Talk:Borel algebra
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[edit] Countably generated
The concept of countably generated is in http://en.wikipedia.org/wiki/Countably_generated How to prove that Borel sets are countably generated? Jackzhp 20:44, 6 November 2006 (UTC)
[edit] from open sets?
A subset of X is a Borel set if and only if it can be obtained from open sets by using a countable series of the set operations union, intersection and complement.
Is this true in general? I know that for metrisable spaces, every Borel set is of the form F-sigma, G-delta, F-sigma-delta, G-delta-sigma, F-sigma-delta-sigma, etc., etc. But I don't think every Borel set is necessarily of this form in a general topological space. (Why else would metrisability be part of the hypothesis of the theorem I read?)
- I was quite wrong here. I read the symbol for the first uncountable ordinal as the symbol for the first infinite ordinal. Whoops! Revolver 07:03, 16 Dec 2004 (UTC)
In any case, I'm not sure it's clear precisely what is meant by "using a countable series of the set operations union, intersection, and complement". Certainly this is meant to include the F-sigma, G-delta, F-sigma-delta, G-delta-sigma, F-sigma-delta-sigma, etc., etc., but what about taking countable unions/intersections of sets lying somewhere in this hierarchy? I'm not even sure if this gives more sets, but regardless, the way it's worded, it's not clear if this construction is intended or not. Revolver
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- I believe The article is true as it now stands with your correction. I may have been responsible for the original sloppy wording.CSTAR 13:14, 16 Dec 2004 (UTC)
I noticed that this article is virtually identical to [1]. Did they copy from us, or the other way around? If it's the latter, it would seem to be a violation. Revolver
Sorry...it was copied from us! We're given credit at the bottom of the page. My bad. Revolver
[edit] generation of Borel sets
I didn't mean iteration to any countable ordinal. I meant that any Borel set could be created by iteration to a countable ordinal, and that this countable ordinal may be arbitrarily large depending on the Borel set. Revolver 19:31, 14 Jun 2005 (UTC)
- I just didn't want to give the impression that there is a Borel set whose existence requires one to iterate uncountably many times. To get the whole algebra, you must go to uncountably many times, but not for a fixed Borel set. Revolver 19:40, 14 Jun 2005 (UTC)
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- Ah, yes.--CSTAR 20:17, 14 Jun 2005 (UTC)
[edit] Serious mistake in article about the Borel algebra
Remark: Originally posted on Oleg Alexandrov's Talk page by anonymous user 129.70.85.106
In the article "http://en.wikipedia.org/wiki/Borel_algebra" it is claimed, the borel algebra is the
"minimal σ-algebra containing the compact sets".
This is not correct. E.g. if you take some set E equipped with the indescrete topology (i.e. the only open sets are the empty set and E itself), then all sets are compact. Thus the minimal σ-algebra containing the compact sets is the powerset, while the borel algebra only consists of E and the empty set.
Other examples are
- any other non-Hausdorff space - an uncountable space equipped with the discrete topology (every subset is open) --anon (129.70.85.106)
Very correct. However, please note that the article states at the beginning that
- By Borel algebra one means either the the minimal σ-algebra containing the open sets or the minimal σ-algebra containing the compact sets.
And below that, it says that the two structures are not in general equivalent. So, you are very correct with the example above. But there is no mistake in the article. One can mean two different things by Borel algebra. In R^n those defintions are equivalent I think, in general they are not, and the article does state that. Oleg Alexandrov 15:29, 22 July 2005 (UTC)
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- Indeed the Borel algebra defined as generated by the compact sets is basically due to Halmos' book, Measure Theory. All differences disappear for locally compact sperable metric spaces. Maybe we should clarify the statement and say that these are two different definitions.--CSTAR 15:49, 22 July 2005 (UTC)
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BTW, Oleg did you answer your own question, or was this someone else's objection?--CSTAR 15:53, 22 July 2005 (UTC)
Yes, it was somebody else's objection. I agree that maybe the statement would need clarification. Oleg Alexandrov 16:08, 22 July 2005 (UTC)
[edit] Construction of a Borel isomorphism
Can someone tell me how to construct an isomorphism between such Polish spaces as the unit ball in L^2[0,1] and the real line with the natural topology? Leocat 12 Oct 2006
- Uh oh, construct? Do I know you? :) --CSTAR 19:24, 12 October 2006 (UTC)
- You better worry about whether you know the answer to my question.Leocat 15 Oct 2006
- Good answer. What I do know is that Kuratowski's theorem says there exists such an isomorphism. Whether this isomorphism is constructible, I don't know. Should I worry sone more? What me, worry?...about that? --CSTAR 14:52, 15 October 2006 (UTC)
[edit] Simple example
I would like to see a simple example what is and what is not a Borel set. TomyDuby 19:31, 31 December 2006 (UTC)
