Virgil
Posts:
8,833
Registered:
1/6/11


Re: Mathematics in brief
Posted:
Dec 9, 2012 10:17 PM


In article <5b78f4cae683408181388351faec8466@10g2000yqo.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 9 Dez., 21:16, Zuhair <zaljo...@gmail.com> wrote: > > On Dec 9, 10:59 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > > > > > > > On 9 Dez., 20:37, Zuhair <zaljo...@gmail.com> wrote: > > > > > > > Only a countable subset can be represented by the Binary Tree. The > > > > > reason is that no path is really actually infinite. > > > > > > Then you are not addressing what Cantor was speaking about, he is > > > > speaking about reals represented by ACTUALLY infinite sequences (paths > > > > in your case). It is clear that the set of all reals represented by > > > > FINITE sequences is countable, but those are just a very small subset > > > > of the set of all reals. > > > > > > If one assumes Actual infinity, then it is easy to recover the > > > > diagonal path from any bijection between the reals and the set of all > > > > paths of the infinite binary tree, and this will be a path that is not > > > > present in the tree of course. > > > > > Then you are wrong from the scratch. Every real number has a > > > representation by an infinite sequence (= infinite path of nodes in > > > the tree). > > > > Why you don't just prove that statement. > > Because it has been proven by Cantor. Can you imagine any diagonal > number that has not a representation in one and the same Binary Tree?
Only in complete infinite binary trees in which every node has two child nodes. But none of WM's trees can have this property as it demonstrably produces uncountably many paths, which WM's trees are not allowed to have. > > Please don't completely forget your ability to think when adhering to > the "experts" who require "proofs" (of that silly kind they prefer) > for 2 + 2 = 4 and theorems like that.
WM recks not his own Rede. > > > Of course this is a clear > > retreat from what you've just said before, where you said that no path > > is actually infinite, anyhow. > > That is the reason why Cantor is wrong. But the Binary Tree that I > assume has an actual infinity of levels. Every path has an actual > infinity of nodes. Nevertheless I construct this actually infinite > Binary Tree by means of a countable set of paths.
The alleged countability of WM's set of paths, requires by its very definition the existence of a list containing all paths in that set.
But then Wm must have a lot of nodes with less than two children, as if all do, there are provably no possible complete lists of all such paths, which there would have to be for WM to be correct. > > > Possibly you are referring to potential > > infinity when you say infinite path of nodes. But by argument of > > potential infinity you cannot have something called "infinite" path of > > nodes, all what you can have is 'finite' paths. And again clearly what > > you are addressing is something quite different from what Cantor is > > speaking about. Cantor is speaking about Reals that are represented by > > ACTUALLY INFINITE paths. > > Nevertheless, every node lies at a finite distance from the root node. > This is just the Tree that I construct!
It is true in any infinite tree, but in an infinite tree there can be no longest path. > > > And so far nobody have succeeded to > > demonstrate any contradiction involved with this concept. > > It is ridiculous! I have shown you the contradiction.
You alleged contradiction does not hold water, or anything else.
In order to have a COMPLETE infinite binary tree, every node in that tree must have two child nodes, and thus every EVERY infinite sequence of child nodes must occur.
But Cantor's diagonal proof shows that the set of all such sequences cannot be listed and thus cannot be countable.
> You escaped into > the completely nonsensical idea, saying: "You will need uncountably > many infinite binary trees to recover all the reals."
Actually that is nonsense. One only needs one COMPLETE infinite binary tree to recover all the reals with countably many paths left over. > > You cannot recognize that I have a contradiction if I can force you to > utter such nonsense?
It is only your own nonsense that creates those alleged contradictions. 

