14 The theory of higher randomness

Higher randomness theory studies algorithmic complexity problems of randomness using techniques and ideas from hyperarithmetic theory and set theory. Since second order arithmetic provides a broader, though sometimes coarser, view of the reals, the picture one gets of randomness is quite different from that for first order theory.

14.1 Higher Kurtz randomness

14.1.1 image and image-Kurtz randomness

Definition 14.1.1. Let x ∈ 2ω. Given a class Γ of sets of reals, x is Γ-Kurtz random if it is not a member of any closed nullset in Γ.

We focus on image, image and image-Kurtz randomness. Recall (§ 2.7.1) that BC denotes the set of Borel codes and {Az | z ∈ BC} is the set of reals generated by the codes.

Proposition 14.1.2.

(i)x is image-Kurtz random if and only if x does not belong to any Π1(HYP)-nullset.

(ii)The set of image-Kurtz random reals is image but not image.

Proof. Note that (i) follows immediate from Theorem 2.7.4. We give a self-contained proof here.

By Theorem 2.7.2, every Π1(HYP)-set is a image-set. Hence no image-Kurtz random real belongs to a Π1(HYP)-nullset. Conversely, let P be a image-nullset. Then image image is image. Since P is closed and the set of hyperarithmetic reals is dense, we have

image

By Corollary 2.2.12, U is also image and hence image. Thus PΠ1(HYP). It follows that x does not belong to a Π1(HYP)-nullset, and hence image. In other words, x is image-Kurtz random.

To prove (ii), observe that the set

image

is image, where Az is as defined in § 2.7.1. By Proposition 13.1.6, the set

image

is image. Hence the set

image

is a image-nullset. Moreover by (i),

image

Hence the set of image-Kurtz random reals is image. Now μ(E)= 1, and since no hyperarithmetic real is image-Kurtz random, by Theorem 13.1.13 one concludes that E is not image.

Theorem 14.1.3 (Kjos-Hanssen, Nies, Stephan and Yu [68]). image-Kurtz randomnessimage-Kurtz randomness = image-Kurtz randomness.

Proof. It is obvious that image-Kurtz randomness ⊆ image-Kurtz randomness ⊆ image-Kurtz randomness. We show that the first inclusion is proper. Let

image

By Proposition 13.1.6, R is image and by Theorem 4.2.1, there is a image-function f : ω → 2<ω uniformizing R. Let

image

Then C is a closed image-set and μ(C)≥ 2−1. Note that if z1Σ1 ∩ HYP and μ(Az1)> p for some p, then there is a finite zz1 in Σ1 ∩ HYP such that μ(Az)> p. By Proposition 14.1.2, every real in C is image-Kurtz random. Since C is closed and image, it can be presented by a image-tree. By Theorem 11.1.2, there is an xC such that xh image. Let e and image be such that image. Then

image

In other words, x is a image-singleton which is therefore not image-Kurtz random.

We now show that every image-Kurtz random real is image-Kurtz random. Let A be a image-open set of measure 1. Define x ={σ ∈ 2<ω |(∀y)(yσyA)}. Then x is a image-real coding A (i. e. yA if and only if there is a σx such that yσ). Hence there is a recursive function f : 2<ωω such that σx if and only if f(σ)∈ image. Define a image-relation Rω × ω so that (k, n)∈ R if and only if nimage and μ(⋃{[σ]|(∃mimagen)(f(σ)= m)}) > 1 − image. Clearly R is a image-relation which can be uniformized by a image-function f. Since μ(A)= 1, f is a total function. Thus the range of f is bounded by a notation nimage. Define B ={y |(∃σ)(yσf(σ)∈ imagen)}. Then BA is a image-open set with measure 1. It follows that every image-open co-nullset has a image-open co-null subset. Hence image-Kurtz randomness = image-Kurtz randomness.

The following result clarifies the connection between image and image-Kurtz randomness.

Proposition 14.1.4. If image = image, then x is image-Kurtz random if and only if x is image-Kurtz random.

Proof. Suppose that image = image and x is image-Kurtz random. If A is a image-closed nullset such that xA, then by the Spector–Gandy Theorem 3.7.3, there is an arithmetic formula φ(z, y) such that the formula (∃z ∈ HYP)φ(z, y) defines A. Since image = image, there is a recursive ordinal β such that

image

Define T = {σ ∈ 2<ω |(∃yB)(yσ)}. Clearly B ⊆ [T]. Since B is image, [T] is image. Since A is closed, we have [T] ⊆ A. Also A is null implies that [T] is null as well. By Lemma 13.1.10 and a routine application of image-uniformization, there is a total image-function f : ω → HYP such that for any k, f(k) ∈ Σ1 ∩ HYP, Af(k) ⊆ 2ω [T] and μ(Af(k))> 2k. Let

image

Then C is a image-closed nullset. However, xC and this is a contradiction.

14.1.2 Base of a cone of Kurtz random reals

From the proof of Theorem 14.1.3, one sees that every hyperdegree above that of image contains a image-Kurtz random real. This fails for image-Kurtz randomness. We say that a hyperdegree a is a base of a cone for Γ-Kurtz randomness if for every hyperdegree ba, b contains a Γ-Kurtz random real.

The hyperdegree of image is a base of a cone for image-Kurtz randomness as proved in Theorem 14.1.3. Not every nonzero hyperdegree is a base of a cone for image-Kurtz randomness (see Lemma 14.6.1). Does there exist a base of a cone for image-Kurtz randomness? If so, such a base, say b, is not hyperarithmetically reducible to a image-singleton. This means that a hyperdegree which serves as a base must be fairly complicated.

Lemma 14.1.5. For any reals x and zT x, there is an x-Kurtz-random real yT z.

Proof. Let zT x′ and fix an enumeration of x-r. e. open sets image. We construct an increasing sequence {σs}s<ω of strings as follows.

Let σ0 = image.

Stage s + 1. Let l0 = 0, l1 = |σs|, and image for n > 1. For each such n, let

image

Then

image

In other words,

image

Case 1. There is an m > l1 + 1 such that image. Let n = m + 1. Then ln+1 − 1 − ln > 2 and ln > m. Then there is a image and a τ image σ such that [τ] ⊆ image and τ ∈ 2m.

Let image.

Case 2. Otherwise. Let image.

This completes the construction at stage s + 1. Let image. Obviously the construction is recursive in z and so yT z. Moreover, if image is of measure 1, then Case 1 holds at stage s + 1. Hence y is x-Kurtz random.

Let l0 = 0 and let image for n > 0. To compute z(n) from y, we y-recursively find the n-th lm such that for all i, j with lmi < j < lm+1, y(i) = y(j). Then z(n) = y(lm).

We call Aω × 2ω auniversal image-closed set if (i) it is image, (ii) for every n, An = {x |(n, x)∈ A} is a image-closed set, and (iii) every image-closed set is An for some n. By Theorems 13.1.3 and 13.1.4, the real x0 = {n | μ(An) = 0} is image. Let

image

Then c may be viewed as a image-real. Since every image-null closed set is image(c), every c-Kurtz-random real is image-Kurtz random.

Theorem 14.1.6. The hyperdegree of c is the base of a cone for image-Kurtz randomness.

Proof. Lemma 14.1.5 implies that the Turing degree of any yT c′ contains a member which is c-Kurtz random and therefore image-Kurtz random. Thus c is a base for image-Kurtz randomness.

We analyze the complexity of c. Since every image-singleton is recursive in image Moreover, the set {x | x is a image-singleton} is image(c). In other words, {x | x is a image-singleton} image. Hence one has image > image. Applying the same argument as that in Proposition 4.4.4, the reals in image are precisely those that are image and so c is not image. Moreover, since c is image, it is Σ1 definable over image and hence cimage. Indeed, all image-reals belong to image. Thus:

c has the largest hyperdegree among all image reals.

14.1.3 Exercises

Exercise 14.1.1. Prove that neither the set of image nor image-Kurtz random reals are image

Exercise 14.1.2. Prove that there is a image-Kurtz random real which is not Schnorr random.

Exercise 14.1.3. Prove that the set of image-Kurtz random reals is not image.

Exercise 14.1.4. Prove that image.

14.2 image and image-Martin-Löf randomness

14.2.1 image-randomness

Definition 14.2.1. Let Γ be a definable class of sets of reals. We say that x is Γ-random if it does not belong to any Γ-nullset.

In this section, we study the key properties of image-randomness.

Proposition 14.2.2. Suppose that A is a image-nullset. There is a image-set Vω × 2<ω such that (∀n)(μ(Vn) = 2n) and A ⊆ ⋂n<ω Vn, where Vn = {σ |(n, σ)∈ V)} is a image-open set.

Proof. As before, {Az} is the collection of sets of reals generated by Borel codes z2.7.1). Since A is a image-nullset, by Lemma 13.1.10 there is a image-function f : ωΣ1 ∩ HYP such that for every n < ω,

μ(Af(n)) < 2n, and

AAf(n)

Then Af(n) = [Un] for some image-set Un ⊆ 2<ω. It now suffices to show that {Un}n<ω may be expanded to a image-collection {Vn}n<ω such that UnVn and μ(Vn) = 2n. Since the set U = {(n, σ) | σUn} is image, there is an mimage such that UT Hm. Note that the set image is r. e. Then by a standard technique in classical randomness theory relativized to image, there is a Vω × 2<ω recursive in image such that for every n, μ(Vn) = 2n and UnVn. Then image

By Proposition 14.2.2, image-randomness may be viewed as a higher analog of Schnorr randomness.

Proposition 14.2.3 (Martin-Löf [97]). The set of image-random reals is image but not image.

Proof. Define a image-set

image

By Proposition 14.2.2, A is precisely the set of image-random reals. Since A does not contain a hyperarithmetic member, by Theorem 13.1.13, it is not image.

14.2.2 image-Martin-Löf randomness

A sequence of open sets {Un}n<ω is a Martin-Löf test (ML test) if μ(Un) < 2n for all n. Given a class Γ of sets of reals (e. g. image), {Un}n<ω is a Γ-ML test if {(n, σ)|σ ∈ 2<ω ∧[σ]∈ Un}∈ Γ.

Definition 14.2.4. Let Γ be a class of sets of reals. Then x is Γ-ML random if image image for any Γ ML test {Un}n<ω.

Proposition 14.2.5 (Hjorth and Nies [51]). There is a universal image-ML test {Un}n<ω, i. e. for every image-ML test image

Proof. Let Ûω × 2<ω be a universal image-set, i. e. Û is image and every image-set image is Ûn = {σ |(n, σ)∈ Û} for some n. Then there is a recursive function f : ω × 2<ωω such that (n, σ)∈ Û ⇔ f(n, σ)∈ image. Let

image

In other words, (n, σ) is enumerated into ̂V if the measure of ̂Un up to stage f(n, σ) is less than 2n. Let

image

Then U is image as well. Setting Un = {σ |(n, σ)∈ U}, we have

image

Hence {Un}n<ω is a image-ML test. By the universality of Û, {Un}n<ω is a universal image-ML test

Corollary 14.2.6 (Hjorth and Nies [51]).

(i)There is a image-closed set of positive measure of image-ML random reals.

(ii)For any zh image, there is a image-ML random xh z.

Proof. (i) follows immediately from Proposition 14.2.5, while (ii) is a consequence of (i) and Theorem 11.1.2.

The following results says that every image-closed set of positive measure essentially contains all the image-ML random reals.

Proposition 14.2.7. Suppose that T ⊆ 2<ω is a image-tree with μ([T]) > 0. Then for any image-ML random x, there is a z ∈[T] such that z = x, i. e. (∃m)(∀nm)(x(n) = z(n)).

Proof. Let T ⊆ 2<ω be as given. For any n ≥ 0, let

image

Then {Un}n<ω is uniformly image. Moreover, μ(U0) = μ(2ω [T]) = 1 − μ([T]). For any n, μ(Un+1]) = μ(Un)⋅(1 − μ([T])) = (1 − μ([T]))n+2. Let image. Then μ(Vn) ≤ ∑mn (1 − μ([T]))m+1 and hence {Vn}n<ω can be viewed as a image-ML test. It follows that there is a least n such that x image Vn. If n = 0, then x ∈ [T]. Otherwise there is a σ0σ1 σn−1x so that x = σ0σ1 σn−1z for some z ∈ [T]. Hence x = z.

14.2.3 Separating image-Martin-Löf randomness from image-randomness

Define a notion of forcing image = 〈D, ≤〉 as follows:

(i)PD if and only if P ⊆ 2ω is image-closed and of positive measure;

(ii)for any P, QD, PQ if and only if PQ.

Suppose Uω × 2<ω is image and Un = {σ |(n, σ)∈ U}. Assume limn→∞ μ(Un) = 0 and let

image

Lemma 14.2.8. imageU is dense.

Proof. Let PD. There is an n such that μ(Un) < μ(P)/2. Now the complement P0 = 2ω Un has measure greater than or equal to 1 −(μ(P)/2). Then P0P is a image-closed set and has measure greater than or equal to μ(P)/2. Thus PP0D.

For any image set G, let image = {P | PDPG = image}.

Lemma 14.2.9. If G is a image-set consisting only of image-random reals, then image is dense in image

Proof. Let PD. Then there is a hyperarithmetic x such that P is image(x). Without loss of generality, we may assume that for any σ, if [σ]∩ P ≠ 0, then μ([σ]∩ P)> 0 (since we may assume that P contains only 1-x-random reals). By Theorem 13.1.13, there is a hyperarithmetic z in P. Hence z image G. Since G is closed, there is an n such that [zn]∩ G = 0. Let Q = [zn]∩ P. Then Qimage.

Theorem 14.2.10 (Chong, Nies and Yu [12]). image-ML randomnessimage-randomness.

Proof. By Proposition 14.2.2, image-ML randomness ⊆ image-randomness. By Proposition 14.2.5, let {Un}n<ω be a universal image-ML test. For n < ω, let

image

By Lemma 14.2.9, image is dense. By Lemma 14.2.8, there is a image-random x such that for any n, there is a Pimage with xP. Hence image, i. e. x is not image-ML random.

We remark that image is an analog of random forcing ℝ discussed in § 6.2.2. However, ℝ is c. c. c. and so preserves all cardinals while image is not c. c. c. in image. By the proof of Theorem 14.2.10 and Corollary 14.3.2 below, any image-generic x collapses image and hence does not preserve admissible ordinals.

14.2.4 Strong image-Martin-Löf randomness

A sequence of open sets {Un}n<ω is a generalized Martin-Löf test (ML test) if limn→∞ μ(Un) = 0. Given a class Γ of sets of reals (e. g. image), {Un}n<ω is a generalized Γ-ML test if {(n, σ)| σ ∈ 2<ωσUn}∈ Γ.

Definition 14.2.11. Given a class Γ of sets of reals, x is strongly Γ-ML random if x image image for any generalized Γ-ML test {Un}n<ω.

Obviously every strongly image-ML random real is image-ML random.

Lemma 14.2.12. If x is the leftmost path of a image-closed set of reals, then x is not strongly image-ML random.

Proof. Let P be a image-closed set. Then there is a image-tree T ⊆ 2<ω such that P = [T]. By the Gandy–Spector Theorem (3.7.1), there is a Σ0-formula φ(σ, β) such that

image

For β < image, let

image

For n < ω and β < image, let

image

Define

image

and

image

The following facts are straightforward to verify: For n < ω and β < image,

(1) Un+1,βUn,β;

(2) μ(Un,β) < 2n;

(3) x ∈ ⋂n<ω Un;

(4) for any z, if zUn,<β Un,β, then z image Un,γ for any γβ.

Now suppose that image. Then there is a σ0 such that

image

Let n0 = |σ0|+ 2. Then there is a β0 < image such that

image

By (2),

image

By (1) and (4),

image

Hence there is a σ1σ0 such that

image

Let n1 = |σ1|+ 2. There is a β1 < β0 such that

image

Repeating the process, one obtains an infinite descending sequence β0 > β1 > … of ordinals which is a contradiction. It follows that {Un}n<ω is a generalized image-ML test and image.

Theorem 14.2.13 (Chong, Nies and Yu). There is a real which is image-ML random but not strongly image-ML random.

Proof. By Corollary 14.2.6, there is image-closed set P of positive measure consisting only of image-ML random reals. By Lemma 14.2.12, there is an xP which is not strongly image-ML random.

14.2.5 Exercises

Exercise 14.2.1 (Sacks [123]). Prove that every image-random real is image-random.

Exercise 14.2.2. Prove that for any image-random x and recursive ordinal β, x(β)T x⊕0(β).

Exercise 14.2.3 (Yu [171]). Prove that the set of image-random reals is not image.

Exercise 14.2.4. Prove that for any image-random x and recursive function f :2ω → 2ω, if f is almost everywhere differentiable, then f is differentiable at x.

Exercise 14.2.5 (Bienvenu, Greenberg and Monin). Prove that for any x which is the leftmost path of a image-closed set of reals, there is a generalized image-Martin-Löf test {Un}n<ω such that image image is countable.

Exercise 14.2.6 (Chong and Yu). Prove that for any xh image, there is a zh x which is image-Martin-Löf random but not strongly image-Martin-Löf random.

14.3 image-randomness

14.3.1 The largest image-nullset

The notion of image-randomness was introduced in Hjorth and Nies [51] and is unique to the study of higher randomness. There is no analog of this notion in the subject of classical randomness theory.

Theorem 14.3.1 (Kechris [64]; Stern [161]; Hjorth and Nies [51]). There is a largest image-nullset.

Proof. Define

image

and

image

Define

image

We show that image is the largest image-nullset. By Theorem 13.1.1, image is image. Moreover, μ(imagen) = 0 for all n < ω. By Theorem 13.1.4, μ(image) = 0.

If A is a image-nullset, then, by Corollary 3.7.4, there is an arithmetic formula φ( ̇x,z) such that if image = image, then xA ⇔ A(image, x) image (∃y)φ(x, y). Hence if image = image, then xA ⇔ A(image, x) image (∃yβ)φ(x, yβ) for some β < image. Since A is null, we have

image

Thus Aimage.

Corollary 14.3.2. For any x ∈ 2ω, the following are equivalent:

(i)x is image-random;

(ii)x is strongly image-ML random and image = image;

(iii)x is image-ML random and image = image;

(iv)x is image-random and image = image.

Proof. Suppose that x is image-random. Since {z | image > image} is a image-nullset, we have image = image and hence (i) ⇒ (ii) ⇒ (iii).

By Theorem 14.2.10, (iii) implies (iv). We now show (iv) ⇒ (i). By the proof of Theorem 14.3.1, if image = image and ximage, then x imagen for some n. Since Qn is a image-nullset, if image = image and x is image-random, then it is image-random.

The equivalence between (i) and (iv) was also obtained by Stern [161].

14.3.2 Domination and image-randomness

Lemma 14.3.3. If x is image-random, then it is image-dominated.

Proof. By the proof of Theorem 13.3.2, for each eω and nimage, the set

image

has measure 1. Note that Ae,n is image. By Corollary 13.1.11, there is a image-set e,nAe,n such that μ(Ae,n) = 1. Hence if x is image-random, then xAe,n. By Corollary 14.3.2, image = image. Thus if gh x, then image for some e and nimage, proving the lemma.

Proposition 14.3.4 (Kjos-Hanssen, Nies, Stephan and Yu [68]). For any x ∈ 2ω, the following are equivalent:

(i)x is image-random;

(ii)x is strongly image-ML random and image-dominated;

(iii)x is image-ML random and image-dominated;

(iv)x is image-random and image-dominated;

(v)x is image-Kurtz random and image-dominated.

(vi)x is image-Kurtz random and image-dominated.

Proof. By Corollary 14.3.2 and Lemma 14.3.3, we have (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi).

To prove (vi) ⇒ (i), it is sufficient to prove (vi) ⇒ (iv) according to Corollary 14.3.2. Suppose x is image-dominated and image-Kurtz random. Let {Un}n<ω be a image-ML test. Then there is a recursive function f : ω × 2<ωω such that for any pair (n, σ), σUn if and only if f(n, σ)∈ image. Since x is image-dominated, image = image. Define a image(x) relation Rω × ω such that

image

where image. By the image-Uniformization Theorem (4.2.1), there is a partial function p uniformizing R. Since ximage Un, p is a total function and so image(x). Since image = image, there is an m0image such that p(n)∈ image for every n. Define a image-Martin-Löf test {̂Un}n<ω so that σ̂Un if and only if f(n, σ)∈ image. Then ximage ̂Un. Let

image

be a image(x)-function. Then there is a image-function f dominating ̂f. For n < ω, define

image

Then P = ⋂n<ω Vn is a image-closed set and xP. Hence x is not image-Kurtz random, a contradiction.

14.3.3 Hyperdegrees of random reals

By Corollary 14.2.6, we know that the hyperdegrees of the set of image-random reals include the cone of hyperdegrees with base image. This raises the natural question of whether the hyperdegrees of image-random reals are closed upwards.

Proposition 14.3.5. If x is image-random and image = image, then there is a yh x such that no zh y is image-random.

Proof. Suppose that x is image-random and image = image. Let

image

Then F(x) is image(x). Since imagexF(x), F(x) is not empty. By Gandy’s Basis Theorem relativized to x, there is a yF(x) with ωy1 = image = image. By Corollary 14.3.2 and Proposition 14.3.4, no zh y is image-random.

On the other hand, the hyperdegrees of image-random reals are closed downwards.

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