|P(Z*)| > |Z*|

Let us explain Cantor's proof about |P(Z*)| > |Z*|.

At the first step of his proof, Cantor shows that there is injection between P(Z*) and Z*, so after this step we know
that |P(Z*)| > = |Z*|, or in other words, |P(Z*)| cannot be less than |Z*|.

In order to clearly show that |P(Z*)| > |Z*|, we have to show that |P(Z*)| not= |Z*|.

If we show that |P(Z*)| not= |Z*|, than and only than we have no choice but to conclude that |P(Z*)| > |Z*|.

It can be done only if we can show that there is no bijection (no no_surjection but no bijection) between P(Z*) and Z*.

Cantor tried to show, by using S definition, that any mapping between some arbitrary z in Z* and S, leading us to contradiction, and since this is the case, we can clearly conclude that there is no such z in Z* that is mapped with S (because of S definition) and we prove that |P(Z*)| > |Z*|.

But I show that S is based on a definition that is a self-contradiction, and therefore S does not exist.

My argument goes like this:

1) Any set has unique members, for example: {a,a,b,c,…} = {a,b,c,…}, so S has unique members.

2) Any z in Z* is mapped once and only once with some P(Z*) member.

3) By (1) and (2), when t (which is some arbitrary z in Z*) map is checked against S, we know that t is not in S, but by S definition, t must be in S, but if t is in S than by S definition t must not be in S, … etc.

4) By S definition itself, this set MUST include ALL of z in Z* members that are paired with P(Z*) members that do not contain them (and S is a P(Z*) member; therefore by (1) and (2), it MUST include t within it, but it cannot be done because of the same S definition).

5) If t is in S, then by S definition we can conclude that the term ALL = more_than_ALL.

6) If t is not in S, then by S definition we can conclude that the term ALL = less_than_ALL.

7) By (5) and (6) we can conclude that the term ALL = not_ALL, which is a contradiction in Excluded-Middle logical reasoning.

Since (by (7)) S definition is based on the term ALL = not_ALL, S definition does not exist, and Cantors proof by contradiction does not hold.

Please pay attention that in this post I did not use the Hierarchy of dependency argument (which is another way to show why S does not exist).

In this post I used http://www.geocities.com/complementarytheory/Russell... , which clearly shows that Russell's first paradox is not a paradox in Excluded-Middle logical reasoning.