Equivalence of Definitions of Sigma-Algebra

Theorem
The following definitions of a $\sigma$-algebra are equivalent:

Definition 1 implies Definition 2
Let $\mathcal R$ be a system of sets on a set $X$ such that:
 * $(1): \quad X \in \mathcal R$
 * $(2): \quad \forall A, B \in \mathcal R: \complement_X \left({A}\right) \in \mathcal R$
 * $(3): \quad \displaystyle \forall A_n \in \mathcal R: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \mathcal R$

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 $\mathcal R$.

So by definition 2 of $\sigma$-ring it follows that $\mathcal R$ is a $\sigma$-ring with a unit.

Thus $\mathcal R$ is a [Definition:Sigma-Algebra|$\sigma$-algebra]] by definition 2.

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

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

From definition 2 of $\sigma$-ring, $\mathcal R$ is:
 * $(1) \quad$ closed under set difference.
 * $(2) \quad$ closed under countable union

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 a $\sigma$-algebra by definition 1.