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 $< \kappa$.


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!


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

  • $\begingroup$ 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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