Algebra of Sets is Closed under Set Difference
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Theorem
Let $S$ be a set.
Let $\RR$ be an algebra of sets on $S$.
Then:
- $\forall A, B \in S: A \setminus B \in \RR$
Proof 1
By definition $2$ of Algebra of Sets:
- $\RR$ is a ring of sets with a unit.
By definition $2$ of Ring of Sets:
\((\text {RS} 2_2)\) | $:$ | Closure under Set Difference: | \(\ds \forall A, B \in \RR:\) | \(\ds A \setminus B \in \RR \) |
$\blacksquare$
Proof 2
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:
\(\ds A, B\) | \(\in\) | \(\ds \RR\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap \relcomp S B\) | \(\in\) | \(\ds \RR\) | Definition of Algebra of Sets: $\text {AS} 3$ and Algebra of Sets is Closed under Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \setminus B\) | \(\in\) | \(\ds \RR\) | Set Difference as Intersection with Relative Complement |
$\blacksquare$