Proof of Proposition 1.2 (i) If ${{\mathrm {cof}}(\lambda )}\leq \delta $ , then the desired statement follows directly from the closure of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ under $\delta $ -sequences. Hence, we may assume that $\kappa ={{\mathrm {cof}}(\lambda )}$ is a weakly compact cardinal greater than $\delta $ . Pick a cofinal sequence $\langle {\gamma _\alpha }~\vert ~{\alpha <\kappa }\rangle $ in $\lambda $ and fix an unbounded subset A of $\lambda $ such that $A\cap \gamma \in {\mathrm {Ult}}({{\mathrm {V}}},{U})$ for all $\gamma <\lambda $ . Given $\alpha <\kappa $ , fix functions $f_\alpha $ and $g_\alpha $ with domain $\delta $ such that $[f_\alpha ]_U=\gamma _\alpha $ and $[g_\alpha ]_U = A\cap \gamma _\alpha $ (recall that we are identifying ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ with its transitive collapse). Let ${c}:{[\kappa ]^2}\longrightarrow {U}$ denote the unique function with the property that

$$ \begin{align*}c(\{\alpha,\beta\}) ~ = ~ \{{\xi<\delta}~\vert~{f_\alpha(\xi)<f_\beta(\xi), ~ g_\alpha(\xi)=f_\alpha(\xi)\cap g_\beta(\xi)}\}\end{align*} $$

holds for all $\alpha <\beta <\kappa $ . In this situation, since $\kappa>\delta $ is weakly compact, we know that $\vert {U}\vert =2^\delta <\kappa $ and hence the weak compactness of $\kappa $ yields an unbounded subset H of $\kappa $ and an element X of U with the property that $c[H]^2=\{X\}$ . Pick a function g with domain $\delta $ and the property that $g(\xi )=\bigcup \{{g_\alpha (\xi )}~\vert ~{\alpha \in H}\}$ holds for all $\xi \in X$ . This construction ensures that $[g]_U\cap \gamma _\alpha =[g_\alpha ]_U$ holds for every $\alpha \in H$ and we can conclude that $[g]_U=A$ . In particular, the set A is not fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

(ii) Fix a ${<}(2^\delta )^+$ -closed ultrafilter F on $\lambda $ that contains all cobounded subsets of $\lambda $ and assume, towards a contradiction, that A is an unbounded subset of $\lambda $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Given $\eta <\lambda $ , fix functions $f_\eta $ and $g_\eta $ with domain $\delta $ that satisfy $[f_\eta ]_U=\eta $ and $[g_\eta ]_U=A\cap \eta $ . Moreover, given $\eta <\lambda $ and $X\in U$ , we set

$$ \begin{align*}A_{\eta,X} ~ = ~ \{{\zeta\in(\eta,\lambda)}~\vert~{X=\{{\xi<\delta}~\vert~{f_\eta(\xi)<f_\zeta(\xi), ~ g_\eta(\xi)=g_\zeta(\xi)\cap f_\eta(\xi)}\}}\}.\end{align*} $$

Then

$$ \begin{align*}\bigcup\{{A_{\eta,X}}~\vert~{X\in U}\} ~ = ~ (\eta,\lambda)\end{align*} $$

holds for every $\eta <\lambda $ . By our assumptions on F, there exists a sequence $\langle {X_\eta }~\vert ~{\eta <\lambda }\rangle $ of elements of U with the property that $A_{\eta ,X_\eta }\in F$ holds for all $\eta <\lambda $ . Given $X\in U$ , we now define

$$ \begin{align*}E_X ~ = ~ \{{\eta<\lambda}~\vert~{X_\eta=X}\}.\end{align*} $$

Since we now have

$$ \begin{align*}\bigcup\{{E_X}~\vert~{X\in U}\} ~ = ~ \lambda,\end{align*} $$

our assumptions on F yield an element $X_*$ of U with $E_{X_*}\in F$ .

Proof of the Claim Pick $\rho \in A_{\eta ,X_*}\cap A_{\zeta ,X_*}\in F$ . Then $\alpha <f_\rho (\xi )$ and

$$ \begin{align*} g_\eta(\xi)\cap\alpha ~ = ~ g_\rho(\xi)\cap\alpha ~ = ~ g_\zeta(\xi)\cap\alpha. \end{align*} $$

⊣

Pick a function g with domain $\delta $ and

$$ \begin{align*}g(\xi) ~ = ~ \bigcup\{{g_\eta(\xi)}~\vert~{\eta\in E_{X_*}}\}\end{align*} $$

for all $\xi \in X_*$ . By the above claim, we now know that $g(\xi )\cap f_\eta (\xi )=g_\eta (\xi )$ holds for all $\eta \in E_{X_*}$ and all $\xi \in X_*\in U$ . But this directly implies that

$$ \begin{align*}[g]_U\cap\eta ~ = ~ [g_\eta]_U ~ = ~ A\cap\eta\end{align*} $$

for all $\eta \in E_{X_*}$ . Hence, we can conclude that $[g]_U = A\in {\mathrm {Ult}}({{\mathrm {V}}},{U})$ , a contradiction.

Now, assume that there is a strongly compact cardinal $\kappa $ with $\delta <\kappa \leq {{\mathrm {cof}}(\lambda )}$ . Then $2^\delta <\kappa $ and the cobounded filter on $\lambda $ is ${<}\kappa $ -closed. Since the filter extension property of strongly compact cardinals (see [Reference Kanamori14, Proposition 4.1]) ensures the existence of a ${<}\kappa $ -closed ultrafilter F on $\lambda $ that contains all cobounded subsets of $\lambda $ , the above computations directly yield the desired conclusion.⊣

Proof of Proposition 1.3 (i) If $\kappa>\delta $ is the minimal cardinal with ${\mathcal {P}}({\kappa })\nsubseteq {\mathrm {Ult}}({{\mathrm {V}}},{U})$ , then every element of ${\mathcal {P}}({\kappa })\setminus {\mathrm {Ult}}({{\mathrm {V}}},{U})$ is unbounded in $\kappa $ and fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

(ii) Fix a limit ordinal $\lambda $ with ${{\mathrm {cof}}(\lambda )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}=j_U(\delta ^+)$ and pick a strictly increasing, cofinal function ${c}:{j_U(\delta ^+)}\longrightarrow {\lambda }$ in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Since $j_U[\delta ^+]$ is a cofinal subset of $j_U(\delta ^+)$ of order-type $\delta ^+$ , we know that $(c\circ j_U)[\delta ^+]$ is a cofinal subset of $\lambda $ of order-type $\delta ^+$ . In particular, the closure of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ under $\delta $ -sequences implies that every proper initial segment of $(c\circ j_U)[\delta ^+]$ is an element of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Finally, since [Reference Kanamori14, Proposition 22.4] shows that $j_U[\delta ^+]\notin {\mathrm {Ult}}({{\mathrm {V}}},{U})$ , we can conclude that the set $(c\circ j_U)[\delta ^+]$ is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

(iii) First, assume that ${{\mathrm {cof}}(\lambda )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}=\delta ^+$ . Since $2^\delta = \delta ^+$ and $U \notin {\mathrm {Ult}}({{\mathrm {V}}},{U})$ , we have ${\mathcal {P}}({\delta ^+})\nsubseteq {\mathrm {Ult}}({{\mathrm {V}}},{U})$ , and we can use (i) to find an unbounded subset A of $\delta ^+$ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Let $\langle {\gamma _\alpha }~\vert ~{\alpha <\delta ^+}\rangle $ be the monotone enumeration of an unbounded subset of $\lambda $ of order-type $\delta ^+$ in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Set $B=\{{\gamma _\alpha }~\vert ~{\alpha \in A}\}$ . Then B is unbounded in $\lambda $ and it is easy to see that B is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

Now, assume that ${{\mathrm {cof}}(\lambda )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}>\delta ^+$ and fix an unbounded subset A of $\lambda $ of order-type $\delta ^+$ . Then the closure of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ under $\delta $ -sequences implies that A is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof of Theorem 1.8 By [Reference Lambie-Hanson and Lücke17, Theorem 3.4], our assumptions allow us to fix a $\square ^{\mathrm {ind}}(\kappa , \delta )$ -sequence

$$ \begin{align*}\langle{C_{\gamma,\xi}}~\vert~{\gamma < \kappa, ~ i(\gamma) \leq \xi < \delta}\rangle.\end{align*} $$

Given $\gamma \in {\mathrm {Lim}}\cap \kappa $ , let ${f_\gamma }:{\delta }\longrightarrow {{\mathcal {P}}({\gamma })}$ denote the unique function with $f_{\gamma }(\xi )=\emptyset $ for all $\xi <i(\gamma )$ and $f_\gamma (\xi )=C_{\gamma ,\xi }$ for all $i(\gamma )\leq \xi <\delta $ . In this situation, Łos’ Theorem directly implies that for all $\beta ,\gamma \in {\mathrm {Lim}}\cap \kappa $ with $\beta \leq \gamma $ , the set $[f_\gamma ]_U$ is closed unbounded in $j_U(\gamma )$ and $[f_\beta ]_U=[f_\gamma ]_U\cap j_U(\beta )$ holds. Define

$$ \begin{align*}A ~ = ~ \bigcup\{{[f_\gamma]_U}~\vert~{\gamma \in {\text{Lim}} \cap \kappa}\}.\end{align*} $$

By our assumptions on $\kappa $ , we know that $j_U(\kappa )=\sup (j_U[\kappa ])$ and therefore A is a closed unbounded subset of $j_U(\kappa )$ with $A\cap j_U(\gamma )=[f_\gamma ]_U$ for all $\gamma \in {\mathrm {Lim}} \cap \kappa $ .

Assume, towards a contradiction, that A is an element of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Then there is ${f}:{\delta }\longrightarrow {{\mathcal {P}}({\kappa })}$ with $[f]_U=A$ and the property that $f(\xi )$ is a closed unbounded subset of $\kappa $ for all $\xi <\delta $ . Since we have

$$ \begin{align*}\{{\xi<\delta}~\vert~{f_\gamma(\xi)=f(\xi)\cap\gamma\neq\emptyset}\} ~ \in ~ U\end{align*} $$

for all $\gamma \in {\mathrm {Lim}} \cap \kappa $ , we can find $\xi <\delta $ with the property that $\xi \geq i(\gamma )$ and $f(\xi )\cap \gamma =C_{\gamma ,\xi }$ holds for unboundedly many $\gamma $ below $\kappa $ . Pick $\beta \in {\mathrm {Lim}}(f(\xi ))$ and $\beta <\gamma <\kappa $ with $\xi \geq i(\gamma )$ and $f(\xi )\cap \gamma =C_{\gamma ,\xi }$ . Then $\beta \in {\mathrm {Lim}}(C_{\gamma ,\xi })$ and this implies that $\xi \geq i(\beta )$ and $C_{\beta ,\xi }=C_{\gamma ,\xi }\cap \beta =f(\xi )\cap \beta $ . These computations show that there is a closed unbounded subset C of $\kappa $ and $\xi <\delta $ such that $\xi \geq i(\beta )$ and $C\cap \beta =C_{\beta ,\xi }$ holds for all $\beta \in {\mathrm {Lim}}(C)$ . But this contradicts 6 in Definition 3.1.⊣

Proof Fix a $\square _\kappa $ -sequence $\langle {A_\gamma }~\vert ~{\gamma \in {\mathrm {Lim}}\cap \kappa ^+}\rangle $ and a closed unbounded subset C of $\kappa $ of order-type ${{\mathrm {cof}}(\kappa )}$ . Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ , let $\lambda _\gamma ={{\mathrm {otp}}\left (A_\gamma \right )}\leq \kappa $ and let $\langle {\beta ^\gamma _\alpha }~\vert ~{\alpha <\lambda _\gamma }\rangle $ denote the monotone enumeration of $A_\gamma $ . Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $\lambda _\gamma \in {\mathrm {Lim}}(C)\cup \{\kappa \}$ , let $B_\gamma =\{{\beta \in A_\gamma }~\vert ~{{{\mathrm {otp}}\left (A_\gamma \cap \beta \right )}\in C}\}$ . Next, if $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $\lambda _\gamma \notin {\mathrm {Lim}}(C)\cup \{\kappa \}$ and ${\mathrm {Lim}}(C)\cap \lambda _\gamma =\emptyset $ , then we define $B_\gamma =A_\gamma $ . Finally, if $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $\lambda _\gamma \notin {\mathrm {Lim}}(C)\cup \{\kappa \}$ and ${\mathrm {Lim}}(C)\cap \lambda _\gamma \neq \emptyset $ , then we set $\alpha =\max ({\mathrm {Lim}}(C)\cap \lambda _\gamma )<\lambda _\gamma $ and we define $B_\gamma =B_{\beta ^\gamma _\alpha }\cup (A_\gamma \setminus B_{\beta ^\gamma _\alpha })$ .

With the help of Fodor’s Lemma, we can now find a stationary subset E of S and $\lambda <\kappa $ with ${{\mathrm {otp}}\left (B_\gamma \right )}=\lambda $ for all $\gamma \in E$ . Then we have $\vert {{\mathrm {Lim}}(B_\gamma )\cap E}\vert \leq 1$ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ . Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ , define $C_\gamma =B_\gamma $ if ${{\mathrm {otp}}\left (B_\gamma \right )}\leq \lambda $ and let $C_\gamma =\{{\beta \in B_\gamma }~\vert ~{{{\mathrm {otp}}\left (B_\gamma \cap \beta \right )}>\lambda }\}$ if ${{\mathrm {otp}}\left (B_\gamma \right )}>\lambda $ .

This completes the proof of the lemma.⊣

Proof By our assumptions, we can apply Lemma 4.2 to obtain a $\square _\kappa $ -sequence $\langle {C_\gamma }~\vert ~{\gamma \in {\mathrm {Lim}}\cap \kappa ^+}\rangle $ and a stationary subset E of $S^{\kappa ^+}_\delta $ such that ${{\mathrm {otp}}\left (C_\gamma \right )}<\kappa $ and ${\mathrm {Lim}}(C_\gamma )\cap E=\emptyset $ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ . Next, note that Lemma 4.1 implies that $j_U((\nu ^\delta )^+)=(\nu ^\delta )^+<\kappa $ holds for all cardinals $\nu <\kappa $ . This allows us to fix the monotone enumeration $\langle {\kappa _\xi }~\vert ~{\xi <\delta }\rangle $ of a closed unbounded set of uncountable cardinals smaller than $\kappa $ of order-type $\delta $ with the property that $j_U(\kappa _\xi )=\kappa _\xi $ holds for all $\xi <\delta $ . In this situation, the normality of U implies that $[\xi \mapsto \kappa _\xi ]_U=\kappa $ and $[\xi \mapsto \kappa _\xi ^+]_U\leq \kappa ^+$ . Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ , let $\xi _\gamma $ denote the minimal element $\xi $ of $\delta $ with $\kappa _\xi ^+>{{\mathrm {otp}}\left (C_\gamma \right )}$ . Note that $\xi _\gamma \geq \xi _\beta $ holds for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ and $\beta \in {\mathrm {Lim}}(C_\gamma )$ .

In the following, we inductively construct a sequence

$$ \begin{align*}\langle{f_\gamma\in\prod_{\xi<\delta}\kappa_\xi^+}~\vert~{\gamma<\kappa^+}\rangle.\end{align*} $$

The idea behind this construction is that these functions represent a cofinal subset of $\kappa ^+$ and thereby in particular witness that $[\xi \mapsto \kappa _\xi ^+]_U = \kappa ^+$ . We identify each $f_\gamma \in \prod _{\xi <\delta }\kappa _\xi ^+$ with a function with domain $\delta $ in the obvious way and define:

• • $f_0(\xi )=0$ for all $\xi <\delta $ .

• • $f_{\gamma +1}(\xi )=f_\gamma (\xi )+1$ for all $\gamma <\kappa ^+$ and $\xi <\delta $ .

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ bounded in $\gamma $ and $\xi <\delta $ , then

  $$ \begin{align*}f_\gamma(\xi) ~ = ~ \min\{{\rho\in{\text{Lim}}}~\vert~{\rho>f_\beta(\xi) \textit{ for all }\beta\in C_\gamma\setminus\max({\text{Lim}}(C_\gamma))}\}.\end{align*} $$

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ and $\xi <\xi _\gamma $ , then $f_\gamma (\xi )=\omega $ .

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ and $\xi _\gamma \leq \xi <\delta $ , then

  $$ \begin{align*}f_\gamma(\xi) ~ = ~ \sup\{{f_\beta(\xi)}~\vert~{\beta\in{\text{Lim}}(C_\gamma)}\}.\end{align*} $$

Proof of the Claim (i) We prove the statement by induction on $0<\gamma <\kappa ^+$ , where the successor step follows trivially from our induction hypothesis. Now, assume that $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ bounded in $\gamma $ . Since $\delta $ is an uncountable regular cardinal, our induction hypothesis allows us to find $\zeta <\delta $ with the property that $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ holds for all $\beta _0,\beta _1\in C_\gamma \setminus \max ({\mathrm {Lim}}(C_\gamma ))$ with $\beta _0<\beta _1$ and all $\zeta \leq \xi <\delta $ . By definition, we now have

$$ \begin{align*}f_\gamma(\xi) ~ = ~ \sup\{{f_\beta(\xi)}~\vert~{\beta\in C_\gamma\setminus\max({\text{Lim}}(C_\gamma))}\}\end{align*} $$

for all $\zeta \leq \xi <\delta $ . Since $C_\gamma \setminus \max ({\mathrm {Lim}}(C_\gamma ))$ is a cofinal subset of $\gamma $ , the desired statement for $\gamma $ now follows directly from our induction hypothesis. Finally, if $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ , then the desired statement for $\gamma $ follows directly from the definition of $f_\gamma $ and our induction hypothesis.

(ii) We prove the claim by induction on $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ . First, if $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ bounded in $\gamma $ and $\beta =\max ({\mathrm {Lim}}(C_\gamma ))$ , then our definition ensures that $f_\beta (\xi )<f_\gamma (\xi )$ holds for all $\xi <\delta $ and hence the desired statement follows directly from our induction hypothesis. Next, if $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ , then our induction hypothesis implies that $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ holds for all $\beta _0,\beta _1\in {\mathrm {Lim}}(C_\gamma )$ with $\beta _0<\beta _1$ and all $\xi _{\beta _1}\leq \xi <\delta $ . Since $\xi _\gamma \geq \xi _\beta $ holds for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ , this fact together with our definition yields the desired statement for $\gamma $ .

(iii) This statement is a direct consequence of the definition of the sequence $\langle {f_\gamma }~\vert ~{\gamma < \kappa ^+}\rangle $ and statement (ii).⊣

Note that the first part of the above claim in particular shows that we have $[f_\beta ]_U<[f_\gamma ]_U<[\xi \mapsto \kappa _\xi ^+]_U$ for all $\beta <\gamma <\kappa ^+$ . Since we already observed that $[\xi \mapsto \kappa _\xi ^+]_U=(\kappa ^+)^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}\leq \kappa ^+$ holds, we can conclude that $[\xi \mapsto \kappa _\xi ^+]_U=\kappa ^+$ .

Next, notice that the fact that $2^\kappa =\kappa ^+$ holds allows us to fix an enumeration $\langle {h_\alpha }~\vert ~{\alpha <\kappa ^+}\rangle $ of $\prod _{\xi <\delta }{\mathcal {P}}({\kappa _\xi ^+})$ of order-type $\kappa ^+$ . In addition, let $\langle {\gamma _\alpha }~\vert ~{\alpha <\kappa ^+}\rangle $ denote the monotone enumeration of E. We now inductively define a sequence

$$ \begin{align*}\langle{c_\gamma}~\vert~{\gamma\in{\text{Lim}}\cap\kappa^+}\rangle\end{align*} $$

of functions with domain $\delta $ satisfying the following statements for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ :

1. (a) $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ for all $\xi <\delta $ .

2. (b) If $\beta \in {\mathrm {Lim}}\cap \gamma $ , then $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

3. (c) If $\gamma \notin E$ , then $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and $\xi _\gamma \leq \xi <\delta $ .

4. (d) If $\gamma \in E$ and $\alpha <\kappa ^+$ with $\gamma =\gamma _\alpha $ , then $c_\gamma (\xi )\neq h_\alpha (\xi )\cap f_\gamma (\xi )$ for all $\xi _\gamma \leq \xi <\delta $ .

The idea behind this definition is to use the fact that the sequence $\langle {[f_\gamma ]_U}~\vert ~{\gamma <\kappa ^+}\rangle $ is not continuous at ordinals of cofinality $\delta $ to diagonalize against the sequence $\langle {[h_\alpha ]_U}~\vert ~{\alpha <\kappa ^+}\rangle $ of subsets of $\kappa ^+$ in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ in d. The inductive definition of this sequence is straightforward, but we decided to give the details to convince the reader that it works. We distinguish between the following cases:

First, we set $\beta _0=0$ if ${\mathrm {Lim}}(C_\gamma )=\emptyset $ and $\beta _0=\max ({\mathrm {Lim}}(C_\gamma ))$ otherwise. Next, we set $\beta _1=0$ if $\gamma =\omega $ and $\beta _1=\max ({\mathrm {Lim}}\cap \gamma )$ otherwise. We then have $\beta _0\leq \beta _1<\gamma $ and $f_{\beta _0}(\xi )<f_\gamma (\xi )$ for all $\xi <\delta $ . Using our induction hypothesis, we can find $\xi _\gamma \leq \zeta <\delta $ with the property that $f_{\beta _0}(\xi )\leq f_{\beta _1}(\xi )<f_\gamma (\xi )$ holds for all $\zeta \leq \xi <\delta $ and, if $\beta _0>0$ , then $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ . Note that our assumptions imply that ${\mathrm {Lim}}(C_\gamma )$ is bounded in $\gamma $ and hence the definition of $f_\gamma (\xi )$ ensures that ${{\mathrm {cof}}(f_\gamma (\xi ))}=\omega $ holds for every $\xi <\delta $ . Therefore, we can fix a sequence of strictly increasing functions $\langle {{k_\xi }:{\omega }\longrightarrow {f_\gamma (\xi )}}~\vert ~{\xi <\delta }\rangle $ with the property that $k_\xi $ is cofinal in $f_\gamma (\xi )$ for all $\xi <\delta $ , $k_\xi (0)=f_{\beta _0}(\xi )$ for all $\xi <\zeta $ , and $k_\xi (0)=f_{\beta _1}(\xi )$ for all $\zeta \leq \xi <\delta $ . Define

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0=0, \\ c_{\beta_0}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0>0, \\ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_1=0, \\ c_{\beta_1}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_1>0. \end{cases} \end{align*} $$

These definitions ensure that $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ for all $\xi <\delta $ . Moreover, if $\beta _0>0$ , then $c_\gamma (\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ holds for all $\xi <\delta $ . This inductively implies that $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ holds for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and all $\xi _\gamma \leq \xi <\delta $ . Next, if $\beta _1>0$ and $\zeta \leq \xi <\delta $ , then $f_{\beta _1}(\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_{\beta _1}(\xi )=c_{\beta _1}(\xi )$ . This allows us to conclude that for all $\beta \in {\mathrm {Lim}}\cap \gamma $ , we have $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

Since our assumptions imply that ${{\mathrm {cof}}(\gamma )}=\omega $ , there is a strictly increasing sequence $\langle {\beta _n}~\vert ~{n<\omega }\rangle $ cofinal in $\gamma $ such that $\beta _n\in {\mathrm {Lim}}\cap \gamma $ for all $0<n<\omega $ , $\beta _0=0$ in case ${\mathrm {Lim}}(C_\gamma )=\emptyset $ , and $\beta _0=\max ({\mathrm {Lim}}(C_\gamma ))$ in case ${\mathrm {Lim}}(C_\gamma )\neq \emptyset $ . By the regularity of $\delta $ , we can find $\xi _\gamma \leq \zeta <\delta $ such that $f_{\beta _n}(\xi )<f_{\beta _{n+1}}(\xi )<f_\gamma (\xi )$ and $c_{\beta _{n+2}}(\xi )\cap f_{\beta _{n+1}}(\xi )=c_{\beta _{n+1}}(\xi )$ for all $\zeta \leq \xi <\delta $ and all $n<\omega $ and, if $\beta _0>0$ , then $c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ . By the definition of $f_\gamma $ , we then have $f_\gamma (\xi )=\sup \{{f_{\beta _n}(\xi )}~\vert ~{n<\omega }\}$ for all $\zeta \leq \xi <\delta $ . Since the definition of $f_\gamma $ also implies that ${{\mathrm {cof}}(f_\gamma (\xi ))}=\omega $ and $f_{\beta _0}(\xi )<f_\gamma (\xi )$ for all $\xi <\delta $ , we can fix a sequence of strictly increasing functions $\langle {{k_\xi }:{\omega }\longrightarrow {f_\gamma (\xi )}}~\vert ~{\xi <\zeta }\rangle $ with the property that $k_\xi $ is cofinal in $f_\gamma (\xi )$ for all $\xi <\zeta $ and $k_\xi (0)=f_{\beta _0}(\xi )$ for all $\xi <\zeta $ . Define

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0=0,\\ c_{\beta_0}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0>0,\\ \bigcup\{{c_{\beta_n}(\xi)}~\vert~{0<n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_0=0,\\ \bigcup\{{c_{\beta_n}(\xi)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_0>0. \end{cases} \end{align*} $$

Then the set $c_\gamma (\xi )$ is closed and unbounded in $f_\gamma (\xi )$ for all $\xi <\delta $ . In addition, if $\beta _0>0$ , then $c_\gamma (\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ for all $\xi <\delta $ . In particular, we have $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and all $\xi _\gamma \leq \xi <\delta $ . Next, if $0<n<\omega $ and $\zeta \leq \xi <\delta $ , then $f_{\beta _n}(\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_{\beta _n}(\xi )=c_{\beta _n}(\xi )$ . This directly implies that for all $\beta \in {\mathrm {Lim}}\cap \gamma $ , we have $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

Let

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \omega, & \textit{for all }\xi<\xi_\gamma,\\ \bigcup\{{c_\beta(\xi)}~\vert~{\beta\in{\text{Lim}}(C_\gamma)}\}, & \textit{for all }\xi_\gamma\leq\xi<\delta. \end{cases} \end{align*} $$

Fix $\beta _0,\beta _1\in {\mathrm {Lim}}(C_\gamma )$ with $\beta _0<\beta _1$ . Then $\beta _0\in {\mathrm {Lim}}(C_{\beta _1})$ and $\beta _1\notin E$ by the choice of the $\square _\kappa $ -sequence and the stationary set E, because we have $\beta _1 \in {\mathrm {Lim}}(C_\gamma )$ . Moreover, if $\xi _\gamma \leq \xi <\delta $ , then the set $c_{\beta _1}(\xi )$ is unbounded in $f_{\beta _1}(\xi )$ , $\xi \geq \xi _{\beta _1}$ , $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ and $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ . Since $c_\gamma (\xi )=\omega =f_\gamma (\xi )$ holds for all $\xi <\xi _\gamma $ , this shows that $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ for all $\xi <\delta $ , and $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ holds for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and all $\xi _\gamma \leq \xi <\delta $ . Moreover, if $\beta _0\in {\mathrm {Lim}}\cap \gamma $ and $\beta _1\in {\mathrm {Lim}}(C_\gamma )$ with $\beta _0<\beta _1$ , then there is $\xi _\gamma \leq \zeta <\delta $ with $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ and $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ and hence we have $f_{\beta _0}(\xi )<f_\gamma (\xi )$ and $c_{\beta _0}(\xi )=c_\gamma (\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ .

Fix $\alpha <\kappa ^+$ with $\gamma =\gamma _\alpha $ . Let $\langle {\beta _\xi }~\vert ~{\xi <\delta }\rangle $ be the monotone enumeration of a subset of ${\mathrm {Lim}}(C_\gamma )$ of order-type $\delta $ that is closed unbounded in $\gamma $ . Given $\xi _\gamma \leq \xi <\delta $ , we have $f_{\beta _\xi }(\xi )<f_\gamma (\xi )$ and we can therefore pick a closed unbounded subset $C^\gamma _\xi $ of $f_\gamma (\xi )$ with $\min (C^\gamma _\xi )=f_{\beta _\xi }(\xi )$ and $C^\gamma _\xi \neq h_\alpha (\xi )\cap [f_{\beta _\xi }(\xi ),f_\gamma (\xi ))$ . Now, define

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \omega, & \textit{for all }\xi<\xi_\gamma, \\ c_{\beta_\xi}(\xi) ~ \cup ~ C^\gamma_\xi, & \textit{for all }\xi_\gamma\leq\xi<\delta. \end{cases} \end{align*} $$

Then $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ for all $\xi <\delta $ and, if $\xi _\gamma \leq \xi <\delta $ , then $c_\gamma (\xi )\neq h_\alpha (\xi )\cap f_\gamma (\xi )$ . Moreover, if $\xi _\gamma \leq \zeta <\xi <\delta $ , then $\beta _\zeta \in {\mathrm {Lim}}(C_{\beta _\xi })$ , $\beta _\xi \notin E$ , $\xi>\xi _{\beta _\xi }$ , $f_{\beta _\zeta }(\xi )<f_{\beta _\xi }(\xi )<f_\gamma (\xi )$ and

$$ \begin{align*}c_{\beta_\zeta}(\xi) ~ = ~ c_{\beta_\xi}(\xi)\cap f_{\beta_\zeta}(\xi) ~ = ~ f_\gamma(\xi)\cap f_{\beta_\zeta}(\xi).\end{align*} $$

In particular, this shows that for all $\beta \in {\mathrm {Lim}}\cap \gamma $ , we have $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

The above construction ensures that $[c_\gamma ]_U$ is a closed unbounded subset of $[f_\gamma ]_U$ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ . Moreover, we have $[c_\beta ]_U=[c_\gamma ]_U\cap [f_\beta ]_U$ for all $\beta ,\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $\beta <\gamma $ . In particular, there is a closed unbounded subset C of $\kappa ^+$ with $C\cap [f_\gamma ]_U=[c_\gamma ]_U$ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ .

Proof of the Claim First, if $\beta <\kappa ^+$ , then there is $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $[f_\gamma ]_U>\beta $ and

$$ \begin{align*}C\cap\beta ~ = ~ (C\cap[f_\gamma]_U)\cap\beta ~ = ~ [c_\gamma]_U\cap\beta ~ \in ~ {\mathrm{Ult}}({{\text{V}}},{U}).\end{align*} $$

Next, assume, towards a contradiction, that C is an element of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Then there is an $\alpha <\kappa ^+$ with $C=[h_\alpha ]_U$ . Since we have

$$ \begin{align*}[h_\alpha]_U\cap[f_{\gamma_\alpha}]_U ~ = ~ C\cap[f_{\gamma_\alpha}]_U ~ = ~ [c_{\gamma_\alpha}]_U,\end{align*} $$

we know that the set $\{{\xi <\delta }~\vert ~{h_\alpha (\xi )\cap f_{\gamma _\alpha }(\xi )=c_{\gamma _\alpha }(\xi )}\}$ is an element of U. In particular, we can find $\xi _{\gamma _\alpha }\leq \xi <\delta $ with $h_\alpha (\xi )\cap f_{\gamma _\alpha }(\xi )=c_{\gamma _\alpha }(\xi )$ , contradicting the definition of $c_{\gamma _\alpha }$ .⊣

This completes the proof of the theorem.⊣

Proof Set $\lambda _0={{\mathrm {cof}}(\lambda )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}$ . Let A be an unbounded subset of $\lambda _0$ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ and let $\langle {\gamma _\eta }~\vert ~{\eta <\lambda _0}\rangle $ be a strictly increasing sequence that is cofinal in $\lambda $ and an element of ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . In this situation, the set $\{{\gamma _\eta }~\vert ~{\eta \in A}\}$ is unbounded in $\lambda $ and fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof As in the proof of Theorem 4.3, we can apply Lemma 4.1 to find the monotone enumeration $\langle {\kappa _\xi }~\vert ~{\xi <\delta }\rangle $ of a closed unbounded set of uncountable cardinals smaller than $\kappa $ of order-type $\delta $ with the property that $j_U(\kappa _\xi )=\kappa _\xi $ holds for all $\xi <\delta $ . Then normality implies that $[\xi \mapsto \kappa _\xi ]_U=\kappa $ and we can repeat arguments from the first part of the proof of Theorem 4.3 to see that $[\xi \mapsto \kappa _\xi ^+]_U=\kappa ^+$ . By our assumptions, there is a function h with domain $\delta $ , $[h]_U=\lambda $ and the property that $h(\xi )$ is a regular cardinal in the interval $(\kappa _\xi ^+,\kappa )$ for all $\xi <\delta $ . Fix a sequence $\langle {h_\gamma \in \prod _{\xi <\delta }h(\xi )}~\vert ~{\gamma <\kappa ^+}\rangle $ such that the sequence $\langle {[h_\gamma ]_U}~\vert ~{\gamma <\kappa ^+}\rangle $ is strictly increasing and cofinal in $\lambda $ .

Pick a $\square _\kappa $ -sequence $\langle {C_\gamma }~\vert ~{\gamma \in {\mathrm {Lim}}\cap \kappa ^+}\rangle $ with ${{\mathrm {otp}}\left (C_\gamma \right )}<\kappa $ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ . Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ , we let $\xi _\gamma $ denote the minimal element $\xi $ of $\delta $ with $\kappa _\xi ^+>{{\mathrm {otp}}\left (C_\gamma \right )}$ .

We now inductively construct a sequence

$$ \begin{align*}\langle{f_\gamma\in\prod_{\xi<\delta}h(\xi)}~\vert~{\gamma<\kappa^+}\rangle\end{align*} $$

by setting:

• • $f_0(\xi )=0$ for all $\xi <\delta $ .

• • $f_{\gamma +1}(\xi )=\max (f_\gamma (\xi ),h_\gamma (\xi ))+1$ for all $\gamma <\kappa ^+$ and $\xi <\delta $ .

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ bounded in $\gamma $ and $\xi <\delta $ , then

  $$ \begin{align*}f_\gamma(\xi) ~ = ~ \min\{{\rho\in{\text{Lim}}}~\vert~{\rho>f_\beta(\xi)\textit{ for all }\beta\in C_\gamma\setminus\max({\text{Lim}}(C_\gamma))}\}.\end{align*} $$

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ and $\xi <\xi _\gamma $ , then $f_\gamma (\xi )=\omega $ .

• • If $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with ${\mathrm {Lim}}(C_\gamma )$ unbounded in $\gamma $ and $\xi _\gamma \leq \xi <\delta $ , then

  $$ \begin{align*}f_\gamma(\xi) ~ = ~ \sup\{{f_\beta(\xi)}~\vert~{\beta\in{\text{Lim}}(C_\gamma)}\}.\end{align*} $$

As in the proof of Theorem 4.3, we have the following claim.

In particular, this shows that the sequence $\langle {[f_\gamma ]_U}~\vert ~{\gamma <\kappa ^+}\rangle $ is strictly increasing. Since the above definition ensures that $[h_\gamma ]_U<[f_{\gamma +1}]_U<[h]_U$ holds for all $\gamma <\kappa ^+$ , we also know that this sequence is cofinal in $\lambda $ .

Next, we inductively define a sequence $\langle {c_\gamma }~\vert ~{\gamma \in {\mathrm {Lim}}\cap \kappa ^+}\rangle $ of functions with domain $\delta $ such that the following statements hold for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ :

1. (a) $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ with ${{\mathrm {otp}}\left (c_\gamma (\xi )\right )}<\kappa _\xi ^+$ for all $\xi <\delta $ .

2. (b) If $\beta \in {\mathrm {Lim}}\cap \gamma $ , then $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

3. (c) If $\beta \in {\mathrm {Lim}}(C_\gamma )$ and $\xi _\gamma \leq \xi <\delta $ , then $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ .

Our inductive construction distinguishes between the following cases:

We set $\beta _0=0$ if ${\mathrm {Lim}}(C_\gamma )=\emptyset $ and $\beta _0=\max ({\mathrm {Lim}}(C_\gamma ))$ otherwise. Moreover, we set $\beta _1=0$ if $\gamma =\omega $ and $\beta _1=\max ({\mathrm {Lim}}\cap \gamma )$ otherwise. This definition ensures that $\beta _0\leq \beta _1<\gamma $ and $f_{\beta _0}(\xi )<f_\gamma (\xi )$ for all $\xi <\delta $ . We can now find $\xi _\gamma \leq \zeta <\delta $ with the property that $f_{\beta _0}(\xi )\leq f_{\beta _1}(\xi )<f_\gamma (\xi )$ for all $\zeta \leq \xi <\delta $ and, if $\beta _0>0$ , then $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ . Since the definition of $f_\gamma $ implies that ${{\mathrm {cof}}(f_\gamma (\xi ))}=\omega $ holds for all $\xi <\delta $ , we can pick a sequence of strictly increasing functions $\langle {{k_\xi }:{\omega }\longrightarrow {f_\gamma (\xi )}}~\vert ~{\xi <\delta }\rangle $ with the property that $k_\xi $ is cofinal in $f_\gamma (\xi )$ for all $\xi <\delta $ , $k_\xi (0)=f_{\beta _0}(\xi )$ for all $\xi <\zeta $ , and $k_\xi (0)=f_{\beta _1}(\xi )$ for all $\zeta \leq \xi <\delta $ . Let

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0=0,\\ c_{\beta_0}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0>0,\\ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_1=0,\\ c_{\beta_1}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_1>0. \end{cases} \end{align*} $$

Then $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ of order-type less than $\kappa _\xi ^+$ for all $\xi <\delta $ and, if $\beta _0>0$ , then $c_\gamma (\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ for all $\xi <\delta $ . In particular, we know that $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and all $\xi _\gamma \leq \xi <\delta $ . Finally, notice that $\beta _1>0$ implies that $f_{\beta _1}(\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_{\beta _1}(\xi )=c_{\beta _1}(\xi )$ hold for all $\zeta \leq \xi <\delta $ . This shows that for all $\beta \in {\mathrm {Lim}}\cap \gamma $ , we have $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\beta (\xi )=c_\gamma (\xi )\cap f_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

Since the limit points of $C_\gamma $ are bounded in $\gamma $ , we have ${{\mathrm {cof}}(\gamma )}=\omega $ and we can pick a strictly increasing sequence $\langle {\beta _n}~\vert ~{n<\omega }\rangle $ cofinal in $\gamma $ such that $\beta _n\in {\mathrm {Lim}}\cap \gamma $ for all $0<n<\omega $ , $\beta _0=0$ in case ${\mathrm {Lim}}(C_\gamma )=\emptyset $ , and $\beta _0=\max ({\mathrm {Lim}}(C_\gamma ))$ in case ${\mathrm {Lim}}(C_\gamma )\neq \emptyset $ . Fix $\xi _\gamma \leq \zeta <\delta $ such that $f_{\beta _n}(\xi )<f_{\beta _{n+1}}(\xi )<f_\gamma (\xi )$ and $c_{\beta _{n+2}}(\xi )\cap f_{\beta _{n+1}}(\xi )=c_{\beta _{n+1}}(\xi )$ for all $\zeta \leq \xi <\delta $ and all $n<\omega $ , and, if $\beta _0>0$ , $c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ . Then the definition of $f_\gamma $ ensures that ${{\mathrm {cof}}(f_\gamma (\xi ))}=\omega $ and $f_{\beta _0}(\xi )<f_\gamma (\xi )$ for all $\xi <\delta $ . Moreover, it also directly implies that $f_\gamma (\xi )=\sup \{{f_{\beta _n}(\xi )}~\vert ~{n<\omega }\}$ holds for all $\zeta \leq \xi <\delta $ . Fix a sequence of strictly increasing functions $\langle {{k_\xi }:{\omega }\longrightarrow {f_\gamma (\xi )}}~\vert ~{\xi <\zeta }\rangle $ such that $k_\xi (0)=f_{\beta _0}(\xi )$ and $k_\xi $ is cofinal in $f_\gamma (\xi )$ for all $\xi <\zeta $ . Define

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0=0,\\ c_{\beta_0}(\xi) ~ \cup ~ \{{k_\xi(n)}~\vert~{n<\omega}\}, & \textit{for all }\xi<\zeta,\textit{ if }\beta_0>0,\\ \bigcup\{{c_{\beta_n}(\xi)}~\vert~{0<n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_0=0,\\ \bigcup\{{c_{\beta_n}(\xi)}~\vert~{n<\omega}\}, & \textit{for all }\zeta\leq\xi<\delta,\textit{ if }\beta_0>0. \end{cases} \end{align*} $$

Given $\xi <\delta $ , the set $c_\gamma (\xi )$ is closed and unbounded in $f_\gamma (\xi )$ and the regularity of $\kappa _\xi ^+$ implies that ${{\mathrm {otp}}\left (c_\gamma (\xi )\right )}<\kappa _\xi ^+$ . Next, $\beta _0>0$ implies that $c_\gamma (\xi )\cap f_{\beta _0}(\xi )=c_{\beta _0}(\xi )$ for all $\xi <\delta $ , and therefore $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ for all $\beta \in {\mathrm {Lim}}(C_\gamma )$ and all $\xi _\gamma \leq \xi <\delta $ . Finally, we have $f_{\beta _n}(\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_{\beta _n}(\xi )=c_{\beta _n}(\xi )$ for all $0<n<\omega $ and $\zeta \leq \xi <\delta $ , and hence for all $\beta \in {\mathrm {Lim}}\cap \gamma $ , we have $f_\beta (\xi )<f_\gamma (\xi )$ and $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ for coboundedly many $\xi <\delta $ .

Let

$$ \begin{align*} c_\gamma(\xi) ~ = ~ \begin{cases} \omega, & \textit{for all }\xi<\xi_\gamma,\\ \bigcup\{{c_\beta(\xi)}~\vert~{\beta\in{\text{Lim}}(C_\gamma)}\}, & \textit{for all }\xi_\gamma\leq\xi<\delta. \end{cases} \end{align*} $$

Given $\beta _0,\beta _1\in {\mathrm {Lim}}(C_\gamma )$ with $\beta _0<\beta _1$ and $\xi _\gamma \leq \xi <\delta $ , the above definition ensures that $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ and $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ . Since $c_\gamma (\xi )=\omega =f_\gamma (\xi )$ holds for all $\xi <\xi _\gamma $ and we have $f_\gamma (\xi )=\sup \{{f_\beta (\xi )}~\vert ~{\beta \in {\mathrm {Lim}}(C_\gamma )}\}$ and ${{\mathrm {otp}}\left (C_\gamma \right )}<\kappa _\xi ^+$ for all $\xi _\gamma \leq \xi <\delta $ , we can conclude that $c_\gamma (\xi )$ is a closed unbounded subset of $f_\gamma (\xi )$ of order-type less than $\kappa _\xi ^+$ for all $\xi <\delta $ , and, if $\beta \in {\mathrm {Lim}}(C_\gamma )$ and $\xi _\gamma \leq \xi <\delta $ , then $c_\gamma (\xi )\cap f_\beta (\xi )=c_\beta (\xi )$ holds. Finally, given $\beta _0\in {\mathrm {Lim}}\cap \gamma $ and $\beta _1\in {\mathrm {Lim}}(C_\gamma )$ with $\beta _0<\beta _1$ , our induction hypothesis yields $\xi _\gamma \leq \zeta <\delta $ with $f_{\beta _0}(\xi )<f_{\beta _1}(\xi )$ and $c_{\beta _0}(\xi )=c_{\beta _1}(\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ , and this ensures that $f_{\beta _0}(\xi )<f_\gamma (\xi )$ and $c_{\beta _0}(\xi )=c_\gamma (\xi )\cap f_{\beta _0}(\xi )$ for all $\zeta \leq \xi <\delta $ .

Given $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ , the properties listed above ensure that $[c_\gamma ]_U$ is a closed unbounded subset of $[f_\gamma ]_U$ of order-type less than $\kappa ^+$ . Moreover, if $\beta ,\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ with $\beta <\gamma $ , then $[c_\beta ]_U=[c_\gamma ]_U\cap [f_\beta ]_U$ . These observations show that there is a closed unbounded subset C of $\lambda $ with $C\cap [f_\gamma ]_U=[c_\gamma ]_U$ for all $\gamma \in {\mathrm {Lim}}\cap \kappa ^+$ and this property directly implies that ${{\mathrm {otp}}\left (C\right )}=\kappa ^+<\lambda $ . Since $\lambda $ is a regular cardinal in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ , this allows us to conclude that the set C is not contained in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ and hence it is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof of Theorem 1.6 Fix a normal ultrafilter U on a measurable cardinal $\delta $ that satisfies the three assumptions listed in the statement of the theorem. By Proposition 1.2, if $\lambda $ is a limit ordinal with the property that the cardinal ${{\mathrm {cof}}(\lambda )}$ is either smaller than $\delta ^+$ or weakly compact, then no unbounded subset of $\lambda $ is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . In the proof of the converse implication, we first consider two special cases.

Proof of the Claim We start by noting that, if ${{\mathrm {cof}}(\kappa )}=\delta ^+$ , then the fact that our assumptions imply that $2^\delta =\delta ^+$ holds allows us to use Proposition 1.3 find a subset of $\kappa $ with the desired properties. Therefore, in the following, we may assume that $\delta ^+<{{\mathrm {cof}}(\kappa )}\leq \kappa $ . Let $\nu \leq \kappa $ be minimal with $\nu ^\delta \geq \kappa $ . By the minimality of $\nu $ , we then have $\mu ^\delta <\nu $ for all $\mu <\nu $ . In particular, the fact that $2^\delta =\delta ^+<\kappa $ implies that $\nu>2^\delta >\delta $ and therefore we know that $\nu ^+=\nu ^\nu \geq \nu ^\delta \geq \kappa \geq \nu $ . These computations show that either ${{\mathrm {cof}}(\nu )}>\delta $ and $\kappa =\nu $ , or ${{\mathrm {cof}}(\nu )}\leq \delta $ and $\kappa =\nu ^+$ .

First, assume that either ${{\mathrm {cof}}(\nu )}>\delta $ and $\kappa =\nu $ , or ${{\mathrm {cof}}(\nu )}<\delta $ and $\kappa =\nu ^+$ . Then Lemma 4.1 shows that $j_U(\kappa )=\kappa $ holds in both cases. Moreover, since ${{\mathrm {cof}}(\kappa )}$ is a regular cardinal greater than $\delta ^+$ and

$$ \begin{align*}j_U({{\text{cof}}(\kappa)}) ~ = ~ {{\text{cof}}(j_U(\kappa))}^{{\mathrm{Ult}}({{\text{V}}},{U})} ~ = ~ {{\text{cof}}(\kappa)}^{{\mathrm{Ult}}({{\text{V}}},{U})},\end{align*} $$

the fact that ${{\mathrm {cof}}(\kappa )}$ is not weakly compact allows us to use Theorem 1.8 to find an unbounded subset of ${{\mathrm {cof}}(\kappa )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}$ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . In this situation, we can then apply Proposition 5.1 to obtain an unbounded subset of $\kappa $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

Finally, assume that ${{\mathrm {cof}}(\nu )}=\delta $ and $\kappa =\nu ^+$ . In this situation, we know that $\nu $ is a singular cardinal of cofinality $\delta $ with $2^\nu =\nu ^+$ and the property that $\mu ^\delta <\nu $ holds for all $\mu <\nu $ . Since the assumptions of the theorem guarantee the existence of a $\square _\nu $ -sequence, we can apply Theorem 4.3 to find an unbounded subset of $\kappa $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof of the Claim First, if $\lambda $ is a cardinal, then we can use the above claim to directly derive the desired conclusion. Hence, we may assume that $\lambda $ is not a cardinal.⊣

Subclaim. There is a cardinal $\kappa $ of cofinality $\delta $ such that

$$ \begin{align*}\lambda ~ \in ~ (\kappa^+,j_U(\kappa)] ~ \cup ~ \{j_U(\kappa^+)\}\end{align*} $$

and $\kappa>\delta $ implies that $\mu ^\delta <\kappa $ for all $\mu <\kappa $ .

Proof of the Subclaim Let $\theta = |\lambda |$ . Then our assumptions imply that

$$ \begin{align*}\delta ~ < ~ {{\text{cof}}(\lambda)} ~ \leq ~ \theta ~ < ~ \lambda ~ < ~ \theta^+.\end{align*} $$

Moreover, we have ${{\mathrm {cof}}(\theta )}\neq \delta $ , because otherwise $\theta $ would be a singular strong limit cardinal of cofinality $\delta $ and our assumptions would allow us to repeat the argument from the first part of the proof of Theorem 4.3 to show that $\theta ^+=(\theta ^+)^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}$ , contradict our assumption that $\lambda $ is a cardinal in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . In addition, we know that there is some $\nu <\theta $ satisfying $\nu ^\delta \geq \theta $ , because otherwise Lemma 4.1 would imply that

$$ \begin{align*}j_U(\theta) ~ = ~ \theta ~ < ~ \lambda ~ < ~ \theta^+ ~ = ~ j_U(\theta^+) ~ = ~ (j_U(\theta)^+)^{{\mathrm{Ult}}({{\text{V}}},{U})},\end{align*} $$

which again contradicts the assumption that $\lambda $ is a cardinal in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Let $\rho <\theta $ be the minimal cardinal with the property that $\rho ^\delta \geq \theta $ holds. Then the minimality of $\rho $ implies that $\mu ^\delta <\kappa $ holds for all $\mu <\rho $ .

First, assume that $\rho =2$ . Then $\delta <\theta \leq 2^\delta =\delta ^+$ and therefore $\theta = \delta ^+$ . Since Lemma 4.1 implies that $j_U(\delta ^{++})=\delta ^{++}$ , we know that $j_U(\delta ^{++})=\delta ^{++}=\theta ^+>\lambda $ and, as above, we can conclude that $\lambda $ is not contained in the interval $(j_U(\delta ^+),\delta ^{++})$ . Moreover, since our assumptions on $\lambda $ directly imply that $\lambda $ is not contained in the interval $(j_U(\delta ),j_U(\delta ^+))$ , we can conclude that $\lambda $ is an element of the set $(\delta ^+,j_U(\delta )]\cup \{j_U(\delta ^+)\}$ in this case. In particular, this shows that the statement of the subclaim holds for $\kappa =\delta $ .

Next, assume that $\rho>2^\delta $ . Then our cardinal arithmetic assumptions and the minimality of $\rho $ imply that ${{\mathrm {cof}}(\rho )}\leq \delta $ and $\theta =\rho ^+$ . But then we already know that ${{\mathrm {cof}}(\rho )}=\delta $ , because otherwise we could apply Lemma 4.1 to conclude that $j_U(\theta ) = \theta < \lambda < \theta ^+ = j_U(\theta ^+)$ . Since our assumptions imply that $(\rho ^+)^\delta =\rho ^+$ , Lemma 4.1 implies that $j_U(\rho ^{++})=\rho ^{++}=\theta ^+$ and this shows that $\lambda $ is not contained in the interval $(j_U(\rho ^+),\rho ^{++})$ . Since $\lambda $ is also not contained in the interval $(j_U(\rho ),j_U(\rho ^+))$ , we can conclude that $\lambda $ is contained in the set $(\rho ^+,j_U(\rho )]\cup \{j_U(\rho ^+)\}$ . This allows us to conclude that the statement of the subclaim holds for $\kappa =\rho $ in this case.⊣

First, assume that $\kappa =\delta $ holds. By our assumptions, Lemma 4.1 shows that $\delta ^{++}=j_U(\delta ^{++})>j_U(\delta ^+)$ . Since we know that $\delta ^+<\lambda \leq j_U(\delta ^+)$ and ${{\mathrm {cof}}(\lambda )}> \delta $ , this implies that ${{\mathrm {cof}}(\lambda )}=\delta ^+$ , and hence we can use Proposition 1.3 to find an unbounded subset of $\lambda $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

Next, assume that $\kappa>\delta $ and $\lambda =j_U(\kappa )$ . Then

$$ \begin{align*}\delta ~ < ~ {{\text{cof}}(\lambda)} ~ \leq ~ {{\text{cof}}(\lambda)}^{{\mathrm{Ult}}({{\text{V}}},{U})} ~ = ~ j_U({{\text{cof}}(\kappa)}) ~ = ~ j_U(\delta) ~ < ~ \delta^{++}\end{align*} $$

and we can conclude that ${{\mathrm {cof}}(\lambda )}=\delta ^+$ . Another application of Proposition 1.3 now yields the desired subset of $\lambda $ .

Now, assume that $\kappa>\delta $ and $\lambda =j_U(\kappa ^+)$ . Then our assumptions ensure the existence a $\square (\kappa ^+)$ -sequence and therefore we can apply Theorem 1.8 to find an unbounded subset of $\lambda $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .

Finally, we assume that $\kappa>\delta $ and $\kappa ^+<\lambda <j_U(\kappa )$ . Then we know that $\mu ^\delta <\kappa $ holds for all $\mu <\kappa $ .

Subclaim. ${{\mathrm {cof}}(\lambda )}=\kappa ^+$ .

Proof of the Subclaim Assume, towards a contradiction, that ${{\mathrm {cof}}(\lambda )}\neq \kappa ^+$ . Since $\kappa $ is singular and ${{\mathrm {cof}}(\lambda )}<\lambda <j_U(\kappa )<\kappa ^{++}$ , this implies that ${{\mathrm {cof}}(\lambda )}<\kappa $ . In this situation, we can repeat an argument from the first part of the proof of Theorem 4.3 to find a monotone enumeration $\langle {\kappa _\xi }~\vert ~{\xi <\delta }\rangle $ of a closed unbounded subset of $\kappa $ of order-type $\delta $ such that $\kappa _0>{{\mathrm {cof}}(\lambda )}$ , $[\xi \mapsto \kappa _\xi ]_U=\kappa $ and $[\xi \mapsto \kappa _\xi ^+]_U=\kappa ^+$ . Fix a function f with domain $\delta $ such that $[f]_U=\lambda $ holds and $f(\xi )$ is a regular cardinal in the interval $(\kappa _\xi ^+,\kappa )$ for all $\xi <\delta $ . Pick a sequence $\langle {f_\alpha }~\vert ~{\alpha <{{\mathrm {cof}}(\lambda )}}\rangle $ of functions with domain $\delta $ such that $f_\alpha (\xi )<f(\xi )$ holds for all $\alpha <{{\mathrm {cof}}(\lambda )}$ and all $\xi <\delta $ , and the induced sequence $\langle {[f_\alpha ]_U}~\vert ~{\alpha <{{\mathrm {cof}}(\lambda )}}\rangle $ is strictly increasing and cofinal in $\lambda $ . Given $\xi <\delta $ , the fact that $f(\xi )$ is a regular cardinal greater than ${{\mathrm {cof}}(\lambda )}$ then yields an ordinal $\gamma _\xi <f(\xi )$ with $f_\alpha (\xi )<\gamma _\xi $ for all $\alpha <{{\mathrm {cof}}(\lambda )}$ . But then $[f_\alpha ]_U<[\xi \mapsto \gamma _\xi ]_U<\lambda $ for all $\alpha <{{\mathrm {cof}}(\lambda )}$ , a contradiction.⊣

By the above computations, we now know that $\kappa $ is a singular cardinal of cofinality $\delta $ with the property that $\mu ^\delta <\kappa $ holds for all $\mu <\kappa $ , and $\lambda $ is a limit ordinal of cofinality $\kappa ^+$ with $\kappa ^+<\lambda <j_U(\kappa )$ that is a regular cardinal in ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Since our assumptions guarantee the existence of a $\square _\kappa $ -sequence, we can use Theorem 5.2 to show that there also exists an unbounded subset of $\lambda $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ in this case.⊣

To conclude the proof of the theorem, fix a limit ordinal $\lambda $ with the property that the cardinal ${{\mathrm {cof}}(\lambda )}$ is greater than $\delta $ and not weakly compact. Set $\lambda _0={{\mathrm {cof}}(\lambda )}^{{\mathrm {Ult}}({{\mathrm {V}}},{U})}$ . By [Reference Jech10, Lemma 3.7.(ii)], we then have ${{\mathrm {cof}}(\lambda _0)}={{\mathrm {cof}}(\lambda )}$ . Hence, we can use the previous claim to find an unbounded subset of $\lambda _0$ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Using Proposition 5.1, we can conclude that there is an unbounded subset of $\lambda $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof of Theorem 1.7 We argue that Jensen-style extender models without subcompact cardinals satisfy the statements (a)–(c) listed in Theorem 1.6. First, notice that the ${\mathrm {GCH}}$ holds in all of these models and hence statement (a) is satisfied. Next, recall that [Reference Schimmerling and Zeman24, Theorem 15]Footnote ³ shows that, in Jensen-style extender models, a $\square _\nu $ -sequence exists if and only if $\nu $ is not a subcompact cardinal. In particular, we know that, in Jensen-style extender models without subcompact cardinals, $\square (\nu ^+)$ -sequences exist for all infinite cardinals $\nu $ . Since [Reference Zeman28, Theorem 0.1] yields the existence of $\square (\kappa )$ -sequences for inaccessible cardinals $\kappa $ that are not weakly compact in the relevant models, we can conclude that statement (b) holds in these models. Finally, the validity of statement (c) in Jensen-style extender models without subcompact cardinals again follows from [Reference Schimmerling and Zeman24, Theorem 15].⊣

Proof By our assumptions, we can use the results of [Reference Gitik6] to show that $2^\delta =\delta ^+$ holds. Set $\kappa =\delta ^{++}$ . Then our assumptions imply that $\kappa $ is not weakly compact in ${\mathrm {L}}[U]$ . In this situation, we can construct a tail of a $\square (\kappa )$ -sequence $\langle {C_\nu }~\vert ~{\xi < \nu < \kappa , ~ \nu \in {\mathrm {Lim}}}\rangle $ in ${\mathrm {L}}[U]$ above some ordinal $\xi> \delta ^+$ with $\xi < \kappa $ , using the argument in [Reference Jensen11, Section 6] for ${\mathrm {L}}$ . A consequence of this proof, published by Todorčević in [Reference Todorčević27, 1.10], but probably first noticed by Jensen (see [Reference Schimmerling22, Theorem 2.5] for a modern account), is that the sequence $\langle {C_\nu }~\vert ~{\xi < \nu < \kappa , ~ \nu \in {\mathrm {Lim}}}\rangle $ remains a tail of a $\square (\kappa )$ -sequence in ${\mathrm {V}}$ . We can now easily extend this sequence to a $\square (\kappa )$ -sequence $\langle {C_\nu }~\vert ~{\nu \in {\mathrm {Lim}} \cap \kappa }\rangle $ in ${\mathrm {V}}$ . Since $2^\delta =\delta ^+$ holds, Lemma 4.1 shows that $j_U(\kappa )=\kappa $ and hence we can use Theorem 1.8 to find an unbounded subset of $\kappa $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof Assume, towards a contradiction, that there is a condition p in ${\mathbb {P}}$ and a ${\mathbb {P}}$ -name $\dot {f}$ for a function with domain $\delta $ with the property that, whenever G is ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ with $p\in G$ , then $[\dot {f}^G]_U=A$ holds in ${\mathrm {V}}[G]$ . As ${\mathbb {P}}$ is $(\delta +1)$ -strategically closed, there is a condition $p_1$ in ${\mathbb {P}}$ below p and a subset X of $\delta $ with the property that, whenever G is ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ with $p_1\in G$ , then $X=\{{\xi <\delta }~\vert ~{\dot {f}^G(\xi )\in {\mathrm {V}}}\}$ .

Proof of the Claim If such a pair of conditions does not exist, then it is easy to check that the condition q forces $\dot {f}(\check {\xi })$ to be equal to the set

contradicting our assumption that $\xi $ is not an element of X.⊣

Proof of the Claim Assume, towards a contradiction, that X is not an element of U. Fix a winning strategy $\sigma $ for Player Even in the game $G_{\delta +1}({\mathbb {P}})$ , some sufficiently large regular cardinal $\theta $ and an elementary submodel M of ${\mathrm {H}}(\theta )$ of cardinality $\delta $ satisfying $(\delta +1)\cup \{\lambda ,\sigma ,\dot {f},p_1,A,U,X,{\mathbb {P}}\}\subseteq M$ and ${}^{{<}\delta }M\subseteq M$ . We define $\eta =\sup (\lambda \cap M)<\lambda $ and fix a function h with domain $\delta $ such that $[h]_U=A\cap \eta $ .

Note that, given a partial run of $G_{\delta +1}({\mathbb {P}})$ of even length less than $\delta $ that consists of conditions in M and was played according to $\sigma $ by Player Even, the given sequence is an element of M and Player Even responds to it with a move in M. Therefore, if $\tau $ is a strategy for Player Odd in $G_{\delta +1}({\mathbb {P}})$ that answers to sequences of conditions in M by playing a condition in M and $\langle {p_\xi }~\vert ~{\xi \leq \delta }\rangle $ is a run of $G_{\delta +1}({\mathbb {P}})$ played according to $\sigma $ and $\tau $ , then $p_\xi \in M$ for all $\xi <\delta $ . Moreover, the previous claim allows us to use elementarity to show for every $\xi \in \delta \setminus X$ and every condition $q\in M\cap {\mathbb {P}}$ with $q\leq _{\mathbb {P}} p_1$ , there is $\gamma \in M\cap \lambda $ and a condition $r\in M\cap {\mathbb {P}}$ with $r\leq _{\mathbb {P}} q$ and

(1)

Now, pick a strategy $\tau $ for Player Odd in $G_{\delta +1}({\mathbb {P}})$ with the following properties:

• • $\tau $ plays the condition $p_1$ in move $1$ .

• • Given $\xi \in X$ , if Player Even played a condition $q\in M\cap {\mathbb {P}}$ in move $(2+2\cdot \xi )$ , then $\tau $ responds by also playing the condition q in the next move.

• • Given $\xi \in \delta \setminus X$ , if Player Even played a condition $q\in M\cap {\mathbb {P}}$ in move $(2+2\cdot \xi )$ , then $\tau $ responds by playing a condition $r\in M\cap {\mathbb {P}}$ with $r\leq _{\mathbb {P}} q$ such that the equivalences of (1) hold true for some $\gamma \in M\cap \lambda $ .

Let $\langle {p_\xi }~\vert ~{\xi \leq \delta }\rangle $ be the run of $G_{\delta +1}({\mathbb {P}})$ played according to $\sigma $ and $\tau $ . By the above remarks, we then have $p_\xi \in M$ for all $\xi <\delta $ . In particular, for every $\xi \in \delta \setminus X$ , there exists $\gamma _\xi <\lambda $ with

Let G be ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ with $p_\delta \in G$ . Then the closure properties of ${\mathbb {P}}$ imply that $[h]_U=A\cap \eta $ holds in ${\mathrm {V}}[G]$ . Since $A=[\dot {f}^G]_U$ holds in ${\mathrm {V}}[G]$ , we know that the set

$$ \begin{align*}Y ~ = ~ \{{\xi<\delta}~\vert~{h(\xi)\textit{ is an initial segment of }\dot{f}^G(\xi)}\}\end{align*} $$

is an element of U. But then there is some $\xi \in Y\setminus X = Y \cap (\delta \setminus X)$ and our construction ensures that the ordinal $\gamma _\xi $ is contained in the symmetric difference of $h(\xi )$ and $\dot {f}^G(\xi )$ , a contradiction.⊣

Now, let G be ${\mathbb {P}}$ -generic over ${\mathrm {V}}$ with $p_1\in G$ . By the previous claim and the closure properties of ${\mathbb {P}}$ , we can find a function f with domain $\delta $ in ${\mathrm {V}}$ such that $[f]_U=[\dot {f}^G]_U=A$ holds in ${\mathrm {V}}[G]$ . Since forcing with ${\mathbb {P}}$ adds no new functions from $\delta $ to the ordinals, we can conclude that $[f]_U=A$ also holds in ${\mathrm {V}}$ , a contradiction as A was chosen to be fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ .⊣

Proof Assume, towards a contradiction, that there is an unbounded subset A of $\kappa $ that is fresh over ${\mathrm {Ult}}({{\mathrm {V}}},{U})$ . Note that, in ${\mathrm {V}}$ , our assumptions imply that $\mu ^\delta =\mu $ and hence Lemma 4.1 implies that $j_U(\kappa )=\kappa $ . In particular, for every $\gamma <\kappa $ , there is a function $f\in {\mathrm {H}}(\kappa )$ with domain $\delta $ and $[f]_U=\gamma $ . By our assumptions, there exists $G {\mathrm {Col}}({\mu },{{<}\kappa })^{\mathrm {W}}$ -generic over ${\mathrm {W}}$ with ${\mathrm {V}}={\mathrm {W}}[G]$ and hence we know that ${}^{{<}\mu }{\mathrm {W}}\subseteq {\mathrm {W}}$ . Moreover, since ${\mathrm {Col}}({\mu },{{<}\kappa })$ satisfies the $\kappa $ -chain condition in ${\mathrm {W}}$ , there exist ${\mathrm {Col}}({\mu },{{<}\kappa })$ -nice names $\dot {A}$ and $\dot {F}$ in ${\mathrm {W}}$ such that $\dot {A}^G=A$ and $\dot {F}^G$ is a function with domain $\kappa $ and the property that for all $\gamma <\kappa $ , the set ${\dot {F}^G(\gamma )}:{\delta }\longrightarrow {{\mathrm {H}}(\kappa )\cap {\mathcal {P}}({\kappa })}$ is a function with $[\dot {F}^G(\gamma )]_U=A\cap \gamma $ .

Work in ${\mathrm {W}}$ and pick an elementary submodel M of ${\mathrm {H}}(\kappa ^+)$ of cardinality $\kappa $ such that ${}^{{<}\kappa }M\subseteq M$ and $(\kappa +1)\cup \{\dot {A},\dot {F},U\}\subseteq M$ . In this situation, the weak compactness of $\kappa $ yields a transitive set N with ${}^{{<}\kappa }N\subseteq N$ and an elementary embedding ${j}:{M}\longrightarrow {N}$ with critical point $\kappa $ (see [Reference Hauser9, Theorem 1.3]).

Now, let $H_0$ be ${\mathrm {Col}}({\mu },{[\kappa ,j(\kappa ))})$ -generic over ${\mathrm {V}}$ . Then there is $H\in {\mathrm {V}}[H_0]$ that is ${\mathrm {Col}}({\mu },{{<}j(\kappa )})$ -generic over ${\mathrm {W}}$ with ${\mathrm {V}}[H_0]={\mathrm {W}}[H]$ and $G\subseteq H$ . In this situation, standard arguments (see [Reference Cummings, Foreman and Kanamori3, Proposition 9.1]) allow us to find an elementary embedding ${j_*}:{M[G]}\longrightarrow {N[H]}$ with $j_*\restriction M=j$ . Set $f=j_*(\dot {F}^G)(\kappa )$ . For any $\gamma < \gamma ^\prime < \kappa $ ,

$$ \begin{align*}\{{\xi<\delta}~\vert~{\dot{F}^G(\gamma)(\xi)\textit{ is an initial segment of } \dot{F}^G(\gamma^\prime)(\xi)}\}\in U.\end{align*} $$

So, given $\gamma <\kappa $ , elementarity implies that the set