Talk:Borel algebra
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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
- 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] (http://www.tutorgig.com/encyclopedia/getdefn.jsp?keywords=Borel_algebra). 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
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)
- Ah, yes.--CSTAR 20:17, 14 Jun 2005 (UTC)