Talk:Infinity-Borel set

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[edit] (Formerly) Disputed

Um, so actually I wrote this article, and no one else has touched it, so really I'm arguing with myself here. But I thought about it and I'm not sure my definition is right. On the other hand, if it is right, it's (I think) easier to follow than any alternative I can think of, so I don't want to jump the gun. But I also don't want to mislead anyone.

Here's the issue; maybe someone can help me out: The ∞-Borel sets are the ones that have ∞-Borel codes. An ∞-Borel code is something that codes up the way you build an ∞-Borel set by starting with open sets (each basic open set gets a code), and then iterating the operations of complementation and wellordered union (there's a code for the complement of a coded set, and a code for the union of a sequence of coded sets).

Problem is--is the class of thus-coded sets in fact closed under wellordered union? To prove it, you'd seem to need AC_α, if your sequence of sets to be unioned up has length α, in order to choose a Borel code for each of the sets in the sequence, so you can form the new Borel code.

Any help appreciated. --Trovatore 04:21, 15 July 2005 (UTC)

[edit] Fixed

So the definition now in place is harder to follow, and it's got a lot of LaTeX that simply cannot be displayed if anyone is expected to read the thing. But it does have the advantage of being correct. --Trovatore 02:20, 16 July 2005 (UTC)

[edit] Alternative definition

So I thought of another approach: Dispense with the codes, and perform the transfinite iteration directly on the sets. Something like this:

For each ordinal α define by transfinite recursion B_α as follows:

1. B₀ is the collection of all open subsets of X.
2. For a given even ordinal α, B_α+1 is the union of B_α with the set of all complements of sets in B_α.
3. For a given even ordinal α, B_α+2 is the set of all wellordered unions of sets in B_α+1.
4. For a given limit ordinal λ, B_λ is the union of all B_α for α<λ

It follows from the Burali-Forti paradox that there must be some ordinal α such that B_β equals B_α for every β>α. For this value of α, B_α is the collection of ∞-Borel sets.

Whaddaya think? Is it clearer than the definition with codes? I did leave more out in this alternative version (e.g. I didn't really say what a wellordered union is, and it takes a little argument to see that clause 3. above gives you a set at all, rather than a proper class). Also the code notion is important. --Trovatore 02:49, 16 July 2005 (UTC)

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Whoops. It is clearer, but it's wrong. Can't prove without AC that every set as defined above has an ∞-Borel code; same error as before. --Trovatore 03:06, 16 July 2005 (UTC)