Hi everyone I´d like if someone could say me if the following is correct. Thanks in advance

Rotate the unit circle by a fixed angle $\alpha$, say $R: C \rightarrow C$; $(1,\theta)\mapsto (1,\theta+\alpha)$

a) If $\alpha/\pi$ is rational, show that each orbit of $R$ is a finite set.

b) If $\alpha/\pi$ is irrational, show that each orbit is infinite and has closure equal to $C$.

For a). Suppose that $\alpha/\pi=m/n$ in lowest terms, clearly $\alpha =\pi m/n$. By induction is easy to show that $R^kp=(1,\theta+k\alpha)$ for any $k$. Then $R^{2n}=(1,\theta+2\pi m)$ which is the initial point $p$.

For b). Suppose that the orbit of a point $p$ in the unit circle is finite. So there is some positive integer such that $R^np=p$, in other words $x=\cos \theta=\cos(\theta+n\alpha)$, where $\theta$ is the angle of the point. Thus $n\alpha=2\pi k$, i.e., $\alpha/\pi=2k/n\in \mathbb{Q}$.

If $\alpha/\pi\in \mathbb{Q}^c$ and $p$ is a point in the unit circle. We have to show that the orbit of $p$ is dense in $C$. Suppose not. So, there is an open interval $I$ on the cicle such that no point of the orbit of $p$ is in $I$. Let $r=\text{diam} \,I$. It follow that no point in the orbit of $p$ is in a distance $<r$. Since otherwise there would be eventually in $I$. Then we have that the orbit of $p$ it does have at most $r^{-1}$ elements, contrary to $\alpha/\pi$ being irrational.

  • $\begingroup$ It looks good. And more words just to add the one more character that the system needs to let me post this comment. $\endgroup$ – Lee Mosher Aug 3 '14 at 16:32
  • $\begingroup$ Hello can you explain to me how the last sentence follows from "Since otherwise there would be eventually in $I$." I get everything up to that point. $\endgroup$ – Sandeep Silwal Aug 3 '14 at 16:57
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    $\begingroup$ @SandeepSilwal Do you refer to the point: "it does have at most 1/r elements"? If so, since the distance of any element in the orbit of $p$ is $\ge r$ we cannot have infinitely many of them, this would imply that $\alpha/\pi$ must be rational. $\endgroup$ – Jose Antonio Aug 3 '14 at 17:27
  • $\begingroup$ Oh I get it now. Thanks! $\endgroup$ – Sandeep Silwal Aug 3 '14 at 17:30

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