On 17 Jun., 14:56, 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." > > A: actually infinite paths exist, > B: the infinite tree contains a path p that can be > distinguished from every path of P. > > You agree > > A ==> B > > So if A is true your claim is false.
A ==> B
should go undisputed, like: If pi is rational, then it can be represented by p/q. Nobody will claim that pi is rational but cannot be represented by p/ q.
The only conceivable instance to lay claim to pi's rationality without its being representable as a fraction is, if Cantor had said that pi is rational. Well, then further arguing would be hopeless and in vain.
> > You want to show > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > by proving ~B > > (Note that assuming ~A is circular)
Of course.
> As we have seen we can create a finite subset > by adding a single path.
That is of no relevance because only allowed subsets can be of interest. Why do you want to raise the impression your arguing with respect to irrelevant subsets would be relevant?
> However, > as previously noted, > to create a path requires creating an infinite subset.
To create a path requires to add that path.
> To do this we > need to add a subset of paths. > > E.g. The subset of paths added is > > 1000... > 11000... > 111000... > ... > > The subset of nodes that is now in the tree is > > t={1,11,111, ...} > > Note that t is not an element of P, so the subset > t is never added to the tree.
If it has not been added, then it cannot be in tree. Otherwise, if it could be in the tree without having been added, why then should it not be in the list without having been added? Why then should it not be in P without being in P? Can you explain that?
