6
$\begingroup$

(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.

Improve this question
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Noah Schweber
    Commented Apr 4, 2016 at 5:41
  • $\begingroup$ The $E^+_\kappa$ definition sounds suspiciously close to the Chang Conjecture. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 4, 2016 at 6:19
  • 1
    $\begingroup$ 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 7:49
  • 2
    $\begingroup$ $E_\kappa$ 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 $\kappa \leq \mu^+.$ $\endgroup$
    – Mohammad Golshani
    Commented Apr 4, 2016 at 9:17
  • 1
    $\begingroup$ Also note that for any cardinal $\lambda \leq \kappa,$ there exists $\mu \in E_\kappa$ with $|\mu|=\lambda.$ $\endgroup$
    – Mohammad Golshani
    Commented Apr 4, 2016 at 9:22

1 Answer 1

Reset to default
6
$\begingroup$

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 $L$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ It's only a contradiction if you assume ZFC is consistent! ;-P $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 4, 2016 at 15:14
  • $\begingroup$ @AsafKaragila That isn't right. My proof did not use Con(ZFC). I proved under Noah's assumption that there is a measurable cardinal in L. Scott proved that there isn't one. Contradiction. Right? $\endgroup$
    – Joel David Hamkins
    Commented Apr 4, 2016 at 15:17
  • $\begingroup$ Well, if ZFC is inconsistent, then it already proved that there is a measurable cardinal in L, so you're fine. :) But you're right. I should have said that it's only a problem if you assume that ZFC is consistent. :) $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 4, 2016 at 15:18
  • $\begingroup$ 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 $E_\kappa$? $\endgroup$
    – Noah Schweber
    Commented Apr 4, 2016 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.