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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.

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Does this help? –  Martin Sleziak May 30 '12 at 17:20
    
@MartinSleziak maybe,I don't know... –  Jr. May 30 '12 at 17:25
    
Hint: $n$ $\text{mod} 2\pi$ is dense in $[0,2\pi].$ –  bobobinks May 30 '12 at 17:29
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2 Answers

up vote 7 down vote accepted

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.

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+1 Concise, perhaps a little too much, but precise. –  DonAntonio May 30 '12 at 18:04
    
@rannousse I think $|r|=1$, did you mean $|n-m|< \epsilon$? –  Jr. May 31 '12 at 21:35
    
@rannousse How do I know that some power of $r$ will be at the same part of the choosen point? –  Jr. May 31 '12 at 21:48
    
@Jr You are right about $|r|=1$. I fixed the proof. –  ranousse May 31 '12 at 22:06
    
@Jr For your second question, multiply by $r$ until $r^n$ is before the considered point $z$ and $r^{n+1}$ after it. Then $|r^n-z|<\varepsilon$ (notice that each time you multiply by $r$ you advance of $\theta$ along the circle) –  ranousse May 31 '12 at 22:12
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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.

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