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Topic: Another AC anomaly?
Replies: 280   Last Post: Dec 27, 2009 3:53 PM

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Virgil

Posts: 870
Registered: 7/27/09
Re: Another AC anomaly?
Posted: Dec 19, 2009 3:14 PM
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In article
<2fac8bb1-4c90-4421-b559-1ea7f0301d4f@e27g2000yqd.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
>

> >  > > Aha, you are clearly a mindreader.  Well, as far as I know mindreading
> >  > > is not part of mathematics.  Anyhow, I can think of numbers larger
> > than
> >  > > that path.
> >  > > But that is completely irrelevant.  I am able to think about a set
> > that
> >  > > contains all natural numbers, you apparently are not.
> >  >
> >  > How do you know that without confirming it by thinking the last too?
> >
> > Why need I to think about a last one (which there isn't) to be able to
> > think
> > about a set that contains all natural numbers?  Apparently you have some
> > knowledge about how my mind works that I do not have.

>
> Yes. A very convincing and often required proof of completenes of a
> linear set is to know the last element.


Knowing the allegedly last element in a set does not work unless one has
imposed a linear ordering on the set, which is in no wise necessary for
sethood. Which is the "last" point of the set of points on a circle?


> One of the rebuttals has meanwhile been changed. Peter Webb
> recognized: It is true that you cannot show pi as a finite decimal,
> but you can't show 1/3 as a finite decimal either.
>
> Just what I said.

> >
> >  > >  > The tree contains all paths that can be constructed by nodes, using
> >  > >  > the axiom of infinity. Which one would be missing?
> >  > >
> >  > > The infinite paths because you stated a priori that your tree did not
> >  > > contain infinite paths.  So it is impossible to construct in your tree
> >  > > infinite paths by the axiom of infinity.
> >  >
> >  > The axiom of infinity establishes the set N from finite numbers.
> >
> > It establishes the *existence* of a set N of finite numbers.

>
> What else should be established?

> >
> >  > It establishes the infinite paths as well in my tree from finite
> >  > paths.
> >
> > No.  That is impossible because you stated that the paths were finite.
> > What it *does* establish is the extistence of a set P of finite paths.

>
> It is rather silly to argue about the uncountability of the set of
> paths.


Then stop doing it. There is a perfectly adequate proof that one cannot
have a list of paths that contains all paths, or , equivalently, one
cannot have a list of all subset of N.
Since being capable of being listed is a necessary requirement for
countability, that proof eliminates countability for the set of all
paths s well as for the power set of any infinite set.

> Only minds completely disformed by set theory could try to
> defend the obviously false position that there were uncountably many
> paths.


Minds thus "deformed" by set theory are often capable of creating valid
proofs that WM not only cannot create but also cannot even follow.
>
> But "10 Questions" will give you the answer why there are not
> uncountably many paths. There are no infinite decimal expansions of
> real numbers. There are not due paths in the tree.


1 = 1.000... is a real number with an infinite decimal expansion, and in
ZF there are infinitely many more of them.

I know not what a "due path" is, but in my infinite binary tree, whose
nodes are the members of N, every path is an infinite subset of N, and
there are more of them than even WM can count.

>
> It is as impossible to express any real number by an infinite decimal
> expansion as it is impossible to express 0 by a unit fraction. The
> rest will be explained in "10 Questions". Therefore I will stop with
> this topic here.


Then in WM's world, 1.000... is not a real number.

Odd of him, but he has always been odd.


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Read Re: Another AC anomaly?
Virgil

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