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This question already has an answer here:

I have a question which I'm deeply confused about. I was trying to do some problems my professor gave us so we could practice for exam, one of them says:

Give a partition of ω in ω parts, everyone of them of cardinal ω.

I know that $ ω=\left \{ 1,2,3,...,n,n+1,.... \right \}$ , but I thought that |ω|=ω.

Can somebody help me?

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marked as duplicate by Asaf Karagila elementary-set-theory Nov 4 '14 at 3:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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There are many ways to do it, since $|\omega\times\omega|=\omega$. Here’s one. Every $n\in\Bbb Z^+$ can be written uniquely in the form $n=2^km$, where $m$ is odd, and you can start by letting $S_k=\{2^km:m\in\omega\text{ is odd}\}$: $S_0$ is the set of odd positive integers, $S_1$ the set of even positive integers that are not divisible by $4$, and so on. That isn’t quite a partition of $\omega$, since it doesn’t cover $0$, so throw $0$ into one of the parts: let $P_0=\{0\}\cup S_0$, say, and $P_k=S_k$ for $k>0$, and $\{P_k:k\in\omega\}$ is then a partition of the kind that you want.

Another way is to use the pairing function, which is a bijection $\varphi:\omega\times\omega\to\omega$, and set $P_k=\varphi[\{k\}\times\omega]$ for each $k\in\omega$.

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For example, the part $P_n$ could consist of all numbers of the form $2^{n-1} q$, where $q$ ranges over the odd numbers. So $P_1$ consists of all odd natural numbers, $P_2$ consists of all natural numbers that are divisible by $2$ but not by $4$, $P_3$ consists of all natural numbers that are divisible by $4$ but not by $8$, and so on.

Since the cardinality of the set of odd positive integers is $\omega$, each of the $P_n$ has cardinality $\omega$.

Remark: Since your post identifies $\omega$ with $\{1,2,3,\dots\}$, we have assumed that that is the usage in your course. The more usual thing is to let $\omega$ be $\{0,1,2,3,\dots\}$. If we use the more ordinary convention, let $f(x)=x-1$, and use $f$ to carry the partition of $\mathbb{N}$ that we have produced above into a partition of $\{0,1,2,3,\dots\}$.

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You could for example write up the natural numbers in a kind of staircase pattern

1

...

1 3

2

...

1 3 6

2 5

4

etc (insired of course by the diagonal argument used to prove countability of the rationals) then take every column as one of your sets.

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