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Topic: Answer to Dik T. Winter
Replies: 441   Last Post: Feb 5, 2013 6:25 AM

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 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Answer to Dik T. Winter
Posted: Jun 11, 2009 3:14 PM

On 11 Jun., 15:46, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> In article <8d9c63c5-b605-4a98-81f2-815875aae...@21g2000vbk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
>  > On 4 Jun., 04:04, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
>  > >  > That is the logic of potential infinity, i.e., of incomplete sets.
>  > >
>  > > Eh?  As far as I know logic is *not* about sets.  It is about axioms,
>  > > statements and inference rules.
>  >
>  > Logic has been obtained from the bahviour of parts of reality that can
>  > be called sets.
>
> O.  Still, as far as I know logic is *not* about sets.
>
>  > >  > In Cantor's diagonal argument you can use the same logic : There is no
>  > >  > last line, therefore there is always a line beyond the checked lines,
>  > >  > But there you don't.
>  > >
>  > > Because you do not check the lines in order.  It is always your basic
>  > > assumption that you first check the first line and after that the next
>  > > line.  That is wrong.
>  >
>  > That is necessary because you cannot find the n-th line unless you
>  > know the line number n - 1 or some equivalent mark.
>
> You are wrong.  A list is a mapping from N to the elements of the list.
> Through that list, given a number n, you find the n-th element of the list
> without referring to any previous elements of the list.

But you cannot find number n without referring to the numbers less
than n.

> To give an example,
> let's have a list of positive rational numbers through the mapping given a
> lnong time ago by David Tribble.  To get the 51st element of the list, we
> calculate the mapping: write 51 in binary, create from it a continued fraction
> as described on
>   <http://homepages.cwi.nl/~dik/english/mathematics/mueck/mapping.html>
> which is [0, 1, 2, 3], calculate it and we find 7/9.  So we know the 51st
> element without knowing the 50th element.  Where did I come at the 47th
> rational in that list (which is 7/11)?

How can you write 51 without knowing what it is? Of course you must
count. In unary this is more difficult than in decimal or binary, but
the principle is the same. How do you obtain this number
1111111111111111111111111111111111111
unless you count the digits?
>
>  > > You simply give a definition of a new number so it is clear
>  > > that it will be different from each of the omega lines, without checking
>  > > directly any of them.
>  >
>  > And in my binary tree I give a definition of an end of a path that
>  > contains aleph_0 nodes.
>
> I have not seen such a definition.  What is the definition of "end of a
> path"?

Virgil calls it tail. That is a good name.
I begin the construction with one path p_0. Then I construct another
path p_1. All nodes that p_1 does not have in common with p_0, is the
tail of p_1. All nodes of the tail are mapped on p_1. Then p_2 is
constructed. All nodes of p_2 that differ from p_0 and p_1 are mapped
on p_2. And so on. You see the ratio of paths and nodes per path at
any stage during the construction is 0.
http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#382,45,Folie 45
>
>  >                         Every end of a path contains aleph_0 nodes.
>  > These nodes can be mapped on that path. Every node will eventually be
>  > mapped on one path. Therefore all nodes are used up for constructing a
>  > countable set of paths.
>
> Perhaps right, depends on how you actually do define things.  But are there
> nodes mapped to all paths?  That is what you assert.

Unless there is at least one node occupied by the path p_n that is not
occupied by the paths p_0 to p_n-1, p_n is not a new path.

>
>  >                                        And the reverse of being
>  > identical is being not identical.
>
> Not in this case bacause you apply the words to different things.  *Unless*
> you assume that what is valid in finite cases also is valid in infinite
> cases.  But let's see:
>     sum{i = 0 .. n} 1/(i!)
> is a rational number.  So according to your logic:
>     e = sum{i = 0 .. n} 1/(i!)
> is also a rational number.

I don't see a difference. But I can assuer you, there is no decimal
expansion for irrational numbers.
>
>  > > Ok.  The logic of finite complete sums of rational elements gives that
>  > > the sum is rational.  Using your logic we get that the complete (i.e. in
>  > > the limit) sum of rational elements is also rational.  And so by that
>  > > logic, e, pi and whatever are rational, and all numbers we do use are
>  > > rational.
>  >
>  > In fact there are no binary expansions of irrational numbers.
>
> What is the relevance?  Where am I talking about binary expansions?

Binary, decimal, whatever. It does not exist.
>
>  > >  >                                Either those sets obey that logic
>  > >  > or they do not exist in a science that is subject to the application
>  > >  > of logic.
>  > >
>  > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist.  Still I think you use at least
>  > They exist as ideas but not as sums of series.
>
> So the equation  e = sum{i = 0 .. oo} 1/(i!)  is invalid in your logic?

sum{i = 0 .. oo} 1/(i!) is a finite word with about 12 symbols. I
suffices to communicate what you mean. But what you think to have been
abbreviating is not abbreviated by this.
>
>  > >  > Small wonder. There cannot be a complete list, because the existence
>  > >  > of a complete infinite linear set like N contradicts logic.
>  > >
>  > > Apparently your logic.  What are the rules of inference in your logic?
>  >
>  > The union of a complete set of linear sets is one of the sets.
>
> Eh?  That is not a rule of inference.

But it is true.
>
>  > > And, also apparently, in your logic the length of the diagonal of a square
>  > > with sides with size 1 does not exist.
>  >
>  > It exists, but not as binary or decimal expansion.
>
> Eh?  Now suddenly sqrt(2) does exist.

Not suddenly. It accept that it exists. But it has no binary or
decimal expansion.
>
>  > >  > That is a blatant lie.
>  > >
>  > > Eh?  What is the lie?  I just state that in ZF there are no non-static
>  > > sets.
>  >
>  > Then every element should be accessible. Why do you start always with
>  > a diminishingly small one?
>
> Every element is accessible.  And you start always at the beginning because
> there is no last one.

But you claim that there is always a larger one than you atrt with.
Why that?
>  >
>  > A potentially infinite set is a set that has no last element.
>
> Ok.  Than N is a potentially infinite set.

That is true. But as such a set is never exhausted, Cantor waits and
waits and waits ...
>
>  > >  >                                   Then for every element that
>  > >  > you choose there is a larger one. If it has been there all time, why
>  > >  > the hell did you not start with this one ?
>  > >
>  > > Because in ZF a set does not need to have a last element.  So whatever
>  > > element you chose, there is *always* a larger one.
>  >
>  > That is potential infinity.
>
> Ok.  Than N is a potentially infinite set that actually does exist in ZF.

If one rejects actual infinity, then there is not "a continuous
function nowhere differentiable". The actual infinity is a necessary
condition of Cantor's diagonal proof
of the power-set theorem. If one rejects actual infinity, then the
theorem becomes unprovable and therefore there are not any cardinals
greater than aleph_0.

[Prof. Alexander A. Zenkin, Doctor of Physical and Mathematical
Sciences, Leading Research Scientist of the Computer Center of the
Russian Academy of Sciences.1937 - 2006]
http://www.math.rutgers.edu/~zeilberg/fb68.html

Regards, WM

Date Subject Author
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Dik T. Winter
5/27/09 mueckenh@rz.fh-augsburg.de
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