Finishing a proof that the points $e^{ixn}$ are dense in $S^1$ for $x\notin \pi \mathbb{Q}$.

Question

I'm trying to prove that for $x\notin \pi \mathbb{Q}$ the set $\left\lbrace e^{ixn}, n\in \mathbb{N} \right\rbrace$ is dense in $S^1$. This is where I'm at:

Since $x$ is not a rational multiple of $\pi$ it follows that for $n,n'\in \mathbb{N}$ s.t. $n\neq n'$ we have $e^{ixn}\neq e^{ixn'}$. Let $s\in S^1$ and $\epsilon>0$ be given, we can partition $S^1$ in finitely many closed arcs of length $\epsilon$, since there are infinitely many points $e^{ixn}$, by the pigeonhole principle, we can find $m,n\in \mathbb{N}$, $m<n$ s.t. $\left| e^{ixn}-e^{ixm}\right|<\epsilon$, thus $\left|e^{ixl}-1\right|<\epsilon,\; (l=n-m)$. Furthermore, notice that if $n_k=kl$ then we have $\left| e^{ixn_{k+1}} - e^{ixn_k}\right|<\epsilon$ as well. Now I'm not entirely sure how to finish the proof. So far the reasoning is fairly similar to what can be found here Dense set in the unit circle- reference needed but I don't quite understand how they end their proof, going from what I have here, to there being some $k$ s.t. $\left| e^{ixn_k}-s\right|<\epsilon$. I feel like I'm missing something fairly obvious but I just don't see it, any help is appreciated.

Answer 1

In brief: the key is to consider the points $(e^{ixl})^k = e^{ix(lk)}$ (with $k \in \Bbb N$), which break up $S^1$ into $\epsilon$-balls.

Begin with any point $z \in S^1$. To show that your set is dense, we need to show that for any $\epsilon > 0$, there exists an $n \in \Bbb N$ such that $|e^{ixn} - z| < \epsilon$. You've found that for any $\epsilon > 0$, there exists an $l \in \Bbb N$ such that $|e^{ixl} - 1| < \epsilon$. From there, let $0 \leq x_0 < 2 \pi$ be such that $e^{ix_0} = e^{ixl}$ (that is, $x_0 \equiv xl \pmod {2\pi}$).

Suppose first that $\operatorname{Im}(e^{ixl}) = \sin(xl) > 0$. Let $0 \leq y < 2 \pi$ be such that $z = e^{iy}$. Note that for any $0 \leq t \leq x_0$, it holds that $|e^{it} - e^{ixl}| < \epsilon$. Similarly, for any $k \in \Bbb N$, it holds for $kx_0 \leq t \leq (k+1)x_0$, that $|e^{it} - e^{ix[l(k+1)]}| < \epsilon$. Now, there must exist a $k \in \Bbb N$ such that $kx_0 \leq y \leq (k+1)x_0$. Thus, with $n = (k+1)l$, we have $|e^{ixn} - e^{iy}| < \epsilon$.

On the other hand, suppose that $\operatorname{Im}(e^{ixl}) < 0$. Let $0 \leq y \leq 2 \pi$ be such that $z = e^{-iy}$, and outline a similar argument to conclude the proof.

Answer 2

The set of real numbers $x\not\in\pi\mathbb{Q}$ contains $\mathbb{Q}\setminus\{0\}$, and is therefore dense in $\mathbb{R}$. The map $\mathbb{R}\to S^{1}, t\mapsto e^{it}$ is a continuous surjection. The image of a dense set under a continuous surjection is dense in the codomain.