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I proved that $x\sim y$ iff $x-y\in \mathbb Z$ is an equivalence relation on $\mathbb R$. I'd like to know if $[x]=\{x+n:n\in\mathbb Z\}$ is an equivalence class for every $x\in \mathbb R$ (if it is well written).

Now, I proved that $(x_1,y_1)\sim(x_2,y_2)$ iff $ x_1=x_2$ is an equivalence relation on $\mathbb R^2$ and know that $[(x,y)]$ is the set of points of the form $(x_0,y)$ for some fixed $x_0$, but don't know how to write it down. I appreciate your help.

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For the first question, yes you are correct. If $x\sim y$ then $x-y=n$ for some $n\in\Bbb Z$, so $y=x-n$; and of course if $y=x+n$ then $x-y\in\Bbb Z$. So $[x]$ is exactly the set $\{x+n\mid n\in\Bbb Z\}$.

For the second question, note that if $(x,y)\sim(x',y')$ then $x=x'$. So the equivalents of $(x,y)$ are those where $x$ is fixed, and $y$ varies over $\Bbb R$. Namely, $$[(x,y)]=\{(x,y')\in\Bbb R^2\mid y'\in\Bbb R\}.$$

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  • $\begingroup$ Would it be redundant to write $\{(x,y)\in \mathbb R^2:y\in \mathbb R, \text{ $x$ is fixed}\}$? I was trying to write something similar, but kind of got confused. $\endgroup$ – Vladimir Vargas Jan 17 '15 at 23:25
  • $\begingroup$ The $(x,y)\in\Bbb R^2$ is not that redundant. Writing that $x$ is fixed is redundant. $\endgroup$ – Asaf Karagila Jan 17 '15 at 23:32
  • $\begingroup$ But if one writes $A=\{(x,y)\in\mathbb R^2:y\in\mathbb R\}$ then $(1,2)\in A$ and $(2,1)\in A$, but $1\neq 2$ so there would be a contradiction. Am I wrong? $\endgroup$ – Vladimir Vargas Jan 17 '15 at 23:35
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    $\begingroup$ No, the expression just has a missing piece of information, what is $x$. But if you write $[(x,y)]=\{(x,y')\mid y'\in\Bbb R\}$, then we know that is $x$. It has an assigned value outside the set. (I opted to $y'$ in the definition of the set, to make it clearer that it's a different $y$.) $\endgroup$ – Asaf Karagila Jan 17 '15 at 23:39
  • $\begingroup$ Great!, cheers. $\endgroup$ – Vladimir Vargas Jan 17 '15 at 23:41

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