I am trying to describe the set of equivalence classes $\mathbb{R}/\mathbb{Z}$, the quotient by the relation on $\mathbb{R}$ defined by $x \sim y$ if and only if $x - y \in \mathbb{Z}$. Here is my attempt.
The set of equivalence classes $\mathbb{R}/\mathbb{Z}$ are of the form $$ \mathbb{R}/\mathbb{Z} = \left\{x + \mathbb{Z} \mid x \in [0,1) \right\}. $$ As every real number $y \in \mathbb{R}$ can be written as $y = \lfloor y \rfloor + c$ for some $c \in [0,1)$, these sets are exhaustive. Furthermore, each set is in fact an equivalence class. Indeed, if $x,y \in \mathbb{R}$ live in the same class $t + \mathbb{Z}$ for some $t \in [0,1]$, then $x = t + z$ and $y = t + z'$ for some $z,z \in \mathbb{Z}$, so $x - y = (t + z) - (t + z') = z - z' \in \mathbb{Z}$, so $x \sim y$. So each class $t + \mathbb{Z}$ is an equivalence class. Furthermore, if $t \neq t'$, then $t + \mathbb{Z} \neq t' + \mathbb{Z}$, because if $x \in t + \mathbb{Z}$ and $y \in t' + \mathbb{Z}$, then $x - y \not \in \mathbb{Z}$.
I wanted to show dual containment, and I believe I've done so. Every real number is in its own equivalence class, $[x]$, so by showing that every $y$ can be written as a sum of its integer and fractional part, I've shown that the LHS is a subset of the RHS. Next I showed that every set $x + \mathbb{Z}$ is in fact an equivalence class and that they don't overlap.
How does this look? Is there an error in the logic?