Zermelo's Well-Ordering Theorem

Theorem
Every set is well-orderable.

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

By the Axiom of Choice, there is a choice function $c$ defined on $\mathcal P \left({S}\right) - \left\{{\varnothing}\right\}$.

We will use this choice function and transfinite induction to define a bijection between $S$ and some ordinal.

Intuitively, we start by pairing $c \left({S}\right)$ with $0$, and then keep extending the bijection by pairing $c \left({S - X}\right)$ with $\alpha$, where $X$ is the set of elements we've dealt with already.


 * Base case: $\alpha = 0$

Let $s_0 = c \left({S}\right)$.


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

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

Otherwise, define:
 * $s_\alpha := c \left({S - \left\{{s_\beta: \beta < \alpha}\right\} }\right)$.

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

Alternative Name
This result is also known as Zermelo's Theorem, for Ernst Zermelo.

Under this name it can often be seen worded:


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

In fact, the Well-Ordering Theorem is equivalent to the Axiom of Choice.