What is the exact reason why the class of all ordinals, $\textbf{ORD}$ is not a set?

Question

In this question, some explanations about why the class of all ordinals is not a set are given. I can also find similar explanations in many books. However, all those explanations and proofs just outline how we can reach a contradiction by assuming $\textbf{ORD}$ is a set, without saying where exactly things go wrong. To be more specific, one way to prove that $\textbf{ORD}$ is NOT a set is the following:

Assume it is a set. Then it is well-ordered and transitive. Hence $\mathbf{ORD}\in \mathbf{ORD}$, a contradiction.

The problem I have is this: Although I see the contradiction in the end, it is not at all obvious what's going on in the middle. I̶t̶'̶s̶ ̶n̶o̶t̶ ̶a̶t̶ ̶a̶l̶l̶ ̶o̶b̶v̶i̶o̶u̶s̶ ̶t̶h̶a̶t̶ ̶$\textbf{ORD}$ ̶i̶s̶ ̶w̶e̶l̶l̶ ̶o̶r̶d̶e̶r̶e̶d̶.̶ ̶I̶t̶'̶s̶ ̶n̶o̶t̶ ̶a̶t̶ ̶a̶l̶l̶ ̶o̶b̶v̶i̶o̶u̶s̶ ̶t̶h̶a̶t̶ $\textbf{ORD}$ ̶c̶o̶n̶t̶a̶i̶n̶s̶ ̶a̶ ̶m̶i̶n̶i̶m̶a̶l̶ ̶e̶l̶e̶m̶e̶n̶t̶.̶ EDIT: sorry for the confusion; by "not at all obvious", I just mean it is not obvious why the assumption of being a set is relevant. The facts (such as transitivity) themselves are obvious to me. In the end, I reach the conclusion, but I just don't feel I know what's going on.

Question: What unique property of sets (by unique I mean the property does not apply to proper classes) help us prove that $\textbf{ORD}$ well-ordered while assuming it is a set?

Sorry if the above is not clear enough. I will do my best to explain what I mean.

Answer 1

There is no such property--the proof that $\mathbf{ORD}$ is well-ordered has nothing to do with it being a set. Indeed, you can prove in ZF that $\mathbf{ORD}$ is a well-ordered class, meaning that every nonempty subset of $\mathbf{ORD}$ has a least element. Namely, given a nonempty subset $X\subseteq\mathbf{ORD}$, pick an element $\alpha\in X$. If $\alpha$ is the least element of $X$, we're done. Otherwise, $\alpha\cap X$ is nonempty, and so has a least element $\beta$ since $\alpha$ is an ordinal. Then $\beta$ is also the least element of $X$, since any $\gamma\in X\setminus\alpha$ must be greater than or equal to $\alpha$ and thus greater than $\beta$. (Here I assume you already know that the ordinals are totally ordered by $\in$, which takes some work to prove.)

The relevance of $\mathbf{ORD}$ being a set is that $\mathbf{ORD}$ is by definition the class of all sets that are transitive and well-ordered by $\in$. So, to conclude that $\mathbf{ORD}\in\mathbf{ORD}$, you need to know that $\mathbf{ORD}$ is a set.

Note that essentially the same argument can be made in other contexts where everything is a set, which you may find less confusing. For instance, if you define $\omega$ as the set of finite ordinals, then you can prove by this argument that $\omega$ must be infinite as follows. First, you prove that $\omega$ is an ordinal. Then, if $\omega$ were finite, you conclude that $\omega$ would be a finite ordinal and hence an element of $\omega$. But then $\{\omega\}$ would be a subset of $\omega$ with no least element, which is a contradiction.

Answer 2

The class $\bf Ord$ is in fact the union of all the von Neumann ordinals. It is in fact very obvious that it is well-ordered, you're just not looking at the right corner of the room.

If $\alpha,\beta$ are two ordinals, then either $\alpha\in\beta$ or $\beta\in\alpha$, because we know that's just the way we build the ordinals ($\varnothing$, $\alpha\cup\{\alpha\}$, and the unions at limit steps). And $\in$ is well-founded, or if you prefer without using Regularity, if $A$ is a set of ordinals, then set $\alpha=\bigcup A$, then $A\subseteq\alpha+1$. Since $(\alpha+1,\in)$ is in fact a well-order, that means $A$ has a minimum there, which is also the minimum in $\bf Ord$.

---

The assumption that $\bf Ord$ is a set is irrelevant to the proof that it is well-ordered by $\in$ or that it is transitive. The assumption that $\bf Ord$ is a set simply tells us that $\bf Ord$ is in fact a von Neumann ordinal, which is a problem, because then $\bf Ord\in Ord$, in which case $\in$ is no longer a well-order, since it is not irreflexive.

Answer 3

I will explain this in layman's terms.

If one assumes that the class of all ordinals ($Ord$) is a set and is well ordered, then since every (well ordered) set has an order type for example the set of natural numbers has an order type of $\omega$, which means that order types are ordinal numbers. But if you assume that $Ord$ is a set, that would mean there is an ordinal (specifically its order type) outside the set of all ordinals, and hence would imply, that $Ord\in Ord$ which basically implies that $Ord$ is smaller than itself and is incomplete. And since no set can have such properties under $\mathsf{ZFC}$, $Ord$ is not a set. This is the heart of the Burali-Forti paradox.

If one takes up $\mathsf{NBG}$ set theory the class of all ordinals $Ord$ would be considered a proper class.