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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!

Martin Sleziak
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CWcx
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  • 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. – CWcx Aug 31 '12 at 20:27

1 Answers1

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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 $[0,1)$.

Brian M. Scott
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    it seems you are assuming that there are infinitely many distinct values of $nr \mod 1$, but how do you know this? – CWcx Aug 31 '12 at 20:57
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    @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
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    Oh I see. Thanks for this nice answer. I definitely didn't think of using pigeonhole principle! – CWcx Aug 31 '12 at 21:02
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    excellent use of the pigeonhole principle – Kirk Boyer Aug 31 '12 at 21:59
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    The idea of proof is excellent. I only want to make a complement. For $ir=u+q_1, jr=v+q_2$ ($u,v\in\mathbb{Z}, q_1,q_2\in[0,1)$), then $(i-j)r \bmod 1=q_1-q_2$ if $q_1\geqslant q_2$, $(j-i)r \bmod 1=q_2-q_1$ if $q_1\leqslant q_2$. Note that $0.01\bmod 1=0.01, -0.01\bmod 1=0.99$. – bianzhiyu Oct 22 '20 at 10:53
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    @bianzhiyu: That is why I used absolute values in the first part of the last sentence; in the last part of the sentence it doesn’t matter whether $(j-i)r$ is positive or negative, since we’re just concerned with the set of integer multiples of it $bmod 1$. – Brian M. Scott Oct 22 '20 at 17:09
  • I don't understand the use of the pigeonhole principle here. Can someone explain. – alexanderyaacov Jan 24 '21 at 02:43
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    @alexanderyaacov: There are $m+1$ numbers $ir\bmod 1$ for $i\in{1,\ldots,m+1}$. For each of them there is exactly one $k\in{0,\ldots,m-1}$ such that $$\frac{k}m\le ir\bmod 1<\frac{k+1}m,,$$ where $k\in{0,\ldots,m-1}$. Since there are $m+1$ possible values of $i$ and only $m$ possible values of $k$, there must be two different values of $i$ that are in the same interval $\left[\frac{k}m,\frac{k+1}m\right)$; in the answer those two values of $i$ are called $i$ and $j$. – Brian M. Scott Jan 24 '21 at 02:49
  • what does it mean "every point of $[0,1)$ is within $\frac1m$ of the set ${n(j-i)r\bmod 1:n\in\Bbb Z}$" can you explain what you mean by that – shangq_tou Apr 09 '21 at 13:40
  • ??????????????? – shangq_tou Apr 09 '21 at 13:51
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    @winic92046: Exactly what I said: if $x\in[0,1)$, there is some $n\in\Bbb Z$ such that $$\left|x-\big(n(j-i)r\bmod 1\big)\right|<\frac1m,.$$ – Brian M. Scott Apr 09 '21 at 19:36
  • @BrianM.Scott Hi, could you explain what it means by "That it’s false for rational r is obvious...". What is false here if r is rational number? could you elaborate a bit why is false if r is a rational number. Also could you elaborate little more how to conclude: so every point of [0,1) is within $\frac{1}{m}$ of the set ..... How do we know every number in [0,1) is within $\frac{1}{m}$ of distance from a number in the set {$n(j-i)r$ $mod 1: n \in Z$}? Thank you. – john_w Apr 29 '21 at 01:26
  • @john_w: If $r$ is rational, ${nr\bmod 1:n\in\Bbb Z}$ is easily seen not to be dense in $[0,1)$: it is in fact a finite set. The points of the set ${n(j-i)r\bmod 1:n\in\Bbb Z}$ are less than $\frac1m$ apart, so the gaps between adjacent points of this set are less than $\frac1m$ in length. – Brian M. Scott Apr 29 '21 at 01:33
  • @BrianM.Scott Hi, can you explain why {$n(j−i)r$ $mod1:n∈Z$} covers the range [0,1]? also say we have a point 0.0001, then it will fall within $\frac{1}{m}$ from say $k_7(j-i)r mod1: k_7 \in Z$ right? and say another point 0.7928 will fall within $\frac{1}{m}$ say from $k_{13}(j-i)r mod1: k_{13} \in Z$ right?, but how do we know the spread of the positions of $k_1(j-i)r mod1$, $k_2(j-i)r mod1$,..., $k_{912302392}(j-i)r mod1$, $k_{912302393}(j-i)r mod1$, $k_{...}(j-i)r mod1$ will spread out so that each of the number in [0,1] will be less than $\frac{1}{m}$ from one of the above? – john_w Apr 29 '21 at 02:36
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    @john_w: Let $a=|(j-i)r|\bmod 1$; $0<a<\frac1m$. Now you’re just looking at the numbers $0,a,2a,3a,4a,\ldots$ (among others). Since $a$ is irrational, there is some $n\in\Bbb Z^+$ such that $na<1<(n+1)a$. The numbers $0,a,2a,3a,\ldots,na$ clearly divide $[0,1)$ into $n+1$ intervals of length $a$, where $a<\frac1m$. Every $x\in[0,1)$ is in one of those intervals and therefore less than $\frac1m$ away from one of those numbers. – Brian M. Scott Apr 29 '21 at 18:07
  • @BrianM.Scott Thank you for the detailed explanations, I think I understand now. Also when you said "Since a is irrational there is some n $\in Z^+$ such that na<1<(n+1)a". Does a has to be irrational in order for the bound to happen? or a can be any real number and we still have the statement that "there is some n∈Z+ such that na<1<(n+1)a" being true? Can we replace "Since a is irrational, there is some n∈Z+ ..." with "Since a is real number, there is some n∈Z+ ..."? – john_w Apr 29 '21 at 20:48
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    @john_w: If $a$ is rational, we can still say that there is some $n\in\Bbb Z^+$ such that $na<1\color{red}{\le}(n+1)a$; the irrationality of $a$ simply ensures that no integer multiple of $a$ is equal to $1$. – Brian M. Scott Apr 30 '21 at 01:50