Algebra of Sets contains Empty Set/Proof 2
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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 $1$ 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 |
$\blacksquare$