# Why do $2^{\lt\kappa}=\kappa$ and $\kappa^{\lt\kappa}=\kappa$ when the Generalized Continuum Hypothesis holds?

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 $\kappa$?

-

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

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}

-
Thanks again, Asaf. –  rankled Dec 13 '11 at 13:23
@rankled: No problems :-) –  Asaf Karagila Dec 13 '11 at 22:22