Let $S$ be a set and let $\mathcal{A} = \{A\subseteq S: \text{A is countable or $A^c$ is countable}\}$.
Exercise: Prove that $\mathcal{A}$ is an $\sigma$-algebra.
I know that $\mathcal{A}$ is called an $\sigma$-algebra if the following properties hold:
i) $\emptyset, S \in \mathcal{A}$.
ii) $A\in \mathcal{A} \Rightarrow A^c \in \mathcal{A}$.
iii) $A_1, A_2, ..., \in \mathcal{A} \Rightarrow \bigcup\limits_{i = 1}^{\infty}A_i \in \mathcal{A}.$
My approach:
i) Quite straightforward: if $A = \emptyset$ then $A \in \mathcal{A}$, because $\emptyset$ is countable, so $\emptyset \in \mathcal{A}$. If $A = S$ we have that $A^c = S\backslash S = \emptyset$ is countable so we know that $S\in \mathcal{A}$.
ii) Pick an arbitrary set $A \in \mathcal{A}$. We know that $A$ is countable or $A^c$ is countable. Assume that $A$ is countable. $A^c = S\backslash A$ is in $\mathcal{A}$, because $(S\backslash A)^c = A$ is countable. Now assume that $A^c$ is countable. $A \in \mathcal{A}$ because $A^c$ is countable.
iii) I'm not sure what to do here. I feel like the solution might be rather obvious because we know that one of the $A_i = S$. Hence $\bigcup\limits_{i = 1}^{\infty}A_i = S \in \mathcal{A}$.
Question: How do I solve part iii)?