Thread: sigma-algebra

Hi!

Trying to get a grasp on .
Let´s say I have a the set .

Now, if I want to find the smallest that includes both 1 and 2. Would this be:

?

The empty set is there, and also its complement . Both 1 and 2 are there and their complement 3. Also the union of 1 and 2 are is there.

Thanks!

Thats not quite right.

First, here is a result for algebras that is quite useful to know:

If is a set and is a -algebra on then
is closed under countable intersections.

Proof:

Let for Then note that for each we have that where is the complement of It follows that


But other than that I can point out that are not themselves sets and can not possibly be in the -algebra. However, I think you meant , which is correct.

You are however, missing some sets in this. Consider this corollary to the previous result: if then This means that for example along with a few other sets.

I think I understood your claim and proof. You say that if we have a sequence of sets in a -algebra their intersection must also lie in . Because all the complements of the sets must lie in , we get that the union of all complements must lie in , and because of this the complement of the union of complements must lie in . And this is equal to the intersection of our original sets.

I don´t think I understand your corollary though.

I updated:

Originally Posted by ecnanif

I don´t think I understand your corollary though.

I updated:

The corollary stating: If means
That since , by the corollary this implies .
He was just showing you how you could get your missing sets in your -algebra. Your updated set is now correct.

Ah okay =)
Thank you!