An assignment reads : 

> 
> Let $\Omega$ be a non-countable collection and let $\mathscr{A}=\{A\in\Omega\ |\ A\text{ is countable or }A^C \text{ is countable}\}$
>  Prove that $\mathscr{A}$ is a sigma-algebra.
> 
> 
> 

It's only necessary to prove that (1) $\Omega\in\mathscr{A}$, (2) every complement of $A\in\mathscr{A}$ is in $\mathscr{A}$, and (3) every countable union $\cup\_{n\in\mathbb{N}}A\_n$ of elements of $\mathscr{A}$ is in $\mathscr{A}$, too.

The first two conditions hold because $\emptyset$ is countable so $\emptyset^C=\Omega\in\mathscr{A}$, and because (2) is implied by the very defintion of $\mathscr{A}$.

The solution of the assignment then proceeds to show that (3) holds, too. If all the $A\_n$ in the countable union are countable, then the countable union is countable too. If one of the $A\_n$, say, $A\_m$ is not countable, then $(\cup\_{n\in\mathbb{N}}A\_n)^C=\cap\_{n\in\mathbb{N}}A\_n^C$. Because $A\_m^C$ is countable (otherwise $A\_m$ would not be an element of $\mathscr{A}$), it follows that $\cap\_{n\in\mathbb{N}}A\_n^C$ is countable because $\cap\_{n\in\mathbb{N}}A\_n^C\subseteq A\_m^C$.

I do not understand that very last claim, namely that $\cap\_{n\in\mathbb{N}}A\_n^C\subseteq A\_m^C$. Can anyone explain?