Zorn's Lemma Implies Zermelo's Well-Ordering Theorem
Theorem
Zorn's Lemma implies Zermelo's Well-Ordering Theorem.
Proof
Let $X$ be a set.
If $X = \O$ the theorem holds vacuously.
Assume $X$ is not empty.
Let $\WW$ be the collection of pairs $\tuple {W, \preceq}$ such that:
- $W \subseteq X$
- $\preceq$ well-orders $W$
Next, define the partial ordering $\preccurlyeq$ on $\WW$ by $\tuple {W, \preceq} \preccurlyeq \tuple {W', \preceq'}$ if and only if:
- $W \subseteq W'$
- $\preceq$ is the restriction of $\preceq'$ to $W$
- For all $w \in W$ and $w' \in W' \setminus W$: $w \preceq' w'$
To apply Zorn's Lemma, we need to show that every chain in $\WW$ has an upper bound.
Let $\CC \subseteq \WW$ be such a chain.
Then we claim $\bigcup \CC \in \WW$ and $\bigcup \CC$ is an upper bound for $\CC$, where we define $\bigcup \CC$ as:
- $\ds \bigcup \CC := \tuple { \bigcup_{ \tuple{W, \preceq} \in \CC } W, \bigcup_{ \tuple{W, \preceq} \in \CC } \preceq }$
First to show $\bigcup \CC \in \WW$.
By Union of Subsets is Subset: Set of Sets:
- $\ds \bigcup_{\tuple{W, \preceq} \mathop \in \CC} W \subseteq X$
Next, let $S \subseteq \bigcup_{\tuple {W, \preceq} \mathop \in \CC} W$ be non-empty.
Fix $s \in S$.
Then for some $\tuple {W, \preceq} \in \bigcup \CC$, we have $s \in W$.
Let $w$ be the smallest element of $S \cap W$ with respect to $\preceq$.
Now let $s' \in S$ be arbitrary.
If $s' \in W$, then $w \preceq s'$ and therefore $w \preceq_{\bigcup \CC} s'$.
If $s' \notin W$, then $s' \in W'$ for some $\tuple {W', \preceq'} \in \bigcup \CC$.
But since $\CC$ is a chain, we have:
- $\tuple {W, \preceq} \preccurlyeq \tuple {W', \preceq'}$
and therefore, $w \preceq' s'$ since $w \in W$ and $s' \in W' \setminus W$.
Then, by definition, $w \preceq_{\bigcup \CC} s'$.
Thus $w$ is the smallest element of $S$ and we conclude $\bigcup \CC$ is well-ordered.
Therefore $\bigcup \CC \in \WW$.
Now to show $\bigcup \CC$ is the sought upper bound of $\CC$.
We observe, for all $\tuple {W, \preceq} \in \CC$:
- $W \subseteq \bigcup_{\tuple {W, \preceq} \mathop \in \CC} W$ by Set is Subset of Union
- $\preceq$ is the restriction of $\preceq_{\bigcup \CC}$ to $W$
- For all $w \in W$ and $w' \in \bigcup \CC \setminus W$, $w \preceq_{\bigcup \CC} w'$
and conclude that:
- $\ds \tuple {W, \preceq} \preccurlyeq \bigcup \CC$
That is, $\ds \bigcup \CC$ is an upper bound of $\CC$.
Thus the hypotheses of Zorn's Lemma hold and we can conclude that $\WW$ has a maximal element.
Let $\tuple {E, \preceq}$ be the maximal element of $\WW$.
Suppose that $E \ne X$.
Then there exists $x_0 \in X \setminus E$.
Define an ordering $\preceq'$ on $E \cup \set {x_0}$ as follows:
- $y \preceq' x$ if and only if $x = x_0$ or $x, y \in E$ and $y \preceq x$.
This is a well-order on $E \cup \set {x_0}$ with:
- $\struct {E, \preceq} \preccurlyeq \struct {E \cup \set {x_0}, \preceq'}$
This contradicts that $\tuple {E, \preceq}$ is the maximal element of $\WW$.
So $E = X$.
Hence $X$ is well-orderable.
$\blacksquare$
Sources
- 1984: Gerald B. Folland: Real Analysis: Modern Techniques and their Applications: $(\text P.3)$