Finding an order isomorphism from $\text{On}\times\text{On}$ to $\text{On}$

Question

Let $\text{On}$ be the class of all ordinals and let $\leq_{\text{c}}$ be the canonical well-ordering on $\text{On}\times\text{On}$. More specifically, $\preceq$ is defined as follows.

Let $\left(\alpha,\beta\right),\left(\gamma,\zeta\right)\in\text{On}\times\text{On}$. Then $\left(\alpha,\beta\right)\leq_{\text{c}}\left(\gamma,\zeta\right)$ if and only if one of the following hold:

1. $\alpha\cup\beta<\gamma\cup\zeta$.

2. $\alpha\cup\beta=\gamma\cup\zeta$ and $\alpha<\gamma$.

3. $\alpha\cup\beta=\gamma\cup\zeta$, $\alpha=\gamma$, and $\beta\leq\zeta$.

Here, $\leq$ is the standard ordering on ordinals.

I have shown that every proper segment of this well-ordered class is a set. This allows me to define a function $\Gamma:\text{On}\times\text{On}\rightarrow\text{On}$ by $$\Gamma\left(\left(\alpha,\beta\right)\right)=\text{Ord}\left\{ \left(\zeta,\eta\right)\in\text{On}\times\text{On}:\left(\zeta,\eta\right)<_{\text{c}}\left(\alpha,\beta\right)\right\} $$ for all $\left(\alpha,\beta\right)\in\text{On}\times\text{On}$.

Here, $\text{Ord}$ is the function which maps each well-ordered set to the unique ordinal which is order isomorphic to it. I want to show that $\Gamma$ is an order isomorphism. I have shown that it is strictly increasing. Now I just have to show that it is surjective. This is the problem I am facing. I can't seem to show that it is surjective. I have tried using induction but that got me nowhere. Any ideas?

Answer 1

This is actually much simpler than you'd expect, if you pay attention to the right things. In fact, you've almost proved this already.

Since $\Gamma$ is a strictly increasing function, its range cannot be a set of ordinals. So it has to be unbounded. But now simply note that if $\Gamma(\alpha,\beta)=\gamma$ and $\delta<\gamma$, then there is a unique initial segment below $(\alpha,\beta)$ which is isomorphic to $\delta$. And by well-ordering, there is a least $(\alpha',\beta')$ below $(\alpha,\beta)$ such that $\Gamma(\alpha',\beta')=\delta$.

In other words, we show that the image must be downwards closed, and not bounded (i.e., not a set). Therefore it has to be an isomorphism.