# Fine structure question: when do levels of $L$ look "a lot" like each other?

(Everything is assuming $V=L$.)

Fix an uncountable regular cardinal $\kappa$, and let $$E_\kappa=\{\mu<\kappa: \mbox{there is an elementary substructure of L_\kappa isomorphic to L_\mu}\},$$ and let $$E_\kappa^+=\{\mu<\kappa: \mbox{\exists an elementary substructure of (L_{\kappa^+}, L_\kappa) which is isomorphic to (L_\kappa, L_\mu)}\}.$$ (In each definition $\mu$ ranges over ordinals.) Here "$(L_\alpha, L_\beta)$" denotes the structure $(L_\alpha; \in)$ augmented by a predicate for $L_\beta$.

Clearly $E_\kappa^+\subseteq E_\kappa$. In the course of a problem I'm working on, I assumed that $E_\kappa$ and $E_\kappa^+$ are quite different, but thinking about it in more detail this seems unjustified. My question is:

Is $E_\kappa^+$ necessarily a proper subset of $E_\kappa$?

I would also be interested in specific cases, e.g. $\kappa=\omega_1$.

EDIT: Oh dear, this is what I get for doing math late at night: the question above has a terrible typo - $E_\kappa^+$ is supposed to be the set of $\mu$ such that for some $\alpha<\kappa$ we have $(L_\alpha,L_\mu)$ is isomorphic to an elementary substructure of $(L_{\kappa^+}, L_\kappa)$. This completely changes the meaning of the question, of course.

I'm leaving the question as is and accepting Joel's answer, since it is a good answer; but I've asked a corrected version of the question at Levels of L resembling each other, take 2.

• I'm slightly worried that this is actually a really easy problem and I'm just missing something obvious; if that's the case, my apologies and please feel free to vote to migrate to math.stackexchange. Commented Apr 4, 2016 at 5:41
• The $E^+_\kappa$ definition sounds suspiciously close to the Chang Conjecture. Commented Apr 4, 2016 at 6:19
• If $\mu$ is the minimum of $E_\kappa$, then $L_\mu$ is the pointwise definable. However, no point in $E_\kappa^+$ has this property. Commented Apr 4, 2016 at 7:49
• $E_\kappa$ seems to be a club of $\kappa.$ But for $E^+_\kappa$, note that the structure $(L_{\kappa^+, \kappa})$ thinks the ordinal height of $L_\kappa$ is the largest cardinal, so it seems if there is any $\mu \in E^+_\kappa,$ then $\kappa \leq \mu^+.$ Commented Apr 4, 2016 at 9:17
• Also note that for any cardinal $\lambda \leq \kappa,$ there exists $\mu \in E_\kappa$ with $|\mu|=\lambda.$ Commented Apr 4, 2016 at 9:22

Under your assumptions, the set $E_\kappa^+$ is empty. Indeed, we don't even need the predicates that you mention. (In any case, since $\kappa$ is definable in $L_{\kappa^+}$, the predicate for $L_\kappa$ in $\langle L_{\kappa^+},\in,L_\kappa\rangle$ would be definable and therefore offer no additional expressive power.)
What I claim is that if $\kappa$ is a cardinal, then there is no elementary substructure of $\langle L_{\kappa^+},\in\rangle$ in $L$ that is isomorphic to $\langle L_\kappa,\in\rangle$.
To see this, suppose that $\kappa$ is a cardinal and there is an elementary substructure $X\prec L_{\kappa^+}$ in $L$ that is isomorphic to $L_\kappa$. This amounts to an elementary embedding $j:L_\kappa\to L_{\kappa^+}$. Since $\kappa$ is definable in the target structure, we must have $j(\mu)=\kappa$ for some $\mu<\kappa$, and so $j$ has a critical point $\delta\leq\mu$. Since all the subsets of $\delta$ in $L$ appear in $L_\kappa$, we may define in $L$ a measure on $\delta$ via $X\in U\iff \delta\in j(X)$. So $\delta$ is a measurable cardinal in $L$, contradicting Scott's theorem that there are no measurable cardinals in $L$.
• This is a great answer - unfortunately, I stated my own question wrong (see my edit and linked new question). This argument does not apply to the correct version; is there a way to modify it to still show that $E_\kappa^+$, correctly defined, is strictly smaller than $E_\kappa$? Commented Apr 4, 2016 at 17:11