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?
Is there a normal function where infinite cardinals are exactly the fixed points
cardinalsordinalsset-theory
Best Answer
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
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.