Sigma-Algebra of Countable Sets

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Theorem

Let $X$ be a set.

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


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


Proof

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


Axiom $(1)$

By Relative Complement with Self is Empty Set:

$\complement_X \left({X}\right) = \varnothing$

In particular, it is countable.

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

$\Box$


Axiom $(2)$

The relative complement of a countable set is by definition co-countable.

The converse holds by Relative Complement of Relative Complement.

Hence:

$E \in \Sigma \implies \complement_X \left({E}\right) \in \Sigma$

$\Box$


Axiom $(3)$

Let $\left({E_n}\right)_{n \mathop \in \N} \in \Sigma$ be a collection.


Suppose that all the $E_n$ are countable.

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

$\displaystyle \complement_X \left({\bigcup_{n \mathop \in \N} E_n}\right) \subseteq \complement_X \left({E_n}\right)$

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

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


Hence the result, from Proof by Cases.

$\blacksquare$


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