I don't know how to prove this exercise, actually I don't even know how to start :
Let is separable metric space. Prove that its borel -algebra is generated by system .
I would appreciate any hint.
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?
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?
Let be the set of all open sets of . Let be the power set of .
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 .