For what kind of infinite subset A of $\mathbb Z$ and irrational number $\alpha$, is $\{e^{k\alpha \pi i}: k\in A \}$ dense in $S^1 $?

There is a well-known result saying that $\{e^{k\alpha \pi i}: k\in \mathbb Z \}$ is dense in $S^1$. By density, we can select an infinite subset $A$ of $\mathbb Z$ such that $\{e^{k\alpha \pi i}: k\in A \}$ is not dense in $S^1$ any more. So there is a natural question:

For what kind of infinite subset A of $\mathbb Z$ and irrational number $\alpha$, is $\{e^{k\alpha \pi i}: k\in A \}$ dense in $S^1 $ ?

I feel like if A contains a collection of an infinite Arithmetic sequence(like 1,3,5,...), then $\{e^{k\alpha \pi i}: k\in A \}$ should be dense in $S^1$(I don't know how to prove this). But this doesn't look like a necessary condition. I think such a "basic" problem must be studied before, so every solution or partial solution(for specific types of A) or reference will be appreciated!

Solution 1:

It is certainly not necessary that $A$ contains an infinite arithmetic sequence, nor even an arbitrarily long arithmetic sequence. In fact, the set $A$ can be as sparse as you like in the following sense: for any irrational $\alpha$ and any sequence of natural numbers $n_1 < n_2 < n_3 < ...$ we can find a subset $A$ of the natural numbers with its elements listed in increasing order as $A=\{a_1,a_2,a_3,...\}$ such that $a_j \ge n_j$ for all $j$ and such that $\{e^{k \alpha\pi i} : k \in A\}$ is dense.

To prove this, choose $B_1,B_2,...$ to be a countable base for the topology on $S^1$. Then define the elements $a_j$ of $A$ by induction subject to the stated conditions plus the additional condition that $e^{a_j \alpha \pi i} \in B_j$. The choice is always possible because at each stage $j$ of the induction, only finitely many values of $k$ are disallowed by the previous stages, and yet what remains after removing those finitely many values of $e^{k \alpha \pi i}$ is still a dense subset of $S^1$ and hence there are infinitely many such values to choose from in $B_j$.