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 $\omega$?
The real cofinality of singular cardinals in $L$ under $0^\#$
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set-theory
large-cardinals
1 Answers
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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.
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0$A$s I am not the OP this is not my decision where to post it, however I feel that these are borderline questions. In MO they might be appreciated but at least at this point they are certainly not research level questions. Since I am trying to get some volume into the advanced set-theory related tags on MSE I always prefer it first for these borderline questions. – 2011-07-28