Zermelo's Well-Ordering Theorem/Proof 3

Proof
Let $S$ be a non-empty set.

Let $C$ be a choice function for $S$.

Let $A$ be an arbitrary set.

Let the mapping $h$ be defined as:


 * $\map h A = \begin {cases}

\map C {S \setminus A} & : A \subsetneqq S \\ x & : \text {otherwise} \end {cases}$

where $x$ is an arbitrary element such that $x \notin S$.

The latter is known to exist from Exists Element Not in Set.

From the Transfinite Recursion Theorem: Formulation $5$, there exists a mapping $F$ on the class of all ordinals $\On$ such that:


 * $\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$

It remains to show the following:


 * $(1): \quad \exists \delta \in \On: \map F \delta \notin S$


 * $(2): \quad$ Let $\beta$ be the smallest ordinal such that $\map F \beta \notin S$. Then:
 * $F \sqbrk \beta = S$
 * and:
 * $F \restriction \beta$ is injective
 * and so $\beta$ can be put into a one-to-one correspondence with $S$.

Hence $S$ can be well-ordered.