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Let $X$ be a set of ordinals.

If $X$ has no largest element, then \[ \sup X \notin X \subseteq \sup X, \] and $\sup X$ is the smallest ordinal $\alpha$ such that $X \subseteq \alpha$.

On the other hand, if $X$ has a largest element $\max X$, then \[ \max X = \sup X \in X \nsubseteq \sup X, \] and the smallest ordinal $\alpha$ such that $X \subseteq \alpha$ is $\sup X + 1$.

Is there any way to express "the smallest ordinal $\alpha$ such that $X \subseteq \alpha$" that works in both cases?

One possibility would be \[ \sup \{ \beta + 1 : \beta \in X \}, \] but I'm looking for something more concise or elegant than that.

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I've seen $\mathrm{ssup}(X)$, the strict supremum of $X$. I'm not a fan though. –  François G. Dorais Dec 26 '12 at 6:40
    
I have seen $\sup^+(X)$ used in this context. –  Asaf Karagila Dec 26 '12 at 7:43
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1 Answer

up vote 4 down vote accepted

$\mathrm{rank}(X)$

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Wow. That's certainly concise and elegant! –  Ari Brodsky Dec 26 '12 at 8:59
    
Very efficient, but, unlike the less efficient possibility in the question, this depends on the specific representation of ordinals by sets. It works for the usual von Neumann ordinals, but not for other possible representations. –  Andreas Blass Dec 26 '12 at 16:42
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