# 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.

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

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 $T$:

- $\neg \left({a \in T}\right)$

Thus since $T = f \left({a}\right)$:

- $\neg \left({a \in \map f a }\right)$

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

Suppose instead that:

- $\neg \left({a \in \map f a}\right)$

Then by the definition of $T$:

- $a \in T$

Thus since $T = f \left({a}\right)$:

- $a \in \map f a$

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

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.

$\blacksquare$

## Sources

- 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