Virgil
Posts:
4,660
Registered:
1/6/11
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Re: Mathematics in brief
Posted:
Dec 10, 2012 3:07 AM
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In article
<5a2d9b2e-c558-446a-908f-1a5f24d3f2cc@r14g2000vbd.googlegroups.com>,
WM <mueckenh@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:
> >
> > 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.
What definition is that? I know of no such definition.
While it is certainly possible to surject the set of paths of a truly
complete binary tree, of the sort that is forbidden to grow in WM's
garden, none of those reals really IS a path.
> Every binary sequence that is not in the BinaryTree
Only in Wolkenmuekenheim are there any.
Elsewhere Every binary sequence is in any complete infinite 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.
Right for once!
>
> >
> > 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 ending that I do not publish).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.
And Cantor's counterproof says that no such list contains all paths and
provides an unambiguous way of determining from any such list ( which
exists as a consequence of WM's claim of countability) any finite number
of the uncountably many paths which are missing.
>
> > 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!
List the ones you have used (and until you have listed them , or at
least prove you can list them, your claim of countability is
notstablished) and then it is trivial to find others.
> >
> > 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?
In large part because those like WM who argue against it here are so
obviously unable to make their case.
> 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.
Don't look now but you still are!
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