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

Ordinals:

  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?
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    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).2012-09-02

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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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    But don't *real* mathematicians start counting at $0$?2012-09-02
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    @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.2012-09-02
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    @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.)2012-09-02
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    @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$2012-09-02
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    @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.2012-09-02
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    @AndréNicolas: Ah the shortcomings of the written language. So much non-verbal communication is lost. Actually under the "Ordinals" and "Usefulness in proofs" sections above, I consider it a **strength** of the first definition that $\text{rank}(\alpha)=\alpha+1$ and every rank is a successor ordinal. Those conditions are universal. I think a hybrid definition would sacrifice the strengths of both definitions.2012-09-02
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    @Travis: Limit points are never a good place to add information in. Limit points are where you are suppose to regroup and ask yourself "What have I got so far?".2012-09-02
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    @AsafKaragila: lol, that's the opposite of what I was asking. I know and understand that. I was asking how does that relate to def 2? How does def 2 seem to introduce new info at limit points?2012-09-03
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    @Travis: Picture yourself in a boat on a river, and the river is doing a transfinite induction on the rank of sets. It's a river, so it needs your help. Using the first definition the limit stages are trivial and require no work -- you only added new information at successor stages. The second definition requires *some* work, because there are *new* sets whose rank is a limit ordinal.2012-09-03
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    @AsafKaragila: I guess I don't see what you mean by there are _new_ sets using the second definition. When I construct $V_\omega$, $\omega$ is not yet a set so I wouldn't be asking what its rank is. When I construct $V_{\omega+1}$ I now have the set $\omega$ and it's a subset of $V_\omega$ so it's rank is $\omega$. So I don't ever have a set whose rank has not been defined yet. It looks like a "labeling" difference to me. I don't see a hierarchical or structural difference.2012-09-26