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Let $\kappa$ ba weakly inaccessible cardinal. Why are there $\kappa$ regular cardinals $\lambda < \kappa$? I've tried a recursive construction, but I don't know what to do in the limit step. Supremum does not work, since then we loose regularity.

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Suppose that $\kappa$ is weakly inaccessible. Thus, it is a regular limit cardinal. So $\kappa=\aleph_\beta$ for some ordinal $\beta$. Since $\kappa$ is a limit cardinal, it must be that $\beta$ is a limit ordinal. Since $\kappa$ is regular, it cannot be that $\beta<\kappa$. So $\kappa=\aleph_\kappa$. Thus, there are $\kappa$ many cardinals below $\kappa$. All the successor cardinals $\aleph_{\beta+1}$ for $\beta<\kappa$ are regular, and there are $\kappa$ many of these.

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Ah, of course! Thanks. –  Martin Brandenburg Sep 2 '10 at 22:20

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