# Multiples in sets of positive upper density

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?

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.
Besicovitch's result mentioned by Seva is quite hard, while in your question you may simply take $A=\cup [n!+1,2n!]$.
• Lacunary sets like this were the first thing I thought of - but for your construction, $A=[2,2]\cup[3,4]\cup[7,12]\cup[25,48]\cup\dotsb$, and there are lots of elements of $A$ divisible by each other? – Seva Jul 22 '17 at 9:51
• But for each fixed $k$, $M_A(k)=A\cap \frac1kA$ is finite – Fedor Petrov Jul 22 '17 at 10:18