It is well-known that if the Erdős cardinal $\kappa(\omega_1)$ exists, then $0^\sharp$ exists, but what if $\kappa(\lambda)$ exists for a limit ordinal $\omega_1^L\leq \lambda<\omega_1$? Does this still impliy that $0^\sharp$ exists?

Or, alternatively: it is known that the existence of $\kappa(\lambda)$ for all countable $\lambda$ is consistent with $0^\sharp$ does not exist (with $V=L$, in fact), but, does it imply that $\omega_1^L=\omega_1$?

  • 2
    $\begingroup$ If $0^\#$ exists, consider $L$ as your new universe. Then in $L$, there are arbitrarily high countable-Erdos cardinals, but also $V=L$. $\endgroup$
    – Asaf Karagila
    Oct 10 '14 at 1:04
  • $\begingroup$ Thank you, but that does not answer the question. A negative answer would require a model in which $0^\sharp$ does not exist and there exists $\kappa(\lambda)$ for $\omega_1^L\leq \lambda<\omega_1$. $L$ does not satisfy the latter. $\endgroup$
    – Carlos
    Oct 10 '14 at 8:59

Similarly to Asaf's comment, let $\kappa = \kappa(\omega_1^L)$ (we assume that it exists). Then in $L_\kappa$, there is $\alpha$-Erdős cardinal for every $L$-countable $\alpha$ (since being $\alpha$-Erdős cardinal is downward absolute between transitive models for countable ordinals $\alpha$). So $L_\kappa$ is a model for $\forall \lambda < \omega_1,\, \kappa(\lambda)$ exists and $\omega_1 = \omega_1^L$.

It is interesting to note that even when $\omega_1$ is larger then $\omega_1^L$, still the existence of $\lambda$-Erdős cardinal for every $\lambda < \omega_1$ does not imply the existence of $0^{\#}$. For example, the existence of $\kappa(\omega_1^L)$-Erdős cardinals has only mild effect on the cardinals of $L$: Although it does imply that $\omega_1^L < \omega_1$ (otherwise $0^{\#}$ exists, and then $\omega_1^L < \omega_1$), it is consistent with $\omega_1 = \omega_2^L$. The reason is the if $V \models \kappa(\omega_1^L)$ exists, and $\omega_2^L$ is countable, then one can construct in $V$ a $L$-generic filter for $Col(\omega, \omega_1^L)$, $G$, and since $L[G] \subseteq V$ and both agree that $\omega_1^L$ is countable, the fact that $\kappa(\omega_1^L)$ is $\omega_1$-Erdős in $V$, implies that it is also $\omega_1^L$-Erdős in $L[G]$, but $\omega_1^{L[G]} = \omega_2^L$.

The same argument shows that if in $V$, for every ordinal $\lambda < \omega_2^L$ we had a $\lambda$-Erdős cardinal, then there is a model in which $\omega_1 = \omega_2^L$, and for every countable ordinal $\alpha$ there is a $\alpha$-Erdős cardinal.

  • $\begingroup$ Thank you! I insist that the first paragraph does not answer the question, by the same reason that in Asaf's case, but the second does! In $L[G]$ $0^\sharp$ does not exist because $\omega_2^L=\omega_1$, but $\kappa(\omega_1^L)$ exists. $\endgroup$
    – Carlos
    Oct 10 '14 at 10:05

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