The "why is that?" is simple. Your post is nonsense. You are stuck on a falsehood that you simply made up. Cantor's theorem does do what you claim. The mappings you've presented as bijective aren't.
Here. Since you can't seem to find a copy of it, here's a proof for Cantor's Theorem. Please point out the construction method you claim it has for making these mappings that aren't bijective.
Cantor's Theorem: For any set, S, and its power set, P(S):
|S| < |P(S)|
where the cardinality ordering is based on mapping functions: |A| <= |B| iff there exists an injective mapping from A to B, and |A| = |B| iff there exists a bijective (injective and surjective) mapping from A to B.
Proof:
There is a trivial injective mapping, f(X)->{X}, from S to P(S), so we know |S| <= |P(S)|. It is sufficent, then, to show there is no surjective mapping, g(), from S to P(S).
1. Assume g() exists. That is, for any B in P(S) there is an A in S such that g(A) -> B.
2. Let C = { X in S | X not in g(X) }.
3. C is a subset of S, and therefore C is in P(S).
4. Since g() is surjective from S to P(S), there must exist some Y in S such that g(Y) -> C.
5. Either Y is in C or Y is not in C. If Y is in C then Y is in g(Y) and, by definition of C, Y cannot be in C. So, Y cannot be in C. Then, Y is not in g
, and therefore, by definition of C, Y must be in C.
6. The existence of Y leads to a contradiction. Therefore, the assumption that g() exists leads to a contradiction. Therefore, g() does not exist.
7. Since there is no surjective mapping from S to P(S), |S| must be strictly less than |P(S)|.
