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Can you partition an infinite set, into an infinite number of infinite sets?

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When you say infinite set, do you mean unbounded or do you mean having an infinite amount of elements? –  Raskolnikov Dec 1 '10 at 15:33
    
@Raskolnikov: Having an infinite number of elements –  utdiscant Dec 1 '10 at 15:47
    
See also: math.stackexchange.com/questions/12213/… –  Aryabhata Dec 1 '10 at 16:20
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4 Answers 4

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Yes.

First, note that it's enough to do it for countably infinite sets. For if $X$ is uncountable and $Y$ is a countably infinite subset, then $Z = X-Y$ is also infinite. Then, if we divide $Y$ into infintely many infinite sets, we've divided $X$ into infinitely many infinite sets.

So, let's consider a countable set $Y$. Of course, we may as well label the elements of $Y$ as $0,1,2,...$. In short, I'm going to break $\mathbb{N}$ into infinitely many infinite sets.

First note that there are infinitely many prime numbers. Further, if $p$ and $q$ are prime numbers, and if $p^a = q^b$, then we must have $p=q$ and $a=b$. This follows from unique factorization of numbers.

So, for each prime $p$, consider the set $Y_p = \{ p^a$ with $a>0\in\mathbb{N}\}$. Then, e.g., $Y_2 = \{2,4,8,16,32,64,128,...\}$ and $Y_3 =\{3,9,27,81,243,...\}$.

Clearly, each $Y_p$ is infinite. Further, if $Y_p$ and $Y_q$ have anything in common, then by what we said before, we must have that $p=q$. In other words, for different primes $p$ and $q$, the sets $Y_p$ and $Y_q$ have no overlap.

Since there are infinitely many primes, there are infinitely many $Y_p$ with no overlap. Finally, let $Z$ be all the naturals which are NOT powers of a prime number. $Z$ is also infinite since, for example, it contains $2*3, 2*3^2, 2*3^3, 2*3^4,...$

Thus, we've dividied $\mathbb{N}$ into infinitely many infinite sets: $Z$ and each of the $Y_p$.

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Certainly. Look at the Cantor pairing function that shows a correspondence between the naturals and pairs of naturals. You have $f:\mathbb{N}\mapsto \mathbb{N\times N}$ which is a byjection. All sets of the form (x,y) for a given x are infinite. So the sets of n that go to (x,y) for a given x are infinite.

On second thought, this is too complicated. $(x,y) \in \mathbb{N \times N}$ is infinite. The sets of the form $(x,1)$ are infinite, as are $(x,2)$, as are $\ldots$. So we have divided an infinite set into an infinite number of infinite sets. The pairing function just projects this onto $\mathbb{N}$.

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For another example with a somewhat different flavour, chop the real line up at integers, dividing it into countable union of unit intervals, each of which has uncountably many points:

$$ \mathbb{R} = \bigcup_{n \in \mathbb{Z}} [n,n+1)$$

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A simple way to split the set of natural numbers.Take the sets $S_n$ of numbers with sum of digits $n$ for each all $n$. Obviously no sets overlap and each $S_n$ contain numbers of the form $1111..0000...$ where the string of 1's contain $n$ of them, and as many zeros as you wish

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