Empty Set and Set form Algebra of Sets

Theorem

Let $S$ be any non-empty set.

Then $\set {S, \O}$ is (trivially) an algebra of sets, where $S$ is the unit.

Proof

From Set Union is Idempotent:

$S \cup S = S$

and

$\O \cup \O = \O$

Then from Union with Empty Set:

$S \cup \O = S$

So $\set {S, \O}$ is closed under union.

From Relative Complement of Empty Set:

$\relcomp S \O = S$

and from Relative Complement with Self is Empty Set:

$\relcomp S S = \O$

so $\set {S, \O}$ is closed under complement.

Hence the result, by definition of algebra of sets.

$\blacksquare$