Algebra of Sets contains Underlying Set/Proof 2

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Theorem

Let $S$ be a set.

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

Then:

$S \in \RR$


Proof

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

\((\text {AS} 2)\)   $:$   Closure under Union:      \(\ds \forall A, B \in \RR:\) \(\ds A \cup B \in \RR \)      
\((\text {AS} 3)\)   $:$   Closure under Complement Relative to $S$:      \(\ds \forall A \in \RR:\) \(\ds \relcomp S A \in \RR \)      


We have that $\RR \ne \O$ and so:

$\exists A \subseteq S: A \in \RR$


Then:

\(\ds \relcomp S A\) \(\in\) \(\ds \RR\) Definition of Algebra of Sets: $\text {AS} 3$
\(\ds \leadsto \ \ \) \(\ds \relcomp S A \cup A\) \(\in\) \(\ds \RR\) Definition of Algebra of Sets: $\text {AS} 2$
\(\ds \leadsto \ \ \) \(\ds S\) \(\in\) \(\ds \RR\) Union with Relative Complement

$\blacksquare$