In article <3f7ebb76-23f8-494a-9247-284a14720b07@3g2000yqk.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 12 Jun., 03:49, "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.
And so your statement that they are isomorphic is wrong. Do you agree?
> > 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.
I have not seen a disprove of that theorem.
> > 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.
No, that is not a valid axiom.
> > > > > 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".
So you did use consciously an ambiguous word without clarifying what you did mean?
> > > 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 that still does not show that all nodes are covered by a countable set of paths. It only shows that every node is covered by a set of paths that contains a countable subset, not that the set itself is countable.
> > > > > 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.
Yes, they are in the tree, so you have to prove that they are also used in the covering, which you have not done. You do not have to show that each node is covered, you have to show that with your covering you use each path. You did show the former.
> > 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 is not an answer to my question. The path is in the tree, but does it contain a node that is not covered by a path that after some node always goes to the right?
> > > > > 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.
Right. But the number of required paths is the minimal number of paths needed to cover all the nodes. That is *not* the number of all paths. Again: consider the set of paths where each path after some point always goes to the right. This set of paths covers each node. Consider the path that alternates going right and left continuously. Can you point to a node in this alternating path that is not covered by one of the paths in the earlier set? But you did agree slightly higher up that that path was in the tree.
> > > > > 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.
Oh. So in an informal discussion you can tell nonsense whatever you like and when challenged you do not need to back it up? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
