Date: Jan 29, 2013 4:09 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 203
On 29 Jan., 09:54, William Hughes <wpihug...@gmail.com> wrote:

> On Jan 29, 9:33 am, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > "All" and "every" in impredicative statements about infinite sets.

>

> > Consider the following statements:

>

> > A) For every natural number n, P(n) is true.

> > B) There does not exist a natural number n such that P(n) is false.

> > C) For all natural numbers P is true.

>

> > A implies B but A does not imply C.

>

> Which is the point. Even though A

> does not imply C we still have

> A implies B.

>

> Let L be a list

> d the antidiagonal of L

> P(n), d does not equal the nth line of L

>

> We have (A)

>

> For every natural number n, P(n) is true.

>

> This implies (B)

>

> There does not exist a natural number n

> such that P(n) is false.

>

> In other words, there is no line of L that

> is equal to d.

And how can C be correct nevertheless? Because "For all" is

contradictory.

There is no natural number that finishes the set N.

There is no finished set N.

There is, in the list of all reminating decimals, no anti-diagonal,

that differs from all terminatig decimals at digits belonging to at

least one of these terminating decimals. Reason: The list is complete.

If you don't believe, consider the Binary Tree constructed from all

finite paths only.

Again, the only solution is, there is no complete set Q.

There is, in the construction of the complete Binary Tree, no node

that adds more than one path to the tree. Nevertheless the completely

constructed tree contains uncountably many paths. No reason to be

taken aback, at least a little bit?

Nevertheless, the steps of construction can be enumerated and

therefore can be considered as a list. In no line you find any

infinite path. But the complete list contains uncountably many

infinite paths - if such exist in the complete construction.

Regards, WM