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 element 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 element for $\left({\mathcal P \left({S}\right), \cap}\right)$.