Disclaimer: This question is part of: Borel Measures: Atoms (Summary)
Given a sigma algebra $\Sigma$ over a countable space $\#\Omega\leq\aleph_0$.
Does it admit a finest measurable partition: $$\Omega=\bigsqcup_{k=1}^\infty E_k\quad E_k\in\Sigma$$ (I guess Zorn's lemma might solve this but not sure about it.)
Yes, it does!
First, fix an element $\omega\in\Omega$.
Then every proper chain has length at most countably infinite: $$\Omega\supsetneq E_1\supsetneq E_2\supsetneq\ldots\quad\omega\in E_k\in\Sigma$$ and therefore admits a lower bound: $$\omega\in\bigcap_{k=1}^\infty E_k\in\Sigma$$ Hence, there is a minimal element by Zorn's lemma: $$\omega\in A_\omega\in\Sigma$$
Now, consider distinct elements $\omega\neq\omega'$.
Assume $A_\omega\cap A_{\omega'}\neq\varnothing$ and $A_\omega\neq A_{\omega'}$.
Also either $\omega\in A_\omega\cap A_{\omega'}$ or $\omega\in A_\omega\cap A_{\omega'}^c$.
So w.l.o.g. $A_\omega\supsetneq A_\omega\cap A_{\omega'}$ with $\omega\in A_\omega\cap A_{\omega'}\in\Sigma$.
But that contradicts minimality.
Concluding, there is a finest measurable partition modulo repetitions: $$\Omega=\bigsqcup_{\omega\in\Omega}A_\omega\quad A_\omega\in\Sigma$$ Especially, it is countable!
