In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> If actually infinite paths exist, then there is a path p that can be > distinguished from every path of P. > And: > There is no path p that can be distinguished from every path of P. > > This kind of proof is called proof by contradiction or modus tollens: > ((A ==> B) & ~B) ==> ~A
But WM has not proved either. If the node set of a maximal infinite binary tree tree is actually infinite and P is any countable or finite set of paths (or other sets of nodes) of that tree (where a path is any maximal linear subset of the set of nodes under the ancestor relation) then there are paths not members of P.
And any such non-member path can be distinguished from all members of P.
> > > > Irrelevant, you distinguish p from every element of P > > Not after they have been written in form of tree.
The set P does not contain, as a member, after a construction of a tree anything that P did not contain, as a member, before that construction.
So that if P starts either finite countably infinite, it remains so, but the set of ALL paths in a maximal infinite binary tee, regardless of how it may have been "constructed", is uncountably infinite, a la Cantor. > > > > > Every path of the tree is is from P. > > > > Nope. Every *node* of the tree is from P. > > Every path of the tree is from P. I explicitly forbid every other path > to enter my tree.
Then your tree is not maximal, and may not even be a tree in the sense that there will be in it linearly ordered sets of nodes which are not subsets of any path.
> But, surprise, there is no other path willing to do > so.
Maybe not in those quasi-trees that WM "constructs" but if WM's the member paths of WM's P are countable in cardinality, then there will be sets of nodes satisfying all the requirements of pathhood in the constructed tree that are not in WM's P.
> What kind of path should apply, after I have every node of the > tree covered with a path of P?
All those uncountably many maximal linearly-ordered sets of nodes that are not members of P. > > > However, there is a *subset of nodes* in the tree that is not > > contained in one element of P. > > Wrong. Every node and every subset of nodes is in a path of P.
It is WM who is dead wrong about that, apparently because he does not even understand what paths are.
There are all kinds of "subsets of nodes" which cannot be members of any path. For example, no path can contain both children of any parent node.
And if P is countable, then its member paths can be listed and the Cantor construction produces a path not in P
> ideas you talk about are not realized by digits or bits or nodes
They are "realized" in the imagination of those whose imaginations are not as crippled s WM's, which is all that is needed in mathematics.