In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 12 Apr., 19:37, Dan <dan.ms.ch...@gmail.com> wrote: > > > >> 3) exists k , forall n , d_n == q_kn > > >No. That negation is valid only for all n =< k. > > > > Why ? This is NOT how logic works . > > But it is how mathematics works.
Only the sort of mathematics that disdains logic, like WMytheology. > > If "=<" or "=<" doesn't appear > > in the affirmation , then there is no reason for it to appear in the > > negation . Either way, a contradiction doesn't exist > > A contradiction does exist *if* d is not more than all d_n.
Since d is only a sequence of d_n's, it is, at last in one sense, is nothing more than "all d_n". But in any case, WM's "contradictions" exist almost entirely inside Wolkenmuekenheim, and never get out. > > Forall n: (d_1, ..., d_n) leaves no digit out.
Out of what?
> Assume (sets of nodes = paths in the Binary Tree)
Not all sets of nodes are paths in most trees, including any trees having more than two nodes.
> > > > The Cantor set doesn't have your required structure . >
NOt necessarily in Wolkenmuekenheim, where WM does it far too regularly. > > > > Unless you've worked out exactly WHERE a contradiction occurs in > > Cantor's argument > > > By the same token unless you've worked out WHERE and HOW my argument > > is wrong, > > Your argument is wrong if d is nothing but the union of all its finite > initial segments (d_1, ..., d_n), since for each of them the diagonal > argument (which is only valid for the first k lines of the list) fails > for all following lines.
What is the value of k immediately after which the diagonal argument fails?
I'll bet WM cannot name it, nor even prove that there is a k for which it works while for k+1 it does not work. > > If you claim, however, that d is more than all its finite initial > segments, then I will present another argument.
Don't bother, as it will be even more wrong that the ones you have been boring us with so far. > > So do you claim that d is not in the list but all its finite initial > segments (d_1, ..., d_n) are in the list?
As ALL of the members of a list in the Cantor proof are infinite sequences, there are NO finite initial segments listed. --