Power Set is Closed under Intersection

Theorem
Let $$S$$ be a set.

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

Then:
 * $$\forall A, B \in \mathcal{P} \left({S}\right): A \cap B \in \mathcal{P} \left({S}\right)$$

Proof
Let $$\forall A, B \in \mathcal{P} \left({S}\right)$$.

Then by the definition of power set, $$A \subseteq S$$ and $$B \subseteq S$$.

From Intersection Subset we have that $$A \cap B \subseteq A$$.

It follows from Subsets Transitive that $$A \cap B \subseteq S$$.

Thus $$A \cap B \in \mathcal{P} \left({S}\right)$$.