Das Kalenderblatt 100227

26/02/2010 - 12:08 von WM | Report spam
A mistake in Cantor’s proof on cardinality of power sets is shown.

[...] Theorem. The set P(A) of all subsets of a set A has a larger
cardinality than A.
Proof. Suppose they have the same number of elements.
Let b : A --> P(A) be a bijection between A and P(A).
(1) Let T = {x in A|x is not element of b(x)}.
Since T is a subset of P(A) and b is onto,
(2) T = b(t) for some t.
Thus t is in b(t) iff (by 2) t is in T iff (by 1) t is not in b(t).
This is a contradiction. Since we cannot do a one-to-one
correspondence between A and P(A), and since P(A) cannot be smaller
than A, the only possible conclusion is that P(A) must be larger than
A.

[...] Let “x is a set that is not member of x” and O the set that it
determines, set O will be member of O if and only if it satisfies the
function that determines O, that is,if and only if O is not member of
O, what is a contradiction.
This contradiction is known as the paradox of the sets or, more
generally, as Russell’s paradox. To examine the contradiction we will
use the usual definitions:
Definition 3.1 A set is ordinary if it is not member of itself.
Definition 3.2 An extraordinary set is a set that belongs to itself.
Let us consider the following work hypothesis: given any property,
there is a set of all things that have this property. Let the property
of being an ordinary set. Applying the work hypothesis, we have a set
of all ordinary sets; let us denominate H the set. Let us now consider
the following argument:
(a) If H does not belong to H, then H is an ordinary set, but then H
does not contain all the ordinary sets.
(b) If H belongs to H, then H is an extraordinary set, but then H
contains a set that is not ordinary.
There are two necessary conditions for H: (i) to contain all ordinary
sets; and (ii) not to contain any extraordinary set. From (a) and (b),
we see that the conditions are not satisfied: the set H cannot belong
to H to exclude extraordinary set in H, and H cannot exclude H to
contain all the ordinary sets of U. If H exists, then S c H where S is
an ordinary set. A non-empty set cannot satisfy both conditions (i)
and (ii). Therefore H does not exist.

[...] Let us consider a non-empty proper subset of A, denoted by B,
such that their elements satisfy “x is not in b(x)” where b : B --> W
and W is a subset of P(A). If T of Cantor’s theorem exists, then B c T
and, therefore, T is not empty.
The set T is the set of the elements x of A such that “x is not in
b(x)” (the elements of T are the elements of A which separately
satisfy the property “x is not in b(x)”) and
“T contains the element a satisfying b(a) = T and T does not contain
the element a of A satisfying b(a) = T” [*]
A non-empty set cannot satisfy the condition [*]. Therefore T does not
exist. As T does not exist, the contradiction of Cantor’s theorem
[...] does not exist and we cannot conclude |P(A)| > |A|.

[J. C. Ferreira: "On power sets", (2001)]
http://arxiv.org/PS_cache/math/pdf/...1291v3.pdf

Gruß, WM
 

Lesen sie die antworten

#1 Roalto
27/02/2010 - 13:08 | Warnen spam
On Fri, 26 Feb 2010 03:08:17 -0800 (PST), WM
wrote:
[Snip]

[J. C. Ferreira: "On power sets", (2001)]
http://arxiv.org/PS_cache/math/pdf/...1291v3.pdf


War ja klar. Wieder was aus der GM-Müllkippe.
Lieber, guterr Herr Professor, mal ne Frage:

Diese Zahl 0,1212.. (..bedeutet Periode(12), also 4/33 in
Dezimalschreibweise)
enthàlt wieviel 1sen und wieviel 2en?

Gruß, WM


Viel Spass weiterhin
Rolf
Wo Frauen geehrt werden,
sind die Götter zufrieden.

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