Rotation map by an irrational on $S^1$

Question

$$\newcommand{\ob}{\operatorname{orbit}} $$ Let $S^1=[0,1)~~mod~~1$. Let $0<\alpha<1$ be an irrational and let $\tau$ be the rotation by $\alpha$ and $\beta \in \ob(\alpha)$. Is it true that if $0\leq x<\tau^{-m}(\beta)<x'<\beta$ then $0<\tau^m(x)<\beta<\tau^m(x')<1$ for all $\beta \in \ob(\alpha)$ or some $\beta \in \ob(\alpha)$. I tried to do one part by taking $\beta=n\alpha-[n\alpha]$, where $[]$ is the greatest integer function, but finally I am getting as $0<\tau^m(x)<\beta+1$.

Answer 1

The orbit of $\alpha$ is dense due to its irrationality. Thus you can easily choose $m>0$ and $\beta$ from the orbit of $\alpha$ such that $0<\tau^{-m}(\beta)<\beta<1/2 \mod 1$. We get the second train of inequalities from the first one by rotating all the considered points by $0< m\alpha \leq 1/2 \mod 1$. The inequalities are preserved because $\beta+m\alpha<1/2+1/2=1$; this is all nicely visible when drawn on the circle. So it is definitely true for some $\beta$ (independently from whether it is from the orbit of $\alpha$ or not).
But it is not true for every $\beta$ from the orbit of $\alpha$. Just contemplate the situation when $\beta-\tau^{-m}(\beta)>1/2$.

To the original problem - one can show, in fact, more:
For any $\beta \in (0,1)$ and any $x,x' \in \mathbb S^1$ there is $m \geq 0$ such that $\tau^m(x) \in [0,\beta]$ and $\tau^m(x') \in (\beta,1)$.
One way how to see this is the following. Thanks to irrationality of $\alpha$ we know that some iterate of $\tau$ is a rotation by a "really small angle". For our purposes this means that for arbitrary fixed $x<x' \in \mathbb S^1$, we take $n>0$ such that $\tau^n =$ rotation by $\gamma$ with $0<\gamma<\min\{\beta,1-\beta,x'-x\}$.
Now, take any $k \geq 0$ such that $\tau^k(x) \in [0,\beta]$ and $\tau^k(x)<\tau^k(x')$ (again, this is possible due to irrationality of $\alpha$). If $\tau^k(x') \in (\beta,1)$, we're done. If not, look at $\tau^{k+n}(x)$ and $\tau^{k+n}(x')$, $\tau^{k+2n}(x)$ and $\tau^{k+2n}(x')$, etc. until the first ocurrence of $\tau^{k+ln}(x') \in (\beta,1)$. Because of how $\gamma$ was chosen, $\tau^{k+ln}(x') \in [0,\beta]$. Obviously, $m=k+ln$ solves the problem. (I switch between the circle and the interval relatively freely hoping that it doesn't cause misunderstandings.)