TITLE: Multiples in sets of positive upper density
QUESTION [11 upvotes]: Suppose we are given $A \subseteq \mathbb{N}$ with $\lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} > 0$. For $k\in \mathbb{N}, k\geq 2$ we set $$M_A(k) = \{a\in A: ka \in A\}.$$
Does there exist $k\in \mathbb{N}, k\geq 2$ such that $M_A(k)$ is infinite?

REPLY [15 votes]: Not necessarily: you can in fact have $M_k(A)=\varnothing$ for all integer $k\ge 2$. This was shown by Besicovitch ("On the density of certain sequences of integers", Math. Ann. 110 (1935), no. 1, 336–341) who has constructed a set (of positive integers) of positive upper density such that none of the elements of the set is divisible by any other element.

REPLY [10 votes]: Besicovitch's result mentioned by Seva is quite hard, while in your question you may simply take $A=\cup [n!+1,2n!]$.