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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 t$h$in$k$ 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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    @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