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(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?, 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.

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    $\begingroup$ If $\mu$ is the minimum of $E_\kappa$, then $L_\mu$ is the pointwise definable. However, no point in $E_\kappa^+$ has this property. $\endgroup$
    – Yizheng Zhu
    Commented Apr 4, 2016 at 17:15
  • $\begingroup$ @YizhengZhu Ah, I missed your comment to the same effect on my previous question. I don't see either piece immediately, though: why is $L_\mu$ pointwise definable, and why is no point in $E_\kappa^+$ pointwise definable? (This may be obvious, constructibility is not my forte.) $\endgroup$
    – Noah Schweber
    Commented Apr 4, 2016 at 17:16
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    $\begingroup$ The minimum of $E_\kappa$ is pointwise definable, since the collection of definable elements of any $L_\mu$ is an elementary substructure, since there are definable Skolem functions. So that elementary substructure must collapse to $L_\mu$ by minimality and hence everything in $L_\mu$ is definable. $\endgroup$
    – Joel David Hamkins
    Commented Apr 4, 2016 at 17:20
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    $\begingroup$ But no $\mu\in E_\kappa^+$ has pointwise definable $L_\mu$, since $L_{\kappa^+}$ can see that $L_\kappa$ is not pointwise definable, and so $L_\alpha$ will also see that about $L_\mu$. $\endgroup$
    – Joel David Hamkins
    Commented Apr 4, 2016 at 17:22
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    $\begingroup$ @YizhengZhu You should post your comment as an answer! $\endgroup$
    – Joel David Hamkins
    Commented Apr 4, 2016 at 17:23

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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.)

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