Let $\alpha$ a irrational number, and $R:S^{1}\rightarrow S^{1}$ the irrational rotation, i.e., $[x]\rightarrow[x+\alpha]$. I need to prove that, for all $[x]\in S^{1}$, the set $\{R^{n}([x])\}$ is dense in $S^{1}$.
First, I can write $R$ by
$$R(e^{2\pi ix})=e^{2\pi i (x+\alpha)}.$$
So, I can write $R^{n}(x)$, $n\in\mathbb{Z}$ by
$$R^{n}(e^{2\pi i x})=e^{2\pi i(x+n\alpha)} $$
I need to prove that, for all $e^{2\pi i x}\in S^{1}$ and for all $[y]=e^{2\pi i y}\in S^{1}$, every neighborhood $V$ of $[y]$ contains a point $[z_{y}]=e^{2\pi i z}$ such that $[z_y]=R^{n}([x])$ for some $n\in\mathbb{Z}$. That is,
$e^{2\pi i z}=e^{2\pi i (x+\alpha n)}\Rightarrow 2\pi iz=2\pi i(x+\alpha n)+2ik\pi\;\textrm{for some}\;k\in\mathbb{Z}\Rightarrow z=x+\alpha n+k.$
Is my way correct? If it does, how can I proceed now? If it doesn't, what I need to do?
I don't think this question is duplicate. I'm showing my attempt to proof, that is different from other proofs.
