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.
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.
Forall n: (d_1, ..., d_n) leaves no digit out. Assume (sets of nodes = paths in the Binary Tree) that d is the union of all paths (d_1, ..., d_n). If forall n: (d_1, ..., d_n) is in the list, then d is in the list.
Therefore this is a contradiction: Forall n: d =/= q_n. d is not in the list. Forall n: (d_1, ..., d_n) is in the list,. d is in the list.
. > > >That depends on the structure of the set. In linearly ordered sets > >like chains of mother-child we have > > The Cantor set doesn't have your required structure .
It has. Every linearly ordered set has that structure.
> 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.
If you claim, however, that d is more than all its finite initial segments, then I will present another argument.
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? If so, what is the difference d \ U(d_1, ..., d_n) d and (d_1, ..., d_n) understood as sets of nodes of paths in decimal tree.