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Definition 5. Let $X$ be a nonempty set. By a partition $P$ of $X$ we mean a set of nonempty subsets of $X$ such that

(a) If $A, B \in \mathscr P$ and $A \neq B$, then $A \cap B = \emptyset$,
(b) $\bigcup\limits_{C \in \mathscr P} C=X$

Example 6. Let $m$ be any fixed positive integer. For each integer $j$, $0\le j \lt m$, let $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$. Then the set $$\{\Bbb Z_0,\Bbb Z_1, \Bbb Z_2 ,\cdots, \Bbb Z_{m-1}\}$$ forms a partition of $\Bbb Z$.

Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.

$\Bbb Z_0, \Bbb Z_1, \Bbb Z_2 ,\cdots, \Bbb Z_{m-1}$ are subsets of $\Bbb Z$, and each are different, so $\{\Bbb Z_0, \Bbb Z_1, \Bbb Z_2,\cdots, \Bbb Z_{m-1}\}$ satisfies the condition (a) in definition 5, but does it also satisfy the condition (b) $\bigcup\limits_{C \in \mathscr P}=X$? I don't think it does because the union of them is not $\Bbb Z$. That is the finite set $\Bbb Z_0 \bigcup \Bbb Z_1 \bigcup \Bbb Z_2 \bigcup \cdots \bigcup \Bbb Z_{m-1} \neq \Bbb Z$ since $\Bbb Z$ is an infinite set. So why $\{\Bbb Z_0, \Bbb Z_1, \Bbb Z_2,\cdots, \Bbb Z_{m-1}\}$ forms a partition of $\Bbb Z$? Isn't it insufficient to be a partition of $\Bbb Z$?

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    $\begingroup$ $\Bbb Z_0 \bigcup \Bbb Z_1 \bigcup \Bbb Z_2 \bigcup \cdots \bigcup \Bbb Z_{m-1}$ isn't finite - it's a finite union of infinite sets. $\endgroup$
    – πr8
    Jan 11, 2016 at 10:30
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    $\begingroup$ For example, if $m=2$, this describes the partition of $\Bbb Z$ in to odd and even numbers ... There are just two parts, but every integer is (either) odd or even $\endgroup$ Jan 11, 2016 at 10:41
  • $\begingroup$ @πr8 $m$ is a fixed positive integer. As for $\Bbb Z_0=\{x \in \Bbb Z\,|\, x=km \text{ for some } k \in \Bbb Z\}$, "for some" in "for some k" is an existential quantifier, and m is fixed. So $\Bbb Z_0$ is a finite set, and similarly so are the others. $\endgroup$
    – buzzee
    Jan 11, 2016 at 10:44
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    $\begingroup$ $\mathbb{Z}_0$ is the set of multiples of $m$. Why should it be finite? $\endgroup$
    – πr8
    Jan 11, 2016 at 10:54
  • $\begingroup$ @πr8 Shoudn't it be $\Bbb Z_0=\{x \in \Bbb Z\,|\, x=km \text{ for all } k \in \Bbb Z\}$ to be an infinite set? $\endgroup$
    – buzzee
    Jan 11, 2016 at 11:05

5 Answers 5

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The Euclidean Division Theorem:

Given two integers $a$ and $b$, with $b ≠ 0$, there exist unique integers $q$ and $r$ such that $$a = bq + r$$ and $$0 ≤ r < |b|,$$ where $|b|$ denotes the absolute value of $b$.

So given a integer $a$, there exist a $r$ such that $a = mq + r$ with $0 ≤ r < |b|$. By the definition of $Z_i$,$i=0,\dots,m-1$, $a\in Z_r$. So $$\bigcup_{i=0}^{m-1}Z_i = Z$$

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Hint. The relation $a=b \mod m$ is an equivalence relation and $\mathbb{Z}_i, i=0 \ldots m-1 $ are the equivalence classes with respect to the relations. Then use a theorem that the set of all equivalence classes form a partition of the set $\mathbb{Z}.$

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  • $\begingroup$ A one-line hint that's nothing more than a reformulation of the question. Good job. $\endgroup$ Jan 11, 2016 at 10:02
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Condition (b) is also satisfied, since any integer has a remainder in $\;\{0,1,\dots, m-1\}\;$ by the Euclidean division theorem.

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Why don't you think it's not a partition?

It's clear that the union is a subset of $\mathbb Z$, so the only way it would not become $\mathbb Z$ is that there's an integer not in the union.

So assume that $z\in \mathbb Z$, then by Euclid division you have $z = qm + r$ where $0\le r< m$. Therefore $z\in \mathbb Z_r$ and therefore in the union.

Note that your motivation that (a) is fulfilled is a bit sloppy. It's not enough that they're different, they have to be pairwise disjoint as well. That too follows from Euclid division (the remainder in the interval is unique).

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Their union is $\mathbb{Z}$ as every consecutive point in $\mathbb Z_j$ for some $j$ belonging to $(0,1,...,m-1)$ is equispaced with a distance $m$, so each $\mathbb Z_j$ completes $1/m$th of the $\mathbb Z$ by its $m$ spaced motion along the real line and as they are non intersecting so their union completes $m$ times $1/m$th of $\mathbb Z$, i.e. the whole $\mathbb Z$.

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