$\newcommand{\IR}{\mathbb R}$
$\newcommand{\IT}{\mathbb T}$
$\newcommand{\w}{\omega}$
$\newcommand{\e}{\varepsilon}$


[Taras Banakh](https://mathoverflow.net/users/61536/taras-banakh) and me proceed a long quest answering a [question](https://math.stackexchange.com/questions/1203722/approximate-vanishing-in-pontryagin-dual) of ougao at Mathematics.SE. Recently we encountered a notion of a remote sequence. We are interested whether it was studied before and want to know a solution of a problem below.


Recall that a circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$, endowed with the operation of multiplication of complex numbers and the topology inherited from $\mathbb C$ is a topological group. Let $(r\_n)\_{n\in\w}$ be an increasing sequence of natural numbers. The sequence $(r\_n)\_{n\in\w}$ is called *remote* if there exists $z\in\IT$ such that $\inf\_{n\in\w}|z^{r\_n}-1|>0$. A *minimum growth rate* of $(r\_n)\_{n\in\w}$ is a number $\inf\_{n\in\w} r\_{n+1}/r\_{n}$.


**Problem.** Find the maximal set $M\subseteq [1,\infty)$ such that if the minimum growth rate of $(r\_n)\_{n\in\w}$ belongs to $M$ then $(r\_n)\_{n\in\w}$ is remote.


*Our try.* It is easy to see that $1\not\in M$. On the other hand, we can show that any real number $m>2$ belongs to $M$ as follows. Let $m$ be the minimum growth rate of $(r\_n)\_{n\in\w}$. Find $\e>0$ such that $2+\frac{\e}{1-\e}\le m$. Then $\frac{r\_{n+1}}{r\_n}\ge 2+\frac\e{1-\e}=\frac{2-\e}{1-\e}$ and hence $\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ for every $n\in\w$. Let $W=\{e^{it}:|t|<\pi \e\}$ and for every $n\in\w$ consider the set $U\_n=\{z\in\IT:z^{r\_n}\in W\}$. Consider the exponential map $\exp:\IR\to \IT$, $\exp:t\mapsto e^{2\pi t i}$, and observe that for every $n\in\w$ any connected component of the set $\exp^{-1}[U\_n]$ is an open interval of length $\frac{\e}{r\_n}$ and every connected component of the set $\IR\setminus \exp^{-1}[U\_n]$ is a closed interval of length $\frac{1-\e}{r\_n}$. Since $\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ and $$\frac{1-\e}{r\_{n+1}}+\frac\e{r\_{n+1}}+\frac{1-\e}{r\_{n+1}}=\frac{2-\e}{r\_{n+1}}=\frac{1-\e}{r\_n}\cdot\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le \frac{1-\e}{r\_n},$$every connected component of the set $\IR\setminus \exp^{-1}[U\_n]$ contains a connected component of the set $\IR\setminus\exp^{-1}[U\_{n+1}]$. Then for every $n\in\w$ we can choose a connected component $I\_n$ of the set $\IR\setminus\exp^{-1}[U\_n]$ such that $I\_{n+1}\subseteq I\_n$. By the compactness of the set $I\_0$, the intersection $\bigcap\_{n\in\w}I\_n$ contains some real number $t$. Then the point $z=\exp(t)$ does not belong to $\bigcup\_{n\in\w}U\_n$, which implies that $z^{r\_n}\notin W$ for every $n\in\w$. The definition of the neighborhood $W$ ensures that $\inf\_{n\in\w}|z^{r\_n}-1|>0$, that is the
sequence $(r\_n)\_{n\in\w}$ is remote.


Thanks.



A property of rapid sequences of natural numbers
A sequence is called *lacunary* if, in your terminology, its minimum growth rate is strictly greater than $1$. The following articles prove that every lacunary sequence is remote. If I understand your question correctly, this means that the (only) maximal $M$ you seek is the interval $(1,\infty)$.


*Pollington, Andrew D.*, On the density of sequences $\{n\_k\xi\}$, Ill. J. Math. 23, 511-515 (1979). [ZBL0401.10059](https://zbmath.org/?q=an:0401.10059).


*de Mathan, B.*, [**Numbers contravening a condition in density modulo 1**](http://dx.doi.org/10.1007/BF01898138), Acta Math. Acad. Sci. Hung. 36, 237-241 (1980). [ZBL0465.10040](https://zbmath.org/?q=an:0465.10040).


