Algebra of Sets contains Empty Set/Proof 1

Theorem

Let $S$ be a set.

Let $\RR$ be an algebra of sets on $S$.

Then:

$\O \in \RR$

where $\O$ denotes the empty set.

Proof

By definition $2$ of Algebra of Sets:

$\RR$ is a ring of sets with a unit.

By definition $2$ of Ring of Sets:

\((\text {RS} 1_2)\)

$:$

Empty Set:

\(\ds \O \in \RR \)

$\blacksquare$