On Thursday, November 7, 2013 9:50:35 PM UTC, David Hartley wrote: > In message <7vKtBqDeeAfSFwPl@212648.invalid>, David Hartley > > <email@example.com> writes > > >Replace X and X' by X_0 and X_1 from [B(4)]^r such that (X,X') = > > >(X_0,X_1) and f(X) = f(X_0), f(X') = f(X_1)(as in Rado's proof). Let > > >X_0^sigma = b_4p. Form X_2 from X_0 by replacing X_0^sigma by b_(4p+2). > > >Since sigma is not in L, f(X_0) = f(X_2). X_1^rho = X_0^sigma is not in > > >X_2 so the lemma can be applied to X_2 and X_1, giving f(X_2) =/= > > >f(x_1) and so f(X_0) =/= f(X_1). > > > > > >Hope there's no mistakes. > > > > Well I've spotted one mistake already. We can only select X_0 and X_1 to > > get (X,X') = (X_0,X_1) not to also have f(X) = f(X_0), f(X') = f(X_1). > > But it doesn't matter, that's enough for f(X_0) =/= f(X_1) to imply f(X) > > =/= f(X') >
Thanks. I did check this carefully and it seems fine. Much better than in the paper.
Finally, I won't need to be sceptical if someone accosts me in the street and tells me. "Verily I say unto you: The Canonical Ramsey theorem is absolutely true."