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?
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asked Jul 22, 2017 at 8:49
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2 Answers
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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.
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$\begingroup$ That's an amazing paper, thanks for making me aware of it! $\endgroup$ Commented Jul 22, 2017 at 13:57
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Besicovitch's result mentioned by Seva is quite hard, while in your question you may simply take $A=\cup [n!+1,2n!]$.
answered Jul 22, 2017 at 9:37
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$\begingroup$ 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? $\endgroup$– SevaCommented Jul 22, 2017 at 9:51
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6$\begingroup$ But for each fixed $k$, $M_A(k)=A\cap \frac1kA$ is finite $\endgroup$ Commented Jul 22, 2017 at 10:18