With von Neumann's definition of $\Omega$ , the ordinal set , we have $[0,\omega)=\omega \; (\forall \omega \in \Omega)$ . In fact with von Neumann approach ,
$\Omega$ cannot be considered as a set ( since $\Omega$ is itself well-
ordered and belong to itself .
With the "modern" definition of $\Omega$ that consist in defining the
ordinals as the equivalence classes of well-ordered sets ( two well ordered
sets being equivalent if there exist a monotonous bijective mapping between
them ) , do we still have this property ? Apparently yes because the
demonstration by transfinite induction is still valid .
Best Answer
Chris Eagle's comment and Pietro Majer's answer are correct, but let me add the following. There is a standard way to overcome the difficulty mentioned by Chris (that each individual equivalence class, except 0, fails to be a set). This method, known as "Scott's trick" (invented by Dana Scott), is to replace each equivalence class by the set of its members of lowest rank (in the usual cumulative hierarchy of ZF set theory). With this trick, individual ordinals (in the older sense that the question calls "modern") become sets, and it becomes sensible to ask whether the collection of all these ordinals is a set. The answer is no. The reason is essentially the Burali-Forti paradox: If the class of ordinals were a set, then, since it is well-ordered (by the relation of "embeddability as initial segment" between well-ordered sets), its order-type would be an ordinal $\alpha$, and it would be order-isomorphic to the collection of ordinals strictly below $\alpha$. That is impossible, because no well-ordered set is isomorphic to a proper initial segment of itself.