Set $T^{\mathbb N}x$ dense in $\mathbb S^1$ (Poincaré recurrence theorem)

Question

Let $Ω =\mathbb S^1$ be the unit circle in $\mathbb R^2 = \mathbb C$, and let $T : Ω → Ω$ be multiplication by $e^{i\alpha}$. For $α \notin π\mathbb Q$ and every $x ∈ Ω$, is the set $T^{\mathbb N}x$ dense in $Ω$?

I think the answer is yes and I was trying to prove it with the poincaré recurrence theorem but I don´t know how could I finish the proof.

Answer 1

Without loss of generality, take $x=1$: indeed, if $A=e^{2\pi\alpha\mathbb N}$ is dense in $\mathbb S^1$, so it is $Ax=e^{2\pi\alpha\mathbb N}x$.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, take a sequence $q_n\in\mathbb{Q}$ such that $\alpha-q_{n}\to0$.

Let $q_n=\frac{a_{n}}{b_{n}}$, $b_{n}\in\mathbb{N}$, we want to prove that $T^{2b_n}\to1$. Indeed, since $e^{2\pi i\theta}=e^{2\pi i(\theta+z)}$, $\forall z\in\mathbb{Z}$, we have \begin{align}T^{2b_n}&=e^{2\pi i\,\alpha b_n}=e^{2\pi i(\alpha b_n-a_{n})}=\\ &=\underbrace{e^{2\pi i\, b_n}}_{=1}e^{2\pi i\left(\alpha-\frac{a_{n}}{b_n}\right)}=e^{2\pi i(\alpha-q_{n})}\to e^{2\pi i\,0}=1.\end{align}

In the same fashion, given $p\in\mathbb{q}$, considering $p_n=p-q_n$, you can prove that $T^{c_n}\to e^{2\pi i p}$ for a suitable choice of $c_{n}\in\mathbb{N}$.

You can conclude by remarking that $e^{\pi\mathbb{Q}}$ is dense in $e^{\pi\mathbb{R}}$.