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.
> > You believe in infinite paths. But you cannot name any digit that > > underpins your belief. Every digit that you name belongs to a finite > > path. > > Right. But there is no finite path that contains them all. I believe in > a path that contains them all, and that is an infinite path.
There is a finite path that contains larger numbers than you can ever think of. > > > 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.
> > > 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? > > > > > 1/3 does not exist as a path. But everything you can ask for will be > > > > found in the tree. > > > > Everything of that kind is in the tree. > > > > > > This makes no sense. Every path in the tree (if all paths are finite) > > > is a rational with a power of 2 as the denominator. So 1/3 does not exist > > > as a path. In what way does it exist in the tree? > > > > 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. Then I appended an infinite tail. When the tree is completed, then the tail becomes invisible, because every sequenece of nodes has been constructed. But if you don't believe me, then look at the tree: You can see and admire every node and its connection to the root and the continuing paths downwards. So let me know what you think is missing. > > > > In what way do numbers like 1/3 exist in your tree? Not as a path, > > > apparently, but as something else. > > > > Isn't a path a sequence of nodes, is it? > > Apparently not in your tree. In your tree a path is a finite sequence of > nodes.
The tree is the union of all paths. There is no end. > > > Everey node of 1/3 (that you > > can prove to belong to 1/3) is in the tree. > > Right, but there is no path that denotes 1/3. > > > > Similar for 'pi' and 'e'. > > > > Yes. Every digit is available on request. > > Right, but there is no path that denotes either 'pi' or 'e'.
Which node is missing? > > > > So when you state that > > > the number of paths is countable that does not mean that the number of > > > real numbers is countable because there are apparently real numbers in > > > your tree without being a path. > > > > Wrong. Not only "apparantly" but provably (on request): > > There is no proof needed. Apparently there are real numbers in your tree > without being a path, because each path is finite (by your own definition).
You are wrong. The paths in the complete tree aer unions of all finite paths. > > > Every digit of > > every real number that can be shown to exist exists in the tree. > > But not every real number is represented in the tree by a path.
Every real number is there, that can be represented by digits. > > > Or would you say that a number, every existing digit of which can be > > shown to exist in the tree too, is not in the tree as a path? > > Yes, by your own admissions. You state (explicitly) that every path is > finite and it is easy to prove that every number that is represented by > such a path is a rational number with a denominator that is a power of 2. > So there are apparently real numbers of which every digit is in the tree > that are not represented as a path, like 1/3.