In article <email@example.com> WM <firstname.lastname@example.org> writes: > On 17 Dez., 14:44, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: ... > .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 axiomsbeing a ring is not rubbish? > > > > > > Yes, exactly that is true. > > > > And why, except by opinion? > > Look here: > http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb5da79d#
What is the relation?
> > 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?
Why need I to think about a last one (which there isn't) to be able to think about a set that contains all natural numbers? Apparently you have some knowledge about how my mind works that I do not have.
> > > > 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. > > To live with that axiom does not create uncountability. See the proof > here: > http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb5da79d#
Where is the proof there? I see only you writing a bit of nonsense and two rebuttals.
> > > 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 *existence* of a set N of finite numbers.
> It establishes the infinite paths as well in my tree from finite > paths.
No. That is impossible because you stated that the paths were finite. What it *does* establish is the extistence of a set P of 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.
How is it possible to yield an infinite path if by your definition paths are finite? That is like saying lim(n -> oo) is a natural number.
> > > > 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, ...
Right, but that is not a path by your definition.
> > > > > 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.
But that is not a path by your statement. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/