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Given a subset $S$ of the positive integers $\mathbf{N}$, let $\mathrm{d}^\star(S)$ be its upper asymptotic density, that is, $$ \mathrm{d}^\star(S)=\limsup_{n\to \infty}\frac{|S \cap [1,n]|}{n}. $$ Also, for each integer $k \ge 0$, set $S+k:=\{s+k:s \in S\}$.

Question. Do there exist $X,Y \subseteq \mathbf{N}$ such that $\mathrm{d}^\star(X)>0$, $\mathrm{d}^\star(Y)>0$, and $$ \mathrm{d}^\star(X \cap (Y+k))=0 \,\,\,\text{ for all }k\ge 0? $$


Ps. The answer is positive (constructively, modulo my mistakes) for the following finite-analogue: "Do there exist infinite sets $X,Y \subseteq \mathbf{N}$ such that $X \cap (Y+k)$ is finite for all $k\ge 0$?"

To this aim, let $X$ be the set of squares and $Y$ be the set of cubes which are not also squares. Then, by a result of Siegel, the equation $|x-y|=k$ has finitely many solutions with $x \in X$ and $y \in Y$ for every fixed nonzero integer $k$.

Edit: I just discovered an excellent proof of the finite-analogue without translation invariance in Albiac-Kalton "Topics in Banach Space Theory" (lemma 2.5.3), namely:

"There exists an uncountable family $(A_i)$ of infinite sets of $\mathbf{N}$ such that $A_i \cap A_j$ is finite for all $i\neq j$."

Proof:Identify $\mathbf{N}$ with $\mathbf{Q}$ and, for each irrational number $\theta$ let $(a_{\theta,n})$ be a sequence of rationals converging to $\theta$. Then the uncountable family $$\{\{a_{\theta,n}:n\in \mathbf{N}\}: \theta \in \mathbf{R}\setminus \mathbf{Q}\}$$ satisfies the hypothesis.

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  • $\begingroup$ When you mention the construction of almost disjoint family from Albiac-Kalton, you mean that for any two sets from this family you have that $X\cap Y+k$ has zero density, right? Whether or not all these sets have positive upper density probably depends on the choice of the sequences and the choice of the bijection between $\mathbf N$ and $\mathbf Q$. $\endgroup$ Mar 31, 2018 at 14:34
  • $\begingroup$ @MartinSleziak You are right in one point: I didn't check translations and the result depends on the choice of the bijection. On the other hand, I was referring the finite analogue, hence we'd need only $X \cap Y+k$ finite for each $k$. I don't know whether this can be fixed $\endgroup$ Mar 31, 2018 at 14:39

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The answer is "yes".

Let $X = \bigcup [4^n, 2 \cdot 4^n)$ and $Y = \bigcup [2 \cdot 4^n, 3 \cdot 4^n)$. Then $d^\star(X), d^\star(Y) \geqslant \tfrac{1}{4}$, but $X \cap (Y + k)$ is a finite set for every $k \geqslant 0$.

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  • $\begingroup$ That was completely elementary, thank you Mateusz $\endgroup$ Mar 9, 2018 at 13:48
  • $\begingroup$ @PaoloLeonetti: Still, it took me a while. You're welcome! $\endgroup$ Mar 9, 2018 at 13:52

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