# How can we express the smallest ordinal $\alpha$ such that $X \subseteq \alpha$?

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.

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

$\mathrm{rank}(X)$
For a given set $$X$$ of ordinals, the following formulas all refer to the same ordinal --- namely, the smallest ordinal larger than every element of $$X$$: $$\min\{\alpha \mid X \subseteq\alpha\} \\ \{ \gamma \mid \gamma \leq \beta \text{ for some } \beta \in X \} \\ \sup \{ \beta+1 \mid \beta \in X \}$$