# A cardinal $\kappa$ such that $2^{\lambda}<\kappa$ for all $\lambda<\kappa$ is regular?

A cardinal $\kappa$ such that $2^{\lambda}<\kappa$ for all $\lambda<\kappa$ is regular?

I would appreciate very much an answer

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## 1 Answer

No. This is just the definition of a strong limit cardinal. It does not have to be regular.

For example, $\beth_\omega$. The $\beth$ numbers are defined as:

$\beth_0=\aleph_0$; $\beth_{\alpha+1}=2^{\beth_\alpha}$; and for a limit $\beta$, $\beth_\beta=\sup\{\beth_\alpha\mid\alpha<\beta\}$. It is not difficult to see that for any limit ordinal $\delta$, $\beth_\delta$ is a strong limit cardinal.

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In fact, the "strong" in "strongly inaccessible" is shorthand for "strong limit". Regularity is a separate issue. –  Andres Caicedo Nov 8 '12 at 23:07