Show that $\sigma(\mathcal{F})$ coincides with the countable-cocountable $\sigma$-algebra. Let $S$ be a set and let $\mathcal{F} = \{\{s\}:s\in S\}$ be the collection consisting of all sets which contain one element of $S$.
Let $\mathcal{A} = \{A\subseteq S:A \text{ is countable or $A^c$ is countable}\}$ be the countable-cocountable $\sigma$-algebra for $S$.
Question How do I show that $\sigma(\mathcal{F})$, the smallest $\sigma$-algebra containing $\mathcal{F}$, coincides with the countable-cocountable $\sigma$-algebra $\mathcal{A}$?
I assume coincides means 'is' here, please correct me if I'm wrong. The idea of $\sigma(\mathcal{F})$ was just introduced to me and I'm not so sure how to use it. I know that $\mathcal{A}$ is in fact an $\sigma$-algebra and I understand that $\mathcal{F}$ is contained by $\mathcal{A}$, but I don't know how to show that there is no smaller $\sigma$-algebra that contains $\mathcal{F}$.
Thanks in advance! 

Suppose $\mathcal{S}$ is a $\sigma$-algebra and $\mathcal{F} \subseteq \mathcal{S}$.
We need to see that $\mathcal{A} \subseteq \mathcal{S}$, and we have minimality and so  $\sigma(\mathcal{F}) = \mathcal{A}$.
So let $A$ be countable, then enumerate $A$ as $A = \{a_n: n \in \mathbb{N}\}$
Then for all $n$, $\{a_n\} \in \mathcal{F}$ so $\{a_n\} \in \mathcal{S}$ as well.
And as the latter is a $\sigma$-algebra, $A = \cup_n \{a_n\} \in \mathcal{S}$.
Now, if $A$ is cocountable, then $X\setminus A$ is countable so $X \setminus A \in \mathcal{S}$ by the above paragraph. But $\mathcal{S}$ is closed under complements so $X \setminus (X\setminus A) = A \in \mathcal{S}$ as well.
So all members of $\mathcal{A}$ are in $\mathcal{S}$ as required.
