I'd like a hint to prove the above assertion. My idea was to find a convergent sequence, of points of $X$, to each point $z \in \mathbb{S}^1$, but I don't think it's right.
$X=\{\cos n + i\sin n : n \in \mathbb{N}\}$ is dense in $\mathbb{S}^1 \subset \mathbb{C}$
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3Possible duplicate of [Sine function dense in $[-1,1]$](http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1) – 2016-01-19
2 Answers
First solution (longer but more general): I propose to show that the set $G=\{n+2k\pi: n\in Z, k\in Z\}$ is dense in R. You can use the fact that the subgroups of R are cyclic or dense. G is not cyclic else $\pi$ would be a rational. So it is dense.
Direct and easier solution: For $\varepsilon>0$ divide the unit circle in parts of equal size whose length does not exceed $\varepsilon$. For $m\ne n$, $e^{im}\ne e^{in}$ (else $\pi$ would be rational). Hence there is an infinite number of $e^{in}$. As there is only a finite number of parts, one can find $m
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0thanks for the explanation! – 2012-06-01
Hints: 1) Show the $n$-th roots of unity on $\,S^1\,$ are the vertices of a regular n-gon (i.e., they're
equally distributed over the circle)
2) since $\,\displaystyle{\lim_{n\to\infty}\frac{2\pi}{n}=0}\,$ , there exists a root of unity whose arc distance from $\,1=e^{2\pi i}\,$ is arbitrarily
small (try some $\epsilon > 0\,$ stuff here)
3) Deduce now that any element $\,z=e^{it}\,\,,2k\pi\leq t<2(k+1)\pi\,,\,k\in\mathbb{Z}\,$ in $\, S^1\, $ is close enough to
some root of unity.