If an infinite set $S$ of positive integers is equidistributed, is $S+S$ also equidistributed?

Solution 1:

$S+S$ need not be equidistributed.

Let $(a_n)_n$ be a sequence of even positive integers such that $$a_{n+1} > 3+a_n+a_{n-1}+\dots+a_2+a_1$$ for each $n \ge 1$ and such that $\{\frac{a_n}{2} : n \ge 1\}$ is equidistributed. It should be clear that such a sequence exists; let me know if you want details. Let $$S = \cup_{n=1}^\infty \{a_n,a_n+1\}.$$ Then $S$ is equidistributed: it is clearly equidistributed mod $2$, and that $\{\frac{a_n}{2} : n \ge 1\}$ is equidistributed implies $S$ is equidistributed mod $n$, for any $n \ge 3$.

Now, $S+S = \cup_{1 \le n \le m < \infty} \{a_n+a_m,a_n+a_m+1,a_n+a_m+2\}$. Note the union is a disjoint one, since $a_{n+1} > 3+a_n+\dots+a_1$ for each $n \ge 1$. Therefore, $S+S$ is not equidistributed mod $2$, since $2$ out of $3$ of $a_n+a_m, a_n+a_m+1,a_n+a_m+2$ (namely, the first and the third) are even.