In message <email@example.com>, Paul <firstname.lastname@example.org> writes >Thanks for your contributions. I don't see things quite the same way. >I think that, as before, the author meant exactly what he said. All >indexing is fine and beginning at b_0 is fine. You may need sufficient >largeness of the x terms where largeness is defined in relation to the >indexing of the b terms. I disagree with you that we have enough >B(r^(s-1)) elements between the members of B(r ^ s). We sometimes >(unless we argue further) want r instead of r-1.
It is crucial that X_sigma0 and X_0 be sunsets of B. Then we know X_sigma0^rho0 has an even index, 2pi, and that b_(2pi+1) will not be in X_0 or X_sigma0. That's the whole reason for using B
There's nothing in the existing specification to stop you choosing X_0 and X_1 with all odd indices (unless r is even of course). That shows immediately that using the B(t) as stated can't work.
Starting with b_0 is not fine. If the smallest member of X_1 is b_0 and the smallest of X_0 is larger then there is no way to construct X_2. Starting B(t) at b_2t is one way around that. Another would be to specify that rho_0 is the smallest index at which the x_i and x'_i differ.
You are quite right about not having enough room to do the construction. All the elements of X_1 could lie between two successive elements of X_0. Then all of X_2 would have to be between the corresponding elements of X_1 and there may be only (r-1) available.
>Whenever we see a problem in the construction, we move around it by >using the Blass-Epstein argument to say that, because we're working >inside B', the problem (not being able to find enough B(r^(s-1) >elements) is inconsistent with the assumed inequality f(x...) =/= >f(x'...).
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 rho0-indexed element of X_sigma0 to the one-higher 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.