Algebra of Sets contains Underlying Set/Proof 1

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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:

\((\text {AS} 1)\)   $:$   Unit:    \(\ds S \in \RR \)      

The result is hence immediate.

$\blacksquare$