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Define the equivalence relation $\sim$ on $\mathbb{R}$ as follows:

$$\forall a,b\in\mathbb{R},\ a\sim b\ \Leftrightarrow\ b-a\in\mathbb{Z}$$

I can prove that this is an equivalence relation, but I saw a claim that the partition $\mathbb{R}/\sim$ can be thought of as the interval $[0,1)$. To be clear, is that the set of equivalence classes $\{[a]_{\sim}:a\in[0,1)\}$?

Additionally, given a partition $\mathscr{P}_{\sim}$ on a set $S$, could I define the equivalence relation $\sim$ on $S$ by taking the cartesian product of each piece of the partition with itself?

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The map $i:(\mathbb{R}/\sim) \to [0,1)$ defined (awkwardly) by $ \{ i([x]) \} = [x] \cap [0,1)$ is a bijection.

Notation: $[x] = \{ y \mid x \sim y \}$.

A partition defines an equivalence class in the manner you described, that is, if $P_1,...,P_n$ is the partition, then the equivalence class is $\cup_k (P_k \times P_k)$.

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  • $\begingroup$ Sorry, but I'm new to this stuff, could you explain the notation $i[x]$? $\endgroup$ – chs21259 Mar 10 '14 at 2:40
  • $\begingroup$ I added a comment above explaining... $\endgroup$ – copper.hat Mar 10 '14 at 2:41
  • $\begingroup$ That makes sense now, thanks! $\endgroup$ – chs21259 Mar 10 '14 at 2:43
  • $\begingroup$ I've made a small change, I hope you don't mind. $\endgroup$ – dtldarek Mar 10 '14 at 2:44
  • $\begingroup$ @dtldarek: Clarifications always welcomed! $\endgroup$ – copper.hat Mar 10 '14 at 2:45
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$[0,1)$ is what we call a set (or system) of distinct representatives - that is, for every $x$ there is exactly one $y\in[0,1)$ such that $x\sim y$.

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  • $\begingroup$ So essentially, if there is an $x\in\mathbb{R}$ such that $x\sim y$ where $y\in[0,1)$ then we have $[x]_{\sim}=[y]_{\sim}$? $\endgroup$ – chs21259 Mar 10 '14 at 2:46
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    $\begingroup$ That's backwards. If there is an $x$ then there exists a $y\in [0,1)$. What you've stated is trivially true, since $x\sim y$ means $[x]=[y]$. $\endgroup$ – Thomas Andrews Mar 10 '14 at 2:47
  • $\begingroup$ Ok that makes sense, and the bijection copperhat posted above makes use of your original fact, where the intersection "picks out" the unique element of the equivalence class that lives in $[0,1)$? $\endgroup$ – chs21259 Mar 10 '14 at 3:04
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Any real number can be written as $n+f$ uniquely with $n$ an integer $f$ the positive fractional part. (for negative numbers such as $-8.25$ you take fractional part to be $+0.75$ and integer part to be $-9$, rather than as $-.25 $ and $-8$.

Two real numbers $a,b$ are equivalent if their difference is an integer which is the same as saying their $f$ parts are equal. So this fractional part determines the equivalence class completely.

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  • $\begingroup$ So, say I want to do the same thing in $\mathbb{R}^2$, and define the same relation on each coordinate of $(a_1,a_2)$. Then I would get the partition being equivalent to the square $[0,1)\times [0,1)$ correct? Because every coordinate an be expressed as $(n+f_1,m+f_2)$? $\endgroup$ – chs21259 Mar 10 '14 at 4:15

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