In some sense $Ord$ is a "proper class" ordinal. Unfortunately the notion of a proper class ordinal is not a straight forward generalization of the notion of "set" ordinals because the proper classes cannot be members of each other. But there are some hopes to define the notion of a "proper class" ordinal using a different view to the usual meaning of a "set" ordinal. For example if we find a property $P$ which is well defined on both sets and proper classes, and then discover a "theorem" like this:

"A set $s$ is an ordinal (in the usual sense) iff $s$ has the property $P$"

Then we can give a "definition" for the notion of a "proper class" ordinal which is compatible with the usual notion of "set" ordinals as follows:

"A class $C$ called a class-ordinal iff $C$ has the property $P$"

As an inexact suggestion in this direction note that the relations $\in$ and $\subsetneq$ are same on set ordinals, i.e.

$\forall \alpha,\beta \in Ord~~~~~\alpha\in \beta \Longleftrightarrow \alpha \subsetneq \beta$

And the relation $\subsetneq$ in well defined between classes. Now the question is:

Question (1): Is there a known generalization of the notion of set ordinals to proper classes?

If the answer of the above question is positive, then:

Question (2): Is there a proper class order type for any proper class well ordering using reasonable assumptions like the Axiom of Global Choice?

If the answer of the above question is positive too, then:

Question (3): What is the order type of $L$ with Godel's well ordering?

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    $\begingroup$ $\mathsf{Ord}$. $\endgroup$ Oct 6, 2013 at 16:17
  • $\begingroup$ @Andres: Dear Andres, I am not sure about using text shape font for mathematical "symbols" and "abbreviations" like $Ord$ and $ZFC$. It seems these should be written as any other symbol in TeX. Is there any official manual to show what we should do in these cases? Can you guide me? $\endgroup$
    – user36136
    Oct 6, 2013 at 17:20
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    $\begingroup$ Dear Ali, I think that @Andres was merely answering the question in the title. The order type of the canonical ordering of $L$ is $\sf Ord$. $\endgroup$
    – Asaf Karagila
    Oct 6, 2013 at 17:26
  • $\begingroup$ @ Asaf: Dear Asaf. What a strange misunderstanding! Thanks. $\endgroup$
    – user36136
    Oct 6, 2013 at 17:45
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    $\begingroup$ Ali, that's just typography. One can use Ord or \mathrm{Ord} or \mathsf{Ord} or similarly ON or ORD. It's just important that the reader is clear on the meaning and that the notation is consistent. $\endgroup$
    – Asaf Karagila
    Oct 6, 2013 at 18:15

1 Answer 1


The first question can be answered by taking the idea of definable well-orderings which are not set-like. That is, we can consider the formula $\varphi(x,y)$ which states that $x,y$ are distinct ordinals and either $y=0$ or $x\neq 0$ and $x\in y$. It is not hard to see that $\varphi$ defines a well-order of order type $\sf Ord+1$ on the ordinals. This can go on, and we can ask (e.g. What is $\omega_1^{CK}(\mathsf{Ord})$?):

Given $M$ a transitive model of $\sf ZFC$, what is the least ordinal $\alpha$ such that $\alpha$ is not a first-order definable over $M$?

To the second question, the answer is in fact the the following things are equivalent:

  1. The axiom of global choice.
  2. Every class can be well-ordered.
  3. There exists a well-ordering of $V$.

The implications $(3)\implies(2)\implies(1)$ is trivial. The proof that $(1)\implies(3)$ is about the same as the proof that the existence of a choice function on $\mathcal P(X)\setminus\{\varnothing\}$ implies that $X$ can be well-ordered.

Finally, the well-ordering of $L$ is of order type $\sf Ord$. To see this, simply note that if $x\in L_\alpha$ then $x$ does not appear before $L_\alpha$ in the order. Therefore we have a cofinal class of sets, which have only set-many predecessors. This is enough to show that there is no point which has class-many predecessors so the only order type fitting this is $\sf Ord$ itself.


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