Power Set with Intersection is Commutative Monoid

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

Then $$\left({\mathcal{P} \left({S}\right), \cap}\right)$$ is a commutative monoid whose identity is $$S$$.

The only invertible element of this structure is $$S$$.

Proof

 * First we show that $$\left({\mathcal{P} \left({S}\right), \cap}\right)$$ is closed.

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

Thus as $$A \cap B \subseteq A$$ from Intersection Subset it follows that $$A \cap B \subseteq S$$ from Subsets Transitive.

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


 * Next, we have that, from Set System Closed with Intersection is Semigroup, $$\left({\mathcal{P} \left({S}\right), \cap}\right)$$ is a commutative semigroup.


 * Next we need to do now is show that $$\left({\mathcal{P} \left({S}\right), \cap}\right)$$ has an Identity.

From Subset Equivalences, we have $$A \subseteq S \iff A \cap S = A = S \cap A$$. Thus we see that $$S$$ acts as the identity.


 * Finally, we show that only $$S$$ has an Inverse:

For $$T \subseteq S$$ to have an inverse under $$\cap$$, we require $$T^{-1} \cap T = S$$. From this it follows that $$T = S = T^{-1}$$.

The result follows by definition of monoid.