OPEN
This is open, and cannot be resolved with a finite computation.
Does there exist $B\subset\mathbb{N}$ which is not an additive basis, but is such that for every set $A\subseteq\mathbb{N}$ of Schnirelmann density $\alpha$ and every $N$ there exists $b\in B$ such that\[\lvert (A\cup (A+b))\cap \{1,\ldots,N\}\rvert\geq (\alpha+f(\alpha))N\]where $f(\alpha)>0$ for $0<\alpha <1 $?
The Schnirelmann density is defined by\[d_s(A) = \inf_{N\geq 1}\frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N}.\]
Erdős
[Er36c] proved that if $B$ is an additive basis of order $k$ then, for any set $A$ of Schnirelmann density $\alpha$, for every $N$ there exists some integer $b\in B$ such that\[\lvert (A\cup (A+b))\cap \{1,\ldots,N\}\rvert\geq \left(\alpha+\frac{\alpha(1-\alpha)}{2k}\right)N.\]It seems an interesting question (not one that Erdős appears to have asked directly, although see
[35]) to improve the lower bound here, even in the case $B=\mathbb{N}$. Erdős observed that a random set of density $\alpha$ shows that the factor of $\frac{\alpha(1-\alpha)}{2}$ in this case cannot be improved past $\alpha(1-\alpha)$.
This is a stronger property than $B$ being an essential component (see
[37]). Linnik
[Li42] gave the first construction of an essential component which is not an additive basis.
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This page was last edited 16 September 2025.
Additional thanks to: Terence Tao and Desmond Weisenberg
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