Prev: integral problem
Next: Prime numbers
From: Virgil on 9 Jul 2006 15:20
In article <1152450425.487453.77870(a)35g2000cwc.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Dik T. Winter schrieb:
>
> > In article <pTsU8Z33pqrEFw3G(a)212648.invalid> David Hartley
> > <me9(a)privacy.net> writes:
> > ...
> > > (WM's fallacy is his claim that the (order-preserving) indexation by the
> > > naturals carries over to the limit.)
> >
> > Indeed. The most basic standard fallacy he uses.
>
> Indeed. The most basic standard fallacy Cantor uses.

Except that in Cantor's case, no sequence processing or limit
processing need be invoked, as each comparison of a Cantor number with
a member of the list can be carried out independently of any other such
comparison.

And, in fact, all of the Cantor comparisons are proved simultaneously.
From: Virgil on 9 Jul 2006 15:24
In article <1152450584.362072.151080(a)m73g2000cwd.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:


> Indeed, if you consider 1.000... - 0.999... = 0 and if you claim that
> all positions are enumerated by natural numbers, then there must be
> natural number n with 10^(-n) = 0. Hence, this assumption is obviously
> wrong.

That depends entirely n what one means by 0.999..., or, in "mueckenh"'s
case, whether one has the delusion that there is a largest natural
number.
From: Virgil on 9 Jul 2006 15:29
In article <1152450704.938408.69690(a)h48g2000cwc.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Virgil schrieb:
>
> > What we have said, and what is quite true, is that for every natural
> > there is a larger natural.
>
> Of course. And therefore it is impossible to exhaust all of them or to
> find a set which is larger than all naturals together.

Except in such systems as ZF, ZFC and NBG where there is an axiom
requiring that such sets exist.

And until "mueckenh", or someone else, can find some internal
contradiction within one of these systems, they remain eminently usable,
and will be the most common standards on which set theories are based.
From: mueckenh on 9 Jul 2006 17:11

Virgil schrieb:

> > > >
> > > > |
> > > > o
> > > > / \
> > > > o o
> > > > / \ / \
> > > >
> > >
> > > When every path contains more than any finite number of edges, then
> > > there are "more" paths than edges.
>
> >
> > Then there are many edges, of course. But why should there be many more
> > paths?
>
> Because one can biject the set of edges with the naturals and biject the
> set of paths with the power set of the naturals, and there are "more"
> members in any power set that in the set itself.

Please note: That fact (if it was a fact) would show nothing but a
contradiction of set theory as you may obtain from he diagram above.

No two paths can be generated without two new edges. Stop to run
through the world of sets with your eyes wide shut.
>

> It is only possible to have more edges than paths when one cannot biject
> the set of paths with the power set of the set of edges. No finite tree
> represents what happens in an infinite tree.

What happens in an any part of the infinite tree is precisely described
in the diagram. No two paths can be generated without two new edges.
Stop to run through the world of sets with your eyes wide shut.

> If "mueckenh" wishes to claim more edges than paths, or even as many
> edges as paths for such ->infinite<- trees, he must either construct an
> explicit injection from the set of paths to the set of edges or a
> surjection in the reverse direction

Here it is. Every edge is mapped on that path which runs through it.
When the path splits in two paths, one half of the former edge and the
new edge are mapped on each of the new paths. Continuing, we have two
edges mapped on every infinite path. This is a surjective mapping of
edges on paths. Or are you unable to do mathematics with fractions? If
not: The number of edges per path is calculated by 1 + 1/2 + 1/4 + ...
= 2.
>
> Then "mueckenh" is claiming a surjection from edges to paths.
>
> While this is quite possible for finite binary trees in which there is a
> finite maximal length to each path, and each path has a ->last<- edge,
> it is not for infinite binary trees in which each path contains
> infinitely many edges and no path contains any last edge.

Do you believe that the infinite paths exist there without consisting
of edges? Do you believe that above diagram becomes invalid somewhere?
This kind of collective naive blindness is required only by
set-theorists.

Regards, WM

From: mueckenh on 9 Jul 2006 17:13

Virgil schrieb:

> > If Cantor's proof fails once, it fails once and for all, because it is
> > an impossibility proof. Due to the claim it is impossible to have the
> > diagonal in the list.
>
> Precisely where does the Cantor diagonal proof fail?

In this and infinitely many infinite lists like this:
0.0
0.1
0.11
0.111
....

In an infinitely long list the diagonal number 0.111... is not
distingiushed from all the list numbers.

It is by far more safe to believe that there are more edges than paths
in the tree than to believe that 0.111... is distinct from every number
of he infinite list.

Regards, WM

First  |  Prev  |  Next  |  Last
Pages: 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111
Prev: integral problem
Next: Prime numbers