Define: $\beth_0=\omega$
$\beth_{\alpha+1}=|\mathcal{P}(\alpha)|=2^{\beth_{\alpha}}$
$\beth_{\lambda}=\sup_{\alpha<\lambda}{\beth_{\alpha}}$ for $\lambda$ a limit ordinal.
(a) Show that $\beth_{\alpha}$ is an initial ordinal for each ordinal $\alpha$.
(b) Show that there exists a $\beth$ fixed point, i.e. a limit ordinal $\kappa$ such that $\beth_{\kappa}=\kappa$.
For (a), I am doing transfinite induction on $\alpha$. $\alpha=0$ is clear, since $\omega$ is an initial ordinal. Now, for $\alpha$ successor, I have the following question: our IH is that $\beth_{\alpha}$ is initial, and we have to show that $\beth_{\alpha+1}$ is initial too. It is clear that $\beth_{\alpha+1}=|\mathcal{P}(\beth_{\alpha})|>|\beth_{\alpha}|=\beth_{\alpha}$ by Cantor's theorem and the IH, since $|\beth_{\alpha}|$ is the unique initial ordinal equipotent to $\beth_{\alpha}$, and by IH, $\beth_{\alpha}$ is an initial ordinal. However, why can't it be the case that there exists $\beth_{\alpha}<\beta<\beth_{\alpha+1}$ and $\beta$ equipotent to $\beth_{\alpha+1}$? Because clearly $\beth_{\alpha}<\beth_{\alpha}+1<\beth_{\alpha+1}$, so indeed there are ordinals between those two (though in this case obviously they are not equipotent).
Same happens for the limit case: IH: For all $\gamma<\alpha$, $\beth_{\gamma}$ is an initial ordinal. Then why is $\beth_{\alpha}=\bigcup_{\lambda<\alpha}\beth_{\lambda}$ not equipotent to $\beta$ for any $\beta<\beth_{\alpha}$?
Maybe there is something that I'm missing.
For (b) I have proven in class that a normal function has arbitrarily high fixed points, and I think the Beth function is normal. But I was wondering if there was a concrete example I could give to illustrate this. I thought about $\beth_{\beth_{\omega}}$, but I think that one is not a fixed point (though I do not know how to prove it). Maybe applying that process infinitely?
(a) comes from the definition of cardinal, since $\beth_\alpha$ is always defined as "the cardinal $\kappa$ such that [something to do with bijections to the function set $^{(\beth_\beta)}2$ if $\alpha=\beta+1$, or $\bigcup_{\beta<\alpha}\beth_\beta$ if $\alpha$ is limit]".
(b) comes from the fixed point lemma that you proved.
Firstly, $\beth_{\beta+1}>\beth_\beta$ for all $\beta$, because Cantor's theorem implies that the function set $^{(\beth_\beta)}2$ is of greater cardinality than $\beth_\beta$. Also, $\beth$ is monotonic: if $\beta\leq\alpha$, then $\beth_\beta\leq\beth_\alpha$, since $^{(\beth_\beta)}2$ is a subset of $^{(\beth_\alpha)}2$.
Then $\beth$ is a normal function as you suspect: