OPEN
This is open, and cannot be resolved with a finite computation.
- $500
Let $A\subset (1,\infty)$ be a countably infinite set such that for all $x\neq y\in A$ and integers $k\geq 1$ we have\[ \lvert kx -y\rvert \geq 1.\]Does this imply that $A$ is sparse? In particular, does this imply that\[\sum_{x\in A}\frac{1}{x\log x}<\infty\]or\[\sum_{\substack{x <n\\ x\in A}}\frac{1}{x}=o(\log n)?\]
Note that if $A$ is a set of integers then the condition implies that $A$ is a primitive set (that is, no element of $A$ is divisible by any other), for which the convergence of $\sum_{n\in A}\frac{1}{n\log n}$ was proved by Erdős
[Er35], and the upper bound\[\sum_{n<x}\frac{1}{n}\ll \frac{\log x}{\sqrt{\log\log x}}\]was proved by Behrend
[Be35]. This $O(\cdot)$ bound was improved to a $o(\cdot)$ bound by Erdős, Sárkőzy, and Szemerédi
[ESS67].
In
[Er73] and
[Er77c] Erdős mentions an unpublished proof of Haight that\[\lim \frac{\lvert A\cap [1,x]\rvert}{x}=0\]holds if the elements of $A$ are independent over $\mathbb{Q}$.
Over the years Erdős asked for various different quantitative estimates, for example\[\liminf \frac{\lvert A\cap [1,x]\rvert}{x}=0\]or even (motivated by Behrend's bound)\[\sum_{\substack{x <n\\ x\in A}}\frac{1}{x}\ll \frac{\log x}{\sqrt{\log\log x}}.\]In
[Er97c] he offers \$500 for resolving the questions in the main problem statement above.
This was partially resolved by Koukoulopoulos, Lamzouri, and Lichtman
[KLL25], who proved that we must have\[\sum_{\substack{x <n\\ x\in A}}\frac{1}{x}=o(\log n).\]See also
[858].
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This page was last edited 06 April 2026. View history
Additional thanks to: Zachary Chase and Desmond Weisenberg
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #143, https://www.erdosproblems.com/143, accessed
