Equivalence of Definitions of Algebra of Sets

Theorem
The following definitions of an algebra of sets are equivalent:

Definition 1 implies Definition 2
Let $\mathcal R$ be a ring of sets with a unit $X$.

By definition, $X \in \mathcal R$.

From definition 2 of ring of sets, $\mathcal R$ 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 $X \setminus A = \complement_X \left({A}\right)$.

So $\mathcal R$ is an algebra of sets by definition 1.

Definition 2 implies Definition 1
Let $\mathcal R$ be a system of sets such that $\forall A, B \in \mathcal R$:
 * $(1): \quad A \cup B \in \mathcal R$
 * $(2): \quad \complement_X \left({A}\right) \in \mathcal R$

Let $A, B \in \mathcal R$.

From the definition:
 * $\forall A \in \mathcal R: A \subseteq X$.

Hence from Intersection with Subset is Subset:
 * $\forall A \in \mathcal R: A \cap X = A$

Hence $X$ is the unit of $\mathcal R$.

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

From the definition of symmetric difference:
 * $A * B = \left({A \setminus B}\right) \cup \left({B \setminus A}\right)$

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

So by definition 1 of ring of sets, it follows that $\mathcal R$ is indeed a ring of sets.