Paul
Posts:
507
Registered:
7/12/10


Re: Surprise at my failure to resolve an issue in an elementary paper by Rado
Posted:
Nov 5, 2013 4:13 AM


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. > >  > > David Hartley
With Andreas Blass's help, I am now past the initial blockage that began this thread. Needless to say, the author did _not_ err in defining L. The two definitions of L are equivalent. Proof that the two definitions are equivalent is an easy exercise but not absolutely immediate. By definition of B', the truth or falsity of a statement of the form f(x...) = f(y...) is uniquely determined by the relevant positions of any 2r tuple which contains all the x elements and all the y elements. The nonimmediacy stems from the fact that there is no 2r tuple staring us in the face. We want to move from relations of the form f(x0, x1, x2) = f(x0, x1', x2) to relations of the form f(y0, y1, y2) = f(y0, y1', y2). However, with no disjointedness guaranteed, we don't necessarily have 6 elements to work with in proving the relationship. The solution is to add r1 (2 for this example) large integers that are larger than any of the other numbers involved. (The infinitude of B' allows us to do this).
Considering (x0, x1, x1', x2, 1000000, 100000000) and (y0, y1, y1', y2, 1000000, 100000000) the membership of B' says exactly that f(x0, x1, x2) = f(x0, x1', x2) implies f(y0, y1, y2) = f(y0, y1', y2).
I'm delighted to get past my initial blockage, and I look forward to mastering the rest of the proof.
I think that you're making an error in your conception of g. No one said that all the alpha_i indices are distinct. alpha_0 < ... < alpha_r1 and alpha_r < alpha_r+1...< alpha_2r1. But this doesn't imply by any means that all the alpha terms are distinct.
Many many thanks for thinking about this paper with me. I work outside academia so I don't generally get a chance to discuss these topics.
Paul Epstein

