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Let $\lambda$ be an ordinal. A cardinal $\kappa$ is $\lambda$-strong iff there is some inner model $M$ and an elementary embedding $$ j \colon V \rightarrow M $$ s.t. $crit(j) = \kappa$ and $V_\lambda \subseteq M$.

I am given the following exercise:

Let $\alpha < \beta < \kappa$ be ordinals, where $\kappa$ is $(\kappa+\beta)$-strong. Then $$ \{ \mu < \kappa \mid \mu \text{ is } (\kappa+\alpha) \text{-strong} \} $$ has size $\kappa$.


I'd like to show that $$ M \models \kappa \text{ is } (j(\kappa)+\alpha) \text{-strong}, $$ but I'm not sure where to begin...

(It seems that I have to "stretch" a $(\kappa, \kappa+\alpha)$-extender $E \in V$ to a $(\kappa, j(\kappa)+\alpha)$-extender $\tilde E \in M$, but how? Is this even the right approach?)


edit:

Consider the case $\alpha =0 $ and recall that a cardinal $\mu$ is $\mu$-strong iff it is measurable.

Now, if $\kappa$ is $\kappa+2$-strong, we may fix a witnessing measure $U$ on $\kappa$. Then $U \in V_{\kappa+2} \subseteq M$ and some basic observations yield $$ M \models \kappa \text{ is measurable} $$ Thus for any fixed $\xi < \kappa$ we get $$ M \models \exists \mu \colon j(\xi) < \mu < j(\kappa) \wedge \mu \text{ is measurable} $$ (take $\mu = \kappa$) and elementarity yields that there is some measurable $\mu \in (\xi, \kappa)$. This proves that there are $\kappa$ many $\mu$-strong cardinals $\mu < \kappa$ and also suggest that we should be more careful and require something like $\alpha+n < \beta < \kappa$ for some $0 < n < \omega$. (Given the kind of proof I have for the edited question below, the exact value of $n$ is then easily calculated.)

I think that this exercise contains a typo and should read $$ \{ \mu < \kappa \mid \mu \text{ is } (\mu+\alpha) \text{-strong} \} $$ instead. I will ask my professor and in the case that this is what he was asking for, I will tidy up my post and provide an answer for future reference.

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  • $\begingroup$ At which level $V_\alpha$ can we find an extender witnessing that there is an embedding $j:V\to M$ with critical point $\kappa$ and such that $V_{\kappa+\tau}\subset M$? $\endgroup$ Jul 4, 2015 at 19:23
  • $\begingroup$ @Andres Such an extender is an element of $V_{\kappa + \tau +1}$ and an argument similar to the one above yields that $\{ \mu < \kappa \mid \mu \text{ is } (\mu+\alpha) \text{-strong} \}$ has size $\kappa$. I still fail to see how I can improve this result to $\kappa$ many $(\kappa+\alpha)$-strong cardinals below $\kappa$. $\endgroup$ Jul 4, 2015 at 19:38
  • $\begingroup$ Actually, it may be $V_{\kappa+\tau+3}$... (using flat pairing functions it seems that we could get away with $+2$ at the end)? $\endgroup$ Jul 4, 2015 at 19:42
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    $\begingroup$ This is false, notice that when $\kappa$ is measurable, then it is $(\kappa+1)$-strong as we can get a transitive class $M$ and an elementary embedding $j:V\rightarrow M$ with $crit(j)=\kappa$ and $M^\kappa\subseteq M$. If what you're trying to prove was true, you would always get $\kappa$ many measurables below any measurable $\kappa$. $\endgroup$ Jul 5, 2015 at 14:29
  • $\begingroup$ @Camilo Yeah, I already figured that, but I somehow managed to delete the relevant part of a previous edit where I said that for this reason we should be more carefully and require $\alpha + 1 < \beta < \kappa$ (or maybe even a little more, but this will fall out of a careful observation of a proof). $\endgroup$ Jul 5, 2015 at 14:36

1 Answer 1

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We can show that if $\kappa$ is $(\kappa+\alpha+1)$-strong, then the set of $(\kappa+\alpha)$-strong cardinals below $\kappa$ has size $\kappa$. First off, as is easy to verify, $M\vDash \kappa\text{ is }(\kappa+\alpha)\text{-strong}$. Let $k: M\prec N$ witness this. Let $e=k\circ j$. Then $e: V\prec N$ witnesses that $\kappa$ is $(\kappa+\alpha)$-strong with target $\lambda\gt j(\kappa)+\alpha$, and let $E$ be the $(\kappa,e(\kappa))$-extender defined by $X\in E_a$ if and only if $X\in Y$ and $a\in e(X)$, where $Y=\{X\in M|X\subseteq [\kappa]^{|a|}\}$. Then $j_E^M: M\prec \text{Ult}_E(M)$ witnesses, in $M$, the $(j(\kappa)+\alpha)$-strongness of $\kappa$, because $e(\kappa)\gt |V_{(j(\kappa)+\alpha)}|^M$ if and only if $k(\kappa)\gt |V_{\kappa+\alpha}|$. To see this, note that $k(\kappa)\gt |V_{\kappa+\alpha}|^M$, because $k(\kappa)\gt |V_{\kappa+\alpha}|^N$, and since $M\vDash k(\kappa)\times V_{\kappa+\alpha}\subseteq N$, $|V_{\kappa+\alpha}|^M=|V_{\kappa+\alpha}|^N$, and since $k(\kappa)\times V_{\kappa+\alpha}\subseteq M$, $|V_{\kappa+\alpha}|^M=|V_{\kappa+\alpha}|$. To establish $E$ is an extender, condition 1. is easy. Coherence follows from the fact that all the functions, if they exist in $M$, are $\Delta_0$. Normality is also simple. Finally, it is easy to establish that the direct limit is well-founded. The rest follows by standard reflection arguments.

All of this aside, I do think it was probably a typo. This process is rather involved (For an exercise), and the result that $\{\mu\lt\kappa|\mu\text{ is }(\mu+\alpha)\text{-strong}\}$ is much simpler and easier. Plus, I didn't even realize you wrote $\kappa+\alpha$ instead of $\mu+\alpha$, which indicates the author might have as well.

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  • $\begingroup$ Why should $M$ see $j_E^M$? $E$ is defined using $j$. If your argument worked, there could be no least measurable cardinal, as pointed out by Camilo Arosemena-Serrato, and thus no measurable cardinal. $\endgroup$ Nov 8, 2021 at 21:15
  • $\begingroup$ Also, $k(\kappa) = j(\kappa)$ since $M \vDash \text{$j(\kappa)$ is inaccessible and greater than $\kappa+\alpha$}$, so $k(j(\kappa)) = j(\kappa)$ $\endgroup$ Nov 8, 2021 at 21:23
  • $\begingroup$ I meant $e(\kappa)=j(\kappa)$. $\endgroup$ Nov 8, 2021 at 21:36

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