Talk:Borel set

message: 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)
thread_title:  Countably generated 

message: 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)
replies: text:  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)
thread_title:  from open sets? 

message: 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)
replies: text:  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)
replies: text:  BTW, Oleg did you answer your own question, or was this someone else's objection?--CSTAR 15:53, 22 July 2005 (UTC)
thread_title:  Serious mistake in article about the Borel algebra 

message: The following statement appears in the references section but is immediately followed by a list of 4 references.
An excellent exposition of the machinery of Polish topology is given in Chapter 3 of the following reference:
I seriously doubt it applies to all 4.  Does anyone know which one(s) its talking about? Timhoooey 00:40, 14 October 2007 (UTC)
thread_title: Ambiguous statement in the references section

message: "if m is an uncountable limit ordinal, Gm is closed under countable unions"
How is this proven without Countable Union Condition?
How is
Aleph_1 x 2^Aleph_0 = 2^Aleph_0
proven without AC?  —Preceding unsigned comment added by JumpDiscont (talk • contribs) 03:39, 20 November 2009 (UTC)
replies: text: Who said it is?  Choice is the default assumption; we don't have to mention it.  If you want to discuss what can happen in models of ~AC, that might be an interesting sidelight, but it would come later in the article. --Trovatore (talk) 03:47, 20 November 2009 (UTC)
thread_title:  Choice 

message: why is the above part of the wikipedia article true and can it be changed to ... OR ... --demus wiesbaden (talk) 14:44, 4 December 2009 (UTC)
replies: text: Yes. — Carl (CBM · talk) 18:47, 4 December 2009 (UTC)
thread_title:  "any measure defined on open sets AND closed sets must also be defined on all Borel sets" 

message: I'm not sure if the paragraph
In the construction by transfinite induction, it can be shown that, in each step, the number of sets is, at most, the power of the continuum. So, the total number of Borel sets is less than or equal to \aleph_1 \times 2 ^ {\aleph_0}\, = 2^{\aleph_0}\,.
In the example section is meant for the borel algebra on the reals. If it is, then obvious the number of Borel sets is exactly continuum, because for any real number x, the set {x} is in the borel algebra (take the countable intersection of all open intervals (x-q,x+q) for q a positive rational number), and the defines an injection from the reals to the Borel algebra. Money is tight (talk) 09:02, 12 March 2010 (UTC)
replies: text: There are exactly 2^{\aleph_0} Borel sets of reals, yes.  Every Borel set can be coded by a real, which gives you the surjection. --Trovatore (talk) 09:06, 12 March 2010 (UTC)
replies: text:  Isn't this a consequence of axiom of countable choice? (In absence of any form of axiom of choice, it is consistent that all sets of real numbers are Borel sets.)  - Mike Rosoft (talk) 21:33, 16 January 2019 (UTC)
replies: text:  Um.  Probably.  I'd have to think about it, but yes, really really pathological things can happen if countable choice fails.  But choice (especially countable choice) is part of the default context.  For a while it was fashionable to be careful about calling out its dependencies specifically, but that is no longer true. --Trovatore (talk) 21:39, 16 January 2019 (UTC)
thread_title:  Cardinality 

message: I revised the section Standard Borel spaces and Kuratowski theorems to give an unambiguous and modern definition of "Borel space".
The previous version said that Mackey used the term "Borel space" to mean what we now call a "measurable space", but that this is a misleading usage. However, it did not clearly state whether it was adopting Mackey's definition, nor did it offer another meaning for the term "Borel space". Yet it then proceeded to use the term several times! 
The usage was consistent with the "obvious" definition of "Borel space", namely a set equipped with the sigma-algebra of Borel sets (of some topology). So I made this the official definition, at the top of the paragraph.
If someone thinks this is ahistorical, or not even correct usage, then I would suggest removing Mackey's no-longer-used definition (or putting it in a footnote) and also removing all usages of "Borel space", replacing them by "measurable space" (suitably qualified), except in the fixed expression "standard Borel space". 
178.38.78.40 (talk) 10:07, 3 May 2015 (UTC)
thread_title:  Revised ambiguous use of "Borel space"