proof of Zermelo’s well-ordering theorem

Let $X$ be any set and let $f$ be a choice function on $𝒫(X)\setminus \{\mathrm{\varnothing }\}$. Then define a function $i$ by transfinite recursion on the class of ordinals as follows:

(the function is undefined if either of the unless clauses holds).

Thus $i(0)$ is just $f(X)$ (the least element of $X$), and $i(1)=f(X-\{i(0)\})$ (the least element of $X$ other than $i(0)$).

Define by the axiom of replacement $\beta =i^{-1}[X]=\{\gamma \mid i(\gamma )=x\text{ for some }x\in X\}$. Since $\beta$ is a set of ordinals, it cannot contain all the ordinals (by the Burali-Forti paradox).

Since the ordinals are well ordered, there is a least ordinal $\alpha$ not in $\beta$, and therefore $i(\alpha )$ is undefined. It cannot be that the second unless clause holds (since $\alpha$ is the least such ordinal) so it must be that , and therefore for every $x\in X$ there is some such that $i(\gamma )=x$. Since we already know that $i$ is injective, it is a bijection between $\alpha$ and $X$, and therefore establishes a well-ordering of $X$ by .

The reverse is simple. If $C$ is a set of nonempty sets, select any well ordering of $\bigcup C$. Then a choice function is just $f(a)=$ the least member of $\mathrm{a}$ under that well ordering.