In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 21 Mrz., 03:41, Virgil <vir...@ligriv.com> wrote: > > > > > > A needed number of lines has a first element. > > > > Nope!
If that need is to have a set of lines whach covers all of |N, then any such set has to satisfy both of the two needs: 1. The seet of line must cintin a first line, and 2, For each line in the set of lines there must be a 'next' line. > > > > Any sufficient set of lines will have a first element, but as there are > > pairs of such sufficient sets which are disjoint, there is no first > > element common to all sufficient sets. > > I am not interested in sufficient sets but in necessary sets.
No particular set of lines is necessary but there are two necessary conditions on sets of lines, for it to contain all of |N: 1. Such a set of lines must have first line, and 2. For each of its lines it must also have a longer line.
> > It is a pity. You and your ilk are unable to apply induction
Induction works fine when one is allowed to have an infinite set like |N which has a first members and which for every member has a unique but different from all predecessors next member, which means that such induction cannot not work in WMytheology. ====================================================================
WM claims to know how to map bijectively the set of infinite binary sequences, B, linearly to the set of reals and then map that image set of reals linearly ONTO the set of all paths, P, of a Complete Infinite Binary Tree.
But each binary rational in |R is necessarily the image of two sequences in B but that one rational can then only produce one image in P, so the mapping cannot be the bijection WM claims.
SO that WM is, as usual with things mathematical, wrong. --