Cantor's Theorem

Theorem
There is no surjection from a set $$S$$ to its power set for any set $$S$$.

Proof
Suppose $$S$$ is a set with a surjection $$f: S \to \mathcal P \left({S}\right)$$.

$$ $$

Now by the Law of the Excluded Middle, there are two choices for every $$x \in S$$:


 * $$x \in f \left({x}\right)$$
 * $$x \notin f \left({x}\right)$$

Let $$T = \left\{{x \in S: x \notin f \left({x}\right)}\right\}$$.

As $$f$$ is supposed to be a surjection, $$\exists a \in S: T = f \left({a}\right)$$.

Thus:
 * $$a \in f \left({a}\right) \implies a \notin f \left({a}\right)$$
 * $$a \notin f \left({a}\right) \implies a \in f \left({a}\right)$$

This is a contradiction, so the initial supposition, i.e. that there is such a surjection, must be wrong.

Comment
Compare this with Russell's Paradox.