Power Set is Sigma-Algebra

Theorem
The power set of a set which is countably infinite is a sigma-algebra.

Proof
Let $$S$$ be a countably infinite set, and let $$\mathcal P \left({S}\right)$$ be its power set.

We have that a power set is an algebra of sets, and so:


 * 1) $$\forall A, B \in \mathcal P \left({S}\right): A \cup B \in \mathcal P \left({S}\right) \ $$
 * 2) $$\mathcal{C}_S \left({A}\right) \in \mathcal P \left({S}\right) \ $$

Now, suppose $$\left \langle {A_i}\right \rangle$$ be a countably infinite sequence of sets in $$\mathcal P \left({S}\right)$$.

Consider an element of the union of all the sets in this sequence:
 * $$x \in \bigcup_{i \in \N} A_i$$

By definition of union, $$\exists i \in \N: x \in A_i$$.

But $$A_i \in \mathcal P \left({S}\right)$$ and so by definition $$A_i \subseteq S$$.

By definition of subset, it follows that $$x \in S$$.

Hence, again by definition of subset, $$\bigcup_{i \in \N} A_i \subseteq S$$.

Finally, by definition of power set, $$\bigcup_{i \in \N} A_i \in \mathcal P \left({S}\right)$$.

So, by definition, $$\mathcal P \left({S}\right)$$ is a sigma-algebra.