define an order relation with respect to which ....... Let $(S;<)$ and $(T;<')$ be two well-ordered sets. On $S \times T$, define an order relation with respect to which it becomes a well-ordered set. Give proofs for your answer.
I was thinking that since S, T are well-ordered than they must have a least element. S and T is the Cartessian product of S and T. So i should find a the least element of $S \times T$ such that it becomes a well ordered set. I was thinking that it may have a partially or a totally ordered set.
 A: Hint: Think about the order in the dictionary.
Some details: Among another things you wish to prove that for all non-empty subsets $S'\times T'$ of $S\times T$, there exists a least element in $S'\times T'$. Two things can happen:


*

*$\forall (s_1,t_1), (s_2, t_2)\in S'\times T'\left(s_1=s_2\right)$ or

*$\exists (s_1, t_1), (s_2, t_2)\in S'\times T'(s_1\neq s_2)$.


I'll consider the first case and leave the second one to you.
Suppose $\forall (s_1,t_1), (s_2, t_2)\in S'\times T'\left(s_1=s_2\right)$ and consider $\min \left(T'\right)$ (why does this minimum exist?). 
Since our hypothesis is that the first coordinate of the elements of $S'\times T'$ are all equal, one has $S'\times T'=\{s\}\times T'$, for some $s\in S$. 
Consider $\left(s, \min\left(T'\right)\right)$ which is an element of $S'\times T'$. I will prove that $\forall (u,t)\in S'\times T'(\left(s, \min\left(T'\right)\right)\leq (u,t))$. Let $(u,t)\in S'\times T'$. From previous considerations it follows that $u=s$. By definition of $\min\left(T'\right)$ one also has $\min \left(T'\right)\leq t$. Thus, by definition of the lexicographic order $\leq _\text{Lex}$, it follows that $\left(s, \min\left(T'\right)\right)\leq_\text{Lex} (u,t)$, as wished.
This means that $\left(s, \min\left(T'\right)\right)=\min \left(S'\times T'\right).$
