Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
4  
$\mathsf{Ord}$. –  Andres Caicedo Oct 6 '13 at 16:17
    
@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? –  Ali Sadegh Daghighi Oct 6 '13 at 17:20
3  
Dear Ali, I think that @Andres was merely answering the question in the title. The order type of the canonical ordering of $L$ is $\sf Ord$. –  Asaf Karagila Oct 6 '13 at 17:26
    
@ Asaf: Dear Asaf. What a strange misunderstanding! Thanks. –  Ali Sadegh Daghighi Oct 6 '13 at 17:45
1  
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. –  Asaf Karagila Oct 6 '13 at 18:15

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.