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

$\blacksquare$