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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 8:14 PM
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In article
<0b77d31b-1160-4c36-8935-58859e094d1c@j19g2000yqk.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 19 Dez., 21:14, Virgil <Vir...@home.esc> wrote:
>

> > > 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?

>
> I am interested in linear sets.


Every set with more than one element has more than one "linearity" or
ordering possible, and any set non-empty set can be ordered so as to
have a first and a last member according to that ordering, though for
one-element sets they are not distinct.
>
> > > 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.

>
> You are wrong. The "perfectly adequate proof" is wrong.


You keep claiming that but have never come up with convincing reasons
for your claim.
>
> > > 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.

>
> There is one simple proof: Every node of the binary tree splits one
> path into two paths. the number of paths cannot surpass the number of
> nodes.


First of all, every node of a COMPLETE INFINITE BINARY TREE splits one
infinite set of infinite paths into two disjoint but equally infinite
subsets of equally infinite paths. Anything less and the original tree,
as is usual with WM's trees, cannot have been a complete infinite binary
tree.


> Nobody would ever have doubted that without the unfortunate
> development of mathematics induced by Cantor.


You have just conceded that it was a MATHEMATICAL developement, so it is
now, and ever will remain, a part of mathematics.
> >
> >
> >

> > > 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.

>
> From the sequence of digits you cannot obtain 1. First you must be
> sure that there never will appear a digit other than 0. But unless you
> know the last digit, you cannot be sure. And you cannot know the last
> digit.


I can know that there ISN'T a last digit, at which point I can and do
know that all of the zero digits are redundant.
>
> Of course you can know a formula saying digit d_n = 0 for all n in N.
> But that is a finite formula.


It may be a finite formula, but it defines infinitely many digit
positions quite satisfactorily, at least for mathematicians.

> It is not an infinite sequence of digits
> that informes you about the due number.


Numbers can exist quite satisfactorily in mathematics without any
particular form of numeral being able to name them all.

WM's peculiar theory of numerals is irrelevant to mathematics.


Date Subject Author
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