# The real cofinality of singular cardinals in $L$ under $0^\#$

Suppose that $0^\#$ exists, is there a relatively simple way to show that for any ordinal $\lambda$, if $\lambda$ is a singular cardinal in $L$ then its real cofinality is $\omega$?

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The statement is not true. Let $\kappa$ be any regular cardinal in $V$, and let $\lambda=(\aleph_{\kappa+\kappa})^L$. This is a singular cardinal in $L$, because $L$ can see the sequence $\langle\aleph_{\kappa+\alpha}^L\mid\alpha\lt\kappa\rangle$ converging to $\lambda$. But the true cofinality of $\lambda$ in $V$ is $\kappa$, since it is the supremum of an increasing $\kappa$ sequence of ordinals and $\kappa$ is regular.
Joel, that is interesting. The OP and myself have an assignment (we have taken the same course) in which a question is given: Assume $0^\#$ exists, if $\alpha$ is a cardinal in $L$ and its true cofinality is uncountable then it is a Silver indiscernible. From since Silver indiscernibles are inaccessible in $L$, since this means that that if $\alpha$ is a singular cardinal in $L$ and its true cofinality is uncountable, then it is inaccesible in $L$ and therefore singular, it means that the true cofinality must be $\omega$. So what you are saying implies that the question is flawed? – Asaf Karagila Jul 28 '11 at 7:52
This question as stated is flawed, but it can be repaired if you add that $\alpha$ is a cardinal in $V$. In other words, the true statement would be: if $\alpha$ is a cardinal in $V$ and singular in $L$, then it must have true cofinality $\omega$, for otherwise it would be a Silver indiscernible and therefore inaccessible in $L$ and therefore regular in $L$. – JDH Jul 28 '11 at 8:50
@Asaf, of course that is fine. By the way, for the future, I would usually think of questions about $0^\sharp$ and similar topics as rising to MathOverflow-level. – JDH Jul 28 '11 at 12:33