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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.

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You mean GCH, not CH, as CH itself talks only about $\aleph_0$ and you meant to say that it holds for all cardinals (or at least all finite cardinals). –  Asaf Karagila Jul 1 '12 at 20:31

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The answer lies in $\beth$ numbers.

The definition is as follows:

  1. $\beth_0=\aleph_0$.
  2. $\beth_{\alpha+1}=2^{\beth_\alpha}$.
  3. 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 $\alpha$.)

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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.

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