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 2
First, $(\text {SA} 1')$ is the same as $(\text {SA} 1)$.

Secondly, $(\text {SA} 3')$ is a special case of $(\text {SA} 3)$.

It remains to show $(\text {SA} 2')$.

Let $A, B \in \Sigma$ be arbitrary.

Observe:

By $(\text {SA} 2)$:
 * $\relcomp X A \in \Sigma$

Furthermore by $(\text {SA} 3)$:
 * $\relcomp X A \cup B \in \Sigma$

Finally, by $(\text {SA} 2)$:
 * $A \setminus B = \relcomp X {\relcomp X A \cup B} \in \Sigma$

Hence $(\text {SA} 2')$ is verified.

Definition 2 implies Definition 1
First, $(\text {SA} 1)$ is the same as $(\text {SA} 1')$.

Secondly, $(\text {SA} 2)$ is a special case of $(\text {SA} 2')$, since for all $A \subseteq X$:
 * $\relcomp X A = X \setminus A$

by definition of relative complement.

It remains to show $(\text {SA} 3)$.

Let $A_n \in \Sigma$ be arbitrary for $n = 1, 2, \ldots$.

Let $F_1 := A_1$.

For all $n \ge 2$, let recursively:
 * $F_n := A_n \setminus \paren {F_1 \sqcup \cdots \sqcup F_{n - 1} }$

Sublemma
For all $n = 1, 2, \ldots$ we have:
 * $F_n \in \Sigma$

and:
 * $A_1 \cup \cdots \cup A_n = F_1 \sqcup \cdots \sqcup F_n$

Proof of Sublemma
We shall prove by induction.

For $n = 1$ the claim is trivial, as $F_1 = A_1$.

Suppose that the claim is true for $n = k - 1$.

In particular:
 * $F_1 \sqcup \cdots \sqcup F_{k - 1} \in \Sigma$

Therefore by $(\text {SA} 2')$:
 * $F_k = A_k \setminus \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \in\Sigma$

Furthermore:

Therefore:
 * $\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigsqcup_{n \mathop = 1}^\infty F_n$

where the belongs to $\Sigma$ by $(\text {SA} 3')$.

Hence $(\text {SA} 3)$ is verified.

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.