and proceed to construction of T ki+1.

If no such leaf λ0 exists, set ξki = ξk−1i and T ki = T k−1i , and set σk(i) = 0,and proceed to construction of T ki+1.

Constructing the new tree T kk Finally, we dene ξkk = ξkk−1, σ

k(k) = 0 and let

T kk be the downward closure of ξkk .This concludes the construction of the trees T ki for i ≤ k.

5 Verication

It now remains to be seen that the construction carried out above will satisfyour requirements, namely that for each i, the ξki converge to some ξi, and thatthe ξi are amongst the initial segments of some real X. Furthermore we needX ≤T ∅′ and dimP (X) > 0, but whenever ΦXi is a total reduction we requiredimP (ΦXi ) < 1.

Controlling Eective Packing Dimension of ∆02 Degrees 11

Theorem 5.1 For each i, there is some L such that when L < k1 < k2, wehave ξk1i = ξk2i , σk1(i) = σk2(i), equal to either 0 or 1, and there is some

string ρ ∈ T ki−1 such that |ρ| ≥ i, ρ ξki , and K(ρ) ≥ 14 |ρ| − 1. In addition,

we may choose L so that if k > L, then for s ≥ k we have T k,si = T si . Now

dene

Ti =⋃k

T ki

and

ξi = limkξki

Then Ti is a c.e. pruned clumpy tree, and there is a path X through Ti such thatfor each k, each ξkj is an initial segment of X for each j ≤ k. In particular,

ξi is an initial segment of X.

Proof In proving this theorem, it will be useful to prove several other prop-erties to hold throughout the induction. Therefore we will also show that ateach stage k, we dene ξki ∈ T ki with the properties that ξki ξ

k+1i , and either

ξki−1 ξki or vice versa. In addition, T k−1i ⊆ T ki and T ki ⊆ T ki−1. We will also

see that whenever A = A ∩ ρ2≤|ρ| is a pruned clump on T ki for k > L, andA's root is an initial segment of a path through Ti, A contains at least 2qρ

many leaves of τ2≤|ρ|, where qρ is the least integer greater than or equal to

(1−∑ij=−1 2−2j−4)|ρ|.

Notice that for the case i = −1, ξk−1 and T k−1 are chosen specically sothat they meet the requirements of the theorem, as well as the additionalproperties mentioned above.

Also notice that at each stage of the construction we preserve the require-ment that T k−1i ⊆ T ki and that T ki ⊆ T ki−1 for each i < k, and ensure that

T kk ⊆ T kk−1 by choice, so that these conditions will hold of all the trees we

build. In addition, at each stage of the construction we ensure that ξkj ξkior ξki ξkj for each i, j < k, and dene ξkk = ξkk−1, so that comparability of

the strings is preserved at every stage. Furthermore we always set ξk−1i ξki .But this means that if we let τk be the longest of the ξki at stage k, then wehave τk τk+1 at each stage k, so it is clear that there is a path through allof the τk which lies in T−1.

Now let i ≥ 0 and x L so large that the statements in the theorem andthe additional hypotheses mentioned hold for each j < i at stage L of theconstruction.

Lemma 5.2 For suciently large M , whenever k > M , ξki = ξMi and

σk(i) = σM (i) 6= −1. Furthermore there is some string ρ with |ρ| ≥ i, ρ ξki ,and K(ρ) > 1

4 |ρ| − 1. In particular, |ξki | ≥ i.In the case that k > M and σk(i) = 1, there is some x such that each string

ρ ∈ Ti−1 which extends ξki has Φρi (x) ↑.On the other hand, if k > M and σk(i) = 0 then for each string λ ∈ Ti

which is the root of a pruned clump on Ti with ξki λ, and each leaf ρ of that

clump, there is a string ρ ∈ Ti with ρ ρ and Φρ(x) ↓ for each x < 2−2i−4|λ|.Furthermore, for any two extensions ρ1, ρ2 of λ on Ti, and each x < 2−2i−4

we have Φρ1i (x) = Φρ1i (x).

12 J. Stephenson

Proof By assumption, there is a path through Ti−1 which has ξj as an initialsegment for each j < i. Thus we may choose M0 large enough to satisfy allof the hypotheses imposed for j < i, and such that there is a pruned clumpA which is on each T ki−1 for stages k > M0, and whose root extends ξj forj < i, and is of length greater than i. Assume that for no j < i do we haveσk(j) = −1 for any k > M0.

Now suppose that for some k > M0, σk−1(i) = −1. Notice that for each

j > i we also have σk(j) = −1, and ξkj = ξki , because we only allow the least

i with σ−1(i) = −1 to make any changes at stage k. The construction of T kiproceeds by searching on T k−1i for a string ρ with |ρ| ≥ i and K(ρ) > 1

4 |ρ|−1,

and which is consistent with our choices ξkj for j < i and ξk−1j for i < j < k.

There is some string ρ ∈ A with the property that K(ρ) > 14 |ρ|−1 by Lemma

2.2 and the inductive hypothesis, and so we will set ξki = ρ for some such ρ.Because of our choice of M0, we know that there is no stage k > M0 at

which σk(j) changes for any j < i. Therefore once ξki has an initial segmentsatisfying the complexity condition discussed above, the only way that we can

have a stage k′ > k for which ξk′−1i 6= ξk

′

i is if σk′(i) = 1 but σk

′−1(i) = 0.

But in this case we choose ξk′

i to be a string which extends ξk′−1i , and thus

inherits satisfaction of the complexity condition. Also notice that if we neverset σk1(i) = 1 at any stage k1 > k, then every time we add a pruned clump to

T k1i with root λ, it is added in a single step, and we add extensions ρ of each

leaf ρ of the pruned clump so that for each x < 2−2i−4|λ| we have Φρi (x) ↓,and so that for any two such extensions ρ1, ρ2 of λ on Ti, if Φρ1i (x) ↓ and

Φρ2i (x) ↓ for some x < 2−2i−4 we have Φρ1i (x) = Φρ1i (x). This agreement ofhalting computations will obviously be preserved under extensions, and sincethe construction only ever extends strings which are already leaves, will holdof all extensions of λ in Ti.

Finally, notice that if σk1(i) = 1 at some stage k1 > M0, and we set τ to

be the longest string in the set ξk1−1, ξk10 , ξk11 , · · · ξk1i−1, ξ

k1−1i , · · · , ξk1−1k1−1, then

we know that there is some leaf λ0 on T k1−1i which extends τ , and which has

an extension λ ∈ T k1i−1 such that for some x < 2−2i−4|λ| every extension ρ of

λ in T k1,si−1 for any s has Φρi (x) ↑. In this case we have set ξk1i = λ, and T k1i is

the tree obtained by adding the downward closure of λ to T k′−1

i .

By the assumption on L it follows from the inductive hypothesis on the

trees T k,si−1 that the strings which are in T k,si−1 for some s are precisely thosestrings in Ti−1. From this we may therefore conclude that there is some x

such that every string ρ on Ti−1 which extends ξk′

i now has the property thatΦρi (x) ↑, as desired.

It clearly suces to take M > M0 to be some stage such that for k > M ,ξki = ξMi and such that if σk(i) = 1 for any k > M , this is true of everyk > M .

Lemma 5.3 For suciently large M , whenever k > M , and s ≥ k,

T si = T k,si . Thus Ti is the union of the T k,si at such stages.

Controlling Eective Packing Dimension of ∆02 Degrees 13

Proof Choose M > L larger than the M of the previous lemma, so that ξkjand σk(j) are constant for all j ≤ i and k > M . Then notice that by inductive

assumption, for s ≥ k we have T k,si−1 = T si−1. Fix some k > M . Notice that we

have T k,ki = T ki by denition. In addition, because σs(j) and ξsj are xed forj ≤ i and s ≥ k, it follows that the construction of T si for s ≥ k either alwaysproceeds via application of S or always by applications of T , and that the

algorithm which constructs T k,si will use the same algorithm, taking as input

T k,s−1i and T k,si−1 rather than Ts−1i and T si−1. In either case, for each s ≥ k we

know that T k,si−1 = T si−1, and it will follow inductively that T si = T k,si .

Lemma 5.4 For suciently large M , whenever k0 > M , for each leaf λ0 of

T k0i which is extended by some path through Ti, there is some k > k0 and a

pruned clump on T ki whose root λ is an extension of λ0. Furthermore this

clump has at least 2qλ many leaves of τ2≤|λ|, where qλ is the least integer

greater than or equal to (1−∑ij=−1 2−2j−4)|λ|.

Proof Take M to be large enough to satisfy each of the previous lemmas,and k0 > M .

Fix some leaf λ0 of T k0i which is an initial segment of a path through Ti.

Notice that this means that we must have ξk0i λ0, since after stage k0, we

only ever add strings to Ti which extend ξk0i . Now Ti−1 is a pruned clumpytree and λ0 lies on a path through it (since Ti ⊆ Ti−1), so we know that thereis a pruned clump on Ti−1 with a root extending λ0, and that we may choosethe root to be as long a string as we like (but the form of T−1 forces all prunedclumps to have roots whose lengths are powers of 4). Thus, for large enough k1we may note that λ0 has an extension λ ∈ T k1i−1 which is the root of a pruned

clump on Ti−1, and such that 4N = |λ| > 4|λ0| for some N > i + 2. Byassumption, the pruned clump on Ti−1 has at least 2qλ many leaves of τ2≤|λ|,

where qλ is the least integer greater than or equal to (1−∑i−1j=−1 2−2j−4)|λ|.

Now suppose that we have σk(i) = 0; then we know that at no stagek > k0 do we set σk(i) = 1. But this means that at each stage k > k0of the construction, we only ever add extensions to T ki if they are given byrunning S to search for them in T ki−1. Choose k1 > k large enough that λ0 is

extended by a pruned clump A with root λ λ0 on T k1i−1 whose length is of

form 4N > 4|λ0| · 22i+4 for some N > i + 2. Also assume that for each leaf

on A we have some extension ρ ∈ T k1i−1 with Φρi,k1(x) ↓ for each x < 2−2i−4|λ|(if this is not true at any stage k1, then we will never extend λ0 in Ti, so it isnot on a path through that tree, contrary to assumption). At the rst suchstage k1 of the construction, the algorithm S will detect the extensions of λ,and will add a pruned clump to T k1i .

The pruned clump B on T k1i with root λ is given as follows: The stringswhich are added to B are chosen from amongst those of the pruned clump A.

For each leaf ρ of B we will have Φρi (x) ↓ whenever x < 2−2i−4|λ|. Further-more any two such leaves will output the same 2−2i−4|λ|-many bits in thiscomputation; we take B to have the maximum possible number of leaves ex-

tending λ and satisfying this condition. Notice that there are 22−2i−4|λ| binary

strings of length 2−2i−4|λ|, and that the pruned clump A on T k1i−1 with root

14 J. Stephenson

λ has at least 2qλ many leaves, where qλ is the least integer greater than or

equal to (1 −∑i−1j=−1 2−2j−4)|λ|. From this it follows by a simple count that

B will have at least 2q′λ leaves, where q′λ is the least integer greater than or

equal to (1−∑ij=−1 2−2j−4)|λ|.

If σk(i) = 1, then there is a pruned A ⊆ T ki with root λ λ0, sinceat this stage all extensions of ξki on T ki−1 are added to T ki . Notice that thenumber of leaves on this pruned clump is at least 2qλ , where qλ is the least

integer greater than or equal to (1−∑i−1j=−1 2−2j−4)|λ|, which is greater than

(1−∑ij=−1 2−2j−4)|λ|.

This concludes the proof of the inductive claims on the Ti.Dene ξi = limk ξ

ki for each i. Notice that we have ξi ξj or vice-versa for

every i and j, because this is true at each stage of the construction. Noticethat each tree Ti contains ξj for every j, since ξ

kj ∈ T ki whenever j ≤ k. But

we have |ξj | ≥ j for every j, and so it follows that there is a unique path Xthrough Ti which has each of the ξj as initial segments.

The real X with initial segments given by the ξj is the real which we desiredto construct.

Lemma 5.5 With X as constructed, dimP (X) ≥ 14 .

Proof For each j, there is some initial segment ρ of X with K(ρ) > 14 |ρ| − 1

and |ρ| ≥ j. Therefore it follows that dimP (X) = lim supnK(X n)

n≥ 1

4 .

Now we must show that the conditions given on the trees are sucient toguarantee that if Y is a real computed by X, then dimP (Y ) < 1.

6 Combinatorial Estimate

Notice that if σk(i) has limit 1, then for some x, every string ρ on Ti hasΦρi (x) ↑, and so in particular Φi cannot compute any reals from X. Thereforeto show that for each i such that ΦXi computes a real, dimP (ΦXi ) < 1, weneed only consider i such that σk(i) has limit 0.

Thus, throughout this section, suppose that we are in the case that σk(i)has limit 0. In this case, every time we added a pruned clump to Ti with aroot λ extending ξi, we made sure that for each leaf ρ of that pruned clump,

we also added some extension ρ of ρ with the property that Φρi (x) ↓ for eachx < 2−2i−4|λ|, and furthermore, for any extensions ρ1, ρ2 of λ such that for

some x < 2−2i−4|λ|, Φρ1i (x) ↓ and Φρ2i (x) ↓, we have Φρ1i (x) = Φρ2i (x).The goal of this section is to provide an upper bound on the Kolmogorov

complexity of the strings in ΦTii which are initial segments of ΦXi , and thusto show that this path has eective packing dimension less than 1. We shalldo so by dening two prex-free machines M1 and M2 which will providedescriptions of strings in ΦTii , using the construction of Ti to provide shortdescriptions.

The computation presented here is perhaps less elegant than that presentedby Conidis in[3], but more general. Its directness is hoped to oer some

intuition as to why the complexities of initial segments of paths through ΦTii

Controlling Eective Packing Dimension of ∆02 Degrees 15

are bounded away from 1, as well as indicating roughly where the strings of(comparatively) high complexity might be found on those paths.

Throughout what follows, we will regard Ti as a computably enumerableset of strings, with the enumeration given by

Ti =⋃s≥k

T ki ,

where k is some stage chosen to be suciently large that T si = T k,si for eachs ≥ k.

In order to have a record of what order clumps were added to the tree Ti,number the pruned clumps on T ki via a computable numbering N as follows:If A is the unique pruned clump on Ti with root ρ ξi, and there is nopruned clump on Ti with root ρ such that ξi ≺ ρ ≺ ρ, then set N(A) = 1.Supposing that we have dened the pruned clumps A for which N(A) = j,dene N(B) = j + 1 for a pruned clump B exactly when the root ρ of Bextends the root ρ0 of some pruned clump A with N(A) = j, but such thatfor no pruned clump C with root ρ do we have ρ0 ≺ ρ ≺ ρ. It should be notedthat if A is a pruned clump on Ti and N(A) = j, and we are given some leaf λon A such that some pruned clump on Ti has a root extending λ, it is possibleto computably identify the unique pruned clump B with N(B) = j + 1 andwhose root extends λ. This is because the enumeration of Ti will add all of Bat some stage, and this will be the rst time it adds any extension of λ.

Denition 6.1 Let Aj be a pruned clump on Ti for 1 ≤ j ≤ n, and for eachj let ρj be the root of Aj . Suppose that we have ρj ≺ ρj+1 for each j < n,and that N(Aj) = j for each j. Then we call A1, A2, · · · , Aj a sequence of

adjacent pruned clumps.

The idea is that if we have a string τ ∈ ΦTii which is an extension ofξi then for some string ρ ∈ Ti we have Φρi = τ . We will let the sequenceA1, A2, · · · , An of adjacent pruned clumps consist of the pruned clumps on Tiwhose roots ρi have ρi τ . Notice that each pruned clump on Ti with a rootof length l has at most 2l leaves, and so each can be identied with a stringof length l by assigning those strings according to the lexicographical orderof the leaves. This assignment is computable, since all of the leaves on such apruned clump are added simultaneously at some stage of the enumeration ofTi.

The rst of the two prex-free machines,M1, will be used to compute thoseinitial segments τ of the real ΦXi which have the property that3(1 + 2−2i−4)|ρ| ≤ |τ | ≤ 2−2i−4|ρ|, where ρ is a root of a pruned clump Aon Ti and the root ρ of a pruned clump B with N(B) = N(A) + 1 extends

ρ; in this case there are extensions λ of ρ on Ti that have the property that

Φλi (x) = τ(x) for each x < |τ |. In this case the computation used will ndsome extension of ρ on Ti which gives a halting computation at least as longas τ , and then to note that extensions of ρ giving such a computation all givethe same computation, so that τ has been computed. In this case we willhave used approximately 2

3 |τ | many bits in the computation.More specically,M1, on input of a binary string σ, attempts to decompose

σ into the form 0n1θ1θ2 · · · θn0l1µ as follows:

16 J. Stephenson

Step 1 M1 takes |θ1| to be the length of the root of the unique pruned clumpon Ti with N(A) = 1, and sets A1 = A, unless σ is too short to parse inthis way, in which case M1 does not halt. Suppose that k < n, and M1 hascomputed θ1, · · · , θk, and a sequence of adjacent pruned clump A1, · · · , Aksuch that for j ≤ k, Ak has root ρk with |ρk| = |θk|. Then M1 interprets θkas corresponding to some r ∈ ω with r < 2|θk| as a binary expansion (possiblywith some leading zeroes), and searches for a pruned clump Ak+1 on Ti whichextends the rth leaf of Ak, and such that N(Ak+1) = k + 1. Notice that ifsuch a pruned clump exists it is unique. If no such pruned clump exists, or ifAk does not have r leaves, M1 does not halt.

Otherwise, letting ρk+1 be the root of Ak+1, M1 will set θk+1 to be the|ρk+1| bits of σ which follow those we have labelled θk. If the string σ doesnot have another |ρk+1|-many bits, then M1 does not halt.

Step 2 OnceM1 has found θ1, · · · θn and A1 · · ·An it then identies the numberl for which the bits immediately following θn are of the form 0l1, and then,if there are precisely l bits following those in the string σ, interprets those lbits as the expansion µ of a binary number m < 2l (if the number of bits inµ is not correct, M1 does not halt).

At this point, M1 nds the string ρ ∈ Ti which is the leaf on An whoselexicographic position is given by θn (we interpret the position as an n-bitbinary number, possibly with some leading zeroes). M1 then searches fora pruned clump An+1 on Ti with N(An+1) = n + 1 and whose root ρn+1

satises ρn+1 ρ. If no such pruned clump is found, M1 will not halt. If sucha pruned clump is found, notice that there is some extension λ ρn+1 on

Ti such that Φλi (x) ↓ for each x < 2−2i−4|ρn+1|, and that all strings λ which

extend ρn+1 have the property that if Φλi (x) ↓ for some x < 2−2i−4, then

Φλi (x) = Φλi (x). This means that we can uniformly compute an unambiguousvalue Φλi (x) 2−2i−4|ρn+1| which is equal for all such λ. At this point M1

checks whether m < 2−2i−4|ρn+1|. If not, then M1 will not halt.Finally, we are ready to give the output of M1: assuming that all of the

computations given above are carried out correctly, and M1 has not yet beenstated not to halt, we setM1(σ) = Φλi (x) 2−2i−4m, with λ as in the previousparagraph.

Notice that M1 is a prex-free machine, since any prex of a string onwhich M1 halts will at some point not have the correct syntactical form andwill be rejected.

In what follows the next two lemmas will prove vital in providing theestimates we require:

Lemma 6.2 If A1 · · ·An+1 is a sequence of pruned clumps, where Aj's rootis ρj for each j ≤ n+ 1, then for each j ≤ n we have |ρj+1| ≥ 4 · 2−2i−4|ρj |.

Proof The construction of Ti ensures that if ρj ρj+1 and N(Aj) = j,and N(Aj+1) = j + 1, then when we added the clump Aj to Ti, we addedsome extension λ of the leaf of Aj which ρj+1 extends, by applying the al-gorithm S. Because of the way this algorithm operates, we must then have|ρj+1| ≥ 4 · 2−2i−4|λ| > 4 · 2−2i−4|ρj |.

Controlling Eective Packing Dimension of ∆02 Degrees 17

Lemma 6.3 Let τ be a string in ΦTii that is an initial segment of the path ΦXithrough ΦTii , and assume that A1, · · · , An+1 is a sequence of adjacent pruned

clumps with ρj the root of Aj, that some extension ρ of ρn+1 has Φρi (x) = τ(x)for each x < |τ |, and that 2−2i−4|ρn| < |τ | ≤ 2−2i−4|ρn+1|. Then we have

K(τ) ≤ n+

n∑j=1

|ρj |+ 2 log2(|τ |) + C

for some C, and in particular if |τ | ≥ 3(1 + 2−2i−4)|ρn|, then we have

K(τ) ≤ 2

3|τ |+D(log2(|τ |))

for some D.

Proof The rst result follows immediately: let σ be given by 0n1θ1θ2 · · · θn0l1µ,where |θj | = |ρj |, and θj tells us which leaf of the pruned clump with root ρjhas the root of Aj+1 on it, and µ is the binary expansion for |τ |, of length l.Then we have M1(σ) = τ .

In the case where we have |τ | ≥ 3(1 + 2−2i−4)|ρn| we use the fact that theconstruction of Ti ensures that |ρj+1| ≥ 4 · 22i+4|ρj | for each j; therefore wehave

n∑j=1

|ρj | ≤n−1∑j=1

(2−2i−4

4

)n−j|ρn|+ |ρn|

≤3

2|ρn|

≤2

3|τ |

But then we have K(τ) ≤ 23 |τn|+ n+ 2 log2 |τ |+C, and noting that we have

|τ | ≥ |ρn| ≥ 4n, it follows that n ≤ log2(|τ |), and the result follows.

Now I give a second prex-free machine M2 which will bound the complexityof other strings on ΦTii . This machine will be used to compute initial segmentsof ΦXi in the cases which M1 was unable to provide adequate bounds for.

The machine M2 proceeds as follows: given a string σ, M2 tries to parseσ to be of the form 0n1θ1θ2 · · · θn0l1µ01ν in a similar manner to M1. It rstmimics the behaviour of M1's step 1, to check whether the initial segment ofthe string is of the form 0n1θ1θ2 · · · θn, where the θj correspond to a sequenceof adjacent pruned clumps and specify which leaf of Aj is to be extended tond the root of Aj+1. In this way it is either able to compute the root of thepruned clump An+1, or does not halt.

After this, M2 checks whether µ is of length 2l, and that for each j < l, the2jth and 2j + 1st bits of µ are equal. If not, M2 will not halt. If the propertydoes hold of µ, M2 then deletes every second bit of µ to obtain a string oflength l; it interprets this string as a binary expansion of a number L, andthen checks whether L is the length of the string ν. If so, then M2 searchesfor an extension λ of the root ρn+1 of An+1 such that Φλi (x) ↓ for everyx < 2−2i−4|ρn+1|. It then takes τ to be the string such that |τ | = 2−2i−4|ρn+1|and Φλi (x) = τ(x) for each x < 2−2i−4|τ |.

Finally, we have M2(σ) = τ ν, i.e. the concatenation of the two strings.

18 J. Stephenson

Lemma 6.4 Let τ be a string in ΦTii that is an initial segment of the path ΦXithrough ΦTii , and assume that A1, · · · , An+1 is a sequence of adjacent pruned

clumps with ρj the root of Aj, that some extension ρ of ρn+1 has Φρi (x) = τ(x)for each x < |τ |, and that 2−2i−4|ρn| < |τ | ≤ 2−2i−4|ρn+1|. Then we have

K(τ) ≤ n+

n−1∑j=1

|ρj |+ 3 log2(|τ |) + |τ | − 2−2i−4|ρn|+ C

for some C, and in particular if |τ | ≤ 3(1 + 2−2i−4)|ρn|, then we have

K(τ) ≤(

1− 2−2i−4

6(1 + 2−2i−4)

)|τ |+D log2(|τ |)

for some D.

Proof Once again, the rst result follows very easily: Let σ be given by0n1θ1θ2 · · · θn−10l1µν, where θj tells us which leaf of the pruned clump withroot ρj has the root of Aj+1 on it, and µ is the doubled binary expansionfor |ν|, which is of length l. Then we have M2(σ) = τ . The 3 log2(|τ |) termcomes from the fact that |µ| ≤ log2(τ) and so |0l1µ| ≤ 3 log2(τ) + 1, whereas|τ | − 2−2i−4|ρn| is the length of ν.

Suppose that we have |τ | ≤ 3(1 + 2−2i−4)|ρn|. Notice that we haven−1∑j=1

|ρj | ≤n−1∑j=1

(2−2i−4

4

)n−j|ρn|

≤2−2i−4

2|ρn|.

Now we once again bound all of the smaller terms by a single term which islogarithmic in size, and apply our assumption on |τ |:

K(τ) ≤2−2i−4

2|ρn|+ |τ | − 2−2i−4|ρn|+D log2(|τ |)

=|τ | − 2−2i−4

2|ρn|+D log2(τ)

≤|τ | − 2−2i−4

6(1 + 2−2i−4)|τ |+D log2(|τ |)

=

(1− 2−2i−4

6(1 + 2−2i−4)

)|τ |+D log2(|τ |)

which provides the bound we sought.

Now for each string τ which lies on the path ΦXi through ΦTii we have a se-quence of pruned clumps A1, · · · , An+1 with roots ρj such that some extensionρ of ρn+1 has Φρi (x) = τ(x) for every x, and that if τ is suciently long wemay choose the sequence so that 2−2i−4|ρn| < |τ | ≤ 2−2i−4|ρn+1|, simply bychoosing a sequence of adjacent pruned clumps which is of suitable length.This is sucient for our needs regarding the reals computed by X.

Controlling Eective Packing Dimension of ∆02 Degrees 19

Lemma 6.5 For some D, all suciently long initial segments τ of ΦXi have

the property that

K(τ) ≤(

1− 2−2i−4

6(1 + 2−2i−4)

)|τ |+D log2(|τ |)

We have now completed all of the work which is required to prove the maintheorem (Theorem 1.2):

Proof of Theorem 1.2 If X is the real dened by taking the ξi as initial seg-ments, then X ≤T ∅′ by construction. In addition we have dimP (X) ≥ 1

4 ,

as seen in Lemma 5.5. Furthermore, for any i such that ΦXi = Y is a totalreduction, we have dimP (Y ) < 1.

Finally, we consider a result on extraction of Kolmogorov complexity:

Theorem 6.6 (Fortnow, Hitchcock, Pavan, Vinodchandran, and Wang[7], as stated by

Bienvenu, Doty, Stephan [2]) For any X ∈ 2ω with dimP (X) > 0 and any

ε > 0, there is some Z ∈ 2ω such that Z ≡T X and (Z) ≥ 1− ε.

This result, applied directly to the Turing degree of X, yields Corollary1.3.

7 Future Directions

Many results which prove the existence of real numbers possessing propertiesexpressed in terms of Kolmogorov complexity, eective packing dimension,and eective Hausdor dimension are able to be eectivised to show that suchreals exist below any array noncomputable degree. See for instance papersof Downey and Greenberg [5], Downey and Ng [6], and Kummer [9]. Thiscommon trend suggests that the obvious analogue may hold in the currentcase.

Question 7.1 Given an array noncomputable degree a, is there a real X ≤T awhich has nonzero eective packing dimension, but which cannot compute any

real of eective packing dimension 1?

Given the suggestive pattern noted above, it seems reasonable to supposethat the answer is very likely yes. Indeed, the author feels that the proof islikely to proceed somewhat as follows: noting that the only oracle questionswhich we are asking are about either the prex-free Kolmorogov complexity ofa string, or asking whether some string forces divergence of an oracle machine,rewrite the list of requirements for the construction so that the oracle questionsasked of ∅′ are given as a uniformly computable list of questions which are wtt-computable by ∅′. But we know that any array noncomputable degree willanswer innitely many of these oracle questions correctly, and so it shouldsuce to nd strings of high complexity if given a sequence of pruned clumps,and likewise to nd strings which force divergence of computations. However,none of the technical details which this argument would rely on have beenchecked.

It should be noted that if a real X has nonzero eective packing dimension,but cannot compute any real of eective packing dimension 1, then X is

20 J. Stephenson

necessarily itself array computable, as an immediate corollary of a result ofDowney and Greenberg:

Theorem 7.2 (Downey, Greenberg[5]) Every array noncomputable degree acomputes a set A with eective packing dimension 1.

From this it is clear that the array noncomputable degrees cannot be theset of degrees which compute a real X as above. However, it was shown byKummer (and restated in the form given below by Downey and Greenberg[5])that amongst the c.e. degrees, array computability may be characterised asfollows:

Theorem 7.3 (Kummer[9]) A c.e. degree is array noncomputable if and only

if it computes a real Y with dimP (Y ) > 0.

This implies that no degree whose members are computed by an arraycomputable c.e. degree can have nonzero eective packing dimension. Fromthis we see that the only c.e. degrees below which it is possible to carryout a construction of a degree with the properties set out in this paper arethe array noncomputable ones. As suggested above, it seems likely that thisis precisely the set of c.e. degrees below which such a construction can oc-cur.

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Acknowledgments

The author would like to thank Rod Downey for suggesting the problem, andfor many useful conversations.This article is work towards a Ph.D. in Mathematics at the University ofChicago under Professors Robert Soare and Denis Hirschfeldt, to whom thanksare also extended.

Department of Mathematics,The University of Chicago,5734 S. University Avenue,Chicago, Illinois 60637,USA jrs@math.uchicago.eduhttp://math.uchicago.edu/ jrs/