Date: Jan 29, 2013 4:36 AM Author: mueckenh@rz.fh-augsburg.de Subject: Re: Matheology § 203 On 29 Jan., 10:18, William Hughes <wpihug...@gmail.com> wrote:

> On Jan 29, 10:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:

>

>

>

>

>

> > 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.

>

> B: There is no line of L that is equal to d

>

> does not imply

>

> C: For all n, line n is not equal to d.

>

> B correct does not mean "C correct nevertheless"-

But we know of cases where C is correct nevertheless. I quoted four of

them in the § 203. Or do you disagree to one of them?

In case you have forgotten the old discussion concerning the

configurations of the Binary Tree construction, here it is repeated:

The complete infinite binary tree is the limit of the sequence of its

initial segments B_k:

____________________

B_0 =

a0.

____________________

B_1 =

a0.

/

a1

____________________

B_2 =

a0.

/ \

a1 a2

____________________

...

____________________

B_k =

a0.

/ \

a1 a2

/ \ / \

...

....a_k

___________________

...

The structure of the Binary Tree excludes that there are any two

initial segments, B_k and B_(k+1), such that B_(k+1) contains two

complete infinite paths both of which are not contained in B_k.

Nevertheless the limit of all B_k is the complete binary tree

including all (uncountably many) infinite paths. Contradiction. There

cannot exist more than countably many infinite paths.

Alternative consideration: Obviously every B_k is finite. None does

contain any infinite path. The infinite paths come into the play only

after all B_k with k in N (by some unknown mechanism). If that is

possible, however, this mechanism can also act in Cantor's diagonal

proof such that the anti-diagonal enters the list only after all lines

at finite positions.

Regards, WM