```Date: Jan 30, 2013 5:15 PM
Author: Virgil
Subject: Re: Matheology � 203

In article <d9df6d19-b0a2-4c52-83a9-456b88acffde@f6g2000yqm.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:> On 30 Jan., 10:31, William Hughes <wpihug...@gmail.com> wrote:> > > For a potentially infinite list L, the> > antidiagonal of L is not a line of L.> > Of course. Every subset L_1 to L_n can be proved to not contain the> anti-diagonal> >> > Does this imply> >> > There is no potentially infinite list> > of 0/1 sequences, L, with the property that> > any 0/1 sequence, s, is one of the lines> > of L.> > Do you mean potentially infinite sequences?> Look, everything Cantor does, concerns only finite initial segments.> You could cut off the sequences behind the digonal digit.But the anti-diagonal does not nave any merely "diagonal" digit, so must be endless.> > The only thing not terminating, then could be the diagonal itself. But> then you would claim that the diagonal differs from every entry,> because it has more digits. In the original argument, the diagonal> differs at the same places that also exist in the entries. Therefore> the argument with the diagonal "being longer" is wrong.Any anti-diagonal differs from each listed entry of the list from which it is derived in AT LEAST one digit position, so is not a listed entry of that list. Thus those. like WM, who claim existence of a complete lists of functions from |N to any set of more than one member are wrong.> > So in fact, Cantor shows that the countable set of all terminating> decimals is uncountable.WM claims to be able to prove it but his "proof" is only invalid in his wild weird world of in Wolkenmuekenheim.What Cantor actually showed was that given a list of functions each having domain |N and a codomain of at least two members there is no surjection from |N to that set of functions. Thus any such set satisfies the standard definition of uncountability.That WM does not like it is his problem, not ours.--
```