I am trying to prove the following, but I'm not sure every step works.
Let $a$ be an irrational real number. Show that the set of numbers of the form $e^{2πina}$, $n \in \mathbb{Z}$, is dense in $S^1$.
Here $S^1$ is the unit circle, parametrized as $\{ e^{2\pi i x}: x \in \mathbb{R}\}$
I am considering $S^1$ as a metric space with arc length as the metric: $d(e^{2\pi i r}, e^{2 \pi i s}) = |r -s|$. Let $O_a = \{e^{2\pi i na}\}$ and fix $ \epsilon > 0 $.
Then it suffices to show that given $n$ and $\epsilon$ there exists $\omega \in \mathbb{R}$ such that $d(e^{2 \pi i n a}, e^{2 \pi i \omega}) = |na - \omega| < \epsilon$.
By the Archimedean property, pick $m \in \mathbb{N}$ such that $\frac{1}{m} < \epsilon$.
Given $m$, by the Dirichlet Approximation Theorem, there exist $p, q \in \mathbb{Z}$ such that $|p(na) - q| < \frac{1}{m}$.
Therefore, pick $\omega = (1-p)na + q$.
Then $|na - \omega| = |na - (1-p)na-q| = |pna - q| < \frac{1}{m} < \epsilon.$
So given an element of $O_a$, specified by $n \in \mathbb{Z}$ and $\epsilon > 0$, there exists an element $e^{2\pi i \omega} \in S^1$ such that $e^{2\pi i\omega}$ is in the $\epsilon$ neighborhood of $e^{2 \pi i n a}$ according to the chosen metric. Therefore, each element of $O_a$ is a limit point of $S^1$, and so $O_a$ is dense in $S^1$.
I'm stuck on how to show $e^{2 \pi i \omega} \neq e^{2 \pi i n a}$ which is necessary for the limit point argument. When I pick $\omega$, it's clearly an irrational number, so I can't rule out these being equal. I'm not sure if there are other problems in the proof attempt.
Thanks!
