**Theorem**
Let $\displaystyle \lambda\in\mathbb{R}$ and consider:

$\displaystyle T_{\lambda}:S^1\rightarrow S^1,\quad T_{\lambda}(\theta)=\theta +2\pi\lambda$

Then, each orbit of $\displaystyle T_{\lambda}$ is dense in $\displaystyle S^1$ if $\displaystyle \lambda$ is irrational

*Proof*
Let $\displaystyle \theta\in S^1$ . The points on the orbit of $\displaystyle \theta$ are distinct for if $\displaystyle T_{\lambda}(\theta)^n=T_{\lambda}^m(\theta)$ we would have $\displaystyle (m-n)\lambda\in \mathbb{Z}$, so that $\displaystyle n=m$ .

__Any infinite set of points on the circle must have a limit point. Thus, given any $\displaystyle \epsilon >0$ there must be integers $\displaystyle n$ and $\displaystyle m$ for which $\displaystyle |T_{\lambda}^n(\theta)-T_{\lambda}^m(\theta)|<\epsilon$ .__ Let $\displaystyle k=n-m$ . Then $\displaystyle |T_{\lambda}^k(\theta)-\theta|<\epsilon$ .

Now $\displaystyle T_{\lambda}$ preserves lengths in $\displaystyle S^1$ . Consequently, $\displaystyle T_{\lambda}^k$ maps the arc connecting $\displaystyle \theta$ to $\displaystyle T_{\lambda}^k(\theta)$ to the arc connecting $\displaystyle T_{\lambda}^k(\theta)$ and $\displaystyle T_{\lambda}^{2k}(\theta)$ which has length less than $\displaystyle \epsilon$ . In particular it follows that the points $\displaystyle \theta,T_{\lambda}^k(\theta),T_{\lambda}^{2k}(\the ta),\ldots$ partition $\displaystyle S^1$ into arcs of length less than $\displaystyle \epsilon$ . Since $\displaystyle \epsilon$ was arbitrary, this completes the proof.

Fernando Revilla