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My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:

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Since the authors refer to Theorem 2.2, I post it here for reference:

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My question concerns Part (b) of Theorem 3.13

Let $\kappa$ be a strongly inaccessible cardinal. If each $X\in S$ has cardinality $< \kappa$ and $|S| < \kappa$, then $\bigcup S$ has cardinality $< \kappa$.

Proof:

Let $\lambda = |S|$ and $\mu = \sup \{|X| \mid X \in S\}$. Then (by Theorem 2.2(a)) $\mu < \kappa$ because $\kappa$ is regular, and $|\bigcup S| \le \lambda \cdot \mu <\kappa$.

In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:

Let $\lambda = |S|$ and $\mu = \sup \{|X| \mid X \in S\}$. We have $\forall X \in S:|X| < \kappa \implies \mu \le \kappa$.

I claim that $\mu < \kappa$. If not, $\mu = \kappa$ and thus $\{|X| \mid X \in S\}$ is cofinal in $\kappa$. It follows that $\operatorname{cf}(\kappa) \le |\{|X| \mid X \in S\}| \le |S|< \kappa$ and thus $\operatorname{cf}(\kappa) < \kappa$. Then $\kappa$ is singular, which contradicts the fact that $\kappa$ is regular.

Hence $\mu < \kappa$. We have $|\bigcup S| \le \lambda \cdot \mu =\max\{\lambda, \mu\} <\kappa$.

Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?

Thank you for your help!

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Your way is just the way they use the theorem, all of $X\in S$ are bounded by $\gamma_X$, if $\sup|X|$ is unbounded, the sequence generated from $\gamma_X$ is unbounded, but the sequence is of length $\lambda<\kappa$, which is contradiction

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  • $\begingroup$ I seem to got it. Please check my reasoning! Assume the contrary that $\mu = \sup \{|X| \mid X \in S\} =\kappa$ or equivalently $\{|X| \mid X \in S\}$ is unbounded in $\kappa$. Then by every unbounded subset of $\kappa$ has cardinality $\kappa$ from Theorem 2.2(a), we have $|\{|X| \mid X \in S\}|=\kappa$. On the other hand, $|\{|X| \mid X \in S\}| \le |S|< \kappa$. Thus $\kappa < \kappa$, which is a contradiction. $\endgroup$ – Le Anh Dung Dec 30 '18 at 6:51
  • $\begingroup$ @LeAnhDung yes, correct $\endgroup$ – Holo Dec 30 '18 at 6:52
  • $\begingroup$ Thank you so much for your prompt reply! $\endgroup$ – Le Anh Dung Dec 30 '18 at 6:53

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