Cantor's Theorem/Proof 1

Proof
$S$ is a set with a surjection $f: S \to \powerset S$.

Then:

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


 * $x \in \map f x$
 * $x \notin \map f x$

Let $T = \set {x \in S: x \notin \map f x}$.

As $f$ is supposed to be a surjection, $\exists a \in S: T = \map f a$.

Thus:
 * $a \in \map f a \implies a \notin \map f a$
 * $a \notin \map f a \implies a \in \map f a$

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