# continuity points of elementary embeddings from $0^\sharp$

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?

• $j$ is an ultrapower embedding by an $L$-$\alpha_0$-complete $L$-ultrafilter on $\alpha_0$ using functions in $L$ so it is continuous at all ordinals whose $L$-cofinality is not $\alpha_0$. – Gabe Goldberg Apr 25 '18 at 16:49
• I don't have much experience with $0^{\#}$ but it may be relevant that usually this property is associated with strong compactness at $\delta$, and there are no inner models (yet) for such large cardinals. – Stamatis Dimopoulos Apr 26 '18 at 7:37
• Derive the ultrapower embedding $j': L \to L$ from $j$ using $\alpha_0$ as a seed. Note that $j' \leq j$ pointwise on the ordinals since $j'$ factors into $j$. But $j$ is the pointwise minimum order-preserving function on the indiscernibles that moves $\alpha_0$. So $j'$ and $j$ agree on the indiscernibles. The indiscernibles generate $L$, so $j' = j$. – Gabe Goldberg Apr 26 '18 at 13:56
• The image of the Silver indiscernibles under $j'$ is a club class of order indiscernibles for $L$ and hence is contained in the Silver indiscernibles. – Gabe Goldberg Apr 26 '18 at 15:05
• More generally, the elementary embeddings of $L$ are parameterized by the order embeddings of the Silver indiscernibles. See for example Schindler's Set Theory, Corollary 10.44 (5). – Gabe Goldberg Apr 26 '18 at 15:13

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.)