3.8 Therorem Let us assume the Generalized Continuum Hypothesis. If $\aleph_\alpha$ is a regular cardinal, then $$\aleph_\alpha^{\aleph_\beta}=\begin{cases}
\aleph_\alpha&\text{if }\beta<\alpha\\\aleph_{\beta+1}&\text{if }\beta\ge\alpha\end{cases}$$
My textbook presents the theorem and its proof as follows:
I would like to ask if my understanding of the below statement is correct. $$B=\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)\text{ be the collection of all bounded subsets of } \omega_\alpha$$
Thank you for your help!
- $\delta<\omega_\alpha \implies \delta$ is bounded
If not, there exists $\delta<\omega_\alpha$ such that $\sup \delta=\omega_\alpha$. Let $(\beta_\xi\mid\xi<\lambda)$ be an increasing enumeration of $\delta$. Then $|\lambda|=|\delta|\le\delta<\omega_\alpha$ and $\lim_{\xi\to\lambda}\beta_\xi=\sup \delta=\omega_\alpha$. It follows that $\aleph_\alpha$ is singular, which contradicts the fact that $\aleph_\alpha$ is regular.
- $X$ is a bounded subset of $\omega_\alpha$ $\implies X\subseteq\delta$ for some $\delta<\omega_\alpha$
Since $X$ is a bounded subset of $\omega_\alpha$, $\sup X<\omega_\alpha$. We have $X\subseteq \{\gamma \in\text{ Ord} \mid \gamma \le \sup X\}$. It is clear that $\{\gamma \in\text{ Ord} \mid \gamma \le \sup X\}$ is a proper initial segment of $\omega_\alpha$ and thus an ordinal. Then $\{\gamma \in\text{ Ord} \mid \gamma \le \sup X\} =\delta$ for some $\delta<\omega_\alpha$.
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$B$ is the collection of all bounded subsets of $\omega_\alpha$ $\implies B=\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$
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$X\in B \implies$ $X$ is a bounded subsets of $\omega_\alpha$ $\implies$ $X\subseteq\delta$ for some $\delta<\omega_\alpha$ $\implies$ $X\in\mathcal{P}(\delta)$ for some $\delta<\omega_\alpha$ $\implies$ $X\in\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$.
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$X\in\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$ $\implies$ $X\in\mathcal{P}(\delta)$ for some $\delta<\omega_\alpha$ $\implies$ $X\subseteq\delta$ for some $\delta<\omega_\alpha$ $\implies$ $\sup X \le \sup \delta < \omega_\alpha$ [Since $\delta$ is bounded] $\implies$ $X$ is bounded.
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Best Answer
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.