In the Wikipedia page about $0^\sharp$, we have the following equivalence:
Silver showed that the existence of an uncountable set of indiscernibles in the constructible universe is equivalent to the existence of $0^\sharp$.
My question is why the uncountable condition is needed? If a countable set of indiscernibles exist, why we cannot (or not always) obtain the collection of true formulas defining $0^\sharp$ from this set?
Suppose $0^\#$ exists. Then working in $L[G]$, where $G$ collapses a Silver indiscernible $\kappa$ to $\omega$, you can find a countable set of $L_\kappa$-indiscernibles (use $\Sigma^1_1$-absoluteness and indiscernibility of $\kappa$), which are therefore actually $L$-indiscernibles (as $L_\kappa\preceq L$, and by ``$L$-indiscernibles'' I just mean in the model-theoretic sense). But these can't give the true $0^\#$, since it can't be added in this way to $L$. (In fact, already in $L$, there are countable ordinals $\alpha$ and a countable set of indiscernibles for $L_\alpha$, by $\Sigma^1_2$-absoluteness.)
