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Topic: Matheology § 246
Replies: 246   Last Post: Apr 26, 2013 2:37 AM

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 246
Posted: Apr 13, 2013 4:05 AM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 12 Apr., 23:29, Dan <dan.ms.ch...@gmail.com> wrote:
> > > No, but it contains all decimal fractions, i.e., all terminating
> > > decimals. There is no chance to represent 1/3 or sqrt(2) by digits.
> > > There is no chance to apply Cantor's argument other than to FISs.
> > > Every d_n is part of a FIS.

> >
> > You can represent by an infinite sequence of digits, at least in an
> > abstract sense. That's why we can talk about pi, and the questions
> > like "what is the 1000000'th digit in the decimal expansion of pi?"
> > have a valid unambiguous answer .

>
> Yes, we hav a formula that tells us the n-th digit of pi. But the
> *decimal representation of pi* does never leave the domain of FISs. A
> decimal string would not allow us to recognize pi. Therefore there is
> no decimal representation of pi and of other actually infinite
> sequences. Therefore the anti-diagonal d, when only presenting its
> digits d_n, cannot differ from all FISs and therefore cannot prove the
> uncountability of irrational numbers.

The original "antidiagonal" did not contain any digits, nor did the list
of sequences from which is was to be constructed contain any.

Thus any argument based on the presence of digits is irrelevant.
>
> Do you agree that
> {1} U {1, 2} U {1, 2, 3} U ... = |N
> but the infinite sequence
> {1}, {1, 2}, {1, 2, 3}, ...
> does not contain |N?
>
> What is the difference?

An infinite union contains every member of each of the sets being
unioned, and every natural is in at least one FISON, so the union of all
FISONs MUST contain all naturals.

But a strictly increasing sequence, like {1}, {1, 2}, {1, 2, 3}, ...,
can never contain its limit, |N.

Those not mired in the wilds of Wolkenmuekenheim are already aware of
the difference between the two.

> >
> > > The Binary Tree is constructed by countably many nodes. So it is
> > > impossible to distinguish uncountably many paths *by nodes*.

But quite possible by sets of nodes.
> >
> > Yes . The binary tree has countable nodes , and uncountable paths .
> > (paths sequences of nodes, ie. subsets of the set of nodes , therefore
> > members of the power-set P(nodes) ) .
> > And as Cantor showed , |P(nodes)| > |nodes|

>
> All what Cantor showed concerns FISs.

Where in either the statement of Cantor's diagonal argument, or in his
proof of it, is any FIS even mentioned. The terms of sequence, but never
the FISs of one!

>
> In fact there is no decimal representation of any irrational number.

There are, in fact, formulae that allow the determination any and every
digit of certain transcendental numbers.
>
> > You have to use a whole sequence of FIS to represent it .

Nope! One can use a formula for each digit.
>
> That can only be done by a finite definition - not by gathering
> infinitely many FISs, obviously.

And has been done by finite definition.
>
> > Real numbers in general are fully represented only by 'infinite'
> > paths , so they are uncountable .
> > There are uncountable ways of combining your "finite-length
> > paths" (that are countable in number) , by union, into "infinite-
> > length paths" (that are uncountable in number) .

>
> Not in the Binary Tree. All paths that are possible there (and are
> represented by sets of nodes) belong to a countable set.

What WM is speaking of are properly only FISONs of such a tree, not
paths. A path is usually required to be a maximal sequence of
parent-child linked nodes, which no FISON can be, or could be defined to
be any maximal set of FISONs ordered by inclusion.

> And it is
> impossible to apply the diagonal argument to the Binary Tree

Only someone as messed up as WM would even think of trying to apply a
"diagonal argument" to something that as obviously has no "diagoanal" as
a binary tree. Why does WM not try to apply the diagonal argument to
elliptic integrals as well?

A list of binary sequences, to which the diagaonal argument DOES apply,
and a Complete Infinite Binary Tree are not at all the same thing.
> >
> > You say that all paths are unions of the "finite paths" , and because
> > the "finite paths" are countable , so are all paths .
> > But that does not follow .

>
> This construction of the Binary Tree shows that it is impossible to
> discern uncountably many paths by nodes.

But not as sets of nodes, which s what paths are.

WM keeps ignoring that given any set, it has more subsets then members,
so there are far more sets of nodes than nodes.

> (But Cantor's argument
> concerns nodes (like d_n) only!)

Cantor's diagonal argument only concerns lists of binary sequences
>
> The Binary Tree contains only those paths that are constructed step by
> step of its initial segments

A path in a Complete Infinite Binary Tree can only be constructed from a
nested and infinite set of such initial segments, each being order
isomorphic to the set of all Finite Initial Segments Of Naturals
(FISONs).
--

Date Subject Author
4/12/13 mueckenh@rz.fh-augsburg.de
4/12/13 fom
4/12/13 JT
4/12/13 fom
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4/12/13 dan.ms.chaos@gmail.com
4/12/13 mueckenh@rz.fh-augsburg.de
4/12/13 fom
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4/12/13 Virgil
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4/12/13 dan.ms.chaos@gmail.com
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4/12/13 mueckenh@rz.fh-augsburg.de
4/12/13 fom
4/12/13 Virgil
4/12/13 dan.ms.chaos@gmail.com
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4/12/13 dan.ms.chaos@gmail.com
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4/13/13 Virgil
4/13/13 mueckenh@rz.fh-augsburg.de
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4/13/13 dan.ms.chaos@gmail.com
4/13/13 mueckenh@rz.fh-augsburg.de
4/13/13 dan.ms.chaos@gmail.com
4/13/13 mueckenh@rz.fh-augsburg.de
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