Hi,

I don't know how to prove this exercise, actually I don't even know how to start (Thinking):

Let is separable metric space. Prove that its borel -algebra is generated by system .

I would appreciate any hint.

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- Oct 3rd 2013, 12:23 AMdrabisigma-algebra in separable space
Hi,

I don't know how to prove this exercise, actually I don't even know how to start (Thinking):

Let is separable metric space. Prove that its borel -algebra is generated by system .

I would appreciate any hint. - Oct 3rd 2013, 03:42 AMPlatoRe: sigma-algebra in separable space
This notation uses vocabulary that is hard for me to follow. I think it means that you are to show that the set of balls. forms a

**basis**for the metric topology.

The countable part comes from the given separable. So perhaps you need to post some more about the terms here. Does "its borel -algebra " simply mean the metric generated topology?

Do you have a definitions of**basis**? - Oct 5th 2013, 07:47 AMdrabiRe: sigma-algebra in separable space
Hi,

thanks for answer.

I'm sorry for the vocabulary, I'll try to explain:

borel -algebra means -algebra generated by open subsets of .

So I'm supposed to show, that it's ok to take only the sets of open balls instead of all open subsets.

It is clear now? - Oct 5th 2013, 08:36 AMPlatoRe: sigma-algebra in separable space
- Oct 5th 2013, 09:46 AMSlipEternalRe: sigma-algebra in separable space
Let be the set of all open sets of . Let be the power set of .

Let

Then by definition, we know the Borel -algebra .

Next, let be the -algebra generated by the sets . In other words, it is the intersection of all -algebras containing the sets .

Now, we want to show that . How do you show that two sets are equal? You show that and .

For any and any , is an open set of , so it is an element of . Since , and is the smallest -algebra containing all sets , it must be that .

Next, how do we show that ? We take an arbitrary open set . Then we need to show that . If this is true for any arbitrary , then since is the smallest -algebra containing , it must be that .

So, question: What exactly is the system ? Is fixed? Is fixed? Are neither fixed? If neither is fixed, then it is easy to find a union of sets of the form that will contain every point of and you don't even need complements. If one is fixed, but not another, this becomes a little more difficult. If both are fixed, then you most definitely do not get .

Edit: If is fixed, but is arbitrary, then you do not get either. Alternately, if is arbitrary, but is fixed, you might still be able to show ... I would have to think about that.

Edit 2: If is fixed, but is arbitrary, then if for all , , then you can show .