# Weakly inaccessible cardinals and Discovering Modern Set Theory

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?

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?

• Thanks, I guess you are dubious of the difficulty rating system if you think the second is easier. – James Jun 6 '13 at 2:40
• I am; but also the second one is just a trick. And once you get that trick it's very easy to solve; but the first one relies on no tricks. It's just verifying the definitions and that can be a pain in the backside area which connects the thighs with the lower back. – Asaf Karagila Jun 6 '13 at 2:41
• Clearly $\lambda_\omega$ has $\lambda_\omega = \aleph_{\lambda_\omega}$, but I think the 'R' rating there comes from the fact that it's not immediately obvious that it's the smallest cardinal of that form, or even that there isn't an inaccessible smaller than it. (It's pretty straightforward, but still takes a bit more arguing, I think.) – Steven Stadnicki Jun 6 '13 at 2:45
• I kinda like the trick, it feels like approximating what you want and taking the limit, which I suppose is precisely what you are doing. – James Jun 6 '13 at 2:52
• @James: It's not quite that trick, this is the trick we use in many other areas (e.g. forcing). I can't quite describe this trick in nice words, though. Maybe it's just getting late... – Asaf Karagila Jun 6 '13 at 2:53

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 $\alpha$.