Partially ordered sets and ordinals

Question

Could you please help me with this problem?

Given the partially ordered set $(A,\le_A)$, where $A=\omega \times \omega$ and $$(a,b)\le_A(a',b')\iff a\le a' \land b \le b'$$ with $\le$ being the usual order on $\omega$. Show that every total order that extends $\le _A$ is a well-order and find all the ordinals isomorphic to an arder that extends $\le _A$.

I'm ok with the first part of the question and I already found a way to obtain $\omega, \omega \cdot n$, and $\omega ^2$, and obviously only limit ordinals are allowed. My claim is that ordinals greather than $\omega^2$ can't be obtained (and showing that the exercise is finished), but I don't know how to show it.

Answer 1

Hint: If $\leq$ is a total order that extends $\leq_A$ which has order type greater than $\omega^2$, there is some $(a,b)\in A$ such that $S=\{(a',b')\in A:(a',b')<(a,b)\}$ has order type $\omega^2$. But now note that $S\subseteq (a+1)\times\omega\cup \omega\times (b+1)$ (why?) and $(a+1)\times\omega\cup \omega\times (b+1)$ can be written as a finite union of subsets of order type $\omega$ (why?). So, it suffices to show that the ordinal $\omega^2$ cannot be covered by finitely many subsets of order type $\omega$.

More details for this last step are hidden below.

Suppose $\omega^2$ is a union of finitely many sets $A_1,\dots,A_n$. For each $a<\omega$, some $A_i$ must contain infinitely many elements of the form $\omega\cdot a+b$ for $b<\omega$. Thus some $A_i$ must contain infinitely many elements $\omega\cdot a+b$ for infinitely many different values of $a$. This $A_i$ then would have order type $\omega^2$. So if $\omega^2$ is a union of finitely many sets, at least one of them must have order type $\omega^2$, and in particular they cannot all have order type $\omega$.