Ordinal set $\Omega$ : von Neumann definition and modern definition

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 .

• With what you call the modern definition, the ordinals themselves are not sets (except $0$), so it doesn't even make sense to ask if the class of ordinals is a set. May 31 '11 at 13:19

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.

• @ChrisEagle's comment and @‍PietroMajer's answer. Oct 6 '20 at 20:13

note that:

• In any case, the class of all ordinals is a proper class.

• The approach of defining ordinals as equivalence classes is not "modern" compared with von Neumann's definition, which is on the contrary a clever and elegant way of avoiding the quotient, and referring more directly to the axioms of set theory (namely, to the relation $\in$). You may like the beautiful pages by P. J. Cohen in the introductory chapters of his book Set Theory and the Continuum Hypothesis.

• Indeed. Ordinals as equivalence classes is the original definition by Georg Cantor, the inventor of both ordinal numbers and set theory. May 31 '11 at 13:42
• It's a helpful answer, but to speak of "the" axioms of set theory is misleading. May 6 '12 at 14:28
• Cohen's book is indeed really beautiful. It's almost the sole reason I have any chance of any understanding of many of the logic discussions on MO. Sep 12 '12 at 21:37
• Oct 6 '20 at 20:15

The "modern definition" you are referring to is indeed classical and predates von Neumann's innovation. The equivalence approach is implicit in the work of Cantor, but it was made explicit by Russell (and is based on Frege's similar treatment of cardinals as equivalence classes).

Chris Eagle pointed out in his comment that with the equivalence relation definition of ordinals, no ordinal is a set except 0. This is indeed the case in Zeremlo-style set theories. But in set theories such as Quine's $NF$ or Quine-Jensen's $NFU$ [in which there is a universal set], the equivalence approach leads to sets. The consistency of $NF$ relative to a $ZF$-style set theory is still open, but Jensen showed the consistency of $NFU$ relative to a small fragment of $ZF$ in 1969, see here for more detail.

P.S. $NFU$'s ability to handle "large objects" makes it an interesting vehicle for the foundations of category theory. Solomon Feferman has done significant work in this area, if interested, see his paper, and talk.