Cantor's Theorem

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Theorem

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


Proof 1

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

Then:

\(\displaystyle \forall x \in S: \map f x\) \(\in\) \(\displaystyle \powerset S\) $\quad$ By Hypothesis $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall x \in S: \map f x\) \(\subseteq\) \(\displaystyle S\) $\quad$ Definition of Power Set $\quad$


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.

$\blacksquare$


Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle.

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

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.


Proof 2

Let $S$ be a set.

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

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.

Aiming for a contradiction, suppose 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.

It follows that $f$ is not a surjection.

$\blacksquare$


Also see

Compare this with Russell's Paradox.


Source of Name

This entry was named for Georg Cantor.


Sources