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 $A, B \in \mathcal S$:


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

where $\complement_X \left({A}\right)$ 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:
 * $(1) \quad$ closed under set union
 * $(2) \quad$ 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 $X - A = \complement_X \left({A}\right)$.

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

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

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.