Equivalence of Definitions of Algebra of Sets

Theorem
There are two definitions given for an algebra of sets:


 * An algebra of sets is a ring of sets with a unit;


 * Given a set $$X \ $$ and a collection of subsets of $$X \ $$, $$\mathcal{S} \subset \mathcal{P} \left({X}\right) \ $$, $$\mathcal{S} \ $$ is called an algebra of sets if, given that $$A, B \in \mathcal{S} \ $$,


 * 1) $$A \cup B \in \mathcal{S} \ $$
 * 2) $$\mathcal{C}_X \left({A}\right) \in \mathcal{S} \ $$

where $$\mathcal{C}_X \left({A}\right)$$ is the relative complement of $$A \ $$ in $$X$$.

These definitions are equivalent.

Ring is Algebra
Let $$\mathcal{R}$$ be a ring of sets with a unit $$X$$.

From Ring of Sets Closed under Various Operations, it is immediate that $$\mathcal{R}$$ is:
 * 1) closed under set union;
 * 2) closed under set difference.

From Unit of System of Sets is Unique, we have that:
 * $$\forall A \in \mathcal{R}: A \subseteq X$$

from which we have that $$X - A = \mathcal{C}_X \left({A}\right)$$.

So $$\mathcal{R}$$ is an algebra of sets as defined.

Algebra is Ring
Let $$\mathcal{S} \ $$ be an algebra of sets such that $$\forall A \in \mathcal{S}: \mathcal{C}_X \left({A}\right) \in \mathcal{S}$$.

Let $$A, B \in \mathcal{S}$$.

From the definition, $$\forall A \in \mathcal{S}: A \subseteq X$$.

Hence from Intersection with Subset is Subset, $$\forall A \in \mathcal{S}: A \cap X = A$$.

Hence $$X$$ is the unit of $$\mathcal{S} \ $$.

From Properties of Algebras of Sets, we have that $$\mathcal{S} \ $$ is closed under set intersection.

We have from the definition of symmetric difference that $$A * B = \left({A - B}\right) \cup \left({B - A}\right)$$.

Since both set union and set difference are closed operations, it follows that symmetric difference is also closed.

So by the definition of ring of sets, it follows that $$\mathcal{S} \ $$ is indeed a ring of sets.