Any infinite set partitioned into a set of countably infinite sets?

Question

Prove that if $s$ is infinite, then it can be partitioned into a set of countably infinite sets $\mathcal{A}$. That is:

• $\bigcup \mathcal{A}=s$

• $\forall a\in \mathcal{A}, a$ is countably infinite

• $\forall a_1, a_2 \in \mathcal{A},\quad a_1=a_2 \implies a_1 \cap a_2 = \emptyset$

I'm not sure how to get going on this one. I'm under the presumption that I should use Zorn's lemma, though I'm not entirely sure how...

Is it true that $\mathcal{A}$ should have the same cardinality as $s$?

Any help appreciated.

Answer 1

HINT: Use Zorn's lemma to show that there is a bijection between $s$ and $s\times\Bbb N$. Use the fact that if $s$ is infinite, then there is at least one $s'\subseteq s$ such that $|s'|=\aleph_0$.

Answer 2

Assuming ZFC, we can take $s$ as an aleph number.

• If $s=\aleph_0$, then $\{s\}$ is a partitioning of $s$ with the required property. So we're done.

• If $s = \aleph_\beta$ for some ordinal $β>0$, then we can define a mapping $f:s→\mathcal{P}(s)$ by writing $f(α)=[ωα,\omega(\alpha+1))$. The image of $f$ is the required partitioning.