For $x,y\in\mathbb{R}$ define x ~ y to mean that $x-y\in\mathbb{Z}$. Prove that ~ is an equivalence relation on $\mathbb{R}$. Describe its equivalence classes.
I've successfully proved x ~ y relation is reflexive, symmetric and transitive.
What I'm not able to do is to describe the equivalence classes. In my view it can be
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$\mathbb{Z}$ as every integer $x,y$ is $x~R~y$ as demonstrated above
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or more specifically even $\mathbb{R}$ given that $x-y\in\mathbb{Z}$.
Am I missing something?
Best Answer
Each equivalence class is a subset of $\mathbb R$ and therefore the set of all equivalence classes must be a set of subsets of $\mathbb R$.
In your case, for each $x\in[0,1)$, consider the set $x+\mathbb Z$. There you have it. The set of all equivalence classes is$$\{x+\mathbb Z\mid x\in[0,1)\}.\tag1$$Note that if $x\in\mathbb R$, then $x-\lfloor x\rfloor\in[0,1)$, that $x\sim x-\lfloor x\rfloor$ and that $x-\lfloor x\rfloor$ is the only $y\in[0,1)$ such that $x\sim y$. Therefore, $(1)$ describes all equivalence classes and each equivalence class appears there only once.