Cantor's Theorem/Proof 3

Theorem

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

That is, $S$ is strictly smaller than its power set.

Proof

Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Let $f: S \to \powerset S$ be a mapping.

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

The set $T$ exists by the Axiom of Specification.

Then:

$T \subseteq S$

so $T \in \powerset S$ by the definition of power set.

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

Aiming for a contradiction, suppose:

$\exists a \in S: T = \map f a$

This means that $f$ maps $a$ to $T$.

We claim that $a \notin T$.

Aiming for a contradiction, suppose:

$a \in T$

By definition of $T$:

$a \notin \map f a$

This is a contradiction, since $T = \map f a$.

Thus by Proof by Contradiction, we conclude that:

$(1): \quad a \notin T$

Since $T = \map f a$, we get:

$a \notin \map f a$

By definition of $T$:

$(2): \quad a \in T$

By Principle of Non-Contradiction, $(1)$ and $(2)$ cannot both be true.

We have reached a contradiction.

Thus by Proof by Contradiction, the supposition that $\exists a \in S: T = \map f a$ must be false.

It follows that $f$ is not a surjection.

$\blacksquare$