In article <257f6992-e801-43b8-a5df-30be71217db4@z9g2000yqi.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 12 Jun., 22:23, Virgil <virg...@nowhere.com> wrote: > > > > > > Sorry, that is impossible. Every node of path p and every set of nodes > > > of path p is covered by one or more paths of P. > > > > Is P a set of paths or merely a set of nodes? > > It is a set of paths. > > > > If it is a set of paths, then it must contain all uncountably many paths > > or there will be a path which is not member of it. > > Wrong. All countably many terminating paths are sufficient to cover > all nodes of the tree.
But paths are not mere nodes, they are sets of nodes, and while every node many be a member of a member of P, that does not imply that every any particular subset of the set f nodes is a member of P. That does not even hold for finite sets.
The set {{a},{b},{c}} contains every member of {a,b,c} as a member of a member but does NOT contain as a member every subset of {a,b,c}. For example {a,b} and {a,c} and {b,c} are not members nor subsets of {{a},{b},{c}}.
WM needs to go back to set theory kindergarten, and brush up.
> > > > > > > > > > So it is possible to contruct another path, p, which can be > > > > distinguished from every element of P. > > > > > That is impossible. The path p_0 = 0.111... for instance is completely > > > covered by terminating paths, whether or not it had been inserted > > > originally. > > > > If P is only a set of terminating paths then no non-terminating path > > will be a member of it. > > All non-terminating paths are in the tree constructed by P.
But not in P itself. Since P is countable it can be put into a list which necessarily has an anti-diagonal not in P. Actually, at least as many anti-diagonals as members.
