Proof Fix $A \subseteq P_\kappa X$ and $x\in d_0(A)$ . By the above argument that ${\mathcal B}_0$ is a base for $\tau _0$ , it follows that $\kappa _x\ge \omega $ . Suppose that $y\in P_{\kappa _x}x$ . Since $(y,x]$ is an open neighborhood of x, we can choose a $z\in (y,x]\cap A$ with $z\neq x$ . This implies $z\in (y,x)\cap A$ , and hence A is $\prec $ -cofinal in $P_{\kappa _x}x$ . Conversely, suppose A is $\prec $ -cofinal in $P_{\kappa _x}x$ and let $(a,b]$ be a basic open neighborhood of x. Then $a\in P_{\kappa _x}x$ and we may choose some $y\in A$ with $a\prec y\in P_{\kappa _x}x$ . Hence $y\in (a,b]\cap A\setminus \{x\}$ .

Proof First let us show that $d_0(A)$ is $\prec $ -closed. Suppose $d_0(A)$ is $\prec $ -cofinal in $P_{\kappa _x}x$ . Fix $a\in P_{\kappa _x}x$ and let $b\in d_0(A)\cap (a,x)$ . Then A is $\prec $ -cofinal in $P_{\kappa _b}b$ , so we may choose $c\in A\cap (a,b)\subseteq A\cap (a,x)$ and hence $x\in d_0(A)$ .

Let us show that $d_0(A)$ is $\prec $ -cofinal in $P_\kappa X$ . Fix $x\in P_\kappa X$ . We define an increasing chain $\langle x_\eta \mid \eta <\kappa \rangle $ in $P_\kappa X$ as follows. Let $x_0=x$ . Given $x_\eta $ choose $x_{\eta +1}\in A$ with $x_\eta \prec x_{\eta +1}$ . If $\eta <\kappa $ is a limit let $x_\eta =\bigcup _{\zeta <\eta }x_\zeta $ . Then $\langle \kappa _{x_\eta }\mid \eta <\kappa \rangle $ is a strictly increasing sequence in $\kappa $ and the set

$$\begin{align*}C=\{\eta<\kappa\mid (\forall\zeta<\eta)\ \kappa_{x_\zeta}<\eta\}=\{\eta<\kappa\mid\kappa_{x_\eta}=\eta\}\end{align*}$$

is a club in $\kappa $ . Thus, since $\kappa $ is weakly Mahlo, we can fix some regular $\kappa _{x_\eta }=\eta \in C$ . Let us show that $x_\eta \in d_0(A)$ . Suppose $a\in P_{\kappa _{x_\eta }}x_\eta $ . Since $\kappa _{x_\eta }=\eta $ is regular, $|a|<\kappa _{x_\eta }$ implies $a\in P_{\kappa _{x_\zeta }}x_\zeta $ for some $\zeta <\eta $ , and therefore $a\in P_{\kappa _{x_{\zeta +1}}}x_{\zeta +1}$ which entails that $a\prec x_{\zeta +1}\in A$ .

Proof The fact that (1) implies (2) follows from Proposition 3.3. Let us show that (2) implies (1). We assume (2) holds, and that $\kappa $ is weakly inaccessible but not weakly Mahlo. Let $C \subseteq \kappa $ be a club consisting of singular cardinals, and let ${D=\{x\in P_\kappa X\mid x\cap \kappa \in C\}}$ . Then D is $\prec $ -cofinal in $P_\kappa X$ . By (2), there is a ${y\in P_\kappa X}$ such that D is $\prec $ -cofinal in $P_{\kappa _y}y$ . Then $\kappa _y$ is a limit cardinal and C is cofinal in $\kappa _y$ , so $\kappa _y\in C$ is a singular cardinal.

Let us argue that $y\cap \kappa $ is an ordinal less than $\kappa $ . Suppose $\alpha \in y\cap \kappa $ . Since D is $\prec $ -cofinal in $P_{\kappa _y}y$ there is some $z\in D\cap P_{\kappa _y}y$ such that $\{\alpha \}\prec z$ . Thus $\alpha \in z$ and $z\cap \kappa $ is an ordinal, which implies $\alpha \subseteq z\cap \kappa \subseteq y\cap \kappa $ . Hence $y\cap \kappa $ is transitive.

Let $a\subseteq \kappa _y$ be cofinal in $\kappa _y$ with $|a|=\mathop{\mathrm{cf}}(\kappa _y)<\kappa _y$ . Since $y\cap \kappa $ is an ordinal we have $a\subseteq \kappa _y=|y\cap \kappa |\subseteq y\cap \kappa $ and thus $a\in P_{\kappa _y}y$ . However, there is no $x\in D\cap P_{\kappa _y}y$ with $a\prec x$ because for such an x, $\kappa \cap x\in C$ would be an ordinal containing the set a which is cofinal in $\kappa _y$ , and hence $\kappa _x\geq \kappa _y$ .

Proof Suppose otherwise, and let $x \in P_\kappa X$ be a counterexample such that $\kappa _x$ is minimal among all counterexamples. Since $P_{\kappa _x} x$ is $(\kappa _x+1)$ -strongly stationary, it is a fortiori $\kappa _x$ -strongly stationary. Therefore, by the definition of $(\kappa _x+1)$ -strong stationarity, we can find $y \in P_{\kappa _x} x$ such that $P_{\kappa _x} x$ is $\kappa _x$ -strongly stationary in $P_{\kappa _y} y$ . Since $\kappa _x> \kappa _y$ , this implies that $P_{\kappa _y} y$ is $(\kappa _y + 1)$ -strongly stationary, contradicting the minimality of $\kappa _x$ .

Proof Note that, if A is $1$ -strongly stationary in $P_{\kappa _x}x$ , then $\kappa _x$ is a limit cardinal and hence weakly inaccessible. We can thus assume that this is the case. (2) $\implies $ (1) is trivial. Let us now assume that A is $1$ -strongly stationary in $P_{\kappa _x} x$ . By Proposition 3.4, it follows that $\kappa _x$ is weakly Mahlo. To see that A is $1$ -s-strongly stationary in $P_{\kappa _x}x$ , fix sets $S_0,S_1 \subseteq P_\kappa X$ that are both $\prec $ -cofinal in $P_{\kappa _x} x$ . Let T be the set of $y \in P_{\kappa _x} x$ such that $S_0$ and $S_1$ are both $\prec $ -cofinal in $P_{\kappa _y} y$ . We claim that T is $\prec $ -cofinal in $P_{\kappa _x} x$ . To see this, fix an arbitrary $y_0 \in P_{\kappa _x} x$ . Define a continuous, $\prec $ -increasing sequence $\langle y_\eta \mid \eta < \kappa _x \rangle $ in $P_{\kappa _x} x$ as follows. The set $y_0$ is already fixed. Given $y_\eta $ , find $z^0_\eta \in S_0$ and $z^1_\eta \in S_1$ such that, for all $i < 2$ , we have $y_\eta \prec z^i_\eta \prec x$ . Then let $y_{\eta + 1} = z^0_\eta \cup z^1_\eta $ . If $\xi < \kappa _x$ is a limit ordinal, let $y_\xi = \bigcup \{y_\eta \mid \eta < \xi \}$ . The set of $\eta < \kappa _x$ for which $\kappa _{y_\eta } = \eta $ is a club in $\kappa _x$ , so, since $\kappa _x$ is weakly Mahlo, we can fix some regular cardinal $\eta < \kappa _x$ such that $\kappa _{y_\eta } = \eta $ . A now-familiar argument then shows that $S_0$ and $S_1$ are both $\prec $ -cofinal in $y_\eta $ , and hence $y_\eta \in T$ .

Since A is $1$ -strongly stationary in $P_{\kappa _x}x$ , we can find $w \in A$ such that T is $\prec $ -cofinal in $P_{\kappa _w}w$ . It follows immediately that $S_0$ and $S_1$ are both $\prec $ -cofinal in $P_{\kappa _w} w$ ; therefore, A is $1$ -s-strongly stationary in $P_{\kappa _x} x$ .

For the “moreover” clause, assume that $\kappa _x$ is strongly inaccessible and A is $1$ -strongly stationary in $P_{\kappa _x} x$ . The fact that $\kappa _x$ is Mahlo follows from the previous paragraphs. To show that A is strongly stationary in $P_{\kappa _x} x$ , suppose C is a weak club subset of $P_{\kappa _x}x$ . Since A is $1$ -strongly stationary there is some $y\in A$ such that C is $\prec $ -cofinal in $P_{\kappa _y}y$ . Since C is weakly closed we have $y\in A\cap C$ .

Finally, suppose $\kappa _x$ is Mahlo and A is strongly stationary in $P_{\kappa _x}x$ . To show that A is $1$ -strongly stationary in $P_{\kappa _x}x$ , fix a set S which is $\prec $ -cofinal in $P_{\kappa _x}x$ . Since $\kappa _x$ is Mahlo, it follows from Proposition 3.3 that $d_0(S)$ is a weak club in $P_{\kappa _x}x$ . Thus, by Fact 2.1, $d_0(S)\cap A$ is strongly stationary in $P_{\kappa _x}x$ and hence there is a $y\in A\cap P_{\kappa _x}x$ such that S is $\prec $ -cofinal in $P_{\kappa _y}y$ .

Proof First let us use a straightforward inductive argument on $n\geq 2$ to show that whenever $A_0,\ldots ,A_{n-1}$ are $0$ -strongly stationary in $P_{\kappa _x}x$ , the set $d_0(A_0)\cap \cdots \cap d_0(A_{n-1})\cap A$ is $0$ -strongly stationary in $P_{\kappa _x}x$ . Suppose $A_0$ and $A_1$ are $\prec $ -cofinal in $P_{\kappa _x}x$ and note that $\kappa _x$ must be a limit cardinal. To show that $d_0(A_0)\cap d_0(A_1)\cap A$ is $\prec $ -cofinal in $P_{\kappa _x}x$ , fix $y\in P_{\kappa _x}x$ . Then $A_0\cap (y,x)$ and $A_1\cap (y,x)$ are $\prec $ -cofinal in $P_{\kappa _x}x$ . Since A is $1$ -s-strongly stationary in $P_{\kappa _x}x$ there is an $a\in A\cap P_{\kappa _x}x$ such that $A_0\cap (y,x)$ and $A_1\cap (y,x)$ are both $\prec $ -cofinal in $P_{\kappa _a}a$ , and hence $y<a$ . Therefore $a\in d_0(A_0)\cap d_0(A_1)\cap A\cap (y,x)$ . Now suppose the result holds for n, and suppose $A_0,\ldots ,A_{n-1},A_n$ are all $\prec $ -cofinal in $P_{\kappa _x}x$ . By our inductive hypothesis, $d_0(A_0)\cap \cdots \cap d_0(A_{n-1})$ is $\prec $ -cofinal in $P_{\kappa _x}x$ , and by the base case the set $d_0(d_0(A_0)\cap \cdots \cap d_0(A_{n-1}))\cap d_0(A_n)\cap A$ is $\prec $ -cofinal in $P_{\kappa _x}x$ . But

$$ \begin{align*} d_0(d_0(A_0)\cap\cdots\cap d_0(A_{n-1}))\cap d_0(A_n) &\subseteq d_0(d_0(A_0))\cap \cdots\cap d_0(d_0(A_{n-1}))\cap d_0(A_n) \\ &\subseteq d_0(A_0)\cap \cdots\cap d_0(A_{n-1})\cap d_0(A_n). \end{align*} $$

Now we prove the statement of the lemma. Fix sets $A_0,\ldots ,A_{n-1}\subseteq P_\kappa X$ that are $\prec $ -cofinal in $P_{\kappa _x}x$ . To show that $d_0(A_0)\cap \cdots \cap d_0(A_{n-1})\cap A$ is $1$ -s-strongly stationary in $P_{\kappa _x}x$ , fix sets S and T that are $\prec $ -cofinal in $P_{\kappa _x}x$ . By the previous paragraph, it follows that the set

$$\begin{align*}d_0(S)\cap d_0(T)\cap d_0(A_0)\cap\cdots\cap d_0(A_{n-1})\cap A\end{align*}$$

is $\prec $ -cofinal in $P_{\kappa _x}x$ and hence there is some

$$\begin{align*}y\in d_0(S)\cap d_0(T)\cap d_0(A_0)\cap\cdots\cap d_0(A_{n-1})\cap A,\end{align*}$$

which establishes that $d_0(A_0)\cap \cdots \cap d_0(A_{n-1})\cap A$ is $1$ -s-strongly stationary.

Proof Suppose A is $1$ -s-strongly stationary in $P_{\kappa _x}x$ and C is a $0$ -s-weak club in $P_{\kappa _x}x$ . Then $d_0(C)\cap P_{\kappa _x}x\subseteq C\cap P_{\kappa _x}x$ and by Lemma 3.12, $d_0(C)\cap A$ is $1$ -s-strongly stationary in $P_{\kappa _x}x$ . Thus $A\cap C\cap P_{\kappa _x}x\neq \emptyset $ . Conversely, assume that $A \cap C\cap P_{\kappa _x}x\neq \emptyset $ whenever C is a $0$ -s-weak club in $P_{\kappa _x}x$ . Fix sets S and T that are $0$ -s-strongly stationary in $P_{\kappa _x}x$ . Then $d_0(S)\cap d_0(T)$ is a $0$ -s-weak club in $P_{\kappa _x}x$ because $d_0(S)\cap d_0(T)\cap P_{\kappa _x}x$ is $1$ -s-strongly stationary and hence $0$ -s-strongly stationary in $P_{\kappa _x}x$ by Lemma 3.12, and $d_0(S)\cap d_0(T)$ is $0$ -s-closed in $P_{\kappa _x}x$ since

$$\begin{align*}d_0(d_0(S)\cap d_0(T))\subseteq d_0(S)\cap d_0(T)\end{align*}$$

as a consequence of the fact that $d_0$ is the limit point operator of the space $(P_\kappa X,\tau _0)$ .

Proof Suppose A is not $1$ -s-strongly stationary in $P_{\kappa _x}x$ . If $\kappa _x$ is a successor cardinal then x is isolated in $(P_\kappa X,\tau _1)$ by Corollary 3.2 and hence $x\notin d_1(A)$ . Suppose $\kappa _x$ is a limit cardinal. Then there are sets S and T which are $0$ -strongly stationary in $P_{\kappa _x}x$ such that $d_0(S)\cap d_0(T)\cap A\cap P_{\kappa _x}x=\emptyset $ . Then it follows that $d_0(S)\cap d_0(T)\cap (0,x]$ is an open neighborhood of x in the $\tau _1$ topology that does not intersect A in some point other than x. Hence $x\notin d_1(A)$ .

Conversely, suppose A is $1$ -s-strongly stationary in $P_{\kappa _x}x$ . We must show that x is a limit point of A in the $\tau _1$ topology. Suppose $x\in I\cap d_0(A_0)\cap \cdots \cap d_0(A_{n-1})$ , where $I\in {\mathcal B}_0$ and $A_0,\ldots ,A_{n-1}\subseteq P_\kappa X$ . Then the sets $A_0,\ldots ,A_{n-1}$ are all $\prec $ -cofinal in $P_{\kappa _x}x$ , and by Lemma 3.12, the set $I\cap d_0(A_0)\cap \cdots d_0(A_{n-1})\cap A$ is $\prec $ -cofinal in $P_{\kappa _x}x$ , which implies that the open neighborhood $I\cap d_0(A_0)\cap \cdots \cap d_0(A_{n-1})$ of x intersects A in some point other than x.

Proof We have already established that $(1)_\xi $ and $(3)_\xi $ hold for $\xi \leq 1$ and $(2)_0$ holds. Given these base cases, the fact that (1), (2), and (3) hold for all $\xi <\kappa $ can be established by simultaneous induction using an argument which is essentially identical to that of [Reference Bagaria2, Proposition 2.10]. For the reader’s convenience, we include the argument here.

First, suppose $(1)_\zeta $ , $(2)_\zeta $ , and $(3)_\zeta $ hold for all $\zeta $ less than some limit ordinal $\xi <\kappa $ . It is clear that $(1)_\xi $ and $(3)_\xi $ also must hold. Let us prove that $(2)_\xi $ holds. Notice that the backward direction of $(2)_\xi $ easily follows from the definition of $\xi +1$ -s-strong stationarity and the fact that $(1)_\zeta $ holds for $\zeta \leq \xi $ . For the forward direction of $(2)_\xi $ , suppose A is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ . Fix $\zeta \leq \xi $ and a pair $S,T$ of $\zeta $ -s-strongly stationary subsets of $P_{\kappa _x}x$ . To show that $A\cap d_\zeta (S)\cap d_\zeta (T)$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ , fix sets $A,B$ that are $\eta $ -s-strongly stationary in $P_{\kappa _x}x$ where $\eta <\zeta $ . Using the fact that (3) holds for $\zeta $ , we see that $S\cap d_\eta (A)\cap d_\eta (B)$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ . Since A is $\xi +1$ -s-strongly stationary, and applying the fact that $(1)_\zeta $ holds, we have

$$\begin{align*}\emptyset\neq d_\zeta(d_\eta(A)\cap d_\eta(B)\cap S)\cap d_\zeta(T)\cap A.\end{align*}$$

But, by Lemma 3.5,

$$\begin{align*}d_\zeta(d_\eta(A)\cap d_\eta(B)\cap S)\cap d_\zeta(T)\cap A = d_\eta(A)\cap d_\eta(B)\cap d_\zeta(S)\cap d_\zeta(T)\cap A.\end{align*}$$

Thus, $d_\zeta (S)\cap d_\zeta (T)\cap A$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ .

One can show that if $(1)_{\leq \xi }$ , $(2)_{\leq \xi }$ , and $(3)_{\leq \xi }$ hold then, by induction on n, $(3)_{\xi +1}$ must also hold. For the reader’s convenience we provide a proof that $(3)_{\xi +1}$ holds for $n=1$ , the remaining argument is the same as [Reference Bagaria2, Proposition 2.10]. Suppose $n=1$ . To prove that $A\cap d_{\zeta _0}(A_0)$ is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ , fix sets S and T that are $\eta $ -s-strongly stationary in $P_{\kappa _x}x$ for some $\eta \leq \xi $ . By $(1)_{\leq \xi }$ , it will suffice to show that $A\cap d_{\zeta _0}(A_0)\cap d_\eta (S)\cap d_\eta (T)\neq \emptyset $ . If ${\zeta _0}=\eta $ , then by $(2)_{\leq \xi }$ , it follows that the set $A\cap d_{\zeta _0}(A_0)\cap d_\eta (d_\eta (S)\cap d_\eta (T))$ , which is contained in ${A\cap d_{\zeta _0}(A_0)\cap d_\eta (S)\cap d_\eta (T)}$ , is ${\zeta _0}$ -s-strongly stationary in $P_{\kappa _x}x$ , and thus $A\cap d_{\zeta _0}(A_0)\cap d_\eta (S)\cap d_\eta (T)\neq \emptyset $ . If ${\zeta _0}<\eta $ , then by $(3)_\eta $ , if follows that $d_{\zeta _0}(A_0)$ is $\eta $ -s-strongly stationary in $P_{\kappa _x}x$ , and by $(2)_{\xi }$ , the set $A\cap d_\eta (d_{\zeta _0}(A_0))\cap d_\eta (d_\eta (S)\cap d_\eta (T))$ , which is contained in $A\cap d_{\zeta _0}(A_0)\cap d_\eta (S)\cap d_\eta (T)$ , is $\eta $ -s-strongly stationary in $P_{\kappa _x}x$ . If $\zeta _0>\eta $ then by $(2)_\xi $ the set $A\cap d_{\zeta _0}(A_0)$ is $\zeta _0$ -s-strongly stationary in $P_{\kappa _x}x$ and thus $A\cap d_{\zeta _0}(A_0)\cap d_\eta (S)\cap d_\eta (T)\neq \emptyset $ .

Let us prove that if $(1)_{\leq \xi }$ , $(2)_{\leq \xi }$ , and $(3)_{\leq \xi +1}$ hold then $(1)_{\xi +1}$ holds (this argument is similar to that of Proposition 3.14). Suppose A is not $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ . Then by $(1)_{\leq \xi }$ , for some $\zeta \leq \xi $ there are sets S and T which are $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ such that $A\cap d_\zeta (S)\cap d_\zeta (T)=\emptyset $ . Thus $d_\zeta (S)\cap d_\zeta (T)\cap (0,x]$ is an open neighborhood of x in the $\tau _{\xi +1}$ topology that does not intersect A in some point other than x. Conversely, suppose A is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ . To show that $x\in d_{\xi +1}(A)$ , let U be an arbitrary basic open neighborhood of x in the $\tau _{\xi +1}$ topology. By Lemma 3.6, we can assume that U is of the form

$$\begin{align*}U=I\cap d_\zeta(A_0)\cap\cdots\cap d_\zeta(A_{n-1}),\end{align*}$$

where $I\in {\mathcal B}_0$ , $n<\omega $ , $\zeta <\xi +1$ and $A_i\subseteq P_\kappa X$ for $i<n$ . Since $x\in U$ it follows from $(1)_\zeta $ that each $A_i$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ , and thus by $(3)_{\xi +1}$ we see that $A\cap d_\zeta (A_0)\cap \cdots \cap d_\zeta (A_{n-1})$ is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ , and thus U intersects A in some point other than x.

Finally, we prove that if $(1)_{\leq \xi +1}$ , $(2)_{\leq \xi }$ , and $(3)_{\leq \xi +1}$ hold, then $(2)_{\xi +1}$ must also hold. Suppose A is $\xi +2$ -s-strongly stationary in $P_{\kappa _x}x$ . By $(2)_{\leq \xi }$ , it suffices to show that whenever S and T are $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ , the set ${A\cap d_{\xi +1}(S)\cap d_{\xi +1}(T)}$ is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ . So, fix Y and Z which are $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ for some $\zeta \leq \xi $ . By $(3)_{\xi +1}$ , it follows that ${S\cap d_\zeta (Y)}$ and $T\cap d_\zeta (Z)$ are $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ , and hence by the $\xi +1$ -s-strong stationarity of A and $(1)_{\xi +1}$ we have $A\cap d_{\xi +1}(S\cap d_\zeta (Y))\cap d_{\xi +1}(T\cap d_\zeta (Z))\neq \emptyset $ . But

$$\begin{align*}d_{\xi+1}(S\cap d_\zeta(Y))\cap d_{\xi+1}(T\cap d_\zeta(Z))=d_{\xi+1}(S)\cap d_{\xi+1}(T)\cap d_\zeta(Y)\cap d_\zeta(Z)\end{align*}$$

by Lemma 3.5, and thus $A\cap d_{\xi +1}(S)\cap d_{\xi +1}(T)$ is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ . The backward direction of $(2)_{\xi +1}$ follows easily from $(1)_{\leq \xi }$ .

Proof Fix $\zeta '$ with $\zeta \leq \zeta ' \leq \xi $ . It follows from Theorem 3.16(3) that $d_\zeta (A)$ is $\xi $ -s-strongly stationary and hence $\zeta '$ -s-strongly stationary in $P_{\kappa _x} x$ . Furthermore, $d_\zeta (A)$ is $\zeta '$ -s-closed below $P_{\kappa _x}x$ since $d_{\zeta '}(d_\zeta (A))\subseteq d_\zeta (d_\zeta (A))\subseteq d_\zeta (A)$ .

Proof For the forward direction, suppose that $P_{\kappa _x} x$ is not $\xi $ -s-strongly stationary. Then there is $\zeta < \xi $ and sets $S,T \subseteq P_{\kappa _x}x$ such that S and T are both $\zeta $ -s-strongly stationary in $P_{\kappa _x} x$ but there is no $y \prec x$ such that S and T are both $\zeta $ -s-strongly stationary in $P_{\kappa _y} y$ . Then, by Theorem 3.16(1), we have $d_\zeta (S) \cap d_\zeta (T) = \{x\}$ , so x is isolated in $(P_\kappa X, \tau _\xi )$ .

For the converse, suppose that $P_{\kappa _x} x$ is $\xi $ -s-strongly stationary, and fix an interval $I \in {\mathcal B}_0$ , an $n < \omega $ , ordinals $\xi _0, \ldots , \xi _{n-1} < \xi $ , and sets $A_0, \ldots , A_{n-1} \subseteq P_{\kappa _x} x$ such that

$$\begin{align*}x \in U := I \cap d_{\xi_0}(A_0) \cap \dots \cap d_{\xi_{n-1}}(A_{n-1}). \end{align*}$$

Let $\zeta := \max \{\zeta _i \mid i < n\} < \xi $ . By Corollary 3.17, each of I, $d_{\xi _0}(A_0)$ , …, $d_{\xi _{n-1}}(A_{n-1})$ is a $\zeta $ -s-weak club in $P_{\kappa _x} x$ . By Corollary 3.19, U is also $\zeta $ -s-weak club in $P_{\kappa _x} x$ . In particular, $U \neq \{x\}$ ; hence, x is not isolated in $P_{\kappa _x} x$ .

Proof We only provide a proof of (1) since the proof of (2) is essentially identical. Suppose A is $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ . Fix $\zeta <\xi $ and assume that $C\subseteq P_{\kappa _x}x$ is a $\zeta $ -s-weak club in $P_{\kappa _x}x$ . Since C is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ there is some $y\in d_\zeta (C)\cap A$ , but since $d_\zeta (C)\subseteq C$ we have $y\in C\cap A$ . Conversely, suppose that for all $\zeta <\xi $ and every $C\subseteq P_{\kappa _x}x$ that is a $\zeta $ -s-weak club in $P_{\kappa _x}x$ we have $A\cap C\neq \emptyset $ . To show that A is $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ , suppose S and T are $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ for some $\zeta <\xi $ . Then, since we are assuming that $P_{\kappa _x}x$ is $\xi $ -s-strongly stationary, it follows by Theorem 3.16(3) that $d_\zeta (S)\cap d_\zeta (T)$ is $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ . Furthermore,

$$\begin{align*}d_\zeta(d_\zeta(S)\cap d_\zeta(T))\subseteq d_\zeta(d_\zeta(S))\cap d_\zeta(d_\zeta(T))\subseteq d_\zeta(S)\cap d_\zeta(T),\end{align*}$$

which implies that $d_\zeta (S)\cap d_\zeta (T)$ is a $\zeta $ -s-weak club in $P_{\kappa _x}x$ . Thus $A\cap d_\zeta (S)\cap d_\zeta (T)\cap P_{\kappa _x}x\neq \emptyset $ , and hence A is $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ as desired.

Proof Suppose $P_{\kappa _x}x$ is $0$ -s-strongly stationary. Then ${\mathop{\mathrm{NS}}}_{\kappa _x,x}^0$ is the ideal $I_{\kappa _x,x}$ consisting of all subsets A of $P_{\kappa _x}x$ such that there is some $y\in P_{\kappa _x}x$ with ${A\cap (y,x)=\emptyset} $ . Clearly this is a nontrivial ideal since $P_{\kappa _x}x\notin I_{\kappa _x,x}$ .

Now suppose $\xi>0$ . Let us show that ${\mathop{\mathrm{NS}}}_{\kappa _x,x}^\xi $ is an ideal. Suppose A and B are both not $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ . By Corollary 3.19, there are sets $C_A,C_B\subseteq P_{\kappa _x}x$ such that $C_A$ is a $\zeta _A$ -s-weak club in $P_{\kappa _x}x$ for some $\zeta _A<\xi $ , $C_B$ is a $\zeta _B$ -s-weak club in $P_{\kappa _x}x$ for some $\zeta _B<\xi $ , such that $C_A\cap A=\emptyset $ and ${C_B\cap B=\emptyset }$ . Then $d_{\zeta _A}(C_A)\cap d_{\zeta _B}(C_B) \cap (A\cup B)=\emptyset $ where $d_{\zeta _A}(C_A)\cap d_{\zeta _B}(C_B)$ is a $\zeta $ -s-weak club in $P_{\kappa _x}x$ for $\zeta =\max \{\zeta _A,\zeta _B\}$ because $d_{\zeta _A}(C_A)\cap d_{\zeta _B}(C_B)\cap P_{\kappa _x}x$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ by Theorem 3.16(3) and furthermore

$$\begin{align*}d_\zeta(d_{\zeta_A}(C_A)\cap d_{\zeta_B}(C_B))\subseteq d_\zeta(C_A)\cap d_\zeta(C_B).\\[-34pt] \end{align*}$$

Proof For the forward direction, suppose that (2) fails, and let $\zeta $ , x, and A form a counterexample, with $\zeta $ minimal among all such counterexamples. Note that we must have $\zeta> 0$ .

Claim 3.22. $P_{\kappa _x} x$ is not $\zeta $ -s-strongly stationary.

Proof Suppose otherwise. We will show that A is in fact $\zeta $ -s-strongly stationary, contradicting our choice of A. By Corollary 3.19, it suffices to show that, for all $\eta < \zeta $ and every $\eta $ -s-weak club C in $P_{\kappa _x} x$ , we have $A \cap C \neq \emptyset $ . Fix such $\eta $ and C. Then C is $\eta $ -s-strongly stationary in $P_{\kappa _x} x$ and hence, by the minimality of $\zeta $ , $\eta $ -strongly stationary in $P_{\kappa _x} x$ . Thus, since A is $\zeta $ -strongly stationary, there is $y \in A$ such that C is $\eta $ -strongly stationary in $P_{\kappa _y} y$ and hence, again by the minimality of $\zeta $ , $\eta $ -s-strongly stationary in $P_{\kappa _y} y$ . But then, since C is an $\eta $ -s-weak club in $P_{\kappa _x} x$ , we have $y \in C \cap A$ , as desired.

Proof We proceed by induction on $\xi $ . We let $\Phi _0(R,S,T)$ be the $\Pi ^1_0$ formula

$$\begin{align*}(\forall y\in S) (\exists x\in R)\ (y,x)\in T\end{align*}$$

so that $\Phi _0[A,P_{\kappa _x}x,\prec ]$ expresses that A is $0$ -s-strongly stationary (i.e., $\prec $ -cofinal) in $P_{\kappa _x}x$ over the structure $(V_{\kappa _x}(\kappa _x,x),\in ,A,P_{\kappa _x}x,\prec )$ .

Suppose $\xi $ is a limit ordinal. It is easy to see that $\Phi _\xi = \bigwedge _{\zeta <\xi }\Phi _\zeta $ is as desired.

Suppose $\xi =\zeta +1$ . Let $\Phi _\zeta $ be the $\Pi ^1_\zeta $ -formula obtained from the induction hypothesis. Then for all $x\in P_\kappa X$ a set $A\subseteq P_{\kappa _x}x$ is $\zeta $ -s-strongly stationary in $P_{\kappa _x}x$ if and only if

$$\begin{align*}(V_{\kappa_x}(\kappa_x,x),\in,A,P_{\kappa_x}x,\prec_x)\models\Phi_\zeta[A,P_{\kappa_x}x,\prec_x].\end{align*}$$

For $x\in P_\kappa X$ we see that $A\subseteq P_{\kappa _x}x$ is $\xi $ -s-strongly stationary in $P_{\kappa _x}x$ if and only if

$$\begin{align*}\Phi_\zeta[A,P_{\kappa_x}x,\prec_x] \land (\forall S\subseteq P_{\kappa_x}x)(\forall T\subseteq P_{\kappa_x}x)[\Phi_\zeta[S,P_{\kappa_x}x,\prec_x]\land\Phi_\zeta[T,P_{\kappa_x}x,\prec_x]\longrightarrow\end{align*}$$

$$\begin{align*}(\exists y\in A) \Phi_\zeta[S\cap P_{\kappa_y}y,P_{\kappa_y}y,\prec_y]\land \Phi_\zeta[T\cap P_{\kappa_y}y,P_{\kappa_y}y,\prec_y]]\end{align*}$$

holds in $(V_{\kappa _x}(\kappa _x,x),\in ,A,P_{\kappa _x}x,\prec _x)$ . It is easy to check that the previous formula is equivalent to a $\Pi ^1_\xi $ formula, hence the desired formula $\Phi _\xi (R,S,T)$ exists.

Proof To show that A is $\xi +1$ -s-strongly stationary in $P_{\kappa _x}x$ , fix sets $S,T\subseteq P_{\kappa _x}x$ that are $\zeta $ -s-strongly stationary where $\zeta \leq \xi $ and let $\Phi _\zeta $ be the $\Pi ^1_\zeta $ -formula obtained from Lemma 3.26. Since A is $\Pi ^1_\xi $ -indescribable, it is $\Pi ^1_\zeta $ -indescribable and the fact that

$$\begin{align*}(V_{\kappa_x}(\kappa_x,x),\in,S,T,P_{\kappa_x}x,\prec_x)\models\Phi_\zeta[S,P_{\kappa_x}x,\prec_x]\land\Phi_\zeta[T,P_{\kappa_x}x,\prec_x]\end{align*}$$

implies that there is some $y\in A\cap P_{\kappa _x}x$ with $\kappa _y=y\cap \kappa $ such that the structure