In my textbook Introduction to Set Theory by Hrbacek and Jech, there is a theorem:

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and its corresponding proof:

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I would like to ask if my understanding of the proof in case $\color{blue}{\aleph_\beta < \operatorname{cf}(\aleph_\alpha)}$ is correct.

Let $S=\{X \subseteq \omega_\alpha \mid |X|=\aleph_\beta\}$. We next prove that $X\in S \implies X$ is bounded. Assume the contrary that there exists $X' \in S$ such that $X'$ is unbounded and thus $\sup X'=\omega_\alpha$. Let $(\delta_\xi \mid \xi<\lambda)$ be an increasing enumeration of $X'$. It follows that $|\lambda|=|X'|=\aleph_\beta$ and $\lim_{\xi \to \lambda}\delta_\xi=\sup X'=\omega_\alpha$. By definition of cofinality, $\operatorname{cf}(\aleph_\alpha) \le \lambda$.

We have $\operatorname{cf}(\aleph_\alpha) \le \lambda \implies |\operatorname{cf}(\aleph_\alpha)| \le |\lambda| \implies \operatorname{cf}(\aleph_\alpha) \le |\lambda| = \aleph_\beta \implies \operatorname{cf}(\aleph_\alpha) \le \aleph_\beta$. This contradicts the fact that $\aleph_\beta < \operatorname{cf}(\aleph_\alpha)$.

Thank you for your help!

  • $\begingroup$ Yes, your understanding is correct $\endgroup$ – Holo Dec 29 '18 at 12:17
  • $\begingroup$ Thank you so much for your verification @Holo! $\endgroup$ – Le Anh Dung Dec 29 '18 at 12:40

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