In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 12 Jun., 03:49, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article > > <0f2e464e-9791-4085-a381-0e428be84...@b9g2000yqm.googlegroups.com> WM > > <mueck...@rz.fh-augsburg.de> writes: > > > On 11 Jun., 14:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > ... > > > > > > And you are deluded. An axiom is a statement of something that > > can > > > > > > not be proven, neither disproven using the remainder of the > > theory. > > > > > > > > > > The axiom can be contradicted. Simple example: The axiom could be: > > The > > > > > binary tree has uncountably many paths. > > > > > > > > Perhaps, although in ZF it is not an axiom. > > > > > > It is, because the paths of the tree are isomorphic with the real > > > numbers in [0, 1] > > > > First off, the paths of the tree are not isomorphic with the real number. > > There is no 1-1 mapping that is order preserving. > > There are two paths for every terminating rational number. The rest is > a simple 1 to 1 mapping. > > > > Moreover, the statement "the binary tree has uncountably many paths" is a > > theorem, not an axiom. It can be proven. > > And it can be disproven.
Until WM provides us with a reasonably complete listing of his fundamental truths, on which he claims to base all his conclusions, he cannot prove anything to anyone. > > > > In ZF there is no axiom that uses the word "uncountable". It is defined > > and > > all statements using it are theorems. > > > > But apparently you still do not understand what an axiom is. > > said: I The axiom *could be*: The binary tree has uncountably many > paths.
Axioms are accepted without proof. The the maximal infinite binary tree has uncountably many paths is a theorem, not an axiom, and is provable from the axioms of ZF plus relevant definitions. > > > > > > > I show that the end of each > > > > > path p of the set P can be mapped on a node, and that all paths p > > of > > > > > P cover all nodes of the tree. > > > > > > > > Ignoring that in ZF the paths do not have an end. > > > > > > The paths of the tree have no end. > > > > So your statment "I show that the end of each path p of the set P can be > > mapped on a node" is blatant nonsense. > > I used the word end, that can have at least two meanings in the sense > of "tail". > > > > > But it can be shown for every node > > > that it gets covered and that all nodes get covered by a countable set > > > of paqths. > > > > You have not shown it. You only show it by assuming that there are > > countably > > many paths. > > If every terminating path of the form 111...1000... is used, then > every node is covered by the last 1.
But when one has the set of all nodes having only finitely many 1 branchings between them and the root node, suddenly one also has all paths with infinitely many one branchings in that same tree.
> > Right. But do you use up *all* paths? *That* is what you have to prove. > > You are always going to it from the wrong way. It is right that all nodes > > can be covered by countably many paths. The set of paths where each path > > from some node always goes right is an example. It does indeed cover all > > nodes. But it is not the set of all paths, because there are paths that > > do not satisfy the condition that from some node onwards it always goes > > to the right. > > These path are in the tree too, after having completed the > construction. > > > > Or can you find a node that is *not* covered by a path that after some node > > to the right? Can you find a node that is covered by a path that > > alternates > > going left and right that is not covered by a path that after some node > > always goes to the right? > > I see that every such alternating path with all its nodes is in the > tree, after construction.
Then there are uncountably many of them, as Cantor proved, and as no one, including WM, has been able to disprove. ssume it is false, it is easy to prove it is false. > > > > > > I do not assume that the number of nodes is countable, but I count > > > them. > > > > Darn. Misreeading *again*. The number of nodes *is* countable. You do > > assume the number of paths is countable. Why are you always misreading > > what people do write? > > Perhaps because they do not express them precisely enough? But I do > not always misread.
You do it often enough, and consistently enough, to make discussions with you dificult.
> We can state: The number of nodes is countable. The number of paths > required to cover all nodes is countable too.
Fair enough, but that does not mean that EVERY set of paths which covers all nodes needs to be countable. The set of ALL paths, not all being required, can be, and is enough larger than any of WM's "required to cover" sets as to be uncountable. > > > > > > That means, you are willing to believe in what the Vatican says? > > > > > > > > Well, no, because that "dogma" is not a valid "axiom". But can you > > tell > > > > me where that "dogma" actually is stated the way you say? > > > > > > Sorry, I only read it some time ago somwhere. But I think the set of > > > dogmas must be in the net for those who are interested. I am not. > > > > So you just state something without being able to back it up. > > That is a ridiculous requirement in an informal discussion, in > particular if not mathematics is concerned.
If that statement is so irrelevant to mathematics, why did you bring it up in a discussion of mathematics?