Symmetric Difference with Union does not form Ring

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

• Symmetric Difference with Intersection forms Ring

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': $\text {(d)}$