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 $\{\kappa^\theta:\theta<\lambda\land\theta\text{ a cardinal}\}=\kappa^{\aleph_\eta}$.


2 Answers 2


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

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.
  • $\begingroup$ One additional verification is needed for step 5, namely that the number of sets in the union does not exceed the sup in the previous line. The number of sets in the union is at most $\lambda$ and it's easy to check that $\lambda$ does not exceed the sup, by checking separately the case of $\lambda$ a successor cardinal and a limit cardinal. $\endgroup$
    – PatrickR
    Mar 2, 2022 at 8:03

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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