Cantor's Theorem/Proof 1

Proof
Aiming for a contradiction, suppose $S$ is a set with a surjection $f: S \to \mathcal P \left({S}\right)$.

Now by Law of 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 that there is such a surjection must be false.