Well ordering on the ordinal numbers

Question

Can someone please explain what it means for an ordinal $\alpha$ to be less than an ordinal $\beta$? and give an example of such... I have also read that $\alpha < \beta$ implies $\alpha \in \beta$. Finally, is there a simple proof to show that the ordinals are well ordered? and what relation are they well ordered with respect to?

Answer 1

An ordinal is a transitive set which is well ordered by the $\in$ relation. The class of ordinals is the class of all ordinals, and it too is a transitive class which is well ordered by $\in$.

This requires proof, of course, and depends on what you know or not know. But since $\in$ is the order relation on ordinals, it follows that every ordinal is also a set of ordinals. Therefore, given a nonempty class of ordinals, pick an element there $\alpha$; either it was minimal, or else intersecting the class with $\alpha$ gives a nonempty subset of $\alpha$, which has a minimum since $\alpha$ is well ordered.

Answer 2

Asaf will probably murder me in my sleep for this. His answer is the correct one, but I wanted to expound a bit on what an ordinal "really is". This whole answer is in the spirit of "in maths, we don't say that the ordered pair $\langle a,b \rangle$ is precisely and exactly $\{\{a\},\{a,b\}\}$ unless we really need to do set theory on it".

I find it can be helpful to think of the ordinals as abstract objects. There are several ways you can do this: for example, as "the smallest well-ordered things", or as equivalence classes of "well-ordered sets" under the relation "there is an order-preserving bijection". (This is just as one can view $\mathbb{N}$ as equivalence classes of "finite sets" under the relation "there is a bijection".) Note that there is no set of all finite sets, just as there is no set of all well-ordered sets, so this is purely a conceptual device.

Then by a slight abuse of notation, we can think of the ordinal $\alpha$ as being a well-ordering, just as we can think of the equivalence class which contains $\{a\}, \{1\}, \{2\}, \{\mathbb{N}\}$ and so on as being $1$. We take the universe of all sets, and we extract the ones which are $\alpha$-ish, and declare that this collection embodies $\alpha$-ness.

Then $\alpha < \beta$ if and only if the well-ordering $\alpha$ is an initial segment of the well-ordering $\beta$, just as $m < n$ if and only if the set $m$ is contained in the set $n$.

Now, when we cast this into the language of set theory, we need to pick some actual objects that really and truly exist: it's no good saying that $2$ is defined to be "all two-element sets", as my handwaving above would suggest, because then it's provable that $2$ doesn't exist at all! Instead, it turns out that a structure which does behave the right way is the set-theoretic ordinals (i.e. the transitive sets) together with the order relation of inclusion; in the same way, a structure which behaves the right way for the naturals is the finite set-theoretic ordinals ordered by inclusion. Now, for each $\alpha$ of the handwavy "equivalence class of well-ordered sets" variety, we can prove that the ordinal $\alpha$ exists in set theory as a transitive set with just the right number of smaller ordinals below it.