Sigma-Algebra of Countable Sets

Theorem
Let $X$ be a set.

Let $\Sigma$ be the set of countable and co-countable subsets of $X$.

Then $\Sigma$ is a $\sigma$-algebra.

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

Axiom $(1)$
By Relative Complement with Self is Empty Set:
 * $\relcomp X X = \O$

In particular, it is countable.

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

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

The converse holds by Relative Complement of Relative Complement.

Hence:
 * $E \in \Sigma \implies \relcomp X E \in \Sigma$

Axiom $(3)$
Let $\family {E_n}_{n \mathop \in \N} \in \Sigma$ be a family of elements of $\Sigma$.

Suppose that all the $E_n$ are countable.

Then by Countable Union of Countable Sets is Countable, $\ds \bigcup_{n \mathop \in \N} E_n$ is also countable.

Hence in $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$.

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

Then by Set Complement inverts Subsets and Set is Subset of Union: General Result:


 * $\ds \relcomp X {\bigcup_{n \mathop \in \N} E_n} \subseteq \relcomp X {E_n}$

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

Thus, by Subset of Countably Infinite Set is Countable, it follows that $\ds \bigcup_{n \mathop \in \N} E_n$ is co-countable.

Hence:
 * $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$

Hence the result, from Proof by Cases.