Showing a $\alpha\times\beta$ is well-ordered for $\alpha,\beta$ ordinals I am concerned with showing the least element existence in every subset of $\alpha\times\beta$. Below is my attempt. My teacher has used similar argument to show $\mathbb{N}\times\mathbb{N}$ is a well-ordered by some 'given' ordering.
QUESTION:
Let $\alpha,\beta$ be ordinals. Show that $\alpha\times\beta$ with the ordering
$(\gamma,\delta)\triangleleft(\lambda,\kappa)\iff[\gamma\cup\delta<\lambda\cup\kappa\vee(\gamma\cup\delta=\lambda\cup\kappa\space\land\space(\gamma<\lambda\vee(\gamma=\lambda\space\&\space\delta<\kappa)))]$ is well-ordered. 
SOLUTION:
Now I show that for every non-empty subset $b$ of $\alpha\times\beta$ we have a least element. We proceed as follows:-
If $\emptyset\neq b\subseteq\alpha\times\beta$ , define $k$ to be the least ordinal in $\{x\cup y\mid (x,y)\in b\}$ with respect to ordering $\in_{Ord}$. The set $\{z\in {Ord}\mid (z,k\backslash z)\in b\}$ is non-empty and if $z_0$ is its least element (again w.r.t $\in_{Ord}$ then $(z_0,k\backslash z_0)$ is the $\triangleleft$ -least element of $b$. 
Am I right?
 A: The way I would think about it is to take a nonincreasing sequence $(\alpha_i,\beta_i)$ and prove that it is eventually constant. Nonincreasing means $(\alpha_{i+1},\beta_{i+1}) \leq (\alpha_i,\beta_i)$ for all $i$ where $\leq$ means $\triangleleft$ or equal. 
First, if we define $\gamma_i = \alpha_i \cup \beta_i$, then the fact that the original sequence is nonincreasing means that $(\gamma_i)$ is non-increasing in the usual ordering of the ordinals. Hence $(\gamma_i)$ is eventually constant. (In fact, the ordinal that $(\gamma_i)$ eventually equals will be exactly the $k$ from your proof sketch.)
Hence, by removing finitely many terms from the beginning of the sequence $(\alpha_i, \beta_i)$, we can assume $\gamma_i$ is constant. That means that the first clause of the definition of $\triangleleft$ will never apply. So move on to the next clause of the definition of $\triangleleft$, and apply the same technique. This will show that $\alpha_i$ is eventually constant. Then move on to the third clause and show that $\beta_i$ is eventually constant. 
