In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 15 Jun., 16:01, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 15, 9:23 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 15 Jun., 00:19, 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." > > > > You are trying a proof by contradiction of > > "Infinite paths exist" > > > > We have > > > If [actually]infinite paths exist > > there is a path p that can be distinguished > > from every path of P. > > > > We need > > > > If infinite paths exist path > > there is no path p that can be distinguished > > from every path of P. > > No we need and have in fact: There is no path p that can be > distinguished from every path of P.
What WM claims he needs does not make those putative needs actual. > > > > Your putative proof > > > > you are not able to distinguish p from P (by means of digits).
If one has any finite or countably infinite listing of infinite binary sequences (paths), then the Cantor construction finds one not in that listing.
> > Wrong. P contains every subset of nodes of the binary tree. There is > nothing else.
If P is listable (countable) then every listing of its members produces a non-member by the Cantor construction.
> > > > > You know that you are unable to > > > distinguish any binary sequence representing a real of the unit > > > interval from every path of the tree. > > > > Correct the tree contains every subset of nodes, > > however P ( a list of subsets of nodes) > > does not contain every subset of nodes > > As my proof shows, P contains every subset of nodes.
WM has often claimed this, but never been able to prove it without imposing conditions contrary to, say, ZF, or whatever axiom system allows a maximal infintie binary tree to exist at all.
In WM's "system", there is no such tree, but his is not the only system. There is nothing in mathematics that requires everyone to play by WM's rules. And until WM learns that, he will be forever an outsider. > > > You can distinguish p from every element of P by > > subsets of nodes. > > > Wrong.
That depends on P. If P, as a set of infinite binary sequences, is uncountably infinite, it may also be maximal, in the sense of containing every possible infinite binary seqeunce, but Cantor proved (or rather a corollary to his construction proved) once and for all that as soon as a set of such sequences can be listed (counted), it is missing at least as many members as it contains,
The tree contains nothing more than the elements of P.
The tree 'contains' all sorts of things not in P including many sets of nodes which are not paths at all.
The tree, as a set of nodes, has no members which are paths, but only has subsets which are paths. So that WM is attempting to obfuscate things by using "contain" ambiguously.
To be clear, one must distinguish between 'contain as a subset' and 'contain as a member'.
While P as a set of paths may contain paths as members, it does not contain any Paths as subsets. So the fact that the union of P (the set of all elements of elements of P) may include all nodes does not in any way entail that P contains all paths.