# Why $2^\kappa=\kappa^{\operatorname{cf}{\kappa}}$, if $\kappa$ is a strong limit cardinal?

On Page 58, Set Theory, Thomas Jech(2006) states the following fact without details.

Another fact worth mentioning is: If $\kappa$ is a strong limit cardinal, $2^\kappa=\kappa^{\operatorname{cf}{\kappa}}$.

My question is how to derive this "fact"?

My first guess is that strong limit cardinals are regular, but it's obviously wrong, since if so, definition of inaccessible cardinal would be redundant.

So another question is what can we say about $\operatorname{cf}{\kappa}$ of a strong limit cardinal $\kappa$?

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## 1 Answer

Let me quote the relevant part of a previously proved theorem:

Theorem 5.16.

$(\rm iii)$ If $\kappa$ is a limit cardinal then $2^{\kappa}=(2^{<\kappa})^{\operatorname{cf}(\kappa)}$

By the fact that $\kappa$ is a strong limit we have that $2^{<\kappa}=\sum_{\lambda<\kappa} 2^{\lambda} = \sup\{2^\lambda\mid\lambda<\kappa\}=\kappa$ and so: $$\kappa^{\operatorname{cf}(\kappa)}\leq\kappa^\kappa=2^\kappa=(2^{<\kappa})^{\operatorname{cf}(\kappa)}=\kappa^{\operatorname{cf}(\kappa)}$$

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Thank you. I shall be more careful to avoid asking this kind of questions. –  Metta World Peace Nov 29 '12 at 0:21
Oh, it's fine. I overlooked a ton of things when I was starting to learn on my own. I still overlook a lot of things, but by now I know that I most likely overlooked them and I try to find out where they are in the paper or whatnot. :-) –  Asaf Karagila Nov 29 '12 at 0:24