In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> 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!
WM has already said that HE can construct a tree with infinitely infinitely long sequences of nodes in it.
So they must exist at least for him. But now he does not want to allow anyone else to be able touse them?
> And that is the assumption we start off with (and that we > disprove).
WM's perpetual claims of his own proofs and disproofs have yet to refer to any actual proof or disproof that is either mathematically or logically valid.
> 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.
Given a maximal infinite binary tree and any finite or countably infinite set of subsets of its node set then there will be paths as node sets not members of that given set. > > > > 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.
If you, or anyone else, claims that they can count (list) all the paths in that tree, then the Cantor diagonal argument refutes you all. > > You cannot distinguish your favourite path from the tree constructed > by means of a countable set of paths.
Sure I can! At least in binary trees with more than two nodes.
The first node of a path in such a tree has at most one child whereas in the tree it has two. That counts as distinguishing a path from a tree. > > What's now? Play the game in order to prove that? > > Thou shalt not belittle my binary tree!
Your form of allegedly maximal infinite binary tree belittles itself by being too little to be a maximal infinite binary tree.