Weakly inaccessible cardinals and Discovering Modern Set Theory

Question

So I've been trying to teach myself some set theory and I've come across some exercises in Just and Weese's Discovering Modern Set Theory. To whit:

Pg. 180

Definition 20: A cardinal $\kappa$ is called weakly inaccessible if $\kappa$ is an uncountable regular limit cardinal.

Exercise 27 (PG): Show that if $\alpha$ is a weakly inaccessible cardinal, then $\alpha=\aleph_\alpha$.

Exercise 28 (R): Show that the smallest ordinal $\alpha$ such that $\alpha=\aleph_\alpha$ is not a weakly inaccessible cardinal.

Ends

For the first I've proved by induction that $\alpha\subseteq\aleph_\alpha$ for all ordinals $\alpha$, obviously I have to use the weakly inaccessible part to go the other way, but I've no idea where to start. Weakly inaccessible gives that every function with cofinal range in $\alpha$ has domain at least $\alpha$ but that is saying that things are big, where as I need that $\aleph_\alpha$ is small.

The second is beyond me, I considered trying to show that ZFC proves the existence of such an ordinal $\alpha$, and then appeal to the fact that the existence of a weakly inaccessible cardinal is independent of ZFC, however this result is yet to be obtained in the text and so I imagine this can't be what is wanted.

Thoughts?

Answer 1

The second one is easier, simply define by recursion $\lambda_0=\mu$, $\lambda_{n+1}=\aleph_{\lambda_n}$, for $n\in\omega$ ($\mu$ is any cardinal). Then calculate what is $\lambda_\omega=\sup\{\lambda_n\mid n\in\omega\}$.

The first one is not much harder either. Because $\aleph_\alpha$ is limit, $\alpha$ is a limit ordinal. Pick a cofinal sequence in $\alpha$, $\langle\delta_i\mid i<\operatorname{cf}(\alpha)\rangle$. Consider the sequence $\aleph_{\delta_i}$. What is its limit? What can you conclude on $\operatorname{cf}(\alpha)$ from the assumption that $\aleph_\alpha$ is regular as well?

Answer 2

For Exercise $28$ let $\alpha_0=0$. Given $\alpha_n$, let $\alpha_{n+1}=\aleph_{\alpha_n}$. Then

$$\alpha=\sup_{n\in\omega}\alpha_n=\sup_{n\in\omega}\alpha_{n+1}=\sup_{n\in\omega}\aleph_{\alpha_n}=\aleph_\alpha\;,$$

but $\operatorname{cf}\alpha=\omega<\alpha$. It’s not hard to see that this $\alpha$ is the smallest ordinal satisfying $\alpha=\aleph_\alpha$.

Added: For Exercise $27$ you’ve already shown that $\alpha\le\aleph_\alpha$. If $\alpha<\aleph_\alpha$, then $\alpha=\aleph_\beta$ for some $\beta<\alpha$, so $\operatorname{cf}\alpha\le\beta<\alpha$, contradicting the regularity of $\alpha$.