
Re: Surprise at my failure to resolve an issue in an elementary paper by Rado
Posted:
Nov 4, 2013 9:43 AM


In message <8933482d892046ce88866402fce0cea2@googlegroups.com>, Paul <pepstein5@gmail.com> writes >Agreed. The redefinition of L leads to problems soon after (3) of page >3. Furthermore, assuming an error in a paper should be a last resort. >An author is far more likely to omit steps of reasoning than to make an >elementary logical error. My initial post merely shows that the >equality I'm complaining about doesn't follow _immediately_. However, >just because an assertion doesn't follow immediately doesn't mean that >it doesn't follow. To show that it doesn't follow, we would need a >counterexample and I haven't seen one.
To avoid all the tedious notation, let's define
D(x,y,i) to mean x and y are subsets of B with r elements which, when ordered by <, differ only at the position indexed by i.
The original definition of L has i in L iff
for all x,y D(x,y,i) > f(x) =/= f(y)
The problematic step in the proof assumes that if i is *not* in L then
for all x,y D(x,y,i) > f(x) = f(y)
Suppose for a counterexample, that B = N and put r = 2 and
f(x) = 2^(1 + min x) if max x is even = 3^(1 + min x) if max x is odd
Then L = {0} with the original definition but is empty with the suggested variation.
It may be that with the specific B defined in the paper the two properties are equivalent but it certainly requires proof.
 David Hartley

