On 30 Jun., 16:12, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > 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. >
I do not argue that there is a bijection with what you dream of as "all real numbers". I show that the number of paths that can be identified by sequences of nodes or bits is countable.
> > 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.
And I have shown that this sequence is covered by finite sequences to the satisfaction of everybody who is using bits to identify a number. > > > 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.
The tree will not differ in any detail from the construced one, if we use all finite paths as before but add the sequence 010101... instead of 000.... > > > > > > 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.
That' swhy you cannot find it. > > > 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.
I showed hat all paths are in the tree that can be identified by nodes.
> > > > > 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.
All that can be identified of 1/3 belongs to a terminating rational. Your "..." has no meaning in rigorous mathematics.
