Given two well-orders $\langle A,R \rangle$ and $\langle B,S \rangle$, one of the following holds.

Question

Let $\langle A,R \rangle$ and $\langle B,S \rangle$ be two well-orders, and let $\text{pred}(A,x,R) := \{y \in A \;|\; yRx\}$ and similarly for $\text{pred}(B,z,S)$.

It is claimed that one of the following must hold:

$1$. $\langle A,R \rangle \cong \langle B,S \rangle$.

$2$. $\exists y \in B\left(\langle A,R \rangle \cong \langle \text{pred}(B,y,S),S\rangle\right)$.

$3$. $\exists x \in A\left(\langle \text{pred}(A,x,R), R \rangle \cong \langle B,S \rangle \right)$.

The proof of this seems like a sketch, giving us $f := \{\langle v,w \rangle\;:\: v \in A \land w \in B \land \text{pred}(A,v,R), R \rangle \cong \langle \text{pred}(B,w,S),S\rangle\}$

and it is claimed that it can not be the case that both initial segments are proper, which I interpret as $\neg (\text{pred}(A,v,R) \subsetneqq A \land \text{pred}(B,w,S) \subsetneqq B)$.

First: Is my understanding of what is claimed correct?

Second: Given that my understanding of what is claimed is correct, I fail to see why this can't be the case. I don't know exactly how to proceed. Any illumination on how to think would be well appreciated.

This question is motivated by the proof of lemma $6.3$, chapter $1$, in Kunen's "Introduction to independence proof's".

Answer 1

What you need to show is

1. $f$ is a function on its domain. In other words, for all $v\in A$, there is at most one $w\in B$ such that $\langle v,w\rangle\in f$. Use lemma 6.1 for this.
2. $f$ is order-preserving- if $v_1<_Rv_2$ in $A$, $v_1,v_2\in\text{dom}(f)$ then $f(v_1)<_S f(v_2)$ in $B$.
3. $\text{dom}(f)$ is an initial segment of $\langle A,R\rangle$ and $\text{ran}(f)$ is an initial segment of $\langle B,S\rangle$.
4. $\text{dom}(f)$ and $\text{ran}(f)$ cannot both be proper initial segments (of $A, B$ respectively) because then you could extend $f$ by one more mapping so $f$ is not maximal.