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Subject: =?ISO-8859-1?Q?Matheology_=A7_111?=
From: WM <mueck...@rz.fh-augsburg.de>
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Matheology =A7 111

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

E =3D (x_1, x_2, =85 , x_nu, =85)

which depend on infinitely many coordinates x_1, x_2, =85 , x_nu, =85, 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  =3D (m, m, m, m, =85 ),
E^II =3D (w, w, w, w, =85 ),
E^III =3D (m, w, m, w, =85 ).

I maintain now that such a manifold M does not have the power of the
series 1, 2, 3, =85, nu, =85.

This follows from the following proposition:

"If E_1, E_2, =85, E_nu, =85 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 =3D (a_1,1, a_1,2, =85 , a_1,nu, =85)
E_2 =3D (a_2,1, a_2,2, =85 , a_2,nu, =85)
...
E_mu =3D (a_mu,1, a_mu,2, =85 , a_mu,nu, =85)
...

where the characters a_mu,nu are either m or w.  Then there is a
series b_1, b_2, =85 b_nu,=85, defined so that b_nu is also equal to m or
w but is different from a_nu,nu.

Thus, if a_nu,nu =3D m, then b_nu =3D w, and if a_nu,nu =3D w, then b_nu =
=3D
m,

Then consider the element

E_0 =3D (b_1, b_2, b_3, =85)

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

E_0 =3D E_mu

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

b_nu =3D a_mu,nu

and so we would in particular have

b_mu =3D 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, =85, E_nu, =85 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: "=DCber 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 =3D (w, m, m, m, m, m, m, ...)
E_2 =3D (m, w, m, m, m, m, m, ...)
E_3 =3D (m, m, w, m, m, m, m, ...)
E_4 =3D (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 =3D 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 =3D (b_1, b_2, b_3, =85) by b_mu =3D m =3D/=3D w =3D 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