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For an infinite cardinal $\kappa$ and an ordinal $\lambda>\kappa,$ $\kappa$ is called $\lambda-$strong, if there is a non-trivial elementary embedding $j: V \rightarrow M$ with $crit(j)=\kappa$ such that $V_\lambda \subset M.$ For any infinite cardinal $\kappa$ let $\lambda_\kappa$ be the least ordinal such that $\kappa$ is not $\lambda_\kappa-$strong.

By Hamkins interesting answer, $\lambda_\kappa$ can not be a weakly compact cardinal.

Question. Can $\lambda_\kappa$ be (consistency) a singular cardinal?

Question. Can $\lambda_\kappa$ be (consistency) the least inaccessible cardinal above $\kappa$?

(I would like to thank Prof. Hamkins for clarifying the question)

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  • $\begingroup$ Isn't it obviously non-empty? Suppose that $\lambda$ is the first ordinal not in $S_\kappa$, then $\lambda+1$ is vacuously in $S_\kappa$. $\endgroup$ – Asaf Karagila Oct 7 '13 at 8:52
  • $\begingroup$ I think the question is not posed in the most illuminating way. For example, your set $S_\kappa$ contains every $\lambda$ that is strictly larger than the degree of strongness of $\kappa$ (the supremum of $\delta$ for which $\kappa$ is $\delta$-strong), simply because these cardinals verify the implication vacuously. These cardinals can be singular or whatever, but they have nothing to do with the strongness of $\kappa$. $\endgroup$ – Joel David Hamkins Oct 7 '13 at 12:10
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    $\begingroup$ ...Rather, it seems to me that the main interest here is to inquire: what kind of ordinal is the degree of strongness of a partially strong cardinal $\kappa$, when the strongness of $\kappa$ does not hold at that limit? My answer shows that this ordinal cannot be weakly compact; you are asking in your edit whether it can be singular. $\endgroup$ – Joel David Hamkins Oct 7 '13 at 12:11
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    $\begingroup$ The edit has changed the meaning of $S_\kappa$, which makes some of the remarks that have been made about it wrong or confusing. In my previous comments, I was suggesting that you don't want a set $S_\kappa$ at all, but rather want to refer to the ordinal $\lambda_\kappa$ that is the degree of strongess of $\kappa$. $\endgroup$ – Joel David Hamkins Oct 7 '13 at 13:26
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Theorem. Every weakly compact cardinal above $\kappa$ is in $S_\kappa$. In other words, if a cardinal $\kappa$ is $\delta$-strong for every $\delta\lt\lambda$, where $\lambda$ is weakly compact, then $\kappa$ is $\lambda$-strong as well.

Proof. Suppose that $\lambda$ is weakly compact and $\kappa$ is $\delta$-strong for every $\delta\lt\lambda$, with $\kappa\lt\lambda$. Let $M$ be a transitive structure of size $\lambda$ with $V_\lambda\subset M\prec H_{\lambda^+}$ and $M^{\lt\lambda}\subset M$. Since $\lambda$ is weakly compact, there is $j:M\to N$ into a transitive set $N$ with critical point $\lambda$. The point now is that while $M$ has all the extenders sufficient to witness that $\kappa$ is $\delta$-strong for all $\delta\lt\lambda$, it follows by elementarity that $N$ has an extender witnessing that $\kappa$ is $\lambda$-strong, and since $V_\lambda\subset N$, it will be right about this. So $\kappa$ is $\lambda$-strong, as desired.QED

I was once at a talk of Sy Friedman, at the logic conference in Wroclaw several years ago, where in a theorem he and Natasha Dobrinen had narrowed the consistency strength of a hypothesis to be between "$\kappa$ is $\lt\lambda$-strong for a weakly compact cardinal" and "$\kappa$ is $\lambda$-strong for a weakly compact $\lambda$," a difference that he had remarked were "within $\epsilon$" and very close together, although they had really wanted to prove an equivalence. I pointed out that actually the two hypotheses were equivalent, by the argument above, and so they had already attained the full equiconsistency they had desired.

The theorem above is can be further improved to weaker related notions than weak compactness, since all we needed was a single embedding $j:M\to N$ with critical point $\lambda$ and $V_\lambda\subset M$ and $M^\omega\subset M$ (in order to know that $M$ is right about the well-founded of the extender ultrapower). This hypothesis is strictly weaker than weak compactness, since every weakly compact cardinal is a limit of such kind of cardinals. It would suffice merely that $V_\lambda\prec_{\Sigma_2} N$ for some transitive set $N$ with $\lambda\in N$.

Update. Regarding your edit, basically anything is possible.

Theorem. If $\kappa$ is $\lambda$-strong and $\kappa\lt\lambda$, then there is a transitive class $M$ in which $\kappa$ is $\lt\lambda$-strong but not $\lambda$-strong, so $\lambda_\kappa=\lambda$ in $M$.

Proof. Let $j:V\to M$ be a $\lambda$-strongness extender embedding for which $j(\kappa)$ is minimal. It follows that $V_\lambda\subset M$ and so $\kappa$ remains $\lt\lambda$-strong in $M$. But $\kappa$ is not $\lambda$-strong in $M$, for if it were, there would be an extender embedding $h:M\to N$ witnessing this, and we could assume that $h(\kappa)\lt j(\kappa)$, since $j(\kappa)$ is inaccessible (and much more) above $\kappa$ and $\lambda$ in $M$; in this case, since $V_\lambda\subset M$ the extender arising from $h$ would have given rise been a $\lambda$-strongness embedding in $V$ with a smaller target for $\kappa$, contrary to the minimality of $j(\kappa)$. QED

So the degree of strongness of $\kappa$ can be singular of cofinality $\omega$, as you desired, or any ordinal at all above $\kappa$.

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  • $\begingroup$ That's really interesting. Can it be some singular cardinal (of countable cofinality)? $\endgroup$ – Mohammad Golshani Oct 7 '13 at 11:58
  • $\begingroup$ So for your second theorem, it should be true that $\lambda$ is not weakly compact in $M$, by your first theorem, even if it is very large in $V$. Is it really the case. I don't see it $\endgroup$ – Mohammad Golshani Oct 7 '13 at 18:03
  • $\begingroup$ Yes, that's right. The point is that we make $\lambda$ definable in $M$ in a way that doesn't reflect, violating weak compactness. Incidentally, this is how one can prove the Laver function for strong cardinals, since if $f$ is the failure-of-strongness function, then $j(f)(\kappa)=\lambda$. $\endgroup$ – Joel David Hamkins Oct 7 '13 at 18:07

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