> The flaw in WM's argument is that he claims that no nonmember can be > found for his "ultimate" open at both ends list, but he is wrong.
I thought he was trying a Cantor like argument to the effect that a non-member is found that should by construction by a member, and then arguing that the contradiction therein had to be resolved by concluding that the set couldn't be listed, even though it was clearly countable.
He was certainly wrong, and that was always apparent, but I prefer a specific identified flaw over a general "must be wrong" type of argument, even if the latter amounts to a proof. OK, sometimes they're not available, but they're what I prefer.
> > Given that "ultimate list" {...a2,a1,a0,q0,q1,q2,...} merely rearrange > it to {q0, a0, q1, a1, q2, a2, ...} and you have comprehensive list of > everything with a findable antidiagonal which is not listed.
