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I'm going to assume GCH here. If that holds, then why do we have the equalities $2^{\lt\kappa}=\kappa$ for every $\kappa$, and $\kappa^{\lt\kappa}=\kappa$ for all regular $\kappa$?

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up vote 6 down vote accepted

Recall the definition of $\lambda^{<\mu} = \sup\{\lambda^\nu\mid\nu<\mu\}$.

Assume GCH, let $\kappa$ be a regular cardinal.

$$2^{<\kappa} = \sup\{2^\lambda \mid \lambda<\kappa\} = \sup\{\lambda^+\mid\lambda<\kappa\} = \kappa$$

Where the last equality holds since if $\kappa$ is a successor cardinal then it is $\lambda^+$ for some $\lambda<\kappa$; and if it is a regular limit cardinal then it is the limit of successor cardinals below it.

$$\kappa^{<\kappa} = \sup\{\kappa^\lambda\mid\lambda<\kappa\} = \kappa$$

The last equality follows from: $$\begin{align} \kappa^\lambda &=\kappa\cdot\sup\{\mu^\lambda\mid\mu<\kappa\}\\ &\le\kappa\cdot\sup\{\max\{2^\mu,2^\lambda\}\mid\mu<\kappa\}\\ &=\kappa\cdot\sup\{\max\{\mu^+,\lambda^+\}\mid\mu<\kappa\}\\ &=\kappa\cdot\kappa=\kappa \end{align}$$

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Thanks again, Asaf. – rankled Dec 13 '11 at 13:23
@rankled: No problems :-) – Asaf Karagila Dec 13 '11 at 22:22
Why do we assume $\nu$ is regular in the definition of $\lambda^{<\mu}$? Especially given that we make no such assumption in the definition of $2^{<\kappa}.$ – goblin May 8 '14 at 13:53
Okay well I'll be interested to hear your thoughts. Pretty sure the definitions do not agree, e.g. consider $\lambda =2$ and $\kappa=\mu = \aleph_{\omega+1}$. Then under GCH, the $\lambda^{<\mu} = \aleph_\omega$, but $2^{<\kappa}$ is $\aleph_{\omega+1}$. – goblin May 8 '14 at 14:18
Yeah, you're right. I have no idea why I wrote that in the definition. Oh well. – Asaf Karagila May 8 '14 at 14:20

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