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.
$\blacksquare$