User:Jshflynn/P-star is Commutative Monoid under Union

Theorem
Let $\Sigma$ be an alphabet and $\mathcal{P}(\Sigma^{*})$ be the P-star of $\Sigma$.

Then $(\mathcal{P}(\Sigma^{*}), \cup)$ is a monoid.

Closure
As $V \subseteq \Sigma^{*} \land W \subseteq \Sigma^{*} \implies (V \cup W) \subseteq \Sigma^{*}$ and $V, W \in \mathcal{P}(\Sigma^{*})$ we have that $\mathcal{P}(\Sigma^{*})$ is closed under $\cup$.

Associativity
Set union is associative.

Identity
The empty language is a language over any alphabet and therefore is an element of $\mathcal{P}(\Sigma^{*})$.

The empty language is equivalent to the empty set and for any set $X$:

$X \cup \emptyset=X$

So the empty language is the identity element.

Commutativity
Set union is commutative.

Hence $(\mathcal{P}(\Sigma^{*}), \cup)$ satisfies all the defining properties of a monoid.