Paul
Posts:
507
Registered:
7/12/10


Re: Surprise at my failure to resolve an issue in an elementary paper by Rado
Posted:
Nov 4, 2013 6:10 PM


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 ErdosRado proof of the same theorem.
I will go back to the original ErdosRado theorem and see if I can patch any gaps there. I hope so  Erdos usually wrote very simply.
Paul Epstein

