Wosets are Isomorphic to Each Other or Initial Segments/Proof Using Choice

Proof
We assume $S \ne \O \ne T$; otherwise the theorem holds vacuously.

Define:


 * $S' = S \cup \text{ initial segments in } S$


 * $T' = T \cup \text{ initial segments in } T$


 * $\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$

We note that $\FF$ is non-empty, because it at least contains a trivial order isomorphism between singletons:


 * $f_0: \set {\text{smallest element in } S} \to \set {\text{smallest element in } T}$

Such smallest elements are guaranteed to exist by virtue of $S$ and $T$ being well-ordered.

By the definition of initial segment, the initial segments of $S$ are subsets of $S$.

For any initial segment $I_\alpha$ of $S$, such a segment has an upper bound by definition, namely, $\alpha$.

Also, no initial segment of $S'$ is the entirety of $S$.

Thus every initial segment of $S'$ has an upper bound, $S$ itself.

Every chain in $S'$ has an upper bound, because it defines an initial segment.

The previous reasoning also applies to $T'$.

By Antilexicographic Product of Totally Ordered Sets is Totally Ordered, $S' \times T'$ is itself a totally ordered set, with upper bound $S \times T$.

Thus the hypotheses of Zorn's Lemma are satisfied for $\FF$.

Let $f_1$ be a maximal element of $\FF$.

Call its domain $A$ and its codomain $B$.

Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$.

Then $f_1$ can be extended by defining $\map {f_1} a = b$.

This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments.

Then precisely one of the following hold:


 * $A = S$, with $S$ order isomorphic to an initial segment in $T$

or:


 * $B = T$, with $T$ order isomorphic to an initial segment in $S$

or:


 * both $A = S$ and $B = T$, with $S$ order isomorphic to $T$.

The cases are distinct by Well-Ordered Class is not Isomorphic to Initial Segment.