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Re: Mathematics in brief
Posted:
Dec 10, 2012 11:35 AM
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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 ending that I do not publish).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 anti-diagonal
of a Cantor-list is not defined unless the Cantor-list 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.hs-augsburg.de/~mueckenh/KB/KB%20201-400.pdf
numbers 211 to 234. The important texts with FOM-representatives are
in English.
Regards, WM
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