Suppose $0^\sharp$ exists, and let $\langle \alpha_i : i \in \text{Ord}\rangle$ be the Silver indiscernibles for $L$. Let $j : L \to L$ be the embedding generated by mapping $\alpha_n$ to $\alpha_{n+1}$ for finite $n$ and fixing the rest of the indiscernibles. Suppose $\alpha_0 < \delta <\alpha_\omega$ and $\delta$ is regular in $L$. Is it the case that $\sup j[\delta] < j(\delta)$?

More generally, pick any order preserving map from the indiscernibles to itself. How does one compute at what points the map is continuous?

Suppose $0^\sharp$ exists, and let $\langle \alpha_i : i \in \text{Ord}\rangle$ be the Silver indiscernibles for $L$. Let $j : L \to L$ be the embedding generated by mapping $\alpha_n$ to $\alpha_{n+1}$ for finite $n$ and fixing the rest of the indiscernibles. Suppose $\alpha_0 < \delta <\alpha_\omega$ and $\delta$ is regular in $L$. Is it the case that $\sup j[\delta] < j(\delta)$?

More generally, pick any order preserving map from the indiscernibles to itself and consider the generated embedding. How does one compute at what points the embedding is continuous?

Suppose $0^\sharp$ exists, and let $\langle \alpha_i : i \in \text{Ord}\rangle$ be the Silver indiscernibles for $L$. Let $j : L \to L$ be the embedding generated by mapping $\alpha_n$ to $\alpha_{n+1}$ for finite $n$ and fixing the rest of the indiscernibles. Suppose $\alpha_0 < \delta <\alpha_\omega$ and $\delta$ is regular in $L$. Is it the case that $\sup j[\delta] < j(\delta)$?

Suppose $0^\sharp$ exists, and let $\langle \alpha_i : i \in \text{Ord}\rangle$ be the Silver indiscernibles for $L$. Let $j : L \to L$ be the embedding generated by mapping $\alpha_n$ to $\alpha_{n+1}$ for finite $n$ and fixing the rest of the indiscernibles. Suppose $\alpha_0 < \delta <\alpha_\omega$ and $\delta$ is regular in $L$. Is it the case that $\sup j[\delta] < j(\delta)$?

More generally, pick any order preserving map from the indiscernibles to itself. How does one compute at what points the map is continuous?