# Do I need that $\mathrm{Ord}$ is well-founded before proving that it is totally ordered?

The ordinals are totally ordered, i.e. for $\alpha,\beta\in\mathrm{Ord}$ we have

$$\alpha\in\beta\quad\vee\quad\beta\in\alpha\quad\vee\quad\alpha=\beta.$$

Each proof I have seen so far started like this: Assume ther are incomparable ordinals. Choose an $\in$-minimal ordinal $\alpha$ so that there is an incomparable ordinal $\beta$.

But how can we do this? How do I know that there is an $\in$-minimal ordinal? I think the axiom of regularity does not help me here, because $\mathrm{Ord}$ is a proper class. Do I have to show that $\mathrm{Ord}$ is well-founded before proving totallity?

• There are other proofs, that don't use well-foundedness. But another way to see an $\in$-minimal such ordinal is to say "Let $\alpha$ be such an ordinal and let $\beta$ be the minimum in $\alpha +1$. Then $\beta$ is $\in$-minimal." Commented May 16, 2017 at 16:37
• What axioms are you starting with? Commented May 16, 2017 at 17:02
• @Asaf ZFC and ordinals are defined as transitive sets with all elements being transitive too. Commented May 16, 2017 at 17:05

You can prove that given any ordinal $\alpha$, the ordinals below $\alpha$ are linearly ordered.
Now if $\alpha$ and $\beta$ are two ordinals, consider $x$ to be the transitive closure of $\{\alpha,\beta\}$. I claim that $x$ is a transitive set (by definition) whose elements are all transitive sets, this is because $$x=\operatorname{tcl}(\{\alpha\})\cup\operatorname{tcl}(\{\beta\})=\alpha\cup\beta\cup\{\alpha,\beta\}.$$
So $x$ is now an ordinal such that both $\alpha$ and $\beta$ are less than $x$. So they are comparable by the general proof given before.
• Then we need the existence of the transitive closure. All proofs I found for this are using $\omega$ or using the existence of recursively defined function which itself is proved using $\omega$. Doesn't this give new problems using the smallest infinite ordinal in such an early phase? Commented May 17, 2017 at 7:45
• That might depend on how you formulate some of the axioms. If you know $\omega$ is a set, you can prove what is missing with ad hoc proofs, where necessary. Then you get the existence of transitive closures. Commented May 17, 2017 at 7:58