A normal function is a class function $f$ from the class of ordinals to itself such that $f$ is strictly increasing and continuous. We can easily show that for every ordinal $\alpha$ there is an ordinal $\beta$ with $ \beta > \alpha$ such that $f(\beta) = \beta$. How can we show that the collection of ordinals $\beta$ such that $f(\beta)=\beta$ forms a proper class?
I imagine we need to show that if they formed a set we would be able to comprehend the set of ordinals, but I don't see how to go about showing that. Thanks
Suppose that $S = \{\beta: f(\beta)=\beta\}$ were a set.
Then so would $\alpha = \bigcup S$ be; and moreover it being a union of ordinals is itself an ordinal.
But then there would be a $\beta > \alpha$ such that $f(\beta)=\beta$. This contradicts the existence of $S$ as a set because for all $\beta' \in S$ we have $\beta' < \alpha < \beta$.