Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that the following space is not Hausdorff. Consider the topological space $S^1$, and let $r$ be an irrational number. Consider the action of $\mathbb{Z}$ on $S^1$ given by $$ S^1\times\mathbb{Z}\to S^1; (e^{ix}, n)\mapsto e^{i(x+2\pi n r)}. $$ Let $S^1/\mathbb{Z}$ denote the orbit space. I want to show that this space is not Hausdorff.

It was suggested to me that I try showing that any orbit under this action is dense in $S^1$, but I am getting stuck on proving that bit. But here is what I was thinking: we know that the topological group $\mathbb{R}/2\pi \mathbb{Z}$ is homeomorphic to $S^1$, as seen by the map $t\mapsto e^{it}$. Denote the following composition of maps $$ \mathbb{R}\to \mathbb{R}/2\pi \mathbb{Z}\simeq S^1\to S^1/\mathbb{Z} $$ by $\phi$. Then if $[e^{ix}]\in S^1/\mathbb{Z}$, it follows that $$ \phi^{-1}([e^{ix}]) = \{x+2\pi(nr+m)\mid n, m\in\mathbb{Z}\}. $$ If I can show that this subset is dense in $\mathbb{R}$, then it follows that the set $[x]$ as a subset of $S^1$ is dense. This is where I am getting stuck, and it is not even clear to me that this is necessarily true.

Any hints or suggestions are appreciated. Thanks!

share|improve this question
    
Can you clarify your action definition? From your later discussion it seems that the action should be something like $(e^{ix},n)\mapsto e^{i(x+2\pi nr)}$. –  Kirk Boyer Aug 31 '12 at 20:15
    
Oh yes your right, there is a typo. I will correct it. –  Ler Aug 31 '12 at 20:27

1 Answer 1

up vote 8 down vote accepted

It suffices to show that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$, where $x\bmod 1=x-\lfloor x\rfloor$. This is true precisely in case $r$ is irrational. That it’s false for rational $r$ is obvious, so assume that $r$ is irrational. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\in\{1,\dots,m+1\}$ and $k\in\{0,\dots,m-1\}$ such that $\frac{k}m\le ir\bmod 1,jr\bmod 1<\frac{k+1}m$. Then $|(j-i)r|\bmod 1<\frac1m$, so every point of $[0,1)$ is within $\frac1m$ of the set $\{n(j-i)r\bmod 1:n\in\Bbb Z\}$, and it follows immediately that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$.

share|improve this answer
    
it seems you are assuming that there are infinitely many distinct values of $nr \mod 1$, but how do you know this? –  Ler Aug 31 '12 at 20:57
    
@leo: If $mr\bmod 1=nr\bmod 1$, there are integers $k,\ell$ such that $mr+k=nr+\ell$. If $m\ne n$, then $r=\frac{\ell-k}{m-n}$ is rational. –  Brian M. Scott Aug 31 '12 at 20:59
    
Oh I see. Thanks for this nice answer. I definitely didn't think of using pigeonhole principle! –  Ler Aug 31 '12 at 21:02
    
excellent use of the pigeonhole principle –  Kirk Boyer Aug 31 '12 at 21:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.