Equivalence of Definitions of Ring of Sets

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Theorem

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

Definition 1

A ring of sets $\RR$ is a system of sets with the following properties:

\((\text {RS} 1_1)\)   $:$   Non-Empty:    \(\ds \RR \ne \O \)             
\((\text {RS} 2_1)\)   $:$   Closure under Intersection:      \(\ds \forall A, B \in \RR:\) \(\ds A \cap B \in \RR \)             
\((\text {RS} 3_1)\)   $:$   Closure under Symmetric Difference:      \(\ds \forall A, B \in \RR:\) \(\ds A \symdif B \in \RR \)             

Definition 2

A ring of sets $\RR$ is a system of sets with the following properties:

\((\text {RS} 1_2)\)   $:$   Empty Set:    \(\ds \O \in \RR \)             
\((\text {RS} 2_2)\)   $:$   Closure under Set Difference:      \(\ds \forall A, B \in \RR:\) \(\ds A \setminus B \in \RR \)             
\((\text {RS} 3_2)\)   $:$   Closure under Union:      \(\ds \forall A, B \in \RR:\) \(\ds A \cup B \in \RR \)             

Definition 3

A ring of sets $\RR$ is a system of sets with the following properties:

\((\text {RS} 1_3)\)   $:$   Empty Set:    \(\ds \O \in \RR \)             
\((\text {RS} 2_3)\)   $:$   Closure under Set Difference:      \(\ds \forall A, B \in \RR:\) \(\ds A \setminus B \in \RR \)             
\((\text {RS} 3_3)\)   $:$   Closure under Disjoint Union:      \(\ds \forall A, B \in \RR:\) \(\ds A \cap B = \O \implies A \cup B \in \RR \)             


Proof

Definition 1 implies Definition 2

Let $\RR$ be a system of sets such that for all $A, B \in \RR$:

$(\text {RS} 1_1): \quad \RR \ne \O$
$(\text {RS} 2_1): \quad A \cap B \in \RR$
$(\text {RS} 3_1): \quad A \symdif B \in \RR$

As $\RR$ is non-empty, there exists some $A \in \RR$.

From Symmetric Difference with Self is Empty Set:

$A \symdif A = \O$

By hypothesis $A \symdif A \in \RR$ and so $\O \in \RR$.

Thus criterion $(\text {RS} 1_2)$ is fulfilled.


From Closure of Intersection and Symmetric Difference imply Closure of Set Difference it follows that criterion $(\text {RS} 2_2)$ is fulfilled.


From Closure of Intersection and Symmetric Difference imply Closure of Union it follows that criterion $(\text {RS} 3_2)$ is fulfilled.

$\Box$


Definition 2 implies Definition 1

Let $\RR$ be a system of sets such that for all $A, B \in \RR$:

$(\text {RS} 1_2): \quad \O \in \RR$
$(\text {RS} 2_2): \quad A \setminus B \in \RR$
$(\text {RS} 3_2): \quad A \cup B \in \RR$

We have that $\O \in \RR$ and so $\RR$ is non-empty.

Thus criterion $(\text {RS} 1_1)$ is fulfilled.


By hypothesis, $\RR$ is closed under $\setminus$ and $\cup$.

Thus:

$\forall A, B \in \RR: \paren {A \setminus B} \cup \paren {B \setminus A} \in \RR$

But by the definition of symmetric difference:

$A \symdif B := \paren {A \setminus B} \cup \paren {B \setminus A}$

Thus:

$\forall A, B \in \RR: A \symdif B \in \RR$

and so $\RR$ is closed under symmetric difference.

Thus criterion $(\text {RS} 3_1)$ is fulfilled.


From Union minus Symmetric Difference equals Intersection:

$\forall A, B \in \RR: \paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$

Thus $\RR$ is closed under set intersection.

Thus criterion $(\text {RS} 2_1)$ is fulfilled.

$\Box$


Definition 2 iff Definition 3

Let $\RR$ be a system of sets such that for all $A, B \in \RR$:

$(\text {RS} 1_2): \quad \O \in \RR$
$(\text {RS} 2_2): \quad A \setminus B \in \RR$
$(\text {RS} 3_2): \quad A \cup B \in \RR$

Criteria $(\text {RS} 1_3)$ and $(\text {RS} 2_3)$ are fulfilled immediately.


Consider $A, B \in \RR: A \cap B = \O$.

Then as $A, B \in \RR$ it follows by $(\text {RS} 3_2)$ that $A \cup B \in \RR$ and so $(\text {RS} 3_3)$ is fulfilled.


Now let $\RR$ be a system of sets such that for all $A, B \in \RR$:

$(\text {RS} 1_3): \quad \O \in \RR$
$(\text {RS} 2_3): \quad A \setminus B \in \RR$
$(\text {RS} 3_3): \quad A \cap B = \O \implies A \cup B \in \RR$

Again, criteria $(\text {RS} 1_2)$ and $(\text {RS} 2_2)$ are fulfilled immediately.


Let $A, B \in \RR$.

Then from Set Difference Union Second Set is Union:

$A \cup B = \paren {A \setminus B} \cup B$

From Set Difference Intersection with Second Set is Empty Set:

$\paren {A \setminus B} \cap B = \O$

Thus from $(\text {RS} 3_3)$:

$A \cup B = \paren {A \setminus B} \cup B \in \RR$

$\blacksquare$


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