TITLE: Zorn's Lemma And Axiom of Choice
QUESTION [25 upvotes]: How can I prove Zorn's lemma is equivalent to Axiom of choice?

REPLY [32 votes]: Assume the axiom of choice. Let $(P,\le)$ a partially ordered set that every chain has an upper bound. Let $f$ to be a choice function from all non-empty subsets of $P$, and let $P_a = \{x\in P\mid a<x\}$ then $P_a=\varnothing$ if and only if $a$ is a maximal element.
We define by using transfinite induction:
Let $a_0$ be an element of $P$, if it is maximal then we have finished. Otherwise $P_0 = \{x\in P\mid a_0<x\}$ is non-empty, let $a_1 = f(P_0)$.
Suppose $a_\alpha$ was defined, if it is maximal then we are done, otherwise $P_\alpha=\{x\in P\mid a_\alpha<x\}$ is non-empty and $a_{\alpha+1} = f(P_\alpha)$.
If $\alpha$ is a limit ordinal, and for all $\beta<\alpha$ we have chosen $a_\beta$, then $\{a_\beta\mid \beta<\alpha\}$ is a chain without a maximal element in $P$, and therefore bounded with an upper bound above all the elements of the chain, thus $\bigcap_{\beta<\alpha} P_\beta\neq\varnothing$, and let $a_\alpha=f(\bigcap_{\beta<\alpha} P_\beta)$.
We argue that this must stop at some point, otherwise we have an injection from the proper class of the ordinals into the set $P$. Why did the process stop? It can only stop if we have reached a maximal element at some $a_\gamma$, as wanted.

Assume Zorn's lemma, and let $X$ be a non-empty collection of non-empty sets. Let $(C,\subseteq)$ be the set of all choice functions from some elements of $X$.
This is a non-empty set, since we can always choose from finitely many sets. Given a chain of choice functions, the union is indeed a function. So the condition of Zorn's lemma is satisfied. We have a maximal element $f$.
If $f$ is not a choice function from all the members of $X$ then we can extend it to one more element, which contradicts the maximality.