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.
Best Answer
The $sup$ is ok for me. Thanks