Let $f$ be a mapping, $\beta$ be an ordinal, $X=\{\alpha\mid f(\alpha)\le \beta\}$, and $\gamma=\sup X$. Is $\gamma\in X$?

Question

Let $f:\operatorname{Ord}\to\operatorname{Ord}$ be a mapping consisting of only addition, multiplication, and exponentiation operations, and $\beta$ be an ordinal. Let $X=\{\alpha\mid f(\alpha)\le \beta\}$ and $\gamma=\sup X$.

1. I would like to ask if $\gamma\in X$ or not?

2. Is there certain conditions on $f$ to make sure that $\gamma\in X$?

When I deal with simple mapping such as $f(\alpha)=\delta\cdot\alpha$, I found that $\gamma\in X$. I don't know if we can generalize it.

Thank you so much!

Answer 1

The property you are asking for is continuity. More succinctly written, continuity for $\operatorname{Ord}\to\operatorname{Ord}$ functions is when

$$f(\gamma)=\sup_{\alpha\in\gamma}f(\alpha)=\sup\{f(\alpha)~|~\alpha\in\gamma\}$$

for every limit ordinal $\gamma$. It is notable that the functions you are talking about are increasing functions, so if $\gamma\in X$, then $f(\gamma)\ge f(\delta)$ for every $\gamma>\delta$. Ordinal functions which are strictly increasing and continuous are called normal functions.

Determining when your expressions are normal is not so hard, as it is the case that $\alpha+\beta,\alpha\cdot\beta,$ and $\alpha^\beta$ are all normal in $\beta$, but they are not normal in $\alpha$.