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A related question was posted on MSE (link), although that had some additional stipulations.

Let $S^1=\{w\in \mathbb{C}:|w|=1\}$ be the unit complex circle. Call a set $R\subseteq\mathbb{N}$ good if, for all $x$ irrational, the set $\{e^{2\pi i xr}:r\in R\}$ is dense in $S^1$. Are there more simple equivalent characterizations of good sets $R$?

This is equivalent to asking the question in the title, and I showed in my answer here that it is equivalent to having a sequence $(r_n)_{n=1}^\infty$ exist for each irrational $x$ such that $(r_nx)_{n=1}^\infty$ is uniformly distributed mod $1$. I know that for a fixed $x$, characterizing all sequences satisfying this condition is tricky, but I was hoping that a (simpler) answer would exist for this question. I know that a sequence $(r_n)$ will satisfy $(r_nx)_{n=1}^\infty$ uniformly distributed mod $1$ for almost all $x$ as well, which makes finding a counterexample tricky (or at least, a non-trivial one where $|R|$ is infinite).

Some more specific questions:

  1. If $R$ has positive lower density, is $R$ good? If not, what about if $R$ just has positive upper density? I suspect that the answer to both of these questions is yes. Are there stricter density conditions that would still make $R$ good?
  2. Another answer to the MSE question showed that if there are arbitrarily long arithmetic progressions, then $R$ is good (note that this also means no density conditions are necessary). Does the same hold for arbitrarily long quadratic progressions? Polynomial progressions of fixed degree (if degree can vary, then we could always find a polynomial with those values)?
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    $\begingroup$ I am not sure about your answer on the other site, but it is a landmark result of Furstenberg that for any irrational $x$, the set $\{2^{n}3^{m}.x\}$ is dense mod $1$, but is not (necessarily) equidistributed, i.e. by choosing a proper Liouville number. $\endgroup$
    – Asaf
    Commented Jul 23 at 19:07
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    $\begingroup$ Positive (lower) density is not enough - for instance the set $R=\{n\in\mathbb N\mid \{n\sqrt{2}\}\in[0,1/2]\}$ has density $1/2$ (by Weyl) but by construction the set $R\sqrt{2}$ is not dense mod $1$. On the other hand, if $R$ has (upper?) density $1$, it should hold, as again by Weyl you can argue it will visit every interval infinitely often. $\endgroup$
    – Wojowu
    Commented Jul 23 at 19:30
  • $\begingroup$ @Asaf But there would be an equidistributed sequence with elements all in that set, right? $\endgroup$
    – Varun Vejalla
    Commented Jul 27 at 3:43
  • $\begingroup$ @VarunVejalla , I don't understand your term "equidistributed sequence" in the set. For equidistribution (at least in the standard meaning in number theory), we fix a sequence and then a base point, i.e. the sequence $n^{2}.x$ is equidistributed for all irrational $x$. If you enumerate the elements of $\{2^{n}3^{m}\}$ in increasing order, call the resulting sequence $r_{n}$, there are irrationals $x$ such that $\{r_{n}.x\}$ is not equidistributed. Obviously for almost all $x$ (Lebesgue), the sequence $\{r_{n}.x\}$ will be equidistributed, by ergodicity of $\times 2,\times 3$. $\endgroup$
    – Asaf
    Commented Jul 29 at 14:05
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    $\begingroup$ @Asaf You don't necessarily have to enumerate those elements in increasing order though. I'm just saying that you'd need a sequence $(r_1,r_2,\cdots)$ such that $r_i\in R$ for all $i$ and $(r_n.x)$ is equidistributed. Does that make sense? The sequence can be different for different values of $x$. $\endgroup$
    – Varun Vejalla
    Commented Jul 29 at 22:12

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