In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 19 Jun., 15:42, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > If every terminating path of the form 111...1000... is used, then > > > every node is covered by the last 1. > > > > But that still does not show that all nodes are covered by a countable set > > of > > paths. It only shows that every node is covered by a set of paths that > > contains a countable subset, not that the set itself is countable. > > It is easy to construct a bijection between a countable set of paths > and all nodes. These nodes can even be defined by the paths that lead > to them. That shows that all nodes of the tree can be covered by a > countable set of paths. > > > Yes, they are in the tree, so you have to prove that they are also used in > > the > > covering, which you have not done. You do not have to show that each node > > is covered, you have to show that with your covering you use each path. > > You > > did show the former. > > If every node is covered, then there remains no path to be covered.
Covering a node with a path does not automatically also cover each of the uncountably many paths through that node.
> > If all nodes are covered then all path that can be distinguished by > nodes, i.e., all reals that can be distinguished by digits, are there.
Then WM must mean something other than the usual by "distinguished".
No single digit, nor finite set of digits, distinguishes any one real from all others, just as no one node, nor even any finite set of nodes, distinguishes any one path in a maximal infinite binary tree from all others. > > > Again: > > consider the set of paths where each path after some point always goes to > > the right. This set of paths covers each node. Consider the path that > > alternates going right and left continuously. Can you point to a node in > > this alternating path that is not covered by one of the paths in the > > earlier > > set? But you did agree slightly higher up that that path was in the tree. > > All parts of that path that really exist, really are in the tree.
If any part of that path does not "really exist", one is no longer talking about maximal infinite binary trees at all, but just another of WM's myths. > > > > That is a ridiculous requirement in an informal discussion, in > > > particular if not mathematics is concerned. > > > > Oh. So in an informal discussion you can tell nonsense whatever you like > > and when challenged you do not need to back it up? > > I have read that there is a dogma of the catholic church stating that > the existence of God can be proved by means of scientific reasoning. I > do not remember the source. Does that turn this information into > nonsense?
It is WM who keeps trying to insert his own dogmas in places that they do not fit.
In order to even talk about infinite sets coherently, one must have at least one infinite set, any inductive set as a metaphor for the set of all naturals will do.
And once one has that set, none of WM's objections obtain, since we can define a partial order on any such set which makes it into a maximal infinite binary tree having all those properties WM objects to.