Fact 2.8 (Feng and Jech)

Let $\kappa \ge \omega _2$ be an uncountable regular cardinal. Then a stationary set $S\subseteq [H_\kappa ]^\omega $ is projective stationary iff ${\mathord {\mathbb P}}_S\in \mathsf {SSP} $ .

Fact 2.16. Let $\kappa $ be an uncountable regular cardinal. A stationary set $S\subseteq [H_\kappa ]^\omega $ is spread out iff it is weakly spread out.

Observation 2.18 [Reference Fuchs10, Observation 2.28]

If a set $S\subseteq [D]^\omega $ is spread out in D, with $\omega _1\subseteq D$ , then S is projective stationary in D.

Observation 2.19. Let $\kappa $ be an uncountable regular cardinal, let $S\subseteq [H_\kappa ]^\omega $ be spread out, and let $C\subseteq [H_\kappa ]^\omega $ be club. Then $S\cap C$ is spread out.

Observation 2.20. Let $A\subseteq B\subseteq C$ , and suppose S is spread out in B. Then both $S\mathbin {\downarrow }[A]^\omega $ and $S\mathbin {\uparrow }[C]^\omega $ are spread out.

Observation 3.2. Let $\mathcal {S}$ , $\kappa $ , ${\vec {T}}$ be as in Definition 3.1, and suppose that $ {\langle \vec {Q},{\vec {S}} \rangle } $ is a diagonal chain through $\mathcal {S}$ up to X with respect to ${\vec {T}}$ . Then $:$

1. (1) $\vec {Q}$ is a diagonal chain through $\mathcal {S}$ up to X.

2. (2) If $\bigcup _{i<\omega _1}T_i=\omega _1$ , then $\vec {Q}$ is an exact diagonal chain through $\mathcal {S}$ up to X.

3. (3) If $\kappa \ge \omega _2$ is regular, $ {\langle H_\kappa \cap X,\in \rangle } \prec {\langle H_\kappa ,\in \rangle } $ , ${\vec {T}}\in X$ and $\bigcup _{i<\omega _1}T_i$ contains a club, then there is a diagonal chain $ {\langle {\vec {R}},{\vec {S}} \rangle } $ through $\mathcal {S}$ up to X with respect to some ${\langle \bar {T}_i\;|\;} {i<\omega _1\rangle }$ such that $\bigcup \bar {T}_i=\omega _1$ . Hence, ${\vec {R}}$ is an exact diagonal chain through $\mathcal {S}$ up to X.

Proof We outline the straightforward proof of (3). By elementarity of $H_\kappa \cap X$ , and since ${\vec {T}}\in X$ , it follows that there is a club $C\subseteq \bigcup _{i<\omega _1}T_i$ in X. Hence, the monotone enumeration f of C is also in X. Define for $i<\omega _1$ :

$$\begin{align*}R_i=Q_{f(i)},\ \bar{T}_i=f^{-1}"T_i.\end{align*}$$

It is then easy to check that $\vec {{\bar {T}}}$ is a partition of $\omega _1$ into stationary sets and $ {\langle {\vec {R}},{\vec {S}} \rangle } $ is a diagonal chain through $\mathcal {S}$ with respect to $\vec {{\bar {T}}}$ , as wished. Since $f\in X$ and for all $\alpha <\omega _1$ , $\vec {Q}{\restriction }\alpha \in X$ , it follows that for all $\alpha <\omega _1$ , ${\vec {R}}{\restriction }\alpha \in X$ , as $ {\langle H_\kappa \cap X,\in \rangle } \prec {\langle H_\kappa ,\in \rangle } $ .

Fact 3.4. Let $\kappa $ be an uncountable regular cardinal, $\emptyset \neq \mathcal {S}\subseteq {\mathcal {P}}([H_\kappa ]^\omega )$ a collection of stationary subsets, and ${\vec {T}}$ an $\omega _1$ -sequence of pairwise disjoint stationary subsets of $\omega _1$ , and let ${\mathord {\mathbb P}}={\mathord {\mathbb P}}^{\mathsf {DSRP}}_{\mathcal {S},{\vec {T}}}$ . Then $:$

1. (1) For every countable ordinal $\gamma $ , the set of conditions p with $\delta ^p,\lambda ^p\ge \gamma $ is dense in ${\mathord {\mathbb P}}_S$ .

2. (2) For every $a\in H_\kappa $ , the set of conditions p such that there is an $\alpha \le \delta ^p$ with $a\in Q^p_\alpha $ is dense in ${\mathord {\mathbb P}}$ .

3. (3) For every $S\in \mathcal {S}$ , the set of conditions p such that there is an $i<\lambda ^p$ such that $S^p_i=S$ is dense.

Proof We need some facts before being able to prove this. The first fact is a generalization of a result from [Reference Friedman8].

Fact 1: Let ${\langle A_i\;|\;} {i<\omega _1\rangle }$ be a sequence of stationary subsets of $\omega _1$ , and let $t:\omega _1\longrightarrow \omega _1$ be a function. Then, for any $\beta ,\alpha <\omega _1$ , with $\alpha>0$ , there is a normal (that is, strictly increasing and continuous) function $f:\alpha \longrightarrow \omega _1$ such that for all $\xi <\alpha $ , $f(\xi )\in A_{t(\xi )}$ and $f(0)>\beta $ .

Proof of Fact 1

Let $\gamma ,\alpha ,\sigma $ be countable ordinals, $\alpha $ a limit. Say that $\gamma $ is $\alpha $ -approachable from $\sigma $ if for every $\delta <\gamma $ , there is a normal function $f:[\sigma ,\sigma +\alpha ]\longrightarrow [\delta ,\gamma ]$ such that for all $\xi \in [\sigma ,\sigma +\alpha ]$ , $f(\xi )\in A_{t(\xi ),}$ and $f(\sigma +\alpha )=\gamma $ . We refer to such a function as a nice function from $[\sigma ,\sigma +\alpha ]$ to $[\delta ,\gamma ]$ .

We will prove by induction on limit ordinals $\alpha <\omega _1$ : for every $\sigma <\omega _1$ , the set of $\lambda <\omega _1$ such that $\lambda $ is $\alpha $ -approachable from $\sigma $ is unbounded in $\omega _1$ . This clearly proves Fact 1.

If $\alpha =\omega $ , then let $\sigma <\omega _1$ be given. Fixing any $\beta <\omega _1$ , we have to find a countable $\lambda \ge \beta $ that is $\alpha $ -approachable from $\sigma $ . To this end, let

$$\begin{align*}\lambda\in \left(A_{t(\sigma+\omega)}\cap\bigcap_{\sigma\le\xi<\sigma+\omega}{\mathrm Lim}(A_{t(\xi)})\right)\setminus\beta.\end{align*}$$

Given any $\delta <\lambda $ , it is then easy to define $f:[\sigma ,\sigma +\omega ]\longrightarrow [\delta ,\lambda ]$ recursively so that f is strictly increasing, $f(\sigma )>\delta $ , for $\xi <\omega $ , $f(\xi )\in A_{t(\xi )}$ , and $\sup \{f(\sigma +n)\;| n<\omega \}=\lambda $ . Thus, setting $f(\sigma +\omega )=\lambda $ yields a nice function from $[\sigma ,\sigma +\omega ]$ to $[\delta ,\lambda ]$ , as wished.

Now suppose this has been proven for $\alpha $ . We have to show the claim for $\alpha +\omega $ . To this end, fix $\sigma <\omega _1$ . Given an arbitrary $\beta <\omega _1$ , we have to find a countable $\lambda \ge \beta $ which is $\alpha +\omega $ -approachable from $\sigma $ . Let $D=\{\gamma <\omega _1\;|\;\gamma $ is $\alpha $ -approachable from $\sigma $ }. Inductively, this set is unbounded in $\omega _1$ . Let

$$\begin{align*}\lambda\in\left(A_{t(\sigma+\alpha+\omega)}\cap{\mathrm Lim}(D)\cap\bigcap_{\sigma+\alpha\le\xi<\sigma+\alpha+\omega}{\mathrm Lim}(A_{t(\xi)})\right)\setminus\beta.\end{align*}$$

To see that $\lambda $ is $\alpha +\omega $ -approachable from $\sigma $ , let $\delta <\lambda $ . Let $\gamma \in (D\cap \lambda )\setminus (\delta +1)$ . Since $\gamma $ is $\alpha $ -approachable from $\sigma $ , there is a nice function ${\bar {f}}$ from $[\sigma ,\sigma +\alpha ]$ to $[\delta ,\gamma ]$ . As in the case $\alpha =\omega $ , we can extend ${\bar {f}}$ to a normal and cofinal function ${\bar {f}}':[\sigma ,\sigma +\alpha +\omega )\longrightarrow \lambda $ , such that for each $n<\omega $ , ${\bar {f}}'(\sigma +\alpha +n)\in A_{t(\sigma +\alpha +n)}$ . Since $\lambda =\sup {\mathrm ran} ({\bar {f}}')$ and $\lambda \in A_{t(\sigma +\alpha +\omega )}$ , we can extend ${\bar {f}}'$ to a nice function from $[\sigma ,\sigma +\alpha +\omega ]$ to $[\delta ,\lambda ]$ by specifying that $f(\sigma +\alpha +\omega )=\lambda $ .

Finally, suppose $\alpha $ is a limit of limit ordinals, and the claim has been proven for all limit ordinals below $\alpha $ . Fixing $\sigma ,\beta <\omega _1$ , we have to find a countable $\lambda>\beta $ which is $\alpha $ -approachable from $\sigma $ . Let ${\langle s_n\;|\;} {n<\omega \rangle }$ be increasing and cofinal in $\alpha $ , $s_0=0$ . Let $\sigma _n=\sigma +s_n$ . Let

$$\begin{align*}D_n=\{\gamma<\omega_1\;|\;\gamma\ \text{is }(s_{n+1}-s_n)\text{-approachable from }\sigma_n\}.\end{align*}$$

Inductively, $D_n$ is unbounded in $\omega _1$ , for each $n<\omega $ . Let

$$\begin{align*}\lambda\in\left(A_{t(\sigma+\alpha)}\cap\bigcap_{n<\omega}{\mathrm Lim}(D_n)\right)\setminus\beta.\end{align*}$$

To see that $\lambda $ is $\alpha $ -approachable from $\sigma $ , fix $\delta <\lambda $ . Find an increasing sequence ${\langle \delta _n\;|\;} {n<\omega \rangle }$ cofinal in $\lambda $ such that $\delta _0>\delta $ and $\delta _n\in D_n$ , for all $n<\omega $ . Using the definition of $D_n$ , we can now find a sequence of functions ${\langle f_n\;|\;} {n<\omega \rangle }$ such that:

• • $f_n$ is a nice function from $[\sigma _n,\sigma _{n+1}]$ to $[\delta _n,\delta _{n+1}]$ .

• • $f_{n+1}(\sigma _{n+1})=\delta _{n+1}=f_n(\sigma _{n+1})$ .

Thus, the union $\bigcup _{n<\omega }f_n$ is a function that can be extended to a nice function from $[\sigma ,\sigma +\alpha ]$ to $[\delta ,\lambda ]$ by mapping $\sigma +\alpha $ to $\lambda $ .

Fact 2: If ${\langle S_i\;|\;} {i<\omega _1\rangle }$ is a sequence of stationary subsets of $[H_\kappa ]^\omega $ and ${\langle T_i\;|\;} {i<\omega _1\rangle }$ is a sequence of pairwise disjoint stationary subsets of $\omega _1$ , then for any $\gamma <\omega _1$ , there is a continuous $\in $ -chain ${\langle Q_\alpha \;|\;} {\alpha <\gamma \rangle }$ such that for all $\alpha <\gamma $ , if $\alpha \in T_i$ , then $Q_\alpha \in S_i$ .

Proof of Fact 2

This is a strengthening of [Reference Feng and Jech5, Lemma 1.2], and the proof of that lemma can be adapted to the present situation. Let ${\mathord {\mathbb Q}}$ be the forcing to add a continuous $\in $ -chain of countable subsets of $H_\kappa $ , of length $\omega _1$ , by initial segments of successor length. This forcing is $\sigma $ -closed. Let ${\langle Q_\alpha \;|\;} {\alpha <\omega _1\rangle }$ be a sequence added by ${\mathord {\mathbb Q}}$ , i.e., $\vec {Q}=\bigcup G$ , for some ${\mathord {\mathbb Q}}$ -generic G. In $V[G]$ , every $S_i$ is still stationary, so the set $A_i=\{\alpha <\omega _1\;|\; Q_\alpha \in S_i\}$ is stationary in $V[G]$ . So by Fact 1, applied in $V[G]$ to the function $t:\omega _1\longrightarrow \omega _1$ defined by

$$\begin{align*}t(\xi)=\left\{ \begin{array}{l@{\qquad}l} i, & \text{if }\xi\in T_i,\\ 0, & \text{if }\xi\in\omega_1\setminus\bigcup_{j<\omega_1}T_j, \end{array} \right.\end{align*}$$

there is a normal function $f:\gamma \longrightarrow \omega _1$ such that for all $\alpha <\gamma $ , $f(\alpha )\in A_{t(\alpha )}$ , which means that if $\alpha \in T_i$ , then $f(\alpha )\in A_i$ , and this means that $Q_{f(\alpha )}\in S_i$ . So, the sequence $\vec {Q}'={\langle Q_{f(\alpha )}\;|\;} {\alpha <\gamma \rangle }$ is as wished, and it belongs to V, since ${\mathord {\mathbb Q}}$ is countably distributive.

We can now prove clauses (1), (2), and (3) simultaneously. Fix a condition $p\in {\mathord {\mathbb P}}_S$ , and let $\gamma <\omega _1$ , $a\in H_\kappa $ and $S\in \mathcal {S}$ be given. We may assume that $\gamma>\delta ^p$ . We may also assume that $\delta ^p+1\in \bigcup _{i<\omega _1}T_i$ , for if not, then we may just define ${\vec {T}}'$ to be like ${\vec {T}}$ , except that $\delta ^p+1\in T^{\prime }_0$ , say. Then $p\in {\mathord {\mathbb P}}^{\mathsf {DSRP}}_{\mathcal {S},{\vec {T}}'}$ , and an extension of p in ${\mathord {\mathbb P}}^{\mathsf {DSRP}}_{\mathcal {S},{\vec {T}}'}$ with the desired properties is also an extension of p in ${\mathord {\mathbb P}}^{\mathsf {DSRP}}_{\mathcal {S},{\vec {T}}}$ . So let’s let $i_0$ be such that $\delta ^p+1\in T_{i_0}$ .

Since it is trivial to extend the second coordinate of a condition, we may assume that $\lambda ^p>\gamma $ , that for every $\alpha \le \gamma $ , if $i<\omega _1$ is such that $\alpha \in T_i$ , then $i<\lambda ^p$ , and that there is some $i<\lambda ^p$ such that $S^p_i=S$ , taking care of clause (3). In order to be able to use Fact 2 now, we have to perform a little index translation, shifting by $\delta ^p+1$ . Thus, let’s define a sequence ${\vec {T}}'={\langle T^{\prime }_i\;|\;} {i<\omega _1\rangle }$ by letting $T^{\prime }_i=\{\xi <\omega _1\;|\;(\delta ^p+1)+\xi \in T_i\}$ . Let’s also define ${\vec {S}}={\langle S_i\;|\;} {i<\omega _1\rangle }$ by

$$\begin{align*}S_i=\left\{ \begin{array}{l@{\qquad}l} S^p_i,&\text{if}\ i<\lambda^p,\ i\neq i_0,\\ \{x\in S^p_{i_0}\;|\; \{a, Q^p_{\delta^p}\}\subseteq x\}, & \text{if}\ i=i_0,\\ S^p_0,&\text{if}\ i\ge\lambda^p. \end{array} \right. \end{align*}$$

Clearly, ${\vec {S}}$ is a sequence of stationary subsets of $[H_\kappa ]^\omega $ , and ${\vec {T}}'$ is a sequence of pairwise disjoint stationary subsets of $\omega _1$ , so we may apply Fact 2 to give us a continuous $\in $ -chain ${\langle R_\xi \;|\;} {\xi <{\bar {\gamma }}\rangle }$ , where ${\bar {\gamma }}=(\gamma +1)-(\delta ^p+1)$ , such that for all $i<{\bar {\gamma }}$ , if $i\in T^{\prime }_j$ , then $R_i\in S_j$ . The condition $q= {\langle \vec {Q}^p{{}^\frown }{\vec {R}},{\vec {S}}^p \rangle } $ is then an extension of p with all the desired properties. This is because for $i<{\bar {\gamma }}$ , if $\delta ^p+1+i\in T_j$ , then $i\in T^{\prime }_j$ , so that $Q^q_{\delta ^p+1+i}=R_i\in S_j=S^q_j$ , and in particular, $Q^q_{\delta ^p}=Q^p_{\delta ^p}\in Q^q_{\delta ^p+1}$ , as $Q^q_{\delta ^p+1}\in S_{i_0}$ .

Proof Let ${\mathord {\mathbb P}}={\mathord {\mathbb P}}^{\mathsf {DSRP}}_{\mathcal {S},{\vec {T}}}$ . Let $\theta $ be a sufficiently large regular cardinal, and let $\mathcal {A}= {\langle H_\theta ,\in ,{\mathord {\mathbb P}},\mathcal {S},{\vec {T}},F,<^* \rangle } $ , where F is some function from $H_\theta ^{{<}\omega }$ to $H_\theta $ and $<^*$ is a well-order of $H_\theta $ . Let

$$\begin{align*}A=\{X\in[H_\theta]^{\omega_1}\;|\;\omega_1\subseteq X\ \text{and there is a diagonal chain through }\mathcal{S}\text{ up to }X\}.\end{align*}$$

To show that A is stationary, it suffices to show that there is an $X\in A$ such that $\mathcal {A}|X\prec \mathcal {A}$ and $X\cap \omega _2\in \omega _2$ (see [Reference Jech13, Exercise 38.10]). By the argument of the proof of [Reference Woodin20, Lemma 2.53], it follows from $\mathsf {FA}(\{{\mathord {\mathbb P}}\})$ that there is an $X\in [H_\theta ]^{\omega _1}$ with $\omega _1\subseteq X$ and $\mathcal {A}|X\prec \mathcal {A}$ , such that there is a G which is ( $X,{\mathord {\mathbb P}}$ )-generic (meaning that G is a filter in ${\mathord {\mathbb P}}$ such that for every dense subset $D\subseteq {\mathord {\mathbb P}}$ with $D\in X$ , $G\cap D\cap X\neq \emptyset $ ). Since $\omega _1\subseteq X$ , it follows that $X\cap \omega _2\in \omega _2$ . To see that $X\in A$ , let $\vec {Q}=\bigcup _{p\in G}\vec {Q}^p$ and ${\vec {S}}=\bigcup _{p\in G}{\vec {S}}^p$ . It then follows from Lemma 3.5 that $ {\langle \vec {Q},{\vec {S}} \rangle } $ is a diagonal chain through $ {\langle X,\mathcal {S} \rangle } $ with respect to ${\vec {T}}$ . To see that for $\alpha <\omega _1$ , $\vec {Q}{\restriction }\alpha ={\langle Q_i\;|\;} {i<\alpha \rangle }\in H_\kappa \cap X$ , note that the set $D_\alpha $ of conditions $p\in {\mathord {\mathbb P}}$ with $\delta ^p\ge \alpha $ is dense in ${\mathord {\mathbb P}}$ and an element of X. Hence, there is a $p\in G\cap D_\alpha \cap X$ . The restriction of the first component of p to $\alpha $ is then also in X, and it is the sequence $\vec {Q}{\restriction }\alpha $ .

Proof For the direction from left to right, if A is a stationary subset of D, then by maximality of $\vec {T}$ , there is an $i<\omega _1$ such that $A\cap T_i$ is stationary, so that condition (a) of Definition 4.3 implies that $\{M\in S\;|\; M\cap \omega _1\in A\}$ is stationary, for every $S\in \mathcal {S}$ . Vice versa, if $\mathcal {S}$ is projective stationary on D, then condition (a) of Definition 4.3 follows immediately, and by the remark above, condition (b) is vacuous by the maximality of $\vec {T}$ .

Proof We have to show that, given a sequence $\vec {D}={\langle D_n\;|\;} {n<\omega \rangle }$ of dense open subsets of ${\mathord {\mathbb P}}$ , the intersection $\Delta =\bigcap _{n<\omega }D_n$ is dense in ${\mathord {\mathbb P}}$ . So, fixing a condition $p\in {\mathord {\mathbb P}}$ , we have to find a $q\le p$ in $\Delta $ . We may assume that $S=S^p_0$ is defined.

Let $\lambda $ be a regular cardinal much greater than $\kappa $ , say $\lambda>2^{2^{|{\mathord {\mathbb P}}|}}$ , and consider the model $\mathcal {N}= {\langle H_\lambda ,\in ,<^*,{\mathord {\mathbb P}},\vec {D},p \rangle } $ , where $<^*$ is a well-ordering of $H_\lambda $ .

Since $S'=\{X\in S\;|\; X\cap \omega _1\in T_0\}$ is stationary, we can let $\mathcal {M}=\mathcal {N}|X\prec \mathcal {N}$ be a countable elementary submodel with $X\cap H_\kappa \in S'$ , so that $X\cap \omega _1\in T_0$ .

Since $\mathcal {M}$ is countable, we can pick a filter G which is $\mathcal {M}$ -generic for ${\mathord {\mathbb P}}$ and contains p. Let

$$\begin{align*}\bar{q}= {\langle \vec{Q}^{\bar{q}},\vec{S}^{\bar{q}} \rangle} = {\left\langle \,\bigcup_{r\in G}\vec{Q}^r,\bigcup_{r\in G}\vec{S}^r \right\rangle} .\end{align*}$$

Using items (1) and (2) of Fact 3.4, it follows that $\delta := {\mathrm dom} (\vec {Q}^{\bar {q}})= {\mathrm dom} (\vec {S}^{\bar {q}})=X\cap \omega _1$ , and that $\bigcup _{i<\delta }Q^{\bar {q}}_i=X\cap H_\kappa \in S$ . Thus, if we set $q= {\langle \vec {Q}^{\bar {q}}{{}^\frown }(X\cap H_\kappa ),{\vec {S}}^{\bar {q}} \rangle } $ , then $q\in {\mathord {\mathbb P}}$ , and q extends every condition in G. Moreover, since $D_n\in M$ , for each $n<\omega $ , it follows that G meets each $D_n$ , and hence that $p\ge q\in \Delta $ , as desired.

Proof Let $D=\bigcup _{i<\omega _1}T_i$ , let $t:D\longrightarrow \omega _1$ be defined by $\alpha \in T_{t(\alpha )}$ , and set ${\mathord {\mathbb P}}={\mathord {\mathbb P}}^{\mathsf {DSRP} }_{\mathcal {S},\vec {T}}$ .

(1) $\implies $ (2): Let $A\subseteq \omega _1$ be stationary, $p\in {\mathord {\mathbb P}}$ and $\dot {C}\in \mathrm {V} ^{\mathord {\mathbb P}}$ such that $p\Vdash _{\mathord {\mathbb P}}$ “ $\dot {C}$ is a club subset of $\omega _1$ .” We will find a condition $q\le p$ in ${\mathord {\mathbb P}}$ that forces that $\dot {C}$ intersects $\check {A}$ . Let $\theta $ be a sufficiently large regular cardinal, say $\theta>2^{2^\kappa }$ .

Case 1: There is an $i_0<\omega _1$ such that $A\cap T_{i_0}$ is stationary.

In this case, fix such an $i_0$ . By assumption, for every $S\in \mathcal {S}$ , $\{M\in S\;|\; M\cap \omega _1\in A\cap T_{i_0}\}$ is stationary. By strengthening p if necessary, we may assume that $i_0<\lambda ^p$ . Let $N\prec {\langle H_\theta ,\in ,p,\dot {C},{\mathord {\mathbb P}},\mathcal {S},\vec {T},<^* \rangle } $ be a countable elementary submodel such that $M=N\cap H_\kappa \in S^p_{i_0}$ and $\delta =M\cap \omega _1\in A\cap T_{i_0}$ . Let g be ${\mathord {\mathbb P}}$ -generic over N with $p\in g$ , and let $\vec {Q}=\bigcup _{q\in g}\vec {Q}^q$ and $\vec {S}=\bigcup _{q\in g}\vec {S}^q$ . Then $\vec {Q}$ is a sequence of length $\delta $ , and $M=\bigcup _{i<\delta }Q_i\in S_{i_0}$ . So since $\delta \in T_{i_0}$ , $q= {\langle \vec {Q}{{}^\frown } M,\vec {S} \rangle } $ is a condition that strengthens p and forces that $\delta \in \dot {C}$ , since $\dot {C}^g$ is club in $\delta $ . Since $\delta \in A$ , this means that q forces that $\dot {C}$ intersects $\check {A}$ , as desired.

Case 2: $A\setminus D$ is stationary.

Let $N\prec {\langle H_\theta ,\in ,\mathcal {S},\vec {T},{\mathord {\mathbb P}},p,\dot {C},<^* \rangle } $ be countable with $N\cap \omega _1=\delta \in A\setminus D$ . Let $g\subseteq {\mathord {\mathbb P}}$ be N-generic with $p\in g$ . Let $\vec {Q}=\bigcup _{q\in g}\vec {Q}^q$ and $\vec {S}=\bigcup _{q\in g}\vec {S}^q$ . Since $\delta = {\mathrm dom} (\vec {Q})\notin D$ , it follows that $q= {\langle \vec {Q}{{}^\frown }(N\cap H_\kappa ),{\vec {S}} \rangle } \in {\mathord {\mathbb P}}$ , and since $\dot {C}^g$ is club in $\delta $ , it follows that q forces that $\check {\delta }\in \dot {C}$ , hence that $\check {A}\cap \dot {C}\neq \emptyset $ .

Case 3: Cases 1 and 2 fail.

Then $A\cap D$ is stationary and for all $i<\omega _1$ , $A\cap T_i$ is nonstationary. Fix, for every $i<\omega _1$ , a club $C_i\subseteq \omega _1$ disjoint from $A\cap T_i$ . Let $A^*=A\cap D\cap (\operatorname *{\mathrm {\bigtriangleup }}_{i<\omega _1}C_i)$ . Then $A^*$ is stationary and has the property that for all $\alpha \in A^*$ and all $\beta <\alpha $ , $\alpha \notin T_\beta $ . So $A^*\subseteq Z=\{\alpha \in D\;|\;\forall \beta <\alpha \ \alpha \notin T_\beta \}$ , and by condition (b) of Definition 4.3, $\bigcup \mathcal {S}$ is projective stationary on Z. Thus, we can pick a countable $N\prec {\langle H_\theta ,\in ,{\mathord {\mathbb P}},p,\dot {C},A^*,\mathcal {S},\vec {T} \rangle } $ such that $\delta =N\cap \omega _1\in A^*$ and $M=N\cap H_\kappa \in \bigcup \mathcal {S}$ . Let $S\in \mathcal {S}$ be such that $M\in S$ . Let $g\subseteq {\mathord {\mathbb P}}$ be N-generic for ${\mathord {\mathbb P}}$ . Let $\vec {Q}=\bigcup _{q\in g}\vec {Q}^q$ and $\vec {S}=\bigcup _{q\in g}\vec {S}^q$ . Then $ {\mathrm dom} ({\vec {S}})= {\mathrm dom} (\vec {Q})=\delta \in A^*$ , and it follows that $t(\delta )\ge \delta $ . Thus, $t(\delta )\notin {\mathrm dom} (f)$ , and we can extend ${\vec {S}}$ to some ${\vec {S}}'$ of length $t(\delta )+1$ so that ${\vec {S}}'{\restriction }\delta ={\vec {S}}$ and $S^{\prime }_{t(\delta )}=S$ . We can then let $\vec {Q}'=\vec {Q}{{}^\frown } M$ , resulting in a condition $q= {\langle \vec {Q}',{\vec {S}}' \rangle } $ extending p and forcing that $\delta \in \dot {C}\cap \check {A}^*$ . Note that $A^*\subseteq A$ .

Thus, in each case, we have found an extension q of p forcing that $\dot {C}$ intersects $\check {A}$ . Thus, the stationarity of A is preserved by ${\mathord {\mathbb P}}$ , and since this holds for any stationary subset of $\omega _1$ , ${\mathord {\mathbb P}}$ is stationary set preserving, that is, $ {\langle \mathcal {S},\vec {T} \rangle } $ is $\mathsf {SSP} $ -projective stationary.

(2) $\implies $ (1): Let ${\mathord {\mathbb P}}$ be stationary set preserving. We have to show that $\mathcal {S}$ is projective stationary on $\vec {T}$ . This amounts to proving the two conditions listed in Definition 4.3.

For condition (a), let $i<\omega _1$ , $S\in \mathcal {S}$ , and let $A\subseteq T_i$ be stationary. We have to show that $S_A=\{M\in S\;|\; M\cap \omega _1\in A\}$ is a stationary subset of $[H_\kappa ]^\omega $ . If not, then let $C\subseteq [H_\kappa ]^\omega $ be club with $S_A\cap C=\emptyset $ . Let G be ${\mathord {\mathbb P}}$ -generic over $\mathrm {V} $ , such that G contains a condition p with $i<\lambda ^p$ and $S^p_i=S$ . Let $\vec {Q}=\bigcup _{q\in G}\vec {Q}^q$ and $\vec {S}=\bigcup _{q\in G}\vec {S}^q$ . In $\mathrm {V} [G]$ , A is still stationary, so there is a

$$\begin{align*}\delta\in A\cap\{\alpha<\omega_1\;|\; Q_\alpha\cap\omega_1=\alpha\}\cap\{Q_\alpha\cap\omega_1\;|\; Q_\alpha\in C\}.\end{align*}$$

But then $\delta =Q_\delta \cap \omega _1$ and $Q_\delta \in C$ , and since $\delta \in A\subseteq T_i$ , we have that $Q_\delta \in S_i$ . So $Q_\delta \in S_A\cap C\neq \emptyset $ . Since C was arbitrary, this shows that $S_A$ is stationary, as claimed.

For condition (b), suppose $A\subseteq D$ is stationary in $\omega _1$ and has the property that for all $\alpha \in A$ and all $\beta <\alpha $ , $\alpha \notin T_\beta $ . Letting $S^*=\bigcup \mathcal {S}$ , we have to show that

$$\begin{align*}S^*_A=\{M\in S^*\;|\; M\cap\omega_1\in A\}\end{align*}$$

is stationary. So let $C\subseteq [H_\kappa ]^\omega $ be club. Let G be ${\mathord {\mathbb P}}$ -generic, and let $\vec {Q}=\bigcup _{q\in g}\vec {Q}^q$ and $\vec {S}=\bigcup _{q\in g}\vec {S}^q$ . Let

$$\begin{align*}\delta\in A\cap\{\alpha<\omega_1\;|\; Q_\alpha\cap\omega_1=\alpha\}\cap\{Q_\alpha\cap\omega_1\;|\; Q_\alpha\in C\}.\end{align*}$$

This is possible, because A is stationary in $\mathrm {V} [G]$ . It follows that $\delta =Q_\delta \cap \omega _1\in A\subseteq D$ , so that $t(\delta )$ is defined and $Q_\delta \in S_{t(\delta )}\in \mathcal {S}$ . It follows that $Q_\delta \in S^*_A\cap C$ .

Proof Let $\theta>\kappa $ , and let $\tau $ , A, $X,$ and a be as in Definition 5.1. In particular, $T\in X$ . By elementarity, X sees that T is not stationary, so there is a club set $C\subseteq \omega _1$ in X, disjoint from T. Letting $\delta =X\cap \omega _1$ , it follows that $C\cap \delta $ is unbounded in $\delta $ , and hence that $\delta \in C$ . Thus, $X\cap \omega _1\notin T$ , and hence, the condition stated in Definition 5.1 holds vacuously.

Proof Set $D=\bigcup _{i<\omega _1}T_i$ . For the implication from left to right, fix $S\in \mathcal {S}$ . Let $\theta $ be sufficiently large, and let $H_\theta \subseteq L_\tau ^A=N\models {\mathsf {ZFC} }^- $ . Let X be countable and full, with $N|X\prec N$ , and assume that $\theta , S, D\in X$ . Fix some $a\in X$ . By a version of Fact 2.16 may also assume that ${\vec {T}}\in X$ . But then, $Z=D\setminus \bigtriangledown _{i<\omega _1}T_i$ is also in X, and Z is nonstationary, by assumption. As in the proof of Remark 5.2, it follows that $\delta =X\cap \omega _1\notin Z$ . Now suppose that $\delta \in D$ . Since $\delta \notin Z$ , this means that $\delta \in T_{i_0}$ , for some $i_0<\delta $ . But since S is spread out on $T_{i_0}$ , there are $\pi $ , Y such that $\pi :N|X\longrightarrow N|Y\prec N$ is an isomorphism that fixes a, and such that $Y\cap H_\kappa \in S$ , as wished.

The converse is trivial, because if every $S\in \mathcal {S}$ is spread out on $\bigcup _{i<\omega _1}T_i$ , then it is trivially also spread out on each $T_i$ (we may assume that D belongs to the relevant X). And condition (b) of Definition 5.3 is vacuous, since ${\vec {T}}$ is maximal.

Proof Let $D=\bigcup _{i<\omega _1}T_i$ , let $t:D\longrightarrow \omega _1$ be defined by $\alpha \in T_{t(\alpha )}$ , and let ${\mathord {\mathbb P}}={\mathord {\mathbb P}}^{\mathsf {DSRP} }_{\mathcal {S},\vec {T}}$ . We treat each implication separately.

(1) $\implies $ (2): Assuming that $\mathcal {S}$ is spread out on $T_i$ , for every $i<\omega _1$ , we have to show that ${\mathord {\mathbb P}}$ is $\infty $ -subcomplete. To this end, let $\theta $ be large enough for Definition 5.1 to apply to every $T_i$ , every $S\in \mathcal {S}$ , as well as to $\bigcup \mathcal {S}$ and $D\setminus \bigtriangleup _{i<\omega _1}T_i$ . Let $N=L_\tau ^A\models {\mathsf {ZFC} }^- $ with $H_\theta \subseteq N$ , and let ${\mathord {\mathbb P}}\in X\prec N$ be countable and full. Let a be some member of X, and let $\sigma :{\bar {N}}\longrightarrow X$ be the inverse of the Mostowski collapse of X, ${\bar {N}}$ transitive. Let ${\bar {{\mathord {\mathbb P}}}}=\sigma ^{-1}({\mathord {\mathbb P}}_S)$ , $\bar {a}=\sigma ^{-1}(a)$ , and let ${\bar {G}}\subseteq {\bar {{\mathord {\mathbb P}}}}$ be ${\bar {N}}$ -generic. As usual, we may assume that certain parameters are in X (see [Reference Jensen, Chong, Feng, Slaman, Woodin and Yang17, p. 116, Lemma 2.5]). Here, we will assume that $\mathcal {S},\vec {T}\in X$ . Let ${\bar {\kappa }}=\sigma ^{-1}(\kappa )$ , $\bar {\mathcal {S}}=\sigma ^{-1}(\mathcal {S})$ and $\vec {\bar {T}}=\sigma ^{-1}(\vec {T})$ . It follows from Lemma 3.5 that if we set $\vec {\bar {Q}}=\bigcup \{\vec {Q}^{\bar {p}}\;|\;\bar {p}\in \bar {G}\}$ and $\vec {{\bar {S}}}=\bigcup \{{\vec {S}}^{\bar {p}}\;|\;\bar {p}\in \bar {G}\}$ , then $\vec {\bar {Q}}$ and $\vec {{\bar {S}}}$ are sequences of length $\omega _1^{\bar {N}}$ .

Let $\delta =X\cap \omega _1=\omega _1^{\bar {N}}$ .

Case 1: $\delta \notin D$ .

In this case we define

$$\begin{align*}q=\big\langle{\langle\sigma(\bar{Q}_i)\;|\;} {i<\delta\rangle}{{}^\frown}(X\cap H_\kappa)\ ,\ {\langle\sigma(\bar{S}_i)\;|\;} {i<\delta\rangle}\big\rangle.\end{align*}$$

Then $q\in {\mathord {\mathbb P}}$ , since $X\cap \omega _1\notin D$ . Moreover, q extends every member of $\sigma "\bar {G}$ . Thus, q forces that $\sigma $ itself satisfies the subcompleteness conditions (1)–(4) of Definition 2.13.

Case 2: $\delta \in D$ .

Let $i_0<\omega _1$ be such that $\delta \in T_{i_0}$ , that is, $i_0=t(\delta )$ .

Case 2.1: $i_0<\delta $ .

In this case, let $\bar {S}={\bar {S}}_{i_0}$ . $\bar {S}$ is then in $\bar {\mathcal {S}}$ , and so, $S=\sigma (\bar {S})\in \mathcal {S}\cap X$ . In particular, S is spread out on $T_{i_0}$ . Moreover, $T_{i_0}\in X$ . So, since $\delta =X\cap \omega _1\in T_{i_0}$ , we can choose a $Y\prec N$ with $Y\cap H_\kappa \in S$ and an isomorphism $\pi :N|X\longrightarrow N|Y$ that fixes a, S and ${\mathord {\mathbb P}}$ . Let $\sigma '=\pi \circ \sigma :{\bar {N}}\prec N$ . Let

$$\begin{align*}q=\big\langle{\langle\sigma'(\bar{Q}_i)\;|\;} {i<\delta\rangle}{{}^\frown}(Y\cap H_\kappa)\ ,\ {\langle\sigma'({\bar{S}}_i)\;|\;} {i<\delta\rangle}\big\rangle.\end{align*}$$

Since $Y\cap H_\kappa \in S$ , it follows that $q\in {\mathord {\mathbb P}}$ (note that $X\cap \omega _1=Y\cap \omega _1=\delta \in T_{i_0}$ , and $Y\cap H_\kappa \in S=\sigma ({\bar {S}}_{i_0})=\sigma '({\bar {S}}_{i_0})$ ), and whenever $G\ni q$ is ${\mathord {\mathbb P}}_S$ -generic over $\mathrm {V} $ , then $\sigma '"{\bar {G}}\subseteq G$ . Since $\sigma '(\bar {a})=a$ , the conditions defining $\infty $ -subcompleteness are satisfied.

Case 2.2: $i_0\ge \delta $ .

In this case, $\delta \in Z=D\setminus \bigtriangledown _{i<\omega _1}T_i$ . By assumption, $\bigcup \mathcal {S}$ is spread out on Z. Moreover, $\bigcup {\mathcal {S}}$ and Z are in X. Let Y, $\sigma '$ be such that $\sigma ':N|X\longrightarrow N|Y$ is an isomorphism fixing a, $\mathsf {S}$ , ${\vec {T,}}$ and ${\mathord {\mathbb P}}$ , and such that $Y\cap H_\kappa \in \bigcup \mathcal {S}$ . Let $S\in \mathcal {S}$ be such that $Y\cap H_\kappa \in S$ . Let ${\vec {S}}'$ be a sequence of length $i_0+1$ extending ${\langle \sigma '({\bar {S}}_i)\;|\;} {i<\delta \rangle }$ with $S^{\prime }_{i_0}=S$ and $S^{\prime }_i\in \mathsf {S}$ for every $i\le i_0$ . Let

$$\begin{align*}q=\big\langle{\langle\sigma'(\bar{Q}_i)\;|\;} {i<\delta\rangle}{{}^\frown}(Y\cap H_\kappa),{\vec{S}}'\big\rangle.\end{align*}$$

Then q is a condition, forcing that $\sigma '"\bar {G}\subseteq \dot {G}$ .

(2) $\implies $ (1): To prove that condition (a) of Definition 5.3 holds, fix an $S\in \mathcal {S}$ and an $i_0<\omega _1$ . Let $\theta $ be large enough to verify that ${\mathord {\mathbb P}}$ is $\infty $ -subcomplete. Let $\tau ,a,X,A,N$ be as in Definition 5.1. So $X\prec N=L_\tau ^A$ is countable and full, $S,a,T_{i_0}\in X$ , and suppose that $\delta =X\cap \omega _1\in T_{i_0}$ , that is, $t(\delta )=i_0$ . By (a variation of) Fact 2.16, we may assume that X contains certain parameters we care about. So let us assume that $\mathcal {S},\vec {T}\in X$ . Since $T_{i_0}$ and $\vec {T}$ are in X, it follows that $i_0\in X$ , and hence that $i_0<\delta $ . Let $\sigma :{\bar {N}}\longrightarrow X$ be the transitive isomorph of X, and let $\bar {{\mathord {\mathbb P}}}$ , $\bar {S}$ , $\bar {\mathcal {S}}$ , $\vec {\bar {T}}$ be the preimages of ${\mathord {\mathbb P}}$ , S, $\mathcal {S}$ , $\vec {T}$ under $\sigma $ , respectively.

Let $\bar {G}$ be $\bar {{\mathord {\mathbb P}}}$ -generic over ${\bar {N}}$ , containing a condition $\bar {p}$ such that $S^{\bar {p}}_{i_0}=\bar {S}$ . By assumption, there is a condition $q\in {\mathord {\mathbb P}}$ such that whenever G is ${\mathord {\mathbb P}}$ -generic over $\mathrm {V} $ and contains q, then there is in $\mathrm {V} [G]$ an elementary embedding $\sigma ':{\bar {N}}\longrightarrow N$ with $\sigma '(\bar {a})=a$ , $\sigma '(\bar {S})=S$ , $\sigma '(\bar {\mathcal {S}})=\mathcal {S}$ , $\sigma '(\vec {\bar {T}})={\vec {T}}$ , and such that $\sigma '"\bar {G}\subseteq G$ . Note that for any $S'\in \mathcal {S}$ , and any $i<\omega _1$ , $S'$ is projective stationary on $T_i$ , since $S'$ is even spread out on $T_i$ . This can be easily shown directly. Hence, by Lemma 4.5, ${\mathord {\mathbb P}}$ is countably distributive, so that $\sigma '$ already exists in $\mathrm {V} $ . We have already argued that the union of the first coordinates of conditions in $\bar {G}$ is of the form ${\langle \bar {Q}_i\;|\;} {i<\omega _1^{\bar {N}}\rangle }$ , where $\bigcup _{i<\delta }\bar {Q}_i=H_{\bar {\kappa }}^{\bar {N}}$ , and that the union of the second coordinates is a sequence ${\langle \bar {S}_i\;|\;} {i<\delta \rangle }$ . Now let $r\in G$ be a condition with $\delta ^r\ge \delta $ , and let $Y= {\mathrm ran} (\sigma ')$ . Then

$$\begin{align*}Q^r_\delta=\bigcup_{i<\delta}Q^r_i=\bigcup_{i<\delta}\sigma'"\bar{Q}_i=\sigma'"H_{\bar{\kappa}}^{\bar{N}}=Y\cap H_\kappa.\end{align*}$$

Moreover, $Y\cap \omega _1=X\cap \omega _1\in T_{i_0}$ , so that $Q^r_\delta \in S^r_{i_0}=\sigma '(\bar {Q}_{i_0})=S$ , by clause (3) of Definition 3.3. Thus, $Y\cap H_\kappa \in S$ , and letting $\pi =\sigma '\circ \sigma ^{-1}$ , we have that $\pi :N|X\prec N|Y$ is an isomorphism fixing a, showing that S is spread out on $T_{i_0}$ .

To prove condition (b) of Definition 5.3, we start in the same setup, but we assume that $\delta \in \bigcup _{i<\omega _1}T_i\setminus \bigtriangledown _{i<\omega _1}T_i$ , that is, $\delta \in T_{i_0}$ , where $i_0\ge \delta $ . Let ${\bar {G}}$ be generic over ${\bar {N}}$ for $\bar {{\mathord {\mathbb P}}}$ , and let $q\in {\mathord {\mathbb P}}$ force the existence of a $\sigma ':{\bar {N}}\prec N$ as before, moving $\vec {{\bar {T}}}$ , $\bar {{\mathord {\mathbb P}}}$ , $\bar {\mathcal {S}}$ and $\bar {a}$ the same way as $\sigma $ , and so that if G is ${\mathord {\mathbb P}}$ -generic with $q\in G$ , then $\sigma '"{\bar {G}}\subseteq G$ . As before, it follows that $\sigma '\in \mathrm {V} $ . Let $r\in G$ be such that $\delta ^r\ge \delta $ . It follows that, letting $Y= {\mathrm ran} (\sigma ')$ , $Q^r_\delta =Y\cap H_\kappa $ . So, since $\delta \in T_{i_0}$ , $i_0<\lambda ^r$ and $Q^r_\delta \in S^r_{i_0}\in \mathcal {S}$ . Hence, $Y\cap H_\kappa \in \bigcup \mathcal {S}$ , and $\pi =\sigma '\circ \sigma ^{-1}$ can serve as our wanted isomorphism.

Proof Part (1): if $\Gamma $ is the class of stationary set preserving forcing notions, then $\Gamma $ -projective stationarity is just the usual concept of projective stationarity. So let $\mathcal {S}$ be a nonempty collection of projective stationary sets in $H_\kappa $ . Let ${\vec {T}}$ be a maximal partition of $\omega _1$ into stationary sets. By Remark 4.4, $ {\langle \mathcal {S},{\vec {T}} \rangle } $ is $\mathsf {SSP} $ -projective stationary, so by assumption, $\mathsf {DSRP} (\mathcal {S},{\vec {T}})$ holds. But since ${\vec {T}}$ is a partition of all of $\omega _1$ , this implies $\mathsf {eDSRP} (\mathcal {S})$ , by Remark 3.7.

The case where $\Gamma $ is the class of all $\inf $ -subcomplete forcing notions is handled similarly. This time, $\Gamma $ -projective stationarity means being spread out. Given $\mathcal {S}$ and a partition ${\vec {T}}$ as above, it follows by Remark 5.4 that $ {\langle \mathcal {S},{\vec {T}} \rangle } $ is $\mathsf {\infty \text {-}SC} $ -projective stationary, so that $\mathsf {DSRP} (\mathcal {S},{\vec {T}})$ holds, which again implies $\mathsf {eDSRP} (\mathcal {S})$ , as ${\vec {T}}$ is a partition of $\omega _1$ .

Part (2) is similar. We can work with a maximal partition ${\vec {T}}$ of A into stationary sets now.

Observation 6.4 [Reference Fuchs10, Observation 3.14]

Let $\kappa>\omega _1$ be regular and fix a stationary subset A of $\kappa $ . The following are equivalent $:$

1. (1) Whenever $\mathcal {S}=\{A_i\;|\; i<\omega _1\}$ is a set of stationary subsets of A, there is a partition ${\langle T_i\;|\;} {i<\omega _1\rangle }$ of $\omega _1$ into stationary sets and a normal function $f:\omega _1\longrightarrow \kappa $ such that for every $i<\omega _1$ , $f"T_i\subseteq A_i$ .

2. (2) $\mathsf {eRefl} (\omega _1,A)$ holds.

3. (3) For any set $\mathcal {S}$ of stationary subsets of A that has size $\omega _1$ , $\operatorname {\mathrm {\mathsf {eTr}}}(\mathcal {S})$ is stationary in $\kappa $ .

Fact 6.5 [Reference Fuchs10, Fact 3.15]

Let $\kappa>\omega _1$ be regular, and suppose there is a stationary $A\subseteq \kappa $ such that $\mathsf {eRefl} (\omega _1,A)$ holds. Then $\kappa ^{\omega _1}=\kappa $ .