# Partitioning an infinite set

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

• 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
• – Aryabhata Dec 1 '10 at 16:20
• – Ami Dec 1 '10 at 21:11

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 $Y_p$.

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 $\mathbb{N}$.

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:

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

• This is for an uncountable set though. – PleaseHelp Mar 28 '16 at 13:55

The natural numbers organized in an infinite number of rows and an infinite number of columns:

1  2  4  7 11 16 ..
3  5  8 12 17 ..
6  9 13 18 ..
10 14 19 ..
:  :


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

Here's a quick way to divide the positive integers into infinitely many infinite sets:

Let $U_1$ be the set of positive odd numbers: $$\{1,3,5,7,\dots\}$$ Let $U_2$ be the set of twice the positive odd numbers: $$\{2,6,10,14,\dots\}$$ Let $U_3$ be the set of four times the positive odd numbers: $$\{4,12,20,28,\dots\}$$ …

Let $U_n$ be the set of $2^{n-1}$ times the positive odd numbers.

That is, the numbers in each set are twice the numbers in the last set. Now, the positive integers have been partitioned into the infinite sets $U_n$.

Note: I wrote this up because it is interesting, and 'rounds out' the theory, that we can partition $$X$$ into an infinite number of blocks with each on being countably infinite.

I am marking this as community wiki. If several people downvote this and/or post comments that they disagree with my sentiments, I'll delete this.

Proposition 1: Every infinite set $$X$$ can be partitioned into blocks with each one being countably infinite.
Proof
Choose a well order $$\le$$ on $$X$$ that has no maximal elements (see this) and let $$\sigma$$ denote the successor function on $$X$$. Let $$L$$ denote the elements of $$X$$ that do not have an immediate predecessor. For each $$\alpha \in L$$ define

$$\tag 1 L_\alpha = \{ S^n(\alpha) \, | \, \text{integer } n \ge 0\}$$

This family partitions $$X$$ (see this) and each set must also be countably infinite. $$\quad \blacksquare$$.

If $$L$$ is finite take any $$\alpha \in L$$ and partition the countably infinite set $$L_\alpha$$ into an infinite number of blocks (see the many fine answers in this thread). So we have

Proposition 2: Every infinite set $$X$$ can be partitioned into an infinite number of blocks
with each one being countably infinite.