Equivalence of Definitions of Algebra of Sets

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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.

$\blacksquare$


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