On 22 Jun., 15:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <fd4c6395-51cf-4045-883e-f5c986b91...@z7g2000vbh.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 19 Jun., 15:42, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > > > If every terminating path of the form 111...1000... is used, then > > > > every node is covered by the last 1. > > > > > > 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. The reason is that real numbers with infinite bit sequences do not exist *as bit sequences". 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. > > > > 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. In order to prove that there is such a path p you must show for every path q that there is a node not covered by q. But that is impossible. If you think, you have accomplished it at some node, there will always be another q that is identical with p to that node. That means, you cannot diagonalize the tree. 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. >
> > 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.