Date: Dec 10, 2012 4:37 AM
Subject: Re: Mathematics in brief
On Dec 10, 9:54 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 10 Dez., 06:32, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 10, 12:21 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 9 Dez., 21:19, Zuhair <zaljo...@gmail.com> wrote:
> > > > On Dec 9, 10:59 pm, WM <mueck...@rz.fh-augsburg.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). But as my proff shows I construct the whole Binary Tree by
> > > > > countably many paths. There are not more nodes available to add
> > > > > further paths.
> > > > > >You will need uncountably many infinite
> > > > > > binary trees to recover all the reals.
> > > > > That is purest nonsense. And it has nothing to do with Cantor's
> > > > > diagonal which is of course an infinite sequence of digits
> > > > > corresponding to a path in the Binary Tree.
> > > > Yes corresponding to an ACTUAL infinite path in the Binary Tree, which
> > > > is something that you already refuse to address.
> > > No, you misunderstood. I construct the actually infinite Binary Tree
> > > by actually infinite paths like "every finite path which is appended
> > > by an actually infinite sequence of 000..." or "every finite path
> > > which is appended by an actually infinite sequence of 111..." or ...
> > Ok, now you are clear. And just to make it more clear you are claiming
> > that the actual infinite path 101010... is a path in your actual
> > infinity binary tree, correct.
> > If so what is the proof that ALL reals belong to that tree?
> There is no proof because the Binary Tree contains all reals between 0
> and 1 by definition. Every binary sequence that is not in the Binary
> Tree is not the representation of a real number between 0 and 1.
> Cantor's proof (concerning the binaries with bits w and m) shows that
> not all that are in the tree can be in the list.
> > And what is the proof that the number of paths in that tree is
> > countable?
> 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.
> > 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.
> > 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.
> Regards, WM
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. I know you have different convictions,
but to call all of those as fools? anyhow.