Power Set is Sigma-Algebra

Theorem
The power set of a set is a sigma-algebra.

Proof
Let $S$ be a 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:


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

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:
 * $\displaystyle x \in \bigcup_{i \in \N} A_i$

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:
 * $\displaystyle \bigcup_{i \in \N} A_i \subseteq S$

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

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