Let $R \subseteq \mathbb{N}$. We say that $R$ is adequate if:
$$\forall n \in \mathbb{N} \; \; \forall \varepsilon > 0 \; \; \forall w \in S^1 \; \; \exists r\in R \; \; |w^n - w^r| < \varepsilon$$
where $S^1$ denotes the set $\{|w| = 1\} \subseteq \mathbb{C}$.
In words, given any $w \in S^1$ and any $n \in \mathbb{N}$, we can approximate $w^n$ arbitrarily well by $\{w^r \}_R$. This is equivalently stated as defining that $S$ is adequate if and only if $\{w^n\}$$\subseteq$$\overline{\{w^r\}}$ for every $w \in S^1$. The study of such sets has become relevant to me in some of my work in complex analysis.
Question: Is there an accessible description of the adequate subsets $R \subseteq \mathbb{N}$?
Here are some elementary results I have proved.
We say that $R$ is perfect if it has every residue mod every base; that is to say, $\forall m,n \in \mathbb{N}, \; \exists r \in R$ s.t. $r \equiv m\text{ mod }{n}$.
Lemma: If $R$ is adequate, then it is perfect.
Proof: Consider $e^{2 \pi i q}$ for $q \in \mathbb{Q}$. $\square$
Lemma: If $R$ contains arbitrarily long stretches of consecutive integers, then it is adequate.
Proof (sketch): Let $R$ be as described. This is clearly sufficient for $w = e^{2 \pi i q}$ for $q \in \mathbb{Q}$. Therefore, let $w = e^{2 \pi i t}$ for $t \in \mathbb{R} \setminus \mathbb{Q}$. By the pigeonhole principle, there exists $N \in \mathbb{N}$ such that $0 < |w^N - 1| < \varepsilon$. Let $M >> N \varepsilon^{-1}$ and let $r_1, ..., r_M$ be a sequence of consecutive integers in $R$. Then, there exists $r \in \{ r_1, ..., r_M\} \subseteq R$ such that $|w^n - w^r| < \varepsilon$. $\square$
Corollary: $R$ need not have positive upper density.
I am interested in an answer to the question I stated, or failing that, any examples of such sets $R$ which anyone is able to construct which are non-trivial in the sense that they are not covered by the second lemma above -- or even any perfect sets which are not covered by the second lemma.
Edit: I found a perfect set which does not contain any consecutive integers (in fact, far from it). Consider:
$$(1!), (2!+1), (3!), (4!+1), (5!+2), (6!), (7!+1), (8!+2), (9!+3), (10!), ...$$
I hope the pattern is clear. I find it difficult to imagine that this set is adequate.