Symmetric Difference with Union does not form Ring
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Theorem
Let $S$ be a set.
Let:
- $\symdif$ denote the symmetric difference operation
- $\cup$ denote the set union operation
- $\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cup}$ does not form a ring.
Proof
For $\struct {S, \symdif, \cup}$ to be a ring, it is a necessary condition that $\cup$ be distributive over $*$.
Also, the identity element for set union and symmetric difference must be different.
However:
- $(1): \quad$ the identity for union and symmetric difference is $\O$ for both operations
- $(2): \quad$ set union is not distributive over symmetric difference:
From Symmetric Difference of Unions:
- $\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
The result follows.
$\blacksquare$
Also see
Sources
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(d)}$