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 Axiom of Choice

For every set of non-empty sets, it is possible to provide a mechanism for choosing one element of each element of the set.

$\ds \forall s: \paren {\O \notin s \implies \exists \paren {f: s \to \bigcup s}: \forall t \in s: \map f t \in t}$

That is, one can always create a choice function for selecting one element from each element of the set.

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 $\CC$ be the set of all chains in $X$.

By the premise, each element of $\CC$ has an upper bound in $X$.

Thus by the Axiom of Choice, there is a mapping $g: \CC \to X$ such that for each $C \in \CC$, $\map g C$ is an upper bound of $C$.

Let $p$ be an arbitrary element of $X$.

Define a mapping $h: \On \to X$ by transfinite recursion thus:

 $\ds \map h 0$ $=$ $\ds p$ $\ds \map h {\alpha^+}$ $=$ $\ds \map f {\map h \alpha}$ $\ds \map h \lambda$ $=$ $\ds \map f {\map g {h \sqbrk \lambda} }$ if $\lambda$ is a limit ordinal

when $h \sqbrk \lambda$ is the image set of $\lambda$ under $h$.

Then $h$ is strictly increasing, and thus injective.

Let $h'$ be the restriction of $h$ to $\On \times \map h \On$.

Then ${h'}^{-1}$ is a surjection from $\map h \On \subseteq X$ onto $\On$.

By the Axiom of Replacement, $\On$ is a set.

Thus we conclude that some element of $X$ has no strict successor, and is thus maximal.
$\blacksquare$