Equivalence of Definitions of Sigma-Algebra

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

Let $A, B \in \Sigma$.

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

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

Hence $X$ is the unit of $\Sigma$.

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

Thus $\Sigma$ is a $\sigma$-algebra by definition 3.

Definition 3 implies Definition 1
Let $\Sigma$ be a $\sigma$-ring with a unit $X$.

By definition, $X \in \Sigma$.

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

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

from which we have that $X \setminus A = \relcomp X A$.

So $\Sigma$ is a $\sigma$-algebra by definition 1.

Definition 1 implies Definition 4
Immediate from the definition of algebra along with the added condition of closure under countable unions.

Definition 4 implies Definition 1
By definition $1$ of algebra of sets, an algebra has the properties:

Replacing $(\text {AS} 2)$ with closure under countable unions immediately yields the first definition.