I am currently looking at what Drake calls the Axiom Schema F, "Every normal function defined for all ordinals has a regular fixed point".
In ZFC+(Axiom F), does it hold that there is a club class of inaccessibles? It clear to me that they are unbounded, but I fail to see whether it's necessarily closed. Given a sequence of (0-)inaccessibles, if there is a limit, is it always regular (thus 1-inaccessible)?
For context: it seems to me that this would be a more elegant way to prove second part of Theorem 4.1 in ch.4 in Drake's Set Theory, An Introduction to Large Cardinals. Since $R_0$ (standard Levy's reflection) implies that given formula $\varphi$, there is a club class of $\alpha$s such that $\varphi \iff \varphi^{V_\alpha}$. If what I'm saying above holds, there would easily be a club class of inaccessible $\alpha$s satisfying the same property. But the same proof is very technical in both Levy's Axiom Schemata of Strong Infinity and Drake.
No, due to a simple reason: Let $C$ be a club class and let $\mu$ be any infinite cardinal (e.g. $\mu = \omega$). Then there is some $\kappa \in C$ such that $\operatorname{cf}(\kappa) = \operatorname{cf}(\mu)$.
Proof. Recursively construct a sequence $(\kappa_{i} \mid i \le \mu)$ be fixing any $\kappa_{0} \in C$, further letting $\kappa_{\alpha+1} := \min C \setminus \{ \kappa_{i} \mid i \le \alpha\}$ for successor ordinals $\alpha+1 \le \mu$ and finally, by letting $\kappa_{\lambda} := \sup \{ \kappa_{i} \mid i < \lambda \}$ for limit ordinals $\lambda \le \mu$. Now, by construction and the fact that $C$ is closed, $\{ \kappa_{i} \mid i \le \mu \} \subseteq C$ and $\kappa_{\mu}$ has cofinality $\operatorname{cf}(\mu)$. Q.E.D.
(The latter follows from the general fact that given a normal sequence $(\kappa_{i} \mid i < \mu)$ of ordinals, we have $\operatorname{cf}(sup_{i < \mu} \kappa_{i}) = \operatorname{cf}(\mu)$.)