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I am studying set theory and have recently encountered the concept of ordinals. The way I understand them, we are trying to classify all well ordered sets (which have desirable inductive properties) into equivalence classes w.r.t. a relation that preserves the order structure, and have a canonical representative for each equivalence classes, but since these turn out to be proper classes and not sets, we call them order types.

And it turns out that if we take the von Neumann ordinals $\left(\left\{ \emptyset,\left\{ \emptyset\right\} ,\left\{ \emptyset,\left\{ \emptyset\right\} \right\} ,\ldots,\omega,\omega\cup\left\{ \omega\right\} ,\ldots\right\} \right)$ then we can prove that any well ordered set is isomorphic to exactly one of these, called its order type.

So my question is what's the motivation for the further abstraction of defining an ordinal as a transitive set which is well ordered w.r.t. $\in $ ? Can't we simply use the fact that the von Neumann ordinals possess these properties? Is there any context where we use ordinals that aren't the von Neumann ordinals? Or is my understanding as presented above somehow flawed?

H.Rappeport
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Your misunderstanding is that you think "ordinals" and "von Neumann ordinals" means different things, such that only some of the ordinals in set theory would be von Neumann ordinals.

Since there is really no such distinction in a formal development of set theory, you become confused.

In other words, the sets you're thinking of when you write $$\emptyset,\{\emptyset\},\left\{ \emptyset,\left\{ \emptyset\right\} \right\} ,\ldots,\omega, \omega \cup \left\{ \omega\right\} ,\ldots$$ are exactly the same sets as the transitive sets that are well-ordered by $\in$.

If we want to be a bit more abstract, we could say that an "ordinal number" means "whatever it is that two well-ordered sets have in common when they are order isomorphic". That would be similar to saying that a "natural number" is whatever two finite sets have in common when they have equally many elements.

This is conceptually simple, but not technically convenient. If we want ordinal numbers to be set-theoretic objects (such that we can make them elements of sets, etc), then we need to specify a particular set to represent each of them within set theory.

Von Neumann proposed to represent each isomorphism class of well-orders by the transitive set that is ordered in this particular way by $\in$. This representation is what is known as the "von Neumann ordinals". It's not a particular selection of ordinals; it's a representation that works for all of them.

The von Neumann representation works so well that it has become pretty much universal in set theory.

The reason why we don't just define them as "$\emptyset,\{\emptyset\},\left\{ \emptyset,\left\{ \emptyset\right\} \right\} ,\ldots,\omega, \omega \cup \left\{ \omega\right\} ,\ldots$" is that for rigorous proofs about them we need a definition that doesn't depend on the reader figuring out for himself exactly what is implied by the suggestive notation "$\ldots$" that appears in it.

hmakholm left over Monica
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How are you defining the von Neumann ordinals if not as a transitive set well-ordered by $\in$? We need some precise definition of the notion, even if it's intuitively clear what we mean.

Now there are other definitions of von Neumann ordinal out there, my personal favorite being "hereditarily transitive set." However, it's useful to have multiple equivalent definitions - if you prefer, you can think of this as one definition and then a multi-part characterization theorem - since some are more convenient in different contexts (e.g. when developing forcing, "hereditarily transitive set" often makes arguments much snappier than "transitive set well-ordered by $\in$").

Noah Schweber
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  • My understanding is that the definition of an ordinal is a transitive set well-ordered by $\in $, and then the von Neumann ordinals, defined recursively, are a special case. The question was more about the motivation for the abstract definition, which the latter part of your answer addresses. – H.Rappeport Jun 23 '18 at 20:08
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    @H.Rappeport On the basis of your understanding, as described in your comment on this answer, it might be a good exercise to (1) write out explicitly the recursive definition of the von Neumann ordinals that you refer to and then (2) prove that all transitive sets well-ordered by $\in$ satisfy that recursive definition. – Andreas Blass Jun 23 '18 at 23:23
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    @Noah Like you, I prefer the "hereditarily transitive" definition, and I usually use (and motivate) it when I teach set theory, but that works because I'm usually using ZFC and, in particular, I have the axiom of regularity available. In the absence of regularity, "hereditarily transitive" doesn't work correctly --- a Quine atom, satisfying $x={x}$, is hereditarily transitive. I'd need to either use the definition in the question or supplement "$x$ is hereditarily transitive" with "and every set containing $x$ (as a member) has an $\in$-minimal element." – Andreas Blass Jun 23 '18 at 23:29
  • @Andreas: Or if you have $x={x_n\mid n<\omega}$ where $x_n={x_k\mid k>n}$, which is my other fun example of a hereditarily transitive set. – Asaf Karagila Jun 24 '18 at 10:50
  • @Noah Why not define an ordinal by recursing on well orders instead, isn’t that most intuitive? – Vivaan Daga Mar 11 '23 at 13:37
  • @AndreasBlass I suppose I'll pose the question to you too, why define ordinals as "transitive sets whose elements are also transitive", when you could define them by recursing on well orders? Is there some historical inertia at play here? Since the latter seems much more intuitive. – Vivaan Daga Mar 12 '23 at 10:28
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    @Shinrin-Yoku Hereditary transitivity is easy to express in the primitive language of set theory. Definitions by recursion need a lot of preparatory definitions. – Andreas Blass Mar 12 '23 at 14:06
  • @AndreasBlass That makes sense. Given that one does not need ordinals, for things like transifinite induction, since a "very large" well order suffices, it makes me wonder whether there is any use in set-theory for having a "canonical" well order, other than convenience? Surely there must be, because convinience feels like a very superflous reason, for something that is studied so deeply... – Vivaan Daga Mar 15 '23 at 14:16
  • @AndreasBlass I suppose my question could be modified to ask, is a "canonical" notion of well orders really needed? Similarly one could ask why the ordinal defintion of cardinal numbers is needed when relations between sets based on injections and bijections. – Vivaan Daga Mar 16 '23 at 12:27
  • Noah, @AndreasBlass I ended up asking the question here(as I feel the answer would be longer than what the comment box could fit): https://math.stackexchange.com/questions/4660421/why-do-we-need-canonical-well-orders. – Vivaan Daga Mar 16 '23 at 18:47
  • My question ended up with a few upvotes, but I guess that given no answer was posted ordinals really are superfluous objects. – Vivaan Daga Mar 17 '23 at 16:54
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I want to suggest an answer by posing a more general question.

One way to view what you know is that the von Neumann construction is a set theoretic model that captures the structure of all ordinals (defined abstractly), and that model is universal (perhaps not the right word): any other model is isomorphic.

But the abstract definition is still important. By analogy, why bother defining a complete ordered field when you can just use the field of Dedekind cuts of rationals, or the field of equivalence classes of Cauchy sequences of rationals?

Ethan Bolker
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  • I suppose because these equivalent definitions of the reals allow one to choose which one is most convenient in a given context. Also, these different formulations of the same thing can maybe give different intuitions. So referring to my question, is there a context in set theory where other ordinals are used? Or is it that this abstract definition captures the important aspects of what it means to be an ordinal? – H.Rappeport Jun 23 '18 at 20:04
  • I "answered" only with the analogy because I'm not a set-theorist or logician so don't know of examples where other formulations would be more useful. But the answer from @NoahSchweber provides one. (His answers are always useful!) – Ethan Bolker Jun 23 '18 at 20:08