I am trying to work through the proof of Lemma 3.10 in Jech's Set Theory, and can't figure out why $$ \forall \xi < \lambda: \beta_\xi < \kappa $$ According to the text the negation of the above will contradict the minimality of $\lambda$, so I tried to obtain that contradiction as follows.
If $\gamma = \min\{\xi\in \lambda: \beta_\xi = \kappa\}$, let $h$ be any bijection between $|\gamma|$ and $\gamma$. Then the family $\{ \beta_{h(\xi)}: \xi \in |\gamma|\}$ will consist of subsets of $\kappa$ of strictly smaller cardinality, while also $|\gamma| < \lambda$ . So to get the contradiction it remains to show that $\bigcup_{\xi\in |\gamma|}\beta_{h(\xi)} = \kappa$ , and this is where I struggle. By construction, $ \bigcup_{\xi\in |\gamma|}\beta_{h(\xi)} = \bigcup_{\xi\in \gamma}\beta_{\xi} $ , but if I see it correctly, the original sequence $\{\beta_\xi: \xi < \lambda\}$ doesn't have to be continuous at $\gamma$.
I would be very grateful if someone could help me out.
Recall that $\kappa$ is a cardinal. So by definition any well-ordered set whose cardinality is $<\kappa$, will also have that the order type is $<\kappa$. Which is exactly what Jech is using here.