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?



continuity points of elementary embeddings from $0^\sharp$
**Claim:** *Suppose $i : L \to L$ is an elementary embedding. If $\kappa$ is an $L$-regular cardinal and $\sup i[\kappa] < i(\kappa)$, then $\kappa$ is a Silver indiscernible.* 


The claim answers your more general question: the discontinuity points of $i$ are then precisely the ordinals with $L$-cofinality equal to some Silver indiscernible at which $i$ is discontinuous (since if $i$ is discontinuous at $\alpha$ then $i$ is discontinuous at the $L$-regular cardinal $\text{cf}^L(\alpha)$, which must therefore be a Silver indiscernible).


*Proof of claim.* Let $\kappa' = \sup i[\kappa]$. Consider the hull $H = H^L(i[L]\cup \kappa')\prec L$. Let $k : L\to L$ be the inverse of the transitive collapse of $H$ and let $j = k^{-1}\circ i$. (If you're comfortable with long extenders, it is easier to see $j$ as the extender ultrapower by the $L$-extender of length $\kappa'$ derived from $i$, and $k$ as the factor embedding.)


We claim that $j(\kappa) = \kappa'$. Since $\kappa'\subseteq H$, we have $\text{crit}(k)\geq \kappa'$ and therefore $\sup j[\kappa] = \sup i[\kappa] = \kappa'$. Hence $\kappa' = \sup j[\kappa] \leq j(\kappa)$. Conversely, we must show $j(\kappa) \leq \kappa'$. Suppose $\alpha < j(\kappa)$; we'll show $\alpha < \kappa'$. Then $\alpha = j(f)(\xi)$ for some ordinal $\xi < \kappa'$ and some function $f\in L$. Fix $\bar \xi < \kappa$ be such that $j(\bar \xi) > \xi$ and let $\bar \alpha = \sup (f\restriction \bar \xi)$. Then $\bar \alpha < \kappa$ since $\kappa$ is $L$-regular, and $j(\bar \alpha) \geq \sup j(f\restriction \bar \xi) \geq j(f\restriction \bar \xi)(\xi) = \alpha$. Hence $\alpha < \kappa'$. So $j(\kappa) = \kappa'$. (This is the proof of a well-known extender fact: if $\kappa$ is $M$-regular then any $M$-extender whose constituent $M$-ultrafilters lie on ordinals less than $\kappa$ gives rise to an embedding of $M$ that is continuous at $\kappa$.)


But now $\kappa' = \text{crit}(k)$: we have seen that $\text{crit}(k)\geq \kappa'$. But $k(\kappa') = k(j(\kappa)) = i(\kappa) > \kappa'$, so $\text{crit}(k) = \kappa'$.


It follows that $\kappa'$ is a Silver indiscernible: it is the critical point of an elementary embedding of $L$. But then since $j(\kappa) = \kappa'$, $\kappa$ is also an indiscernible. (Otherwise fix a finite set of ordinals $a\subseteq \kappa$ and a finite set of indiscernibles $b$ above $\kappa$ such that $\kappa$ is definable in $L$ from $a\cup b$. Then $j(\kappa) = \kappa'$ is definable in $L$ from $j(a)\cup j(b)$. But $j(a)\subseteq \kappa'$ and $j(b)$ is a finite set of indiscernibles above $\kappa'$. This contradicts that $\kappa'$ is an indiscernible, since an indiscernible is never definable from ordinals below it and indiscernibles above it.)


