Cofinality of Theta if sharps exist

If ℝ# exists then why is cof(θL(ℝ)) = ω? Also I have the same question for the L(Vλ+1) generalization (if it's actually a different proof; I presume it isn't), i.e. if θ is defined as the sup of the surjections in L(Vλ+1) of Vλ+1 onto an ordinal, then if Vλ+1# exists why is cof(θL(Vλ+1)) = ω?

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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. –  Theo Johnson-Freyd Oct 27 '09 at 17:45
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. –  Richard Dore Oct 27 '09 at 18:35
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. –  Scott Cramer Oct 27 '09 at 20:58
I edited the question for formatting under Richard's guidance. I hope everything is correct. –  Anton Geraschenko Oct 28 '09 at 22:47

2 Answers

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 nth cardinal above continuum. Then the theory of reals with a parameter s_n in L_{a_n+1}(R) is a set of reals A_n. They are Wadge cofinal in Theta, another words the sequence is not in L(R) but each A_n is and that is why you get a singularization.

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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. –  Scott Cramer Oct 27 '09 at 21:01

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. R^# is a mouse over 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.

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