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Suppose $\Omega$ $=$ $\Bbb N$. Verify that the following set of subsets of $\Omega$ is an algebra: $\mathcal A$ $=$ $\{$ $A$ $\subset$ $\Bbb N$ $:$ $A$ is countable or $A$$^c$ is countable $\}$

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    Every subset of N is countable, so you want to prove that intersections and unions of subsets of N are subsets of N them-self. – 2017-01-01
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    Perhaps in the context "countable" means "countably infinite". In that case $\mathcal A$ would be a (Boolean) algebra, but not a $\sigma$-algebra. – 2017-01-01

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Yes, it is even a sigma-algebra.

Clearly $\mathcal{A}$ is closed under complement, and contains the empty set and the whole set. Now assuming $A_1,A_2,\ldots$ are in $\mathcal{A}$, we have

(1) If all the $A_i$'s are countable;

The union $\bigcup A_i$ is also countable, since it is a countable union of countable sets, so we have $\bigcup A_i\in\mathcal{A}$.

(2) If $A_{i}^{c}$ is countable for some $i\geq 1$;

Then the complement of $\bigcup A_i$ is a subset of a countable set, hence itself countable and we also have $\bigcup A_i\in\mathcal{A}$.

Thus $\mathcal{A}$ is a sigma-algebra.

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    If "countable" means "at most countable", then $\mathcal A$ is simply $\mathcal P(\mathbb N)$ ... – 2017-01-01
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    @HenningMakholm: The point here is that this proof works for any set $\Omega$, not just $\mathbb{N}$. – 2017-01-01