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Homeproof of Zermelo's well-ordering theorem

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# proof of Zermelo’s well-ordering theorem

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

$i(\beta)=f(X-\bigcup_{{\gamma<\beta}}\{i(\gamma)\})\text{ unless }X-\bigcup_{{% \gamma<\beta}}\{i(\gamma)\}=\emptyset\text{ or }i(\gamma)\text{ is undefined % for some }\gamma<\beta$ |

(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 $X-\bigcup_{{\gamma<\alpha}}\{i(\gamma)\}=\emptyset$, and therefore for every $x\in X$ there is some $\gamma<\alpha$ 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 $x<_{X}y\leftrightarrow i^{{-1}}(x)<i^{{-1}}(y)$.

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 $a$ under that well ordering.

## Mathematics Subject Classification

03E25*no label found*

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