Proof. Pick a strictly increasing, cofinal sequence $\langle {\kappa _\xi }~\vert ~{\xi <\mu }\rangle $ of measurable cardinals in $\kappa $ with the property that $\mu <\kappa $ implies that $\mu <\kappa _0$ . Given $\xi <\mu $ , fix a normal ultrafilter $U_\xi $ on $\kappa _\xi $ and let

$$ \begin{align*}\langle\langle{N^\xi_\alpha}~\vert~{\alpha\in{\mathrm{On}}}\rangle,\langle{{j^\xi_{\alpha,\beta}}:{N^\xi_\alpha}\longrightarrow{N^\xi_\beta}}~\vert~{\alpha\leq\beta\in{\mathrm{On}}}\rangle\rangle\end{align*} $$

denote the iteration of $\langle {\mathrm {V}},U_\xi \rangle $ . Given $\xi <\mu $ , we then have $j^\xi _{0,\kappa }(\kappa _\xi )=\kappa $ and $j^\xi _{0,\alpha }(\kappa )=\kappa $ for all $\alpha <\kappa $ . In particular, we know that $\vert {{\mathcal {P}}({\kappa })^{N^\xi _\kappa }}\vert =\kappa $ holds for all $\xi <\mu $ . Therefore, we can find $x\in D$ with $x\notin N^\xi _\kappa $ for all $\xi <\mu $ . Given $\xi <\mu $ , we then have $x\neq j^\xi _{0,\kappa }(x)\cap \kappa $ and hence we know that

(1) $$ \begin{align} x\cap j^\xi_{0,\lambda}(\kappa_\xi) ~ \neq ~ j^\xi_{0,\lambda}(x\cap\kappa_\xi) \end{align} $$

holds for all sufficiently large $\lambda <\kappa $ .

By earlier remarks, the pair $\langle {\mathrm {V}},\mathcal {E}\rangle $ is linearly iterable. In the following, we inductively construct systems with the properties listed above while also ensuring that for every $s\in {}^{{<}\mu }\kappa $ , there exists ${{\mathrm {dom}}(s)}\leq \xi <\mu $ with $\kappa _s=\kappa _\xi $ and $U_s=U_\xi $ . Note that this additional property will directly ensure that (iv) holds in the end.

First, we define $I_\emptyset $ to be the trivial linear iteration of $\langle {\mathrm {V}},\mathcal {E}\rangle $ . Moreover, we pick some inaccessible cardinal $\nu _\emptyset <\kappa $ such that $z\in {\mathrm {H}}(\nu _\emptyset )$ and $\mu <\kappa $ implies $\mu <\nu _\emptyset $ .

Next, assume that $\zeta \in {\mathrm {Lim}}\cap \mu $ and the objects $\nu _t$ , $\kappa _t$ , $U_t$ and $I_t$ are defined for all $t\in {}^{{<}\zeta }\kappa $ . Fix $s\in {}^\zeta \kappa $ , and define $I_s$ to be the unique linear iteration of $\langle {\mathrm {V}},\{{U_{s\restriction \eta }}~\vert ~{\eta <\zeta }\}\rangle $ of length $\sup _{\eta <\zeta }{\mathrm {lh}}({I_{s\restriction \eta }}){}<\kappa $ with the property that $U^{I_s}_\alpha =U^{I_{s\restriction \eta }}_\alpha $ holds for all $\eta <\zeta $ and $\alpha <{\mathrm {lh}}({I_{s\restriction \eta }}){}$ . In addition, define $\nu _s$ to be an inaccessible cardinal smaller than $\kappa $ and bigger than both $\sup _{\eta <\zeta }\kappa _{s\restriction \eta }$ and ${\mathrm {lh}}({I_s}){}$ . This setup ensures that ${\mathrm {lh}}({I_s})>1$ implies that ${\mathrm {lh}}({I_s})\in {\mathrm {Lim}}$ , and therefore we know that (v) holds. Moreover, these definitions directly ensure that the relevant parts of (vi) and (vii) hold in this case. In addition, since ${\mathrm {lh}}({I_s})<\nu _s$ and $I_s$ only makes use of ultrafilter on cardinals contained in the interval $(\nu _\emptyset ,\nu _s)$ , the fact that the cofinality of $\kappa $ is not contained in this interval allows us to conclude that (viii) holds in this case. Next, notice that our construction directly ensures that (ix) holds in this case. Moreover, if $\eta <\zeta $ with ${\mathrm {lh}}({I_{s\restriction \eta }}){}<{\mathrm {lh}}({I_s}){}$ , then the fact that (v) and (x) hold for all $\eta <\rho <\zeta $ ensures that

$$ \begin{align*}{\mathrm{H}}(\kappa_{s\restriction\eta})^{M^{I_{s\restriction\eta}}_\infty} ~ = ~ {\mathrm{H}}(\kappa_{s\restriction\eta})^{M^{I_s}_\infty}\end{align*} $$

and

$$ \begin{align*}i^{I_s}_{{\mathrm{lh}}({I_{s\restriction\eta}}){},\infty}\restriction{\mathrm{H}}(\kappa_{s\restriction\eta})^{M^{I_s}_\infty} ~ = ~ {\mathrm{id}}_{{\mathrm{H}}(\kappa_{s\restriction\eta})^{M^{I_s}_\infty}}.\end{align*} $$

By the definition of $I_s$ , this shows that (x) also holds in this case. Finally, pick $t\in {}^{{<}\mu }\kappa $ with ${{\mathrm {dom}}(t)}\leq \zeta $ and $\xi \in {{\mathrm {dom}}(t)}$ with $s\restriction \xi =t\restriction \xi $ and $s(\xi )\neq t(\xi )$ . Set $\rho =\min (\nu _{s\restriction (\xi +1)},\nu _{t\restriction (\xi +1)})$ . Since we know that $\rho <\min (\kappa _{s\restriction (\xi +1)},\kappa _{t\restriction (\xi +1)})$ , we can use (x) and (xi) to show that

$$ \begin{align*}i^{I_s}_{0,\infty}(x)\cap\rho ~ = ~ i^{I_{s\restriction(\xi+1)}}_{0,\infty}(x)\cap\rho ~ \neq ~ i^{I_{t\restriction(\xi+1)}}_{0,\infty}(x)\cap\rho ~ = ~ i^{I_t}_{0,\infty}(x)\cap\rho.\end{align*} $$

By the properties of $\nu _{s\restriction (\xi +1)}$ ensured by our induction hypothesis, these computations show that (xi) also holds in this case.

Now, assume that $\zeta <\mu $ and the objects $\nu _t$ , $\kappa _u$ , $U_u$ and $I_t$ are defined for all $t,u\in {}^{{<}\mu }\kappa $ with ${{\mathrm {dom}}(t)}\leq \zeta $ and ${{\mathrm {dom}}(u)}<\zeta $ . Fix $s\in {}^\zeta \kappa $ , and pick $\zeta \leq \xi <\mu $ with $\kappa _\xi>\nu _s$ . Set $\kappa _s=\kappa _\xi $ and $U_s=U_\xi $ . By Equation (1), there exists a limit ordinal $\kappa _s<\lambda <\kappa $ with the property that

(2) $$ \begin{align} x\cap j^\xi_{0,\lambda}(\kappa_s) ~ \neq ~ j^\xi_{0,\lambda}(x\cap\kappa_s). \end{align} $$

Let $\langle {\lambda _\beta }~\vert ~{\beta <\kappa }\rangle $ denote the unique continuous sequence of ordinals with $\lambda _0=0$ and $\lambda _{\beta +1}=\lambda _\beta +j^\xi _{0,\lambda _\beta }(\lambda )$ for all $\beta <\lambda $ . Since $\kappa $ is a limit of inaccessible cardinals, we know that $\lambda _\beta <\kappa $ holds for all $\beta <\kappa $ . Given $\beta <\kappa $ , define $I_{s^\frown \langle \beta \rangle }$ to be the unique linear iteration of $\langle {\mathrm {V}},\mathcal {E}\rangle $ of length ${\mathrm {lh}}({I_s})+i^{I_s}_{0,\infty }(\lambda _\beta )$ with $U^{I_{s^\frown \langle \beta \rangle }}_\alpha =U^{I_s}_\alpha $ for all $\alpha <{\mathrm {lh}}({I_s}){}$ and $U^{I_{s^\frown \langle \beta \rangle }}_\alpha =i^{I_{s^\frown \langle \beta \rangle }}_{0,\alpha }(U_s)$ for all ${\mathrm {lh}}({I_s}){}\leq \alpha <{\mathrm {lh}}({I_{s^\frown \langle \beta \rangle }}){}$ . That means we linearly iterate $U_s$ on top of what we already have to obtain $I_{s^\frown \langle \beta \rangle }$ . Moreover, for every $\beta <\kappa $ , we define $\nu _{s^\frown \langle \beta \rangle }$ to be the least inaccessible cardinal greater than ${\mathrm {lh}}({I_{s^\frown \langle \beta +1\rangle }}){}$ . These definitions then directly ensure that (v) and (ix) hold. In addition, for all $\beta <\kappa $ , we have

$$ \begin{align*}{\mathrm{lh}}({I_s}) ~ < ~ \nu_s ~ < ~ \kappa_s ~ < ~ \lambda ~ \leq ~ {\mathrm{lh}}({I_{s^\frown\langle\beta+1\rangle}}) ~ < ~ \nu_{s^\frown\langle\beta\rangle}\end{align*} $$

and this can be used to conclude that $i^{I_s}_{0,\infty }(\kappa _s)=\kappa _s$ and $i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }(\nu _{s^\frown \langle \beta \rangle })=\nu _{s^\frown \langle \beta \rangle }$ . This shows that the relevant instances of (vi) and (vii) hold in this case. Moreover, the fact that all linear iterations of the form $I_{s^\frown \langle \beta \rangle }$ with $\beta <\kappa $ have length less than $\kappa $ and only make use of ultrafilters on cardinals contained in the interval $[\kappa _0,\kappa )$ directly implies that (viii) holds in this case as $\mu < \kappa _0$ in case $\mu < \kappa $ . Next, notice that, if $0<\beta <\kappa $ and ${\mathrm {lh}}({I_s}){}\leq \alpha <{\mathrm {lh}}({I_{s^\frown \langle \beta \rangle }}){}$ , then our construction ensures that

$$ \begin{align*}{\mathrm{H}}(\kappa_s)^{M^{I_s}_\infty} ~ = ~ {\mathrm{H}}(\kappa_s)^{M^{I_{s^\frown\langle\beta\rangle}}_{{\mathrm{lh}}({I_s}){}}} ~ = ~ {\mathrm{H}}(\kappa_s)^{M^{I_{s^\frown\langle\beta\rangle}}_\alpha}\end{align*} $$

and

$$ \begin{align*}i^{I_{s^\frown\langle\beta\rangle}}_{{\mathrm{lh}}({I_s}){},\alpha}\restriction{\mathrm{H}}(\kappa_s)^{M^{I_s}_\infty} ~ = ~ {\mathrm{id}}_{{\mathrm{H}}(\kappa_s)^{M^{I_s}_\infty}}.\end{align*} $$

This directly implies that (x) holds in this case. Finally, fix $\beta <\gamma <\kappa $ . Then ${\mathrm {lh}}({I_{s^\frown \langle \beta +1\rangle }}){}<{\mathrm {lh}}({I_{s^\frown \langle \gamma +1\rangle }}){}$ , and hence we know that $\nu _{s^\frown \langle \beta \rangle }\leq \nu _{s^\frown \langle \gamma \rangle }$ .

Proof of the claim

Let

$$ \begin{align*}\langle\langle{N_\alpha}~\vert~{\alpha\in{\mathrm{On}}}\rangle,\langle{{j_{\alpha_0,\alpha_1}}:{N_{\alpha_0}}\longrightarrow{N_{\alpha_1}}}~\vert~{\alpha_0\leq\alpha_1\in{\mathrm{On}}}\rangle\rangle\end{align*} $$

denote the iteration of $\langle M^{I_s}_\infty ,i^{I_s}_{0,\infty }(U_s)\rangle $ . Given $\delta <\kappa $ and $\alpha <i^{I_s}_{0,\infty }(\lambda _\delta )$ , the definition of $I_{s^\frown \langle \delta \rangle }$ ensures that the following statements hold:

• ○ $M^{I_{s^\frown \langle \delta \rangle }}_{{\mathrm {lh}}({I_s})+\alpha }=N_\alpha $ and $M^{I_{s^\frown \langle \delta \rangle }}_\infty =N_{i^{I_s}_\infty (\lambda _\delta )}$ .

• ○ $i^{I_{s^\frown \langle \delta \rangle }}_{0,{\mathrm {lh}}({I_s})+\alpha }=j_{0,\alpha }\circ i^{I_s}_{0,\infty }$ and $i^{I_{s^\frown \langle \delta \rangle }}_{0,\infty }=j_{0,i^{I_s}_{0,\infty }(\lambda _\delta )}\circ i^{I_s}_{0,\infty }$ .

Now, set $M_*=M^{I_{s^\frown \langle \beta \rangle }}_\infty $ , $x_*=i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }(x)$ , $\kappa _*=i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }(\kappa _s)$ , $\lambda _*=i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }(\lambda )$ and $U_*=i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }(U_s)$ . Note that elementarity ensures that

$$ \begin{align*}i_{0,\infty}^{I_s}(j^\xi_{0,\lambda_\beta}(\lambda)) ~ = ~ j_{0,i^{I_s}_{0,\infty}(\lambda_\beta)}(i^{I_s}_{0,\infty}(\lambda)) ~ = ~ i^{I_{s^\frown\langle\beta\rangle}}_{0,\infty}(\lambda) ~ = ~ \lambda_*\end{align*} $$

and this allows us to conclude that

$$ \begin{align*}i^{I_s}_{0,\infty}(\lambda_\gamma) ~ \geq ~ i^{I_s}_{0,\infty}(\lambda_{\beta+1}) ~ = ~ i^{I_s}_{0,\infty}(\lambda_\beta+j^\xi_{0,\lambda_\beta}(\lambda)) ~ = ~ i^{I_s}_{0,\infty}(\lambda_\beta)+\lambda_*.\end{align*} $$

In particular, we know that

(3) $$ \begin{align} {\mathrm{lh}}({I_{s^\frown\langle\beta\rangle}}){}+\lambda_*\leq{\mathrm{lh}}({I_{s^\frown\langle\gamma\rangle}}){}. \end{align} $$

Now, define

$$ \begin{align*}\langle\langle{N^*_\alpha}~\vert~{\alpha\in{\mathrm{On}}}\rangle,\langle{{j^*_{\alpha_0,\alpha_1}}:{N^*_{\alpha_0}}\longrightarrow{N^*_{\alpha_1}}}~\vert~{\alpha_0\leq\alpha_1\in{\mathrm{On}}}\rangle\rangle\end{align*} $$

to be the iteration of $\langle M_*,U_*\rangle $ . Given ordinals $\alpha _0\leq \alpha _1$ , we then have $N^*_{\alpha _0}=N_{i^{I_s}_\infty (\lambda _\beta )+\alpha _0}$ and $j^*_{\alpha _0,\alpha _1}=j_{i^{I_s}_\infty (\lambda _\beta )+\alpha _0,i^{I_s}_\infty (\lambda _\beta )+\alpha _1}$ . In particular, we can use Equation (3) to find an ordinal $\alpha \geq \lambda _*$ with $M^{I_{s^\frown \langle \gamma \rangle }}_\infty =N^*_\alpha $ and $i^{I_{s^\frown \langle \gamma \rangle }}_{0,\infty }=j^*_{0,\alpha }\circ i^{I_{s^\frown \langle \beta \rangle }}_{0,\infty }$ .

By elementarity, the inequality (2) implies that

$$ \begin{align*}x_*\cap j^*_{0,\lambda_*}(\kappa_*) ~ \neq ~ j^*_{0,\lambda_*}(x_*\cap\kappa_*).\end{align*} $$

Moreover, our setup ensures that

$$ \begin{align*}\nu_{s^\frown\langle\beta\rangle} ~> ~ {\mathrm{lh}}({I_{s^\frown\langle\beta+1\rangle}}){} ~ \geq ~ i^{I_s}_{0,\infty}(\lambda_{\beta+1}) ~ \geq ~ i^{I_s}_{0,\infty}(j^\xi_{0,\lambda_\beta}(\lambda)) ~ = ~ \lambda_* ~ > ~ \kappa_*\end{align*} $$

and

$$ \begin{align*}\nu_{s^\frown\langle\beta\rangle} ~ = ~ j^*_{0,\lambda_*}(\nu_{s^\frown\langle\beta\rangle}) ~> ~ j^*_{0,\lambda_*}(\kappa_*).\end{align*} $$

Since for $\alpha \geq \lambda _*$ as above

$$ \begin{align*}i^{I_{s^\frown\langle\gamma\rangle}}_{0,\infty}(x) ~ = ~ j^*_{0,\alpha}(i^{I_{s^\frown\langle\beta\rangle}}_{0,\infty}(x)) ~ = ~ (j^*_{\lambda_*,\alpha}\circ j^*_{0,\lambda_*})(x_*)\end{align*} $$

and

$$ \begin{align*}j^*_{\lambda_*,\alpha}\restriction(j^*_{0,\lambda_*}(\kappa_*)) ~ = ~ {\mathrm{id}}_{j^*_{0,\lambda_*}(\kappa_*)},\end{align*} $$

we know that

$$ \begin{align*} \begin{aligned} i^{I_{s^\frown\langle\gamma\rangle}}_{0,\infty}(x)\cap j^*_{0,\lambda_*}(\kappa_*) ~ & = ~ j^*_{0,\lambda_*}(x_*)\cap j^*_{0,\lambda_*}(\kappa_*) ~ = ~ j^*_{0,\lambda_*}(x_*\cap\kappa_*) \\ & \neq ~ x_*\cap j^*_{0,\lambda_*}(\kappa_*) ~ = ~ i^{I_{s^\frown\langle\beta\rangle}}_{0,\infty}(x)\cap j^*_{0,\lambda_*}(\kappa_*) \end{aligned} \end{align*} $$

and the fact that $\nu _{s^\frown \langle \beta \rangle }> j^*_{0,\lambda _*}(\kappa _*)$ then yields the statement of the claim.

The above claim now shows that (xi) also holds in this case. This completes the proof of the lemma.

Proof. Work in ${\mathrm {V}}[G]$ , pick a function $c\in {}^\lambda \kappa $ with the desired properties and define $I_c$ as above. If ${\mathbb {P}}$ is contained in ${\mathrm {H}}(\kappa _\emptyset )$ , then we can use the Lévy–Solovay theorem to show that for all $\xi <\lambda $ , the set $\{{B\in {\mathcal {P}}({\kappa _\xi })}~\vert ~{\exists A\in U_{c\restriction \xi } ~ A\subseteq B}\}$ is a normal ultrafilter on $\kappa _\xi $ , and therefore we know that $U_{c\restriction \xi }$ itself is a $\sigma $ -complete ${\mathrm {V}}$ -ultrafilter. Since the same conclusion obviously holds true if forcing with ${\mathbb {P}}$ does not add bounded subsets of $\kappa $ , we can apply [Reference Steel49, Theorem 3.3] to conclude that the pair $\langle {\mathrm {V}},\{{U_{c\restriction \xi }}~\vert ~{\xi <\lambda }\}\rangle $ is linearly iterable, and therefore we know that $M^{I_c}_\infty $ is well-founded.

Next, since (x) of Lemma 2.1 ensures that

$$ \begin{align*}{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_c}_{{\mathrm{lh}}({I_{c\restriction\xi}}){}}} ~ = ~ {\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty} ~ = ~ {\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\zeta}}_\infty} ~ = ~ {\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_c}_{{\mathrm{lh}}({I_{c\restriction\zeta}}){}}}\end{align*} $$

and

$$ \begin{align*}i^{I_c}_{{\mathrm{lh}}({I_{c\restriction\xi}}){},{\mathrm{lh}}({I_{c\restriction\zeta}}){}}\restriction{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty} ~ = ~ i^{I_{c\restriction\zeta}}_{{\mathrm{lh}}({I_{c\restriction\xi}}){},\infty}\restriction{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty} ~ = ~ {\mathrm{id}}_{{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty}}\end{align*} $$

hold for all $\xi <\zeta <\lambda $ with ${\mathrm {lh}}({I_{c\restriction \xi }}){}<{\mathrm {lh}}({I_{c\restriction \zeta }}){}<{\mathrm {lh}}({I_c}){}$ , we know that

$$ \begin{align*}{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty} ~ = ~ {\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_c}_\infty}\end{align*} $$

and

(5) $$ \begin{align} i^{I_c}_{{\mathrm{lh}}({I_{c\restriction\xi}}){},\infty}\restriction{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty} ~ = ~ {\mathrm{id}}_{{\mathrm{H}}(\kappa_{c\restriction\xi})^{M^{I_{c\restriction\xi}}_\infty}} \end{align} $$

hold for all $\xi <\lambda $ with ${\mathrm {lh}}({I_{c\restriction \xi }}){}<{\mathrm {lh}}({I_c}){}$ . In particular, it follows that $i^{I_c}_{0,\infty }(z)=z$ and, if $\mu <\kappa $ , then $i^{I_c}_{0,\infty }(\mu )=\mu $ . In addition, for all $\xi <\lambda $ with the property that ${\mathrm {lh}}({I_{c\restriction \xi }}){}<{\mathrm {lh}}({I_c}){}$ , we have $i^{I_c}_{0,{\mathrm {lh}}({I_{c\restriction \xi }}){}}=i^{I_{c\restriction \xi }}_{0,\infty }$ and therefore

(6) $$ \begin{align} i^{I_c}_{0,{\mathrm{lh}}({I_{c\restriction\xi}}){}}(\alpha) ~ < ~ i^{I_c}_{0,{\mathrm{lh}}({I_{c\restriction\xi}}){}}(\kappa_{c\restriction\xi}) ~ = ~ i^{I_{c\restriction\xi}}_{0,\infty}(\kappa_{c\restriction\xi}) ~ = ~ \kappa_{c\restriction\xi} \end{align} $$

for all $\alpha <\kappa _{c\restriction \xi }$ . In particular, a combination of Equations (5) and (6) allows us to conclude that $i^{I_c}_{0,\infty }[\kappa _{c\restriction \xi }]\subseteq \kappa _{c\restriction \xi }$ holds for all $\xi <\lambda $ . If the sequence $\langle {\kappa _{c\restriction \xi }}~\vert ~{\xi <\lambda }\rangle $ is cofinal in $\kappa $ , then this observation directly implies that $i^{I_c}_{0,\infty }(\kappa )=\kappa $ . In the other case, if the above sequence is bounded by $\rho <\kappa $ , then $I_c$ is a linear iteration of length less than $\kappa $ that only uses ultrafilters on measurable cardinals in the interval $[\kappa _\emptyset ,\rho ]$ and, since $\kappa $ is a limit of inaccessible cardinals whose cofinality is not contained in this interval, we also know that $i^{I_c}_{0,\infty }(\kappa )=\kappa $ holds in this case.

Finally, pick functions $c,d\in {}^\lambda \kappa $ whose proper initial segments are all contained in ${\mathrm {V}}$ and $\xi <\lambda $ with $c\restriction \xi =d\restriction \xi $ and $c(\xi )\neq d(\xi )$ . Then Equation (5) implies that

$$ \begin{align*}i^{I_c}_{0,\infty}(x)\cap\kappa_{c\restriction\xi} ~ = ~ i^{I_{c\restriction\xi}}_{0,\infty}(x)\cap\kappa_{c\restriction\xi} ~ = ~ i^{I_d}_{0,\infty}(x)\cap\kappa_{c\restriction\xi}.\end{align*} $$

If we now define

$$ \begin{align*}\rho ~ = ~ \min(\nu_{c\restriction(\xi+1)},\nu_{d\restriction(\xi+1)}) ~ < ~ \min(\kappa_{c\restriction(\xi+1)},\kappa_{d\restriction(\xi+1)}),\end{align*} $$

then statement (xi) of Lemma 2.1 directly implies that Equation (4) holds.

Proof of Theorem 1.1

Let $\mu $ be an infinite regular cardinal, let $\kappa $ be a limit of measurable cardinals with ${{\mathrm {cof}}(\kappa )}=\mu $ , let z be an element of ${\mathrm {H}}(\kappa )$ and let D be a subset of ${\mathcal {P}}({\kappa })$ of cardinality greater than $\kappa $ that is definable by a $\Sigma _1$ -formula with parameters $\kappa $ and z. An application of Lemma 2.2 with the trivial partial order now yields $x\in D$ and systems $\langle {\kappa _s}~\vert ~{s\in {}^{{<}\mu }\kappa }\rangle $ , $\langle {U_s}~\vert ~{s\in {}^{{<}\mu }\kappa }\rangle $ and $\langle {I_c}~\vert ~{c\in {}^\mu \kappa }\rangle $ such that the following statements hold for all $s,t\in {}^{{<}\mu }\kappa $ and all $c,d\in {}^\mu \kappa $ :

• ○ $\kappa _s$ is a measurable cardinal smaller than $\kappa $ .

• ○ $U_s$ is a normal ultrafilter on $\kappa _s$ .

• ○ $I_c$ is a linear iteration of $\langle {\mathrm {V}},\{{U_{c\restriction \xi }}~\vert ~{\xi <\mu }\}\rangle $ with $M^{I_c}_\infty $ well-founded.

• ○ The sequence $\langle {\kappa _{c\restriction \xi }}~\vert ~{\xi <\mu }\rangle $ is cofinal in $\kappa $ .

• ○ $i^{I_c}_{0,\infty }(\mu )=\mu $ , $i^{I_c}_{0,\infty }(\kappa )=\kappa $ and $i^{I_c}_{0,\infty }(z)=z$ .

• ○ If $\xi <\mu $ with $c\restriction \xi =d\restriction \xi $ and $c(\xi )\neq d(\xi )$ , then

  $$ \begin{align*}i^{I_c}_{0,\infty}(x)\cap\kappa_{c\restriction\xi} ~ = ~ i^{I_d}_{0,\infty}(x)\cap\kappa_{c\restriction\xi}\end{align*} $$
  and
  $$ \begin{align*}i^{I_c}_{0,\infty}(x)\cap\rho ~ \neq ~ i^{I_d}_{0,\infty}(x)\cap\rho,\end{align*} $$
  where $\rho =\min (\kappa _{c\restriction (\xi +1)},\kappa _{d\restriction (\xi +1)})$ .

We now define

$$ \begin{align*}{\iota}:{{}^\mu\kappa}\longrightarrow{{\mathcal{P}}({\kappa})};~{c}\longmapsto{i^{I_c}_{0,\infty}(x)}.\end{align*} $$

Then $\iota $ is an injection.

Proof of the claim

Fix $c\in {}^\mu \kappa $ . Pick a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ such that

$$ \begin{align*}D ~ = ~ \{{y\subseteq\kappa}~\vert~{\varphi(\kappa,y,z)}\}.\end{align*} $$

As $x \in D$ , $\varphi (\kappa ,x,z)$ holds in V and hence the properties listed above ensure that $\varphi (\kappa ,\iota (c),z)$ holds in $M^{I_c}_\infty $ . Since the upwards absoluteness of $\Sigma _1$ -statements directly implies that $\varphi (\kappa ,\iota (c),z)$ holds in ${\mathrm {V}}$ , we can conclude that $\iota (c)$ is an element of D.

This completes the proof of the theorem.

Proof. Assume, towards a contradiction, that the above conclusion fails. Then we may work in a model ${\mathrm {V}}$ of ${\mathrm {ZFC}}+{{\mathrm {V}}=M}$ that contains infinitely many measurable cardinals. Let $\kappa $ be the least limit of measurable cardinals, and let G be ${\mathrm {Add}}({\omega },{1})$ -generic over ${\mathrm {V}}$ . An application of (ii) and (iii) in ${\mathrm {V}}[G]$ now yields $z\in {\mathrm {H}}(\kappa )^{\mathrm {V}}$ and a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ with the property that the set $\{{y\in {\mathrm {V}}[G]}~\vert ~{{\mathrm {V}}[G]\models \varphi (\kappa ,y,z)}\}$ is a subset of ${\mathcal {P}}({\kappa })^{\mathrm {V}}$ that has cardinality greater than $\kappa $ in ${\mathrm {V}}[G]$ . In this situation, the homogeneity of ${\mathrm {Add}}({\omega },{1})$ in ${\mathrm {V}}$ ensures that these statements about the class defined by the formula $\varphi $ and the parameters $\kappa $ and z hold in every ${\mathrm {Add}}({\omega },{1})$ -generic extension of ${\mathrm {V}}$ .

Now, let $G_0\times G_1$ be $({\mathrm {Add}}({\omega },{1})\times {\mathrm {Add}}({\omega },{1}))$ -generic over ${\mathrm {V}}$ . Then the above observations show that

$$ \begin{align*}D ~ = ~ \{{y\in{\mathrm{V}}[G_0]}~\vert~{{\mathrm{V}}[G_0]\models\varphi(\kappa,y,z)}\}\end{align*} $$

is a subset of ${\mathcal {P}}({\kappa })^{\mathrm {V}}$ that has cardinality greater than $\kappa $ in ${\mathrm {V}}[G_0]$ . An application of Lemma 2.1 in ${\mathrm {V}}[G_0]$ then yields $x\in D$ and systems $\langle {\nu _s}~\vert ~{s\in {}^{{<}\omega }\kappa }\rangle $ , $\langle {\kappa _s}~\vert ~{s\in {}^{{<}\omega }\kappa }\rangle $ , $\langle {U_s}~\vert ~{s\in {}^{{<}\omega }\kappa }\rangle $ and $\langle {I_s}~\vert ~{s\in {}^{{<}\omega }\kappa }\rangle $ satisfying the statements listed in the lemma with respect to z and some subset of D of cardinality $\kappa ^+$ . Define

$$ \begin{align*}c ~ = ~ \bigcup G_1 ~ \in ~ ({}^\omega 2)^{V[G_0,G_1]}\setminus{\mathrm{V}}[G_0],\end{align*} $$

and let I denote the unique linear iteration of $\langle {\mathrm {V}}[G_0],\{{U_{c\restriction n}}~\vert ~{n<\omega }\}\rangle $ of length $\sup _{n<\omega }{\mathrm {lh}}({I_{c\restriction n}}){}$ in ${\mathrm {V}}[G_0,G_1]$ with $U^I_\alpha =U^{I_{c\restriction n}}_\alpha $ for all $n<\omega $ and $\alpha <{\mathrm {lh}}({I_{c\restriction n}}){}$ . Then Lemma 2.2 shows that $M^I_\infty $ is well-founded. Set $x_*=i^I_{0,\infty }(x)$ . Since Lemma 2.2 ensures that $i^I_{0,\infty }(\kappa )=\kappa $ and $i^I_{0,\infty }(z)=z$ , we can use the elementarity of $i^I_{0,\infty }$ and $\Sigma _1$ -upwards absoluteness to conclude that $\varphi (\kappa ,x_*,z)$ holds in ${\mathrm {V}}[G_0,G_1]$ . Since ${\mathrm {V}}[G_0,G_1]$ is an ${\mathrm {Add}}({\omega },{1})$ -generic extension of ${\mathrm {V}}$ , our earlier observations allow us to conclude that $x_*$ is an element of ${\mathcal {P}}({\kappa })^{\mathrm {V}}$ . If we now pick $n<\omega $ and set $\rho =\min (\kappa _{(c\restriction n)^\frown \langle 0\rangle },\kappa _{(c\restriction n)^\frown \langle 1\rangle })$ , then clause (iii) of Lemma 2.2 shows that $c(n)$ is the unique $i<2$ with

$$ \begin{align*}x_*\cap\rho ~ = ~ i^{I_{(c\restriction n)^\frown\langle i\rangle}}_{0,\infty}(x)\cap\rho.\end{align*} $$

This conclusion implies that c is an element of ${\mathrm {V}}[G_0]$ , a contradiction.

Proof. Pick a strictly increasing sequence $\langle {\kappa _m}~\vert ~{m<\omega }\rangle $ of infinite cardinals that is cofinal in $\kappa $ . In addition, fix an enumeration $\langle {a_\alpha }~\vert ~{\alpha <\kappa }\rangle $ of ${\mathrm {H}}(\kappa )$ . Define U to be the set of all pairs $\langle s,t\rangle $ in ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ with the property that ${\mathrm {lh}}({s})={\mathrm {lh}}({t})$ and $\langle s\restriction l,t\restriction l\rangle \in a_{s(m)}$ for all $l\leq m<{\mathrm {lh}}({s})$ with $s[l]\cup t[l]\subseteq \kappa _m$ . Then it is easy to see that U is a subtree of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ . Assume, towards a contradiction, that there exists a subtree T of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ with $p[T]={}^\omega \kappa \setminus p[U]$ . Pick a function $x\in {}^\omega \kappa $ with the property that $a_{x(m)}={\mathrm {H}}(\kappa _m)\cap T$ holds for all $m<\omega $ .

Now, assume that there is $y\in {}^\omega \kappa $ with $\langle x,y\rangle \in [T]$ . Then $\langle x,y\rangle \notin [U]$ and there exists $l<\omega $ with $\langle x\restriction l,y\restriction l\rangle \notin U$ . Then there exists $l\leq m<\omega $ with $x[l]\cup y[l]\subseteq \kappa _m$ and $\langle x\restriction l,y\restriction l\rangle \notin a_{x(m)}={\mathrm {H}}(\kappa _m)\cap T$ . But, this yields a contradiction because $\langle x\restriction l,y\restriction l\rangle $ is an element of T. This shows that there is $y\in {}^\omega \kappa $ with $\langle x,y\rangle \in [U]$ . Then $\langle x,y\rangle \notin [T]$ and there is $l<\omega $ with $\langle x\restriction l,y\restriction l\rangle \notin T$ . Pick $l\leq m<\omega $ with $x[l]\cup y[l]\subseteq \kappa _m$ . Then the fact that $\langle x,y\rangle \in [U]$ implies that $\langle x\restriction l,y\restriction l\rangle \in a_{x(m)}\subseteq T$ , a contradiction.

Proof. Assume, towards a contradiction, that the set $\{{\|y\|_{\vec {a}}}~\vert ~{y\in B}\}$ is unbounded in $\kappa ^+$ . Pick a subtree S of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ with $p[S]=B$ . By Proposition 4.3, there exists a subtree T of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ with the property that the set $A={}^\omega \kappa \setminus p[T]$ is not a $\mathbf {\Sigma }^1_1$ -subset of ${}^\omega \kappa $ . Given $x\in {}^\omega \kappa $ , set

$$ \begin{align*}T_x ~ = ~ \{{t\in{}^{{<}\omega}\kappa}~\vert~{\langle x\restriction{\mathrm{lh}}({t}),t\rangle\in T}\}.\end{align*} $$

Then $T_x$ is a subtree of ${}^{{<}\omega }\kappa $ for all $x\in {}^\omega \kappa $ and $A=\{{x\in {}^\omega \kappa }~\vert ~{[T_x]=\emptyset }\}$ . By standard arguments (see [Reference Kechris30, Section 2.E]), we now know that a function $x\in {}^\omega \kappa $ is contained in A if and only if there exists an ordinal $\gamma <\kappa ^+$ and a function ${r}:{T_x}\longrightarrow {\gamma }$ with $r(s)>r(t)$ for all $s,t\in T_x$ with $s\subsetneq t$ . Since our assumption implies that for every $\gamma <\kappa ^+$ , there is $y\in B$ with the property that there exists an order-preserving embedding of $\langle \gamma ,<\rangle $ into $\langle \kappa ,\lhd _{\hspace {0.9pt}\bigcup \{{a_{y(\xi )}}~\vert ~{\xi <{{\mathrm {cof}}(\kappa )}}\}}\rangle $ , we know that A consists of all $x\in {}^\omega \kappa $ with the property that there exists $y\in B$ and a function ${f}:{T_x}\longrightarrow {\kappa }$ such that for all $s,t\in T_x$ with $s\subsetneq t$ and all $m<\omega $ with $f(s),f(t)<\kappa _m$ , we have ${\prec }{f(t)},{f(s)}{\succ }\in a_{y(m)}$ .

Below, we aim to derive a contradiction from the above assumption by constructing a subtree U of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ with the property that a pair $\langle x,y\rangle $ in ${}^\omega \kappa \times {}^\omega \kappa $ is an element of $[U]$ if and only if y codes (in some fixed canonical way) functions ${c}:{T_x}\longrightarrow {\omega }$ , ${f}:{T_x}\longrightarrow {\kappa }$ and ${u,v}:{\omega }\longrightarrow {\kappa }$ such that $f(p)<\kappa _{c(p)}$ for all $p\in T_x$ , the pair $\langle u,v\rangle $ is an element of $[S]$ and ${\prec }{f(q)},{f(p)}{\succ }\in a_{u(\max (c(p),c(q)))}$ holds for all $p,q\in T_x$ with $p\subsetneq q$ . Note that if U is a tree with these properties, $\langle x,y\rangle $ is an element of U and c, f, u and v are the functions coded by y, then u is an element of B and the functions f and u witness that x is an element of A. But this means that, if we succeed in constructing such a tree U, then we derive a contradiction because the properties of U ensure that $p[U]=A$ and hence such a tree U witnesses that A is a $\mathbf {\Sigma }^1_1$ -subset of ${}^\omega \kappa $ .

We now show that our assumptions allow us to construct a tree U with the properties described above. For every $s\in {}^{{<}\omega }\kappa $ , we set

$$ \begin{align*}T_s ~ = ~ \{{t\in{}^{{<}\omega}\kappa}~\vert~{{\mathrm{lh}}({t})\leq{\mathrm{lh}}({s}), ~ \langle s\restriction{\mathrm{lh}}({t}),t\rangle\in T}\}.\end{align*} $$

Define U to be the subset of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ consisting of pairs $\langle s,t\rangle $ with ${\mathrm {lh}}({s})={\mathrm {lh}}({t})$ and the property that for all $m<{\mathrm {lh}}({s})$ , there exist $c_m,f_m,u_m,v_m\in {\mathrm {H}}(\kappa )$ such that $a_{t(m)}=\langle c_m,f_m,u_m,v_m\rangle $ and the following statements hold for all $l\leq m$ :

• ○ $\langle u_l,v_l\rangle ,\langle u_m,v_m\rangle \in S$ , ${\mathrm {lh}}({u_m})=m+1$ , $u_l=u_m\restriction (l+1)$ and $v_l=v_m\restriction (l+1)$ .

• ○ ${c_m}:{{\mathrm {H}}(\kappa _m)\cap T_{s\restriction m}}\longrightarrow {\omega }$ with $c_m\restriction {{\mathrm {dom}}(c_l)}=c_l$ .

• ○ ${f_m}:{\{{w\in {\mathrm {H}}(\kappa _m)\cap T_{s\restriction m}}~\vert ~{c_m(w)\leq m}\}}\longrightarrow {\kappa _m}$ with $f_m\restriction {{\mathrm {dom}}(f_l)}=f_l$ and ${\prec }{f_m(q)},{f_m(p)}{\succ }\in a_{u_m(m)}$ for all $p,q\in {{\mathrm {dom}}(f_m)}$ with $p\subsetneq q$ .

Then U is a subtree of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ .

Proof of the Claim

First, fix $\langle x,y\rangle \in [U]$ . Then there are ${c}:{T_x}\longrightarrow {\omega }$ , ${f}:{T_x}\longrightarrow {\kappa }$ and $\langle u,v\rangle \in [S]$ with the property that for all $m<\omega $ , the set $a_{y(m)}$ is equal to the quadruple

$$ \begin{align*}\langle c\restriction({\mathrm{H}}(\kappa_m)\cap T_{x\restriction m}),f\restriction\{{w\in{\mathrm{H}}(\kappa_m)\cap T_{x\restriction m}}~\vert~{c(w)\leq m}\},u\restriction(m+1),v\restriction(m+1)\rangle.\end{align*} $$

Then $u\in B$ and ${\prec }{f(q)},{f(p)}{\succ }\in a_{u(m)}$ holds for all $p,q\in T_x$ with $p\subsetneq q$ and all $m<\omega $ with $f(p),f(q)<\kappa _m$ . By earlier observations, this shows that $x\in A$ .

Now, pick $x\in A$ . Then we can find $\langle u,v\rangle \in S$ and a function ${f}:{T_x}\longrightarrow {\kappa }$ such that for all $p,q\in T_x$ with $p\subsetneq q$ and all $m<\omega $ with $f(p),f(q)<\kappa _m$ , we have ${\prec }{f(q)},{f(p)}{\succ }\in a_{u(m)}$ . Let ${c}:{T_x}\longrightarrow {\omega }$ denote the unique function with $c(p)=\min \{{m<\omega }~\vert ~{f(p)<\kappa _m}\}$ . If we then pick $y\in {}^\omega \kappa $ such that the set $a_{y(m)}$ is equal to the quadruple

$$ \begin{align*}\langle c\restriction({\mathrm{H}}(\kappa_m)\cap T_{x\restriction m}),f\restriction\{{w\in{\mathrm{H}}(\kappa_m)\cap T_{x\restriction m}}~\vert~{c(w)\leq m}\},u\restriction(m+1),v\restriction(m+1)\rangle\end{align*} $$

for all $m<\omega $ , then we can conclude that $\langle x,y\rangle \in [U]$ .

The above computations allow us to conclude that $A=p[U]$ , contradicting the fact that A is not a $\mathbf {\Sigma }^1_1$ -subset of ${}^\omega \kappa $ .

Proof of Theorem 4.1

Assume that there is no inner model with infinitely many measurable cardinals. Then [Reference Koepke31, Theorem 2.14] implies that $0^{\text {long}}$ (as defined in [Reference Koepke31, Definition 2.13]) does not exist. Let $U_{\text {can}}$ denote the canonical sequence of measures, and let ${\mathrm {K}}[U_{\text {can}}]$ denote the canonical core model (as defined in [Reference Koepke31, Definition 3.15]). Then our assumption implies that ${{\mathrm {dom}}(U_{\text {can}})}$ is finite and [Reference Koepke31, Theorem 3.23] shows that there is a generic extension ${\mathrm {K}}[U_{\text {can}},G]$ of ${\mathrm {K}}[U_{\text {can}}]$ by finitely many Prikry forcings with the property that for every ordinal $\tau \geq \omega _2$ and every $X\subseteq \tau $ such that $\vert {X}\vert $ is a regular cardinal smaller than $\vert {\tau }\vert $ , there exists $Z\in {\mathcal {P}}({\tau })^{{\mathrm {K}}[U_{\text {can}},G]}$ with $X\subseteq Z$ and $\vert {Z}\vert ^{{\mathrm {K}}[U_{\text {can}},G]}<\tau $ .

Now, let $\kappa $ be a singular strong limit cardinal of countable cofinality. Then $\kappa $ is singular in ${\mathrm {K}}[U_{\text {can}},G]$ and $\kappa ^+=(\kappa ^+)^{{\mathrm {K}}[U_{\text {can}},G]}$ . Moreover, since forcing with a finite iteration of Prikry forcings preserves all cardinals, we also know that $\kappa ^+=(\kappa ^+)^{{\mathrm {K}}[U_{\text {can}}]}$ . Set $U={U_{\text {can}}}\restriction {\kappa }$ and ${\mathrm {K}}={\mathrm {K}}[U]$ (see [Reference Koepke31, Definition 3.1]). Then [Reference Koepke31, Theorem 3.2] shows that ${\mathrm {K}}$ is an inner model of ${\mathrm {ZFC}}$ . Moreover, we can use [Reference Koepke31, Theorem 3.9.(iii)] to conclude that ${\mathcal {P}}({\kappa })^{{\mathrm {K}}[U_{\text {can}}]}\subseteq {\mathrm {K}}$ and therefore we know that $\kappa ^+=(\kappa ^+)^{\mathrm {K}}$ .

Next, let $<_{\mathrm {K}}$ denote the canonical well-ordering of ${\mathrm {K}}$ given by [Reference Koepke31, Theorem 3.4]). For every $\kappa \leq \gamma <\kappa ^+$ , let ${b_\gamma }:{\kappa }\longrightarrow {\gamma }$ denote the $<_{\mathrm {K}}$ -least bijection between $\kappa $ and $\gamma $ , and set $y_\gamma =\{{{\prec }{\alpha },{\beta }{\succ }}~\vert ~{\alpha ,\beta <\kappa , ~ b_\gamma (\alpha )<b_\gamma (\beta )}\}$ . Finally, we define $D=\{{y_\gamma }~\vert ~{\kappa \leq \gamma <\kappa ^+}\}$ . Then D is a subset of $\mathcal {W}\mathcal {O}_\kappa $ of cardinality $\kappa ^+$ .

In the following, assume, towards a contradiction, that there is a continuous injection ${\iota }:{{}^\omega \kappa }\longrightarrow {{\mathcal {P}}({\kappa })}$ with ${{\mathrm {ran}}(\iota )}\subseteq D$ . Fix a strictly increasing sequence $\vec {\kappa }=\langle {\kappa _m}~\vert ~{m<\omega }\rangle $ of cardinals that is cofinal in $\kappa $ and an enumeration $\vec {a}=\langle {a_\alpha }~\vert ~{\alpha <\kappa }\rangle $ of ${\mathrm {H}}(\kappa )$ . Define T to be the set of all pairs $\langle s,t\rangle $ in ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ such that ${\mathrm {lh}}({s})={\mathrm {lh}}({t})$ , and the following statements hold for all $l\leq m<{\mathrm {lh}}({s})$ :

• ○ $a_{s(m)}\subseteq \kappa _m$ and $a_{s(l)}=a_{s(m)}\cap \kappa _l$ .

• ○ $a_{t(l)},a_{t(m)}\in {}^{{<}\omega }\kappa $ with $l\leq {\mathrm {lh}}({a_{t(l)}})\leq {\mathrm {lh}}({a_{t(m)}})$ , $a_{t(l)}=a_{t(m)}\restriction {\mathrm {lh}}({a_{t(l)}})$ and $\iota (u)\cap \kappa _m=a_{s(m)}$ for all $u\in {}^\omega \kappa $ with $a_{t(m)}\subseteq u$ .

This definition directly ensures that T is a subtree of ${}^{{<}\omega }\kappa \times {}^{{<}\omega }\kappa $ . Pick $\langle x,y\rangle \in [T]$ . Set $u=\bigcup \{{a_{y(m)}}~\vert ~{m<\omega }\}\in {}^\omega \kappa $ and $v=\bigcup \{{a_{x(m)}}~\vert ~{m<\omega }\}\subseteq \kappa $ . By the definition of T, we then have $\iota (u)=v\in D\subseteq \mathcal {W}\mathcal {O}_\kappa $ and this shows that x is an element of $\mathsf {WO}(\vec {\kappa },\vec {a})$ . This shows that $p[T]\subseteq \mathsf {WO}(\vec {\kappa },\vec {a})$ , and therefore Lemma 4.5 yields an ordinal $\gamma <\kappa ^+$ with $\|x\|_{\vec {a}} < \gamma $ for all $x\in p[T]$ .

Since, for every ordinal $\kappa \leq \delta <\kappa ^+$ , there is a unique element y of D with ${{\mathrm {otp}}\left (\kappa ,\lhd _y\right )}=\delta $ , we know that the map

$$ \begin{align*}{i}:{{}^\omega\kappa}\longrightarrow{\kappa^+};~{u}\longmapsto{{{\mathrm{otp}}\ \left(\kappa,\lhd_{\iota(u)}\right)}}\end{align*} $$

is an injection and we can find $u\in {}^\omega \kappa $ with ${{\mathrm {otp}}\left (\kappa ,\lhd _{\iota (u)}\right )}>\gamma $ . Pick $x\in {}^\omega \kappa $ with $a_{x(m)}=\iota (u)\cap \kappa _m$ for all $m<\omega $ . In addition, pick $y\in {}^\omega \kappa $ with the property that for all $l\leq m<\omega $ , we have $a_{y(l)},a_{y(m)}\in {}^{{<}\omega }\kappa $ , $l\leq {\mathrm {lh}}({a_{y(l)}})\leq {\mathrm {lh}}({a_{y(m)}})$ , $a_{y(l)}\subseteq a_{y(m)}=u\restriction {\mathrm {lh}}({a_{y(m)}})$ and $\iota (w)\cap \kappa _m=\iota (u)\cap \kappa _m$ for all $w\in {}^\omega \kappa $ with $u\restriction {\mathrm {lh}}({a_{y(m)}})\subseteq w$ . Note that this is possible as $\iota $ is a continuous injection. Then $\langle x,y\rangle \in [T]$ and $x\in p[T]\subseteq \mathsf {WO}(\vec {\kappa },\vec {a})$ with $\|x\|_{\vec {a}}={{\mathrm {otp}}\left (\kappa ,\lhd _{\iota (u)}\right )}>\gamma $ , a contradiction.

Proof. The statement of the lemma will be a direct consequence of the following two claims:

Proof of the claim

First, the weak homogeneity of ${\mathbb {P}}$ in ${\mathrm {V}}$ ensures that

$$ \begin{align*}U_{can}^{{\mathrm{V}}[G]} ~ \subseteq ~ {\mathrm{K}}[U_{can}]^{{\mathrm{V}}[G]} ~ \subseteq ~ {\mathrm{HOD}}^{{\mathrm{V}}[G]} ~ \subseteq ~ {\mathrm{V}}.\end{align*} $$

In particular, we know that the set $U_{can}^{{\mathrm {V}}[G]}(\kappa )\cap {\mathcal {P}}({\kappa })^{\mathrm {V}}$ is an element of ${\mathrm {V}}$ for every $\kappa \in {{\mathrm {dom}}(U_{can}^{{\mathrm {V}}[G]})}$ . In this situation, we can now use the first claim to inductively show that the definition of the canonical measure sequence ensures that $U_{\text {can}}^{\mathrm {V}}\restriction \xi =U_{can}^{{\mathrm {V}}[G]}\restriction \xi $ holds for all $\xi \in {\mathrm {On}}$ .

Now, let ${\mathbb {P}}$ be a partial order, and let G be ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ . Pick a sufficiently large cardinal $\delta $ such that ${\mathbb {P}}\times {\mathrm {Col}}({\omega },{\delta })$ densely embeds into ${\mathrm {Col}}({\omega },{\delta })$ , and let H be ${\mathrm {Col}}({\omega },{\delta })$ -generic over ${\mathrm {V}}[G]$ . Since ${\mathrm {Col}}({\omega },{\delta })$ is weakly homogeneous in both ${\mathrm {V}}$ and ${\mathrm {V}}[G]$ , we can now use the above claim twice to conclude that ${\mathrm {K}}[U_{can}]^{\mathrm {V}}={\mathrm {K}}[U_{can}]^{{\mathrm {V}}[G,H]}={\mathrm {K}}[U_{can}]^{{\mathrm {V}}[G]}$ .

Proof of Proposition 3.2

Let M denote the class term with the property that ${\mathrm {ZFC}}$ proves the following statements:

• ○ If either $0^{\text {long}}$ exists or $0^{\text {long}}$ does not exist and the model ${\mathrm {K}}[U_{can}]$ contains infinitely many measurable cardinals, then M is equal to the constructible universe ${\mathrm {L}}$ .

• ○ Otherwise, M is equal to ${\mathrm {K}}[U_{can}]$ .

Then the standard results about ${\mathrm {L}}$ together with [Reference Koepke31, Theorem 3.2] show that ${\mathrm {ZFC}}$ proves that M is a transitive model of ${\mathrm {ZFC}}+{{\mathrm {V}}=M}$ that contains all ordinals. Moreover, Lemma 4.6 together with the fact that $0^{\text {long}}$ cannot be added by forcing show that M is forcing invariant.

The above claim now allows us to find a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ with the property that for every uncountable cardinal $\kappa $ , we have ${\mathrm {L}}_{\kappa ^+}=\{{x}~\vert ~{\varphi (\kappa ,\kappa ,x)}\}$ and, if $0^{\text {long}}$ does not exist and ${\mathrm {K}}[U_{can}]$ contains only finitely many measurable cardinals, then there exists $z\in {\mathrm {H}}(\kappa )$ with ${\mathrm {H}}(\kappa ^+)^{{\mathrm {K}}[U_{can}]}=\{{x}~\vert ~{\varphi (\kappa ,x,z)}\}$ . Finally, if the existence of n measurable cardinals is consistent with the axioms of ${\mathrm {ZFC}}$ for some natural number n, then the existence of exactly n measurable cardinals in ${\mathrm {K}}[U_{can}]$ is consistent with ${\mathrm {ZFC}}$ .

Proof of Theorem 1.2

Let $\kappa $ be a singular strong limit cardinal with the property that for every subset D of ${\mathcal {P}}({\kappa })$ of cardinality greater than $\kappa $ that is definable by a $\Sigma _1$ -formula with parameters in ${\mathrm {H}}(\kappa )\cup \{\kappa \}$ , there exists a perfect embedding ${\iota }:{{}^{{{\mathrm {cof}}(\kappa )}}\kappa }\longrightarrow {{\mathcal {P}}({\kappa })}$ with ${{\mathrm {ran}}(\iota )}\subseteq D$ . Assume, towards a contradiction, that there is no inner model with a sequence of measurable cardinals of length ${{\mathrm {cof}}(\kappa )}$ . Then Theorem 4.1 implies that the cofinality of $\kappa $ is uncountable. Moreover, we know that there is no inner model with a measurable cardinal of Mitchell order $1$ and therefore we can construct the canonical core model ${\mathrm {K}}$ as in [Reference Zeman53] (which is Steel’s core model [Reference Steel47] in this easier setting). Note that our hypothesis implies that, in ${\mathrm {K}}$ , the sequence of measurable cardinals below $\kappa $ is bounded below $\kappa $ . In addition, as $\kappa $ is singular in ${\mathrm {V}}$ , weak covering (see [Reference Zeman53, Theorem 7.5.1]) holds for ${\mathrm {K}}$ at $\kappa $ , that is, we have $ (\kappa ^+)^{\mathrm {K}} = \kappa ^+$ . Finally, we know that $\kappa $ is singular in ${\mathrm {K}}$ , because otherwise the fact that $\kappa $ is a singular cardinal of uncountable cofinality would allow us to apply the second part of [Reference Cox10, Theorem 1] to find an inner model in which $\kappa $ has Mitchell order greater than $0$ .

We will now construct a tree of height ${{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ that is an element of ${\mathrm {K}}$ and then argue that this tree does not have a perfect subtree in ${\mathrm {V}}$ . These arguments use ideas from [Reference Hayut and Müller21] that ultimately go back to Solovay’s argument for the consistency strength of the Kurepa hypothesis (see [Reference Jech24, Section 4]). Our tree consists of hulls of initial segments of ${\mathrm {K}}$ of size $\kappa $ , and we will argue that we can obtain such initial segments in a $\Sigma _1$ -definable way with parameters in ${\mathrm {H}}(\kappa ) \cup \{ \kappa \}$ .

In the following, we say a premouse N (in the sense of [Reference Zeman53, Section 4.1]) is good if the following statements hold:

• ○ N is iterable (in the sense of [Reference Zeman53, Section 4.2]).

• ○ $\kappa +1 \subseteq N$ and $\vert {N}\vert = \kappa $ .

• ○ ${{\mathrm {cof}}(\kappa )}^N={{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ .

• ○ $\kappa $ is the largest cardinal in N.

• ○ If $\gamma < \kappa $ is the supremum of the measurable cardinals below $\kappa $ in ${\mathrm {K}}$ , then

  $$\begin{align*}N \vert \gamma^{++} ~ = ~ {\mathrm{K}} \vert \gamma^{++}. \end{align*}$$
  In particular, K and N have the same measurable cardinals and the same total measures below $\kappa $ .

Following [Reference Hayut and Müller21], we say a pair $\langle M, \bar {x}\rangle $ is an active node at $\rho $ for some $\rho < {{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ if there is a good premouse N and some $x \in \left ({}^{{{\mathrm {cof}}(\kappa )}} \kappa \right )^N$ with ${{\mathrm {ran}}(x)} \subseteq \mathrm {Reg}^N$ , the regular cardinals in N, such that the following statements hold:

• ○ x is strictly increasing and cofinal in $\kappa $ .

• ○ M is equal to the transitive collapse of ${{\mathrm {Hull}}}^{N}(x(\rho ) \cup \{x\})$ and $\bar {x} \in M$ is the image of x under the transitive collapse.

• ○ If ${\pi }:{M}\longrightarrow {N}$ is the corresponding uncollapsing map, then ${{\mathrm {crit}}\left ({\pi }\right )} = \bar {x}(\rho )$ .

In addition, we say a pair $\langle M, \bar {x}\rangle $ is an active node if there is some ordinal $\rho <{{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ such that $\langle M, \bar {x}\rangle $ is an active node at $\rho $ .

We now let T denote the unique partial order defined by the following clauses:

1. (i) The elements of T are triples of the form $\langle M, \bar {x}, s\rangle $ satisfying the following properties:

   1. (a) The pair $\langle M, \bar {x}\rangle $ is either an active node or equal to the pair $\langle \emptyset , \emptyset \rangle $ .

   2. (b) s is an element of $({}^{{<}{{\mathrm {cof}}(\kappa )}}\kappa )^{\mathrm {K}}$ with the property that the set

      $$ \begin{align*}\bigcup_{0 < \alpha < \kappa} s^{{-}1}(\{\alpha\})\end{align*} $$
      is finite.

   3. (c) If $\langle M, \bar {x}\rangle $ is an active node at $\rho $ , then ${{\mathrm {dom}}(s)} \geq \rho $ .

2. (ii) The order of T is defined by

   $$ \begin{align*}\langle M_0, x_0, s_0\rangle ~ \leq_{T} ~ \langle M_1, x_1, s_1\rangle\end{align*} $$
   if and only if the following statements hold:

   1. (a) $M_0$ is the transitive collapse of ${{\mathrm {Hull}}}^{M_1}(x_1(\rho ) \cup \{x_1\})$ for some ordinal $\rho $ and $x_0$ is the image of $x_1$ under the transitive collapse, or $M_0 = x_0 = \emptyset $ . In the following, write $\rho $ for the minimal such ordinal and $\rho = -1$ if $M_0 = x_0 = \emptyset $ .

   2. (b) $s_0$ is an initial segment of $s_1$ .

   3. (c) There is no ordinal $\rho '$ between $\rho $ and ${{\mathrm {dom}}(s_0)}$ with the property that $\langle {{\mathrm {Hull}}}^{M_1}(x_1(\rho ') \cup \{x_1\}), x_1\rangle $ transitively collapses to an active node which, in case $\rho \neq -1$ , is not $\langle M_0, x_0\rangle $ .

It is now easy to see that T is a tree of height ${{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ that is contained in ${\mathrm {K}}$ and has the property that each node is splitting into $\kappa $ -many successors. Moreover, each $x \in \left ({}^{{{\mathrm {cof}}(\kappa )}} \kappa \right )^{\mathrm {K}}$ that is strictly increasing and cofinal in $\kappa $ with range contained in $\mathrm {Reg}^N$ naturally gives rise to a cofinal branch $b_x$ through T and two different such elements $x,y \in \left ({}^{{{\mathrm {cof}}(\kappa )}} \kappa \right )^{\mathrm {K}}$ give rise to different branches $b_x$ and $b_y$ . Hence, the fact that the ${\mathrm {GCH}}$ holds in ${\mathrm {K}}$ implies that the set of cofinal branches through T has cardinality at least

$$ \begin{align*}\left(\kappa^{{{\mathrm{cof}}(\kappa)}}\right)^{\mathrm{K}} ~ = ~ (\kappa^+)^{\mathrm{K}} ~ = ~ \kappa^+.\end{align*} $$

Proof of the claim

As the proof of the well-foundedness of $\mathcal {R}_b$ is easier, we focus on the argument that $\mathcal {R}_b \lhd K$ . By our first claim, it suffices to show that $\mathcal {R}_b$ is a good premouse. We obtain iterability for $\mathcal {R}_b$ by reflecting countable elementary substructures of $\mathcal {R}_b$ into models in the tree T, as in [Reference Hayut and Müller21], using the fact that

$$ \begin{align*}{{\mathrm{cof}}({{\mathrm{cof}}(\kappa)}^{\mathrm{K}})} ~ = ~ {{\mathrm{cof}}(\kappa)} ~> ~ \omega\end{align*} $$

(see [Reference Jech25, Lemma 3.7(ii)]). In the following, write $\langle {M_\rho }~\vert ~{\rho < {{\mathrm {cof}}(\kappa )}^{\mathrm {K}}}\rangle $ for the sequence of models appearing in active nodes at $\rho $ along the branch b. Then the definition of T ensures that for every $\xi < \kappa $ , there is some $\rho < {{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ such that if $M_\rho $ is the transitive collapse of ${{\mathrm {Hull}}}^{N_\rho }(x_\rho (\rho ) \cup \{x_\rho \})$ for some good premouse $N_\rho $ and some $x_\rho \in \left ({}^{{{\mathrm {cof}}(\kappa )}} \kappa \right )^{N_\rho }$ , then $x_\rho (\rho )> \xi $ . Therefore, we know that $\kappa \subseteq \mathcal {R}_b$ and elementarity implies that $\kappa +1 \subseteq \mathcal {R}_b$ . Our setup also directly ensures that $\vert {\mathcal {R}_b}\vert = \kappa $ . As $N_\rho \lhd K$ and the critical point of the inverse of the collapse embedding $\pi _\rho \colon M_\rho \rightarrow N_\rho $ is at least $x_\rho (\rho )$ , this also shows that $N_\rho | \gamma ^{++} = K | \gamma ^{++}$ , where $\gamma < \kappa $ is the supremum of the measurable cardinals below $\kappa $ in ${\mathrm {K}}$ and $N_\rho $ . Moreover, we know that $\kappa $ has cofinality ${{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ in $\mathcal {R}_b$ , as witnessed by $x_b$ . Finally, $\kappa $ is the largest cardinal in $\mathcal {R}_b$ by elementarity as $\kappa = \sup ({{\mathrm {ran}}(x_\rho )})$ is the largest cardinal in $N_\rho $ for all $\rho < {{\mathrm {cof}}(\kappa )}$ .

Proof of the claim

It clearly suffices to construct an injection from the set of all active nodes to ${\mathrm {H}}(\kappa )\cap {\mathcal {P}}({\kappa })$ . Let $\langle M,\bar {x}\rangle $ be an active node at some $\rho <{{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ . Since M is the transitive collapse of an elementary submodel of some good premouse $N=\langle \mathrm {J}^{\vec E}_{\alpha },\vec E\rangle $ , we know that M is of the form $\langle \mathrm {J}^A_\epsilon ,A\rangle $ and there is a well-ordering $\lhd $ of M that is definable in M. Let ${\tau }:{\langle M,\lhd \rangle }\longrightarrow {\langle \lambda ,<\rangle }$ denote the corresponding transitive collapse and associate $\langle M,\bar {x}\rangle $ with the element

$$ \begin{align*} \begin{aligned} & \{{{\prec}{0},{{\prec}{\alpha},{\beta}{\succ}}{\succ}}~\vert~{\alpha,\beta<\lambda, ~ \tau^{{-}1}(\alpha)\in\tau^{{-}1}(\beta)}\} \\ & \cup ~ \{{{\prec}{1},{\alpha}{\succ}}~\vert~{\alpha<\lambda, ~ \tau^{{-}1}(\alpha)\in A}\} \\ & ~ \cup ~ \{{{\prec}{2},{\alpha}{\succ}}~\vert~{\alpha<\lambda, ~ \tau^{{-}1}(\alpha)\in\bar{x}}\} \end{aligned} \end{align*} $$

of ${\mathrm {H}}(\kappa )\cap {\mathcal {P}}({\kappa })$ . It is now easy to see that the resulting injection is definable in the desired way.

Proof of the claim

Let ${\mathbb {P}}$ be a countably closed forcing notion, and let G be ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ . Suppose, towards a contradiction, that there is a cofinal branch b through T in ${\mathrm {V}}[G]$ that is not contained in ${\mathrm {V}}$ . By considering the direct limit of the active nodes along b and using the fact that ${{\mathrm {cof}}(\kappa )}^{\mathrm {V}}$ has uncountable cofinality in ${\mathrm {V}}[G]$ , we obtain a pair $\langle \mathcal {R}_b, x_b\rangle $ such that b (modulo some choice of an almost zero sequence s) can be recovered from $\mathcal {R}_b$ and $x_b$ via the transitive collapses of models of the form

$$\begin{align*}{{\mathrm{Hull}}}^{\mathcal{R}_b}(x_b(\rho) \cup \{x_b\}), \end{align*}$$

for $\rho < {{\mathrm {cof}}(\kappa )}^{\mathrm {K}}$ . As ${{\mathrm {cof}}(\kappa )}^{\mathrm {V}}$ has uncountable cofinality in ${\mathrm {V}}[G]$ , the argument in one of our previous claims then shows that $\mathcal {R}_b \lhd {\mathrm {K}}^{V[G]}$ holds in $V[G]$ . By [Reference Zeman53, Theorem 7.4.11], the core model ${\mathrm {K}}$ is forcing absolute, that is, we have ${\mathrm {K}}^{\mathrm {V}} = {\mathrm {K}}^{{\mathrm {V}}[G]}$ . Therefore, we know that $\mathcal {R}_b$ and hence b is already an element of ${\mathrm {V}}$ , a contradiction.

Fix a strictly increasing, cofinal function ${c}:{{{\mathrm {cof}}(\kappa )}}\longrightarrow {{{\mathrm {cof}}(\kappa )}^{\mathrm {K}}}$ , and let $T_*$ denote the unique partial order defined by the following clauses:

1. (i) The elements of $T_*$ are functions t such that ${{\mathrm {dom}}(t)}\in {{\mathrm {cof}}(\kappa )}$ and the following statements hold:

   1. (a) If $\alpha \in {{\mathrm {dom}}(t)}$ , then $t(\alpha )$ is a branch through T of order-type $c(\alpha )$ .

   2. (b) If $\alpha <\beta \in {{\mathrm {dom}}(t)}$ , then $t(\alpha )$ is an initial segment of $t(\beta )$ .

2. (ii) The ordering of $T_*$ is given by inclusion.

It then follows that $T_*$ is a tree of height ${{\mathrm {cof}}(\kappa )}$ with the property that every node has $\kappa $ -many successors. Since the tree T has at least $\kappa ^+$ -many branches, it follows that $T_*$ also has at least $\kappa ^+$ -many branches. Moreover, by using the injection i, it is possible to construct an injection ${i_*}:{T_*}\longrightarrow {{\mathrm {H}}(\kappa )\cap {\mathcal {P}}({\kappa })}$ with $\emptyset \notin {{\mathrm {ran}}(i_*)}$ that is definable by a $\Sigma _1$ -formula with parameters in ${\mathrm {H}}(\kappa )\cup \{\kappa \}$ . Finally, the above computations also imply that forcing with a countably closed partial order does not add a new cofinal branch to $T_*$ .

Define D to be the set of all subsets of $\kappa $ of the form

$$ \begin{align*}y_b ~ = ~ \{{{\prec}{\alpha},{{\prec}{\beta},{\sup(i_*(b\restriction\alpha))}{\succ}}{\succ}}~\vert~{\alpha<{{\mathrm{cof}}(\kappa)}, ~ \beta\in i_*(b\restriction\alpha)}\}\end{align*} $$

for some function b with domain ${{\mathrm {cof}}(\kappa )}$ and the property that $b\restriction \alpha \in T_*$ for all $\alpha <{{\mathrm {cof}}(\kappa )}$ . Since the fact that the tree $T_*$ has at least $\kappa ^+$ -many cofinal branches implies that D has cardinality greater than $\kappa $ and the above computations show that D is definable by a $\Sigma _1$ -formula with parameters in ${\mathrm {H}}(\kappa )\cup \{\kappa \}$ , our assumption yields a perfect embedding ${\iota }:{{}^{{{\mathrm {cof}}(\kappa )}}\kappa }\longrightarrow {{\mathcal {P}}({\kappa })}$ with ${{\mathrm {ran}}(\iota )}\subseteq D$ . Using the fact that ${{\mathrm {cof}}(\kappa )}$ is uncountable, a routine construction now allows us to find

• ○ a system $\langle {u_s}~\vert ~{s\in {}^{{<}{{\mathrm {cof}}(\kappa )}}2}\rangle $ of elements of ${}^{{<}{{\mathrm {cof}}(\kappa )}}2$ ,

• ○ a strictly increasing sequence $\langle {\kappa _\alpha }~\vert ~{\alpha <{{\mathrm {cof}}(\kappa )}}\rangle $ that is cofinal in $\kappa $ , and

• ○ a system $\langle {a_s}~\vert ~{s\in {}^{{<}{{\mathrm {cof}}(\kappa )}}2}\rangle $ of bounded subsets of $\kappa $

such that the following statements hold for all $s,t\in {}^{{<}{{\mathrm {cof}}(\kappa )}}2$ :

1. (i) If ${\mathrm {lh}}({s}){}={\mathrm {lh}}({t}){}$ , then ${\mathrm {lh}}({u_s}){}={\mathrm {lh}}({u_t}){}$ .

2. (ii) $a_s$ is a subset of $\kappa _{{\mathrm {lh}}({s}){}}$ .

3. (iii) If $s\subseteq t$ , then $a_s=a_t\cap \kappa _{{\mathrm {lh}}({s}){}}$ and $u_s\subseteq u_t$ .

4. (iv) $a_{s^\frown \langle 0\rangle }\neq a_{s^\frown \langle 1\rangle }$ and $u_{s^\frown \langle 0\rangle }\neq u_{s^\frown \langle 1\rangle }$ .

5. (v) $\iota [\{{x\in {}^{{\mathrm {cof}}(\kappa )}2}~\vert ~{x\restriction {\mathrm {lh}}({s}){}= s}\}]=\{{y\in {{\mathrm {ran}}(\iota )}}~\vert ~{y\cap \kappa _{{\mathrm {lh}}({s}){}}=a_s}\}$ .

6. (vi) If $\alpha <{\mathrm {lh}}({s}){}$ , then there are $\gamma \leq \delta <\kappa _{{\mathrm {lh}}({s}){}}$ with ${\prec }{\alpha },{{\prec }{\gamma },{\delta }{\succ }}{\succ }\in a_s$ .