# Cardinal Exponentiation $\lim_{\alpha\to\kappa} \alpha^\lambda$

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 $\kappa$.
Thank you for your reply. But I'm not sure I follow your argument. If there exists an unbounded cofinality sequence of functions whose range is $[0, \alpha)$, then the supremum is $[0, \kappa)$. If there is no such sequence, $\lim_{\alpha\to\kappa} \alpha^\lambda < \kappa^\lambda$. – Metta World Peace Nov 27 '12 at 7:29
No, the point is that the union of these function spaces only contain function whose range is bounded below κ So no function in this union could possibly have a cofinal range. Think about the union of all function spaces of functions from $(0,1)$ into $[0,x]$ for $x\in\mathbb R$. Is there any function in this union whose range is unbounded? – Asaf Karagila Nov 27 '12 at 8:00
Functions in the sequence may not be bounded below $\alpha$, but must be bounded below $\kappa$, so their union is bounded below $\kappa$. However, $\kappa^\lambda$ contains unbounded functions, so $\lim_{\alpha\to\kappa} \alpha^\lambda \le \kappa^\lambda$. – Metta World Peace Nov 27 '12 at 8:09