Empty Set and Set form Algebra of Sets

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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.

$\blacksquare$