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We use $\mathsf{AC}$ here. If $0 < \lambda < \operatorname{cf} \kappa$, $f \in \kappa^\lambda$ is bounded so belongs to $\alpha^\lambda$ for an ordinal $\alpha < \kappa$. Thus $\kappa^\lambda = \bigcup_{\alpha < \kappa} \alpha^\lambda$. Here we used set exponentiation, not cardinal exponentiation. Then we do cardinal exponentiation. $\kappa^\lambda \le \kappa \cdot \sup_{\alpha < \kappa} \lvert \alpha^\lambda \rvert = \kappa \cdot \sup_{\alpha < \kappa} \lvert \alpha \rvert^\lambda$, and since $\kappa \le \kappa^\lambda$ and $\sup_{\alpha < \kappa} \lvert \alpha \rvert^\lambda \le \kappa^\lambda$, $\kappa^\lambda = \kappa \cdot \sup_{\alpha < \kappa} \lvert \alpha \rvert^\lambda$. Here I have a question: If $\kappa$ is limit, $\kappa^\lambda = \sup_{\alpha < \kappa} \lvert \alpha \rvert^\lambda$? What I first thought is like this. Since $\kappa^\lambda = \bigcup_{\alpha < \kappa} \alpha^\lambda = \bigcup_{\mu < \kappa} \mu^\lambda$ where $\mu$ is cardinal, $\kappa^\lambda = \sup_{\mu < \kappa} \mu^\lambda$ because in terms of ordinals, union is same as supremum. But $\bigcup_{\mu < \kappa} \mu^\lambda$ is not union of ordinals, is it? Nevertheless it is union of a chain. If we provide a well-ordering for set exponentiation $\kappa^\lambda$ by cardinal exponentiation $\kappa^\lambda$, does set exponentiation $\mu^\lambda$ match to the initial segment by cardinal exponentiation $\mu^\lambda$? I can't think any further about this.

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You ask whether the equality $\kappa^\lambda=\kappa\cdot\sup_{\alpha<\kappa}|\alpha|^\lambda$ can be simplified to $\kappa^\lambda=\sup_{\alpha<\kappa}|\alpha|^\lambda$ for $\kappa$ limit.

Perhaps the easiest way to argue is to note that for any cardinal $\rho<\kappa$, we also have $\rho^+<\kappa$, and $(\rho^+)^\lambda>\rho$, so $\rho<\sup_{\alpha<\kappa}|\alpha|^\lambda$ and, since this holds for all $\rho<\kappa$, then $\kappa\le\sup_{\alpha<\kappa}|\alpha|^\lambda$, and indeed the equality you ask about holds in this case.

In the sketch of your thoughts you present after the question, you correctly note that in the union it is enough to only consider those $\alpha$ that themselves are cardinals (because $\kappa$ is a limit cardinal, so for any $\alpha$ there is a cardinal $\mu$ larger than $\alpha$ and smaller than $\kappa$, and trivially any function with range contained in $\alpha$ has range contained in $\mu$). So, as sets of functions, we indeed have $$ \kappa^\lambda=\bigcup_{\mu<\kappa}\mu^\lambda.$$ Then you ask whether this implies that the cardinals $\kappa^\lambda$ and $\sup_{\mu<\kappa}\mu^\lambda$ coincide, but seem confused abut how to prove this. You can argue as follows: Call $\rho=\sup_\mu\mu^\lambda$. First, $|\bigcup_\mu\mu^\lambda|\le|\bigsqcup_\mu\mu^\lambda|$, where $\sqcup$ denotes disjoint union. For instance, given a function $f$ in $\bigcup_\mu\mu^\lambda$, map it to the copy of $f$ in the copy of $\mu^\lambda$ in $\bigsqcup_\mu\mu^\lambda$, where $\mu$ is least such that $f\in\mu^\lambda$.

But $|\bigsqcup_\mu\mu^\lambda|=\sum_\mu\mu^\lambda$, where the expressions are now cardinals rather than sets of functions, and $\sum_\mu\mu^\lambda\le\sum_\mu\rho=\kappa\cdot\rho$. Since $\kappa\le\rho$, as argued earlier, then this last product simplifies to $\rho$. We have shown that $$\left|\bigcup_{\mu<\kappa}\mu^\lambda\right|\le\sup_\mu\mu^\lambda.$$ The other inequality, should be clear since, as sets of functions, for any $\mu<\kappa$, $\mu^\lambda$ is a subset of $\bigcup_{\mu<\kappa}\mu^\lambda$. This shows that $\sup_{\mu<\kappa}\mu^\lambda=|\bigcup_{\mu<\kappa}\mu^\lambda|$. But we already have that $\bigcup_{\mu<\kappa}\mu^\lambda=\kappa^\lambda$, and we are done.

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  • $\begingroup$ If $x,y$ are infinite cardinals then $xy=\max (x,y).$ If $k$ is a limit cardinal and $\lambda >0$ then $\sup_{a\in k}|a|^{\lambda}\ge \sup_{a\in k}|a|=k,$ so $k^{\lambda}=$ $k\cdot \sup_{a\in k}|a|^{\lambda}=$ $\max (k,\sup_{a\in k}|a|^{\lambda})=$ $\sup_{a\in k}|a|^{\lambda}.$ $\endgroup$ – DanielWainfleet Mar 8 at 5:29

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