My professor stated the following
"For every normal function $f \colon ON \to ON$ and for every ordinal $\beta$ there exist an ordinal $\gamma \ge \beta$ such that $f(\gamma)=\gamma$". (normal function is by definition a continuos and increasing function)
I'm trying to prove it but i don't able to prove one fact:
Let $\gamma=\bigcup \{f^n(\beta) \colon n \in \omega \}$, where $f^0(\beta)=\beta$. I have to prove that this is a limit ordinal...but arguin by absurd i don't find the way to arrive to a contradiction.
HINT: Prove the following general statement: if $A$ is a set of ordinals whose order type is a limit ordinal, then $\sup A$ is a limit ordinal.