Matheology § 111

[WM]

Matheology § 111

Let m and n be two different characters, and consider a set M of
elements

E = (x_1, x_2, … , x_nu, …)

which depend on infinitely many coordinates x_1, x_2, … , x_nu, …, and
where each of the coordinates is either m or w. Let M be the totality
of all elements E.

To the elements of M belong e.g. the following three:

E^I = (m, m, m, m, … ),
E^II = (w, w, w, w, … ),
E^III = (m, w, m, w, … ).

I maintain now that such a manifold M does not have the power of the
series 1, 2, 3, …, nu, ….

This follows from the following proposition:

"If E_1, E_2, …, E_nu, … is any simply infinite series of elements of
the manifold M, then there always exists an element E_0 of M, which is
not equal to any element E_nu."

For proof, let there be

E_1 = (a_1,1, a_1,2, … , a_1,nu, …)
E_2 = (a_2,1, a_2,2, … , a_2,nu, …)
...
E_mu = (a_mu,1, a_mu,2, … , a_mu,nu, …)
...

where the characters a_mu,nu are either m or w. Then there is a
series b_1, b_2, … b_nu,…, defined so that b_nu is also equal to m or
w but is different from a_nu,nu.

Thus, if a_nu,nu = m, then b_nu = w, and if a_nu,nu = w, then b_nu =
m,

Then consider the element

E_0 = (b_1, b_2, b_3, …)

of M, then one sees straight away, that the equation

E_0 = E_mu

cannot be satisfied by any positive integer mu, otherwise for that mu
and for all values of nu

b_nu = a_mu,nu

and so we would in particular have

b_mu = a_mu,mu

which through the definition of b_nu is impossible. From this
proposition it follows immediately that the totality of all elements
of M cannot be put into the sequence: E_1, E_2, …, E_nu, … otherwise
we would have the contradiction, that a thing E_0 would be both an
element of M, but also not an element of M.

[G. Cantor: "Über eine elementare Frage der Mannigfaltigkeitslehre",
Jahresbericht der DMV 1 (1890-91) 75-78]

A proof by contradiction fails, if only one counter example can be
found. Here it is:

Consider the sequence

E_1 = (w, m, m, m, m, m, m, ...)
E_2 = (m, w, m, m, m, m, m, ...)
E_3 = (m, m, w, m, m, m, m, ...)
E_4 = (m, m, m, w, m, m, m, ...)

This matrix formed by the a_mu,nu has no line mu and no column nu with
all characters a_mu,nu = m. (Since such a line would need infinitely
many predecessors, namely all lines with a finite number of m, it
cannot belong to the sequence. It is the limit of the sequence.)

Define E_0 = (b_1, b_2, b_3, …) by b_mu = m =/= w = a_mu,mu.

The first mu characters of E_0 agree with the first mu characters of
all E_nu for all nu > mu. Since there is no last mu and no last nu,
this situation does never change. Otherwise we would have the
contradiction that a matrix has more* characters m on the diagonal
than it has in any line and in any column**.

*) i.e. a number of m before the first w that is larger than every
finite number.
**) where the number of m before the first w is always finite.

Regards, WM

[Uirgil]

In article
<8f500670-73f5-4432...@h16g2000vby.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Matheology § 111
>
> Let m and n be two different characters, and consider a set M of
> elements
>

> E = (x_1, x_2, Š , x_nu, Š)
>
> which depend on infinitely many coordinates x_1, x_2, Š , x_nu, Š, and

> where each of the coordinates is either m or w. Let M be the totality
> of all elements E.
>
> To the elements of M belong e.g. the following three:
>

> E^I = (m, m, m, m, Š ),
> E^II = (w, w, w, w, Š ),
> E^III = (m, w, m, w, Š ).

>
> I maintain now that such a manifold M does not have the power of the

> series 1, 2, 3, Š, nu, Š.

>
> This follows from the following proposition:
>

> "If E_1, E_2, Š, E_nu, Š is any simply infinite series of elements of

> the manifold M, then there always exists an element E_0 of M, which is
> not equal to any element E_nu."
>
> For proof, let there be
>

> E_1 = (a_1,1, a_1,2, Š , a_1,nu, Š)
> E_2 = (a_2,1, a_2,2, Š , a_2,nu, Š)
> ...
> E_mu = (a_mu,1, a_mu,2, Š , a_mu,nu, Š)

> ...
>
> where the characters a_mu,nu are either m or w. Then there is a

> series b_1, b_2, Š b_nu,Š, defined so that b_nu is also equal to m or

> w but is different from a_nu,nu.
>
> Thus, if a_nu,nu = m, then b_nu = w, and if a_nu,nu = w, then b_nu =
> m,
>
> Then consider the element
>

> E_0 = (b_1, b_2, b_3, Š)

>
> of M, then one sees straight away, that the equation
>
> E_0 = E_mu
>
> cannot be satisfied by any positive integer mu, otherwise for that mu
> and for all values of nu
>
> b_nu = a_mu,nu
>
> and so we would in particular have
>
> b_mu = a_mu,mu
>
> which through the definition of b_nu is impossible. From this
> proposition it follows immediately that the totality of all elements

> of M cannot be put into the sequence: E_1, E_2, Š, E_nu, Š otherwise

> we would have the contradiction, that a thing E_0 would be both an
> element of M, but also not an element of M.
>
> [G. Cantor: "Über eine elementare Frage der Mannigfaltigkeitslehre",
> Jahresbericht der DMV 1 (1890-91) 75-78]
>
> A proof by contradiction fails, if only one counter example can be
> found. Here it is:
>
> Consider the sequence
>
> E_1 = (w, m, m, m, m, m, m, ...)
> E_2 = (m, w, m, m, m, m, m, ...)
> E_3 = (m, m, w, m, m, m, m, ...)
> E_4 = (m, m, m, w, m, m, m, ...)

It is trivial that this sequence of sequences does not exhaust the set
of such sequences, as it contains neither
(m, m, m, m, m, m, m, ...) nor (w, w, w, w, w, w, w, ...)

So that WM's counterclaim that a sequence of such sequences can be
exhaustive is still false.