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, 2014 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, 2014 at 8:59

1 Answer 1


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, 2014 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.