Let $m$ be any fixed positive integer. For each integer $j$, $0\le j \lt m$, let $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$.
The set notations for $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ would be as below. Then how do we list elements of $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ from the set notations below?
$\Bbb Z_0= \{x \in \Bbb Z\,|\, x=km \text{ for some } k \in \Bbb Z\}$
$\Bbb Z_1= \{x \in \Bbb Z\,|\, x=km+1 \text{ for some } k \in \Bbb Z\}$
$\Bbb Z_2= \{x \in \Bbb Z\,|\, x=km+2 \text{ for some } k \in \Bbb Z\}$
FYI
"Definition 6 Let F be an arbitrary family of sets. The union of the sets in F, denoted by $\bigcup\mathscr F$, is the set of all elements that are in A for some $A\in\mathscr F$.
$\bigcup\limits_{A \in \mathscr F}A$={$x\in U$|$x \in A$ for some $A\in \mathscr F$}"
"Definition 7 Let F be an arbitrary family of sets. The intersection of sets in F, denoted by $\bigcap\limits_{A\in\mathscr F}A$ or $\bigcap\mathscr F$, is the set of all elements that are in A for all $A \in\mathscr F$.
"
$\bigcap\limits_{A\in\mathscr F}A$={$x\in U$| x∈A for all $A\in \mathscr F$}
Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.
Best Answer
Are you looking for something like this? $$\Bbb Z_0 = \{0, \pm m, \pm2m, \pm3m, \ldots\}\\ \Bbb Z_1 = \{1, \pm m +1, \pm2m + 1, \ldots\}\\ \Bbb Z_2 = \{2, \pm m +2, \pm2m + 2, \ldots\} $$
I'll do a concrete example, so you can see how it works. Say $m = 3$. Then $$ \Bbb Z_0 = \{x \mid x = 3k\text{ for some }k\in \Bbb Z\} $$ It is spelled out "$\Bbb Z_0$ is the set of all $x$ such that $x = 3k$ for some $k \in \Bbb Z$". This means that $15 \in \Bbb Z_0$, since $15 = 3\cdot 5$ (in that case, $k = 5$). However, $43 \notin \Bbb Z_0$, because there is no $k\in \Bbb Z$ so that $43 = 3k$. We get $$ \Bbb Z_0 = \{0, \pm 3, \pm 6, \pm 9, \ldots\} = \{0, 3, -3, 6, -6, 9, -9,\ldots\} $$ On the other hand, we have $$ \Bbb Z_1 = \{x \mid x = 3k+1\text{ for some }k \in \Bbb Z\} $$ which means that a number $x$ is an element of $\Bbb Z_1$ iff there is a $k\in \Bbb Z$ such that $x = 3k+1$. For instance, $43 \in \Bbb Z_1$, since there is a $k\in \Bbb Z$ that makes $43 = 3k + 1$ (it's $14$). However, $15 \notin \Bbb Z_1$, since there is no integer $k$ such that $15 = 3k + 1$. This gives $$ \Bbb Z_1 = \{1, \pm3 + 1, \pm6 + 1, \ldots\} = \{1, 4, -2, 7, -5, \ldots\} $$ Lastly, in exactly the same way, we get $$ \Bbb Z_2 = \{2, \pm3 + 2, \pm 6 + 2, \ldots\} = \{2, 5, -1, 8, -4,\ldots\} $$