On 4 Dez., 23:18, Virgil <Vir...@home.esc> wrote: > In article > <ce07a2f1-cf4c-4e53-b5d5-b1fcd2d80...@s20g2000yqd.googlegroups.com>, > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 4 Dez., 09:20, Virgil <Vir...@home.esc> wrote: > > > In article > > > <949e3b4b-04c5-4d6e-ba64-7f95d37db...@r24g2000yqd.googlegroups.com>, > > > > Actual mathematics would say that while every > > > node and every edge is included in the sequence, there is no member of > > > the sequence having all of them at once. > > > Everey member covers the first node. > > The n-the member covers all nodes including 1 and n. > > So logic forces us (i.e. those who can) to conclude that if all nodes > > are covered then there is a path covering all nodes. > > That WM's particular form of logic is rife with contradictions. > > For every 0 node there is a path in WM's set NOT containing it, > and for every path in WM's set there is a node NOT contained in it, > so why does WM insist that there is a path in your set containing every > node?
That is potential infinity. For every node there is a path surpassing it For every path there is a node surpassing it. There is no "all nodes" and no "all paths".
Actual infinity infinity or set theory, what is the same, claim that there is a path containing all nodes (namely the path 0.000...) and that the union of all paths 0.111... 0.0111... 0.00111... ... contains the path 0.000... but that there is no single path of the list containing all nodes of 0.000...
That is simply impossible. The union of paths of the list contains exactly as many nodes of 0.000... as one of those paths to be unioned. If none of them contains all nodes of 0.000..., then even infinitely many will not accomplish that goal.
The union of paths cannot result in more than any contributing path. The union can be potentially infinite. A union of finite paths cannot yield a path of fixed length the lengths of which surpasses every contributing path.