Power Set with Union is Commutative Monoid

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

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

The only invertible element of this structure is $$\varnothing$$.

Proof

 * First, from Power Set Closed under Union, we have that $$\forall A, B \in \mathcal{P} \left({S}\right): A \cup B \in \mathcal{P} \left({S}\right)$$.


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


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

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

Then from Union with Null: we have $$A \cup \varnothing = A = \varnothing \cup A$$.

Thus we see that $$\varnothing$$ acts as the identity.


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

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

The result follows by definition of monoid.