In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > So it is. > Proof: Let me construct a tree. I will not tell you what kind of paths > I use. For instance I could use terminating paths or paths with tails > 111... or paths with tails pi-3 or paths with tails e-2 or a mixture > of many kinds of paths. I will present you the tree when it is ready. > Would you bet to be able to distinuish your path from the set that I > have used?
If you hide the paths you have used, there is no way, without peeking, that anyone else can Tel wish paths yo have used, but that is immaterial.
The point is that whatever paths you have used, by the time the maximal infinite binary tree finally exists, there are paths in it that you did not use.
> > So let us play the game. Choose a path and try to distinguish it from > the set of paths that completely cover my tree.
Until I know which paths you used, I can only guess, but if I guess slyly enough, and your set of paths is countable, the odds are greatly in my favor. > > > > > Every node of path p and every set of nodes > > > of path p is covered by one or more paths of P. > > > > Indeed, one *or more* paths of P. > > The set of nodes in path p is > > not covered by one path in P. > > Ok. Let us start. I often get spam mails offering some millions of > dollars. I will risk one of them, say 10 million dollars on the claim > that you will not be able to distinguish your path from the set of > paths that I have used for construction. What about your stakes?
Is your set of paths countable (and actually counted or listed, at least algorithmically), and recorded with a neutral referee?
If 'yes' to both, you would almost certainly lose at any stakes.
All your opponent would have to do is use the "Cantor antidiagonal" to your own list of paths, which you cannot ever include in that list.