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

 * Closure: Let $$A, B \in \mathcal{P} \left({S}\right)$$.

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

We also have $$A \cup B \subseteq S \iff A \subseteq S \land B \subseteq S$$ from Union Smallest.

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


 * Associativity:

The operation $$\cup$$ is associative from Union is Associative.


 * Identity:

From Union with Null, we have $$A \cup \varnothing = A = \varnothing \cup A$$. Thus we see that $$\varnothing$$ acts as the identity.


 * Commutativity:

The operation $$\cup$$ is commutative from Union is Commutative.


 * Inverse:

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