Let C be a linearly oredered set and A, B be well-ordered subsets of C then for all $\alpha\in C$ there are finitely many $(s,t)\in A\times B$ such that $s+t=\alpha$
this question was brought as a result of the:1)$A\cup B$ is well ordered. 2)A+B={a+b : $a\in A$,$b\in B$} is well-ordered. however, I know the proof of these statements which is by Zorn's lemma and the axiom of choice I don't get the proof of this statement.
(Note: I'm making assumptions below about the structure of addition e.g. if $a>b\Rightarrow\forall c,a+c>b+c$, etc..)
Assume there are infinitely many. Then let $A'$ be the subset of $A$ such that $s \in A'$ iff there is a $t \in B$ where $s+t=\alpha$. Define $B'$ similarly. $A'$ is infinite iff $B'$ is infinite.
We will show that $A'$ has an infinite strictly decreasing sequence, meaning that it is $A$ has an infinite strictly decreasing sequence, meaning $A$ is not well-ordered. We can construct this sequence using the ordering of $B'$. Since $B'$ is an infinite subset of a linearly ordered set, there is a strictly increasing increasing infinite sequence $t_1, t_2,... \in B'$. From this, we generate $s_1,s_2,... \in A'$ such that $s_i+t_i=\alpha$ for all $i$. Since the $t_i$ are strictly increasing, the $s_i$ are strictly decreasing. So, $A$ is not well-ordered since it has an infinite strictly decreasing sequence.
So we conclude that $A'$ (and thus $B'$) are finite.