Cantor's Theorem/Proof 2

Theorem
Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Then there is no surjection from $S$ onto $\mathcal P \left({S}\right)$.

Proof
Let $f: S \to \mathcal P \left({S}\right)$ be a mapping.

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

Then $T \subseteq S$, so $T \in \mathcal P \left({S}\right)$ by the definition of power set.

We will show that $T$ is not in the image of $f$ and therefore $f$ is not surjective.

Suppose for the sake of contradiction that $\exists a \in S: T = f \left({a}\right)$.

Suppose that $a \in f \left({a}\right)$.

Then by the definition of $T$, $\neg \left({a \in T}\right)$.

Thus since $T = f \left({a}\right)$, $\neg \left({a \in f \left({a}\right) }\right)$.

By Rule of Implication:


 * $(1) \quad a \in f \left({a}\right) \implies \neg \left({ a \in f \left({a}\right) }\right)$

Suppose instead that $\neg \left({a \in f \left({a}\right)}\right)$.

Then by the definition of $T$, $a \in T$.

Thus since $T = f \left({a}\right)$, $a \in f \left({a}\right)$.

By Rule of Implication:


 * $(2) \quad \neg \left({ a \in f \left({a}\right) }\right) \implies a \in f \left({a}\right)$

By Non-Equivalence of Proposition and Negation, applied to $(1)$ and $(2)$, this is a contradiction.

As the specific choice of $a$ did not matter, we derive a contradiction by Existential Instantiation.

Thus by Proof by Contradiction, the supposition that $\exists a \in S: T = f \left({a}\right)$ must be false, so $f$ is not a surjection.