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13.9. DNC functions and e ective Hausdor dimension

619

13.9.1 Dimension in h-spaces

Greenberg and Miller [170] generalized many notions we have seen in this chapter to h-spaces. The following is the natural generalization of the notions of weak and strong s-Martin-L¨of randomness to hω.

Definition 13.9.2. Let s [0, 1].

(i)

A test for weak s-Martin-L¨of randomness is a uniformly c.e. sequence

(ii)

{Vk}k ω of subsets of hsuch that

σ VK μ( σ )s 2−k for all k.

A test for strong s-Martin-L¨of randomness is a uniformly c.e. se-

 

 

 

 

 

 

quence {Vk}k ω of subsets of hsuch that for every k and every

 

prefix-free Vk Vk , we have

σ Vk

μ( σ )s 2−k.

 

 

 

 

 

(iii)A hω is weakly s-Martin-L¨of random if A / k Vk for all tests for weak s-Martin-L¨of randomness {Vk}k ω.

(iv)A hω is strongly s-Martin-L¨of random if A / k Vk for all tests for strong s-Martin-L¨of randomness {Vk}k ω.

Many results proved in Section 13.5 still hold for hω, such as Theorem 13.5.13, which says that if t < s and A is weakly s-random, then A is strongly t-random. Thus, for A hω, the supremum of all s for which A is weakly s-random equals the supremum of all s for which A is strongly s-random, and, by analogy with the 2ω case, can be called the e ective (Hausdor ) dimension of A, denoted by dimh(A).

We can also generalize the notion of a Solovay s-test.

Definition 13.9.3. Let s [0, 1]. A Solovay s-test is a c.e. set S h

such that σ S μ( σ )s < ∞. A set A is covered by this test if A σ for infinitely many σ S.

Theorem 13.5.7 still holds: If A is weakly s-random then A is not covered by any Solovay s-test. If A is covered by some Solovay s-test then A is not weakly t-random for any t > s.

We can also define martingales for h-spaces.

Definition 13.9.4. A supermartingale (for h) is a function d : hR 0 such that for all σ h,

d(σi) h(|σ|)d(σ).8

i<h(|σ|)

If s [0, 1], we say that a supermartingale d s-succeeds on A hω if lim supn d(A n)μ( A n )1−s = .

8Note that this condition is equivalent to d(τ )μ( τ ) d(σ)μ( σ ), where the sum is taken over all immediate successors τ of σ.

620 13. Algorithmic Dimension

As in the 2ω case, A hω is strongly s-Martin-L¨of random iff there is no c.e. supermartingale that s-succeeds on A. The proofs are as before, using in particular the fact that Kolmogorov’s Inequality holds in

hω: For a supermartingale d and

a prefix-free set C h, we have

σ C

d(σ)μ(σ) d(λ). The easiest way to see that this is the case is

 

ω

to think of d(σ)μ(σ) as inducing a semimeasure on h .

It is also worth noting that there is an optimal c.e. supermartingale for h, constructed as usual.

Our goal is to use h-spaces to construct an element of 2ω of minimal degree with e ective dimension 1. To do so, we will need to be able to translate between hω and 2ω in a dimension-preserving way. If h is very

well-behaved then there is no problem. For example, if X 4ω and we let Y (2n) = #X(2n) $ and Y (2n+ 1) = X(n) mod 2, then Y ≡T X, and it is easy to check that dim(Y ) = dimh(X). To handle more complicated h-spaces, it is convenient to work through Euclidean space rather than going directly from hω to 2ω.

There is a natural measure-preserving surjection of hω onto the Euclidean interval [0, 1]. First map strings to closed intervals: Let πh(λ) = [0, 1]. Once πh(σ) = I is defined, divide I into h(|σ|) many intervals I0, I1, . . . , Ih()|−1

ofh equal

length and let πh(σk)

= I

for k < h( σ ). Note that indeed

 

h

 

 

k

 

h

| |

ω

by letting

μ (σ) = μ(π

 

(σ)) for all σ. Now extend π

 

continuously to h

 

πh(X) be the unique element of

n πh(X n). (The fact that h(n) 2 for

all n ensures that this

intersection is indeed a singleton.) The mapping πh

is

 

 

 

 

 

 

 

not quite 1-1, but it is 1-1 if we ignore the (countably many) sequences that are eventually constant. Note that for all X hω, we have X ≡T πh(X).

The theory of e ective dimension can also be developed in the space [0, 1]. We do not have martingales, but we can still, for example, define Solovay tests, as we saw in Definition 13.5.8. Let dim[0,1](X) be the infimum of all s such that there is an interval Solovay s-test that covers X (that is, such that X is in infinitely many elements of the test).

Proposition 13.9.5 (Greenberg and Miller [170]). For all h and X hω, we have dim[0,1](πh(X)) dimh(X).

Proof. Let s > dimh(X) and let S be a Solovay s-test for h that covers X. Since πh is measure-preserving, the image of S under πh is an interval Solovay test covering πh(X).

Equality of these dimensions does not hold in general, but we will see that we do get equality if h does not grow too quickly or irregularly. Suppose that S is an interval Solovay s-test covering some πh(X). We would like to cover X by something like (πh)1(S). The problem, of course, is that the basic sets (closed intervals with rational endpoints) in [0, 1] are finer than the basic sets in hω; not every closed interval is in the range of πh. We thus need to refine S by replacing every I S by finitely many intervals in the range of πh. While we can control the Lebesgue measure of such a

γn+1
|I|

13.9. DNC functions and e ective Hausdor dimension

621

collection, if the exponent s is smaller than 1 then the process of replacing large intervals by a collection of smaller ones may increase the s-weighted sum of the lengths of the intervals significantly. We show that if h does not grow too irregularly and we increase the exponent s slightly, then this sum remains finite.

 

Let In = πh(hn) and let I =

n In = πh(h). Let γn =

1

. Then In

 

|hn|

consists of

|

 

n

|

many closed

intervals, each of length γ

 

.

 

 

 

h

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

For a closed interval I [0, 1], let nI be the unique n such that γn

|I| > γn+1. Let kI

be the largest natural number such that |I|

> kγn+1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h

 

 

 

 

 

 

 

 

Then there is an I

InI +1 of size kI + 1 with I

 

 

I.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

rational endpoints, let

 

For a set S of closed subintervals of [0, 1] with

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

 

 

 

S =

 

I S

I. Thus S I, and if x = π (X) and x is covered by S then X is

covered by (πh

)

1

(S). If S is c.e. then so are S and (π

h

)

1

(S). Furthermore,

 

 

 

 

 

 

 

 

 

 

I S

|I| =

 

σ (πh)1(S) μ(σ) .

 

 

 

 

 

 

 

 

 

 

 

We

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

for h and

 

 

 

express the

regularity of h via the following condition

t > s 0:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h(n)1−s

 

 

 

 

 

 

 

(13.1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

< ∞.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(h(0)

· · ·

h(n))t−s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

Lemma 13.9.6 (Greenberg and Miller [170]). Suppose that (13.1) holds

for t, s, and h, and that S is a

set of closed intervals in [0, 1] such that

 

 

 

 

|

|

I

|

|

 

is also finite.

 

I

 

S

I

s is finite. Then

S

 

I

t

 

 

 

 

 

 

 

 

Proof. Let I be any closed interval in [0, > k and k h(n),

|I|t = (k + 1)γnt +1 2nt−+1s γns+1 = 2k

1]. Let n = nI and k = kI . Since

γns+1

γnt−+1s |I|s

|I|s

 

 

 

 

 

< 2k1−sγnt−+1s |I|s

 

 

 

2h(n)1−s

 

 

 

 

 

 

 

 

 

|I|s.

 

 

 

 

 

 

 

 

t s

 

 

 

 

 

 

 

 

(h(0) . . . h(n))

Thus if we let Sn = {I S : nI = n}, then

 

 

 

 

 

 

 

 

 

2h(n)1−s

 

 

 

 

 

 

 

|I|t =

n

|I|t

n

 

(h(0) . . . h(n))t−s

|I|s

I

 

I Sn

I Sn

s

 

 

 

h(n)1−s

S

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 |I|

 

 

 

 

,

 

 

 

 

 

 

 

n

 

(h(0) . . . h(n))t−s

 

 

 

 

 

 

I S

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which by assumption is finite.

Thus for all X hω, if (13.1) holds for t, s, and h, and dim[0,1](πh(X)) < s, then dimh(X) t. So if (13.1) holds for all t > s 0, then dim[0,1](πh(X)) = dimh(X) for all X hω.

622 13. Algorithmic Dimension

For example, (13.1) holds for all t > s 0 for the constant function h(n) = 2. Thus, as discussed above, dimension in [0, 1] is the same as dimension in 2ω. However, this condition holds for some unbounded functions

has well (for example h(n) = 2n).

The following is a su cient condition for (13.1) to hold for all t > s 0.

Lemma 13.9.7 (Greenberg and Miller [170]). If

 

 

 

 

 

 

 

 

 

 

 

 

lim

 

 

log h(n)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m n log h(m) =h0

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

[0,1]

(π

h

(X)) for

then (13.1) holds for all t > s 0, and so dim

(X) = dim

 

 

all X hω.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Proof. Let f (n) = log h(n). Let t > s

0 and let ε <

t−s

. For su ciently

large n, we have f (n) < ε

 

 

 

 

 

 

 

1−s

 

h(n) and let

 

m n

f (m). Let g(n) = h(0)

· · ·

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ε

 

 

 

 

For su ciently large n, we have h(n) < g(n)

 

,

δ = ε(1 − s) 1(ts− s) < 0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and hence

h(n)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

< g(n)δ . Thus there is an N such that

 

 

 

 

 

 

 

g(n)t−s

 

 

 

 

 

 

 

 

 

 

h(n)1−s

 

 

<

 

g(n)δ =

2δ

log g(n)

,

 

 

 

 

n N

(h(0) . . . h(n))t−s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which is finite because 2δ < 1 and log g(n) n (since h(n) 2).

The regularity condition of Lemma 13.9.7 is not, strictly speaking, a slowness condition, because, for example, h(n) = 2n2 satisfies this condition, yet there is a monotone function that is dominated by h but does not satisfy the condition. However, the condition does hold for all su ciently slow monotone functions.

Lemma 13.9.8 (Greenberg and Miller [170]). If h is nondecreasing and dominated by 2kn for some k, then

 

 

 

 

 

log h(n)

 

 

 

lim

 

= 0,

 

 

n

 

 

m n log h(m)

 

 

 

 

 

 

 

h

[0,1]

 

h

 

 

ω

.

and so dim (X) = dim

 

(π

(X)) for all X h

Proof. If h is bounded then it is eventually constant, and the condition is easily verified, so assume that h is unbounded. Fix c > 0. There is an Nc such that log h(n) > c for all n Nc. For n > Nc,

 

 

 

log h(n)

 

 

 

 

kn

 

 

kn

 

 

 

 

 

 

 

 

.

 

 

 

m n log h(m)

 

 

Nn c log h(m)

c(n − Nc)

 

kn

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

Since limn

 

 

=

 

,

 

 

 

 

 

 

 

 

 

 

c(n−Nc)

c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

log h(n)

 

 

 

 

 

 

 

 

 

 

 

 

lim

 

 

 

 

k

.

 

 

 

 

 

 

m n log h(m)

 

 

 

 

 

 

 

 

n

c

Since c is arbitrary, this limit is 0.

13.9. DNC functions and e ective Hausdor dimension

623

Finally, we get the result we will need to translate between hω and 2ω.

Corollary 13.9.9 (Greenberg and Miller [170]).

dominated by 2kn for some k, then every X hω that dimh(X) = dim(Y ).

If h is nondecreasing and computes a Y 2ω such

13.9.2Slow-growing DNC functions and sets of high e ective dimension

For natural numbers a > b > 0, let Qba be the collection of functions f such that for all n,

1.f (n) {0, . . . , a − 1},

2.|f (n)| = b, and

3.if Φn(n) then Φn(n) / f (n).

By standard coding, Qba can be seen as a computably bounded Π01 subclass of ωω. Note that Q1a is essentially the same as DNCa.

To connect these sets for di erent values of a and b, we will use the concept of strong reducibility of mass problems from Definition 8.9.1. Recall that P ωω is strongly reducible to R ωω if there is a Turing functional Ψ such that Ψf P for all f R. If P is a Π01 class, then we may assume that the Ψ in the above definition is total, and hence the reduction is a truth table reduction, since we can always define ΨX (n) = 0 if there is an s such that X / P [s] and ΨX (n)[s] .

Lemma 13.9.10 (Greenberg and Miller [170]). If a > b > 0 then Qba+1+1 s

Qba, uniformly in a and b. (Uniformity here means that an index for the reduction functional Ψ can be obtained e ectively from a and b.)

Proof. From n and y < a we can compute an mn,y such that

1.for all x < a such that x =y, we have Φmn,y (mn,y ) = x iff Φn(n) = x; and

2.Φmn,y (mn,y ) = y iff either Φn(n) = y or Φn(n) = a.

 

 

 

 

 

 

 

 

 

 

 

 

 

Let f

Qab . Let f (n) =

y<a f (mn,y ). Fix n. For all y < a, if Φn(n) then

Φn(n) / f (mn,y ), so Φn(n) / f (n). Furthermore, if there is some y < a

such that y f (mn,y), then Φn(n) =a, so Φn(n) / f (n) {a}. Finally, if

 

 

 

 

 

 

 

 

 

 

 

 

|f (n)|

= b then f (mn,y) is in fact constant for all y < a, so that y f (mn,y)

for all y f (n). Thus we can define Ψ as follows:

 

 

 

 

 

Ψ

f

 

some subset of f (n) of size b + 1

if

f (n)

 

> b

 

 

(n) =

f (n)

a

 

 

if

|f (n)|

= b.

 

 

 

 

 

 

{ }

 

 

|

 

 

 

 

 

 

 

 

|

 

624 13. Algorithmic Dimension

Corollary 13.9.11 (Greenberg and Miller [170]). If a 2 then Qba+1+b s

DNCa, uniformly in a and b.

For a 2 and c > 0, let Pac be the collection of functions f aω such

that for all n and all x < c, if Φ n,x ( n, x ) then Φ n,x ( n, x ) =f (n). Note that Pa1 s DNCa.

Lemma 13.9.12 (Greenberg and Miller [170]). For all a > b > 0 and c > 0, if c(a − b) < a then Pac s Qba, uniformly in a, b, and c.

Proof. Fix f Qba and n. For all x < c, if Φ n,x ( n, x ) then

Φ n,x ( n, x ) [0, a) \

f ( n, y ).

 

y<c

The set on the right has size at most c(a − b), so if c(a − b) < a then we can choose some x < a not in that set and define Ψf (n) = x.

Corollary 13.9.13 (Greenberg and Miller [170]). If a 2 and c > 0 then

Pcac s DNCa, uniformly in a and c.

Proof. Let b

so P c

c(a 1)+1

so P c

c(a 1)+1

uniform.

= c(a − 1) − a + 1. Then Qab+1+b s DNCa and

 

 

 

 

c((a + b) (b + 1)) = c(a − 1) < a + b,

 

 

 

 

= P c

 

s Qb+1

. Since c > 0, we have c(a

1) + 1

 

ca,

c

a+b

c

c

 

 

a+b

 

 

 

 

 

 

 

 

Pca, and hence Pca s Pc(a−1)+1. All these reductions are

Greenberg and Miller [170] used the classes Pcac to construct sequences of positive e ective dimension.

Theorem 13.9.14 (Greenberg and Miller [170]). Let a 2 and ε > 0. Every f DNCa computes a set of e ective Hausdor dimension greater than 1 − ε, via a reduction that is uniform in a and ε.9

Proof. Fix c > 1. We work in the space (ca)ω . Let d be the universal c.e. supermartingale for this space. By scaling we may assume that d(λ) < 1.

For σ (ca), let Sσ be the set of k < c such that d(σk) a|σ|+1. Note that these sets are uniformly c.e. From σ, we can compute an mσ such that

for each x < c, we have Φ mσ ,x ( mσ , x ) = k if k is the xth element to be enumerated into Sσ , and Φ mσ ,x ( mσ , x ) if |Sσ | < x.

The idea here is that if d(σ) a|σ| then |Sσ| c, by the supermartingale condition, so all elements of Sσ are “captured” by the map e →Φe(e). We can thus use a function g Pcac to avoid all such extensions: given such g, inductively define X (ca)by letting the (n + 1)st digit of X be g(mX n). Then, by induction, d(X n) an for all n.

9That each f DNCa computes such a set is of course not a new fact, since the Turing degree of a bounded DNC function is PA and so computes a 1-random set. The extra information is the uniformity.

13.9. DNC functions and e ective Hausdor dimension

625

Now let s 0 and suppose that d s-succeeds on X, that is, that d(X n)μca(X n)1−s is unbounded. Since μca(X n) = (ca)−n,

 

 

d(X n)μca(X n)1−s an(ca)−n(1−s) = cs−1as n .

Thus, if d s-succeeds on X then cs−1as

> 1, so as > c1−s. Taking the

logarithm of both sides, we have

s

1

 

 

> 1 − s, so s > 1

 

. Thus

loga c

1+loga c

ca

 

1

 

 

 

 

 

 

 

dim

(X) 1

 

. Since limc loga c = , given ε we can find some

1+loga c

c and

X such that dimca

(X) > 1

− ε. By Corollary 13.9.9, X computes a

ω

 

 

 

Y 2

 

such that dim Y > 1 − ε.

 

 

 

 

 

Theorem 13.9.15 (Greenberg and Miller [170]). Let a 2. Every f

DNCa computes a set of e ective Hausdor dimension 1, via a reduction that is uniform in a.

Proof. We combine the constructions of sets of e ective dimensions closer and closer to 1 into one construction. Let h(n) = (n + 1)a. Let d be the universal c.e. supermartingale for hω. Given f DNCa, obtain gn Pnan for all n > 0 uniformly from f .

For σ hn, let Sσ be the set of k n such that d(σk) a|σ|+1 and

compute an mσ such that for each x n, we have Φ mσ ,x ( mσ , x ) = k if k is the xth element to be enumerated into Sσ, and Φ mσ ,x ( mσ , x ) if

|

S

 

we can define X(n) = g

n+1

(m

X n

) and

σ| < x. As in the previous proof,

 

n

 

 

 

 

 

inductively prove that d(X n) a

 

for all n.

 

 

 

 

 

 

h

a−n

0,

 

 

 

 

 

 

 

Now, μ

(X n) = n! , so for s

 

 

 

 

 

 

 

 

d(X n)μh(X n)1−s

asn

.

 

 

 

 

 

 

(n!)1−s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let s < 1. Let k be such that ks−1a2s−1 < 1. For almost all n we have n! > (ka)n, so for almost all n we have

asn

 

asn

= (ks−1a2s−1)n < 1,

(n!)1−s

(ka)n(1−s)

 

 

and hence d cannot s-succeed on X. Thus dimh(X) = 1, so by Corollary 13.9.9, X computes a Y 2ω of e ective dimension 1.

Finally, we can paste together these constructions for all a 2 to get the desired result.

Theorem 13.9.16 (Greenberg and Miller [170]). There is a computable

˜

DNCh˜ computes a set of

order h : N N \ {0, 1} such that every f

e ective Hausdor dimension 1.

Proof. Let h(n) = (n+1)2n, and let d be the universal c.e. supermartingale for h. For σ hn, let Sσ be the set of k < 2n such that d(σk) (n + 1)! and

compute an mσ such that for each x < 2n, we have Φ mσ ,x ( mσ , x ) = k if k is the xth element to be enumerated into Sσ, and Φ mσ ,x ( mσ , x ) if

|Sσ| < x.

626 13. Algorithmic Dimension

n

For n > 0, we have Ph2(n) DNCn+1 uniformly in n, so there is an e ective list of truth table functionals Ψn such that Ψfn Ph2(nn) for all f DNCn+1. Let ψn be a computable bound on the use function of Ψn. Let

mn = 1 + sup{mσ, x : σ hn x < 2n}

and let un = ψn(mn). Let u0 = 0.

For all n > 0, if ρ is a sequence of length un that is a DNCn+1-string (that is, ρ (n+1)un and for all y < un such that Φy (y) , we have Φy(y) =ρ(y); or equivalently, ρ is an initial segment of a sequence in DNCn+1) then Ψn(ρ)

is a P

2n

 

-string (an initial segment of a sequence in P 2n

) of length at least

h(n)

 

 

 

we may assume that u

 

 

h(n)

for all n > 0. Let

m . By increasing ψ

n

n

< u

n+1

˜ n

 

 

 

 

 

 

 

 

 

 

h(k) = n + 1 for k [un−1, un).

 

 

 

 

 

 

 

If f

 

DNC

h˜ then f un is a DNCn+1-string for all n > 0, so

combining

 

 

 

 

n

 

the reductions Ψn, we can define a g T f such that for all n and all σ h

 

,

1.

g(mσ ) < h(|σ|) and

 

 

 

 

 

 

 

2.

for all x < 2n, if Φ mσ ,x ( mσ , x ) then g(mσ) = Φmσ ,x ( mσ , x ).

 

We can now use g to define X hω as in the last two constructions, by letting X(n) = g(mX n). By induction on n, we can show that d(X n) n!. As before, we can do so because if σ hn and d(σ) n!, then there are at most 2n many immediate successors τ of σ such that d(τ ) (n + 1)!, and so they are all captured by the function e →Φe(e) and avoided by g.

Finally, we show that dimh(X) = 1, which by Corollary 13.9.9 implies that X computes a Y 2ω of e ective dimension 1.

Let s < 1. For any σ hn,

μh(σ) =

 

1

 

=

 

 

1

.

 

 

 

 

 

 

 

 

 

2021

· · · 2(n−1)n!

2(n2)n!

 

Thus for all n,

 

 

 

 

 

 

 

 

 

d(X n)μh(X n)1−s

(n!)s

 

 

 

(n!)

,

n

n

 

 

 

2(1−s)(2)

 

2(1−s)(2)

which is bounded (and indeed tends to 0). Thus d does not s-succeed on

X.

Thus, by Theorem 13.9.1, we have the following.

Theorem 13.9.17 (Greenberg and Miller [170]). There is a minimal degree of e ective Hausdor dimension 1.

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