On 13 Jun., 02:16, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 12, 4:36 pm, WM <mueck...@rz.fh-augsburg.de> wrote: >
> > 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. > > > > 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. > > OK. I will give you the list of things you > agreed with. Point out where you made your error. > > There is a subset of nodes that cannot be found in a single > element of P.
That is correct --- iff actually infinitely long sequences of nodes exist! And that is the assumption we start off with (and that we disprove). Consider the set P of terminating paths and p = 1/pi. The nodes of 1/pi are not in a single element of P. > > A subset of nodes is distinguished from an element > q of P if and only if it is not contained in q.
1/pi is not contained in any element of P --- if 1/pi exists! > > A subset of nodes is distinguished from every > element of P if and only if it is not contained in > a single element of P.
Same as above. No error yet. > > There is a subset of nodes > which forms a path, call it p, > that is not contained in a single element of P.
No error detected so far. > > The path p is distinguished from every element > of P.
Ok. > > All of the nodes of path p are in the tree.
That is correct because the tree contains all subsets of nodes.
All information that is available about a number in binary representation is in (one path of) the tree. If I present you the complete tree without saying how I constructed it, then you have exactly the same information that can be applied in Cantor's list argument.
You cannot distinguish your favourite path from the tree constructed by means of a countable set of paths.
