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I know that an infinite set can be partitioned into 2 infinite subsets.

Can one partition an infinite set into finitely many infinite subsets?


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Are you asking that given an infinite set, can that set be partitioned into $n$ infinite sets for all $n$? – ferson2020 Jan 30 '13 at 16:26
In fact, you can also partition it into infinitely many infinite sets. For example, with the set $\mathbb N$, you can let $A_n=\{k\in\mathbb N\mid k=2^nm\text{ with }m\text{ odd}\}$ for $n=0,1,2,\ldots$ – Hagen von Eitzen Jan 30 '13 at 16:30
Let $n$ be given. Partition the integers into those that are multiples of $n$, those that leave a remainder of 1 when divided by $n$, those that leave a remainder of 2 when divided by $n$, ... and those that leave a remainder of $n-1$ when divided by $n$. – MJD Jan 30 '13 at 17:20
up vote 11 down vote accepted

Well, if you can do it with 2, you can keep partitioning one of them.

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Ok, thanks Andreas. – student Jan 30 '13 at 16:26

Of course. We do it by induction.

Theorem: If $A$ is an infinite set and $n>0$ is a natural number then we can write $A$ as a disjoint union of $n$ infinite sets.

Proof. For $n=1$ this is obvious, so we actually start with $n=2$.

For $n=2$ we can do it because every infinite set can be split into two infinite sets.

Suppose that we can split $A$ into $n$ parts, $A_1,\ldots, A_n$. Each is infinite, split $A_n$ into $B$ and $C$, and so the partition $A_1,\ldots,A_{n-1},B,C$ is a partition of $A$ into $n+1$ parts. $\square$

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