On Jun 17, 5:13 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 16 Jun., 23:44, Virgil <virg...@nowhere.com> wrote: > > > > The tree is constructed by all terminating paths with tails 000... > > > Which is a countable set of paths, and therefore, by Cantor, does not > > include all paths. > > By Jove, it does!
No, it doesn't. As you yourself have already pointed out, this set is easily bijectible with the set of paths terminating with ANY given tail, INCLUDING THE EMPTY tail (which would leave us with the set of all FINITE paths). The set of all FINITE paths will "cover" or "be able to construct" THE WHOLE tree, according to your (foolish) usage of these terms. This does NOT imply that actually infinite paths do not exist. The fact that one set of paths "covers or can construct" the tree goes NOWHERE toward proving that paths outside that set cannot be "contained in" the tree.
