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We know that if $0^\#$ exists then it's not in $L$. For an infinite ordinal $\alpha$, denote by $I_\alpha$ the initial segment of length $\alpha$ of Silver indiscernibles.

Question: For which $\alpha$ (if any) is $I_\alpha$ in $L$?

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No infinite set of Silver indiscernibles is in $L$. (Of course, every finite set of Silver indiscernibles is in $L$ since $L$ contains all finite sets of ordinals.)

To see this, assume $0^\#$ exists in $V$ and suppose $(\eta_i)_{i<\omega}$ is an infinite increasing sequence of Silver indiscernibles. I will show that $0^\#$ is definable from $(\eta_i)_{i<\omega}$ using parameters from $L$. So if $(\eta_i)_{i<\omega}$ were in $L$, then $0^\#$ would be in $L$ too. Since $0^\# \notin L$, it follows that $(\eta_i)_{i<\omega}$ is also not in $L$.

Pick an uncountable $V$-cardinal number $\kappa$ which is greater than all the $\eta_i$. Then $L_\kappa \prec L$. So if $\phi(v_0,\dots,v_{n-1})$ is any formula in the language of set theory, then $${}^\ulcorner\phi{}^\urcorner \in 0^\# \Leftrightarrow L \vDash \phi(\eta_0,\dots,\eta_{n-1}) \Leftrightarrow L_\kappa \vDash \phi(\eta_0,\dots,\eta_{n-1}).$$ Since $\kappa$, $L_\kappa$, and the satisfaction relation for $L_\kappa$ are all in $L$, it follows that $0^\#$ is definable from $(\eta_i)_{i<\omega}$ using parameters from $L$.

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Thank you, Francois. –  ZeroDagger Jul 18 '11 at 22:44
    
No problem! It was a pleasure! –  François G. Dorais Jul 19 '11 at 4:17

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