equivalence of Zorn’s lemma and the axiom of choice

Let $X$ be a set partially ordered by such that each chain has an upper bound. Define .Let $p(X)=\{p(x)\mid x\in X\}$. If $p(x)=\mathrm{\varnothing }$ then it follows that $x$ is maximal.

Suppose no $p(x)=\mathrm{\varnothing }$. Then by the axiom of choice (http://planetmath.org/AxiomOfChoice) there is a choice function $f$ on $p(X)$, and since for each $p(x)$ we have $f(p(x))\in p(x)$, it follows that . Define $f_{\alpha }(p(x))$ for all ordinals $\alpha$ by transfinite induction:

$f_0(p(x))=x$

$f_{\alpha +1}(p(x))=f(p(f_{\alpha }(p(x))))$

And for a limit ordinal $\alpha$, let $f_{\alpha }(p(x))$ be an upper bound of $f_i(p(x))$ for .

This construction can go on forever, for any ordinal. Then we can easily construct an injective function from $Ord$ to $X$ by $g(\alpha )=f_{\alpha }(p(x))$ for an arbitrary $x\in X$. This must be injective, since implies . But that requires that $X$ be a proper class, in contradiction to the fact that it is a set. So there can be no such choice function, and there must be a maximal element of $X$.

For the reverse, assume Zorn’s lemma and let $C$ be any set of non-empty sets. Consider the set of functions $F=\{f\mid \forall a\in \mathrm{dom}(f)(a\in C\wedge f(a)\in a)\}$ partially ordered by inclusion. Then the union of any chain in $F$ is also a member of $F$ (since the union of a chain of functions is always a function). By Zorn’s lemma, $F$ has a maximal element $f$, and since any function with domain smaller than $C$ can be easily expanded, $\mathrm{dom}(f)=C$, and so $f$ is a choice function for $C$.