Algebra of Sets contains Empty Set

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 1

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:

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

$\blacksquare$

Proof 2

By definition $1$ of algebra of sets, we have that:

\((\text {AS} 3)\)

$:$

Closure under Complement Relative to $S$:

\(\ds \forall A \in \RR:\)

\(\ds \relcomp S A \in \RR \)

Thus:

\(\ds S\)

\(\in\)

\(\ds \RR\)

Algebra of Sets contains Underlying Set‎

\(\ds \leadsto \ \ \)

\(\ds \relcomp S S\)

\(\in\)

\(\ds \RR\)

Definition of Algebra of Sets: $\text {AS} 3$

\(\ds \leadsto \ \ \)

\(\ds \O\)

\(\in\)

\(\ds \RR\)

Relative Complement with Self is Empty Set

$\blacksquare$