Question about cardinals without GCH Without assuming the Generalized Continuum Hypothesis, how to show that there exists a uncountable cardinal $\kappa$ such that, for every $\lambda < \kappa$, one have  $2^\lambda < \kappa$. With assumption of GCH, for $\kappa = \aleph_{\omega}$ the affirmation holds. I would appreciate any hint.
 A: The answer lies in $\beth$ numbers.
The definition is as follows:


*

*$\beth_0=\aleph_0$.

*$\beth_{\alpha+1}=2^{\beth_\alpha}$.

*If $\delta$ is a limit ordinal and $\beth_\gamma$ was defined for $\gamma<\delta$ then $\beth_\delta=\sup\{\beth_\gamma\mid\gamma<\delta\}$.


Now you can prove that $\beth_\omega$ is a strong limit, and if $\delta$ is a limit ordinal then $\beth_\delta$ is a strong limit cardinal. You can even show that for every $\mu$ there is a strong limit $\kappa$ such that $\mu<\kappa$.
(Note that GCH is the assertion $\beth_\alpha=\aleph_\alpha$ for all $\alpha$.)
A: HINT: Let $\beth_0=\omega$. Given $\beth_\alpha$ for some ordinal $\alpha$, let $\beth_{\alpha+1}=2^{\beth_\alpha}$. Given $\beth_\alpha$ for all $\alpha<\eta$, where $\eta$ is a limit ordinal, let $\beth_\eta=\sup\{\beta_\alpha:\alpha<\eta\}$. Can you find your answer in there?
These are the beth numbers, a hierarchy similar to that of the aleph numbers, but constructed by cardinal exponentiation instead of successor.
