In article <eda6be39-323e-4b70-b40c-a43519b16e73@33g2000vbe.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 22 Jun., 15:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: ... > > > > 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. > > > > > > It is easy to construct a bijection between a countable set of paths > > > and all nodes. These nodes can even be defined by the paths that lead > > > to them. That shows that all nodes of the tree can be covered by a > > > countable set of paths. > > > > Yes, they can be covered by a countable set of paths. Nothing more. It > > does *not* show a bijection, and not a construction of a bijection. > > I do not argue that there is a bijection with all real numbers.
You did assert that there was a bijection with the set of all paths in the tree. You have not proven that.
> The > reason is that real numbers with infinite bit sequences do not exist > *as bit sequences".
Oh, well, you are apparently a non-standard form of "exist". By the axiom of infinite, the infinite set N exists, and so there does exist a function: f: N -> {0, 1} f(n) = 0 if n is odd = 1 if n is even. I would think that would be an appropriate infinite bit sequence of 1/3.
> pi exists in form of many formulas, Vielta, > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist > asa a subject to Cantor's proof. All existing sequences are in my > tree. And obviously there are only countably many.
I would argue that the infinite bit sequence .0101010101 exists and is in your tree. Your bijection above does not include that bit sequence.
> > > > 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. > > > > > > If every node is covered, then there remains no path to be covered. > > > > What do you mean with "covering a path"? Until now you were talking about > > "covering nodes". And whatever it may mean, in what way does "every node > > is covered" show that there remains "no path to be covered"? > > Try to find a path p that contains a node not yet covered by a path q > used to construct the tree.
Is *that* a definition of "covering a path"? But if your paths cover all nodes, there is no such path.
> It contains > already all paths that can be distinguished by nodes, i.e., all reals > that can be distinguished by digits. And As my construction B shows, > this set is countable.
And the path .0101010101... is not in that bijection, nevertheless there is no node in the tree that is *not* covered by that path. So you did show only a bijection between a subset of the paths with the nodes. Back to square 1 I would say.
> > > No. From that we can obtain that the number 1/3 is also in a list of > > > terminating rationals. There does not exist a sequence 0.010101... > > > that is longer than *every* finite sequence of that form. > > > > Ah, so now you contend that 1/3 is a terminating rational. > > No. It is a rational that has no decimal and no binary representation. > It has a ternary representation though.
Eh? You explicitly did state (as I did quote): "From that we can obtain that the number 1/3 is also in a list of terminating rationals." I would state this to mean that 1/3 is a terminating rational.
But I would also say that what I wrote higher up consists (in view of the axiom of infinity) an infinite bitstring of 1/3. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
