Zermelo's Well-Ordering Theorem implies Hausdorff's Maximal Principle
Theorem
Let Zermelo's Well-Ordering Theorem hold.
Then Hausdorff's Maximal Principle holds.
Proof
Let $X$ be a non-empty set.
Let $X$ contain at least two elements.
Otherwise, any non-empty ordering on $X$ is trivially a maximal chain.
By Zermelo's Well-Ordering Theorem, $X$ can be well-ordered set.
Fix such a well-ordering.
Let $\le$ be any ordering on $X$.
Let $\map P {a, Y}$ be the predicate:
- $a$ is $\le$-comparable with every $y \in Y$.
Here, $a$ and $Y$ are bound variables.
That is, $\map P {a, Y}$ holds if, for every $y \in Y$, $a \le y$ or $y \le a$.
Define the mapping:
- $\map \rho {f: S_x \to \powerset X} = \begin {cases} f \sqbrk {S_x} \cup \set x & : \map P {x, S_x} \\ f \sqbrk {S_x} & : \text {otherwise} \end{cases}$
where $S_x$ is the initial segment in $X$ defined by $x$, and $\map f {\,\cdot\,}$ denotes an image set.
Using the Principle of Recursive Definition for Well-Ordered Sets, we can use $\rho$ to uniquely define:
- $h: X \to \powerset X$:
- $\map h \alpha = \map \rho {h {\restriction_{S_\alpha} } } = \begin {cases} h \sqbrk {S_\alpha} \cup \set \alpha & : \map P {\alpha, S_\alpha} \\ h \sqbrk {S_\alpha} & : \text {otherwise} \end{cases}$
Then $\map h \alpha$ is a $\le$-totally ordered set, by virtue of the construction of $h$ in terms of the predicate $P$.
Similarly, $h \sqbrk {S_\alpha}$ is totally ordered, for the same reason.
Then the union:
- $Z = \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$
admits an ordered sum in terms of $\le$ imposed on each $\map h \alpha$.
By Ordered Sum of Tosets is Totally Ordered Set, this is a totally ordered set.
In particular, it is a chain of $\struct {X, \le}$.
We claim this chain is maximal.
To see this, consider $x_0 \notin \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$.
Then $x_0$ is not $\le$-comparable with the elements of $Z$, because $\neg \map P {x, Z}$.
That is, there are no proper supersets of $Z$ that are totally ordered.
Therefore the ordered sum on $Z$ is a maximal chain in $\struct {X, \le}$.
$\blacksquare$
Also see
Sources
- 2000: James R. Munkres: Topology (2nd ed.) $\S 1.11$ Supplementary Exercise $6$