# Finite and Infinite Arithmetic Progressions Related to Beta-Expansion

Authors: Bing Li; Chao Ma
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Let1<β<2 and ε(x,β) be the β-expansion of x∈[0,1). Denote by Aβ(x) the set of positions where the digit 1 appears in ε(x,β). We consider the sets of points x such that Aβ(x) contains arbitrarily long arithmetic progressions and includes infinite arithmetic progressions, respectively. Their sizes are investigated from the topological, metric, and dimensional viewpoints.


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## Body

## 1. Introduction

LetA be a sequence of natural numbers. The lower asymptotic density ρ_(A) of A (the upper asymptotic density ρ¯(A) of A, resp.) is defined as
(1)lllρ_(A)=liminf⁡N→∞∑n≤N,n∈A1N,(ρ¯(A)=limsup⁡N→∞∑n≤N,n∈A1N,resp.).
If ρ_(A)=ρ¯(A), the common value is called the asymptotic density of A, denoted by ρ(A).The concept of asymptotic density was extensively applied to number theory such as arithmetic progressions, the first of which is the following theorem due to Szemerédi [1].Theorem 1 (see [1]).
Ifρ¯(A)>0, then the sequence A contains arbitrarily long arithmetic progressions.The issue to find what kind of sequenceA can contain arbitrarily long arithmetic progressions is very popular nowadays. For instance, Green and Tao [2] proved that the sequence P of prime numbers contains arbitrarily long arithmetic progressions, which extends Theorem 1 since ρ(P)=0.Assume that the sequenceA contains arbitrarily long arithmetic progressions; a further question is whether A can contain an infinite arithmetic progression. In 1972, Wagstaff Jr. [3] found a sequence of natural numbers which contains arbitrarily long arithmetic progressions but does not have any infinite one. In fact, he even gave more information as the following.Theorem 2 (see [3]).
For eacha,b∈[0,1] with a≤b, there exists a sequence of natural numbers A such that A contains no infinite arithmetic progression and ρ_(A)=a and ρ¯(A)=b.The topics of arithmetic progressions and binary expansion were firstly linked by Šalát and Tomanová [4]. They considered the sequence of the positions where the digit 1 appears in the binary expansion of real numbers in the unit interval. In this paper, we investigate the same questions for β-expansion with 1<β≤2 and generalize the results in [4].Let1<β≤2. The β-transformation Tβ:[0,1)→[0,1) is defined as Tβ(x)=βx  (mod 1) for any x∈[0,1). Consider the greedy β-expansion of x:(2)x=ɛ1(x,β)β+ɛ2(x,β)β2+⋯+ɛn(x,β)βn+⋯,
with ɛ1(x,β)=[βx] and ɛn(x,β)=ɛ1(Tβn-1(x),β) for all n≥2, called β-digits of x. Here [y] denotes the integer part of y∈R. It is clear that all the digits ɛn(x,β)∈{0,1} for any x∈[0,1) and n≥1.Define(3)Aβ(x)={n≥1:ɛn(x,β)=1},
that is, the set of natural numbers with ɛn(x,β)=1. In other words, Aβ(x) is the sequence of positions where the digit 1 appears in the β-expansion of x.Denote(4)Fβ={x∈[0,1):thesequenceAβ(x)containslllllarbitrarilylongarithmeticprogressionsx∈[0,1)Aβ(x)},Iβ={x∈[0,1):thesequenceAβ(x)containslllllinfinitearithmeticprogressionsx∈[0,1)Aβ(x)}.Obviously,Iβ⊂Fβ since an infinite arithmetical progression contains arbitrarily long ones. In Section 2, we will give some introduction to β-expansion. In Sections 3 and 4, we will describe the sets Fβ, Iβ and their complements Fβc, Iβc from the viewpoints of topology, metric, and Hausdorff dimension.


## 2. Preliminary forβ-Expansion

Whenβ=2, any sequence in {0,1}N except countably many elements is a β-expansion of some point x∈[0,1). It is not this case when 1<β<2 is not an integer. Denote
(5)Σβ={ω∈{0,1}N:ωisllllllllllaβ-expansionofsomex∈[0,1){0,1}N}.
A 0-1 sequence ω is said to be admissible if ω∈Σβ and a 0-1 word (ɛ1,…,ɛn) is admissible if there exists one point x∈[0,1) such that ɛ1(x,β)=ɛ1,…,ɛn(x,β)=ɛn. Denote by Σβn the set of all admissible words of length n. Let Sβ be the closure of Σβ and then (Sβ,σ) is a subshift of ({0,1}N,σ), where σ is the shift transformation of the symbolic space {0,1}N. Parry [5] gave a criterion to check if a sequence in {0,1}N is admissible where the β-expansion of the number 1 (defined similarly with (2)) plays an essential role. If the β-expansion of 1, denoted by ɛ(1,β), is finite, that is, ending with zeros, we call the modified periodic sequence ɛ*(1,β):=(ɛ1(1,β),…,ɛn-1(1,β),ɛn(1,β)-1)∞ the infinite β-expansion of 1, where ɛn(1,β)≠0 is the last nonzero element. If ɛ(1,β) is infinite, namely, that there are infinitely many 1s in the sequence ɛ(1,β), then just let ɛ*(1,β)=ɛ(1,β). We use the notation ≺ to represent the lexicographical order in the symbolic space {0,1}N.Theorem 3 (see [5]).
Letω∈{0,1}N. Then
(6)ω∈Σβiffσkω≺ɛ*(1,β),∀k≥0.For the caseβ=2, Sβ={0,1}N; that is, the coding space is the full shift space. When 1<β<2, Sβ is a proper subset of {0,1}N. If ɛ(1,β) is finite, Sβ is a subshift of finite type (SFT) which can be obtained by forbidding finitely many words. The set of β’s with Sβ being SFT is dense in (1,2] (see [5]). If 1<β1<β2, then Σβ1⊂Σβ2 and Sβ1⊂Sβ2, which means that every β1-admissible sequence is β2-admissible.Definition 4.
Let(ɛ1,…,ɛn)∈Σβn. The set
(7)J(ɛ1,…,ɛn):={x∈[0,1):ɛ1(x,β)=ɛ1,…,ɛn(x,β)=ɛn}
is called a cylinder of order n.Any cylinderJ(ɛ1,…,ɛn) is a left-closed and right-open interval and its length
(8)|J(ɛ1,…,ɛn)|≤β-n.
We say that a cylinder J(ɛ1,…,ɛn) is full if TβnJ(ɛ1,…,ɛn)=[0,1); that is, any admissible sequence can be concatenated with the word (ɛ1,…,ɛn). In other words, (ɛ1,…,ɛn,ω)∈Σβ if ω∈Σβ. We remark that |J(ɛ1,…,ɛn)|=β-n if J(ɛ1,…,ɛn) is full. The following lemma, due to Li and Wu [6], will be used to prove Proposition 6.Lemma 5 (see [6]).
Let(ɛ1,…,ɛn)∈Σβn. Then there exists M>0 such that the cylinder J(ɛ1,…,ɛn,0M) is full, where M just depends on n and β; 0M means the word of M consecutive zeros.Whenβ=2, the Lebesgue measure L is a Tβ-invariant measure. If β is not an integer, the Lebesgue measure L is not Tβ-invariant; however Rényi [7] proved that there exists a unique invariant measure μβ equivalent to L. Parry [5] and Gel’fond [8] independently gave the density formula of μβ and L.The functionɛ(·,β):x↦ɛ(x,β) maps x∈[0,1] to its β-expansion. Conversely there is a projection πβ from the subshift Sβ to [0,1] defined as πβ(ω)=∑i=1∞(ωi/βi)‍ for any ω=(ω1,…,ωi,…)∈Sβ. In particular, πβɛ(x,β)=x. We have the following commutative diagram; that is, Tβ∘πβ=πβ∘σ:


## 3. Topological Properties ofFβ and Iβ

LetNn(x,β)=∑k=1nɛk(x,β) be the sum of first n digits of the β-expansion of x∈[0,1]. Define
(9)E={x∈[0,1):lim⁡n→∞Nn(x,β)n=0}.Proposition 6.
The setE is of the first category.Proof.
The setE can be written as
(10)E=⋂k=1∞‍⋃N=1∞‍⋂n=N∞‍{x∈[0,1):Nn(x,β)n<1k}=:⋂k=1∞‍⋃N=1∞‍⋂n=N∞‍Ek,n.
For any k,n∈N, the set Ek,n is the union of some cylinders J(ɛ1,…,ɛn) of order n satisfying ɛ1+⋯+ɛn<n/k. Note that the cylinder J(ɛ1,…,ɛn) is a left-closed and right-open interval; denote by J¯(ɛ1,…,ɛn) its closure, which is obtained by adding the right endpoint of J(ɛ1,…,ɛn). Let
(11)E¯k,n=⋃ɛ1+⋯+ɛn<n/k‍J¯(ɛ1,…,ɛn).
Choose K≥1 large enough such that there exists x0∈[0,1) with lim⁡n→∞(Nn(x0,β)/n)>1/K. Such existence is because the sequence u:=(10m)∞ is admissible where m≥2 with ɛm*(1,β)=ɛ1*(1,β)=1 and lim⁡n→∞(Nn(πβ(u),β)/n)=1/(m+1)>0. Notice that
(12)E⊂⋃N=1∞‍⋂n=N∞‍E¯K,n.
Since the set E¯K,n is closed, ∩n=N∞E¯K,n is a closed set; it suffices to show that the interior of ∩n=N∞E¯K,n is empty. Suppose it is not empty; then there exists one interval inside ∩n=N∞E¯K,n, which implies that there exists some cylinder J(ɛ1,…,ɛm) lying in this interval and then J(ɛ1,…,ɛm)⊂∩n=N∞E¯K,n. Thus there exists M∈N only depending on m and β such that the cylinder J(ɛ1,…,ɛm,0M) is full by Lemma 5. Therefore the sequence (ɛ1,…,ɛm,0M,ɛ1(x0,β),…,ɛi(x0,β),…) is admissible; put
(13)y=πβ(ɛ1,…,ɛm,0M,ɛ1(x0,β),…,ɛi(x0,β),…)∈J(ɛ1,…,ɛm).
Thus y∈∩n=N∞E¯K,n which implies limsup⁡n→∞(Nn(y,β)/n)≤1/K contradicting
(14)lim⁡n→∞Nn(y,β)n=lim⁡n→∞Nn(x0,β)n>1K.
So we complete the proof.Remark 7.
By the similar argument with the proof of Proposition6, we can prove that the set
(15)Eα={x∈[0,1):lim⁡n→∞Nn(x,β)n=α}
is of the first category for any 0≤α≤1. When 1<β<2, the set Eα may be empty. For instance, β=(5+1)/2; the set Σβ can be characterized as that
(16)ω∈Σβifftheword11doesnotappearinω;ωdoesnotendwith101010….
So lim⁡sup⁡n→∞(Nn(x,β)/n)≤1/2 for any x∈[0,1], which indicates Eα=∅ if α>1/2.Theorem 8.
The setFβ is residual in [0,1).Proof.
Firstly we claim thatFβc⊂{x∈[0,1):ρ(Aβ(x))=0}. Indeed, if ρ(Aβ(x))>0, by Theorem 1, we know that the sequence Aβ(x) contains arbitrarily long arithmetical progressions; that is, x∈Fβ.
Secondly, combining the facts that(17)Fβc⊂{x∈[0,1):ρ(Aβ(x))=0}={x∈[0,1):lim⁡n→∞Nn(x,β)n=0}=E
and the set E is of the first category (Proposition 6), we know that the set Fβc is of the first category. So the set Fβ is residual.Theorem 9.
The setIβ is of the first category in [0,1).Proof.
LetAβ(x)={a1(x)<a2(x)<a3(x)<⋯<an(x)<⋯} and
(18)B={{an+1(x)-an(x)}n≥1x∈[0,1):thegap⁡sequencelllll{an+1(x)-an(x)}n≥1isbounded}.
Then Iβ⊆B. Note that B=⋃m=1∞Bm, where
(19)Bm={x∈[0,1):|an+1(x)-an(x)|≤m,foranyn≥1}.
It suffices to show that, for any m≥1, the set Bm is nowhere dense. That is, for any interval I⊂[0,1), there exists an interval I′⊂I such that I′∩Bm=∅. Indeed, for given I⊂[0,1), we can find a cylinder J(ɛ1,ɛ2,…,ɛn) such that J(ɛ1,ɛ2,…,ɛn)⊂I when n is large enough. Put I′=J(ɛ1,ɛ2,…,ɛn,0m+1). Obviously I′⊂I. Since, for any x∈I′, the word of m+1 consecutive zeros appears in the β-expansion of x, we have I′∩Bm=∅. Therefore Bm is nowhere dense.The combination of Theorems8 and 9 implies the following.Corollary 10.
The setFβ∖Iβ is residual in [0,1).


## 4. Metric and Dimensional Properties ofFβ and Iβ

Theorem 11.
L
(
F
β
)
=
1 and L(Iβ)=0.Proof.
Note that∑k=1n‍ɛk(x,β)=∑k=0n-1‍χ[1/β,1)(Tβkx), where χ is the indicator function. By Birkhoff’s ergodic theorem, we have, for μβ-almost every (a.e.) x∈[0,1),
(20)lim⁡n→∞∑k=1nɛk(x,β)n=∫01‍χ[1/β,1)(x)dμβ(x)=μβ([1β,1))>0.
Note that μβ is equivalent to the Lebesgue measure L; then, for L—a.e. x, the asymptotic density of Aβ(x)(21)ρ(Aβ(x))=μβ([1β,1))>0;
that is, L{x∈[0,1):ρ(Aβ(x))=μβ([1/β,1))}=1. By Theorem 1, we know
(22){x∈[0,1):ρ(Aβ(x))=μβ([1β,1))}⊂Fβ.
Thus L(Fβ)=1.
Leta,d≥1 be natural numbers. Define
(23)Iβ(a,d):={x∈[0,1):ɛa+kd(x,β)=1∀k≥0}.
Then Iβ=∪(a,d)Iβ(a,d), where (a,d)∈N×N. To prove that L(Iβ)=0, it suffices to show that L(Iβ(a,d))=0 for any (a,d)∈N×N. In fact, let
(24)Iβ(a,d,m)={x∈[0,1):ɛa+kd(x,β)=1∀0≤k≤m}
and then
(25)Iβ(a,d)=⋂m=0∞‍Iβ(a,d,m).
The set Iβ(a,d,0) consists of finitely many cylinders J(ɛ1,…,ɛa-1,ɛa) with ɛa=1. Now we estimate the length of J(ɛ1,…,ɛa-1,ɛa) by the following two cases.
(
i
) If the word (ɛ1,…,ɛa-1,1) is admissible, then the cylinder J(ɛ1,…,ɛa-1,0) is full by Lemma 2.9(i) in [6] and consequently |J(ɛ1,…,ɛa-1,0)|=β-a. Since |J(ɛ1,…,ɛa-1,0)|+|J(ɛ1,…,ɛa-1,1)|=|J(ɛ1,…,ɛa-1)|, |J(ɛ1,…,ɛa-1)|≤β-(a-1), and 1<β≤2, we have
(26)|J(ɛ1,…,ɛa-1,1)||J(ɛ1,…,ɛa-1)|=1-β-a|J(ɛ1,…,ɛa-1)|≤1-β-aβ-(a-1)=1-1β≤12.
That is, |J(ɛ1,…,ɛa-1,1)|≤(1/2)|J(ɛ1,…,ɛa-1)|.
(
ii
) If the word (ɛ1,…,ɛa-1,1) is not admissible, then J(ɛ1,…,ɛa-1,0)=J(ɛ1,…,ɛa-1), which is not contained in Iβ(a,d,0).
Therefore,(27)L(Iβ(a,d,0))=∑(ɛ1,…,ɛa-1,1)∈Σβa‍|J(ɛ1,…,ɛa-1,1)|≤12∑(ɛ1,…,ɛa-1)∈Σβa-1‍|J(ɛ1,…,ɛa-1)|≤12.
Whenm=1, note that Iβ(a,d,1)⊂Iβ(a,d,0) and Iβ(a,d,1) consists of the cylinders J(ɛ1,…,ɛa-1,1,ɛa+1,…,ɛa+d-1,1) with (ɛ1,…,ɛa-1,1,ɛa+1,…,ɛa+d-1,1)∈Σβa+d. A similar argument with the estimate of L(Iβ(a,d,0)) implies
(28)∑(ɛ1,…,ɛa-1,1,ɛa+1,…,ɛa+d-1,1)∈Σβa+d‍J(ɛ1,…,ɛa-1,1,ɛa+1,…,ɛa+d-1,1)≤12|J(ɛ1,…,ɛa-1,1)|
for any (ɛ1,…,ɛa-1,1)∈Σβa. Therefore,
(29)L(Iβ(a,d,1))≤12L(Iβ(a,d,0))≤122.
Repeating this argument, we obtain
(30)L(Iβ(a,d,m))≤12m+1,
which implies L(Iβ(a,d))=0 together with (25). Then we complete the proof.The following is a consequence of Theorem11.Corollary 12.
ConsiderL(Fβ∖Iβ)=1.Lemma 13.
Let1<β≤2 such that Sβ is a SFT. Then dim⁡HIβ=1.Proof.
SinceSβ is a SFT, there exists N∈N such that it is enough to forbid finitely many words of length N-1. Note that Iβ⊃Iβ(1,d) for any d∈N. Now we construct a Cantor subset F of Iβ(1,d) with large Hausdorff dimension when d is large.
Letd′=d+1 with d>2N and define
(31)F={ɛ2+d-N+kd′…ɛ1+d+kd′=0N∀k≥0x∈[0,1):ɛ1+kd′=1,lllllɛ2+kd′…ɛN+1+kd′=ɛ2+d-N+kd′…ɛ1+d+kd′=0N∀k≥0},
where ɛi means ɛi(x,β) for simplicity. Then the set F can be regarded as a Cantor-like set as the following: the interval F0 of level 0 is [0,1) and the set Fi+1 of level i+1 consists of the cylinders J(1,0N,ɛN+2…ɛd-N+1,0N,…,1,0N,ɛN+2+id′…ɛd-N+1+id′, 0N) for any i≥0, where ɛN+2+jd′…ɛd-N+1+jd′ is any admissible word for 0≤j≤i. Arrange such cylinders from left to right and denote them by Ji,l,1≤l≤iM, where M is the number of admissible words of length d-2N; that is, M=#Σβd-2N. Then
(32)Fi=⋃l‍Ji,l,F=⋂i=0∞‍Fi.
LetJi,l be any cylinder of level i and let Ji-1,l' be its father cylinder of level i-1. Note that both Ji,l and Ji-1,l' are full cylinders; we know that their lengths are β-i(d+1) and β-(i-1)(d+1), respectively. So the ratio between |Ji,l| and |Ji-1,l'| is β-(d+1). Therefore F is a self-similar set. Obviously F satisfies strong separation condition (SSC). By the Hausdorff dimension formula for the self-similar set with SSC (see Falconer [9], Page 130), we know that dim⁡HF is the unique solution s of the equation
(33)∑l=1M‍(β-(d+1))s=1.
That is, dim⁡HF=(log⁡M)/(d+1)log⁡β. Note that M≥βd-2N; then
(34)dim⁡HIβ≥dim⁡HI(1,d)≥dim⁡HF≥d-2Nd+1.
We obtain dim⁡HIβ=1 by letting d→∞.Theorem 14.
Consider(i)
dim⁡HFβc=0;(ii)
dim⁡HIβ=1.Proof.
(
i
) Since Fβc⊂{x∈[0,1):lim⁡n→∞(Nn(x,β)/n)=0}=:E, it suffices to show that the Hausdorff dimension of the right-hand side set is zero. In fact, note that Nn(x,β)=∑k=0n-1‍χ[1/β,1)(Tβkx); by the main theorem in [10], we have
(35)dim⁡HE=sup⁡{hμ(Tβ)log⁡β:μisTβ-invariant,llllllllllllllllllllllllllllllllllllllμ([1β,1))=0hμ(Tβ)log⁡β},
where hμ(Tβ) is the measure-theoretical entropy of Tβ with respect to μ. Since the Tβ-invariant measure μ with μ([1/β,1))=0 is just the Dirac measure at the origin 0, the measure-theoretical entropy hμ(Tβ)=0, which implies dim⁡HE=0. Therefore dim⁡HFβc=0.
(
ii
) Let 1<βn<β be a sequence of numbers satisfying Sβn which are SFTs and βn→β when n→∞. Note that Σβn⊂Σβ; write Hββn=πβ(Σβn). Define the function h:Hββn→[0,1) as h(x)=πβn(ɛ(x,β)) for any x∈Hββn.
We claim thath(Iβ(1,d)∩Hββn)=Iβn(1,d). Indeed, on the one hand, for any x∈Iβ(1,d)∩Hββn, we know ɛ(h(x),βn)=ɛ(x,β) by the definition of the function h. Note that x∈Iβ(1,d); then h(x)∈Iβn(1,d), which leads to h(Iβ(1,d)∩Hββn)⊂Iβn(1,d). On the other hand, for any y∈Iβn(1,d), take x=πβ(ɛ(y,βn))∈Hββn. Clearly h(x)=y and ɛ(x,β)=ɛ(y,βn), which implies x∈Iβ(1,d). So Iβn(1,d)⊂h(Iβ(1,d)∩Hββn).
By Theorem3.1(4) in [11], the function h is (log⁡βn/log⁡β)-Hölder which implies
(36)log⁡βnlog⁡βdim⁡HIβn(1,d)=log⁡βnlog⁡βdim⁡Hh(Iβ(1,d)∩Hββn)≤dim⁡H(Iβ(1,d)∩Hββn).
Since Iβ(1,d)∩Hββn⊂Iβ(1,d), we have
(37)dim⁡HIβ(1,d)≥log⁡βnlog⁡βdim⁡HIβn(1,d),
which concludes the result by letting d→∞ and then n→∞, together with (34) in Lemma 13.



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