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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.

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  • $\begingroup$ I would prefer the naming convention $\mathbb{Z}_{j,m}$ since the definition is clearly dependent on $m$. Then e.g. $\mathbb{Z}_{1,m}=\{km+1 | k\in\mathbb{Z}\}$ which should look familiar ("modulo $m$"). $\endgroup$
    – Dr_Be
    Jan 12, 2016 at 8:52
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    $\begingroup$ In my view it is better to write $m\mathbb Z$, $1+m\mathbb Z$ etc. instead of $\mathbb Z_0$, $\mathbb Z_1$ etc. The notation $\mathbb Z_n$ is allready commonly used for $\{r+n\mathbb Z\mid r=0,1,\dots,n-1\}$. Where did you meet these notations? $\endgroup$
    – drhab
    Jan 12, 2016 at 8:53
  • $\begingroup$ @drhab It's just another curiosity sprang from math.stackexchange.com/questions/1607816/… $\endgroup$
    – buzzee
    Jan 12, 2016 at 9:00

2 Answers 2

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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 $$ \Bbb Z_2 = \{2, \pm3 + 2, \pm 6 + 2, \ldots\} = \{2, 5, -1, 8, -4,\ldots\} $$

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  • $\begingroup$ since it's "for some k $\in Z$", not 'for every k $\in Z$' Shouldn't it be $$\Bbb Z_0 = \{k, 2k, 3k, \ldots\}\\ \Bbb Z_1 = \{k +1, 2k + 1, 3k+1\ldots\}\\ \Bbb Z_2 = \{k +2, 2k+ 2, 3k+2\ldots\} $$? $\endgroup$
    – buzzee
    Jan 12, 2016 at 9:08
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    $\begingroup$ @buzzee No, the $k$ are the numbers I've colored red here: $$ \Bbb Z_1 = \{\color{red}0m + 1, \color{red}{\pm 1} m +1, \color{red}{\pm2}m + 1, \ldots\} $$ $\endgroup$
    – Arthur
    Jan 12, 2016 at 9:12
  • $\begingroup$ Do you regard "for some k $\in Z$" as a meaning of 'k can be any integer'? $\endgroup$
    – buzzee
    Jan 12, 2016 at 9:32
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    $\begingroup$ The notation $$\Bbb Z_j = \{x \mid x = km + j\text{ for some }k \in \Bbb Z\}$$means, more or less by definition, "$x \in \Bbb Z_j$ iff there is some $k \in \Bbb Z$ such that $x = km + j$", so no, "... for some $k$" means "there exists a(t least one) $k$ such that...". $\endgroup$
    – Arthur
    Jan 12, 2016 at 9:42
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    $\begingroup$ @buzzee No. What you have written is "The set of all $x$ such that $x = 3k$ for all $k \in \Bbb Z$". There are no members in that set, since there is no $x$ that is simultaneously equal to all possible different $3k$. You might be thinking of something different, namely $\{3k \mid k \in \Bbb Z\}$. It is the same set, but the notation is used a bit more directly. It is "the set of all numbers of the form $3k$ as $k$ ranges over the integers". $\endgroup$
    – Arthur
    Jan 12, 2016 at 17:52
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Start with another notation: $r+m\mathbb Z:=\{r+mk\mid k\in\mathbb Z\}$.

This instead of $\mathbb Z_r$.


Now define $\mathbb Z/m\mathbb Z:=\{r+m\mathbb Z\mid r=0,1,\dots,m-1\}$.

Then $\mathbb Z/m\mathbb Z$ is a list of the sets you mention.

Note that - defined like this - $\mathbb Z/m\mathbb Z$ is a partition of $\mathbb Z$.

Another commonly used notation for $\mathbb Z/m\mathbb Z$ is $\mathbb Z_m$.

However, that notation is unfortunately allready in use in your question.

Following your notation (I dislike it) we have $\mathbb Z/m\mathbb Z:=\{\mathbb Z_0,\dots,\mathbb Z_{m-1}\}$.

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  • $\begingroup$ Unfortunately the definition of $\mathbb Z_j$ in the question is not the same as this commonly used notation. $\endgroup$ Jan 12, 2016 at 10:12
  • $\begingroup$ @HenningMakholm Yes, and indeed unfortunately. I have chosen for the other option ($\mathbb Z/m\mathbb Z$) now to diminish the chance on confusion. $\endgroup$
    – drhab
    Jan 12, 2016 at 10:20

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