Ring of sets

Remarks

The finite subsets of $\N$ form a ring of sets: the union and the difference of two finite sets are finite. This ring does not contain $\N$, so it is not an algebra of sets. The bounded intervals of $\R$ do not form a ring of sets, since clause 2 fails: $[0, 3] \setminus [1, 2] = [0, 1) \cup (2, 3]$ is not an interval. A ring contains $\varnothing = A \setminus A$ and the intersection $A \cap B = A \setminus (A \setminus B)$. It also contains the symmetric difference $A \mathbin{\triangle} B = (A \setminus B) \cup (B \setminus A)$. With $\triangle$ as addition and $\cap$ as multiplication, a ring of sets is a commutative ring in the sense of algebra, possibly without identity. Every ring of sets is a semiring of sets.