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 $\mathcal A$ 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 $\alpha: \left\{{s_1}\right\} \to \left\{{s_2}\right\}$ such that $\alpha \left({s_1}\right) = s_2$.

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

We can impose an ordering $\le$ on $\mathcal A$ 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 $\mathcal C$ be a chain in $\left({\mathcal A, \le}\right)$.

Then $\displaystyle \bigcup \left\{{\phi \in \mathcal C}\right\}$ is an upper bound of every $\phi \in \mathcal C$, and it lies in $\mathcal A$.

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

Let:
 * $M_1 = \left\{{s_1: \left({s_1, s_2}\right) \in M}\right\}$
 * $M_2 = \left\{{s_2: \left({s_1, s_2}\right) \in M}\right\}$

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

 * Well-Ordering Theorem