Is there a normal function where infinite cardinals are exactly the fixed points?

Question

A normal function is an increasing function $F\colon \mathrm{Ord} \to \mathrm{Ord}$ where for limit $\alpha$, $F(\alpha) = \sup_{\beta < \alpha} F(\beta)$. Every normal function has arbitrarily large fixed points, and enumerating them makes a new normal function. Examples of normal functions are $\alpha \mapsto \omega^\alpha, \alpha \mapsto \epsilon_\alpha, \alpha \mapsto \omega_\alpha$. The first two maps preserves cardinality for infinite ordinals, easily proved with $\mathsf{AC}$ because successor step preserves cardinality, and without $\mathsf{AC}$ because they are absolute between $L$ and $V$. It follows that they fix uncountable cardinals. But there are non-cardinal fixed points, for example $\epsilon_0$ for the first. Is there a normal function where the set of infinite cardinals, or some tail segment of it is exactly the fixed points?

Answer 1

No.

Suppose that $\omega_1$ is a fixed point of some function $F(\alpha)$, then $F(\omega_1)=\omega_1$. Since the function is normal, that means that $\alpha<\omega_1$ implies that $F(\alpha)<\omega_1$ as well.

But now we can show that there is a club of fixed points below $\omega_1$. The same holds for any uncountable cofinality.

One way around this would be to consider a model of $\sf ZF$ in which every infinite ordinal has countable cofinality, but off the cuff that's not going to work either since it will require you to identify a countable cofinal sequence in each ordinal, which seems like a bit more choice than what you'd usually have.¹

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1. That's not to say that it's inconsistent or impossible. It may very well be doable. You'd still need to come up with that definition, though. Probably if $\alpha_n$ is the cofinal sequence up to $\omega_\alpha$, you'd like to have something like $F(\xi)$ to be the least $\alpha_n$ above $\xi$ (in the appropriate interval), and the identity on the cardinals.

   But frankly, that seems like we're trying too hard.