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.
Actually, WM is right for once. Every node (there are only countably many of them after all) can be "covered" by a countable set of paths. Simply pick the nth path so as to contain the nth node in some counting of the nodes.
However often WM is wrong on his own, one should be careful not to assume him always wrong. He is not that consistent.