Sigma-Algebra of Countable Sets

Theorem
Let $X$ be a set.

Let $\mathcal A$ be the collection of countable and co-countable subsets of $X$.

Then $\mathcal A$ is a $\sigma$-algebra.

Proof
Let us verify the axioms of a $\sigma$-algebra in turn.

Axiom $(1)$
By Relative Complement Self Null, $\complement_X \left({X}\right) = \varnothing$; in particular, it is countable.

Hence $X$ is co-countable, and so $X \in \mathcal A$.

Axiom $(2)$
The relative complement of a countable set is by definition co-countable.

The converse holds by Relative Complement of Relative Complement.

Hence $A \in \mathcal A \implies \complement_X \left({A}\right) \in \mathcal A$.

Axiom $(3)$
Let $\left({A_n}\right)_{n \in \N} \in \mathcal A$ be a collection.

Suppose that all the $A_n$ are countable.

Then by Union of Countable Sets, $\displaystyle \bigcup_{n \in \N} A_n$ is also countale, hence in $\mathcal A$.

Now suppose that at least one $A_n$ is only co-countable.

Then by Complements Invert Subsets and Subset of Union, have:


 * $\displaystyle \complement_X \left({\bigcup_{n \in \N} A_n}\right) \subseteq \complement_X \left({A_n}\right)$

By definition of co-countable, the latter is countable.

Hence, by Subset of Countable Set, it follows that $b\displaystyle \bigcup_{n \in \N} A_n$ is co-countable, hence in $\mathcal A$.

Hence the result, from Proof by Cases.