A property of rapid sequences of natural numbers

Question

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

Taras Banakh and me proceed a long quest answering a question 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.

Answer 1

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.

de Mathan, B., Numbers contravening a condition in density modulo 1, Acta Math. Acad. Sci. Hung. 36, 237-241 (1980). ZBL0465.10040.