Re: Possible Proof of the Lusin-Purves Theorem

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 11/01/04 

Date: Mon, 01 Nov 2004 13:03:29 -0600

Acid Pooh wrote:
> poopdeville@gmail.com (Acid Pooh) wrote in message news:<4765002.0410281448.1bfc2b81@posting.google.com>...
>>Mike Oliver <mike_lists@verizon.net> wrote in message news:<2ua4f5F28ose7U1@uni-berlin.de>...
>>>Acid Pooh wrote:
>>>>The Lusin-Purves theorem states that X,Y are standard Borel spaces and
>>>>B is a Borel set in XxY such that each cross-section B_x is countable,
>>>>then the projection of B onto X is a Borel set. I've seen several
>>>>proofs of this fact, but I think I might have come up with a novel
>>>>(though elementary) one. This sounds kind of crankish, especially
>>>>since I don't seem to use the countability of each section in any
>>>>significant way. Constructive criticism/counter-examples (but please,
>>>>be gentle. I haven't slept in a few days)
>>>>
>>>>Pf/ Since X and Y are Standard, XxY is Standard. We may assume that
>>>>B is Borel since we can enlarge the topology on XxY so that B is
>>>>closed and open. Thus B is Polish. The Cantor-Bendixson theorem
>>>>implies that B can be written as B = O union P, where O is countable,
>>>>P perfect, and O intersect P empty. Let a be any point in p and a_n
>>>>any sequence in P converging to a. Since the projection mapping pi_X
>>>>is continuous, if a_n -> a, pi_X(a_n) -> pi_X(a). Since every point
>>>>in P is a limit point, pi_X(P) is closed, hence Borel.
[MO's objection and AP's original response snipped]
> OK, I think I fixed the problem -- obviously, the claim that pi_X[P]
> (from now on denoted by P since the fix makes no reference to the
> projection map) is closed is much too strong. Let cl(P) be the
> closure of P and let \ be the set-minus operation (So A\B = A
> intersect B^c). Here goes:
>
> Let Q = cl(P)\P. The Cantor-Bendixson theorem implies that we can
> write Q = R union S, where R is perfect and S is countable. Note that
> since R is perfect (thus closed), it is Borel and S is a countable
> union of singletons, thus Borel.
>
> Since we have
>
> P = cl(P)\ Q
> = cl(P) \ (S union R)
> = cl(P) intersect (S union R)^c
> = cl(P) intersect (S^c intersect R^c),
>
> P is Borel.

To use Cantor-Bendixson on Q, you have to know Q is closed, which I
don't see how you're going to get without already knowing that P is
very simple (if P were open, or closed, that would do it).

BTW I appreciate attempts to simplify notation, but I found it
difficult to figure out in what topology you were taking the
closure of P, and which P it was.

I don't exactly recall the rest of your argument, but if it really
doesn't need that the sections are countable, it can't be right--and
there's a straightforward procedure for finding the error: Take a
known counterexample to the conclusion, and follow it through the
steps of the proof. See where it breaks down.

In this case you just need a Borel set in the plane whose
projection is complete analytic. For example, let T on
\omega\times\omega be the tree of attempts to find an infinite
descending sequence of a relation on omega. Then [T], the set
of all branches through T, is a closed set in Baire space
cross Baire space, but its projection is complete analytic.

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