Virgil
Posts:
4,661
Registered:
1/6/11
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Re: FAILURE OF THE DISTINGUISHABILITY ARGUMENT. THE TRIUMPH OF CANTOR: THE REALS ARE UNCOUNTABLE!
Posted:
Jan 11, 2013 9:40 PM
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In article
<117f2274-de68-4a54-b90d-f3e423c3dedf@c16g2000yqi.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 11 Jan., 10:39, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <3810bc42-c275-4897-94ba-8280508e9...@10g2000yqk.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 10 Jan., 22:35, Virgil <vir...@ligriv.com> wrote:
> >
> > > > > But there is a striking ground that is more fundamental than any wrong
> > > > > or correct logical conclusion, namely that you cannot find out any
> > > > > real number of the unit interval the path-representation of which is
> > > > > missing in my Binary Tree constructed from countable many paths. At
> > > > > least by nodes, you cannot distinguish further reals, can you?
> >
> > > > We certainly cannot tell which ones are missing until WM tells us which
> > > > ones are present,
> >
> > > In mathematics reals are represented by sequences of digits or bits.
> >
> > They can be, but they are quite often represented by other methods.
>
> Correct. But Cantor's list requires decimals or equivalent
> representations.
Actually, Cantors original list for anti-diagonalization was of
sequences of letters from {m,w}, not digits, and were not interpreted as
numbers.
> >
> > > In the Binary Tree bits correspond to nodes. I can prove that every
> > > path that can be defined by listing its nodes is covered by my
> > > construction, i.e., it is contained in the Binary Tree.
> >
> > We have no problem with all of your paths being in that tree, what we
> > have the problem with is the paths that you have necessarily omitted in
> > any construction that produces only countably many paths.
>
> I have not omitted any path that can be defined by a list of nodes.
Until you have listed all your paths, the claim that the set of them is
countably remains unproven.
> >
> > Every path in a Complete Infinite Binary Tree can be represented by an
> > infinite binary sequence, a mapping from |N to {0,1} and every such
> > mapping necessarily occurs in any representation of ALL paths.
>
> Since all finite paths are in the tree and since |N does not contain
> an infinite n, there cannot be missing anything.
Almost all functions, f, from |N to {0,1} will have infinitely many n
for which f(n) = 0 and infinitely many n for which f(n) = 1.
Thus for each subset S of |N, there is an f with f(n) = 1 if and only if
n is a member of S.
So the set of paths of length |N clearly bijects with the power set of
|N, which is well known not be uncountable.
> If so, let me know
> what is missing - from looking onto the covered nodes.
What is missing is any proof that the set of all functions from N to
{0,1} is countable.
> >
> > But the set of all such representation is, by Cantor's diagonal proof,
> > not countable, since countable = listable.
>
> Cantor's diagonal proof proves that the set of distinguishable or
> definable reals cannot be put in bijection with |N. There are
> countable sets that cannot be in bijection with |N.
Any countable set can be surjected from |N, as the existence of such a
surjection is the definition of being countable.
> Compare the set of
> definable reals.
> >
> > > You will
> > > already understand this, when you know that I use every finite path.
> >
> > There are NO finite paths in a Complete Infinite Binary Tree.
>
> Call the finite initial segments finite paths. I construct the Binary
> Tree by all finite initial segments.
Every such finite initial segment is the initial segment of uncountably
many complete infinite paths.
So that the countability of such finite initial segments is irrelevant.
> >
> > A path is, by definition, as long as possible already, so that no
> > extension of a path is possible
> >
> > > But I append always an infinite extension.
> >
> > No one can do it to a path, which in Complete Infinite Binary Tree is
> > already endless by definition.
>
> No one can find paths missing in the Complete Infinite Binary Tree
> that contains all nodes covered by at least one path. Nevertheless
> many claim to be able (but of course they are not).
No one other than WM makes any such claim.
Aus dem Paradies, das Cantor uns geschaffen, soll uns niemand vertreiben
können.
No one shall expel us from the Paradise that Cantor has created.
David Hilbert
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