The gap structure of a family of integer subsets

In this paper we investigate the gap structure of a certain family of subsets of N which produces counterexamples both to the “density version” and the “canonical version” of Brown’s lemma. This family includes the members of all complementing pairs of N. We will also relate the asymptotical gap structure of subsets of N with their density and investigate the asymptotical gap structure of monochromatic and rainbow sets with respect to arbitrary infinite colorings of N.


Introduction
Let N be the set of all nonnegative integers.The gap of a finite subset A = {a 1 , . . ., a k } of N is the number gap(A) := max{a i+1 − a i : 1 i k − 1}.An infinite subset X of N is piecewise syndetic if it contains arbitrarily large subsets with uniformly bounded gaps.This means that the sequence in k ∈ N defined by d k (X) := min{gap(A) : A ⊂ X and |A| = k + 1}, (1) the electronic journal of combinatorics 21(1) (2014), #P1.47 is bounded.An induction argument in the number of colors shows [2,3] that any finite coloring of N admits a monochromatic piecewise syndetic set.This result is known as Brown's lemma.
Brown's lemma does not admit a density version analogous to Szemerédi's theorem [8], that is, there are subsets X of N with positive density which are not piecewise syndetic.An example of such a subset is given in [1], Theorem 2.8.
Brown's lemma also does not admit a canonical version analogous to the Erdős-Graham canonical version of van der Waerden's theorem [6].In fact, T. Brown [4,5] showed that there is an infinite coloring τ : N → N for which the sequence in k ∈ N defined by is not bounded.The infinite coloring used by T. Brown consists of infinitely many translates of an infinite set, that is, it is a coloring associated to a certain complementing pair of N. Two infinite subsets X 1 and X 2 of N are a complementing pair of N, and we write In this case we can define an infinite coloring τ : In this paper we will investigate the gap structure of a certain family of subsets of N which produces counterexamples both to the "density version" and the "canonical version" of Brown's lemma.This family includes the members of all complementing pairs of N. We will also investigate the asymptotical upper bounds of d k (X) and d k (τ ) when X is a subset of N with positive upper density and τ is an infinite coloring of N.

A familiy of non-piecewise syndetic sets with positive density
We will denote by σ(X) and σ(X), respectively, the upper density and the lower density of X: If σ(X) = σ(X), the density of X is equal to this common value and is denoted by σ(X).
Consider two infinite sequences a n and d n of positive integers, with a 0 = 1.Assume that a n is strictly increasing, d n is nondecreasing and a n+1 an is an integer for each n ∈ N. Fix an integer K > 0. We define recursively an increasing sequence of finite subsets I n := I n (a n , d n , K) of N, with β n := max I n , as follows: I 0 = [0, K] and the electronic journal of combinatorics 21(1) (2014), #P1.47 Hence converges.Moreover, σ(I) = σ(I).
Proof.Taking into account (4) we have This sequence is always convergent and This means that the largest limit of subsequences of |I∩[0,n]| n is attained by x n .Hence which means that σ(I) > 0 if and only if the series (5) converges.If the series (5) diverges, then it is clear that σ(I) = σ(I) = 0. Assume now that the series (5) converges.In this case that is lim n dn an = 0. On the other hand, Since lim n dn an = 0, we have lim x n = lim y n , that is the smallest limit of subsequences of is attained by y n and it is equal to σ(I).
Remark 4. Taking into account its construction, if lim d n = ∞ the subset I is not piecewise syndetic.For example, if a n = 2 n and d n = n, I is not piecewise syndetic but it has positive density K+1 K+2 .This family of subsets is optimal in the following sense.This means that, if d an(K+1) (X) d n+1 , then d k (X) d k (I) for all k, and consequently σ(X) σ(I) = σ(I).

Complementing pairs of N
Complementing pairs of N admit the following characterization (see [9] and the references therein).Given two infinite subsets X 1 and X 2 of N, we have N = X 1 ⊕ X 2 if and only if there exists a sequence m i , with m i 2 for all i ∈ N, such that X 1 is the set of all finite sums i 0 x 2i M 2i and X 2 is the set of all finite sums i 0 x 2i+1 M 2i+1 , where M 0 = 1, M i = i j=1 m j and 0 x i < m i+1 .Let
Proof.To see that X 1 is not piecewise syndetic we only have to check that lim d n = ∞.We can rewrite (6) as Since m i 2 for all i 1, we have M 2i − M 2i−1 1, which means that d n is strictly increasing.
We say that A ⊂ N is a rainbow set with respect to a coloring τ : .
Hence, if gap(A) d for some fixed d and |A| is large enough, from condition (7) we get |τ (A)| < |A|, that is, we can not have arbitrarily large rainbow sequences with bounded gaps.
On the other hand, τ does not admit arbitrarily large monochromatic sequences with uniformly bounded gaps because X 1 is not piecewise syndetic and, for each color n 0 , the monochromatic subset τ −1 (n 0 ) is just the translation copy of X 1 by n 0 .
Remark 9.The infinite coloring used in [5] is the one defined by the complementing pair N = X 1 ⊕ X 2 with X 1 the set of all finite sums i even 2 i and X 2 the set of all finite sums i odd 2 i .In this case, m i = 2 for all i 1, and condition (7) certainly holds.
the electronic journal of combinatorics 21(1) (2014), #P1.47 4 Asymptotical gap structure of positive density sets Not surprisingly, the sequence d k (X) defined by (1) grows at most linearly with k for sets X with positive density.
Proposition 10.Let X be a subset of N with positive lower density σ := σ(X).Then Proof.Given 0 < ǫ < σ, for all sufficiently large n, we must have As the following theorem shows, this asymptotical bound is not optimal.
Theorem 11.Let ̟ : [0, +∞[→ R be a continuous increasing function so that ̟(x)/x 2 decreases with x.Then, if the integral Proof.Let X be a subset of N with ̟(k) = O(d k (X)) and consider the increasing sequences a n and d n defined by a n = 2 n and d n = d 2 n (X).Consider the subset diverges.But, taking the substitution x = 2 y , we get x 2 dx.
Proof.Set a n = 2 n , d n = ⌈̟(2 n )⌉, and consider the subset I = I(a n , d n , 1).If the integral (9) converges, we can apply the integral convergence test, as in the proof of Theorem 11, to conclude that the series (5) converge, and consequently σ(I) > 0. Since ̟ is increasing and, for 2 n < k < 2 n+1 , we have ). Set X = I, and we are done.
Remark 13.In [7], R. Salem and D.C. Spencer studied the influence of gaps in the density of integer subsets.However, a different notion of gap structure is considered there.More precisely, given an positive increasing function ω of the real nonnegative variable x, they were concerned with subsets X of N satisfying the following property: for any closed interval [a, a + l], with a 0 and l > 0, there exists an open interval not less than ω(l) which contains no points of X.For that purpose, they used sequences u(n) defined by where g p is a given sequence of positive integers.For g 0 = 1 and g p 1, these sequences are of the form I(a n , d n , 1), with a n = 2 n and d n = g 0 + g 1 + . . .+ g n .In spite of the different notions of gap structure, the asymptotical bounds given by Theorems 11 and 12 are the same as those given by Theorems I and II in [7].

Asymptotical gap structure and infinite colorings
Next we investigate the asymptotical growth with k of the sequence d k (τ ) defined by (2).
Theorem 14.Given an infinite coloring τ : N → N, we have and define α n = ⌈ n θ(n) ⌉.By the pigeonhole principle, there always exists a monochromatic subset A αn of [1, n] with α n elements.For each n, consider also a rainbow subset B θ(n) of [1, n] with θ(n) elements and θ(n) distinct colors.
the electronic journal of combinatorics 21(1) (2014), #P1.47 Example 15.When τ is the infinite coloring of N associated to the complementing pair N = X 1 ⊕ X 2 , where X 1 is the set of all finite sums x 2i M 2i , with 0 x i < m i+1 , we can give the following asymptotical bounds for d k (τ ).To simplify the discussion, assume further that, for some m 2, we have m i = m for all i 1.In this case, from (6)  This means that gap(A) grows asymptotically as fast as |A| 2 for monochromatic subsets A. From (8) we see that gap(A) is asymptotically bounded below by |A| for rainbow sets A.
and the structure of I 3 is illustrated by the following figure.

1 Lemma 2 .
The subset I(a n , d n , K) := I = n∈N I n of N has positive upper density if and only if the positive series

Lemma 5 .
For each n ∈ N, we have d an(K+1) (I) = d n+1 .Moreover, given X ⊂ N, then σ(X) σ(I) if d an(K+1) (X) d n+1 for each n ∈ N. Proof.The first assertion follows directly from the definitions of I and d k (I).With the respect to the second assertion, observe that, for each k ∈ N, we have d k (I) = d an k (K+1) (I), where n k = max{n : a n (K + 1) k}.

Theorem 8 .
Given a complementing pair N = X 1 ⊕ X 2 , consider the associated infinite coloring τ , as defined in the Introduction section.If lim then there does not exist d ∈ N and arbitrarily large sets A such that gap(A) d and A is either monochromatic or rainbow.. Proof.Observe that the number of colors in each interval of the form J k i = [kM 2i , (k + 1)M 2i ] is precisely the cardinality of the set 2i−1 j=0 x 2j+1 M 2j+1 : 0 x j < m j+1 .Hence, each interval J k i = [kM 2i , (k+1)M 2i ] has exactly M + 2i colors and each color appears exactly M − 2i times.Let A = {b 1 , . . ., b n } be a finite subset of N and choose s minimal so that A ⊆ J k−1 s ∪J k s .We have 2M 2(s−1) b n −b 1 gap(A)n.On the other hand, |τ (A)| 2M + 2s .Then |τ (A)| gap(A)|A|m 2s M − 2(s−1)