
Re: Surprise at my failure to resolve an issue in an elementary paper by Rado
Posted:
Nov 3, 2013 9:26 PM


In message <3c7b369084eb4e4f87d776486e14a27e@googlegroups.com>, Paul <pepstein5@gmail.com> writes >We are given that rho_0 does not belong to L. However, L is defined by >a "for all" statement. So, for rho_0, the forall statement is false >and we can find some yi and yi' to make f(z0, ..., z_r1) = f(y0,..., >y_r1) true. > >But the author is stating something much stronger  that we can deduce >the equality for an arbitrary yi and yi'.
I've never met Ramsey theory before, so I may also be missing something. But as far as I can see you are quite correct: this step is a nonsequitur.
The solution is presumably to redefine L such that the deduction is valid. Ah! I think this ties in with something you've written in the other thread (which I haven't read carefully). You and I are interpreting his definition of L in the same way, and I can't see any other way of reading it. But he presumably intended to define L as the set of numbers rho < r which have the property
there exist two ordered sets of r numbers differ only at the position indexed by rho which do not have the same image under f.
(I'm not writing it in his style, what terrible notation.)
More straightforwardly, L is the set of rho < r which do *not* have the property
whenever two ordered sets of r numbers differ only at the position indexed by rho, then they have the same image under f.
So for rho_0 not in L the step in question is valid.
 David Hartley

