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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?

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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.

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    Did you follow the links he gave? The Wikipedia article does not appear to be referring to the full boolean ring. And, indeed, in the first instance, it is not even a ring. (There are two different definitions of "ring of sets" given in the article...)2012-12-26
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    @Thomas: Yes, of course. Wasn’t it clear that I was responding specifically to the penultimate sentence of his question?2012-12-26
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    I didn't get that, no. Thought you were explaining something else. (In particular, you don't mention "ring of sets" anywhere in your answer.) The fact that the "primary" example of a ring of sets is the Boolean one, and that it is also a "ring" in the algebraic sense, is the core of the answer, and probably deserves specific mention.2012-12-26
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    @Thomas: Okay. Better now?2012-12-26
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    @BrianM.Scott, It's helpful, thank you! What if $X={\Bbb R}$? What is $({\Bbb Z}/2{\Bbb Z})^{|X|}$?2012-12-26
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    @Jack: You’re welcome! It’s simply the direct product of $2^\omega=\mathfrak c=|\Bbb R|$ two-element rings; I don’t know of any special name for it.2012-12-26