Re: Possible Proof of the Lusin-Purves Theorem

From: Acid Pooh (poopdeville_at_gmail.com)
Date: 10/28/04 

Date: 28 Oct 2004 15:48:34 -0700

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.
>
> I don't quite follow how this is supposed to show that pi_X[P]
> is closed. Suppose x_0, x_1,... are elements of pi_X[P] converging
> to x in X. Then you can pick a_0, a_1,... such that x_n = pi_X[a_n],
> but how do you know that the a_n converge? I don't even see why
> they should converge in the original topology on XxY, much less on
> the new strengthened topology.

Yeah, thanks. The graph of f(x) = 1/x in R^2 is a counter-example to
my claim. Ah well, back to the drawing board.

'cid 'ooh

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