Proving \sigma(O) = \sigma(C2) in Sigma Algebras | Open Sets and Complements

[azdang]

I am working on the last step of a proof to show that $$\sigma$$(O) = $$\sigma$$(C₂).
C₂ = {(-$$\infty$$, a): a $$\epsilon$$ R and O = all the open sets in R¹.
I have already showed that $$\sigma$$(C₂) C $$\sigma$$(O).

I am now trying to show the converse, that $$\sigma$$(O) C $$\sigma$$(C₂). To do this, I know I just have to show that O C $$\sigma$$(C₂). This is what I have so far:

For every V in O, V is an open set in R¹.

V = $$\bigcup$$(a_i,b_i) from i=1 to infinity. So, I just have to show that (a,b) is in $$\sigma$$(C₂) really.

So, (a,b) = (-$$\infty$$, b) $$\cap$$ (a, $$\infty$$). Obviously, (-$$\infty$$, b) is in C₂, which means it is in $$\sigma$$(C₂).

But I'm having a hard time showing why (a, $$\infty$$) is in C₂. The complement of this would be (-$$\infty$$, a] but I'm not sure this gets me any closer. Can anyone help me figure out why (a, $$\infty$$) is in C₂? I might be missing something really obvious or just going about it all wrong, as I am trying to follow a model for a very similar problem we did in class. Thank you!

 

[Dick]

You know a sigma algebra is closed under countable unions, right? So the complement of (-infinity,c)=[c,infinity) is in C2. What's the union of all of the sets [c,infinity) for c>a? How can you make that a countable union?

 

[azdang]

Is it something like U[a + $$\frac{1}{n}$$, $$\infty$$) from n=1 to infinity, and that would be equal to (a, $$\infty$$)?

 

[Dick]

Is it something like U[a + $$\frac{1}{n}$$, $$\infty$$) from n=1 to infinity, and that would be equal to (a, $$\infty$$)?

That works.

 

[azdang]

Ooh, thank you so much, Dick! I have a bunch of these to work on, so I could be back for some clarification. Thanks again! :)

 

[azdang]

I have another sigma algebra question. This one may be primarily in part to the fact that I just don't understand what the question is saying. It says:

Let A be a sigma-algebra of subsets of $$\Omega$$ and let B be an element of A. Show that F = {A $$\cap$$ B: A is in A} is a sigma-algebra of subsets of B. Is it still true when B is a subset of $$\Omega$$ that does not belong to A?

So, B would itself be a subset of $$\Omega$$, right? I'm also confused on whether A is a specific element in A or an arbitrary element. And, how we would know what the subsets of B are. Hopefully, once I know what the question is saying, I should be able to figure something out.

 

[Dick]

Let's rewrite this without sort of halfway TeXing it. "Let A be a sigma-algebra of subsets of O and let b be an element of A. Show that F={a intersect b: for all a in A} is a sigma-algebra of subsets of b. Is it still true when b is a subset of O that does not belong to A?". Does that help? There are two cases, b is an element of the sigma algebra A and b is not an element of A. For the first case start trying to check the axioms that define a sigma algebra for F. Then start worrying about the second case.

 

[azdang]

Okay, so basically, the B is fixed, but the A could represent any of the elements in sigma-algebra A. So for F, we could have:

A n B
A^c n B = (A U B^c)^c
B n B = B
B^c n B = {}
(A U B) n B = B
O n B = B
{} n B = <-- A little confused, is this {} or B

I think I'm beginning to see why F is a sigma-algebra of subsets of B. Clearly, B is in F, {} is in F, and it looks like it is closed under complement and countable union in #2, but I'm thinking there must be a more general way to do this because there wouldn't be any way to list out all of the possible intersections.

 

[Dick]

{}nB={}, surely. But now you have to figure out a way to prove it without listing ALL of the elements of the sigma algebra A, which you can't. List the axioms defining a sigma algebra. They are what you have to prove. And let's call 'A' the sigma algebra and 'a' an element of A. Try not to use the same symbol for two different things. It's makes proofs confusing.

 

[azdang]

Yeah, I know. We can blame my book for that, although it is 'script' A for the sigma-algebra, which I tried to show by making it Bold and Italicized.

Well, the first requirement is that B would have to be in F. B could be represented as O intersect B, since O would be an element in A.

The second would be that it is closed under complement. We know $$a \cap B$$ is in F so we can check if the complement is, too. So, $$(a \cap B)^c = a^c \cup B^c.$$ Alright, I'm assuming we have to find a way to write this complement so that it has the form $$a \cap B$$. The only thing I could think to do was actually intersect this complement and B which equals $$a^c \cap B$$, which would be in F, but I'm not sure if that shows it is closed under complement.

Lastly, we'd need to check that it is closed under countable union. What I think we would need to show is this:
$$\bigcup_{i=1}^{\infty}(a_i \cap B)$$ which would equal $$(\bigcup_{i=1}^{\infty}a_i) \cap B$$ just by the distributive law. And then $$(\bigcup_{i=1}^{\infty}a_i)$$ is an element of A since that is closed under countable union as a result of being a sigma-algebra, therefore, this entire intersection is in F. Therefore, closed under countable union. I'm not completely sure about this one, but it seems to make sense.

 

[Dick]

That looks pretty good to me, except for the complement part. For that remember you want to show F is a sigma algebra of subsets of B, not O. The complement of (anB) IN B is B-(anB) (where '-' is the set theory difference). Try doing some set theory stuff on that.

 

[azdang]

Yeah, that was the part I was iffy about. Let's see. So, B-(a$$\cap B$$) would be all the things in B not also in A. Could we say this is the same as B-a? If so, isn't that just $$B \cap a^c$$, which would be in F since $$a^c$$ is an element of A.

 

[Dick]

Yeah, that was the part I was iffy about. Let's see. So, B-(a$$\cap B$$) would be all the things in B not also in A. Could we say this is the same as B-a? If so, isn't that just $$B \cap a^c$$, which would be in F since $$a^c$$ is an element of A.

Correcto.

 

[azdang]

Oh cool! Thank you again. You've been really helpful on these sigma-algebra problems. They are something I have never seen in all my years of math, so it's kinda tricky to get used to. Have a good day!

 

[azdang]

Oops, forgot to tackle part of the question: If F is still a sigma-algebra if B is a subset of Omega that does not belong to A. My thoughts are that it is still a sigma-algebra because I didn't seem to use the fact that B is in A at all in the proof that F is a sigma-algebra. Would that be correct?

 

[Dick]

I agree. I can't see that you used that B is an element of A either. If you believe your proof, then it must be true.