I have a task. Prove that $(\cos(n\alpha), \sin(n\alpha))$, with $n\in\mathbb{N}$ is dense on unit circle.($\alpha$ chosen such that our set is infinite ) We want to show, that for every element's neighbourhood there is an element of that that is in the neighborhood. We know that there all points are different (there is no cycle on the circle). But why can we use pigeonhole principle to prove that if we divide the circle in $k$ arcs then there exists two elements which a closer than $\frac{2\pi}{k}$?

EDIT: Thanks everyone. Now everything is clear. This question is [SOLVED]


marked as duplicate by rtybase, Claude Leibovici, Guy Fsone, Paul Sinclair, user223391 Jan 18 '18 at 19:03

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  • $\begingroup$ Can you rewrite the second sentence? It's wrong the way it is stated. $\endgroup$ – zhw. Jan 18 '18 at 3:17
  • $\begingroup$ As written, the statement is wrong. $\alpha=\pi$ is irrational and $(\cos(\pi n),\sin(\pi n))$ is not dense in the unit circle. $\endgroup$ – Jack D'Aurizio Jan 18 '18 at 10:56
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    $\begingroup$ It's a lot easier to use the theorem saying that if $f$-continuous function and $\left(x_n\right)_{n\in\mathbb{N}}$ is dense, then $\left(f(x_n)\right)_{n\in\mathbb{N}}$ is also dense in $f$'s range. In this case $f(x)=e^{i\cdot \alpha \cdot x}$ and $f$'s range is the unit circle. $\endgroup$ – rtybase Jan 18 '18 at 12:45
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    $\begingroup$ Nope, it will be $\left\{n + m \cdot \frac{2\pi}{\alpha} \mid n,m\in\mathbb{Z} \right\}$ assuming $\frac{2\pi}{\alpha}$ is irrational, as per Kronecker's approximation theorem. And $f(x)=e^{i\cdot \alpha \cdot x}$ is a periodic function. $\endgroup$ – rtybase Jan 18 '18 at 14:36
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    $\begingroup$ @rtybase thanks a lot $\endgroup$ – Kirill Losev Jan 18 '18 at 15:46

The basic idea is that $(\{n\alpha\})$ is dense in $(0, 1)$. This is Weyl's equidistribution theorem: https://en.wikipedia.org/wiki/Equidistribution_theorem

Your statement easily follows from this.

  • $\begingroup$ I want to find not very difficult proof. But every proof that I've found is based on pigeonhole principle. For example link. But I can't understand why we'll have at least two elements in one segment. $\endgroup$ – Kirill Losev Jan 18 '18 at 11:43
  • $\begingroup$ @KirillLosev If you have 10 coins and 9 cups, and every coin is in one of those cups, then some cup must have two or more coins. That is the pigeon hole principle. $\endgroup$ – zhw. Jan 18 '18 at 18:43
  • $\begingroup$ @zwh yep, but I think here we use the idea that there exists cup with more than one coin.(but it's okay if some cups are empty) $\endgroup$ – Kirill Losev Jan 19 '18 at 9:22

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