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.


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 $|s'|=\aleph_0$.

  • $\begingroup$ Hi @Asaf, from your hints, I have figured out a solution that may work. Could you please verify it? We have $|s|=|s\times\Bbb N|$. Thus there is a bijection $f:s\times\Bbb N\to s$. We define a family $(s_i\mid i\in s)$ by $s_i=f[\{i\}\times\Bbb N]$ for all $i\in s$. Since $f$ is bijective, $s_{i_1}\cap s_{i_2}=\emptyset$ for all $i_1\neq i_2\in s$, and $\bigcup_{i\in s}s_i=\bigcup_{i\in s}f[\{i\}\times\Bbb N]=f[\bigcup_{i\in s}\{i\}\times\Bbb N]=f[s\times\Bbb N]=s$. Finally, $|s_i|=|f[\{i\}\times\Bbb N]|=|\{i\}\times\Bbb N|=|\Bbb N|=\aleph_0$ since $f$ is bijective. $\endgroup$
    – Akira
    Sep 15 '18 at 14:48

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.


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