In article <ab2eed5f-d45a-42bf-8293-d6b4db344e6f@s12g2000yqi.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 12 Jun., 21:40, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 12, 2:25 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 12 Jun., 18:31, William Hughes <wpihug...@hotmail.com> wrote: > > > > > > On Jun 12, 12:13 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 12 Jun., 17:42, William Hughes <wpihug...@hotmail.com> wrote: > > > > > > > > On Jun 12, 10:36 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > On 12 Jun., 12:23, William Hughes <wpihug...@hotmail.com> wrote: > > > > > > > > > > On Jun 12, 3:14 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > On 11 Jun., 23:06, William Hughes <wpihug...@hotmail.com> > > > > > > > > > wrote: > > > > > > > > > > > > On Jun 11, 4:38 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 11 Jun., 21:33, William Hughes <wpihug...@hotmail.com> > > > > > > > > > > > wrote: > > > > > > Your claim is that "no possibility exists to construct or to > > > > distinguish by one or many or infinitely many nodes > > > > of the tree another path." > > > > <snip attempt to change P > > after p has been chosen> > > You should give up. There is no attempt to change P. There is a proof > that you cannot distinguish p from the set of paths P that has been > used to construct the tree.
If that list of paths is finite or only countably infinite, then its anti-diagonal is not in it.
So that unless WM's set of paths is uncountable, one can, quite mechanically, find a non-member path. > > > > Note that you have agreed > > > > the binary tree contains a path > > p that can be distinguished from > > every element of P. > > > If I have agreed, then I was in error. The binary tree contains no > path that can be distinguished from the countable set P that has been > used to construct the tree.
If P is countable then for any counting of its members there is an anti-diagonal path NOT included in P.
> Or let me put is this way: Every path that > is present in the tree belongs to a countable set that could have been > unsed to construct the tree.
Which is, as usual, not the same thing at all.
Every path is a member of many countable sets that could be used to "construct" the tree by WM's method. But that toes not imply that there is any one countable set which contains all these paths.
And for every countable set roving it to be countable provides a direct method for proving it incomplete.
> > Wanna raise the stakes?
Whatever stakes you like, since as soon as your list is proved countable it is also proved incomplete.
