Proof For the sake of contradiction, suppose S does not contain a club subset of $\kappa $ in V. Then $T=\kappa \setminus S$ is stationary and we may let G be generic for $P(\kappa )/{\mathop {\mathrm {NS}}}_{\kappa }$ with $T\in G$ . Then $\kappa \in j(T)$ , but this contradicts our assumption that $\kappa \in j(S)$ .

Proof

1. (1) It is easy to see that $D_0$ is closed below $\kappa $ . Let $j:V\to \mathop {\mathrm {Ult}}$ be any generic ultrapower obtained by forcing with $P(\kappa )/{\mathop {\mathrm {NS}}}_{\kappa }$ . Then $j(f^{\kappa }_{\xi })(\kappa )=\xi $ is a limit ordinal and hence $\kappa \in j(D_0)$ . By Proposition 2.2, $D_0$ is unbounded in $\kappa $ .

2. (2) Using that $j(f^{\kappa }_{\xi +1})(\kappa )=\xi +1$ , it follows that $\kappa \in j(D_1)$ , and the result thus follows by Proposition 2.2.

Observation 2.5. If $\varphi $ is a $\Pi ^1_{\xi }$ - or $\Sigma ^1_{\xi }$ -formula over $V_{\kappa }$ , and $\alpha <\kappa $ is regular, then whenever $\varphi \mathrm {|}^{\kappa }_{\alpha }$ is defined, it is a $\Pi ^1_{f^{\kappa }_{\xi }(\alpha )}$ - or $\Sigma ^1_{f^{\kappa }_{\xi }(\alpha )}$ -formula over $V_{\alpha }$ respectively.

Observation 2.7. Let $\kappa $ be a regular uncountable cardinal, and let $\xi <\kappa ^+$ . Let $\vec \pi =\langle \pi ^{\kappa }_{\xi ,\alpha }\mid \alpha <\kappa \rangle $ . Then $j(\vec \pi )_{\kappa }^{-1}=j\upharpoonright \xi $ .

Proof Since each $\pi ^{\kappa }_{\xi ,\alpha }$ is the transitive collapse of $F^{\kappa }_{\xi }(\alpha )$ , it follows by elementarity that $j(\vec \pi )_{\kappa }$ is the transitive collapse of $j(F^{\kappa }_{\xi })(\kappa )=j"\xi $ . Hence, the image of $j(\vec \pi )_{\kappa }$ is $\xi $ , and its inverse is thus clearly identical to $j\upharpoonright \xi $ .

Proof Regularity of $\kappa $ in $\mathop {\mathrm {Ult}}$ is needed so that $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }$ could possibly be defined in $\mathop {\mathrm {Ult}}$ . The proof proceeds by induction on $\xi <\kappa ^+$ . The case when $\xi <\kappa $ is easy, for then by Remark 2.6(1),

$$\begin{align*}j(\varphi(A_1,\ldots,A_n))=\varphi(j(A_1),\ldots,j(A_n)), \end{align*}$$

and thus $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }=\varphi $ by the definition of the restriction operation in this case. Successor steps above $\kappa $ are easily treated as well, for by Remark 2.6(2), in this case,

$$\begin{align*}j(\forall X\psi(X))=\forall X j(\psi(X)). \end{align*}$$

At limit steps $\xi \ge \kappa $ , if $\varphi =\bigwedge _{\zeta <\xi }\psi _{\zeta }$ is a $\Pi ^1_{\xi }$ -formula, let $\vec \psi =\langle \psi _{\zeta }\mid \zeta <\xi \rangle $ , and let $\vec \pi =\langle \pi ^{\kappa }_{\xi ,\alpha }\mid \alpha <\kappa \rangle $ . Then, by Remark 2.6(3), $j(\varphi )=\bigwedge _{\zeta <j(\xi )}j(\vec \psi )_{\zeta }$ , and therefore

$$\begin{align*}j(\varphi)\mathrm{|}^{j(\kappa)}_{\kappa}=\bigwedge_{\zeta\in f^{j(\kappa)}_{j(\xi)}(\kappa)}j(\vec\psi)_{j(\vec\pi)_{\kappa}^{-1}(\zeta)}\mathrm{|}^{j(\kappa)}_{\kappa}=\bigwedge_{\zeta\in\xi}(j(\psi_{\zeta}))\mathrm{|}^{j(\kappa)}_{\kappa}=\varphi,\end{align*}$$

using that $f^{j(\kappa )}_{j(\xi )}(\kappa )=j(f^{\kappa }_{\xi })(\kappa )=\xi $ by our choice of canonical functions at the level of $j(\kappa )$ in $\mathop {\mathrm {Ult}}$ , and by Observation 2.7.

The case when $\varphi $ is a $\Sigma ^1_{\xi }$ -formula is treated in exactly the same way.

Proof Note that $j(\Phi )(\kappa )=j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }=\varphi $ by Lemma 2.8.

Proof Assume for a contradiction that the conclusion of the lemma fails. This means that there is a stationary set T consisting of regular and uncountable cardinals $\alpha $ such that $\varphi \mathrm {|}^{\kappa }_{\alpha }$ is not defined. Assume that $T\in G$ . Then, $\kappa \in j(T)$ , and therefore $\kappa $ is regular in $\mathop {\mathrm {Ult}}$ ; however $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }$ is not defined in $\mathop {\mathrm {Ult}}$ . But, by Lemma 2.8, $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }=\varphi $ , which clearly yields a contradiction.

Proof By induction on $\xi <\kappa ^+$ . This is immediate in case $\xi <\kappa $ , for then by Remark 2.6(1), $j(\varphi (A_1,\ldots ,A_n))=\varphi (j(A_1),\ldots ,j(A_n))$ , and thus $j(\varphi )\mathrm {|}^{j(\kappa )}_{\alpha }=\varphi \mathrm {|}^{\kappa }_{\alpha }$ by the definition of the restriction operation in this case. It is also immediate for successor steps above $\kappa $ , for then by Remark 2.6(2), $j(\forall \vec X\psi )=\forall \vec X j(\psi )$ .

At limit steps $\xi \ge \kappa $ , if $\varphi =\bigwedge _{\zeta <\xi }\psi _{\zeta }$ is a $\Pi ^1_{\xi }$ -formula, let $\vec \psi =\langle \psi _{\zeta }\mid \zeta <\xi \rangle $ , and let $\vec \pi =\langle \pi ^{\kappa }_{\xi ,\alpha }\mid \alpha <\kappa \rangle $ . Then, by Remark 2.6(3), $j(\varphi )=\bigwedge _{\zeta <j(\xi )}j(\vec \psi )_{\zeta }$ , and therefore, assuming for now that $j(\varphi )\mathrm {|}^{j(\kappa )}_{\alpha }$ is defined,

$$\begin{align*}j(\varphi)\mathrm{|}^{j(\kappa)}_{\alpha}=\bigwedge_{\zeta\in j(f^{\kappa}_{\xi})(\alpha)}j(\vec\psi)_{j(\vec\pi)_{\alpha}^{-1}(\zeta)}\mathrm{|}^{j(\kappa)}_{\alpha}=\bigwedge_{\zeta\in j(f^{\kappa}_{\xi})(\alpha)}j(\psi_{j^{-1}(j(\vec\pi)_{\alpha}^{-1}(\zeta))})\mathrm{|}^{j(\kappa)}_{\alpha},\end{align*}$$

using that $j(\vec \pi )_{\alpha }^{-1}[j(f^{\kappa }_{\xi })(\alpha )]=j(F^{\kappa }_{\xi })(\alpha )\subseteq j(F^{\kappa }_{\xi })(\kappa )=j"\xi $ . By our inductive hypothesis, for each $\gamma \in \xi $ and every regular $\alpha <\kappa $ , $j(\psi _{\gamma })\mathrm {|}^{j(\kappa )}_{\alpha }=\psi _{\gamma }\mathrm {|}^{\kappa }_{\alpha }$ . Thus,

$$\begin{align*}j(\varphi)\mathrm{|}^{j(\kappa)}_{\alpha}=\bigwedge_{\zeta\in j(f^{\kappa}_{\xi})(\alpha)}\psi_{j^{-1}(j(\vec\pi)_{\alpha}^{-1}(\zeta))}\mathrm{|}^{\kappa}_{\alpha}. \end{align*}$$

Now,

$$\begin{align*}\varphi\mathrm{|}^{\kappa}_{\alpha}=\bigwedge_{\zeta\in f^{\kappa}_{\xi}(\alpha)}\psi_{(\pi^{\kappa}_{\xi,\alpha})^{-1}(\zeta)}\mathrm{|}^{\kappa}_{\alpha}. \end{align*}$$

Since $\alpha <\kappa $ we have $j(f^{\kappa }_{\xi })(\alpha )=f^{\kappa }_{\xi }(\alpha )$ , and furthermore

$$\begin{align*}(\pi^{\kappa}_{\xi,\alpha})^{-1}[f^{\kappa}_{\xi}(\alpha)]=F^{\kappa}_{\xi}(\alpha)=(j^{-1}\circ j(\vec\pi)_{\alpha}^{-1})[j(f^{\kappa}_{\xi})(\alpha)], \end{align*}$$

showing the above restrictions of $\varphi $ and of $j(\varphi )$ to be equal,Footnote ⁵ and thus in particular also showing that $j(\varphi )\mathrm {|}^{j(\kappa )}_{\alpha }$ is defined, as desired.

The case when $\varphi $ is a $\Sigma ^1_{\xi }$ -formula is treated in exactly the same way.

Proof Assume for a contradiction that the conclusion of the proposition fails. By Lemma 2.10, this means that there is a stationary set T consisting of regular and uncountable cardinals $\alpha $ such that the set $D_{\alpha }$ has stationary complement $E_{\alpha }\subseteq \alpha $ . Using Lemma 2.10 once again, we may assume that $(\varphi \mathrm {|}^{\kappa }_{\alpha })\mathrm {|}^{\alpha }_{\beta }$ is defined for every $\alpha \in T$ and every $\beta \in E_{\alpha }$ . Let $\vec E$ denote the sequence $\langle E_{\alpha }\mid \alpha \in T\rangle $ . Assume that $T\in G$ . Then, $\kappa \in j(T)$ , and thus $j(\vec E)_{\kappa }$ is stationary in $\mathop {\mathrm {Ult}}$ . But,

$$\begin{align*}j(\vec E)_{\kappa}=\{\beta<\kappa\mid(j(\varphi)\mathrm{|}^{j(\kappa)}_{\kappa})\mathrm{|}^{\kappa}_{\beta}\ne j(\varphi)\mathrm{|}^{j(\kappa)}_{\beta}\}.\end{align*}$$

Note that by Lemma 2.9, $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }=\varphi $ . But then, by Lemmas 2.10 and 2.11, $j(\vec E)_{\kappa }$ is nonstationary in $\mathop {\mathrm {Ult}}$ , which gives our desired contradiction.

Proof Suppose for a contradiction that the conclusion of the lemma fails. Using Lemma 2.10, this means that there is a stationary set $T\subseteq H$ such that for every $\alpha \in T$ ,

$$\begin{align*}\mathrm{code}^{\kappa}(\varphi)\cap\alpha\ne\mathrm{code}^{\alpha}(\varphi\mathrm{|}^{\kappa}_{\alpha}). \end{align*}$$

Assume $T\in G$ . Then, $\kappa \in j(T)$ , and hence in $\mathop {\mathrm {Ult}}$ , $\kappa $ is regular and

$$\begin{align*}\mathrm{code}^{j(\kappa)}(j(\varphi))\cap\kappa\ne\mathrm{ code}^{\kappa}(j(\varphi)\mathrm{|}^{j(\kappa)}_{\kappa}). \end{align*}$$

But by Lemma 2.8, this means that in $\mathop {\mathrm {Ult}}$ ,

(1) $$ \begin{align} \mathrm{code}^{j(\kappa)}(j(\varphi))\cap\kappa\ne\mathrm{code}^{\kappa}(\varphi). \end{align} $$

Let us show that (1) is false, thus yielding our desired contradiction. First, let us consider the case in which $\xi <\kappa $ . We have $j(\varphi (A_1,\ldots ,A_n))=\varphi (j(A_1),\ldots ,j(A_n))$ in this case, and by our choice of bijections we see that in $\mathop {\mathrm {Ult}}$ , $b^{j(\kappa )}\supseteq b^{\kappa }$ and hence first order parameters are coded in the same way. Furthermore, by our choice of reasonable coding, we observe that $\mathrm {code}^{j(\kappa )}(j(\varphi ))\cap \kappa $ and $\mathrm {code}^{\kappa }(\varphi )$ have the same slices and hence $\mathrm { code}^{j(\kappa )}(j(\varphi ))\cap \kappa = \mathrm {code}^{\kappa }(\varphi )$ .

Let us inductively look at the cases when $\xi \ge \kappa $ . The successor ordinal case is immediate, comparing all slices of $\mathrm {code}^{j(\kappa )}(j(\varphi ))\cap \kappa $ and of $\mathrm {code}^{\kappa }(\varphi )$ .

Assume now that $\xi $ is a limit ordinal. We will again be comparing the slices of $\mathrm {code}^{j(\kappa )}(j(\varphi ))\cap \kappa $ and of $\mathrm { code}^{\kappa }(\varphi )$ . The slices with index $0$ clearly agree. Our assumption, which we haven’t used yet, yields a formula $\psi $ of complexity $\xi $ such that in $\mathop {\mathrm {Ult}}$ ,

$$\begin{align*}\mathrm{code}^{\kappa}(\psi)=\mathrm{code}^{j(\kappa)}(j(\varphi))\cap\kappa.\end{align*}$$

The slices with index $1$ thus agree between $\mathrm { code}^{\kappa }(\psi )$ and $\mathrm {code}^{\kappa }(\varphi )$ , for they are coding the same well-ordering, and hence they also agree for our desired formulas. The remaining slices agree inductively, contradicting (1) as desired.

Proof Suppose $S\subseteq \kappa $ is subtle, $\vec {S}$ is an S-list, and suppose for a contradiction that $S\setminus A$ is subtle. Using that the set of inaccessible cardinals below $\kappa $ is in the subtle filter on $\kappa $ , we may assume that all elements of S are inaccessible. We will also assume our canonical functions at $\kappa $ to be based on the bijections $b_{\kappa ,\xi }$ used to define the coding at $\kappa $ in the above.

Suppose $\beta \in S\setminus A$ . Let $B_{\beta }=\{\alpha \in S\cap \beta \mid S_{\alpha }=S_{\beta }\cap \alpha \}$ . Since $B_{\beta }\cup \{\beta \}$ is homogeneous for $\vec {S}$ , it follows that for some limit ordinal $\xi _{\beta }$ with $\beta \le \xi _{\beta }<\beta ^+$ , the set $B_{\beta }$ is not $\Pi ^1_{\xi _{\beta }}$ -indescribable in $\beta $ . Let $\varphi ^{\beta }$ thus be a $\Pi ^1_{\xi _{\beta }}$ sentence over $V_{\beta }$ such that $V_{\beta }\models \varphi ^{\beta }$ , and for all $\alpha \in B_{\beta }$ , we have that $V_{\alpha }\not \models \varphi ^{\beta }\mathrm {|}^{\beta }_{\alpha }$ whenever $\varphi ^{\beta }\mathrm {|}^{\beta }_{\alpha }$ is defined.

For each $\beta \in S\setminus A$ , let $E_{\beta }$ code the pair

$$\begin{align*}\langle S_{\beta},\mathrm{code}^{\beta}(\varphi^{\beta})\rangle \end{align*}$$

as a subset of $\beta $ in a natural way. This defines an $(S\setminus A)$ -list $\vec {E}=\langle E_{\beta }\mid \beta \in S\setminus A\rangle $ . By Theorem 3.1, there is a Mahlo cardinal $\beta \in S\setminus A$ and a stationary set $H\subseteq (S\setminus A)\cap \beta $ such that $H\cup \{\beta \}$ is homogeneous for $\vec {E}$ . Let $\varphi =\varphi ^{\beta }$ , and let $\xi =\xi _{\beta }$ .

Recall that for each $\alpha \in S\setminus A$ , we have $(\mathrm {code}^{\alpha }(\varphi ^{\alpha }))_1=\Gamma [R^{\alpha }_{\xi _{\alpha }}]$ , and thus the homogeneity of $H\cup \{\beta \}$ implies that for all $\alpha \in H$ , we have

$$\begin{align*}R^{\beta}_{\xi}\cap(\alpha\times\alpha)=R^{\alpha}_{\xi_{\alpha}}, \end{align*}$$

and therefore

$$\begin{align*}\xi_{\alpha}=\mathop{\mathrm{ot}}\nolimits(\alpha,R^{\alpha}_{\xi_{\alpha}})=\mathop{\mathrm{ot}}\nolimits(\alpha,R^{\beta}_{\xi}\cap(\alpha\times\alpha))=\mathop{\mathrm{ ot}}\nolimits(b_{\beta,\xi}[\alpha])=f^{\beta}_{\xi}(\alpha). \end{align*}$$

Thus, for each $\alpha \in H$ , we have a $\Pi ^1_{f^{\beta }_{\xi }(\alpha )}$ -formula $\varphi ^{\alpha }$ over $V_{\alpha }$ such that

$$\begin{align*}\mathrm{code}^{\alpha}(\varphi^{\alpha})=\mathrm{code}^{\beta}(\varphi)\cap\alpha. \end{align*}$$

Therefore, by Lemma 3.3, there is a club $C\subseteq \beta $ such that $\varphi \mathrm {|}^{\beta }_{\alpha }$ is defined, and

$$\begin{align*}\mathrm{code}^{\beta}(\varphi)\cap\alpha=\mathrm{code}^{\alpha}(\varphi\mathrm{|}^{\beta}_{\alpha}) \end{align*}$$

for all $\alpha \in H\cap C\ne \varnothing $ . Fix some $\alpha \in H\cap C$ . We have

$$\begin{align*}\mathrm{code}^{\alpha}(\varphi^{\alpha})=\mathrm{code}^{\beta}(\varphi)\cap\alpha=\mathrm{code}^{\alpha}(\varphi\mathrm{|}^{\beta}_{\alpha}), \end{align*}$$

and hence $\varphi ^{\alpha }=\varphi \mathrm {|}^{\beta }_{\alpha }$ . However, since $S_{\beta }\cap \alpha =S_{\alpha }$ as well by the homogeneity of H, we have $\alpha \in B_{\beta }$ , and hence $V_{\alpha }\not \models \varphi \mathrm {|}^{\beta }_{\alpha }$ , contradicting that $V_{\alpha }\models \varphi ^{\alpha }$ .

Proof The only nontrivial property is normality. Note first that ${\mathcal I}(I)\supseteq {\mathcal I}([\kappa ]^{<\kappa })\supseteq {\mathop {\mathrm {NS}}}_{\kappa }$ , where the latter is due to an easy argument that may be found within the proof of [Reference Baumgartner3, Theorem 2.3].

Assume now that $A\in {\mathcal I}(I)^+$ . Let C be the club set of all ordinals below $\kappa $ that are closed under the Gödel pairing function $\Gamma $ . By the above, it follows that $A\cap C\in {\mathcal I}(I)^+$ , and we may thus assume that every element of A is closed under Gödel pairing.

Assume that $f\colon A\to \kappa $ is regressive, and let $A_{\alpha }=f^{-1}(\alpha )$ for every $\alpha <\kappa $ . Assume for a contradiction that $A_{\alpha }\in {\mathcal I}(I)$ for every $\alpha <\kappa $ . Thus, for every $\alpha <\kappa $ , we may fix an $A_{\alpha }$ -list $\vec A_{\alpha }=\langle a^{\alpha }_{\beta }\mid \beta \in A_{\alpha }\rangle $ which has no homogeneous set in $I^+$ .

Let $\vec A=\langle a_{\beta }\mid \beta \in A\rangle $ be an A-list defined by letting, for every $\beta \in A$ , $a_{\beta }$ code both $\{f(\beta )\}$ and $a^{f(\beta )}_{\beta }$ , using Gödel pairing. Since $A\in {\mathcal I}(I)^+$ , there is $H\in I^+$ that is homogeneous for $\vec A$ . It follows that H is homogeneous for f, and we let $\alpha $ be the value of f on H. It then follows that H is homogeneous for $\vec A_{\alpha }$ . This yields our desired contradiction.

Observation 3.7. For any cardinal $\kappa $ , the subtle ideal on $\kappa $ is contained in the weakly ineffable ideal on $\kappa $ .

Proof Assume that $A\subseteq \kappa $ is not an element of the weakly ineffable ideal ${\mathcal I}([\kappa ]^{<\kappa })$ on $\kappa $ , that $\vec a$ is an A-list, and let C be a club subset of $\kappa $ . By Lemma 3.6, $A\cap C\in {\mathcal I}([\kappa ]^{<\kappa })^+$ . It follows that we can find $H\subseteq A\cap C$ that is homogeneous for H. This shows that A is subtle.

Proof We proceed by induction on $\gamma <\kappa ^+$ . When $\gamma =0$ , the result follows directly from Theorem 3.4, because the subtle ideal is contained in the weakly ineffable ideal ${\mathcal I}([\kappa ]^{<\kappa })$ , and $f^{\kappa }_0(\alpha )=0$ for all $\alpha <\kappa $ .

Suppose $\gamma =\delta +1<\kappa ^+$ is a successor ordinal, and suppose for a contradiction that $S\setminus A\in {\mathcal I}^{\delta +2}([\kappa ]^{<\kappa })^+$ . Let $C=\{\alpha <\kappa \mid f^{\kappa }_{\delta +1}(\alpha )=f^{\kappa }_{\delta }(\alpha )+1\}$ be the club subset of $\kappa $ obtained from Lemma 2.3. Then, the set

$$\begin{align*}E=\{\alpha\in S\setminus A\mid \alpha\text{ is inaccessible}\}\cap C \end{align*}$$

is in ${\mathcal I}^{\delta +2}([\kappa ]^{<\kappa })^+$ . For each $\alpha \in E$ , let $B_{\alpha }=\{\beta \in S\cap \alpha \mid S_{\beta }=S_{\alpha }\cap \beta \}$ . Since $B_{\alpha }\cup \{\alpha \}$ is homogeneous for $\vec {S}$ and $\alpha \in S\setminus A$ , there is an ordinal $\xi _{\alpha }<\alpha ^+$ such that $B_{\alpha }\in {\mathcal I}^{f^{\kappa }_{\delta }(\alpha )+1}(\Pi ^1_{\xi _{\alpha }}(\alpha ))$ , and hence we may fix a $B_{\alpha }$ -list $\vec {B}^{\alpha }=\langle b^{\alpha }_{\beta }\mid \beta \in B_{\alpha }\rangle $ such that $\vec {B}^{\alpha }$ has no homogeneous set in ${\mathcal I}^{f^{\kappa }_{\delta }(\alpha )}(\Pi ^1_{\xi _{\alpha }}(\alpha ))^+$ .

For $\alpha \in E$ , let $E_{\alpha }$ code the triple $\langle S_{\alpha },B_{\alpha },\vec {B}^{\alpha }\rangle $ as a subset of $\alpha $ in a natural way. This defines an E-list $\vec {E}=\langle E_{\alpha }\mid \alpha \in E\rangle $ . Since $E\in {\mathcal I}^{\delta +2}([\kappa ]^{<\kappa })^+$ , we may fix $H\in \mathcal P(E)\cap {\mathcal I}^{\delta +1}([\kappa ]^{<\kappa })^+$ which is homogeneous for $\vec {E}$ . It follows that H is homogeneous for $\langle S_{\alpha }\mid \alpha \in E\rangle $ , $\langle B_{\alpha }\mid \alpha \in E\rangle $ , and $\langle \vec {B}^{\alpha }\mid \alpha \in E\rangle $ . We let $D=\bigcup _{\alpha \in H} S_{\alpha }$ , $B=\bigcup _{\alpha \in H}B_{\alpha }$ , and $\vec {B}=\bigcup _{\alpha \in H}\vec {B}^{\alpha }=\langle b_{\alpha }\mid \alpha \in B\rangle $ . Since $B=\{\alpha <\kappa \mid S_{\alpha }=D\cap \alpha \}$ , it follows that $H\subseteq B$ .

Let $A_0$ be the set of all ordinals $\alpha \in H$ such that

$$\begin{align*}\exists X\subseteq H\cap\alpha\left[(\forall\xi<\alpha^+\ X\in {\mathcal I}^{f^{\kappa}_{\delta}(\alpha)}(\Pi^1_{\xi}(\alpha))^+) \land (X\cup\{\alpha\}\text{ is hom. for }\vec{B})\right].\end{align*}$$

By our inductive hypothesis, $H\setminus A_0\in {\mathcal I}^{\delta +1}([\kappa ]^{<\kappa })$ , and hence $A_0\in {\mathcal I}^{\delta +1}([\kappa ]^{<\kappa })^+$ . Thus, we may fix an $\alpha \in A_0$ . Since $\alpha \in H$ , it follows by homogeneity that $\vec {B}\upharpoonright (H\cap \alpha )=\vec {B}^{\alpha }\upharpoonright H$ . But by the definition of $A_0$ , and since $\xi _{\alpha }<\alpha ^+$ , there is some $X\in \mathcal P(H\cap \alpha )\cap {\mathcal I}^{f^{\kappa }_{\delta }(\alpha )}(\Pi ^1_{\xi _{\alpha }}(\alpha ))^+$ which is homogeneous for $\vec {B}^{\alpha }$ , which is a contradiction.

Now let us suppose $\gamma <\kappa ^+$ is a limit ordinal, and suppose again for a contradiction that $S\setminus A\in {\mathcal I}^{\gamma +1}([\kappa ]^{<\kappa })^+$ . Since by Lemma 2.3, the set

$$\begin{align*}C=\{\alpha<\kappa\mid f^{\kappa}_{\gamma}(\alpha)\text{ is a limit ordinal and } f^{\kappa}_{\gamma}(\alpha)=\bigcup_{\delta\in F^{\kappa}_{\gamma}(\alpha)} f^{\kappa}_{\delta}(\alpha) \} \end{align*}$$

is in the club filter on $\kappa $ , it follows that the set

$$\begin{align*}E=\{\alpha\in S\setminus A\mid \alpha \text{ is inaccessible}\}\cap C \end{align*}$$

is in ${\mathcal I}^{\gamma +1}([\kappa ]^{<\kappa })^+$ .

For each $\alpha \in E$ , let $B_{\alpha }=\{\beta \in S\cap \alpha \mid S_{\beta }=S_{\alpha }\cap \beta \}$ . Since $B_{\alpha }\cup \{\alpha \}$ is homogeneous for $\vec {S}$ , and $\alpha \in S\setminus A$ , there is some $\xi _{\alpha }<\alpha ^+$ such that $B_{\alpha }\in {\mathcal I}^{f^{\kappa }_{\gamma }(\alpha )}(\Pi ^1_{\xi _{\alpha }}(\alpha ))$ . Since $\alpha \in C$ , we have

$$\begin{align*}B_{\alpha}\in{\mathcal I}^{f^{\kappa}_{\gamma}(\alpha)}(\Pi^1_{\xi_{\alpha}}(\alpha))=\bigcup_{\delta\in F^{\kappa}_{\gamma}(\alpha)}{\mathcal I}^{f^{\kappa}_{\delta}(\alpha)}(\Pi^1_{\xi_{\alpha}}(\alpha))=\bigcup_{\beta<\alpha}{\mathcal I}^{f^{\kappa}_{b_{\kappa,\gamma}(\beta)}(\alpha)}(\Pi^1_{\xi_{\alpha}}(\alpha)). \end{align*}$$

Using that $f^{\kappa }_{\gamma }(\alpha )$ is a limit ordinal once again, we may choose an ordinal $g(\alpha )<\alpha $ such that

$$\begin{align*}B_{\alpha}\in{\mathcal I}^{f^{\kappa}_{b_{\kappa,\gamma}(g(\alpha))}(\alpha)+1}(\Pi^1_{\xi_{\alpha}}(\alpha)).\end{align*}$$

This defines a regressive function $g:E\to \kappa $ , and by the normality of ${\mathcal I}^{\gamma +1}([\kappa ]^{<\kappa })^+$ that follows from Lemma 3.6, there is an $E^*\in \mathcal P(E)\cap {\mathcal I}^{\gamma +1}([\kappa ]^{<\kappa })^+$ and some $\beta _0<\kappa $ such that $g(\alpha )=\beta _0$ for all $\alpha \in E^*$ . Let $\nu =b_{\kappa ,\gamma }(\beta _0)$ and notice that for all $\alpha \in E^*$ ,

$$\begin{align*}B_{\alpha}\in{\mathcal I}^{f^{\kappa}_{\nu}(\alpha)+1}(\Pi^1_{\xi_{\alpha}}(\alpha)). \end{align*}$$

For each $\alpha \in E^*$ , we fix a $B_{\alpha }$ -list $\vec {B}^{\alpha }=\langle b^{\alpha }_{\beta }\mid \beta \in B_{\alpha }\rangle $ such that $\vec {B}^{\alpha }$ has no homogeneous set in ${\mathcal I}^{f^{\kappa }_{\nu }(\alpha )}(\Pi ^1_{\xi _{\alpha }}(\alpha ))^+$ .

Now we define an $E^*$ -list by letting $E^*_{\alpha }$ code the triple $\langle S_{\alpha },B_{\alpha },\vec {B}^{\alpha }\rangle $ as a subset of $\alpha $ in a natural way, for all $\alpha \in E^*$ . This defines $\vec {E}^*=\langle E^*_{\alpha }\mid \alpha \in E^*\rangle $ . Since $E^*\in {\mathcal I}^{\gamma +1}([\kappa ]^{<\kappa })^+$ , we may fix an $H\in \mathcal P(E^*)\cap {\mathcal I}^{\gamma }([\kappa ]^{<\kappa })^+$ which is homogeneous for $\vec {E}^*$ . Then, H is homogeneous for $\langle S_{\alpha }\mid \alpha \in S\rangle $ , $\langle B_{\alpha }\mid \alpha \in E^*\rangle $ and $\langle \vec {B}^{\alpha }\mid \alpha \in E^*\rangle $ . We let $D=\bigcup _{\alpha \in H}S_{\alpha }$ , $B=\bigcup _{\alpha \in H}B_{\alpha }$ and $\vec {B}=\bigcup _{\alpha \in H}\vec {B}^{\alpha }=\langle b_{\alpha }\mid \alpha \in B\rangle $ . Since $B=\{\alpha <\kappa \mid S_{\alpha }=D\cap \alpha \}$ , it follows that $H\subseteq B$ .

Now since $\nu <\gamma $ , we have $H\in {\mathcal I}^{\gamma }([\kappa ]^{<\kappa })^+\subseteq {\mathcal I}^{\nu +1}([\kappa ]^{<\kappa })^+$ , and we may apply the inductive hypothesis to the H-list $\vec {B}\upharpoonright H$ . Let $A_0$ be the set of all ordinals $\alpha \in H$ such that

$$\begin{align*}\exists X\subseteq H\cap\alpha\left[(\forall\xi<\alpha^+\ X\in{\mathcal I}^{f^{\kappa}_{\nu}(\alpha)}(\Pi^1_{\xi}(\alpha))^+)\land (X\cup\{\alpha\}\text{ is hom. for }\vec{B}\upharpoonright H)\right]. \end{align*}$$

It follows that $H\setminus A_0\in {\mathcal I}^{\nu +1}([\kappa ]^{<\kappa })$ , which implies that $A_0\in {\mathcal I}^{\nu +1}([\kappa ]^{<\kappa })^+$ . Fix $\alpha \in A_0$ . Since $\alpha \in H$ , homogeneity implies that $\vec {B}\upharpoonright (H\cap \alpha )=\vec {B}^{\alpha }\upharpoonright H$ . But, by the definition of $A_0$ , and the fact that $\xi _{\alpha }<\alpha ^+$ , it follows that there is some $X\in \mathcal P(H\cap \alpha )\cap {\mathcal I}^{f^{\kappa }_{\nu }}(\Pi ^1_{\xi _{\alpha }}(\alpha ))^+$ which is homogeneous for $\vec {B}^{\alpha }$ , a contradiction.

Proof Suppose $S\setminus A\in {\mathcal S}([\kappa ]^{<\kappa })^+$ . Let M be a $\kappa $ -model with $S\setminus A,f\in M$ and let $U\subseteq [\kappa ]^{\kappa }$ be a $\kappa $ -amenable M-normal M-ultrafilter such that $S\setminus A\in U$ . Let $j:M\to N$ be the usual elementary embedding obtained from U such that N is transitive.

Since U is $\kappa $ -amenable, it follows that for every $B\subseteq \kappa ^n\times \kappa $ in M, the set $\{\vec {\alpha }\in \kappa ^n\mid B_{\vec {\alpha }}\in U\}$ is in M where $B_{\vec {\alpha }}=\{\beta <\kappa \mid \vec {\alpha }\mathbin {{}^{\smallfrown }}\beta \in B\}$ . Thus, we can define the product ultrafilters $U^n$ on $P(\kappa ^n)^M$ by induction as follows. For $B\subseteq \kappa ^n\times \kappa $ , we let $B\in U^{n+1}=U^n\times U$ if and only if $B\in M$ and $\{\vec {\alpha }\in \kappa ^n\mid B_{\vec {\alpha }}\in U\}\in U^n$ . For each $n\in \omega \setminus \{0\}$ , we let $j_{U^n}:M\to N_{U^n}$ be the ultrapower of M by $U^n$ and note that, it follows from [Reference Gitman10, Proposition 2.32] that $N_{U^n}$ is well-founded. Furthermore, by [Reference Gitman10, Lemma 2.31], we have $j_{U^{n+1}}=j_{j_{U^n}(U)}\circ j_{U^n}$ where $j_{j_{U^n}(U)}$ is the ultrapower of $N_{U^n}$ by $j_{U^n}(U)$ (for more details on product ultrafilters one may consult [Reference Gitman10, Chapter 2] or [Reference Gitman11]).

For each $n\in \omega \setminus \{0\}$ , let $f_n=f\upharpoonright [S]^n$ . Since f is regressive, it follows by elementarity that the ordinal $\gamma _1=j_{U^1}(f)(\{\kappa \})$ is less than $\kappa $ . Furthermore, for each $n<\omega $ the ordinal $\gamma _{n+2}=j_{U^{n+2}}(f)(\{\kappa ,j_U(\kappa ),j_{U^2}(\kappa ),\ldots ,j_{U^{n+1}}(\kappa )\})$ is less than $\kappa $ . Fix $n\in \omega \setminus \{0\}$ . Let $A_n=\{\vec {\alpha }\in [S]^n\mid f_n(\vec {\alpha })=\gamma _n\}$ . By [Reference Gitman10, Lemma 2.34], there is a $B_n\in U$ such that for all $\beta _1<\cdots <\beta _n$ in $B_n$ we have $(\beta _1,\ldots ,\beta _n)\in A_n$ , that is, $f_n(\beta _1,\ldots ,\beta _n)=\gamma _n$ . Hence $B_n$ is homogeneous for $f_n$ .

Clearly $B=\bigcap _{n<\omega }B_n$ is homogeneous for f, and since M is a $\kappa $ -model and U is M-normal, it follows that $B\in U$ and hence $\kappa \in j(B)$ . Now we have $B\in N$ and furthermore, N thinks that B is homogeneous for f. But, since $\kappa \in j(S\setminus A)$ , it follows that N thinks that there are no subsets of S which are both $\Pi ^1_{\eta }$ -indescribable in $\kappa $ for all $\eta <(\kappa ^+)^N$ and homogeneous for f. Hence, in N, there must be some $\eta <(\kappa ^+)^N$ such that B is not $\Pi ^1_{\eta }$ -indescribable in $\kappa $ . Working in N, fix a $\Pi ^1_{\eta }$ -sentence $\varphi $ over $V_{\kappa }$ that is true in $V_{\kappa }$ such that for all $\alpha \in B$ we have $V_{\alpha }\models \lnot \varphi \mathrm {|}^{\kappa }_{\alpha }$ . Since M and N are both $\kappa $ -models and since $P(\kappa )^M=P(\kappa )^N$ , it follows that $(V_{\kappa }\models \varphi )^M$ and

$$\begin{align*}((\forall\alpha\in B)\ V_{\alpha}\models\lnot\varphi\mathrm{|}^{\kappa}_{\alpha})^M. \end{align*}$$

Hence by elementarity,

$$\begin{align*}((\forall\alpha\in j(B))\ V_{\alpha}\models\lnot j(\varphi)\mathrm{|}^{j(\kappa)}_{\alpha})^N. \end{align*}$$

But this is a contradiction because $\kappa \in j(B)$ and $(V_{\kappa }\models \varphi )^N$ where $j(\varphi )\mathrm {|}^{j(\kappa )}_{\kappa }=\varphi $ (see the proof of Lemma 2.8).