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The following comes from Wikipedia.

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It looks very strange to me since this "ring" seems to have nothing to do with the "ring" in abstract algebra. $(\cal R, \bigcup)$ is not an Abelian group (no inverse element if empty set is chosen as the identity), and $(\cal R, \backslash)$ is not a monoid (does not satisfy associativity). Am I right?

Then why this thing is called a ring?

Thanks!

Tony
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1 Answers1

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In fact, it is a ring of sets under the operations

  1. Symmetric difference as addition
  2. Intersection as multiplication.

Under these operations, a ring of subsets is a ring, even an algebra over $\mathbb F_2$. To see this, consider a universal set $X$, then we have a bijection $$A\mapsto 1_A$$ with $A$ a subset of $X$ and $1_A$ its characteristic function. Then the operations above become the normal sum and product of functions (over $\mathbb F_2$).

Quang Hoang
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  • I made a blog post that addressed a very similar idea while proving that you can put a ring structure on any set. Great answer $+1$. – Cameron Williams Oct 06 '15 at 02:17
  • Thanks, the idea is not new though. They are known as Boolean Algebra – Quang Hoang Oct 06 '15 at 02:23
  • Definitely not but it's still a good answer! – Cameron Williams Oct 06 '15 at 02:24
  • @QuangHoang Thank you for your answer. I have a further question. What is the identity for intersection? – Tony Oct 06 '15 at 03:12
  • That'd be the set with characteristic function equal to $0$. Can you guess? – Quang Hoang Oct 06 '15 at 03:13
  • @QuangHoang Thanks for answering. I still don't get it. Do you mean the empty set? – Tony Oct 06 '15 at 03:39
  • Oh, sorry, you asked for intersection. That'd be the set with characteristic function $1$, i.e. the universal set $X$. It makes sense as $A\cap X=A$. – Quang Hoang Oct 06 '15 at 03:50
  • @QuangHoang Yes. Thanks. That's the problem where I am puzzled. To claim $X$ is the identity, we need first show $X$ is in $\cal R$ given that the $\cal R$ is closed under symmetric difference and intersection. You can check all three definitions here https://proofwiki.org/wiki/Definition:Ring_of_Sets (suppose the system of sets $\cal R$ is a subset of the power set $2^X$). I cannot see why the whole set is in the ring. – Tony Oct 06 '15 at 03:55
  • @QuangHoang Or maybe "ring of set" is not a "ring" in the algebra sense when it does not include the whole set? So the two concepts, although strongly connected, are still somewhat different? – Tony Oct 06 '15 at 04:08
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    A ring in Abstract Algebra need not a multiplicative identity (see Wikipedia ). The axioms for Ring of Sets translate exactly to ring axioms by the identification mentioned in my answer. – Quang Hoang Oct 06 '15 at 04:20
  • @QuangHoang Ok. Right. I found there is conflict among definitions of Ring in different places. Some require the inverse element, some do not. Anyway your answer is very helpful. Thank you again. – Tony Oct 06 '15 at 20:25