
Re: The uncountability infinite binary tree.
Posted:
Dec 16, 2012 4:37 AM


On Dec 16, 11:43 am, WM <mueck...@rz.fhaugsburg.de> wrote: > 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
The CIBT (complete infinite binary tree) is defined in the following manner:
CIBT is a Tree The root node of CIBT is labeled 0 Each node of CIBT has two child nodes each having a different label from the other Any node of ClBT receives only ONE label that is either 0 or 1.
The ClBT is COMPLETE, in the sense that all possible binary paths belong to it. I already said that.
But YOUR construction which is done by appending infinite paths to countably many finite stumps of the of the ClBT is NOT the CIBT itself, it is another tree, at best it is a proper substree of the CIBT, actually it can even be not a subtree of CIBT, however I'll take that the appending process is respective of the binary structure of CIBT, so this constructed tree of yours would be as I said just a PROPER subtree of the CIBT. And the countability of your constructed tree is not important since it is not the CIBT itself. It is very easy to show a missing path from your constructed tree that of course belongs to the CIBT. As I said again and again just apply the diagonal argument on your so constructed tree and you'll see the missing path.
Can you tell me what is your proof that the tree you've constructed by the appending process is the CIBT??? try to prove it, and you will see that you cannot.
All you have done up till now is to provide a blind assertion that the tree constructed by the appending process is the CIBT itself? I didn't see any proof of that, you didn't present any? so if you have a proof of that then show it, so that one can see if it is valid or not?
YOU (and anyone who agree to your claims) is the one who must supply a proof that the tree constructed by the appending process is the complete infinite binary tree itself. But do you have such a proof? I simply think you CANNOT have any. But lets see if you have.
Again, supply the proof that your constructed tree is the CIBT?
Zuhair

