Bernoulli decomposition and arithmetical independence between sequences

In this paper we study the following set\[A=\{p(n)+2^nd \mod 1: n\geq 1\}\subset [0.1],\] where $p$ is a polynomial with at least one irrational coefficient on non constant terms, $d$ is any real number and for $a\in [0,\infty)$, $a \mod 1$ is the fractional part of $a$. By a Bernoulli decomposition method, we show that $A$ must have box dimension equal to one.


INTRODUCTION
This paper is a self-contained sequel of [Y18a]. We use the idea of Bernoulli decomposition to consider a number theoretic problem. Given two sequences x = {x n } n≥1 , y = {y n } n≥1 in [0, 1], it is often of great interest to study their independence. In terms of sequences with dynamical background, this can be also understood as the disjointness between dynamical systems, see [F67] for more details. Intuitively, we want to say that two sequences x, y are independent if {(x n , y n )} n≥1 is in some sense close to the product set X × Y, where X, Y are the sets of numbers in the sequence x, y respectively. We give two natural ways of expressing this idea. Definition 1.1. Let x = {x n } n≥1 , y = {y n } n≥1 be two sequences in [0,1]. We write X, Y to be the sets of numbers in the sequence x, y respectively. Then we say that x and y are: • topologically independent if X × Y is equal to the closure of the set of elements in the sequence {(x n , y n )} n≥1 . • arithmetically independent if the set H(x, y) of numbers in the sequence {x n + y n } n≥1 attains the maximal box dimension, namely, dim B H(x, y) = min{1, dim B X + dim B Y }.
As an easy example we see that {nα} n≥1 and {nβ} n≥1 are topologically and arithmetically independent if 1, α, β are linearly independent over the field Q. It is also possible to study the independence between {nα} n≥1 and {n 2 β} n≥1 based on Weyl's equidistribution theorem. Naturally, a next question is to ask about the independence between {nα} n≥1 and {2 n d} n≥1 , where d is any real number. For a polynomial p with degree k with real coefficients, we write p(n) = k i=0 a i n i . We say that p is irrational if at least one of the numbers a 1 , . . . , a k is an irrational number. In this paper, we show the following result. See Section 2.2 for a clarification of the notation that appear below. Theorem 1.2. Let p be an irrational polynomial and let d be any real number. Then the sequences {p(n) mod 1} n≥1 and {2 n d mod 1} n≥1 are arithmetically independent.
We note that there is a curious connection between sequences of form {p(n)+2 n d mod 1} n≥1 and αβ-sequences. Let α, β be two real numbers, an αβ-sequence {x n } n≥1 is such that x 1 = 0 and for each i ≥ 1 we can choose x i+1 = x i + α mod 1 or x i+1 = x i + β mod 1 freely. We have the following interesting problem. Conjecture 1.3. Let α, β be such that 1, α, β are independent over the field of rational numbers. Then any αβ-sequence has full box dimension.
This conjecture is related to affine embeddings between Cantor sets, symbolic dynamics and Diophantine approximation, see [K79], [FX18] and [Y18b]. A lot of ideas for proving Theorem 1.2 appeared in [Y18b] for αβ-sets. For this reason, we can consider Theorem 1.2 as a cousin of Conjecture 1.3. Although the Bernoulli decomposition method we introduce in this paper cannot be used directly for αβ-sequences, it still sheds some lights on Conjecture 1.3.
We also consider here a number theoretic result which is closely related to the independence of sequences. Let m be an odd number and we consider the ring R[m] of residues modulo m. This is the finite set {0, . . . , m−1} together with the integer multiplication and addition modulo m. In this setting we can also consider the sequence {2 n + cn mod m} n≥0 where c is an integer such that gcd(c, m) = 1. On one hand the +c mod m action on R[m] can be seen as uniquely ergodic, which is analogous to +α mod 1 action on the unit interval with an irrational number α. On the other hand, {2 n mod m} n≥0 is an orbit under the ×2 mod m action. A analogy of Theorem 1.2 would be that {2 n + cn mod m} n≥0 is large in R[m]. We prove the following result which confirm this intuition. We remark that the method for proving the following result shares some strategy for proving Theorem 1.2. This is a special case of Problem 6 in the third round of the 27-th Brazilian Mathematical Olympiad, [27BMO]. For this reason, we will not provide the solution for the fully generalized case. Theorem 1.4. Let m ≥ 3 be an odd number and c be such that gcd(c, m) = 1. Let D(m) be the number of residue classes visited by {2 n + cn mod m} n≥0 . Then D(m) = m. In other words, for each r ∈ R[m], there is an integer n r such that 2 nr + cn r ≡ r mod m.

PRELIMINARIES
2.1. Dimensions. We list here some basic definitions of dimensions mentioned in the introduction. For more details, see [F05,Chapters 2,3] and [M99,Chapters 4,5]. We shall use N (F, r) for the minimal covering number of a set F in R n with balls of side length r > 0.
Then for all δ > 0 we define the following quantity The g-Hausdorff measure of F is Similarly the lower box dimension of F is If the limsup and liminf are equal, we call this value the box dimension of F and we denote it as dim B F.

2.2.
The unconventional fractional part symbol. For a real number α, it is conventional to use {α} for its fractional part. It is unfortunate that {.} is also used to denote a set or a sequence as well. For this reason we will use mod 1 for the fractional part. More precisely, for a real number x we write x mod 1 to denote the unique number a in [0, 1) such that a−x is an integer.
2.3. Sets and sequences. We write {x n } n≥1 for the sequence x 1 x 2 x 3 . . . . Sometimes it is convenient to use {x n } n≥1 to denote the following set Thus {x n } n≥1 and dim B {x n } n≥1 should be understood in this way.
2.4. Filtrations, atoms and entropy. Let X be a set with σ-algebra X . A filtration of σ-algebras is a sequence F n ⊂ X , n ≥ 1 such that Given a measurable map S : X → X and a finite measurable partition A of X, we denote S −n A to be the following finite collection of sets (notice that S might not be invertible) Then we use ∨ n−1 i=0 S −i A to be the σ-algebra generated by In this sense ∨ n−1 i=0 S −i A is generated by a finite partition A n−1 of X which is finer than A. Let µ be a probability measure, then we define the entropy of µ with respect to a finite partition A as follows We define the entropy of S as follows Here we implicitly used Sinai's entropy theorem, see [PY98,Lemma 8.8] 2.5. Dynamical systems and factors. A measurable dynamical system is in general denoted as (X, X , S, µ) where X is a set with σ-algebra X and measure µ and a measurable map S : X → X. In case when X is clear in context (for example Borel σ-algebra in Borel spaces) then we do not explicitly write it down. Given two dynamical systems (X, X , S, µ), (X 1 , X 1 , S 1 , µ 1 ), a measurable map f : X → X 1 is called a factorization map and (X 1 , X 1 , S 1 , µ 1 ) is called a factor of (X, X , S, µ) if µ 1 = f µ and f • S = S 1 • f.
Then we take a σ-algebra on Ω generated by cylinder subsets. A cylinder subset Z ⊂ Ω is such that Z = i∈N Z i and Z i = Λ for all but finitely many integers i ∈ N. We construct a probability measure µ on Ω by giving a probability measure µ Λ = {p λ } λ∈Λ on Λ and set µ = µ N Λ . We require here that p λ = 0 for all λ ∈ Λ. Then this system is weak-mixing and has entropy h(S, µ) = λ∈Λ −p λ log p λ . We call this system a Bernoulli system. We can also introduce a metric topology on Ω by defining d(ω, ω ′ ) = #Λ − min{i∈N:ω i =ω ′ i } . This turns Ω into a compact and totally disconnected space. For ω ∈ Ω and r ∈ (0, 1), we use B(ω, r) to denote the r-ball around ω with radius r with respect to the metric d constructed above.

A MATHEMATICAL OLYMPIAD PROBLEM
We first illustrate a short proof of Theorem 1.4 which provides us some motivation.
Proof of Theorem 1.4. Let l = ord(2, m) be the order of 2 in the multiplication group (Z/mZ) * . This can be done because gcd(2, m) = 1. For convenience we consider c = 1 and note that other cases can be shown with the same method. Since l = ord(2, m) we consider the following sequence {2 nl + nl mod m} n≥0 . We see that 2 nl ≡ 1 mod m for all n ≥ 0. However H = {nl mod m} n≥0 is a subgroup of Z/mZ of order m/gcd(l, m). For convenience we write ∆ = gcd(l, m). This ∆ plays the same role of the entropy in the proof of Theorem 1.2. If ∆ = 1 then D(m) = m follows automatically. We consider the case when ∆ > 1. Now for each integer r we consider the following sequence This sequence forms a coset of H. More precisely it is 2 r + r + H. Now if {2 r + r mod ∆} r≥0 would visit all residue classes modulo ∆, then 2 r + r + H, r ≥ 0 would visit all cosets of H in Z/mZ and {2 n + n} n≥1 would visit all residue classes modulo m. Since ∆ is an odd number as well we see that we have reduced the problem for m to the problem for ∆ which is strictly smaller than m. We can iterate this reduction procedure. Since we are considering positive integer set, either we eventually obtain ∆ = 1 or else we can consider further gcd(∆, ord(2, ∆)) < ∆. The later can not happen infinitely often. This concludes the proof.

EQUIDISTRIBUTED SEQUENCES
In this section, we prove some results for equidistributed sequences. We are particularly interested in taking subsequences.
Definition 4.1. Let H = {x n } n≥1 be an equidistributed sequence in [0, 1] with respect to the Lebesgue measure. Let K ⊂ N = {k 1 , k 2 , . . . } be an integer sequence. We introduce the following sequence We will be interested mostly in the closure of the set of numbers in A K (H). For convenience we denote C K (H) to be the closure of the set of numbers in A K (H). Observe that C N (H) = [0, 1]. C K (H) does not need to contain any intervals. Indeed, let {y n } n≥1 be an enumeration of rational numbers in (0, 1). For each n ≥ 1, let I n ⊂ [0, 1] be the open interval centred at y n with length ǫ n > 0. Assume further that n ǫ n = 0.5. One thing to observe is that I = ∪ n I n has Lebesgue measure at most 0.5. Since H equidistributes, for each n ≥ 1, {k : x k ∈ I n } has natural density equal to ǫ n . Then we see that has natural density at most 0.5 for each integer N ≥ 1. We construct a sequence K ⊂ N in the following way. First, for N = 1, we choose an integer k 1 and a subset K 1 ⊂ {1, . . . , k 1 } such that {x n } n∈K 1 ∩ I 1 = ∅. We can achieve further that #K 1 ≥ 0.49k 1 . Now let N = 2, similar as the above step, we can choose an integer k 2 > 1 and a subset K 2 ⊂ {k 1 + 1, k 1 + k 2 } such that {x n } n∈K 2 ∩ (I 1 ∪ I 2 ) = ∅.
Continuing this manner, for each N ≥ 1, we obtain a non-empty set K N ⊂ N such that {x n } n∈K N ∩ (I 1 ∪ · · · ∪ I N ) = ∅.
Then it is possible to see that for each integer n ≥ 1, x k ∈ I n for at most finitely many k ∈ K. Thus C K (H) either does not contain y n or contains y n as a isolated point. Since {y n } n≥1 is dense in [0, 1], C K (H) can not contain any intervals. We can even achieve that K has upper natural density at least 0.49.
The above argument shows that we can not say much about topological structures of C K (H). In turn, we focus on the Lebesgue measure. Proof. We first examine a simple case. Let I ⊂ [0, 1] be an interval with length larger than 1 − ρ.
Then since H equidistributes we see that has natural density larger than 1 − ρ. Therefore K ∩ K ′ cannot be empty. This implies that C K (H) ∩ I = ∅. Now let I be any open set with Lebesgue measure larger than 1 − ρ. By [SS05,Theorem 1.3], I can be written as a countable union of closed intervals with disjoint interiors, say I = ∪ n≥1 I n . Then for a large enough integer N the Lebesgue measure of ∪ n≤N I n is greater than 1 − ρ. Now we want to study the following set, It is tempting to claim that K ′′ natural density greater than 1 − ρ. The difficulty here is that the intervals I n , n ≥ 1 are not really disjoint. In fact for I n ∩ I n ′ can contain at most one point. For this reason we shrink each interval I n to I ′ n by fixing their centres and reducing their lengths to (1 − ǫ)I n . By choosing ǫ small enough, ∪ n≤N I ′ n has Lebesgue measure greater than 1 − ρ. Now we see that has Lebesgue measure at least ρ.
The above lemma is essentially sharp. We can not hope to bound the Lebesgue measure of C K (H) to be larger than the upper natural density of K. Indeed, let I = [0, 0.5] and we take Then C K (H) = I and it has Lebesgue measure 0.5. Because of the equidistribution property of H, K has natural density 0.5. However, it is possible to obtain better bounds for specially chosen sequences K. Lemma 4.3. Let H, K as in Definition 4.1. Let (Ω, S, ν) be a Bernoulli system and let Ω ′ ⊂ Ω be a measurable set with positive ν measure. For each ω ∈ Ω, we define Proof. Let I ⊂ [0, 1] be a closed interval. For each ω we are interested in the following limit, Of course we need to require that n i=1 ½ Ω ′ (S n (ω)) is not zero for large enough n. This happens for almost all ω. We see that almost surely This can be seen by using the ergodicity of Bernoulli systems. Now we want to estimate the following sum For a fixed interval I with length |I| we know that the following sequence has natural density |I|. If we are able to show that almost surely then we will see that for almost all ω, For convenience we write X n (ω) = ½ Ω ′ (S n (ω)). If Ω ′ is a ball of radius 2 −r , we decompose the integer set with modulo r. For each j ∈ {0, . . . , r − 1} we denote (2), . . . } in increasing order. By the law of large numbers we see that for ν almost all ω, We can apply this to each j with #K j = ∞. Now we see that, If #K j = ∞, we saw that for ν almost all ω, If #K j < ∞ then we see that for all ω, We want to sum up the above bounds over j ∈ {0, . . . , r − 1}. As there are finitely many terms to sum we see that the little-o parts sum up to and we have Dividing n in the above expression we see that for ν almost all ω, Taking n → ∞ we see that This proves ( * ) for Ω ′ being a ball of radius 2 −r for each r ≥ 1. Suppose that Ω ′ satisfies ( * ) then Ω ′ c satisfies ( * ) as well because for all ω and all n ≥ 1. A cylinder set in Ω can be written as a finite disjoint union of balls in Ω.
Let Ω ′ 1 , . . . , Ω ′ k be k ≥ 2 pairwise disjoint balls in Ω. Then we know that ( * ) holds for each Ω ′ i , i ∈ {1, . . . , k}, for almost all ω ∈ Ω. Thus we can deduce that for almost all ω ∈ Ω, ( * ) holds simultaneously for all Ω ′ 1 , . . . , Ω ′ k . This implies that ( * ) holds for ∪ k i=1 Ω ′ i as well. Thus we have shown that ( * ) holds for all cylinder subsets of Ω. We want to upgrade this result for the σ-algebra generated by cylinder sets, which will be sufficient since this is the σ-algebra we are considering. Let Ω ′ be an element in this σ-algebra with ν-measure p. Then for each ǫ > 0 we can find a cylinder set Ω ′′ such that ν(Ω ′ ∆Ω ′′ ) ≤ ǫ. Then we see that For ν almost all ω, the first summation is ν(Ω ′′ )|I|n + o N (n) (by what we have shown in above) and the second can be bounded by this is because the Bernoulli system in consideration is ergodic. This implies that for almost all ω (in a manner that depends on ǫ), We apply the above with a sequence ǫ i ↓ 0, as a result for almost all ω, (1) For the other direction, observe that With similar argument as above we see that, We apply the above with ǫ ↓ 0 again and we see that for almost all ω, (2) Combining (1), (2) we see that ( * ) holds for the σ-algebra generated by cylinder sets. Now we apply the above argument to intervals with rational end points, as there are countably many of them, we see that almost surely ( * ) holds for all I with rational end points. This concludes the proof.
In particular, the above lemma says that if we choose a subset K of integers by choosing each integer with probability p, then almost surely C K (H) = [0, 1]. This happens no matter how small p is, as long as p > 0. We see that almost surely, K has natural density p and if we would apply Lemma 4.2 we only obtain that C K (H) has Lebesgue measure at least p, which is much weaker than the result of Lemma 4.3. Now we finish this section by combining the above lemmas and obtain the following theorem.
Theorem 4.4. Let H be a equidistributed sequence in [0, 1]. Let K be an integer sequence with lower natural density ρ > 0. Let (Ω, S, µ) be a Bernoulli system. Let Ω i , i ∈ I be a finite collection of pairwise disjoint events (measurable subsets of Ω). Suppose that i∈I µ(Ω i ) = 1 − ǫ, where ǫ < ρ. Then there is a full measure set Ω ′ ⊂ Ω( not depending on K) such that for each ω ∈ Ω ′ , there is an i = i(ω, K) such that has Lebesgue measure at least ρ − ǫ. Here K Ω i is defined as follows Proof. We see that almost surely, K Ω i has natural density µ(Ω i ) for each i ∈ I. By Lemma 4.3, we see that almost surely, Let Ω ′ ⊂ Ω be a full measure set in which the above two properties hold for each i ∈ I. Then for each ω ∈ Ω ′ we see that Since K has lower natural density ρ > ǫ, we see that K ∩ ∪ i∈I K Ω i has lower natural density at least ρ − ǫ > 0. Notice that K Ω i , i ∈ I are pairwisely disjoint. For each i ∈ I we write We remark here that the above inequality may not hold if ρ i , i ∈ I are replaced by lower natural densities. We see that there is an i ∈ I such that We enumerate K Ω i as k 1 < k 2 < . . . . Then K i ⊂ K, viewed in this enumeration, has upper natural density at least ρ i /µ(Ω i ) which is at least ρ − ǫ. Now let H ′ = A K Ω i (H) and let K ′ i be the following sequence has Lebesgue measure at least ρ − ǫ. This concludes the proof.

BERNOULLI FACTORS AND A COMBINATORIAL SINAI FACTOR RESULT
In this section we closely follow [Y18a,Section 9]. In order to increase the readability and self-containing, we present here full details.
When (Ω, S B , ν) is a factor of (X, S, µ) with the same entropy, then intuitively all the complicities are carried by (Ω, S B , ν) and therefore the fibres of f should not be too complicated with respect to the map S. The following result expresses this intuition in a clear way. The following result is known as Rohlin's disintegration theorem, and we adopt the version in [S12].
Definition 5.2 (Simmons). Let f : X → Y be a measurable map between two measurable spaces and let µ be a measure on X with projection µ Y = f µ on Y . We call a collection of measures {µ y } y∈Y a system of conditional measures if the following properties hold, 1 : For all y ∈ Y , µ y is a measure supported on f −1 (y) and for µ Y almost all y ∈ Y , µ y is a probability measure. 2 : We have the law of measure disintegration. For all Borel set B ⊂ X, we have If X, Y are also metric spaces (f need not to be continuous) we require further that the following holds for µ Y almost all y ∈ Y .
3 : µ y = lim r→0 µ f −1 (B(y,r)) , where the limit is in the weak* sense and µ f −1 (B(y,r)) is the conditional measure of µ on f −1 (B(y, r)), namely, for any Borel set B ⊂ X with positive µ measure, Theorem 5.3 (Rohlin, Simmons). Let f : X → Y be a measurable map between two metric spaces with corresponding Borel σ-algebra. Then there exists a system of conditional measures.
Then we have the following result due to [W16,Lemma 6.4] which is a direct consequence of conditional Shannon-McMillan-Breiman theorem, Egorov's theorem and the Portmanteau theorem.
Theorem 5.4 (Wu). Let (X, S, µ) be an ergodic dynamical system with X being a Borel space. Let A be a finite partition of X such that ∨ ∞ i=0 S −i A generates the σ-algebra of X. For each x ∈ X not on the boundaries of sets in If µ does not positive measures to boundaries of S −i A for all i ∈ N and h(S, µ) > 0 then there exist a Bernoulli factor (Ω, S B , ν) with measurable factorization map f : X → Ω and for each δ > 0 there exist a X δ ⊂ X and a constant C δ with the following properties, 1 :µ(X δ ) > 1 − δ.
2 :For all x ∈ X δ and n ≥ 1, µ f (x) (A n (x)) ≥ C δ 2 −nδ and µ f (x) is a probability measure. 3 :For all integers n ≥ 1, there exists a measurable set B n δ ⊂ Ω with ν(B n δ ) ≥ 1 − δ and a r = r(δ, n) > 0 such that for all ω ∈ B n δ and all atoms A n we have From the above results we can obtain the following combinatorial version of Sinai's factor theorem.
Definition 5.5 (Combinatorial Bernoulli factor). Let (X, S, µ) be an ergodic system. Let A be a finite partition generating the Borel σ-algebra of X. For δ > 0, we say that (X, S, µ) is δ-Bernoulli if the following statements hold: There is a number c δ > 0 and for each integer n ≥ 1, there is an integer N (n) and a measurable decomposition D n = {D n (1), . . . , D n (N (n))} of X such that (D N n , πS, πµ) is a Bernoulli system where π : X → D is defined by taking π(x) to be the sequences of sets D n ∈ D n such that S n (x) ∈ D n . For each i ∈ {1, . . . , N (n)}, we writeM n (i) to be the collection of atoms of A n intersecting D n (i). For each i, there is a subcollection M n (i) ⊂M n (i) consisting at most c δ 2 nδ many elements such that the union of atoms in ∪ i M n (i) has µ measure at least 1 − 2δ.
We say that (X, T, µ) satisfies the weak Bernoulli property if it is δ-Bernoulli for each δ > 0.
The following result is a variant of [W16, Theorem 6.1] and it is closely related to Austin's theorem on the weak Pinsker property, see [A18].
Theorem 5.6 (weak Bernoulli property). We adopt the conditions in Theorem 5.4. Then (X, S, µ) satisfies the weak Bernoulli property. Moreover, let ǫ > 0 be arbitrarily chosen in (0, 1) and H = {h k } k≥1 be an equidistributed sequence in [0, 1]. For each δ ∈ (0, 1), there is a constant c δ > 0 and X ′ δ with full µ measure such that for all n ≥ 1, all x ∈ X ′ δ and all K ⊂ N with lower natural density at least ρ > 2δ + ǫ, there is a collection M n = M n (x, K) of at most c δ 2 nδ atoms of A n with the following property: Denote the union of elements in M n as M n . We construct the following sequence Then the following set has Lebesgue measure at least ǫ C K∩K ′ (x) (H).
Proof. We use Theorem 5.4 to find a set X δ with µ(X δ ) > 1 − δ. Then for each integer n ≥ 1 we can find B n δ with ν(B n δ ) ≥ 1 − δ and r = r(δ, n) > 0. Without loss of generality we shall assume that r = d −k where d is the number of digits of the Bernoulli system and k is an integer. For each ω ∈ B n δ we have Now because of the topology we chose for Ω, we see that B(ω, r) consists all sequences in Ω with the same first k digits as ω. In particular if ω ′ ∈ B(ω, r) then B(ω ′ , r) = B(ω, r). This property reflects the fact that Ω is an ultrametric space. Notice that for any Bernoulli system (Ω, S B , ν), any ball of positive radius has positive ν measure. In particular µ(f −1 (B(ω, r))) > 0 and by properties (2)(3) in Theorem 5.4, for each ω ′ ∈ B n δ ∩ B(ω, r) we have On the other hand we clearly have atoms An Since there are only finitely many r balls in Ω we see that as ω varies in B n δ there are finitely many different sets of form Y (ω). Denote the collection of these sets as {Y 1 , . . . , Y N ′ (n) } where N ′ (n) is an integer. For each i ∈ I = {1, . . . , N ′ (n)}, let Ω(i) ⊂ B n δ be the set of form B(ω, r) ∩ B n δ such that Y i = X δ ∩ f −1 (Ω(i)). We notice here that the union of all Y i is a rather large subset of X, more precisely we have the following result, For each i ∈ I we write the collection of atoms intersecting Y i as M n (i) and write their union as M n (i). Then we saw that #M n (i) ≤ 2 nδ (1 − δ)C δ and the following collection of atoms in In order to finish the proof for the weak Bernoulli property, we need to specify D n . Denote the number of r-balls of Ω as N (n), which is in general larger than N ′ (n). We enumerate them as B 1 , . . . , B N (n) . Then precisely N ′ (n) of them intersect B n δ . We assume that these are the first N ′ (n) elements. We take D n (i) to be f −1 (B i ) for each i. For each i ∈ {1, . . . , N (n)}, we takẽ M n (i) to be the collection of atoms in A n intersecting D n (i). For each i ∈ {1, . . . , N ′ (n)} we already constructed M n (i) in above. For i ∈ {N ′ (n) + 1, N (n)} we can choose M n (i) to be empty and concludes the proof of the first part. Now we consider the following sequence for x ∈ X, by the ergodic theorem we see that for µ almost all x ∈ X, K(x) has natural density at least 1 − δ. For each i ∈ I and x ∈ X we construct the following set , Ω(i))).
Here K(f (x), Ω(i)) = {k ∈ N : S k • f (x) ∈ Ω(i)}. By Lemma 4.3, we claim that for µ almost all x ∈ X and any sequence K with lower natural density at least 2δ + ǫ there exists an i ∈ I such that ,Ω(i)) (H) has Lebesgue measure at least ǫ. Indeed, we know that K ∩ K(x) has lower natural density at least δ + ǫ for µ almost all x ∈ X and i∈I ν(Ω(i)) = ν(B n δ ) ≥ 1 − δ. By Lemma 4.3, there is a full ν measure set Ω ′ with the property that for all sequence K 0 with lower natural density at least δ + ǫ, for each ω ′ ∈ Ω ′ there is a i ∈ I such that has Lebesgue measure at least ǫ. As Ω ′ has full ν measure, there is at least one (in fact, µ-almost all) x ∈ X such that f (x) ∈ Ω ′ . Now if x ∈ X is such that K(x) ∩ K has lower density at least δ + ǫ and at the same time f (x) ∈ Ω ′ , then there is i ∈ I such that has Lebesgue measure at least ǫ. However, such choices of x ∈ X ranges over a full µ-measure set, which proves the claim. The second part of the theorem follows since the above argument holds for all n ≥ 1 and we can find a full measure set X ′ δ ⊂ X which satisfies all our requirements.

ON SEQUENCES {p(n) + 2 n d mod 1} n≥1
Given any irrational number c and any number d we consider the sequence {nc + 2 n d mod 1} n≥1 . If d is a rational number then 2 n d takes only finitely many values. From here we know that in this case {nc + 2 n d mod 1} n≥1 have box dimension 1, in fact it must be dense in [0, 1]. If d is randomly picked according to the Lebesgue measure then intuitively nc and 2 n d are somehow independent and {nc + 2 n d mod 1} n≥1 can be viewed as an equidistributed sequence with a random pertubation and it is plausible that {nc + 2 n d mod 1} n≥1 should have box dimension 1.
Theorem 6.1. Let c be an irrational number and d be any real number. We have Proof. Let Or 2 (d) be the following orbit closure Or 2 (d) is then a closed ×2 mod 1 invariant subset of [0, 1] and therefore there exist ×2 mod 1 invariant probability measures whose supports are contained in Or 2 (d). By taking ergodic components we can find an invariant measure µ whose support is contained in Or 2 (2) and the ×2 mod 1 action is ergodic with respect to µ. Our idea is to take a µ-typical point d ′ and study the sequence {nc + 2 n d ′ mod 1} n≥1 . In order to set up the link between {nc + 2 n d mod 1} n≥1 and {nc + 2 n d ′ mod 1} n≥1 we first observe that d ′ ∈ Or 2 (d) and therefore there exists a sequence {i k } k≥1 such that 2 i k d mod 1 → d ′ as k → ∞. Consider the points i k c mod 1, without loss of generality we shall assume that i k c mod 1 → c * as k → ∞, where c * ∈ [0, 1]. Consider the points (i k +1)c+2 i k +1 d for integers k. We see that this sequence of points converge to c * +c+2d ′ mod 1. Similarly for each integer M we see that (i k + M )c+ 2 i k +M d converges to c * + M c+ 2 M d ′ mod 1. As this holds for any integer M we see that Therefore we see that it is enough to show that {nc + 2 n d ′ } n≥1 has full lower box dimension. Now consider the ergodic dynamical system ([0, 1], T, µ) where T is the ×2 mod 1 action. Consider the partition [0, 0.5] ∪ [0.5, 1] which we denote as A. It is easy to show that T −n A generates the Borel σ-algebra as n → ∞. Then we have h(T, µ) = h(T, µ, A). If µ supports on boundaries of T −n A for integers n then as those boundaries are rational numbers we see that we can pick d ′ ∈ Q and in this case {nc + 2 n d ′ } n≥1 = [0, 1]. Now assume that µ does not have any boundary supports. Our further argument can be split into two cases based on whether or not h(T, µ) is 0.
6.1. Zero entropy. When h(T, µ) is equal to zero, then for each integer n we have 2 n many disjoint dyadic intervals of length 2 −n covering [0, 1]. Denote this collection of intervals as D n . We see that Suppose that the above sum is bounded from above by ǫ ′ n for each large enough n. Then for each δ ′ > 0, we define the following subset If v / ∈ V δ ′ then − log µ(v) > nδ ′ log 2 and therefore Then we see that On the other hand because v∈V µ(v) = 1 then we see that we see that for all large enough n, at least 1−10δ ′ portion of µ measure is supported in a collection of at most 2 nδ ′ many dyadic intervals in D n . Let d ′ be a µ-typical number which is generic with respect to µ. Let δ > 0 be a small number. Let M be a large integer and there is a collection of at most 2 M δ many dyadic intervals in D M which support at least 1 − 10δ portion of µ measure. Denote this collection of dyadic intervals as M M and write their union as S M . As d ′ is generic with respect to µ we see that for large enough integers n, {2 n d ′ mod 1} spends at most 1 − 20δ portion of time in S M . More precisely, A M = {n ∈ N : 2 n d ′ mod 1 ∈ S M } is an integer sequence with natural density at least 1 − 20δ. By Lemma 4.2, the closure E of {nc mod 1} n∈A M has Lebesgue measure at least 1 − 20δ. We will need at least (1 − 20δ)2 M many dyadic intervals in D M to cover E for otherwise the Lebesgue measure of E is smaller than 1 − 20δ. Denote this collection of dyadic intervals as F M . We say that the pair (s M , f M ) ∈ M M ×F M related (or s M is related to f M ) if there is an integer n ∈ A M such that 2 n d ′ mod 1 ∈ s M and nc mod 1 ∈ f M . Now each s M ∈ S M is related to at least one of f M in F M . Then by the pigeonhole principle we see that there exists at least one f M which is related to at least (1 − 20δ)2 M /2 M δ many elements in M M . This implies that in order to cover {nc+2 n d ′ mod 1} n∈A M with dyadic intervals in D M , we need at least 0.01(1 − 20δ)2 M /2 M δ many of them. As the above argument holds for all large enough integer M we see that dim B {nc + 2 n d ′ mod 1} n≥1 ≥ 1 − δ. As δ can be arbitrarily small, this concludes the case when h(T, µ) = 0. Having established the result for {nc + 2 n d mod 1} n≥1 it is natural to ask what is the situation for {p(n) + 2 n d mod 1} n≥1 where p : Z → R is a polynomial with at least one irrational coefficient. In fact we can treat the polynomial case with a similar method. If we know that d is a generic point with respect to an ergodic measure of the doubling map, then the argument in the above proof can be applied to show that {p(n) + 2 n d mod 1} n≥1 has full box dimension. The only thing we need here is that {p(n) mod 1} n≥1 equidistributes in [0, 1]. If d is not such a generic point we want to choose another number d ′ in Or 2 (d). This step is very crucial. In the case when p(n) = nc for a irrational number c, we can use some sort of translational invariance of the sequence {nc mod 1} to set up the link between {nc + 2 n d mod 1} and {nc + 2 n d ′ mod 1}. We can not directly use this argument for the general polynomial situation. Some modifications are needed. As an illustration, we first deal with the sequence {n 2 c + 2 n d mod 1}.
Again we have to re-select a number d ′ in Or 2 (d) and we want to set up a link between {n 2 c + 2 n d mod 1} and {n 2 c + 2 n d ′ mod 1}. Let i k → ∞ be a sequence of integers such that 2 i k d → d ′ as i k → ∞. Then we consider points i 2 k c, (i k + 1) 2 c, (i k + 2) 2 c. It is of no loss of generality to assume that there exist three numbers c 1 , c 2 , c 3 ∈ [0, 1] such that the fraction parts of i 2 k c, (i k + 1) 2 c, (i k + 2) 2 c converge to c 1 , c 2 , c 3 respectively. Indeed, we only need to take a subsequence out of a subsequence three times to ensure all numbers converge as required. Now, observe the following relation which holds for all integer k ≥ 1, . This implies that (by treating the integer part and fractional part in the above expression separately) we have c 3 − c 2 mod 1 = c 2 − c 1 + 2c mod 1. Now for each integer M we consider the M + 1 points i 2 k c, . . . , (i k + M ) 2 c. We can still assume that they converge ( mod 1) respectively to c 0 , c 1 , . . . , c M . Arguing as above we see that for each integer m ∈ {1, . . . , M }, c m+1 − c m mod 1 = c m − c m−1 + 2c mod 1. This implies that {c 0 , c 1 , . . . c M } is actually {c 0 + n(c 1 − c 0 ) + n(n − 1)c} n=M n=1 . It is evident that along the sequence i k → ∞ we have 2 i k +m d mod 1 → 2 m d ′ mod 1 for each m ∈ {0, . . . , M }. This implies that {c 0 + n(c 1 − c 0 ) + n(n − 1)c + 2 n d ′ } n=M n=1 ⊂ {n 2 c + 2 n d} n≥1 . Now for each integer M we obtained a number c 0 (M ) such that {c 0 (M ) + n(c 1 (M ) − c 0 (M )) + n(n − 1)c + 2 n d ′ } n=M n=1 ⊂ {n 2 c + 2 n d} n≥1 . We can assume that for a subsequence M k → ∞ both c 0 (M k ), c 1 (M k ) converges. Therefore there exist numbers c * 0 , c * 1 ∈ [0, 1] such that, {c * 0 + nc * 1 + n(n − 1)c + 2 n d ′ } n=∞ n=1 ⊂ {n 2 c + 2 n d} n≥1 . This means that {n 2 c + 2 n d} n≥1 contains a translational copy of {nc * 1 + n(n − 1)c + 2 n d ′ } n≥1 and this is exactly the result we want. Notice that {nc * 1 + n(n − 1)c mod 1} equidistributes in [0, 1].
Theorem 6.2. Let c be an irrational number and d be any real number. We have dim B {n 2 c + 2 n d mod 1} n≥1 = 1.
The above method can be iterated to obtain the result for general polynomials. We omit here the full detail and only state the result. Theorem 6.3. Let d be any real number and p be a polynomial with at least one irrational coefficient of non constant term. We have dim B {p(n) + 2 n d mod 1} n≥1 = 1.
Proof. Our task is to show that if we choose a number d ′ inside the orbit closure Or 2 (d) then there is an equidistributed sequence {x n } n≥1 such that {x n + 2 n d ′ } ⊂ {p(n) + 2 n d mod 1} n≥1 . As we can choose d ′ to be generic with respect to an ergodic measure of the doubling map, the argument in the proof of Theorem 6.1 can be applied to show the result of this theorem. If c is a rational number, it is easy to see that for each k ≥ 1, the fractional part of n k c for integers n can attain at most finitely many values. Therefore we can assume that p has degree at least two and the coefficient of the highest term is a irrational number.
Given m ≥ 2 many real numbers x 1 , . . . , x m . We define the m-th order difference inductively as follows. For m = 2 we write Having defined the (k − 1)-th order difference ∆(x 1 , . . . , x k−1 ) we can define the k-th order difference as For example we see that If p has degree m ≥ 2, then ∆ m (p(n), p(n + 1), . . . , p(n + m − 1)), viewed as a function of n, is a polynomial of degree 0, namely, a constant c ′ . On the other hand, for any given numbers x 1 , . . . , x m−1 , there is a unique way to extend these numbers to an infinite sequence x 1 , . . . , x n , . . . such that ∆ m (x n , . . . , x n+m−1 ) = c ′ for all n ≥ 1. It is also possible to see that x n = p 1 (n) mod 1 for a polynomial p 1 with degree m and the coefficient of the highest order term is the same as that of p. Then the same argument as we have shown in the case when p(n) = n 2 c can be applied here. As a result we see that there is a irrational polynomial p 0 , which is in general different than p such that From here we can use the entropy argument in the proof of Theorem 6.1 and finish the proof of this Theorem.
It is interesting to see whether we can replace {p(n) mod 1} with an arbitrary equidistributed sequence x n in [0, 1]. The difficulty is that in this case we cannot easily shift d in {x n + 2 n d} into d ′ . The above results seem to suggest the plausibility the following statement.
Let H = {x n } n≥1 be an equidistributed sequence in We note that the above problem is closely related to the results in [HS12]. In fact Or 3 (d ′ ) = {3 n d ′ mod 1} n≥1 and Or 2 (d ′′ ) = {2 n d ′′ mod 1} n≥1 are ×3, ×2 invariant sets respectively. Then by [HS12,Theorem 1.3], for each projection π : (x, y) ∈ R 2 → ux + vy ∈ R with u, v ∈ R and uv = 0, the projected image π(Or 3 (d ′ ) × Or 2 (d ′′ )) always attains the maximal dimension, namely, dim H π(Or 3 (d ′ ) × Or 2 (d ′′ )) = min{1, dim H Or 2 (d ′′ ) + dim H Or 3 (d ′ )}. In particular if we choose u = v = 1 in above we see that the sum set Or 3 (d ′ ) + Or 2 (d ′′ ) attains the maximal dimension. This argument almost gives an answer to Question 6.5. The difficulty of Question 6.5 is that we are not taking the sum set of Or 2 (d ′′ ) and Or 3 (d ′′ ) but sums between individual elements in the sequences {2 n d ′′ mod 1} n≥1 and {3 n d ′ mod 1} n≥1 . 7. FURTHER QUESTIONS 7.1. Hausdorff dimension. So far we considered the box dimension sequences of form {p(n)+ 2 n d mod 1} n≥1 . A natural question is what about the Hausdorff dimension of its closure? One observation is that if {2 n d mod 1} has full Hausdorff dimension, then because it is a closed ×2 mod 1 invariant set, it must be [0, 1]. In particular we have 0 ∈ {2 n d mod 1}. By the argument in the proof of Theorem 6.1 we see that there is another polynomial with at least one irrational coefficient such that {p ′ (n) + 2 n 0} is contained in the closure of {p(n) + 2 n d mod 1} n≥1 . This implies that {p(n) + 2 n d mod 1} n≥1 itself must be dense in [0, 1]. On the other hand, if {2 n d mod 1} has Hausdorff dimension strictly smaller than 1, then observe that the following difference set {p(n) + 2 n d mod 1} n≥1 − {2 n d mod 1} n≥1 is dense in [0, 1]. Since {2 n d mod 1} is ×2 mod 1 invariant we see that the box dimension of {2 n d mod 1} is equal to its Hausdorff dimension which is strictly smaller than 1. This implies that {p(n) + 2 n d mod 1} n≥1 must have positive Hausdorff dimension. Thus we proved the following theorem.
Theorem 7.1. Let d be any real number and p be a polynomial with at least one irrational coefficient of non-constant term. We have dim H {p(n) + 2 n d mod 1} n≥1 > 0.
The above result and argument is from D-J. Feng and we thank him for permission to include it here. It is interesting to see whether stronger results can hold. We pose here the following problem.
Question 7.2. Let d be any real number and p be a polynomial with at least one irrational coefficient of non constant term. Is it true that there is a absolute constant a > 0 such that dim H {p(n) + 2 n d mod 1} n≥1 ≥ a.
Can a be chosen to be 1?
In the beginning of Section 4, we see that a subsequence of an equidistributed sequence may not be anywhere dense. Motivated by this result we ask here the following question. Question 7.3. Let d be any real number and p be a polynomial with at least one irrational coefficient of non constant term. Can {p(n) + 2 n d mod 1} n≥1 be nowhere dense in [0,1]?
A negative answer to the above question would affirmatively answer Question 7.2.

ACKNOWLEDGEMENT
HY was financially supported by the University of St Andrews.