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I was looking for an equivalence relation between the points of the real line such that the equivalence classes are $1$ unit long segments.

I found two so far

1) Let $x_1,x_2\in\mathbb{R}$ and $$ x_1\sim x_2\iff \lfloor x_1\rfloor =\lfloor x_2\rfloor $$

2) Let $x_1,x_2\in\mathbb{R}$ and

$$ x_1\sim x_2\iff \lceil x_1\rceil =\lceil x_2\rceil. $$

Since both of these felt a bit boring I tried to look for some others. One idea was to think about $x_1$ and $x_2$ as coordinates and try to set up an equivalence relation that way, but didn´t get anywhere.

Is there a "neater" way to partition the real line into $1$ unit long segments?

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  • $\begingroup$ Sure. Let $0 < m < 1$ then $x_1 \equiv x_2 \iff $ there is an integer $n$ so that $x_1, x_2 \in [n+m, n+1 + m)$. $\endgroup$ – fleablood Mar 2 '19 at 19:17
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Sure.

$x_1 \equiv x_2 \iff x_1, x_2 \in I_n$ where $I_n$ is a unit long segment.

In you cases, Case $1$ is the unit segments $I_n = [n, n+1): n\in \mathbb Z$ and in Case $2$ the unit segments are $I_n =(n, n+1]: n\in \mathbb Z$.

For and $w \in (0,1)$ we can have $I_n =[n+w, n+w + 1): n\in \mathbb Z$ and $I_n = (n+w, n+w + 1]: n\in \mathbb Z$.

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For every $c\in[0,1)$ there are also the equivalence relations $$x_1\sim x_2\qquad \iff\qquad \lfloor x_1+c\rfloor=\lfloor x_2+c\rfloor,$$ $$x_1\sim x_2\qquad \iff\qquad \lceil x_1+c\rceil=\lceil x_2+c\rceil.$$ The only possible variation on this is to move the endpoints of the intervals into the adjacent equivalence class, giving another countably infinitely many variations on these two classes of equivalence relations.

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