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I think the title of the question says it all. I unfortunately did not seem to conclude anything. Some ideas I had:

It is easy to show that (given $T$ is the rotation) $\{T^n(\theta)\}$ is a set of distinct points. Furthermore, given that the circle is a compact metric, it must have a limit point $x$. By continuity of the rotation function, $T^n(x)$ is a limit as well since taking $T$ of every term yields the same sequence (with only the first term removed). By induction, we have infinitely many distinct limit points $\{T^n(x)\}$.

That's all I could come up with! It was also easy to show that the orbit is infinite. I still don't seem to be able to get close to the required result however.

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Please, use $..$ to write your math formulas. – Sigur Jan 19 '13 at 17:11
@MR1992: You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Jan 19 '13 at 17:13
I have typeset the formulas in your question. Please double-check that I transcribed correctly, note the use of dollar signs, and see the previous comment for more resources on formatting mathematics. – Neal Jan 19 '13 at 17:48
up vote 3 down vote accepted

First, let's get rid of the $\pi$ and talk about fractions of a circle and regular irrational numbers instead.

Let $a<1$ be an irrational number, and assume that the orbit that you get by going $a$ of the circle (say, clockwise) $n$ times around a point is not dense in the circle, that is, there exists an open interval $I$ on the circle of size $\varepsilon$ which it does not cross.

Assume there were $m>n$, such that the points corresponding to rotations by $na$ and $ma$ have a distance $d<\varepsilon$. This would mean that rotation by $m-n$ gives a new point, which is of distance $d$ from the previous point, and because $d<\varepsilon$, this means that if this rotation is done repeatedly, eventually a point will be in $I$.

So there can be no such $m,n$, meaning that points in the orbit are at least $\varepsilon$ apart, which means that there are at most $\frac{1}{\varepsilon}<\infty$ of them, contradicting the fact that $a$ is irrational.

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