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

Theorem
Let $\left({S, \preceq_S}\right)$ and $\left({T, \preceq_T}\right)$ be well-ordered sets.

Then precisely one of the following hold:


 * $\left({S, \preceq_S}\right)$ is order isomorphic to $\left({T, \preceq_T}\right)$

or:


 * $\left({S, \preceq_S}\right)$ is order isomorphic to an initial segment in $\left({T, \preceq_T}\right)$

or:


 * $\left({T, \preceq_T}\right)$ is order isomorphic to an initial segment in $\left({S, \preceq_S}\right)$

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

Define:


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


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


 * $\mathcal F = \left\{ { f:S' \to T' \ \vert \ f \text{ is an order isomorphism} } \right\}$

We note that $\mathcal F$ is not empty, because it at least contains a trivial order isomorphism between singletons:


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

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

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

Let $f_1$ be a maximal element of $\mathcal F$. 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 $f_1\left({a}\right) = b$.

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 No Isomorphism from Woset to Initial Segment.