Cantor's Theorem/Proof 2

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.

Let $S$ be a set.

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

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

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

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:

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

Suppose that:

$a \in \map f a$

Then by the definition of $T$:

$\neg \paren {a \in T}$

Thus since $T = \map f a$:

$\neg \paren {a \in \map f a}$

$(1): \quad a \in \map f a \implies \neg \paren {a \in \map f a}$

Suppose instead that:

$\neg \paren {a \in \map f a}$

Then by the definition of $T$:

$a \in T$

Thus since $T = \map f a$:

$a \in \map f a$

$(2): \quad \neg \paren {a \in \map f a} \implies a \in \map f a$

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 $\exists a \in S: T = \map f a$ must be false.

It follows that $f$ is not a surjection.

$\blacksquare$

2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $2.10$
2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text{A}$: Set Theory: Cantor's Theorem