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The article on sigma-ideals in wikipedia claims they are a special kind of ideals:

http://en.wikipedia.org/wiki/Sigma-ideal

But, unfortunately, no explanation to that regard is offered (not at least that I can identify as such). Could anyone explain it in an easy way?

Thanks in advance.

Javier Arias
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  • Compare the two definitions. Countable unions of elements of the $\sigma$-ideal are in the $\sigma$-ideal. It follows that finite unions are. – André Nicolas May 23 '15 at 00:37
  • Yep, but still no definition of the sigma ideal as ideal is provided....... – Javier Arias May 23 '15 at 00:43
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    I just looked, and the definition is provided. The property I used is called (iii). In the definition of ideal in an algebra, or $\sigma$-algebra, you will find a corresponding clause, except that one only asks for finite unions to be in. – André Nicolas May 23 '15 at 00:48
  • But in order for something to be in the ideal, did it not have to be defined with respecto to the product between that something and and element of the ring? Where is the product here? I do not get it. – Javier Arias May 23 '15 at 00:51
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    Product in this case is intersection. Maybe look under Boolean ring. – André Nicolas May 23 '15 at 00:54

1 Answers1

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A $\sigma$-ideal is an ideal (with additional conditions) in the representation of the $\sigma$-algebra $\Sigma$ as a Boolean ring.

Every Boolean algebra can become a Boolean ring by taking the ring addition to be $A\oplus B=(A\sqcap \overline B)\sqcup(\overline A\sqcap B)$ and the ring multiplication to be $A\otimes B=A\sqcap B$.

In case of a subalgebra of the subset algebra (which a $\sigma$-algebra is), this works out as the ring operations in the Boolean ring being symmetric difference (with $\varnothing$ as additive identity) and intersection (with $X$ itself as multiplicative identity).

A ring ideal must be closed under multiplication (that is, intersection) with arbitrary $\sigma$-algebra elements, and this is exactly being closed under taking subsets.

A ring ideal must also be closed under finite unions, because $A\cup B=A\oplus B\oplus (A\cap B)$. A $\sigma$-ideal is additionally closed under countable unions, which cannot be expressed in purely ring-theoretic language.

hmakholm left over Monica
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  • Makholm. Do you have some references on the topic? Some paper one might quote and have a look at as first overview? I would be glad to know about it. – Javier Arias Jun 05 '15 at 13:12
  • @JavierArias: No, all I have are the pertinent Wikipedia articles and some long-forgotten lecture notes in measure theory and general topology. – hmakholm left over Monica Jun 05 '15 at 13:29
  • Ok, thanks anyway....I will try to come to grips with this very difficult matter to me..... – Javier Arias Jun 05 '15 at 13:34
  • Makholm Does this relate to your explanation? https://en.wikipedia.org/wiki/Coarse_structure – Javier Arias Jun 25 '15 at 10:59
  • @JavierArias: Hm, I've never met that concept before. From a cursory glance I'd say probably not, but there's plenty of room for me to be wrong. – hmakholm left over Monica Jun 25 '15 at 13:30
  • Thanks for your reply. I just thought you might just have had that into account when giving your answer.....never mind....forget it :) By the way, maybe you can help me with this out:http://math.stackexchange.com/questions/1288275/visual-representation-of-right-or-left-ideals-of-a-ring – Javier Arias Jun 25 '15 at 17:48