$X=\{\cos n + i\sin n : n \in \mathbb{N}\}$ is dense in $\mathbb{S}^1 \subset \mathbb{C}$

Question

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.

Answer 1

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 $m<n$ such that $e^{im}$ and $e^{in}$ are note equal and in the same part. Hence there exists $\theta\ne0$ such as $|\theta|<\varepsilon$ and $e^{i(n-m)}=e^{i\theta}$. Then $r=e^{i(n-m)}\in X$ and using powers of $r$ you can find an element of $X$ at distance less then than $\varepsilon$ for any point on the circle.

Answer 2

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.