Weakly inaccessible cardinal equivalent to regular aleph fixed point?

Question

Are the following propositions equivalent ?

• $\kappa$ is a weakly inaccessible cardinal
• $\aleph_\kappa = \kappa \land cof(\kappa) = \kappa$

Answer 1

The answer is yes.

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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.

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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 $\kappa=\aleph_\kappa$.

Answer 2

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

If $k=\aleph_k$ then $k$ is a limit cardinal by def'n of $\aleph_k.$ So if also $cof(k)=k$ then $k$ is a regular limit cardinal.