# Is my understanding of a proof from textbook Introduction to Set Theory by Hrbacek and Jech correct?

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. $$B=\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)\text{ be the collection of all bounded subsets of } \omega_\alpha$$

1. $$\delta<\omega_\alpha \implies \delta$$ 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.

1. $$X$$ 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 $$\delta<\omega_\alpha$$.

1. $$B$$ is the collection of all bounded subsets of $$\omega_\alpha$$ $$\implies B=\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$$

• $$X\in B \implies$$ $$X$$ is a bounded subsets of $$\omega_\alpha$$ $$\implies$$ $$X\subseteq\delta$$ for some $$\delta<\omega_\alpha$$ $$\implies$$ $$X\in\mathcal{P}(\delta)$$ for some $$\delta<\omega_\alpha$$ $$\implies$$ $$X\in\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$$.

• $$X\in\bigcup_{\delta<\omega_\alpha}\mathcal{P}(\delta)$$ $$\implies$$ $$X\in\mathcal{P}(\delta)$$ 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.

1. This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $$\sup\delta \le \delta < \omega_\alpha.$$ And this has nothing to do with regularity.
• If you don't know already, a good thing to think about next would be where in the proof regularity of $\aleph_\alpha$ is used. – spaceisdarkgreen Dec 29 '18 at 6:17
• I have fixed 1. as follows: $\forall\gamma\in\delta:\gamma<\delta \implies \sup \delta \le \delta < \omega_\alpha \implies \delta$ is bounded. I guess you have some implicit assumption when writing $\color{blue}{\sup\delta = \delta < \omega_\alpha}$. I have a counter-example: $5=\{0,1,2,3,4\}$ and thus $\sup 5= 4 \neq 5$. I think the regularity of $\aleph_\alpha$ is used in the statement every $X\in S$ is a bounded subset of $\omega_\alpha$. Please have a check on my above reasoning. Thank you for your help! – Le Anh Dung Dec 29 '18 at 7:39