Sigma-Algebra of Countable Sets

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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 $X \in \Sigma$.

$\Box$


Axiom $(2)$

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

The converse holds by Relative Complement of Relative Complement.

Hence:

$E \in \Sigma \implies \relcomp X E \in \Sigma$

$\Box$


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.

$\blacksquare$


Sources

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