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

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> g is sufficient to give the desired result if P and P' are disjoint. E

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> is presumably included to extend that to the case where they intersect,

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> but I can't see how it does. g still only addresses disjoint subsets of

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> P u P' u E. I think that can be worked round by altering the definition

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> of g. Instead of listing the ways in which each set of size 2r can be

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> split into disjoint sets of size r with the same image under f, it

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> should give the ways in which it can yield two not necessarily disjoint

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> sets of size r with the same image under f. The sets B' and B derived

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> from this new g by Ramsey's theorem will still have any properties

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> required for the rest of the proof, as the old g will be constant on any

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> set that the new g is constant on.

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>

>

> And now I think I may see a way around the original problem.

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>

>

> Suppose D(x,y,i) but fx = fy. By the P/Q lemma, fx=fz for any other z

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> such that D(x,z,i) [P = P' = x, Q = y, Q' = z]

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>

>

> There's a possible problem if x_i < y_i but z_i < x_i. The lemma as

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> stated won't apply but the proof of it will extend to cover this case.

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>

>

> Now looking back at a rho not in L. By the definition

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>

>

> there exist distinct x,y such that D(x,y,i) and f(x) = f(y)

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>

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> By the P/Q lemma, for all x,y such that D(x,y,i), f(x) = f(y)

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>

>

> which is what we needed.

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>

>

>

>

> So it seems the problem step can be justified using a lemma from later

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> in the paper and, unless I've gone astray, that lemma can be proved if

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> 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