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Let X $=\mathbb{R}$

Let $x \sim y \leftrightarrow x - y \in \mathbb{Z}$.

It is intuitively obvious why this would have an inifinite number of equivalence classes. Is there a rigorous way of proving this?

1123581321
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1 Answers1

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Say $x,y\in [0,1)$ with $x\neq y$. Can $x-y\in\mathbb{Z}$?

Reveillark
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  • In that case $0 < |x - y| < 1$, so $x - y$ is not an integer. – 1123581321 Mar 27 '19 at 03:35
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    That's assuming $y<x$, in general you may have $-1<x-y<1$. But that's not the point. The conclusion here is that every number $x\in [0,1)$ gives rise to a different equivalence class. – Reveillark Mar 27 '19 at 03:37