Denseness of a certain subset of the unit circle I am trying to show that the set of numbers of the form $e^{2\pi ina}$, where $a$ is irrational and $n\in\mathbb Z$ , is dense in the unit circle.
Solution: Noting that elements of the unit circle can be written as $z=e^{2\pi i\theta}$ with $\theta\in\mathbb R$, and using the standard topology of $\mathbb C$;
\begin{align*}
\left|e^{2\pi ian}-e^{2\pi i\theta}\right| &= \big|[\cos(2\pi an)-\cos(2\pi\theta)]-i[\sin(2\pi an)-\sin(2\pi\theta)]\big|\\
 &= \left|-2\sin[\pi(na-\theta)]\sin[\pi(na+\theta)]-2i\sin[\pi(na-\theta)\cos[\pi(na+\theta)]\right|\\
&=2|\sin[\pi(na-\theta)]||\sin[\pi(na+\theta)]+i\cos[\pi(na+\theta)]|\\
&=2|\sin[\pi(na-\theta)]|\leq 2\pi|na-\theta|<2\pi\epsilon_1=\epsilon.
\end{align*}
I have used the following: $$\cos(P)-\cos(Q)=-2\sin\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right),~ \sin(P)-\sin(Q) = 2\cos\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right)$$ and $|\sin x|\leq |x|$.
The last inequality follows from the fact that there is always a real number ($\theta$) arbitrarily close to any rational number ($na$).
This last argument makes me feel like I have cheated (a bit). Is there a way to complete this exercise without the prior knowledge of the denseness of the rationals on the real line? Furthermore, please recommend alternative approaches, if there are any.

As Leonid observed, your proof is incorrect. Since you asked for the alternative approach, this is my try.

*

*Let $a_n:=a\cdot n-\lfloor a\cdot n\rfloor$. Then $e^{2\pi i\cdot a\cdot n} = e^{2\pi i\cdot a_n}$.

*It sufficies to show that $\{a_n:n\in\Bbb N\}$ is dense in $[0,1]$. Indeed, for any $z\in S$ (unit circle) there exists $\theta\in [0,1]$ such that $z=e^{2\pi i \theta}$. If $a_n$ is close to $\theta$ then $e^{2\pi i\cdot a_n}$ is close to $z$.

*Observe that irrationality of $a$ implies that $a_m\neq a_n$ for $m\neq n$.

*Take any $N\in\Bbb N$ (big number). Consider $N$ boxes $[0,1/N],[1/N,2/N],\ldots,[1-1/N,1]$ covering $[0,1]$.

*One of the boxes contains infinitely many numbers $a_n$. Particularly, there are $m,n\in\Bbb N$, $m<n$, such that $|a_m-a_n|\leq 1/N$.

*Assume $0<a_n-a_m\leq 1/N$ (the other case is similar, as we consider indices from $\Bbb Z$, not only $\Bbb N$). Then $0<a_{k}\leq 1/N$ for $k:=n-m$.

*Observe that $a_{2k}=2a_{k}$, $a_{3k}=3a_{k}$ and generally $a_{jk}=ja_{k}$ as long as $ja_{k}<1$. These points form an arithmetic progression with the difference $d=a_{k}\leq 1/N$.

*Therefore for any $\theta\in[0,1]$ there exists $j$ such that $|a_{jk}-\theta|\leq 1/N$.

*Since $N$ can be arbitrarily large, the set $\{a_n:n\in\Bbb N\}$ is dense in $[0,1]$.

