In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 13 Jun., 08:30, Virgil <virg...@nowhere.com> wrote: > > > > > 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. > > > > > How do they come in? > > > > They are a natural and inevitable consequence of the tree becoming a > > maximal infinite binary tree. As soon as all the required nodes are > > there, so, automatically, are all those unused paths. As soon as one has > > the set of all nodes, every possible subset is a subset, and all those > > paths are merely paticular subsets. > > All paths that I have used contain all possible subsets.
Outside of WM's world of MathUnrealism, if one has a maximal infinite binary tree and defines in it a path to be any maximal set of nodes totally ordered by the ancestor relation, then there are uncoutably many of paths.
WM must be using some other definition of path in order to have so few of them. > > > > > The tree is constructed from a countable set P of > > > terminating paths. > > > > If the union of that set P contains every node of the maximal infinite > > binary tree as a member, then it contains every set of nodes as a > > subset, therefore contains every path as a subset. > > It does. And it is a countable union.You remember a proof by Feferman > and Levy? They showed that the statement that the set of all real > numbers is the union of a denumerable set of denumerable sets cannot > be refuted.
Did they say it could be proved? And didn't their proof require denial of the axiom of choice? > > > > > If you have a Cantor list filled with those paths, > > > then you claim to be able to construct another path p. So this path > > > does not creaqpo intio the list. > > > > But it is in a new list. But P is not just a list, it must be a set of > > sets of nodes and the union of P (the set of all members of members of > > P) must then be the set of ALL nodes, at least if the tree is to be > > constructed from it. > > > > And once one has a set of all nodes, one has as subsets all those paths > > WM doesn't like hearing about. > > > > > But it does creap into the tree? > > > > Do you mean "creep" > > probably. Why do you English always speek such that one cannot > distinguish between weak and week? > > > > > When > > > does it creap? Where does it lurk and who tells it when to start? > > > > If the subsets of P cover all nodes then the union of P is the set of > > nodes and the subsets of that union include all those paths that upset WM > > so badly, so I guess one would have to say that when building your P, as > > soon as what you have built so far covers all nodes, you have all paths. > > And if the right hand side of the tree is not yet ready, but the left > hand side? Is there a border?
The "border" appears to be between having all those nodes which precede and/or follow some particular node, and not having them.
As soon as one has them, one also has uncountably many paths. > > > > > > > > > That is nonsense! There is no actually infinite path. So nothing needs > > > to creap and superstition can be expelled from mathematics. > > > > That is WM's nonsense. > > > > WM can hardly build his own infinite paths, as he has been doing here > > for some time, and then deny infinite paths to everyone else. > > I can assume its existence and then find that this assumption was > wrong.
You cannot find the assumption wrong without assuming that we do not grant.
> 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. > > > > > But if I trick you out and decide only after you have guessed? > > > > That is not the game originally proposed, so that WM finds he needs to > > cheat. > > > > > Or if I > > > am honest, as is my character, and use only every path that can be > > > named? Then your odds are zero if not less. > > > > For any finite list of more than one name, the concatenation of all its > > names cannot be included within it as a name. And WM declares that > > absolutely no infinite lists of names are possible, so he loses. > > The list of everything is shown below. > > > > > > > > > > > > > > > > > > > > > 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. > > > > > Yes to both. Here is the algorithmus generating the words than label > > > my paths including the dictionary and grammar of all languages that > > > are able to identify paths: > > > > > 0 > > > 1 > > > 00 > > > 01 > > > 10 > > > 11 > > > > None of the above are paths, or even sets of nodes, so are not allowable. > > All paths that you can tink of and anything else is encoded in th > elist above.
Nope. Only nodes are encoded. To encode a path one would need an infinite binary sequence whose nodes are the finite initial sequences in your list. Either that or infinite subsets of members of your list. In either case, there are more that countably many of them.