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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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If $A$ has a largest member $\alpha=\max A$, then $\sup A=\alpha$; if $A$ has no largest member, then $\sup A$ is a limit ordinal. –  Brian M. Scott May 14 '12 at 10:17
    
Thanks. I knew it was trivial ... –  Marc Moretti May 14 '12 at 14:06
1  
@Marc: You can (and should) write an answer to your own question and accept it. –  Asaf Karagila May 16 '12 at 15:55

1 Answer 1

up vote 1 down vote accepted

The $sup$ is ok for me. Thanks

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