Zermelo's Well-Ordering Theorem

Theorem
Let the Axiom of Choice be accepted.

Then every set is well-orderable.

Proof
Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

By the Axiom of Choice, there is a choice function $c$ defined on $\powerset S \setminus \set \O$.

We will use $c$ and the Principle of Transfinite Induction to define a bijection between $S$ and some ordinal.

Intuitively, we start by pairing $\map c S$ with $0$, and then keep extending the bijection by pairing $\map c {S \setminus X}$ with $\alpha$, where $X$ is the set of elements already dealt with.

Basis for the Induction
$\alpha = 0$

Let $s_0 = \map c S$.

Inductive Step
Suppose $s_\beta$ has been defined for all $\beta < \alpha$.

If $S \setminus \set {s_\beta: \beta < \alpha}$ is empty, we stop.

Otherwise, define:
 * $s_\alpha := \map c {S \setminus \set {s_\beta: \beta < \alpha} }$

The process eventually stops, else we have defined bijections between subsets of $S$ and arbitrarily large ordinals.

Now, we can impose a well-ordering on $S$ by embedding it via $s_\alpha \to \alpha$ into the ordinal $\beta = \ds {\bigcup_{s_\alpha \mathop \in S} \alpha}$ and using the well-ordering of $\beta$.

Also see

 * Well-Ordering Theorem is Equivalent to Axiom of Choice