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

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

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?

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
• 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. – Le Anh Dung Dec 30 '18 at 6:51