Equivalence of Definitions of Algebra of Sets

Theorem

The following definitions of the concept of Algebra of Sets are equivalent:

Definition 1

Let $X$ be a set.

Let $\powerset X$ be the power set of $X$.

Let $\mathcal R \subseteq \powerset X$ be a set of subsets of $X$.

Then $\mathcal R$ is an algebra of sets over $X$ if and only if the following conditions hold:

 $(AS \, 1)$ $:$ Unit: $\displaystyle X \in \mathcal R$ $(AS \, 2)$ $:$ Closure under Union: $\displaystyle \forall A, B \in \mathcal R:$ $\displaystyle A \cup B \in \mathcal R$ $(AS \, 3)$ $:$ Closure under Complement Relative to $X$: $\displaystyle \forall A \in \mathcal R:$ $\displaystyle \relcomp X A \in \mathcal R$

Definition 2

An algebra of sets is a ring of sets with a unit.

Proof

Definition 1 implies Definition 2

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.

$\Box$

Definition 2 implies Definition 1

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.

$\blacksquare$