Subset Product/Examples/Congruence Modulo Initial Segment of Natural Numbers

Example of Subset Product
Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:
 * $\N_{<m} = \set {0, 1, \ldots, m - 1}$

Let $\RR_m$ denote the equivalence relation:
 * $\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$

For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$:
 * $\eqclass a m := \set {a + z m: z \in \Z}$

Let $\Z_m$ be the set defined as:
 * $\Z_m := \set {\eqclass a m: a \in \N_{<m} }$

Let $+_\PP$ denote the operation induced on $\powerset \Z$ by integer addition.

Then the algebraic structure $\struct {\Z_m, +_\PP}$ is closed in the sense:
 * $\forall \eqclass a m, \eqclass b m \in \Z_m: \eqclass a m +_\PP \eqclass b m \in \Z_m$

Proof
From Equivalence Class Example: Congruence Modulo $\N_{<m}$:
 * $\RR_m$ is an equivalence relation
 * $\eqclass a m$ is the equivalence class of $a \in \N_{<m}$ under $\RR_m$.

Let $\eqclass a m, \eqclass b m \in \RR_m$ be arbitrary.

Then we have:

It also follows that:
 * $\forall a, b \in \N_{<m}: \eqclass a m = \eqclass b m \iff a = b$