On 17 Dez., 14:44, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <ba9f2c85-caa2-491d-9263-bea1bed3a...@p19g2000vbq.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 16 Dez., 03:31, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > Why that? Group, ring and field are treated in my lessons. > > > > > > You think that something that satisfies the ZF axioms being a collection > > > of sets is rubbish, while something that satisfies the ring axioms being > > > a ring is not rubbish? > > > > Yes, exactly that is true. > > And why, except by opinion?
> Aha, you are clearly a mindreader. Well, as far as I know mindreading is not > part of mathematics. Anyhow, I can think of numbers larger than that path. > But that is completely irrelevant. I am able to think about a set that > contains all natural numbers, you apparently are not.
How do you know that without confirming it by thinking the last too? > > > > > Every digit that is on the diagonal of Canbtor's list is a > > > > member of a finite initial segment of a real number. > > > > > > Right, but there is no finite initial segment that contains them all. > > > > That is pure opinion, believd by the holy bible (Dominus regnabit in > > aeternum > > et ultra. [2. Buch Moses: Exodus 15 Vers 18]) or forced upon us by the > > men-made axiom of infinity. > > Sorry, I have no knowledge of the bible. But live without that axiom when > you can't stomach it. And do not attack mathematicians who live with that > axiom.
> > > > > You can only argue about such digits. And all of them (in form of > > > > bits) are present in my binary tree. > > > > > > Right, but your tree does not contain infinite paths, as you explicitly > > > stated. > > > > The tree contains all paths that can be constructed by nodes, using > > the axiom of infinity. Which one would be missing? > > The infinite paths because you stated a priori that your tree did not > contain infinite paths. So it is impossible to construct in your tree > infinite paths by the axiom of infinity.
The axiom of infinity establishes the set N from finite numbers. It establishes the infinite paths as well in my tree from finite paths.
> You can construct infinite > sequences of nodes, but as you stated *explicitly* that your tree did > not contain infinite paths, those infinite sequences of nodes are > apparently not paths within your terminology.
They are not constructed, but it might happen that the come in by the union of some finite paths: The sequence of finite paths 1, 11, 111, ... may yield an infinite path 111... as a limit.
However, there is never more than one limit of a sequence and every convergent sequence hat at least one finite term. > > > > > It exists in that fundamentally arithmetical way: You can find every > > > > bit of it in my binary tree constructed from finite paths only. You > > > > will fail to point to a digit of 1/3 that is missing in my tree. > > > > Therefore I claim that every number that exists is in the tree. > > > > > > In that case you have a very strange notion of "existing in the tree". > > > Apparently you do *not* mean "existing as a path". So when you say that > > > the number of (finite) paths is countable, I agree, but 1/3 is not > > > included in that, because it is not a path according to your statements. > > > > It is. I constructed a finite path from the root node to each other > > node. > > Yup, you constructed a finite path, and that does not represent 1/3.
It 1/3 can be represented by a bit sequence, then it is represented in my tree by the union of the node sets representing 0.0, 0.01, 0.010, ... > > > Then I appended an infinite tail. > > Whatever that may be, it is *not* a path according to your explicit statement > that the tree did not contain infinite paths.
You have misread my construction. I use finite paths and then I append an infinite tail.