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 $p(x)=\{y\in X\mid x. Let $p(X)=\{p(x)\mid x\in X\}$. If $p(x)=\emptyset$ then it follows that $x$ is maximal.

Suppose no $p(x)=\emptyset$. 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 $x. 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 $i<\alpha$.

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 $\alpha<\beta$ implies $g(\alpha). 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 $F=\{f\mid\forall a\in\operatorname{dom}(f)(a\in C\wedge f(a)\in a)\}$ 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, $\operatorname{dom}(f)=C$, and so $f$ is a choice function for $C$.

Title equivalence of Zorn’s lemma and the axiom of choice EquivalenceOfZornsLemmaAndTheAxiomOfChoice 2013-03-22 12:59:04 2013-03-22 12:59:04 Henry (455) Henry (455) 10 Henry (455) Proof msc 03E25