I've just identified that the definition we used for the rank of a set in my set-theory class (1.) is different than the one I commonly find on the web (2.).

  1. $\text{rank}(A)=\min\{\alpha\mid A\in{V}_\alpha\}$
  2. $\text{rank}(A)=\min\{\alpha\mid A\subseteq{V}_\alpha\}$

There are a couple of key differences.

Empty set:

  1. $\emptyset\in\{\emptyset\}={V}_1\implies\text{rank}(\emptyset)=1$
  2. $\emptyset\subseteq\emptyset={V}_0\implies\text{rank}(\emptyset)=0$


  1. $\alpha\in{V}_{\alpha+1}\implies\text{rank}(\alpha)=\alpha+1$
  2. $\alpha\subseteq{V}_\alpha\implies\text{rank}(\alpha)=\alpha$

Usefulness in proofs:

  1. $\text{rank}(A)$ is always a successor ordinal
    • $\not\exists{A}\left(\text{rank}(A)=\omega\right)$
    • $\not\exists{A}\left(\lim\left(\text{rank}(A)\right)\right)$
  2. $\text{rank}(A)$ is the smallest ordinal greater than the rank of every member of $A$
    • $\forall\alpha\exists{A}\left(\text{rank}(A)=\alpha\right)$

So here's what I'm wondering:

  • How common is the first definition (the less common one)?
  • Are there any other advantages or subtleties in either that I'm missing?
  • 2
    $\begingroup$ I think you've pretty much said all there is to be said. The first definition is a little strange in that the rank is never a limit ordinal, which seems unnatural. But in practice, the aesthetic difference is all there is to it. I've yet to come across the first one being used, but it does not make for much of a difference (as it is basically the second one +1). $\endgroup$ – tomasz Sep 2 '12 at 14:40

There is a very good merit for the first definition (which is the one I am accustomed to, actually).

You don't want any new information to be added in limit stages. You want all the information to be added in successor steps, and the limit points are just accumulation points, they accumulate everything you had so far.

In reality there is absolutely no difference between the two definitions, because it is not hard to see that the second is just taking a step back from the first.

Philosophically speaking, sets exists when they are elements, not subsets. This gives, in my opinion, a stronger foundation for the first definition. Note that $\varnothing$ is empty and this means that we essentially begin with nothing, therefore it is appropriate that no set should have rank zero.

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  • 1
    $\begingroup$ But don't real mathematicians start counting at $0$? $\endgroup$ – André Nicolas Sep 2 '12 at 14:52
  • $\begingroup$ @Andre, of course. But I still see it is fitting not to assign any set rank zero. The rank is the point where the set came to existence, a birthday -- if you will. No sets are born on the zeroth day. $\endgroup$ – Asaf Karagila Sep 2 '12 at 15:13
  • $\begingroup$ @AsafKaragila: The part about sets existing once they are elements instead of subsets (hence subclasses) makes sense. Could you expound upon how the second definition implies (or seems to imply) new information is added at limit stages? (Not the philosophy - not why you want limits to be accumulations instead of introducing new material - that I already understand, but how the definition relates to this philosophy.) $\endgroup$ – Travis Bemrose Sep 2 '12 at 16:22
  • $\begingroup$ @AndréNicolas: Before there is the empty set, there is nothing. The empty set is something - it's that nothingness put into a set. You can think of is as zero is the rank of nothingness: $\text{rank}(\ \ \ )=0$ $\endgroup$ – Travis Bemrose Sep 2 '12 at 16:32
  • $\begingroup$ @Travis: My comment to Asaf Karagila was meant as a joke. But we can make a hybrid of the two definitions, letting the empty set have rank $0$ and yet making all other ranks successor ordinals. The whole distinction between the several ways of defining rank doesn't matter, all we want is a complexity measure to do induction on. $\endgroup$ – André Nicolas Sep 2 '12 at 16:37

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