I'm reading ch 7 of Nadkarni's book Spectral Theory of Dynamical Systems and I've come across this statement in a proof I'm currently trying to understand:
If $z\in S^1$ (where $S^1$ denotes the Circle Group) and $z^{n_k}$, with $k\in\mathbb{N}$, has only a finite set of limit points of the form $e^{2\pi i \frac{p}{q}}$ (where $p$ and $q$ are integers and $q>0$), then for some integer $p_0$, $z^{-p_0n_k}\rightarrow 1$ as $k\rightarrow\infty$.
I don't really understand, why this holds and would very much appreciate any input on this.
+1 I’m more than twenty years dealing with topological groups and I also don’t understand this (unless the trivial case $p_0=0$). Indeed, let $S^1_Q=\{z\in S^1: z=e^{2\pi r}\mbox{ for some }r\in \Bbb Q \}$ and $z\in S^1\setminus S^1_Q$ be any point. Then $\{z^n:n\in\Bbb N\}$ is dense in $S^1$. Therefore for any point $t\in S^1\setminus S^1_Q $ there exists a sequence $\{z^{n_k}: k\in\Bbb N\}$ of natural powers of $z$ converging to $t$. Then $\{z^{n_k}\}$ has no other limit points, but for each non-zero integer $p_0$ the sequence $\{z^{-p_0n_k}\}$ converges to $z^{-p_0}\ne 1$.