The cardinal inequality question that I would like to get some help about is a crucial inequality used in the proof of Lemma 8.14 (using Lemma 8.15) in Jech.
If we assume $\aleph_\alpha^{\aleph_1}<\aleph_{\omega_1}$ for all $\alpha<\omega_1$ and $\aleph_{\alpha}^{\text{cf }\aleph_\alpha}=\aleph_{\alpha+1}$ for all $\alpha\in S$ where $S\subset \omega_1$ is a stationary set, then can we conclude that there exists a stationary set $S_0\subset\omega_1$ such that $\aleph_\alpha^{\aleph_1}=\aleph_{\alpha+1}$ for all $\alpha\in S_0$? Jech notes on Page 98 that we can take $S_0$ to be the set $$S_0:=\{\alpha<\omega_1:\aleph_\alpha>2^{\aleph_1}\text{ and }\aleph_\alpha^{\aleph_0}=\aleph_{\alpha+1}\}$$
I can see why $S_0$ is a stationary subset of $\omega_1$ (since $S\cap C$ where $C:=\{\text{limit ordinals less than }\omega_1\}$ is a stationary subset), but I couldn't really figure out why $S_0$ is such that $\alpha\in S_0$ implies $\aleph_\alpha^{\aleph_1}=\aleph_{\alpha+1}$. Please advise me!
I have came up with a possibly wrong proof, so please correct me if I got anything wrong. Also, please feel free to post any alternative solution! Any help is appreciated!
If we show that $\aleph_\alpha^{\aleph_0}=\aleph_\alpha^{\aleph_1}$ for all limit ordinals $\alpha<\omega_1$ with $\aleph_\alpha>2^{\aleph_1}$, then we are basically done. Indeed, $$T:=S\cap\{\alpha<\omega_1:\aleph_{\alpha}>2^{\aleph_1}\text{ and }\alpha\text{ is a limit ordinal}\}$$ is a stationary subset of $\omega_1$ and $\alpha\in T$ implies $\aleph_\alpha^{\aleph_1}=\aleph_\alpha^{\aleph_0}=\aleph_{\alpha+1}$.
We will prove the above claim by induction.
Base case: Let $\theta<\omega_1$ be the first limit ordinal such that $\aleph_\theta>2^{\aleph_1}$. We have $\aleph_1^{\aleph_1}=\max\{2^{\aleph_1},\aleph_1\}<\aleph_\theta$. And if $\aleph_\nu^{\aleph_1}=\max\{2^{\aleph_1},\aleph_\nu\}$, then $\aleph_{\nu+1}^{\aleph_1}=\max\{2^{\aleph_1},\aleph_{\nu+1}\}<\aleph_\theta$. If $\gamma$ is a limit ordinal such that $\gamma<\theta$ (meaning that $\aleph_\gamma\leq 2^{\aleph_1}$), then $\aleph_\gamma^{\aleph_1}=\max\{2^{\aleph_1},\aleph_\gamma\}$ by the fact that $\aleph_\gamma\leq 2^{\aleph_1}$. If $\gamma=\theta$ and $\nu<\gamma$ implies $\aleph_\nu^{\aleph_1}=\max\{2^{\aleph_1},\aleph_\nu\}$, then $\aleph_\nu^{\aleph_1}<\aleph\gamma$ for all $\nu<\gamma$, hence $\aleph_\gamma^{\aleph_1}=\aleph_\gamma^{\text{cf } \aleph_\gamma}=\aleph_\gamma^{\aleph_0}$. Since $\gamma=\theta$, the base case is just proven.
Induction step: Assume $\zeta<\omega_1$ is a limit ordinal such that $\aleph_\zeta>2^{\aleph_1}$ and such that for all limit ordinals $\nu<\zeta$ with $\aleph_\nu>2^{\aleph_1}$, we have $\aleph_\nu^{\aleph_1}=\aleph_\nu^{\aleph_0}$. We can show by induction that $\aleph_\nu^{\aleph_1}=\max\{\aleph_\nu,\aleph_\nu^{\aleph_0}\}=\aleph_\nu^{\aleph_0}$ for all $\theta\leq\nu<\zeta$ (where $\theta$ is as defined in the base case). If there exists $\nu<\zeta$ such that $\aleph_\nu^{\aleph_1}=\aleph_\nu^{\aleph_0}\geq \aleph_\zeta$, then $\aleph_\zeta^{\aleph_1}=\aleph_\nu^{\aleph_1}=\aleph_\nu^{\aleph_0}\leq \aleph_\zeta^{\aleph_0}\leq \aleph_\zeta^{\aleph_1}$.
If $\aleph_\nu^{\aleph_1}=\aleph_\nu^{\aleph_0}<\aleph_\zeta$ for all $\nu<\zeta$, then $\aleph_\zeta^{\aleph_1}=\aleph_\zeta^{\text{cf }\aleph_\zeta}=\aleph_\zeta^{\aleph_0}$.