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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.

flag	asked Dec 26 at 1:01 Ari Brodsky 3●1

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

Amit Kumar Gupta · Accepted Answer · 2012-12-26 01:24:24Z UTC

4

$\mathrm{rank}(X)$

answered Dec 26 at 1:24

Amit Kumar Gupta
2,909●10●26

	Wow. That's certainly concise and elegant! – Ari Brodsky Dec 26 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 at 16:42

Is there a name for the smallest ordinal $\alpha$ such that $X \subseteq \alpha$

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