
Re: The uncountability infinite binary tree.
Posted:
Dec 16, 2012 3:43 AM


On 16 Dez., 07:18, Zuhair <zaljo...@gmail.com> wrote: > On Dec 16, 12:12 am, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > On 15 Dez., 21:27, Zuhair <zaljo...@gmail.com> wrote: > > > > On Dec 15, 10:22 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > What do you mean I appended the sequence 000..., Can you explain that > > > in details. I mean the full detail of how did you construct this tree > > > by this appending. How do you prove the the resulting constructed tree > > > is the infinite binary tree. DETAILS please. At least refer me to an > > > article that has all the details about that alleged construction if > > > there is any. > > > The finite paths are the following: > > 0. > > 0.0 > > 0.1 > > 0.00 > > 0.01 > > 0.10 > > 0.11 > > ... > > > Each of these paths now is equipped with an infinite tail 000... > > > 0.000... > > 0.0000... > > 0.1000... > > 0.00000... > > 0.01000... > > 0.10000... > > 0.11000... > > ... > > > Some paths appear more than once. Some nodes are constructed more than > > once. But that does not matter. This set of paths is nevertheless > > countable. > > And there is no "diagonal" that *at a finite level n* differs from all > > paths.
> > Ok but the resulting construction is not the INFINITE binary tree > we've already defined.
Liar! The resulting construction is what you accepted as the complete infinite Binary Tree CIBT. Should I
The set of paths that I used is countable. And you will see it when I tell you what I used. But before I unveil my construction let me know whether the constructed Binary Tree is complete in your opinion. Here it is:
0. 0 1 0 1 0 1 ...
Every level starts with 0, is alternating between 0 an 1, and has twice as many nodes as the level above.
You said: This tree has ALL possible binary paths as paths of it. In other words it is COMPLETE.
> For example the path representing the decimal > expansion of pi is not among the paths of your constructed tree.
First: pi is not a real number of the unit interval Second: Every finite initial segment of pi  3 is in the CIBT.
And even pi  3 is there because I cheated. I did not use the tails 000... but the tails with the bitstring of pi3.
>Of > course the tree you've constructed that way is COUNTABLE, I agree, but > it is not a complete infinite binary tree, it is not even near.
You are dreaming. But let us play again. I construct a CIBT by using tails that I do not publish. You have to find out what paths are missing.
> Now > even if you add any tail of your choice, I mean suppose you have a > countable set of tails T of your choice, and you append those tails to > countably many finite stumps of the infinite binary tree, still the > resulting construction is countable, but it is not the complete > infinite binary tree, why because simply and I say simply apply the > method I've showed you in the head post on your so constructed tree > and you will recover a path of the complete binary tree that is not > among the paths of your constructed tree. I showed you that, but you > keep refusing to understand that part.
If you are really unable to recognize it yourself, then listen to others like George Greene: The CIBT is complete. It is impossible to find a digit sequence defined by nodes that is not contained in it.
Regards, WM

