If ℝ^{#} exists then why is cof(θ^{L(ℝ)}) = ω? Also I have the same question for the L(V_{λ+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(V_{λ+1}) of V_{λ+1} onto an ordinal, then if V_{λ+1}^{#} exists why is cof(θ^{L(Vλ+1)}) = ω?
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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 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. 


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. 

