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. > > 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. > > > > > 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. > > > > > Therefore after having completed the > > > > covering of the whole tree, there remains no node that could be used > > > > to construct a path that does not belong to P. > > > > > > This is the wrong way around. You assume that you can cover this way the > > > whole tree (I think with this you mean each path in the tree). But that > > > is what you have to prove. > > > > There is not much to prove. Append a tail of a path to every node. > > Then every node is covered by at least one path, hence it does not > > remain uncovered. > > 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. > > > > > This disproves the > > > > mentioned axiom. > > > > > > Indeed, when you assume 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. We can state: The number of nodes is countable. The number of paths required to cover all nodes is countable too.
> > > > 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.
