Symmetric Difference with Intersection forms Ring
Theorem
Let $S$ be a set.
Let:
- $\symdif$ denote the symmetric difference operation
- $\cap$ denote the set intersection operation
- $\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain.
Proof 1
From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, \symdif}$ is an abelian group, where $\O$ is the identity and each element is self-inverse.
From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
Also Intersection Distributes over Symmetric Difference.
Thus $\struct {\powerset S, \cap}$ is a commutative ring with a unity which is $S$.
From Intersection with Empty Set:
- $\forall A \in \powerset S: A \cap \O = \O = \O \cap A$
Thus $\O$ is indeed the zero.
However, from Set Intersection Not Cancellable, it follows that $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
$\blacksquare$
Proof 2
From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.
Hence $\powerset S$ is a ring of sets, and hence a commutative ring.
From Intersection with Subset is Subset, we have $A \subseteq S \iff A \cap S = A$.
Thus we see that $S$ is the unity.
Also during the proof of Power Set with Intersection is Monoid, it was established that $S$ is the identity of $\struct {\powerset S, \cap}$.
We also note that set intersection is not cancellable, so $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
The result follows.
$\blacksquare$
Also see
Sources
- 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.3$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $6 \ \text{(iii)}$