Equivalence of Definitions of Ring of Sets
Theorem
The following definitions of the concept of Ring of Sets are equivalent:
Definition 1
A system of sets $\RR$ is a ring of sets if and only if $\RR$ satisfies the ring of sets axioms:
\((\text {RS} 1_1)\) | $:$ | 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 system of sets $\RR$ is a ring of sets if and only if $\RR$ satisfies the ring of sets axioms:
\((\text {RS} 1_2)\) | $:$ | 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 system of sets $\RR$ is a ring of sets if and only if $\RR$ satisfies the ring of sets axioms:
\((\text {RS} 1_3)\) | $:$ | 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 $(\text {RS} 3_3)$:
- $A \cup B = \paren {A \setminus B} \cup B \in \RR$
$\blacksquare$
Sources
- 1970: Avner Friedman: Foundations of Modern Analysis ... (previous) ... (next): $\S 1.1$: Rings and Algebras