Logical question while trying to prove a theorem about comparing well-ordered sets I'm reading Set theory of Jech. I try to prove Thm 2.8 for myself but I'm not sure if the first step of my proof is correct. So the theorem affirms that if $W_1$ and $W_2$ are two well-ordered sets, then exactly one of the three cases holds：i) $W_1$ is isomorphic to $W_2$, ii) $W_1$ is isomorphic to an initial segment of $W_2$, iii)$W_2$ is isomorphic to an initial segment of $W_1$.
I want to define a function that maps the least element $x_1$ of $W_1$ to least element $y_1$ of $W_2$, and then the least element $x_2$ of $W_1\backslash \{x_1\}$ to least element $y_2$ of $W_2\backslash \{y_1\}$, and continue like this until of one the sets $W_1$ and $W_2$ is exhausted. 
But I'm not sure if the above definition works if both of the sets $W_1$ and $W_2$ are uncountable (otherwise I could use induction). That is, I don't know if I can really reach the last step where either all elements of $W_1$ are taken or all elements of $W_2$ are taken.
Could you tell me if this is correct and why? Thanks!
 A: You're right that the procedure you describe may fail to exhaust all elements of $W_1$ and $W_2$ if they are infinite (they don't even have to be uncountable). Instead, how about something like this:
Call a partial function from $W_1$ to $W_2$ "good" if it is an order isomorphism from an initial segment of $W_1$ to an initial segment of $W_2$.
Prove that any two good functions agree on the intersection of their domains. (Namely, there cannot be a first element that they disagree on).
The union of all good maps is therefore a partial function $W_1\to W_2$. Prove that it is itself a good map. Call it $f$.
Now either the domain of $f$ is all of $W_1$ or its range is all of $W_2$. (Otherwise we could map the first unused element of $W_1$ to the first unused element of $W_2$ and get a larger good function, contradicting the fact that $f$ is the union of all good functions).
But this means exactly that we're in one of the three situations in your claim.
A: Define $f:W_1\to W_2$ recursively by setting
$$ f(\min W_1) = \min W_2 \quad\text{and}\quad f(x) = \min (W_2\setminus f(I_x)),\ x\in W_1,$$ where $I_x:=\{y\in W_1 \mid y<x\}$ is the initial segment of $x$. From here there are a few things to worry about:
1) How do we know that $W_2\setminus f(I_x)$ is nonempty? We should first assume that $W_2$ is not isomorphic to an initial segment of $W_1$.
2) How do we know that defining a function recursively like this is valid? It's certainly intuitive, but the details can get a bit messy (especially when we jump to transfinite recursion).
3) Once we accept that $f$ is a well-defined function, then we need to show that it is an order isomorphism.

If you assume choice (thanks Henning Makholm), then I recommend trying an argument by Zorn's lemma. The ideas you have with using transfinite induction are great, but the beauty of Zorn's lemma is that you can avoid these obstructions.
A: First, a few preliminary results:  If $<_A$ is a well-order on A then (1).There is no order-isomorphism $f$ from $A$ onto a proper initial segment of $A.$ (Consider the least $a\in A$ such that $f(a)<a.$ )... (2). The only isomorphism from $A$ onto $A$ is $id_A.$ (Consider the least $a\in A$ such that $f(a)\ne a$ and consider $f^{-1}a.$)... (3). If $(A,<_A)$ and $(B,<_B)$ are well-orders and $f,g$ are isomophisms from $A$ onto $B$ then $f=g.$ (Consider that $g^{-1}f$ is an isomorphim of $A$ onto $A$ and apply (2).)
Let $(A,<_A)$ and $(B,<_B)$ be well-orders. 
For $a\in A$ let $(-,a]_A=\{a'\in A: a'<_A a \lor a'=a\}.$
For $b\in B$ let $(-,b]_B=\{b'\in B: b'<_B b \lor b'=b\}.$
Let $A'$ be the set of those $a\in A$ for which there exists an order-isomorphism  $f_a:(-,a]_A\to (-,b]_B$ for some $b\in B.$
Let $B'$ be the set of those $b\in B$ for which there exists an order-isomorphism $g_b:(-,b]_B\to (-,a]_A$ for some $a\in A.$
We can readily show that $A'=A$ or $A'=(-,a]_A\setminus \{a\}$ for some $a\in A,$ and that $B'=B$ or $B'=(-,b]\setminus \{b\}$ for some $b\in B.$ 
From (1),(2), and (3) if $a\in A'$ then $f_a$ is unique, and if $b\in B'$ then $f_b$ is unique. So by the Axiom of Replacement we obtain the sets $F=\{f_a(a):a\in A'\}$ and $G=\{g_b(b): b\in B'\}.$
We can  readily show $h(a)=f_a(a)$ is an isomorphism from $A'$ to $F$ and that $j(b)=g_b(b)$ is an isomorphism from $B'$ to $G.$  
I will leave the rest to you.   
