(Everything below 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 $$E\_\kappa^+=\{\mu<\kappa: \mbox{$\exists M\prec (L\_{\kappa^+}, L\_\kappa)$ with $M\cong (L\_\alpha, L\_\mu)$ for some $\alpha<\kappa$}\}.$$ (In each definition $\mu$ ranges over ordinals.) Here "$(L\_\alpha, L\_\beta)$" denotes the structure $(L\_\alpha; \in)$ augmented by a predicate for $L\_\beta$.


My question is:



> 
> Is $E\_\kappa^+$ always a proper subset of $E\_\kappa$?
> 
> 
> 


*Note: This is a corrected version of [Fine structure question: when do levels of $L$ look "a lot" like each other?](https://mathoverflow.net/questions/235233/fine-structure-question-when-do-levels-of-l-look-a-lot-like-each-other), in which $E\_\kappa^+$ was defined incorrectly (it was late and I was tired). I suspect the answer is the same - that $E\_\kappa^+$ is much, much smaller than $E\_\kappa$ - but the situation is significantly different: for one thing, Joel's answer to the lined question shows that $E\_\kappa^+$ as defined there is in fact **empty**, which it is definitely not here.*



Levels of L resembling each other, take 2
If $\mu$ is the minimum of $E\_\kappa$, then $L\_\mu$ is the pointwise definable. However, if $\mu \in E\_\kappa^+$, then $L\_\mu$ is not pointwise definable. (See Joel's comment above. I am too lazy to type.)


