
Re: Mathematics in brief
Posted:
Dec 10, 2012 11:35 AM


On 10 Dez., 10:37, Zuhair <zaljo...@gmail.com> wrote:
> > This proof is given by constructing the whole Binary Tree by countably > > many actually infinite paths (i.e. finite paths with infinite endings > > like 010101... or 000... or 111... or 001001001... or any desired > > ending that I do not publish). My proof rests upon your inability to > > find out what paths are missing. > > But those are not all the reals, there can be reals that do not end > with a tail that is a repeated segment, > what you are speaking about is > actually the rationals which are known to be countable. > Anyhow I don't know the details of how did you construct your binary > tree, can it for example represent irrational numbers like the square > root of 2 (which doesn't end by a repeated segment tail) > or the transcendental reals.
Of course I can use also tails that consist of the bits of sqrt(2) or of pi or I can mix arbitrary tails. The point is, that you cannot define a real number or an infinite tail by following its nodes. It is simply impossible. You can only give a finite definition of infinite tails (like I did above  who could follow all bits of pi?). But there are only countably many finite definitions.
> > > I can see that the number of paths in any FINITE binary > > > tree is less than the number of its nodes? but can that feature > > > survive at infinite level? and what is the proof? I do see that the > > > number of nodes in your tree is countable. But would it follow that > > > the number of paths must be so at infinite level? > > > Find a path that I have not used! > > That is easy, you claim that there is a bijection between N and the > set of all infinite paths in your tree, correct! Then simply apply the > diagonal argument of Cantor and you will get a linear graph > that is not among the paths of your tree, and this will correspond to > a real, since it reflects a binary sequence, so your tree is NOT of > all paths that reals corresponds to. In other words your tree if > countable then it is incomplete.
No. The idea that real numbers could be defined by infinite sequences of nodes without a finite definition defining them, is simply wrong. > > > > How I see matters is that if I assume that there is a bijection > > > between N and the set of all infinite paths in your tree, then I can > > > easily construct a diagonal using Cantor's argument, and this diagonal > > > would provably be a path that is not in that Tree. So either your tree > > > must have uncountably many paths (with countably many nodes) or your > > > tree has countably many paths but is incomplete, i.e. there are > > > infinite binary paths that are not paths of it. > > > Or the idea of countable and uncountable sets is humbug. Why do you > > refuse to take into account this possibility? Because you cannot > > believe that many thousands of mathematicians have behaved like fools > > in the last hundred years? No reason to be ashamed. I have been among > > them myself for a long time. > > To be honest my answer is YES. I find it hard to believe that many > thousands of mathematicians behaved as you stated for a whole century > to time, not only that, among those thousands are people who are > considered geniuses of all times like Harvey Friedman for example, > Frege, Tarski, Godel, Hilbert, Von Neumann, Quine, and many many > others, who are authorities by known standards. To go say that they > acted as fools etc... is insulting really. How can you account for > such a claim. To be honest I don't see your claim to be reasonable. > Anyhow everything in this strange world is possible, but for one to > hold of such claim, then he must demonstrate a solid I mean really > really solid argument to verify his stance, since it is way against > what one should expect really.
Answer these questions honestly to yourself. You need not publish the answers. Have you ever realized that 0.111... is *not* an infinite sequence but only a finite expression allowing you to determine every digit of an infinite sequence? Have you ever realized that a real number defined as the antidiagonal of a Cantorlist is not defined unless the Cantorlist has a finite definition like, for instance, 0.1 0.11 0.111 ... ?
> I know you have different convictions, > but to call all of those as fools? anyhow.
This is a long story. I think I am entitled to do so. If you are interested why, you can find the reason here: http://www.hsaugsburg.de/~mueckenh/KB/KB%20201400.pdf numbers 211 to 234. The important texts with FOMrepresentatives are in English.
Regards, WM

