Proof If $U\in \mathrm {V}$ is a measure on $\kappa $ and $\mathbb {P}\in \mathrm {V}$ is a $\le \kappa $ -distributive poset, no subsets of $\kappa $ are added by $\mathbb {P}$ . Then $U\in \mathrm {V}\subseteq \mathrm {V}^{\mathbb {P}}$ is still a measure in $\mathrm {V}^{\mathbb {P}}$ .

Proof Let $j:\mathrm {V}\rightarrow \mathrm {M}$ be a $\delta $ -strong embedding so that $j(\vec {x})=\vec {x}$ , $\delta <j(\kappa )$ , and $\mathrm {M}\mid \delta =\mathrm {V}\mid \delta \vDash "{\varphi (\vec {x})}.\text {"}$ In $\mathrm {M}$ , there is then some level of the cumulative hierarchy below $j(\kappa )$ that satisfies $\varphi (j(\vec {x}))$ and so by elementarity, in $\mathrm {V}$ there’s a level $\mathrm {V}\mid \alpha $ of the cumulative hierarchy with $\operatorname {\mathrm {rank}}(\vec {x})<\alpha <\kappa $ satisfying $\varphi (\vec {x})$ . Because $\delta>\kappa $ , we could have thrown in any fixed ordinal $\beta <\kappa $ as a useless parameter into $\vec {x}$ so the idea above gives an $\alpha>\beta $ with this property. here

Proof Let E be the derived $(\kappa ,\mu )$ -extender: $E=\{\langle r,X\rangle \in [\mu ]^{<\omega }\times \mathcal {P}([\kappa ]^{|r|}):r\in j(X)\}$ . We form the ultrapower $\mathrm {Ult}_E(\mathrm {V})$ with elementary $j_E:\mathrm {V}\rightarrow \mathrm {Ult}_E(\mathrm {V})$ and can factor $j=j_E\circ k_E$ where $\mathrm {cp}(k_E)\ge \mu $ . Note that $|\mathrm {Ult}_E(\mathrm {V})\mid \gamma |^{\mathrm {Ult}_E(\mathrm {V})}=\nu \le |\mathrm {V}\mid \gamma |<\mu $ for some $\nu $ . In particular, we can code $\langle \mathrm {Ult}_E(\mathrm {V})\mid \gamma ,\in \rangle $ as a subset of $\nu $ . This subset $R\subseteq \nu $ must also be in $\mathrm {M}$ because $\operatorname {\mathrm {cp}}(k_E)\ge \mu $ :

$$\begin{align*}k_E(R)=\{\beta\in k_E(\nu):\beta\in k_E(R)\}=\{\beta\in \nu:k_E(\beta)\in k_E(R)\}=R\text{.}\end{align*}$$

But then in $\mathrm {M}$ , by elementarity, $k_E(R)=R$ codes $\langle \mathrm {M}\mid \gamma ,\in \rangle =\langle \mathrm {V}\mid \gamma ,\in \rangle $ for every $\gamma <\lambda $ . This shows $\mathrm {V}\mid \lambda =\mathrm {Ult}_E(\mathrm {V})\mid \lambda $ , meaning E is $\lambda $ -strong.

Result 1.17. Suppose $\kappa $ is strong and this is not destroyed by $\mathrm {Add}(\kappa ^+,1)$ . Suppose $\kappa ^{+\sharp \sharp }$ exists. Then, there are arbitrarily large $\delta <\kappa $ whose $\delta +2$ -strength is destroyed by $\mathrm {Add}(\delta ^+,1)$ .

Proof Let A be $\mathrm {Add}(\kappa ^+,1)$ -generic over $\mathrm {V}$ . By hypothesis, $\kappa $ is still strong in $\mathrm {V}[A]$ and so we get a sufficiently strong embedding $j:\mathrm {V}[A]\rightarrow \mathrm {M}[j(A)]$ with $\operatorname {\mathrm {cp}}(j)=\kappa $ so that $\lambda =\kappa ^{+\sharp \sharp }$ is still $\lambda +2$ -strong in $\mathrm {M}[j(A)]$ .

Without loss of generality, $\mathrm {M}[j(A)]$ has enough agreement with $\mathrm {V}[A]$ to witness this fact as well. Since $\kappa <\lambda <j(\kappa )$ , this is reflected down into $\mathrm {V}[A]$ : there are arbitrarily large $\delta <\kappa $ such that $\delta $ ’s $\delta +2$ -strength is destroyed by $\mathrm {Add}(\delta ^+,1)$ in $\mathrm {V}[A]$ . It follows that $\mathrm {Add}(\kappa ^+,1)*\mathrm {Add}(\delta ^+,1)$ is $<\delta $ -directed closed and $\le \delta $ -distributive in $\mathrm {V}$ but destroys $\delta $ ’s $\delta +2$ -strength. Since $\mathrm {Add}(\kappa ^+,1)$ is sufficiently distributive, it can’t add to nor destroy $\delta $ ’s $\delta +2$ -strength. So it must be that $\mathrm {Add}(\delta ^+,1)^{\mathrm {V}[A]}$ destroyed it. Since $\mathrm {Add}(\kappa ^+,1)$ is $\le \kappa $ -distributive, $\mathrm {Add}(\delta ^+,1)^{\mathrm {V}[A]}=\mathrm {Add}(\delta ^+,1)^{\mathrm {V}}$ destroyed $\delta $ ’s $\delta +2$ -strength in $\mathrm {V}$ .

Proof (2) always implies (1) since any failure of hyperstrength is reflected to $\mathrm {V}\mid \delta $ for arbitrarily large $\delta $ . So we assume (1) and aim to show (2) by induction. For $\alpha =1$ —i.e., $0$ -hyperstrength or just being strong—this follows by the absoluteness of strength of extenders between levels of the cumulative hierarchy. Extenders witnessing the $\lambda $ -strength of $\kappa $ for $\lambda <\delta $ can be found with length $\mu =|\mathrm {V}\mid \lambda |^+$ by Lemma 1.12 and so are in $\mathrm {V}\mid {\mu +\omega }\subseteq \mathrm {V}\mid \delta $ with $\delta $ a strong limit. So $\kappa $ being strong implies it is strong in $\mathrm {V}\mid \delta $ .

For limit $\alpha $ , this follows inductively: for each $\xi <\alpha $ , $\kappa $ is $<\xi +1$ -hyperstrong iff $\mathrm {V}\mid \delta \vDash $ “ $\kappa $ is $<\xi +1$ -hyperstrong” for all strong limit $\delta>|\kappa +\xi |^+$ closed under $\lambda \mapsto |\mathrm {V}\mid \lambda |^+$ . Hence $\kappa $ is $<\alpha $ -hyperstrong implies $\mathrm {V}\mid \delta $ satisfies this for such $\delta $ greater than $\sup _{\xi <\alpha }|\kappa +\xi |^+\le |\kappa +\alpha |^+$ .

For successor $\alpha $ , assume the results for $<\alpha $ -hyperstrength. In fact, we take the inductive hypothesis as this holding for any transitive model of $\mathsf {ZFC}$ containing $|\kappa +\alpha |^+$ . But without loss of generality, and for the sake of notation, take our model as $\mathrm {V}$ . As notation, for E an extender, let $j_E:\mathrm {V}\rightarrow \operatorname {\mathrm {Ult}}_E(\mathrm {V})$ be the canonical embedding.

Suppose $\kappa $ is $\alpha $ -hyperstrong and let $\delta \ge |\kappa +\alpha |^+$ be an arbitrary strong limit closed under $\lambda \mapsto |\mathrm {V}\mid \lambda |^+$ . Inductively, $\mathrm {V}\mid \delta \vDash "{\kappa \text { is }< \alpha \text {-hyperstrong.}}\text {"}$ Let $\lambda <\delta $ be arbitrary. In $\mathrm {V}$ , there’s a $\lambda $ -strong extender E on $\kappa $ such that $\mathrm {Ult}_E(\mathrm {V})\vDash "{\kappa \text { is }<\alpha \text {-hyperstrong.}}\text {"}$ Consider $\gamma =\sup _{\beta <\delta }j_E(\beta )$ which is a strong limit in $\mathrm {Ult}_E(\mathrm {V})$ because $\delta $ is a strong limit in $\mathrm {V}$ . Moreover,

$$\begin{align*}\gamma \ge \delta \ge|\kappa+\alpha |^+\ge(|\kappa+\alpha |^{+})^{\mathrm{Ult}_E(\mathrm{V})}\text{.}\end{align*}$$

To see that $\gamma $ is closed under $\lambda '\mapsto |\mathrm {V}\mid \lambda '|^+$ in $\mathrm {Ult}_E(\mathrm {V})$ , let $\lambda '<\gamma $ be arbitrary. Then $\lambda '<j_E(\beta )$ for some $\beta <\delta $ . Since $|\mathrm {V}\mid \beta |^+<\delta $ , we have by elementarity

$$\begin{align*}\mathrm{Ult}_E(\mathrm{V})\models"{|\mathrm{Ult}_E(\mathrm{V})\mid\lambda'|\le |\mathrm{Ult}_E(\mathrm{V})\mid j_E(\beta)|=j_E(|\mathrm{V}\mid \beta|)<\gamma}.\text{"}\end{align*}$$

As a strong limit in $\mathrm {Ult}_E(\mathrm {V})$ , $(|\mathrm {V}\mid \lambda '|^+)^{\mathrm {Ult}_E(\mathrm {V})}<\gamma $ . So by the inductive hypothesis on $\alpha $ ,

$$\begin{align*}\mathrm{Ult}_E(\mathrm{V})\mid\gamma \vDash "{\kappa \text{ is }<\alpha \text{-hyperstrong.}}\text{"}\end{align*}$$

Yet $E\in \mathrm {V}\mid \delta $ and taking the ultrapower there yields $\mathrm {Ult}_E^{\mathrm {V}\mid \delta }(\mathrm {V}\mid \delta )=\mathrm {Ult}_E(\mathrm {V})\mid \gamma $ . Hence E witnesses the $\alpha $ -hyperstrength of $\kappa $ in $\mathrm {V}\mid \delta $ .

Proof

1. 1. Suppose $\kappa $ is not hyperstrong. By reflection,

   (*) $$ \begin{align} \mathrm{V}\mid \delta \vDash "{\exists \gamma \ (\kappa \text{ is not }\gamma \text{-hyperstrong})\wedge \forall\alpha\in\mathrm{Ord}\ (\beth_\alpha\text{ and }|\mathrm{V}\mid\alpha|^+\text{ exists})}\text{"} \end{align} $$
   for some $\delta $ . By Lemma 1.11, we get some such that ( $*$ ) holds. Since $\beth _\alpha <\delta $ for every $\alpha <\delta $ , $\delta $ is a strong limit with then $\delta>|\kappa +\gamma |^+$ for any $\gamma $ as in ( $*$ ). Corollary 1.21 therefore implies $\kappa $ is not $\gamma $ -hyperstrong for some , a contradiction.

2. 2. Suppose $\kappa $ is not $\alpha $ -hyperstrong for some . This fact can be reflected to some level $\mathrm {V}\mid \delta $ where $\delta $ is a strong limit, and by Lemma 1.11, to $\mathrm {V}\mid \delta $ for arbitrarily large strong limit . In particular, we get some strong limit $\delta _0$ with such that $\mathrm {V}\mid {\delta _0}\vDash $ “ $\kappa $ is not $\alpha $ -hyperstrong.” But applying Corollary 1.21 in , by the hypothesis, for all limit $\delta>|\kappa +\alpha |^+$ , a contradiction with the hypothesis on $\delta _0$ . Hence $\kappa $ must be $\alpha $ -hyperstrong for each , and so hyperstrong by (1).

Result 1.23. Let $\kappa $ be srs such that exists. Then $\kappa $ is hyperstrong. In fact, any single -srs embedding witnesses all degrees of hyperstrength of $\kappa $ . Moreover, in $\mathrm {V}\mid \kappa $ , there is a proper class of hyperstrongs.

Proof Let $\kappa $ be srs. $\kappa $ is already $0$ -hyperstrong. There is then a -srs embedding $j:\mathrm {V}\rightarrow \mathrm {M}$ with $\operatorname {\mathrm {cp}}(j)=\kappa $ . We proceed by induction on $\alpha $ to show $\kappa $ is $<\alpha $ -hyperstrong in $\mathrm {V}$ and $\mathrm {M}$ , with the base case of $\alpha =1$ already true. The limit case is trivial, so we consider only the successor case: showing $\alpha $ -hyperstrength in $\mathrm {V}$ and $\mathrm {M}$ . By Lemma 1.22 (1), we may assume .

Suppose $\kappa $ is $<\alpha $ -hyperstrong in $\mathrm {V}$ and $\mathrm {M}$ so that $\kappa $ is $\alpha $ -hyperstrong in $\mathrm {V}$ . If there is some $\lambda $ with no $\lambda $ -strong extender in $\mathrm {M}$ witnessing $\alpha $ -hyperstrength of $\kappa $ , then by reflection, some $\mathrm {M}\mid \delta $ (correctly) satisfies there’s no such extender. By Lemma 1.11, we can assume with $\delta $ a strong limit so that $\mathrm {M}\mid \delta =\mathrm {V}\mid \delta $ also has no such extender. Corollary 1.21 tells us $\mathrm {V}$ must satisfy $\kappa $ isn’t $\alpha $ -hyperstrong, a contradiction.

We also can show that $\kappa $ is the limit of hyperstrongs, giving $\mathrm {V}\mid \kappa $ as a model of a proper class of hyperstrongs. Suppose not: $\kappa $ is the least hyperstrong above some $\lambda $ . By elementarity, $j(\kappa )$ is still the least hyperstrong above $j(\lambda )=\lambda $ in $\mathrm {M}$ but So by Lemma 1.22 (2), $\kappa $ is hyperstrong in $\mathrm {M}$ . But $\lambda <\kappa <j(\kappa )$ contradicts that $j(\kappa )$ is the least hyperstrong above $\lambda $ in $\mathrm {M}$ .

Proof Clearly if $\kappa $ is srs, any $\lambda $ -srs embedding $j:\mathrm {V}\rightarrow \mathrm {M}$ has $\kappa $ as strong in both $\mathrm {V}$ and $\mathrm {M}$ , meaning $\kappa $ is $1$ -hyperstrong. For the other direction, let $\lambda>\kappa $ be arbitrary, and let $j:\mathrm {V}\rightarrow \mathrm {M}$ be $\lambda $ -strong where $\kappa $ is strong in $\mathrm {M}$ . If some $\delta $ is strong in $\mathrm {M}$ for $\kappa <\delta <\lambda $ , let $\delta $ be least. Then there is a $\lambda $ -strong embedding $h:\mathrm {M}\rightarrow \mathrm {N}$ with $\mathrm {cp}(h)=\delta $ such that $\delta $ is not strong in $\mathrm {N}$ , e.g., using the Mitchell-least $\lambda $ -strong embedding. It follows that in $\mathrm {N}$ , there are no strongs in the interval $(\kappa ,j(\delta ))\supseteq (\kappa ,\lambda )$ . Hence $h\circ j:\mathrm {V}\rightarrow \mathrm {N} \lambda $ -strong and $\mathrm {V}$ , $\mathrm {N}$ agree that the only strong below $\lambda $ is $\kappa $ , so $\kappa $ is $\lambda $ -srs.