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I'm trying to show that, for $\lambda$ an infinite cardinal and $\kappa$ any cardinal, that $$\kappa^{<\lambda} = \sup\{\kappa^\theta:\theta<\lambda\land\theta\text{ a cardinal}\},$$ where $\kappa^{<\lambda}=|\bigcup\{{}^\alpha\kappa:\alpha<\lambda\}|$ by definition. (Throughout ${}^\alpha\kappa$ denotes all functions from $\alpha\to\kappa$, and $\alpha$ always denotes an ordinal.)

It is easy to show $\geq$. For the other direction, I have the following sketch for $\lambda$ a limit cardinal:

For each infinite successor cardinal $\aleph_{\eta+1}\leq\lambda$, there is an injection from $\bigcup\{{}^\alpha\kappa:\aleph_\eta\leq\alpha<\aleph_{\eta+1}\}$ and ${}^{\aleph_{\eta+1}}\kappa$, induced by the bijection between $\{\alpha:\aleph_\eta\leq\alpha<\aleph_{\eta+1}\}$ and $\aleph_{\eta+1}$. You have to be careful about zero functions, but since there are only $\aleph_{\eta+1}$ many of them in $\bigcup\{{}^\alpha\kappa:\aleph_\eta\leq\alpha<\aleph_{\eta+1}\}$, it isn't a problem. From this, I think I can get that $$|\bigcup\{{}^\alpha\kappa:\alpha<\lambda\}|\leq\sup\{\kappa^\theta:\theta\leq\lambda\land\theta\text{ a successor cardinal}\},$$ but since $\lambda$ is a limit cardinal, the latter is exactly equal to $\sup\{\kappa^\theta:\theta<\lambda\land\theta\text{ a cardinal}\}$. Does this look reasonable?

If $\lambda=\aleph_{\eta+1}$ is a successor cardinal however, it seems like the above argument doesn't work. At best, it shows that $\kappa^{<\lambda}\leq\kappa^\lambda$, while $\{\kappa^\theta:\theta<\lambda\land\theta\text{ a cardinal}\}=\kappa^{\aleph_\eta}$.

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$$\begin{align*} \sup\{\kappa^\theta : \theta < \lambda,\ \theta\mbox{ cardinal}\} &= \sup\{|{}^\theta\kappa| : \theta < \lambda,\ \theta\mbox{ cardinal}\} \\ &= \sup\{|{}^{\leq\theta}\kappa| : \theta < \lambda,\ \theta\mbox{ cardinal}\}\\ &= \sup\{\theta^+\cdot|{}^{\leq\theta}\kappa| : \theta < \lambda,\ \theta\mbox{ cardinal}\}\\ &= \sup\{|{}^{<\theta^+}\kappa| : \theta < \lambda,\ \theta\mbox{ cardinal}\}\\ &= \left|\bigcup\{{}^{<\theta^+}\kappa : \theta < \lambda,\ \theta\mbox{ cardinal}\}\right|\\ &= \left|\bigcup\{{}^\alpha\kappa: \alpha < \lambda\}\right| \end{align*} $$

Each line can be justified as follows:

  1. Definition of exponentiation.
  2. There is an injection $F : {}^{\leq\theta}\kappa \to {}^\theta\kappa$ defined by: $$F(f)(\beta) = \begin{cases}f(\beta)+1, & \beta \in \mathrm{dom}(f)\\\\0, & \mbox{otherwise}\end{cases}$$
  3. Since cardinal multiplication is just the maximum of the two cardinals, and $$|{}^{\leq\theta}\kappa| \geq \kappa^\theta \geq 2^\theta \geq \theta^+$$
  4. For each $\theta < \alpha < \theta^+$, fix a bijection $b_\alpha : \theta \to \alpha$. Then we get an injection $G : {}^{<\theta^+}\kappa \to \theta^+\times{}^{\leq\theta}\kappa$ defined by: $$G(f) = \begin{cases}(\mathrm{dom}(f), f), & \mathrm{dom}(f) \leq \theta\\\\(\mathrm{dom}(f), f\circ b_{\mathrm{dom}(f)}), & \mbox{otherwise}\end{cases}$$
  5. Because $\{{}^{<\theta^+}\kappa : \theta < \lambda,\ \theta\mbox{ cardinal}\}$ forms an increasing $\subseteq$-chain of sets.
  6. Because the two unions are the same set.
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In the case of a successor cardinal, let $\mu^+=\lambda$, then $\kappa^{<\lambda}=\kappa^\mu$. This is already in your proof, we have that there are $\lambda$ ordinals and from each there are at most $\kappa^\mu$ functions, so we have $\kappa^\mu\cdot\lambda$. But $\kappa^\mu\geq2^\mu\geq\lambda$ and therefore $\kappa^{<\lambda}=\kappa^\mu$, which is clearly $\sup\{\kappa^\theta:\theta<\lambda\}$.

For the limit case your proof seems okay.

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