Density of a set of periodic points

Question

Having $f(\theta)=2\theta$ a map of $S^1$, I have to prove that periodic points of $f$ are dense in $S^1$, i.e, $\theta$ is periodic of period $n$ if and only if $2^n\cdot\theta=\theta+2k\cdot\pi$ for some integer $k$, i.e, if and only if $\theta=\frac{2k\cdot\pi}{2^n-1}$. But, do you have any idea how I can prove that the set of periodic points is dense in $S^1$?

Answer 1

Since $x\mapsto\alpha x$ is always a homeomorphism when dealing with Euclidean space (for nonzero $\alpha$) the question:

Is $\{2\pi\cdot\frac{k}{2^n-1}:(k,n)\in\Bbb N\}$ dense in $[0,2\pi)$?

Is equivalent to the question:

Is $\{\frac{k}{2^n-1}:(k,n)\in\Bbb N\}$ dense in $[0,1)$?

Take any $x\in[0,1)$. If $x=0$, all neighbourhoods of $x$ contain intervals of the form $[0,r)$ for $0\lt r\le 1$ and it is straightforward that $k=1$ and $n$ sufficiently large gives $\frac{k}{2^n-1}\in[0,r)$. If $0\lt x\lt1$, neighbourhoods of $x$ contain intervals of the form $(x-r,x+r)$ for sufficiently small $r\gt0$. Choose $n\in\Bbb N$ such that $\frac{1}{2^n-1}\lt2r$. Let $k$ be the largest natural such that $\frac{k}{2^n-1}\lt x-r$ (such $k$ always exists since $n$ is fixed, the maximum of a finite set of naturals will always exist). It follows that $\frac{k+1}{2^n-1}\gt x-r$ (by definition of $k$) but also that $\frac{k+1}{2^n-1}\lt x+r$ (by choice of $n$) hence $\frac{k+1}{2^n-1}\in(x-r,x+r)$.

We conclude that the set is dense.