On Thursday, November 7, 2013 9:50:35 PM UTC, David Hartley wrote: > In message <7vKtBqDeeAfSFwPl@212648.invalid>, David Hartley > > <firstname.lastname@example.org> 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') > > -- >
I think I can offer a tiny further improvement in your simplification as follows. In contrast to Rado, the set on which we show canonicity is B(4) rather than B(2). Your lemma 1 can then be used unchanged because B(4) is a subset of B(2). This approach lets us omit the step: [Replace X and X' by X_0 and X_1 from [B(4)]^r such that (X,X') = (X_0,X_1)]
Does this work as a further simplification or does it simplify too much, and thereby introduce an error?