Zero of Power Set with Intersection

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

Consider the algebraic structure $$\left({\mathcal P \left({S}\right), \cap}\right)$$, where $$\cap$$ denotes set intersection.

Then the empty set $$\varnothing$$ serves as the zero for $$\left({\mathcal P \left({S}\right), \cap}\right)$$.

Proof
First we note that from Empty Set Element of Power Set, we have that $$\varnothing \in \mathcal P \left({S}\right)$$.

Then from Intersection with Null: we have:
 * $$\forall A \subseteq S: A \cap \varnothing = \varnothing = \varnothing \cap A$$

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

So:
 * $$\forall A \in \mathcal P \left({S}\right): A \cap \varnothing = \varnothing = \varnothing \cap A$$

Thus we see that $$\varnothing$$ acts as the zero for $$\left({\mathcal P \left({S}\right), \cap}\right)$$.