Question on Cardinality ..Help a)  Let $n$ be a positive integer. Define a relation on $\mathbb{Z} $, which yields a partition of $\mathbb{Z}$ with $n$ elements; and give the partition.
b) Deduce that $n\omega = \omega$ where $\omega$ is the cardinality if $\mathbb{Z}$.
I was thinking that I can define the mapping $f:  \mathbb{Z} \rightarrow\mathbb{Z}^+ - \lbrace1,2,3\rbrace$ as $f(-n) = 2n$ if $n \in \mathbb{Z}^+ $
$ f(n) = 2n + 1$ if $n \in \mathbb{Z}^+ \cup \lbrace 0 \rbrace$
$f$ is thus a $1-1$ corresponce between the set $\mathbb{Z} $ and the set $\mathbb{Z}^+ - \lbrace1,2,3\rbrace$
but the cardinality of the subset $\mathbb{Z}^+ - \lbrace1,2,3\rbrace$ is not $n$ , thats the problem.That's all I have tried..any help anyone?
 A: Recall that a partition on a set $X$ is a set of non-empty subsets of $X$ such that each element of $X$ appears in exactly one of the subsets in the partition. So, what are some partitions of $\mathbb{Z}$. Well the easiest would be the partition $P_1=\{\mathbb{Z}\}$ which has one element. On the other end of the extreme scale, we could define the partition $$P_{\infty} = \{\ldots \{-2\}, \{-1\}, \{0\}, \{1\}, \{2\} \ldots\}=\{\{n\}\mid n\in\mathbb{Z}\}.$$ This partition has $\omega$ elements.
How might one go about making a partition of $\mathbb{Z}$ with two elements? Well we just need to split $\mathbb{Z}$ into exactly two non-empty, non-overlapping subsets - one way to do this would be the partition $P_2=\{\{0\},\mathbb{Z}\setminus\{0\}\}$, and another would be $P_2'=\{\{n\mid n\mbox{ even}\},\{n\mid n\mbox{ odd}\}\}$. Both $P_2$ and $P_2'$ are partitions and both have cardinality $2$, so either would work, and in fact there are (infinitely) many other ways to get a partition with two elements.
I think from here you might be able to think of a partition with $3$ elements. How about $n$ elements?
This seems to be the part of the question that you're having trouble with so I'll not write anything on part b unless you feel you need some hints for that as well.
