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 $0 < \alpha < 1$?

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 $0 < \alpha < 1$?

Note: here Erdős seems to use a slightly weaker notion of an additive basis (see [Er56] at the top of page 135). In particular, for this problem, a set is an additive basis of order $k$ if every natural number can be written as a sum of _at most_ $k$ elements of the set, rather than as a sum of _precisely_ $k$ elements.

A positive solution was given by GPT 5.5 Pro (prompted by gebyjaff, cleanup by Liam Price); in fact a sparse random set $B$ has this property, with $f(\alpha)\gg \alpha (1-\alpha)^2$.