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') -- David Hartley