# Algebra vs. Sigma Algebra

1. Sep 17, 2009

### LumenPlacidum

I'm trying to understand the difference between these two algebraic structures.

An algebra is a collection of subsets of a set X that is closed under pairwise unions and complements of individual subsets with respect to X.

A sigma algebra is a collection of subsets of a set X that is closed under countably many unions and complements of individual subsets with respect to X.

My trouble is that if an algebra is only closed under pairwise unions, than doesn't the entire collection of unions need to be at most countable? Then, since a sigma algebra is also an algebra, wouldn't the "at most" part be included in it too?

Can someone help me see an example of an algebra that is not a sigma algebra?

Edit: Oh, I have an idea for an example. Please let me know if my concept is incorrect.

If I take for my set X the real numbers, then I decide to place into my algebra the subsets containing exactly one real number, then I think as an algebra, I must get the entire power set of the reals as my algebra, which relies on the pairwise unions not having to be countable.

If I start with the same subsets but try to generate a sigma algebra, then my sigma algebra contains the empty set, the reals, and all the finite and cofinite subsets of the real numbers (i.e. all the sets composed of finitely many reals and all the sets that are all the reals except for finitely many reals). However, I would never obtain any open interval of the real numbers, since that would require uncountably-many unions.

Last edited: Sep 17, 2009
2. Sep 17, 2009

### Preno

it is closed under countable unions, not "countably many unions".
No, you get the algebra of finite and co-finite sets of reals.
No, you will get the sigma algebra of countable and co-countable sets of reals. For example, it will include the set of rationals.
Yes, the smallest sigma algebra over the set of real singletons doesn't include any intervals.