Zero of Power Set with Union

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

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

Then $$S$$ serves as the zero for $$\left({\mathcal P \left({S}\right), \cup}\right)$$.

Proof
We note that from Subset of Itself, $$S \subseteq S$$ and so $$S \in \mathcal P \left({S}\right)$$ from the definition of the power set.

From Union with Superset is Superset‎, we have:
 * $$A \subseteq S \iff A \cup S = S = S \cup A$$.

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

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

Thus we see that $$S$$ acts as the zero.