Axiom of Choice Implies Zorn's Lemma/Proof 2
Theorem
Acceptance of the Axiom of Choice implies the truth of Zorn's Lemma.
Statement of Zorn's Lemma
Let $\struct {X, \preceq}, X \ne \O$ be a non-empty ordered set such that every non-empty chain in $X$ has an upper bound in $X$.
Then $X$ has at least one maximal element.
Proof
Aiming for a contradiction, suppose that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.
By the Axiom of Choice, there is a mapping $f: X \to X$ such that:
- $\forall x \in X: x \prec \map f x$
Let $\mathcal C$ be the set of all chains in $X$.
By the premise, each element of $\mathcal C$ has an upper bound in $X$.
Thus by the Axiom of Choice, there is a mapping $g: \mathcal C \to X$ such that for each $C \in \mathcal C$, $\map g C$ is an upper bound of $C$.
Let $p$ be an arbitrary element of $X$.
Define a mapping $h: \operatorname {Ord} \to X$ by transfinite recursion thus:
\(\ds \map h 0\) | \(=\) | \(\ds p\) | ||||||||||||
\(\ds \map h {\alpha^+}\) | \(=\) | \(\ds \map f {\map h \alpha }\) | ||||||||||||
\(\ds \map h \lambda\) | \(=\) | \(\ds \map f {\map g {f \sqbrk \lambda} }\) | if $\lambda$ is a limit ordinal |
Then $h$ is strictly increasing, and thus injective.
Let $h'$ be the restriction of $h$ to $\operatorname {On} \times \map h {\operatorname{On} }$.
Then ${h'}^{-1}$ is a surjection from $\map h {\operatorname{On} } \subseteq X$ onto $\operatorname{On}$.
By the Axiom of Replacement, $\operatorname{On}$ is a set.
By Burali-Forti Paradox, this is a contradiction.
Thus we conclude that some element of $X$ has no strict successor, and is thus maximal.
$\blacksquare$