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Let $A$ a set of ordinals. We know that $\sup A:=\bigcup A$ is an ordinal. Frequently, in proofs, one use that it is a limit ordinal. I would want to know when it is.

To show that it is limit : let $\beta<\sup A$, there exists an ordinal $\gamma\in A$ so that $\beta<\gamma\leq \sup A$ (if not, every $\gamma\in A$ is least than $\beta$ so $\sup A\leq \beta$, contradiction). So, I want to say that if $\gamma<\sup A$ then $\sup A$ is limit ($|A|$ is it necessary limit ?). For example, $\sup(\omega+2)=\omega+1$ successor

But if the unique $\gamma$ such that $\beta<\gamma\leq \sup A$ is $\gamma=\sup A$, it means that $\sup A=\max A$ and it is limit iff $\max A$ is limit.

If we can't find a $\gamma\in A$, maybe we can find an ordinal $\gamma$ such that $\beta<\gamma<\sup A$, that is, when $\beta\in\bigcup\bigcup A$ so when $\bigcup A=\bigcup\bigcup A$.

Can somebody make the notion of $sup$ clearest for me ? Thanks.

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    @Marc: You can (and should) write an answer to your own question and accept it.2012-05-16

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The $sup$ is ok for me. Thanks

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Just in case you don't completely understand, the supremum of a well-ordered set $(A,<)$ with no maximim (given a larger well-ordered set $(B,<)$) is the least element in $B$ not in $A$.

For limot ordinals, the supremum of a set $A$ is the smallest ordinal not in $A$.

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    If $A$ is a set of ordinals with a maximum element, then the supremum of $A$ is *not* "the smallest ordinal not in $A$".2017-09-18