Algebra of Sets contains Empty Set

From ProofWiki
Jump to navigation Jump to search


Let $S$ be a set.

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


$\O \in \RR$

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:    \(\ds \O \in \RR \)      


Proof 2

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

\((\text {AS} 3)\)   $:$   Closure under Complement Relative to $S$:      \(\ds \forall A \in \RR:\) \(\ds \relcomp S A \in \RR \)      


\(\ds S\) \(\in\) \(\ds \RR\) 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