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On Page 57 of Jech's Set Theory, Lemma 5.19

If $\kappa$ is a limit cardinal, and $\lambda \geq \operatorname{cf}{\kappa}$, then $\kappa^\lambda = (\lim_{\alpha\to\kappa} \alpha^\lambda)^{\operatorname{cf}{\kappa}}$

I have difficulty in understanding one step of proof, which is

$(\lim_{\alpha\to\kappa} \alpha^\lambda)^{\operatorname{cf}{\kappa}} \leq (\kappa^\lambda)^{\operatorname{cf}{\kappa}}$

Here's how far I understand this inequality. $\lim_{\alpha\to\kappa} \alpha^\lambda = \bigcup_{\alpha<\kappa}\alpha^\lambda$ , the later is the union of function spaces $\{f:\lambda \to \alpha \}$ for all $\alpha < \kappa$. Since $\lambda \ge \operatorname{cf}{\kappa}$, it's possible to have $\sup\{f(\xi): \xi<\lambda\}=\kappa \notin \kappa$. Thus we have an element doesn't belong to $\kappa^\lambda$ ,which implies the reverse $\lim_{\alpha\to\kappa} \alpha^\lambda \ge \kappa^\lambda$ for $\lambda \ge \operatorname{cf}{\kappa}$.

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It is the union of function spaces, but the range of all functions is bounded below $\kappa$. While there exists a cofinality sequence of functions there is not function of whose is cofinal.

Think about it this way, if $\alpha<\kappa$ then $\alpha\subseteq\kappa$, then $f\colon\lambda\to\alpha$ is also a function into $\kappa$.

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    Yes, exactly. You're welcome!2012-11-27