Power Set is Algebra of Sets

Theorem
Let $$S$$ be a set.

Let $$\mathcal{P} \left({S}\right)$$ be the power set of $$S$$.

Then $$\mathcal{P} \left({S}\right)$$ is an algebra of sets where $$S$$ is the unit.

Proof
From Power Set Closed under Intersection and Power Set Closed under Symmetric Difference, we have that:


 * 1) $$\forall A, B \in \mathcal{P} \left({S}\right): A \cap B \in \mathcal{P} \left({S}\right)$$;
 * 2) $$\forall A, B \in \mathcal{P} \left({S}\right): A * B \in \mathcal{P} \left({S}\right)$$.

From the definition of power set:
 * $$\forall A \in \mathcal{P} \left({S}\right): A \subseteq S$$

and so $$S$$ is the unit of $$\mathcal{P} \left({S}\right)$$.

Thus we see that $$\mathcal{P} \left({S}\right)$$ is a ring of sets with a unit.

Hence the result, from definition 2 of an algebra of sets.