
Re: Another notcompletelyinsignificant gap in the Rado paper
Posted:
Nov 7, 2013 4:35 PM


In message <mvZqxm9Vg$eSFw0i@212648.invalid>, David Hartley <me9@privacy.net> writes >There is a sense in which that applies here. If X_0 and X_1 are >interleaved in that way, then we can jump straight to the last part of >the argument. The point of the X_i sequence is to get us to the >position where we can change the rho0indexed element of X_sigma0 to >the onehigher indexed b without disturbing the relative orderings with >X_0. If X_0 and X_1 are disjoint then that can't happen anyway and we >can just put sigma0 = 1 and finish.
This suggests a possible alternative proof of part (b).
Suppose {x_0,..,x_(r1)}_< and {x'_0,...,x'_(r1)}_< are subsets of B which differ at an index which is in L. We need to show f(X) =/= f(X').
Lemma 1. If {x_0,..,x_(r1)}_< and {x'_0,...,x'_(r1)}_< c B; rho in L and x'_rho =/= x_i for any i
then f({x_0,..,x_(r1)}) =/= f({x'_0,..,x'_(r1)})
Proof. Put X = {x_0,..,x_(r1)} and X' = {x'_0,..,x'_(r1)}.
x'_rho is in B, so equals b_2p for some p. Let
x"_rho = b_(2r+1), x"_i = x'_i for all other i
and let X" = {x"_0,..,x"_(r1)}
Now (X,X') = (X,X") so if f(X) = f(X') then f(X) = f(X"). But since rho is in L, f(X') =/= f(X"). Hence f(X) =/= f(X')
Now there may not be index fitting the requirements of the lemma. In that case let rho be the largest member of L where X and X' differ. Wlog,
x_rho < x'_rho = x_sigma for some sigma > rho
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.
 David Hartley

