Date: Nov 4, 2013 6:10 PM
Author: Paul
Subject: Re: Surprise at my failure to resolve an issue in an elementary paper<br> by Rado

On Monday, November 4, 2013 11:00:04 PM UTC, David Hartley wrote:
> Working a bit further on I've hit another problem. The proof of the
>
> lemma about the sets P,P',Q and Q' also seems flawed. The definition of
>
> g is sufficient to give the desired result if P and P' are disjoint. E
>
> is presumably included to extend that to the case where they intersect,
>
> but I can't see how it does. g still only addresses disjoint subsets of
>
> P u P' u E. I think that can be worked round by altering the definition
>
> of g. Instead of listing the ways in which each set of size 2r can be
>
> split into disjoint sets of size r with the same image under f, it
>
> should give the ways in which it can yield two not necessarily disjoint
>
> sets of size r with the same image under f. The sets B' and B derived
>
> from this new g by Ramsey's theorem will still have any properties
>
> required for the rest of the proof, as the old g will be constant on any
>
> set that the new g is constant on.
>
>
>
> And now I think I may see a way around the original problem.
>
>
>
> Suppose D(x,y,i) but fx = fy. By the P/Q lemma, fx=fz for any other z
>
> such that D(x,z,i) [P = P' = x, Q = y, Q' = z]
>
>
>
> There's a possible problem if x_i < y_i but z_i < x_i. The lemma as
>
> stated won't apply but the proof of it will extend to cover this case.
>
>
>
> Now looking back at a rho not in L. By the definition
>
>
>
> there exist distinct x,y such that D(x,y,i) and f(x) = f(y)
>
>
>
> By the P/Q lemma, for all x,y such that D(x,y,i), f(x) = f(y)
>
>
>
> which is what we needed.
>
>
>
>
>
> So it seems the problem step can be justified using a lemma from later
>
> in the paper and, unless I've gone astray, that lemma can be proved if
>
> we modify g. That's all I have time for tonight.
>
> --

Thanks a lot.

Ironically, my whole purpose in going to this paper was the belief that it simplified the original Erdos-Rado proof of the same theorem.

I will go back to the original Erdos-Rado theorem and see if I can patch any gaps there. I hope so -- Erdos usually wrote very simply.

Paul Epstein