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 |
$\blacksquare$