Cantor's Theorem/Proof 1

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.

Aiming for a contradiction, suppose $S$ is a set with a surjection $f: S \to \powerset S$.

Then:

$\ds \forall x \in S: \, $

$\ds \map f x$

$\in$

$\ds \powerset S$

$\ds \leadsto \ \ $

$\ds \forall x \in S: \, $

$\ds \map f x$

$\subseteq$

$\ds S$

Definition of Power Set

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.

$\blacksquare$

This proof depends on the Law of the Excluded Middle.

This is one of the logical axioms that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

This in turn invalidates this proof from an intuitionistic perspective.