Zermelo's Theorem (Set Theory)

Theorem
Every set of cardinals is well-ordered with respect to $\le$.

Proof
Let $S_1$ and $S_2$ be sets which are not empty.

Suppose there exists an injection $f: S_1 \to S_2$ and another injection $g: S_2 \to S_1$.

Then by the Cantor-Bernstein-Schröder Theorem there exists a bijection between $S_1$ and $S_2$ and by definition $S_1$ is equivalent to $S_2$.

Let $\AA$ be the set of invertible mappings $\phi: A \to B$ where $A \subseteq S_1$ and $B \subseteq S_2$.

Since $S_1$ and $S_2$ are not empty, $\exists s_1 \in S_1$ and $\exists s_2 \in S_2$.

Thus we can construct the mapping $\alpha: \set {s_1} \to \set {s_2}$ such that $\map \alpha {s_1} = s_2$.

This is trivially an invertible mapping, so $\AA$ is not empty.

We can impose an ordering $\le$ on $\AA$ by letting $\phi_1 \le \phi_2 \iff \phi_1 \subseteq \phi_2$, that is, if $\phi_2$ is an extension of $\phi_1$.

Let $\CC$ be a chain in $\struct {\AA, \le}$.

Then $\ds \bigcup \set {\phi \in \CC}$ is an upper bound of every $\phi \in \CC$, and it lies in $\AA$.

The conditions of Zorn's Lemma are satisfied, so we can find a maximal element $M$ in $\AA$.

Let:
 * $M_1 = \set {s_1: \tuple {s_1, s_2} \in M}$
 * $M_2 = \set {s_2: \tuple {s_1, s_2} \in M}$

We have that $M_1 \subsetneq S_1$ and $M_2 \subsetneq S_2$ both together contradict the fact that $M$ is maximal element.

Thus either $M_1 = S_1$ or $M_2 = S_2$, and possibly both.

Thus either:
 * $M$ is an injection of $S_1$ into $S_2$

or:
 * $M^{-1}$ is an injection of $S_2$ into $S_1$.

Thus either $S_1 \le S_2$ or $S_2 \le S_1$, and the result follows.

Also known as
This is called by some authors the Trichotomy Problem.

Also see

 * Zermelo's Well-Ordering Theorem