In article <88doghFgf9U1@mid.individual.net>, Sylvia Else <sylvia@not.here.invalid> wrote:
> On 23/06/2010 4:31 PM, Virgil wrote: > > > 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.
I do not think that WM has any coherent argument. Though many ambiguities and non-sequiturs that his arguments always include make it difficult to be sure. > > 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. > > I thought that was what I did.
Perhaps so, but since it was not in this post, I wasn't sure.
My newsreader does not make it easy to recover earlier posts in a thread once they have been marked as read. > > Sylvia.
