Let $\Omega$ be a countable set, $\mathcal A$ be a $\sigma$-algebra that seperates points $(\forall \omega_1,\omega_2 \in \Omega, \omega_1 \ne \omega_2, \exists A \in \mathcal A:\omega_1 \in A, \omega_2 \notin A)$. Then $\mathcal P(\Omega)=\mathcal A$.
What I have so far is:
Proof.
$\supseteq:\surd$
$\subseteq:$
$\emptyset,\Omega\in \mathcal A$ is implied by definition.
Let $A \subseteq\Omega, (\ A \ne \emptyset, A \ne \Omega)$, so that $a\in A$, $b \in \Omega \smallsetminus A$. Then $b\in A^c, a\in \Omega \smallsetminus A^c$. Hence $A\in \mathcal A \implies A^c \in \mathcal A$.
Now I am stuck at proving $(A_n)_{n\ge 1} \in \mathcal A \implies \bigcup_{n\in \mathbb N}A_n \in \mathcal A$. Can anyone help me with that?
Since $\mathcal{A}$ separates points and is closed under complement, for any $\omega\in\Omega$ for any $q\in\Omega$ with $q\ne\omega$, there exists $A_{q}$ in $\mathcal{A}$ with $\omega\in A_q$ and $q\notin A_q$. Then the intersection of all $A_q$ for $q\in\Omega\setminus\{\omega\}$ is a countable intersection, thus in $\mathcal{A}$, and equals $\{\omega\}$. Thus $\mathcal{A}$ contains all singletons of $\Omega$. But since $\Omega$ is countable, all of its subsets are countable unions of singletons, so $\mathcal{A} = P(\Omega)$.