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.

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


  • $\begingroup$ Wow. That's certainly concise and elegant! $\endgroup$ Dec 26 '12 at 8:59
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    $\begingroup$ 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. $\endgroup$ Dec 26 '12 at 16:42
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    $\begingroup$ Related: Discussion: Differing definitions for the rank of a set $\endgroup$
    – jeq
    Jun 24 '17 at 14:15

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 $X$: $$ \min\{\alpha \mid X \subseteq\alpha\} \\ \{ \gamma \mid \gamma \leq \beta \text{ for some } \beta \in X \} \\ \sup \{ \beta+1 \mid \beta \in X \} $$

None of these are as concise as Amit's answer, and they also aren't as concise as I had in mind when asking the question, but they may be the best we can hope for if we are concerned about the pitfalls raised in the comments (by Andreas and jeq).


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