"Ring of subsets" and "rings" The wikipedia defines "ring of sets" and "ring" differently. Is there any relationship between the two "ring"s? I cannot even find an Abelian group in the power set of a given nonempty set $X$. Is there a reason or just coincident that "ring" appears in both of these concepts?
 A: Here’s the prototypical example of a ring of sets in all three senses.
For any set $X$, $\wp(X)$ is an Abelian group under the operation $\triangle$ of symmetric difference, $$A\triangle B=(A\cup B)\setminus(A\cap B)=(A\setminus B)\cup(B\setminus A)\;.$$ $\varnothing$ is the identity, and every other element has order $2$, since $A\triangle A=\varnothing$. Such a group is known as a Boolean group. It becomes a Boolean ring when you add intersection ($\cap$) as the multiplication. It’s a unital ring, since $X$ is the multiplicative identity.
Identifying each $A\subseteq X$ with its indicator (characteristic) function $\chi_A$ gives you an isomorphism from $\langle\wp(X),\triangle,\cap\rangle$ to $(\Bbb Z/2\Bbb Z)^{|X|}$ the direct product of $|X|$ many copies of the ring $\Bbb Z/2\Bbb Z$.
The Boolean ring $\langle\wp(X),\triangle,\cap\rangle$ is trivially a ring of sets in both of the senses mentioned in the Wikipedia article, since it includes every subset of $X$. If $\mathscr R\subseteq\wp(X)$ is any unital subring of $\langle\wp(X),\triangle,\cap\rangle$, then $\mathscr R$ is closed under set difference: $A\setminus B=A\cap(X\triangle B)\in\mathscr R$ if $A,B\in\mathscr R$. $\mathscr{R}$ is also closed under union, since $A\cup B=(A\setminus B)\triangle B$. Thus, $\mathscr{R}$ is also a ring of sets in the order-theoretic and measure-theoretic senses.
