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. Each n is finite. But for each n, also each n+1 is included (when using iduction - not when removing any finite set). > > 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.
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.
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? In the latter case, all finite paths could be removed without removing the infinite paths. Only in this case my anti-Cantor-proof (for all d_n) could be rejected with the argument that the anti-diagonal d is more than all its d_n. (Although this rejection would fail, because the limits of sequences need not differ, when the sequences differ.)