# Is my understanding of this proof about cardinality correct?

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

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