Zermelo's Well-Ordering Theorem

Theorem
Every set is well-orderable.

Proof
This proof makes use of the Axiom of Choice. In fact, the Well-Ordering Theorem is equivalent to the Axiom of Choice.

Let $S$ be a set, and let $\mathcal{P}(S)$ be the power set of $S$.

By the Axiom of Choice, there is a choice function $c$ defined on $\mathcal{P}(S)-\{\emptyset\}$. We will use this choice function and transfinite induction to define a bijection between $S$ and some ordinal. Intuitively, we start by pairing $c(S)$ with $0$, and then keep extending the bijection by pairing $c(S - X)$ with $\alpha$, where $X$ is the set of elements we've dealt with already.


 * Base case: $\alpha = 0$

Let $s_0 = c(S)$.


 * Inductive step: Suppose $s_\beta$ has been defined for all $\beta < \alpha$.

If $S - \{s_\beta \mid \beta < \alpha\}$ is empty, we stop.

Otherwise, define $s_\alpha = c(S-\{s_\beta \mid \beta < \alpha\})$.

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 $s_\alpha \mapsto \alpha$ into the ordinal $\beta = \displaystyle{\bigcup_{s_\alpha \in S} \alpha}$ and using the well-ordering of $\beta$.