Let $X \subset \mathbb{N}$ and say that $X$ is super-equidistributed if for all $\alpha \in \mathbb{R} \setminus \mathbb{Z}$ there exists $C(\alpha) > 0$ such that for all $N$ $$ \left| \sum_{x \in X, x \leq N} \exp(2\pi i \alpha x) \right| < C(\alpha). $$

My conjecture is the following: If $X$ is super-equidistributed, then either $X$ is finite or $\mathbb{N} \setminus X$ is finite.

Why should this be true? Well there isn't too much evidence but one can show that if $X$ is super-equidistributed, then for all integers $a,q$ we have $$ \#\{x \in X: x \leq N, x \equiv a \pmod{q}\} = \frac{|X \cap [1,N]|}{q} + O_q(1). $$

This is a strong condition on $X$, but it is not enough. For example a friend of mine suggested the set $S = \bigcup_{n \geq 1} [n!, 2 \cdot n!) \cap \mathbb{Z}$. This satisfies this property mod $q$ for all $q$ but is neither finite nor is $\mathbb{N} \setminus S$. However if we pick $\alpha = e$, one can show that $\sum_{x \in S, x \leq N} \exp(2 \pi i \alpha x)$ is unbounded. To see this one essentially uses that the fractional part of $e \cdot k!$ is very well behaved for any $k$.

Also note that any set $X$ such that $X$ or $\mathbb{N} \setminus X$ is finite is super-equidistributed since $\sum_{x \geq 0} \exp(2 \pi i \alpha x) = (1 - \exp(2 \pi i \alpha))^{-1}$.

I would appreciate any thoughts on this problem. Possible counter-examples, solutions or consequences are all welcome.