Metric decomposability theorems on sets of integers

A set A⊂N$\mathcal {A}\subset \mathbb {N}$ is called additively decomposable (resp., asymptotically additively decomposable) if there exist sets B,C⊂N$\mathcal {B},\mathcal {C}\subset \mathbb {N}$ of cardinality at least two each such that A=B+C$\mathcal {A}=\mathcal {B}+\mathcal {C}$ (resp., AΔ(B+C)$\mathcal {A}\Delta (\mathcal {B}+\mathcal {C})$ is finite). If none of these properties hold, the set A${\mathcal {A}}$ is called totally primitive. We define Z$\mathbb {Z}$ ‐decomposability analogously with subsets A,B,C$\mathcal {A,B,C}$ of Z$\mathbb {Z}$ . Wirsing showed that almost all subsets of N$\mathbb {N}$ are totally primitive. In this paper, in the spirit of Wirsing, we study decomposability from a probabilistic viewpoint. First, we show that almost all symmetric subsets of Z$\mathbb {Z}$ are Z$\mathbb {Z}$ ‐decomposable. Then we show that almost all small perturbations of the set of primes yield a totally primitive set. Further, this last result still holds when the set of primes is replaced by the set of sums of two squares, which is by definition decomposable.


Statement of the results
This article is concerned with sum sets in the integers.Given two subsets A, B of Z, their sum set A + B is the set {a + b : (a, b) ∈ A × B}.We denote 2A = A + A. Further, we denote by A ∼ B the property that A∆B is finite.A set C ⊂ N is called additively decomposable (resp.asymptotically additively decomposable) if there exist sets A, B ⊂ N of cardinality at least two each such that (resp.C ∼ A + B).A set which is not asymptotically additively decomposable is called totally primitive.Similarly, a set C ⊂ Z is called additively Z-decomposable (resp asymptotically additively Z-decomposable) if there exist sets A, B ⊂ Z of cardinality at least two each such that C = A + B (resp.C ∼ A + B).
An old conjecture of Ostmann [11, page 13] asserts that the set P of primes is totally primitive.In spite of serious efforts by numerous authors and notable advances (see [2,3] and the very recent [5] on related problems), the problem remains unsolved.The philosophy supporting this idea is that additive decomposability is a very rare property, so most sets occurring in number theory which are not specifically defined to be a sum set 1 should not have it.A theorem of Wirsing [12] actually asserts that almost all sets are totally primitive, where "almost all" refers to the construction of a random subset of N by selecting each integer into the set with probability 1/2 independently of each other.
On the other hand, Ruzsa recently showed [10] in the Number Theory Web Seminar that the widely believed Hardy-Littlewood prime tuples conjecture implies that the signed set of primes P ∪ (−P) is aymptotically additively Z-decomposable, i.e. there exist sets A, B ⊂ Z such that P ∪ (−P) ∼ A − B. Our first result shows that this property is actually typical.To state it, let (ξ n ) n∈N be a sequence of independent identically distributed Bernoulli variables satisfying P(ξ n = 1) = P(ξ n = 0) = 1/2 for each n ∈ N.
Theorem 1. Almost all symmetric subsets of Z are additively Z-decomposable.More precisely, let D = {n ∈ Z : ξ |n|=1 }.Then P(D is Z-decomposable) = 1.This theorem will follow from a finite tuples property that we will prove.Regarding subsets of N, the same finite tuple property yields the following.A theorem of Granville [4] shows that the Hardy-Littlewood conjecture implies that set of primes contains a sumset A + B where both summands are infinite.Thus what is known of the primes, resp.the signed primes, under the Hardy-Littlewood conjecture, is true of almost every subset of N, resp.almost every symmetric subset of Z. Further, let us observe that a much stronger statement than Theorem 2 holds, namely that every dense subset of N contains a sumset A + B where both summands are infinite; this was conjectured by Erdős and proven by Moreira, Richter and Robertson [8] (see also [7] for a simpler proof).Almost every subset being dense, this statement implies Theorem 2, which may be thought of as a "cheap" version of the theorem of Moreira, Richter and Robertson.
Regarding decomposability of subsets of N, we consider new probability distributions whose mass is concentrated "near" a fixed set of interest such as the set of the primes.First we introduce a standard notational convention: a set (or equivalently an increasing sequence) of integers is denoted by a calligraphic letter (e.g.A), its elements are denoted by the corresponding lower case letter (e.g.A = {a n : n ∈ N}), its counting function by the corresponding upper case letter (e.g. Let S be an infinite set and define a function f = f S by f (x) = x/S(x).We make the following two hypotheses.S1 f (x) tends to infinity as x does.S2 The number of Then we fix a sequence (δ n ) n∈N which has the following properties.
An obvious consequence of D2 is that lim sup n δ n = ∞.Also we may assume that ℓ is even.We will often denote δ −1 n by η n .Moreover we consider a sequence ε n of independent random integers such that n .Thanks to D3, such a sequence exists: we may take ε n to be uniformly distributed on the interval of integers (−(s n , so we require a certain amount of dispersion; otherwise the problem is too close to the deterministic question of whether S itself is asymptotically additively decomposable, which is completely different.
Finally we consider the random sequence c n = ε n + s n and the random set Observe that the definition of ε n ensures that c n < c n+1 for all n, so that the sequence ε n uniquely determines C. The following theorem shows that we can take S to be the set of the primes or the set of the sums of two squares.Basically, it suffices to disturb the nth prime or sum of two squares by a random integer of standard deviation a small power of log log n.Theorem 4. The set P of the primes and the set 2Q of the sums of two squares fulfill the hypotheses S1,S2.When S is either of these two sets and the sequence δ satisfies δ n ≤ (s n+1 − s n−1 )/2 for all n ∈ N and δ n ≫ min((s n+1 − s n−1 )/2, (log log n) ι ) for n ≥ 2 and some ι > 0 arbitrarily small, the properties D1-3 hold.
In this sense, almost every small perturbation of either of these two sets is totally primitive.Regarding sums of two squares, this statement is a kind of inverse Atkin theorem: Atkin [1] proved that for almost every small perturbation Q ′ of the set Q of the squares, the sumset 2Q ′ is dense, in sharp contrast with 2Q.Here we perturb the sumset instead of perturbing the summands, and lose almost surely the decomposability property.

Proof of Theorems 1 and 2
Both results rely on the finite tuples property, which is the property of a set C ⊂ Z such that for any finite H ⊂ Z, there exist infinitely many n ∈ N such that n + H ⊂ C.
Here let (ξ n ) n∈N be a sequence of independent identically distributed Bernoulli variables satisfying P(ξ n = 1) = P(ξ n = 0) = 1/2 for each n ∈ N and let C = {n ∈ N : ξ n = 1}.Lemma 5. Almost all subsets of N, and therefore almost all symmetric subsets of Z, have the finite tuples property.That is, with probability 1, C and therefore C ∪ −C have the finite tuples property.
Proof.Fix a particular finite non empty set H ⊂ Z of cardinality k.Then for any given n ∈ N large enough, the probability that a random subset C ⊂ N is such that n + H ⊂ C is 2 −k .One can extract an infinite sequence n j of integers such that the events n j + H ⊂ C are pairwise independent, for instance n j = j(diamH+1) where diam denotes the diameter (difference between maximum and minimum).Therefore, by the Borel-Cantelli lemma, with probability 1, there exist infinitely many n such that n + H ⊂ C. Now this is true for any particular H, but since there are only countably many finite tuples H ⊂ N, we can take the intersection and conclude.Lemma 6.Any symmetric subset of Z satisfying the finite tuples property is additively Zdecomposable; it is in fact the difference set of two infinite subsets of N.
Proof.Indeed, let D be a symmetric subset of Z satisying the finite tuple property (and therefore infinite), and let D = {d 1 , d 2 , . . .} be an ordering of D. We will construct iteratively increasing sequences

Proof of Theorems 3 and 4
Proof of Theorem 3. Let Dec be the set of additively decomposable subsets of N. Since an asymptotically additively decomposable set differs from an element of Dec by finitely many editions, of which there are countably many, it suffices to show that P(C ∈ Dec) = 0. Let Dec 1 be the set of sets C of the form C = A + B for some A, B ⊂ N satisfying min(|A|, |B|) ≥ 2 and A(x) + B(x) < log 2 2 x f (x) log f (x) for infinitely many integers x.Let Dec 2 be the set of sets C of the form First we show that P(C ∈ Dec 1 ) = 0. We note that Dec . Let x be a large integer and r ≤ x/2.We note that if To see this, note that We turn to Dec 3 .We note that Dec 3 = x 0 ≥1 x≥x 0 Dec 3 (x) where Dec 3 (x) is the set of sets C of the form C = A + B for some infinite sets A, B ⊂ N satisfying A(x) + B(x) ≥ log 2 2 x f (x) log f (x) .Suppose C = A+B where A and B are infinite and A(x)+B(x) ≫ x/(f (x) log f (x)), in particular max(A(x), B(x)) ≥ κx/(f (x) log f (x)) for some constant κ > 0 and every x large enough.Let x be a large integer and assume without loss of generality that A(x) ≥ κx/(f (x) log f (x)).Let b 1 < b 2 < . . .< b ℓ be in B. Then we know that for any n ∈ A the set {n for some j 1 < j 2 < . . .< j ℓ .However the first property of ε n implies easily that either Recall that, by hypothesis, the number of k such that s k+1 ≤ x and By Markov's inequality, we find that In view of equation ( 2) infer that P(C ∈ Dec 3 (x)) = o(1) whence P(C ∈ Dec 3 ) = 0.
Proof of Theorem 4. By the prime number theorem, when S is the set P of the primes, we have f P (x) ∼ log x which proves S1.Further, let π m (x) be the number of primes p ≤ x such that p + m ∈ P. Then by Selberg's sieve π m (x) ≪ x log 2 x p|m (1 + 1/p), where the implied constant is absolute; see for instance [6].Then the number of In particular the number of primes p k ≤ x such that p k+2 −p k ≤ h is O h (x/f (x) 2 ) = o h (x/(f (x) log f (x))) (in fact it is O h (x/f (x) 3 )).Thus S2 holds.
Similarly, for the set S = 2Q of sums of two squares, we have f S (x) ≍ √ log x by a classical result of Landau, which proves S1.Further, let θ m (x) be the number of sums of two squares s ≤ x such that s + m ∈ S. Then again θ m (x) ≪ x log x p|m,p≡3 mod 4 (1 + 1/p), where the implied constant is absolute; this may be achieved via Selberg's sieve, see [9].Arguing like in equation ( 3), we find that the number of s k ≤ x such that s k+1 ≤ s k + M is at most M x log x .In particular the number of sums of two squares s k ≤ x such that s k+2 − p k ≤ h is O h (x/f (x) 2 ) = o h (x/(f (x) log f (x))) (in fact here too, it is O h (x/f (x) 3 )).Thus S2 holds.
Simultaneously this proves D1.D3 holds by definition and D2 may be proven along the same lines as above.

Theorem 2 .
Almost all subsets of N contain a sumset A + B where both summands are infinite.More precisely, let C = {n ∈ N : ξ n = 1}.Then with probability 1, C contains a sumset A + B where both summands are infinite.

Theorem 3 .
The set C is almost surely totally primitive.

Lemma 7 .
a k , b k of positive integers such that D = {a k } − {b k } and d k = a k − b k .To achieve this, start with any pair (a 1 , b 1 ) such that d 1 = a 1 − b 1 .Assuming finite increasing sequences of positive integers a 1 , . . ., a k (forming a set A k ) and b 1 , . . ., b k (forming a set B k ) have already been constructed and satisfy a i − b i = d i and A k − B k ⊂ D, let us construct a k+1 / ∈ A k and b k+1 / ∈ B k such that A k ∪ {a k+1 } − B k ∪ {b k+1 } ⊂ D. Let us look for a positive integer x such that x − B k ⊂ D and x − d k+1 − A k ⊂ D (by symmetry equivalently −x + d k+1 + A k ⊂ D).There exist infinitely many such x, due to the finite tuples property applied to the tuple −B k ∪ −(d k+1 + A k ).So there exists such an x outside of the finite set A k ∪ (d k+1 + B k ), and we pick a k+1 = x and b k+1 = x − d k+1 and we are done.Any subset D of N satisfying the finite tuples property contains a sumset A+B where both summands are infinite.Lemmata 5 and 7 imply Theorem 2. Proof.We will construct iteratively increasing sequences a k , b k of positive integers such that {a k : k ∈ N}+ {b k : k ∈ N} ⊂ D. We start with any pair (a 1 , b 1 ) such that a 1 + b 1 ∈ D. Assuming pairwise distinct a 1 , . . ., a k (forming a set A k ) and b 1 , . . ., b k (forming a set B k ) have already been constructed and satisfy A k + B k ⊂ D, we first select a k+1 to be an integer outside A k satisfying a k+1 + B k ⊂ D, which exists by the finite tuple property.We then set A k+1 = {a k+1 } ∪ A k and take b k+1 ∈ N outside B k such that b k+1 + A k+1 ⊂ D.

2 xf
by definition of S and property D1.Taking r = ⌊ log 2 (x) log f (x) ⌋, we infer that P(C ∈ Dec 1 (x)) = exp(−Ω(x/f (x)) = o(1).This implies that P(C ∈ Dec 1 ) = 0. Now we seek to show that P(C ∈ Dec 2 ) = 0. Note that every set D ∈ Dec 2 satisfies lim sup min(d k+1 − d k , d k − d k−1 ) < ∞.Indeed, if D = A + B where B is finite, let H be the largest gap between two consecutive elements of B; then min(d k+1 − d k , d k − d k−1 ) ≤ H for every k.Further we claim that for any h and k integers, (1)