In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 10 Apr., 17:14, William Hughes <wpihug...@gmail.com> wrote: > > On Apr 10, 4:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 10 Apr., 16:10, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Apr 10, 12:26 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 10 Apr., 08:59, William Hughes <wpihug...@gmail.com> wrote: > > > > > > <snip> > > > > > > > > ... we are agreed that it makes perfect > > > > > > sense to say that any one line (and all its predecessors) > > > > > > can be removed, but the collection of all lines > > > > > > cannot be removed > > > > > > > By a single move. When applying induction, i.e., when n is removed, > > > > > also n+1 is removed, every finite line is removed. Then no finite > > > > > lines remain - only actually infinite lines remain. > > > > > > Nope. You can only use induction to show that a finite > > > > collection of finite lines can be removed. > > > > You cannot use induction to prove > > > > that the collection of all finite lines can be removed. > > > > > That's wrong (even if it is your opinion). Sorry. Induction holds for > > > all n. > > > > Yes and each one of those n is finite. > > Induction does not hold for the collection > > of all n. > > You are wrong.
Technically, WH is not wrong, but quite correct, since induction can only prove that something is true for EACH MEMBER of the relevant inductive set, but does not prove it true true for that set itself. Given a set of natuals, E, such that 2 is a member of E and for each member m , m + 2 is also a member, and that E is a subset of any other set with that property, then one can correctly say that every member of that set is even, but one cannot properly claim that the set is even.
> > Thus, the fact that there is no line (along with > > all its predecessors) that cannot be removed > > is not a contradiction. > > Every line can be placed and can be removed by the same procedure.
Outside of Wolkenmuekenheim, placing and removing require different procedures. > > In principle that does not disturb anybody. Only when the completion > of all finite lines is an infinite line, only when the completion of > all finite paths is an infinite path, then we get a problem.
WM may find himself in trouble but no one else has to. > > Therefore again my question: When the Binary Tree is constructed by > all its nodes (which is possible, since we can determine, for every > node, the step which places that node) does this Binary Tree contain > only all finite paths, or does it contain also infinite paths that can > be distinguished from all finite paths?
Let us consider maximal nested sets of FISONs. If they are the paths then they already exist as subsets of the set of FISONS, otherwise their unions are the paths, and are inherent.
> In the latter case, all finite > paths could be removed without removing the infinite paths.
NOT with either of the above definitions of paths.
> Only in > this case my anti-Cantor-proof (for all d_n) could be rejected
Any and every where outside of Wolkenmuekenheim your anti-Cantor arguments fail to prove anything except your own incompetence.
> with > the argument that the anti-diagonal d is more than all its d_n.
It is more in the sense that it is also provably different from each of the listed sequences from which it was constructed.
> (Although this rejection would fail, because the limits of sequences > need not differ, when the sequences differ.)
Two sequences differ if they differ in at least one position. Any anti-diagonal differs from each line of the list from which it is built in at least one position.
Ergo, such an antidiagonal necessarily differs from every line of the list from which it was built. --