Lexicographic Order forms Well-Ordering on Ordered Pairs of Ordinals

Theorem
The Lexicographic Order, $\operatorname{Le}$, is a strict well-ordering on $\left( \operatorname{On} \times \operatorname{On} \right)$.

Proof 1
This is an instance of Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering.

Total Ordering
Assume $\left(x,y\right) \operatorname{Le} \left(x,y\right)$. Then, $x < x \lor \left( x = x \land y < y \right)$. Both are contradictory, so $\operatorname{Le}$ is irreflexive.

Assume $\left(\alpha,\beta\right) \operatorname{Le} \left(\gamma,\delta\right) \land \left(\gamma,\delta\right) \operatorname{Le} \left(\epsilon,\zeta\right)$. We shall split up into two cases:

If $\alpha < \gamma$, then $\alpha < \epsilon$, so $\left(\alpha,\beta\right) \operatorname{Le} \left(\epsilon,\zeta\right)$.

If $\alpha = \gamma$, then $\alpha < \epsilon \lor \left( \alpha = \epsilon \land \delta < \zeta \right)$.

But also, if $\alpha = \gamma$, then $\beta < \delta$.

Therefore, $\left(\alpha = \epsilon \land \beta < \zeta\right)$. Therefore, $\left(\alpha,\beta\right) \operatorname{Le} \left(\epsilon,\zeta\right)$.

In either case, $\operatorname{Le}$ is transitive. So $\operatorname{Le}$ is a strict ordering.

Strict Total Ordering
Now, assume that $\neg \left(\alpha,\beta\right) \operatorname{Le} \left(\gamma,\delta\right) \land \neg \left(\gamma,\delta\right) \operatorname{Le} \left(\alpha,\beta\right)$.

Then, $\neg \alpha < \gamma$ and $\neg \gamma < \alpha$, so $\alpha = \gamma$.

Similarly, $\neg \beta < \delta$ and $\neg \delta < \beta$, so $\beta = \delta$.

By Equality of Ordered Pairs, $\left(\alpha,\beta\right) = \left(\gamma,\delta\right)$. Therefore, $\operatorname{Le}$ is a strict total ordering.

Well-Ordering
Take any nonempty subset $A$ of $\left( \operatorname{On} \times \operatorname{On} \right)$. We shall allow $A$ to be any class. This isn't strictly necessary, but it will not alter the proof.

Let the mapping $1^{st}$ send each ordered pair $\left(x,y\right)$ to its first member $x$.


 * $\displaystyle 1^{st} = \{ \left( \left( x,y \right) z \right) : z = x \}$

Then, $1^{st} : A \to \operatorname{On}$ is a mapping. Take $\operatorname{Im}\left(A\right)$, the image of $A$ under $1^{st}$.


 * $\displaystyle \operatorname{Im}\left(A\right) \subseteq \operatorname{On}$, so by Subset of Ordinals has Minimal Element, $\operatorname{Im}\left(A\right)$ has a minimal element, which we shall call $\alpha$.

Set $B = \{ y \in \operatorname{On} : \left( \alpha, y \right) \in A \}$. $\alpha$ is a minimal element of $\operatorname{Im}\left(A\right)$, so $\left(\alpha,y\right) \in \operatorname{Im}\left(A\right)$, for some $y \in \operatorname{On}$. Therefore, this class $B$ is nonempty. Furthermore, $B$ is some subset of the ordinals.

By Subset of Ordinals has Minimal Element, we can, again, conclude that $B$ has a minimal element. We will set this minimal element equal to $\beta$. Therefore, $\left(\alpha, \beta\right) \in A$.

Now, assume there is some element of $A$, $\left(\gamma,\delta\right)$ such that $\left(\gamma,\delta\right) \operatorname{Le} \left(\alpha,\beta\right)$.

Then, $\gamma \le \alpha$, but for all ordered pairs in $A$, $\alpha$ is a minimal first element, so $\gamma = \alpha$.

But this implies that $\delta < \beta$ and $\left(\alpha,\delta\right) \in A$. This contradicts the fact that $\beta$ is the minimal element satisfying $\left(\alpha,\beta\right) \in A$. This is a contradiction, so $\left(\alpha,\beta\right)$ is the $\operatorname{Le}$-minimal element of $A$.