# Weakly inaccessible cardinal equivalent to regular aleph fixed point?

Are the following propositions equivalent ?

• $$\kappa$$ is a weakly inaccessible cardinal
• $$\aleph_\kappa = \kappa \land cof(\kappa) = \kappa$$
• Can you please explain why you think this holds? In particular, why the condition $\aleph_\kappa = \kappa$? Commented Oct 29, 2019 at 11:29
• The condition $\aleph_\kappa = \kappa$ expresses the fact that $\kappa$ is an aleph fixed point, i.e. a fixed point of the function $\alpha \mapsto \aleph_\alpha$. $cof(\kappa)=\kappa$ expresses the fact that $\kappa$ is regular. I think this could hold because in mathoverflow.net/questions/64955/… I read that the inaccessible cardinals are precisely the regular fixed points of the beth function, so I wonder if there is something equivalent for weakly inaccessible cardinals. Commented Oct 29, 2019 at 12:12

If $$\kappa=\aleph_\kappa$$, then $$\kappa$$ is not a successor cardinal (if $$\kappa$$ were the successor of $$\aleph_\alpha$$, then we would have $$\kappa=\aleph_\alpha^+=\aleph_{\alpha+1}$$ and the successor ordinal $$\alpha+1$$ is not a cardinal), so it is a weak limit.

When we also assume that $$\mathrm{cf}(\kappa)=\kappa$$, this will together with the previous imply that $$\kappa$$ is regular and a weak limit; that is, $$\kappa$$ is weakly inaccessible.

On the other hand, assume $$\kappa$$ is weakly inaccessible and let $$\kappa=\aleph_\alpha$$ for some $$\alpha\leq\kappa$$. We can see that $$\alpha\neq\beta+1$$ for any ordinal $$\beta$$ (otherwise $$\aleph_\alpha=\aleph_{\beta+1}=\aleph_\beta^+$$ means that $$\aleph_\alpha$$ is a successor cardinal and thus not a weak limit), thus see that $$\alpha$$ must be a limit ordinal. Since $$\kappa$$ is regular and $$\alpha$$ is a limit ordinal, we have $$\kappa=\aleph_\alpha=\mathrm{cf}(\aleph_\alpha)=\mathrm{cf}(\alpha)\leq\alpha\leq\kappa.$$ So $$\alpha=\kappa$$, and thus indeed $$\kappa=\aleph_\kappa$$.

If 𝜅 is a regular limit cardinal, let $$S$$ be the set of cardinals less than 𝜅.
Then $$\sup S=$$ 𝜅 because 𝜅 is a limit cardinal.
So $$S$$ is cofinal in 𝜅, so $$|S|\ge \operatorname{cof}(𝜅)=𝜅.$$
So $$|S|=𝜅,$$ since $$S\subset 𝜅.$$
So $$S$$ is $$\in$$-order-isomorphic to $$𝜅.$$
So $$𝜅$$ is the $$𝜅$$-th cardinal,i.e. $$𝜅=\aleph_𝜅.$$

If $$𝜅=\aleph_𝜅$$ then $$𝜅$$ is a limit cardinal by def'n of $$\aleph_𝜅.$$
So if also $$\operatorname{cof}(𝜅)=𝜅$$ then $$𝜅$$ is a regular limit cardinal.