Normal function on an ordinal

A function $f:\lambda \to \lambda$ on an ordinal is normal when it is strictly increasing and continuous at limit ordinals: $f(\delta )={\sup }_{\alpha <\delta }f(\alpha )$ for every limit $\delta <\lambda$.

Ordinal addition, multiplication and exponentiation are continuous in their right argument by recursion. For fixed $\beta$, $\beta +\alpha$ is normal for every $\beta$, $\beta \alpha$ for $\beta >0$, and ${\beta }^{\alpha }$ for $\beta \ge 2$. Reversing the roles changes the conclusion: $\alpha +\beta$ is not normal for $\beta >0$, while $\alpha \beta$ and ${\alpha }^{\beta }$ are normal only for fixed $\beta =1$. Large principal-ordinal limits expose the failures of continuity; constant zero or one cases fail strict increase.

For a strictly increasing ordinal class function continuous at limits, start with ${\alpha }_0=0$ and iterate ${\alpha }_{n+1}=f({\alpha }_n)$. If the sequence stabilizes, its stable value is a fixed point. Otherwise its supremum $\alpha$ is a limit and continuity gives $f(\alpha )={\sup }_{n}f({\alpha }_n)={\sup }_n{\alpha }_{n+1}=\alpha$. Starting the iteration above any prescribed ordinal gives arbitrarily large fixed points.

If a normal ordinal function defined on all ordinals begins at a countable value and preserves countability from $\Phi (\xi )$ to $\Phi (\xi +1)$, transfinite induction shows it maps all countable ordinals to countable ordinals. At countable limits this uses countable unions of countable sets, hence axiom of countable choice in the usual proof. Iteration from zero either stabilizes at a countable fixed point or has a countable limit fixed by continuity. Applying this to ${\omega }^{\alpha }$ gives the countable epsilon number ${ϵ}_0$.