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

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