If $\mathbb R^\sharp$ exists then why is $\textrm{cof}(\Theta^{L(\mathbb R)})=\omega$? Also I have the same question for the $L(V_{\lambda+1})$ generalization (if it's actually a different proof; I presume it isn't), i.e. if $\Theta$ is defined as the sup of the surjections in $L(V_{\lambda+1})$ of $V_{\lambda+1}$ onto an ordinal, then if $V_{\lambda+1}^\sharp$ exists why is $\textrm{cof}(\Theta^{L(V_{\lambda+1})}) = \omega$?
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1$\begingroup$ In general, and particularly on this post, please give some background. A format I like: ask your question. Then have a section titled "background" that quickly glosses all the notation and definitions. Sometimes finish with a section on why you care. I mean, probably people who will be able to answer the question will know more of the notation than I do, but not everyone's notation is identical. $\endgroup$– Theo Johnson-FreydCommented Oct 27, 2009 at 17:45
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5$\begingroup$ The terms Scott uses are standard for the subject. The only one that might be unclear he defined (what Theta is in L(V_lambda+1)). Your comment seems analogous to complaining that someone posting an algebraic geometry question didn't give a reference to what a stack is. A little more motivation for the question, however, seems pretty reasonable. $\endgroup$– Richard DoreCommented Oct 27, 2009 at 18:35
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3$\begingroup$ You're right, motivation should have been added. While it's not the most interesting question in the world, I've been rather confused attempting to work with sharps and the structure that they impose beyond just covering. I've also been confused about the cofinality of theta and it's impact. This seemed like a simple enough question that might help elucidate both ideas since it gives a connection between them. $\endgroup$– Scott CramerCommented Oct 27, 2009 at 20:58
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1$\begingroup$ I edited the question for formatting under Richard's guidance. I hope everything is correct. $\endgroup$– Anton GeraschenkoCommented Oct 28, 2009 at 22:47
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2 Answers
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This is because the pieces of the sharp singularize $\Theta$. Let $s_n$ be the sequence of the first $n$ cardinals above continuum and let $a_n$ be the $n$th cardinal above continuum. Then the theory of reals with a parameter $s_n$ in $L_{a_n+1}(\mathbb R)$ is a set of reals $A_n$. They are Wadge cofinal in $\Theta$, another words the sequence $\langle A_n: n<\omega\rangle$ is not in $L(\mathbb R)$ but each $A_n$ is and that is why you get a singularization.
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2$\begingroup$ I was a little confused as to why the A_n are Wadge cofinal, but I suppose it's just by contradiction: if not they would all be Wadge reducible to something in L(R), which would imply that R^# is in L(R), which is ridiculous. Thanks. $\endgroup$ Commented Oct 27, 2009 at 21:01
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$\begingroup$ Do we know if this result still holds without excluded middle? $\endgroup$ Commented Sep 20, 2023 at 18:02
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Scott, the best way to think of sharps is via mice. Think of $x^\#$ as a mouse over $x$ with one measure which is iterable. $\mathbb R^\#$ is a mouse over $\mathbb R$ with one measure which is iterable. Things become very easy ones you make the move from sharps as reals or sets of reals or etc to sharps as mice.
