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:

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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!

  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.
  2. Yes, this is correct.
  3. Yes, this is correct.

You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.

  • $\begingroup$ 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. $\endgroup$ – spaceisdarkgreen Dec 29 '18 at 6:17
  • $\begingroup$ 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! $\endgroup$ – Le Anh Dung Dec 29 '18 at 7:39
  • $\begingroup$ @LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist. $\endgroup$ – spaceisdarkgreen Dec 29 '18 at 7:45
  • $\begingroup$ Thank you for all of your dedicated help! $\endgroup$ – Le Anh Dung Dec 29 '18 at 7:57

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