Date: Jan 30, 2013 3:57 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 203
On 30 Jan., 09:40, William Hughes <wpihug...@gmail.com> wrote:

> On Jan 30, 9:28 am, WM <mueck...@rz.fh-augsburg.de> wrote:

>

>

>

>

>

> > On 30 Jan., 00:16, William Hughes <wpihug...@gmail.com> wrote:

>

> > > On Jan 29, 10:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > > > On 29 Jan., 21:28, William Hughes <wpihug...@gmail.com> wrote:

>

> > > <snip>

>

> > > > > It does, however, imply that d in not one

> > > > > of the lines of the list L

>

> > > > For that sake you must check all lines. Can you check what is not

> > > > existing?

>

> > > So now your claim is

>

> > > We can know

>

> > > There does not exist a natural number n

> > > such that d is equal to the nth line

> > > of L

>

> > > but we cannot know

>

> > > d is not one of the lines of L

>

> > You are trying hard to misunderstand!

>

> Do you agree

>

> i. There does not exist a natural number n

> such that d is equal to the nth line

> of L

>

> and

>

> ii. d is one of the lines of L

>

> are mutually exclusive?-

In existing finite sets this is true. In actually infinite sets it is

not true, since there is no chance to prove something for an actually

infinite set, i.e. a number of lines that is larger than every natural

number. You will always fail as in the case tgo produce an actual

infinite sequence as the limit of

0.1

0.11

0.111

...

You can prove something for all natural numbers, but not for a larger

set.

Consider the Binary Tree. There you will find it confirmed: If there

are uncountably many paths, then they don't come into being in one of

the steps constructing the Binary Tree. So, if they are in the tree,

after all (steps), then they have crept in somehow.

Regards, WM