Proof Let $\kappa $ be a VSS cardinal. Then there is an ordinal $\lambda> \kappa $ , a transitive set M such that $V_\lambda \subset M$ , and a generic elementary embedding $j : V_\lambda \to M$ with $\operatorname {\mathrm {crit}}(j) = \kappa $ . (Here we will not need $j(f)(\kappa ) \le \lambda $ for any particular function f.)

Let $\vec {A}$ be a $\kappa $ -sequence of sets such that $\vec {A}(\alpha ) \subset \alpha $ for every ordinal $\alpha < \kappa $ . Then we may define a subset $A \subset \kappa $ by $A = j(\vec {A})(\kappa )$ . We will show that the set

$$ \begin{align*}S = \{\alpha < \kappa : A \cap \alpha = \vec{A}(\alpha)\}\end{align*} $$

is stationary. Letting C be a club set in $\kappa $ we have $\kappa \in j(C)$ , and because

$$ \begin{align*}j(A) \cap \kappa = A = j(\vec{A})(\kappa),\end{align*} $$

we have $\kappa \in j(S)$ also, so it follows that

$$ \begin{align*}\kappa \in j(C) \cap j(S) = j(C \cap S),\end{align*} $$

and by the elementarity of j we have $C \cap S \ne \emptyset $ .

Proof Because $\kappa $ is a limit of inaccessible cardinals we may define a function $g : \kappa \to \kappa $ such that $g(\alpha )$ is the least inaccessible cardinal greater than $\max \{f(\alpha ), \alpha \}$ for all $\alpha < \kappa $ . Because $\kappa $ is VSS with respect to the function $g+1$ defined by $\alpha \mapsto g(\alpha )+1$ , there is an ordinal $\beta> \kappa $ , a transitive set N with $V_\beta \subset N$ , and a generic elementary embedding

$$ \begin{align*}j: V_\beta \to N \text{ with } \operatorname{\mathrm{crit}}(j) = \kappa \text{ and } j(g)(\kappa) < \beta.\end{align*} $$

By the definition of g from f and the elementarity of j it follows that $j(g)(\kappa )$ is the least inaccessible cardinal in N greater than $\max \{j(f)(\kappa ), \kappa \}$ . Because $j(g)$ is a function from $j(\kappa )$ to $j(\kappa )$ we have $j(g)(\kappa ) < j(\kappa )$ . Letting $\lambda = j(g)(\kappa )$ we therefore have $\lambda> \kappa $ and

$$ \begin{align*}j(f)(\kappa) < \lambda < j(\kappa).\end{align*} $$

Because $\lambda < \beta $ and $V_\beta \subset N$ , the inaccessibility of $\lambda $ is absolute from N to V. Define $j_1 = j \restriction V_\lambda $ and M = $j(V_\lambda ) = V_{j(\lambda )}^N$ . Then $j_1 : V_\lambda \to M$ is a generic elementary embedding with $\operatorname {\mathrm {crit}}(j_1) = \kappa $ and $j_1(f)(\kappa ) < \lambda < j_1(\kappa )$ as desired. Because $V_\lambda $ satisfies ZFC it follows by the elementarity of $j_1$ that M satisfies ZFC.

Proof Let $\kappa $ be a VSS cardinal and let $f : \kappa \to \kappa $ be a function in L. Then by Lemma 2.2 there is an inaccessible cardinal $\lambda> \kappa $ , a transitive model M of ZFC with $V_\lambda \subset M$ , and a generic elementary embedding $j : V_\lambda \to M$ with $\operatorname {\mathrm {crit}}(j) = \kappa $ and $j(f)(\kappa ) < \lambda < j(\kappa )$ . Note that $L^{V_\lambda } = L_\lambda $ and $L^M = L_\theta $ where $\theta = \mathrm {Ord}^M$ . Moreover, because $\lambda $ is inaccessible we have $L_\lambda = V_\lambda ^L$ . For the elementary embedding $j_1 = j \restriction V_\lambda ^L$ we have

$$ \begin{align*}j_1 : V_\lambda^L \to L_\theta \text{ and } \operatorname{\mathrm{crit}}(j_1) = \kappa \text{ and } j_1(f) = j(f).\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_\lambda ^L)$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in L[G]$ such that

$$ \begin{align*}j_2 : V_\lambda^L \to L_\theta \text{ and } \operatorname{\mathrm{crit}}(j_2) = \kappa \text{ and } j_2(f) = j(f).\end{align*} $$

We have $V_\lambda ^L = L_\lambda \subset L_\theta $ and $j_2(f)(\kappa ) = j(f)(\kappa ) < \lambda $ , so $j_2$ witnesses the VSS property for $\kappa $ in L with respect to f.

Proof (1) $\!\implies\! $ (2): Assume that $\kappa $ is VSS and let $f: \kappa \to \kappa $ . By Lemma 2.2 there is an inaccessible cardinal $\lambda> \kappa $ , a transitive model M of ZFC with $V_\lambda \subset M$ , and a generic elementary embedding

$$ \begin{align*}j : V_\lambda \to M \text{ with }\operatorname{\mathrm{crit}}(j) = \kappa \text{ and }j(f)(\kappa) < \lambda < j(\kappa).\end{align*} $$

Define $\beta = \max \{j(f)(\kappa ), \kappa +1\}$ , so $\kappa < \beta < \lambda $ and $V_\beta ^M = V_\beta $ . Let $j_1 = j \restriction V_\beta $ and note that

$$ \begin{align*}j_1 : V_\beta \to V_{j(\beta)}^M \text{ and } \operatorname{\mathrm{crit}}(j_1) = \kappa \text{ and } j_1(\kappa) = j(\kappa) \text{ and } j_1(f) = j(f).\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_\beta )$ be a V-generic filter. Because $M[G]$ is wellfounded, by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in M[G]$ such that

$$ \begin{align*}j_2 : V_\beta \to V_{j(\beta)}^M \text{ and } \operatorname{\mathrm{crit}}(j_2) = \kappa \text{ and } j_2(\kappa) = j(\kappa) \text{ and } j_2(f) = j(f).\end{align*} $$

Then we have

$$ \begin{align*}j_2(\operatorname{\mathrm{crit}}(j_2)) = j(\kappa) \text{ and } j(f) \in \operatorname{\mathrm{ran}}(j_2) \text{ and }j(f)(\operatorname{\mathrm{crit}}(j_2)) \le \beta,\end{align*} $$

because $j(f)(\kappa ) \le \beta $ . By the elementarity of j and the definability of forcing there is an ordinal $\bar {\beta } < \lambda $ such that, letting $g \subset \operatorname {\mathrm {Col}}(\omega ,V_{\bar {\beta }})$ be a V-generic filter, there is an elementary embedding $j_3 \in V_\lambda [g]$ with

$$ \begin{align*}j_3 : V_{\bar{\beta}} \to V_{\beta} \text{ and } j_3(\operatorname{\mathrm{crit}}(j_3)) = \kappa \text{ and } f \in \operatorname{\mathrm{ran}}(j_3) \text{ and } f(\operatorname{\mathrm{crit}}(j_3)) \le \bar{\beta}.\end{align*} $$

Therefore statement 2 holds for f.

(2) $\!\implies\! $ (1): Assume that statement 2 holds and let $f: \kappa \to \kappa $ . Note that statement 2 implies the ineffability of $\kappa $ by an argument similar to Proposition 2.1. It follows that $\kappa $ is a limit of inaccessible cardinals, so by increasing the values of f we may assume that for all $\alpha < \kappa $ , $f(\alpha )$ is an inaccessible cardinal greater than $\alpha $ and we may furthermore define $f^+(\alpha )$ to be the least inaccessible cardinal greater than $f(\alpha )$ . Applying statement 2 to the function $f^+: \kappa \to \kappa $ yields ordinals $\bar {\lambda }$ and $\lambda $ and a generic elementary embedding

$$ \begin{align*}j :V_{\bar{\lambda}} \to V_\lambda \text{ with } j(\bar{\kappa}) = \kappa \text{ and } f^+ \in \operatorname{\mathrm{ran}}(j) \text{ and } f^+(\bar{\kappa}) \le \bar{\lambda},\end{align*} $$

where $\bar {\kappa } = \operatorname {\mathrm {crit}}(j)$ . By restricting j if necessary we may assume that $\bar {\lambda }$ is equal to $f^+(\bar {\kappa })$ and is therefore an inaccessible cardinal less than $\kappa $ . Let $\bar {\beta } = f(\bar {\kappa })$ and $\beta = j(\bar {\beta })$ .

Note that $f^+ \in \operatorname {\mathrm {ran}}(j)$ implies $f \in \operatorname {\mathrm {ran}}(j)$ because for all $\alpha < \kappa $ , $f(\alpha )$ is definable in $V_\lambda $ as the largest inaccessible cardinal less than $f^+(\alpha )$ , so we may define $\bar {f} = j^{-1}(f)$ . Then we have $\bar {f} : \bar {\kappa } \to \bar {\kappa }$ and $\bar {\beta } = j(\bar {f})(\bar {\kappa })$ . Note that

$$ \begin{align*} \bar{\kappa} < \bar{\beta} < \bar{\lambda} < \kappa < \beta < \lambda.\end{align*} $$

For the elementary embedding $j_1 = j \restriction V_{\bar {\beta }}$ we have

$$ \begin{align*} j_1 : V_{\bar{\beta}} \to V_\beta \text{ and } \operatorname{\mathrm{crit}}(j_1) = \bar{\kappa} \text { and } j_1(\bar{\kappa}) = \kappa \text{ and } j_1(\bar{f}) = f.\end{align*} $$

Let $g \subset \operatorname {\mathrm {Col}}(\omega , V_{\bar {\beta }})$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in V[g]$ such that

$$ \begin{align*} j_2 : V_{\bar{\beta}} \to V_\beta \text{ and } \operatorname{\mathrm{crit}}(j_2) = \bar{\kappa} \text { and } j_2(\bar{\kappa}) = \kappa \text{ and } j_2(\bar{f}) = f.\end{align*} $$

Letting $M = V_\beta $ we therefore have

$$ \begin{align*} V_{\bar{\beta}} \subset M \text{ and } j_2 : V_{\bar{\beta}} \to M \text{ and } \operatorname{\mathrm{crit}}(j_2) = \bar{\kappa} \text { and } j_2(\bar{f})(\bar{\kappa}) = \bar{\beta}.\end{align*} $$

Because $\bar {\lambda }$ is inaccessible in V and remains so in $V[g]$ , a Skolem hull argument in $V[g]$ yields a transitive set $M' \in V_{\bar {\lambda }}[g]$ and an elementary embedding $j_3 \in V_{\bar {\lambda }}[g]$ such that

$$ \begin{align*} V_{\bar{\beta}} \subset M' \text{ and } j_3 : V_{\bar{\beta}} \to M' \text{ and } \operatorname{\mathrm{crit}}(j_3) = \bar{\kappa} \text { and } j_3(\bar{f})(\bar{\kappa}) = \bar{\beta}.\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega , V_\beta )$ be a V-generic filter. By the elementarity of j and the definability of forcing, there is a transitive set $M'' \in V_{\lambda }[G]$ and an elementary embedding $j_4 \in V_{\lambda }[G]$ with

$$ \begin{align*}V_{\beta} \subset M'' \text{ and } j_4 : V_{\beta} \to M'' \text{ and } \operatorname{\mathrm{crit}}(j_4) = \kappa \text{ and } j_4(f)(\kappa) = \beta.\end{align*} $$

Therefore $\kappa $ is VSS with respect to f.

Proof Assume that $0^\sharp $ exists and let $\kappa $ be a Silver indiscernible. Then there is an elementary embedding $j: L \to L$ with $\operatorname {\mathrm {crit}}(j) = \kappa $ . Let $f : \kappa \to \kappa $ be a function in L and define $\lambda = \max \{j(f)(\kappa ), \kappa +1\}$ . For the elementary embedding $j_1 = j \restriction V_\lambda ^L$ we have

$$ \begin{align*} j_1 : V_\lambda^L \to V_{j(\lambda)}^L \text{ and } \operatorname{\mathrm{crit}}(j_1) = \kappa \text{ and } j_1(\kappa) = j(\kappa) \text{ and } j_1(f) = j(f).\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_\lambda ^L)$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in L[G]$ such that

$$ \begin{align*} j_2 : V_\lambda^L \to V_{j(\lambda)}^L \text{ and } \operatorname{\mathrm{crit}}(j_2) = \kappa \text{ and } j_2(\kappa) = j(\kappa) \text{ and } j_2(f) = j(f).\end{align*} $$

We have $j_2(f)(\kappa ) = j(f)(\kappa ) \le \lambda $ , so $j_2$ witnesses the VSS property for $\kappa $ in L with respect to the function f.

Proof Assume that $\kappa $ is 2-iterable. Then by Gitman and Welch [Reference Gitman and Welch6, Theorem 4.7] there is a transitive model M of ZFC with $V_\kappa \in M$ and an elementary embedding $j : M \to N$ with critical point equal to $\kappa $ where N is a transitive model of ZFC and $M = V_{j(\kappa )}^N$ .

First, we show that $\kappa $ is VSS in N. Let $f : \kappa \to \kappa $ be in N and therefore also in M. Define $\lambda = \max \{j(f)(\kappa ), \kappa + 1\}$ . Because $j(f)$ is a function from $j(\kappa )$ to $j(\kappa )$ we have $\lambda < j(\kappa ) = \mathrm {Ord}^M$ , so $V_\lambda ^M = V_\lambda ^N$ and we have an elementary embedding $j_1 = j \restriction V_\lambda ^M$ with

$$ \begin{align*} j_1 : V_\lambda^N \to V_{j(\lambda)}^N \text{ and } \operatorname{\mathrm{crit}}(j_1) = \kappa \text{ and } j_1(\kappa) = j(\kappa)\text{ and } j_1(f) = j(f).\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_\lambda ^N)$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in N[G]$ such that

$$ \begin{align*} j_2 : V_\lambda^N \to V_{j(\lambda)}^N \text{ and } \operatorname{\mathrm{crit}}(j_2) = \kappa \text{ and } j_2(\kappa) = j(\kappa)\text{ and } j_2(f) = j(f).\end{align*} $$

Because $j_2(f)(\kappa ) = j(f)(\kappa ) \le \lambda $ , this elementary embedding $j_2$ witnesses the VSS property for $\kappa $ in N with respect to f.

Now let C be club in $\kappa $ . Then $\kappa \in j(C)$ , so the model N satisfies the statement “there is a VSS cardinal in $j(C)$ ” and by the elementarity of j it follows that the model M satisfies the statement “there is a VSS cardinal in C.” Let $\bar {\kappa } \in C$ be VSS in M. Because we have $j(\bar {\kappa }) = \bar {\kappa }$ and

$$ \begin{align*}j(V_\kappa) = j(V_\kappa^M) = V_{j(\kappa)}^N = M,\end{align*} $$

it follows by the elementarity of j that $\bar {\kappa }$ is VSS in $V_\kappa $ .

Proof Let $\kappa $ be a VSS cardinal, let $A \subset V_\kappa $ , and assume toward a contradiction that no cardinal less than $\kappa $ is virtually A-extendible in $\langle V_\kappa , V_{\kappa +1}; \mathord {\in }\rangle $ . Then we may define a function $f:\kappa \to \kappa $ such that for every ordinal $\alpha < \kappa $ , $f(\alpha )$ is the least ordinal greater than $\alpha $ such that for every ordinal $\theta < \kappa $ there is no generic elementary embedding from $\langle V_{f(\alpha )}; \mathord {\in },A \cap V_{f(\alpha )}\rangle $ to $\langle V_\theta; \mathord {\in },A \cap V_\theta \rangle $ that has critical point $\alpha $ and maps $\alpha $ above $f(\alpha )$ .

Because $\kappa $ is VSS, by Lemma 2.2 there is an inaccessible cardinal $\lambda> \kappa $ , a transitive model M of ZFC with $V_\lambda \subset M$ , and a generic elementary embedding

$$ \begin{align*}j : V_\lambda \to M \text{ with } \operatorname{\mathrm{crit}}(j) = \kappa \text { and }j(f)(\kappa) < \lambda < j(\kappa).\end{align*} $$

Defining $\beta = j(f)(\kappa )$ , we have

$$ \begin{align*}\kappa < \beta < \lambda < j(\kappa) < j(\beta) < \mathrm{Ord}^M\kern-1.2pt .\end{align*} $$

Let $j_1 = j \restriction V_\beta $ , considered as an elementary embedding whose domain is the structure $\langle V_\beta; \mathord {\in }, j(A) \cap V_\beta \rangle $ . Then we have

$$ \begin{align*}j_1 : \langle V_\beta; \mathord{\in}, j(A) \cap V_\beta\rangle \to \langle V_{j(\beta)}^M; \mathord{\in}, j(j(A) \cap V_\beta)\rangle \text{ and } \operatorname{\mathrm{crit}}(j_1) = \kappa \text{ and } j_1(\kappa) = j(\kappa).\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_\beta )$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in M[G]$ such that

$$ \begin{align*}j_2 : \langle V_\beta; \mathord{\in}, j(A) \cap V_\beta\rangle \to \langle V_{j(\beta)}^M; \mathord{\in}, j(j(A) \cap V_\beta)\rangle \text{ and } \operatorname{\mathrm{crit}}(j_2) = \kappa \text{ and } j_2(\kappa) = j(\kappa).\end{align*} $$

Note that $j_2(\kappa ) = j(\kappa )> j(f)(\kappa ) = \beta $ . By the elementarity of j and the definability of forcing it follows that there is a cardinal $\alpha < \kappa $ such that, letting $g \subset \operatorname {\mathrm {Col}}(\omega ,V_{f(\alpha )})$ be a V-generic filter, there is an elementary embedding $j_3 \in V_\lambda [g]$ with

$$ \begin{align*}j_3 : \langle V_{f(\alpha)}; \mathord{\in}, A \cap V_{f(\alpha)}\rangle \to \langle V_{\beta}; \mathord{\in}, j(A) \cap V_\beta\rangle \text{ and } \operatorname{\mathrm{crit}}(j_3) = \alpha \text{ and } j_3(\alpha)> f(\alpha).\end{align*} $$

Because $V_\lambda \subset M$ we have $j_3 \in M[g]$ , and because $\beta < j(\kappa )$ it then follows by the elementarity of j that there is an ordinal $\theta < \kappa $ and an elementary embedding $j_4 \in V_\lambda [g]$ such that

$$ \begin{align*}j_4 : \langle V_{f(\alpha)}; \mathord{\in}, A \cap V_{f(\alpha)}\rangle \to \langle V_{\theta}; \mathord{\in}, A \cap V_\theta\rangle \text{ and } \operatorname{\mathrm{crit}}(j_4) = \alpha \text{ and } j_4(\alpha)> f(\alpha).\end{align*} $$

The existence of such a generic elementary embedding $j_4$ contradicts the definition of the function f.

Proof (1) $\!\implies\! $ (2): Assume that statement 2 fails, so there is a generic extension $V[G]$ of V by a poset $\mathbb {P}$ and a real $x \in A^{V[G]} \setminus V$ . Letting $\langle T,\tilde {T}\rangle $ be a $\mathbb {P}$ -absolutely complementing pair of trees for A in V we have $x \in \operatorname {\mathrm {p}}[T]^{V[G]} \setminus L[T]$ , so by Mansfield’s theorem in $V[G]$ (see Jech [Reference Jech7, Lemma 25.24]) there is a perfect tree $U \in L[T]$ on $\omega $ such that $[U] \subset \operatorname {\mathrm {p}}[T]$ in $V[G]$ . This implies $[U] \subset A$ in V, so A is not thin.

(2) $\!\implies\! $ (3): Assume statement 2 and let $\mathbb {P}$ be a poset. Then there is a tree $T_{\neg A}$ on $\omega \times \mathrm {Ord}$ such that

$$ \begin{align*}{} \mathbin{\Vdash}_{\mathbb{P}} \operatorname{\mathrm{p}}[T_{\neg A}] = \omega^\omega \setminus A.\end{align*} $$

Let $n< \omega $ . From $T_{\neg A}$ , one can define a tree $T_{\neg A^n}$ on $\omega ^n \times \mathrm {Ord}$ such that

$$ \begin{align*}{} \mathbin{\Vdash}_{\mathbb{P}} \operatorname{\mathrm{p}}[T_{\neg A^n}] = (\omega^\omega)^n \setminus A^n\kern-1.2pt .\end{align*} $$

Now let $B \subset A^n$ . We will show that B is $\mathbb {P}$ -Baire. Take trees $T_B$ and $T_{A^n \setminus B}$ on $\omega ^n \times \left |B\right |$ and $\omega ^n \times \left |A^n \setminus B\right |$ respectively that project to B and $A^n \setminus B$ respectively in every generic extension. (Such trees can be trivially defined for every pointset.) From $T_{\neg A^n}$ and $T_{A^n \setminus B}$ , one can define a tree $T_{\neg B}$ on $\omega ^n \times \mathrm {Ord}$ such that every generic extension satisfies $\operatorname {\mathrm {p}}[T_{\neg B}] = \operatorname {\mathrm {p}}[T_{\neg A^n}] \cup \operatorname {\mathrm {p}}[T_{A^n \setminus B}]$ . Then we have

$$ \begin{align*}{} \mathbin{\Vdash}_{\mathbb{P}} \operatorname{\mathrm{p}}[T_{\neg B}] = (\omega^\omega)^n \setminus B,\end{align*} $$

so the pair of trees $\langle T_B, T_{\neg {B}}\rangle $ witnesses that B is $\mathbb {P}$ -Baire.

(3) $\!\implies\! $ (4): This follows directly from the existence of a wellordering of A given by the axiom of choice.

(4) $\!\implies\! $ (1): Suppose toward a contradiction that some universally Baire set of reals A has a universally Baire wellordering but is not thin. Because A has a perfect subset there is a continuous injection $f : 2^\omega \to \omega ^\omega $ whose range is contained in A. Taking the preimage of a universally Baire wellordering of A under the continuous function $f \times f$ we obtain a wellordering of $2^\omega $ with the Baire property, which leads to a contradiction using the Kuratowski–Ulam theorem (see Kanamori [Reference Kanamori8, Corollary 13.10]).

Proof Suppose toward a contradiction that for every ordinal $\alpha < \kappa $ we have

$$ \begin{align*}\forall A_0 \in \operatorname{\mathrm{uB}}_\kappa^{V[G\restriction\alpha]}\, A_0^{V[G]} \ne A.\end{align*} $$

Then for every ordinal $\alpha < \kappa $ , because $\kappa $ is inaccessible in $V[G \restriction \alpha ]$ and we have

$$ \begin{align*} \operatorname{\mathrm{uB}}_\kappa^{V[G \restriction \alpha]} = \bigcap_{\beta < \kappa} \operatorname{\mathrm{uB}}_\beta^{V[G \restriction \alpha]} \text{ and } \mathbb{R}^{V[G]} = \bigcup_{\beta < \kappa} \mathbb{R}^{V[G\restriction \beta]},\end{align*} $$

it follows that for all sufficiently large ordinals $\beta < \kappa $ we have $\beta> \alpha $ and

$$ \begin{align*}\forall A_0 \in \operatorname{\mathrm{uB}}^{V[G\restriction\alpha]}_{\beta}\, A_0^{V[G \restriction \beta]} \ne A \cap V[G \restriction \beta].\end{align*} $$

Then by the $\kappa $ -chain condition for $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\kappa )$ it follows that there is a function $f : \kappa \to \kappa $ in V such that for all $\alpha < \kappa $ we have $f(\alpha )> \alpha $ and

$$ \begin{align*}\forall A_0 \in \operatorname{\mathrm{uB}}^{V[G\restriction\alpha]}_{f(\alpha)}\, A_0^{V[G \restriction f(\alpha)]} \ne A \cap V[G \restriction f(\alpha)].\end{align*} $$

Because $\kappa $ is a limit of inaccessible cardinals in V we may additionally assume that $f(\alpha )$ is inaccessible in V for every ordinal $\alpha < \kappa $ .

Because $\kappa $ is VSS, by Lemma 2.2 there is an inaccessible cardinal $\lambda> \kappa $ , a transitive model M of ZFC in V with $V_\lambda \subset M$ , and a generic elementary embedding

$$ \begin{align*} j : V_{\lambda} \to M \text{ with } \operatorname{\mathrm{crit}}(j) = \kappa \text{ and } j(f)(\kappa) < \lambda < j(\kappa).\end{align*} $$

Defining $\beta = j(f)(\kappa )$ , we have

$$ \begin{align*}\kappa < \beta < \lambda < j(\kappa) < j(\beta) < \mathrm{Ord}^M\kern-1.2pt .\end{align*} $$

Note that $\beta $ is inaccessible in M by our assumption on f and the elementarity of j, and because $V_\lambda \subset M$ and $\beta < \lambda $ this implies that $\beta $ is also inaccessible in V.

We can extend j to a generic elementary embedding

$$ \begin{align*}{\hat \jmath} : V_{\lambda}[G] \to M[H],\end{align*} $$

where $H\subset \operatorname {\mathrm {Col}}(\omega ,\mathord {<}j(\kappa ))$ is an V-generic filter such that $H \restriction \kappa = G$ . By the fact that $\beta = j(f)(\kappa )$ and the elementarity of ${\hat \jmath }$ it follows that

$$ \begin{align*}\forall A_0 \in \operatorname{\mathrm{uB}}^{M[G]}_{\beta}\,A_0^{M[H \restriction \beta]} \ne {\hat \jmath}(A) \cap M[H \restriction \beta].\end{align*} $$

We will now obtain a contradiction by showing

$$ \begin{align*}A \in \operatorname{\mathrm{uB}}^{M[G]}_{\beta} \text{ and } A^{M[H \restriction \beta]} = {\hat \jmath}(A) \cap M[H \restriction \beta].\end{align*} $$

In $V[G]$ , because A is universally Baire we may take a $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\beta )$ -absolutely complementing pair of trees $\langle T,\tilde {T}\rangle $ on $\omega \times \mathrm {Ord}$ such that $\operatorname {\mathrm {p}}[T] = A$ . We may assume that $\langle T,\tilde {T}\rangle \in V_{\lambda }[G]$ : if necessary, replace T and $\tilde {T}$ by their images under the transitive collapse of an elementary substructure of a sufficiently large rank initial segment of $V[G]$ containing $V_\beta [G] \cup \{\beta ,T,\tilde {T}\}$ and having cardinality $\beta $ . Because $V_\lambda \subset M$ we have $V_\lambda [G] \subset M[G]$ and the pair $\langle T,\tilde {T}\rangle \in V_{\lambda }[G]$ witnesses $A \in \operatorname {\mathrm {uB}}^{M[G]}_{\beta }$ as desired.

We have $\langle {\hat \jmath }(T), {\hat \jmath }(\tilde {T})\rangle \in M[H]$ and it follows by the elementarity of ${\hat \jmath }$ that

$$ \begin{align*}\operatorname{\mathrm{p}}[{\hat \jmath}(T)]^{M[H]} = {\hat \jmath}(A).\end{align*} $$

Mapping branches pointwise by j gives the inclusions

$$ \begin{align*}\operatorname{\mathrm{p}}[T] \subset \operatorname{\mathrm{p}}[{\hat \jmath}(T)] \text{ and } \operatorname{\mathrm{p}}[\tilde{T}] \subset \operatorname{\mathrm{p}}[{\hat \jmath}(\tilde{T})],\end{align*} $$

which are absolute to transitive models containing these trees, so in particular they hold in $M[H]$ . Because the pair $\langle T,\tilde {T}\rangle $ is $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\beta )$ -absolutely complementing in $V[G]$ and the pair $\langle {\hat \jmath }(T), {\hat \jmath }(\tilde {T})\rangle $ is complementing in $M[H]$ , these inclusions become equalities when restricted to the reals of $M[H \restriction \beta ]$ , so

$$ \begin{align*}A^{M[H \restriction \beta]} = \operatorname{\mathrm{p}}[T]^{M[H \restriction \beta]} = \operatorname{\mathrm{p}}[ {\hat \jmath}(T)]^{M[H \restriction \beta]} = {\hat \jmath}(A) \cap M[H \restriction \beta],\end{align*} $$

completing the desired contradiction.

Proof By Lemma 5.1 every universally Baire set of reals in $V[G]$ is definable in a uniform way from V, G, and elements of $V_\kappa $ , namely an ordinal $\alpha < \kappa $ and a $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\alpha )$ -name for a $\kappa $ -universally Baire set of reals. (Recall that the canonical extension of a $\kappa $ -universally Baire set of reals does not depend on any choice of trees.) Because the rank initial segment $V_\kappa $ of the ground model has cardinality $\omega _1$ in $V[G]$ , it follows that $V[G]$ satisfies $\left |\operatorname {\mathrm {uB}}\right | = \omega _1$ .

Now let B be a set of reals in $L(\operatorname {\mathrm {uB}}, \mathbb {R})^{V[G]}$ . Then B is definable in $V[G]$ from a universally Baire set A, a real x, and an ordinal $\xi $ . By Lemma 5.1 there is an ordinal $\alpha < \kappa $ and a tree T on $\omega \times \mathrm {Ord}$ in $V[G \restriction \alpha ]$ such that $A = \operatorname {\mathrm {p}}[T]^{V[G]}$ . Letting $\dot {T} \in V$ be a $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\alpha )$ name for T and letting $y \in V[G]$ be a real coding x and the hereditarily countable set $G \restriction \alpha $ , the set B is definable in $V[G]$ from the parameter $\langle \xi , \dot {T}\rangle \in V$ and the real parameter y. Then B has the three claimed regularity properties by Lemmas III.1.4, III.1.6, and III.1.7 respectively of Solovay [Reference Solovay14].

Proof Assume toward a contradiction that some function $f: \omega _1 \to \omega _1$ fails to have this property. For every ordinal $\lambda> \omega _1$ the set $\{\sigma \in \mathcal {P}_{\omega _1}(\lambda ) : \sigma \cap \omega _1 \in \omega _1\}$ is always club in $\mathcal {P}_{\omega _1}(\lambda )$ , so it follows from our assumption that the set of all $\sigma \in \mathcal {P}_{\omega _1}(\lambda )$ such that $\sigma \cap \omega _1 \in \omega _1$ and $\operatorname {\mathrm {o.\!t.}}(\sigma ) < f(\sigma \cap \omega _1)$ contains a club set in $\mathcal {P}_{\omega _1}(\lambda )$ . (This statement holds for $\lambda \le \omega _1$ as well, since it is weaker for smaller $\lambda $ .) Increasing the values of f if necessary, we may assume that f is a strictly increasing function.

We may consider every real x as coding a structure $\langle \omega; E_x \rangle $ where $E_x$ is a binary relation on $\omega $ . More precisely, for all $m,n \in \omega $ we let $\langle m,n\rangle \in E_x$ if and only if $x(2^m3^n) = 0$ . We may use AC to choose for every countable ordinal $\beta $ a real $x_\beta $ that codes $\beta $ in the sense that $\langle \omega; E_{x_\beta }\rangle \cong \langle \beta , \in \rangle $ . We claim that the set of reals

$$ \begin{align*} A = \{x_\beta : \beta \in \operatorname{\mathrm{ran}}(f)\}, \end{align*} $$

which is uncountable, is also universally Baire and thin. This will contradict our assumption that every universally Baire set of reals has the perfect set property.

Because $\left |A\right | = \omega _1$ we may trivially define a tree T on $\omega \times \omega _1$ from A such that $\operatorname {\mathrm {p}}[T] = A$ in every generic extension of V. To prove the claim, it will therefore suffice to show that for every poset $\mathbb {P}$ there is a tree $\tilde {T}$ on $\omega \times \mathrm {Ord}$ such that the pair $\langle T,\tilde {T}\rangle $ is $\mathbb {P}$ -absolutely complementing, which for this trivial tree T simply means

$$ \begin{align*} {} \mathbin{\Vdash}_{\mathbb{P}} \operatorname{\mathrm{p}}[\tilde{T}] = \omega^\omega \setminus A.\end{align*} $$

This will show that A is universally Baire by definition, and also that A is thin by condition 2 of Lemma 4.1. (Note that our assumption on f is only needed to prove universal Baireness of A. The thinness of A follows from the more general fact that the set $\{x_\beta : \beta <\omega _1\}$ is thin, which can be proved using the boundedness lemma; see Jech [Reference Jech7, Corollary 25.14].)

Fix a poset $\mathbb {P}$ and let $\eta = \max \{|\mathbb {P}|^+,\omega _1\}$ . Using our hypothesis on the existence of club sets, for every ordinal $\lambda $ in the interval $[\omega _1,\eta )$ we may choose a function

$$ \begin{align*}g_\lambda : \lambda^{\mathord{<}\omega} \to \lambda,\end{align*} $$

such that for every $g_\lambda $ -closed set $\sigma \in \mathcal {P}_{\omega _1}(\lambda )$ we have $\sigma \cap \omega _1 \in \omega _1$ and $\operatorname {\mathrm {o.\!t.}}(\sigma ) < f(\sigma \cap \omega _1)$ . (If $\mathbb {P}$ is countable then this interval is empty and there is nothing to do in this step.) Moreover, we can choose $g_\lambda $ to satisfy the additional property that for every $g_\lambda $ -closed set $\sigma \in \mathcal {P}_{\omega _1}(\lambda )$ the set $\sigma \cap \omega _1$ is f-closed. Then because f is strictly increasing, these properties imply that for every $g_\lambda $ -closed set $\sigma \in \mathcal {P}_{\omega _1}(\lambda )$ the order type of $\sigma $ is not in the range of f.

As a first step in defining the tree $\tilde {T}$ , note that there is a tree $\tilde {T}_1$ on $\omega \times \omega $ in V such that in every generic extension of V we have

$$ \begin{align*} \operatorname{\mathrm{p}}[\tilde{T}_1] = \{x \in \omega^\omega: \langle \omega;E_x\rangle \text{ is not a well-ordering}\},\end{align*} $$

because this set of reals is $\Sigma ^1_1$ .

As a second step in defining $\tilde {T}$ , note that for every ordinal $\beta < \omega _1^V$ there is a tree $\tilde {T}_{2,\beta }$ on $\omega \times \omega $ in V such that in every generic extension of V we have

$$ \begin{align*}\operatorname{\mathrm{p}}[\tilde{T}_{2,\beta}] = \{x \in \omega^\omega : \langle \omega;E_x\rangle \cong \langle \beta;\mathord{\in}\rangle \text{ and } x \notin A\},\end{align*} $$

because this set of reals is $\Sigma ^1_1(x_\beta )$ where $x_\beta $ is our chosen code of $\beta $ . (The condition $x \notin A$ in the definition of this set either subtracts the single point $x_\beta $ from the set or leaves it unchanged, according to whether or not $\beta $ is in the range of f.)

As a third step in defining $\tilde {T}$ , note that for every ordinal $\lambda \in [\omega _1^V, \eta )$ there is a tree $\tilde {T}_{3,\lambda }$ on $\omega \times \lambda $ in V such that in every generic extension of V we have

$$ \begin{align*}\operatorname{\mathrm{p}}[\tilde{T}_{3,\lambda}] = \{x \in \omega^\omega : \langle \omega;E_x\rangle \cong \langle \sigma;\mathord{\in}\rangle \text{ for some}\ g_\lambda\text{-closed set}\ \sigma \in \mathcal{P}_{\omega_1}(\lambda)\}.\end{align*} $$

To show this, it suffices to represent the set of all such reals x as the projection of a closed subset of $\omega ^\omega \times \lambda ^\omega $ onto its first coordinate. Using a definable pairing function $\lambda \times \omega \cong \lambda $ , it equivalently suffices to represent the set of all such reals x as the projection of a closed subset of $\omega ^\omega \times \lambda ^\omega \times \omega ^\omega $ onto its first coordinate. An example of such a closed set is the set of all triples $\langle x, h, y\rangle \in \omega ^\omega \times \lambda ^\omega \times \omega ^\omega $ such that h is an embedding $\langle \omega; E_x\rangle \to \langle \lambda ;\mathord {\in }\rangle $ and y is a function telling us how far we have to look ahead in order to verify that the range of h is $g_\lambda $ -closed. To be precise, we require that for all $n < \omega $ the pointwise image of the set $\{ h(i) : i < n \}^{\mathord {<}n}$ under $g_\lambda $ is contained in $\{ h(i) : i < y(n) \}$ .

Now we can define a tree $\tilde {T}$ on $\omega \times \eta $ in V as an amalgamation of these trees, so in every generic extension of V we have

$$ \begin{align*}\operatorname{\mathrm{p}}[\tilde{T}] = \operatorname{\mathrm{p}}[\tilde{T}_1] \cup \bigcup_{\beta < \omega_1^V} \operatorname{\mathrm{p}}[\tilde{T}_{2,\beta}] \cup \bigcup_{\lambda \in [\omega_1^V, \eta)} \operatorname{\mathrm{p}}[\tilde{T}_{3,\lambda}].\end{align*} $$

Let $G \subset \mathbb {P}$ be a V-generic filter and let x be a real in $V[G]$ . We want to show

$$ \begin{align*}x \in A \iff x \notin \operatorname{\mathrm{p}}[\tilde{T}].\end{align*} $$

Assume that $x \in A$ . Then clearly we have $x \notin \operatorname {\mathrm {p}}[\tilde {T}_1]$ and $x \notin \operatorname {\mathrm {p}}[\tilde {T}_{2,\beta }]$ for all $\beta < \omega _1^V$ . By the definition of A we have $x = x_\beta $ for some $\beta $ in the range of f. Now let $\lambda \in [\omega _1^V, \eta )$ . As previously noted, the order type of a $g_\lambda $ -closed set $\sigma \in \mathcal {P}_{\omega _1}(\lambda )^V$ cannot be in the range of f, so we have $x \notin \operatorname {\mathrm {p}}[\tilde {T}_{3,\lambda }]$ in V and equivalently in $V[G]$ . Therefore $x \notin \operatorname {\mathrm {p}}[\tilde {T}]$ .

Conversely, assume that $x \notin A$ . If $\langle \omega ;E_x\rangle $ is not a wellordering then we have $x \in \operatorname {\mathrm {p}}[\tilde {T}_1]$ . If $\langle \omega ;E_x\rangle $ is a wellordering of order type less than $\omega _1^V$ then let $\beta $ be its order type and note that $x \in \operatorname {\mathrm {p}}[\tilde {T}_{2,\beta }]$ . If $\langle \omega ;E_x\rangle $ is a wellordering of order type greater than or equal to $\omega _1^V$ then let $\lambda $ be its order type. Note that $\lambda < \eta $ because $\eta $ was chosen to be large enough that forcing with $\mathbb {P}$ does not collapse it to be countable. Therefore $\lambda \in [\omega _1^V,\eta )$ and we have $x \in \operatorname {\mathrm {p}}[\tilde {T}_{3,\lambda }]$ because $\langle \omega ;E_x\rangle \cong \langle \lambda ;\mathord {\in }\rangle $ and the set $\lambda \in \mathcal {P}_{\omega _1}(\lambda )^{V[G]}$ is trivially $g_\lambda $ -closed. In every case we have shown the desired conclusion that $x \in \operatorname {\mathrm {p}}[\tilde {T}]$ .

Proof Let $\kappa = \omega _1^V$ . First we will show that $\kappa $ is inaccessible in L. Because $\kappa $ is regular in V it is regular in L, so by GCH in L it remains to show that $\kappa $ is a limit cardinal in L. Suppose toward a contradiction that there is a cardinal $\eta < \kappa $ of L such that $(\eta ^+)^L = \kappa $ and define the function $f : \kappa \to \kappa $ in L such that for every ordinal $\alpha < \kappa $ , $f(\alpha )$ is the least ordinal $\beta> \max \{\alpha , \eta \}$ such that $L_\beta \models \left | \alpha \right | \le \eta $ . (Note that $\beta < \kappa $ by Gödel’s condensation lemma.)

By our assumption there is an ordinal $\lambda> \kappa $ such that for a stationary set of $\sigma \in \mathcal {P}_\kappa (\lambda )$ in V we have $\sigma \cap \kappa \in \kappa $ and $\operatorname {\mathrm {o.\!t.}}(\sigma ) \ge f(\sigma \cap \kappa )$ . Because the set of all $X \in \mathcal {P}_\kappa (L_\lambda )$ such that $X \prec L_\lambda $ , $X \cap \kappa \in \kappa $ , and $\eta \cup \{\eta \} \subset X$ is club in $\mathcal {P}_\kappa (L_\lambda )$ , there is such a set X with the additional property that $\bar {\lambda } \ge f(\bar {\kappa })$ where $\bar {\kappa } = X \cap \kappa $ and $\bar {\lambda } = \operatorname {\mathrm {o.\!t.}}(X \cap \lambda )$ . Note that

$$ \begin{align*} \eta < \bar{\kappa} < f(\bar{\kappa}) \le \bar{\lambda} < \kappa < \lambda.\end{align*} $$

By the condensation lemma, X is the range of an elementary embedding

$$ \begin{align*}j : L_{\bar{\lambda}} \to L_\lambda \text{ with } \operatorname{\mathrm{crit}}(j) = \bar{\kappa} \text{ and } j(\bar{\kappa}) = \kappa.\end{align*} $$

By the definition of f we have $L_{f(\bar {\kappa })} \models \left | \bar {\kappa } \right | \le \eta $ , and because $f(\bar {\kappa }) \le \bar {\lambda }$ it follows that $L_{\bar {\lambda }} \models \left | \bar {\kappa } \right | \le \eta $ . Then we have $L_\lambda \models \left | \kappa \right | \le \eta $ by the elementarity of j, contradicting the fact that $\kappa $ is a cardinal in V.

Now because $\kappa $ is inaccessible in L it follows that $L_\beta = V_\beta ^L$ for a cofinal set of ordinals $\beta < \kappa $ . We will use this fact to show that L satisfies statement 2 of Proposition 2.5, which is an equivalent condition for $\kappa $ to be VSS.

Let $f : \kappa \to \kappa $ be a function in L and define the function $g: \kappa \to \kappa $ in L where for every ordinal $\alpha < \kappa $ , $g(\alpha )$ is the least ordinal $\beta \ge \max \{f(\alpha ), \alpha +1\}$ such that $L_\beta = V_\beta ^L$ . Applying our assumption to the function $g + 1$ , we obtain an ordinal $\lambda> \kappa $ such that for a stationary set of $\sigma \in \mathcal {P}_\kappa (\lambda )$ in V we have $\sigma \cap \kappa \in \kappa $ and $\operatorname {\mathrm {o.\!t.}}(\sigma )> g(\sigma \cap \kappa )$ . By increasing $\lambda $ if necessary, we may assume that $L_\lambda = V_\lambda ^L$ .

Because the set of all $X \in \mathcal {P}_\kappa (L_\lambda )$ such that $X \prec L_\lambda $ , $X \cap \kappa \in \kappa $ , and $f \in X$ is club in $\mathcal {P}_\kappa (L_\lambda )$ , there is such a set X with the additional property that $\bar {\lambda }> g(\bar {\kappa })$ where $\bar {\kappa } = X \cap \kappa $ and $\bar {\lambda } = \operatorname {\mathrm {o.\!t.}}(X \cap \lambda )$ . By the condensation lemma, X is the range of an elementary embedding

$$ \begin{align*}j : L_{\bar{\lambda}} \to L_\lambda \text{ with } \operatorname{\mathrm{crit}}(j) = \bar{\kappa} \text{ and } j(\bar{\kappa}) = \kappa.\end{align*} $$

Define $\bar {\beta } = g(\bar {\kappa })$ and $\beta = j(\bar {\beta })$ and $\bar {f} = j^{-1}(f)$ . Note that

$$ \begin{align*} \bar{\kappa} < \bar{\beta} < \bar{\lambda} < \kappa < \beta < \lambda.\end{align*} $$

We have $L_{\bar {\beta }} = V_{\bar {\beta }}^L$ because $\bar {\beta }$ is in the range of the function g. Therefore $L_{\bar {\lambda }} \models L_{\bar {\beta }} = V_{\bar {\beta }}$ and it follows by the elementarity of j that $L_\lambda \models L_\beta = V_\beta $ . Because $L_\lambda = V_\lambda ^L$ , this implies $L_\beta = V_\beta ^L$ . For the elementary embedding $j_1 = j \restriction V_{\bar {\beta }}^L$ we have

$$ \begin{align*} j_1: V_{\bar{\beta}}^L \to V_\beta^L \text{ and } \operatorname{\mathrm{crit}}(j_1) = \bar{\kappa} \text{ and } j_1(\bar{\kappa}) = \kappa \text{ and } j_1(\bar{f}) = f.\end{align*} $$

Let $G \subset \operatorname {\mathrm {Col}}(\omega ,V_{\bar {\beta }}^L)$ be a V-generic filter. Then by the absoluteness of elementary embeddability of countable structures there is an elementary embedding $j_2 \in L[G]$ such that

$$ \begin{align*} j_2: V_{\bar{\beta}}^L \to V_\beta^L \text{ and } \operatorname{\mathrm{crit}}(j_2) = \bar{\kappa} \text{ and } j_2(\bar{\kappa}) = \kappa \text{ and } j_2(\bar{f}) = f.\end{align*} $$

Because $f(\bar {\kappa }) \le g(\bar {\kappa }) = \bar {\beta }$ , this elementary embedding $j_2$ witnesses statement 2 of Proposition 2.5 for $\kappa $ in L with respect to the function f.

Question 5.5. What is the consistency strength of the theory ZFC+ “every set of reals in $L(\mathbb {R},\operatorname {\mathrm {uB}})$ has the Baire property”?

Proof Let A be a universally Baire set of reals in $L[G]$ . Then by Lemma 5.1 there is an ordinal $\alpha < \kappa $ and a $\kappa $ -universally Baire set of reals $A_0$ in $L[G\restriction \alpha ]$ such that $A = A_0^{L[G]}$ . Increasing $\alpha $ if necessary, we may assume that it is a successor ordinal, so $G\restriction \alpha $ is countable in $L[G\restriction \alpha ]$ and therefore $L[G\restriction \alpha ] = L[z]$ for some real $z \in L[G]$ . Take a $\operatorname {\mathrm {Col}}(\omega ,\mathord {<}\kappa )$ -absolutely complementing pair of trees $\langle T, \tilde {T}\rangle $ in $L[z]$ such that $\operatorname {\mathrm {p}}[T]^{L[z]} = A_0$ and therefore $\operatorname {\mathrm {p}}[T]^{L[G]} = A$ . Define the sets of reals

$$ \begin{align*} B_0 &= \operatorname{\mathrm{p}}[\tilde{T}]^{L[z]} = \mathbb{R}^{L[z]} \setminus A_0,\\ B &= \operatorname{\mathrm{p}}[\tilde{T}]^{L[G]} = \mathbb{R}^{L[G]} \setminus A. \end{align*} $$

We claim that for every real $x \in L[G]$ we have $x \in A$ if and only if there is a tree $T' \in L[z]$ such that $x \in \operatorname {\mathrm {p}}[T']$ and $\operatorname {\mathrm {p}}[T'] \cap B_0 = \emptyset $ . To prove the forward direction of the claim, note that if $x \in A$ then we can simply let $T' = T$ .

To prove the reverse direction of the claim, let x be a real in $L[G]$ and assume that $x \in \operatorname {\mathrm {p}}[T']$ for some tree $T' \in L[z]$ such that $\operatorname {\mathrm {p}}[T'] \cap B_0 = \emptyset $ . Because $\operatorname {\mathrm {p}}[T'] \cap B_0 = \emptyset $ and $B_0 = \operatorname {\mathrm {p}}[\tilde {T}]^{L[z]}$ we have

$$ \begin{align*}L[z] \models \operatorname{\mathrm{p}}[T'] \cap \operatorname{\mathrm{p}}[\tilde{T}] = \emptyset.\end{align*} $$

The statement $\operatorname {\mathrm {p}}[T'] \cap \operatorname {\mathrm {p}}[\tilde {T}] = \emptyset $ is absolute between $L[z]$ and $L[G]$ by the absoluteness of wellfoundedness of the tree of all triples $\langle r, s_1, s_2\rangle $ such that $\langle r,s_1\rangle \in T'$ and $\langle r,s_2\rangle \in \tilde {T}$ , so because $x \in \operatorname {\mathrm {p}}[T']$ we have $x \notin \operatorname {\mathrm {p}}[\tilde {T}]$ and therefore $x \in \operatorname {\mathrm {p}}[T] = A$ .

We can use the claim to define A in $L[G]$ without reference to T. Because the class $L[z]$ is $\Sigma _1(z)$ and the statement $y \notin \operatorname {\mathrm {p}}[T']$ (where y is a real) is witnessed by the existence of a rank function for the section tree $T_y$ , it follows from the claim that A is $\Sigma _1(z,B_0)$ . Because $\text {HC} \prec _{\Sigma _1} V$ , this implies that A is $\Sigma _1^{\text {HC}}(z,B_0)$ and is therefore $\Sigma ^1_2(z,b)$ where b is a real coding $B_0$ . A symmetric argument shows that B is ${\boldsymbol {\Sigma }}^1_2$ in $L[G]$ , so A and B are both ${\boldsymbol {\Delta }}^1_2$ in $L[G]$ .

Proof of Theorem 1.3 Assume that there is a $\Sigma _2$ -reflecting VSS cardinal $\kappa $ . Then $\kappa $ is clearly $\Sigma _2$ -reflecting in L and moreover it is VSS in L by Lemma 2.4. Let $G \subset \operatorname {\mathrm {Col}}(\omega ,\mathord {<}\kappa )$ be an L-generic filter. Because $\kappa $ is $\Sigma _2$ -reflecting in L we have ${\boldsymbol {\Delta }}^1_2 \subset \operatorname {\mathrm {uB}}$ in $L[G]$ by the proof of Feng et al. [Reference Feng, Magidor and Woodin2, Theorem 3.3]. On the other hand, because $\kappa $ is VSS in L we have $\operatorname {\mathrm {uB}} \subset {\boldsymbol {\Delta }}^1_2$ in $L[G]$ by Proposition 6.1.

Conversely, assume $\operatorname {\mathrm {uB}} = {\boldsymbol {\Delta }}^1_2$ . Because ${\boldsymbol {\Delta }}^1_2 \subset \operatorname {\mathrm {uB}}$ , the cardinal $\omega _1^V$ is $\Sigma _2$ -reflecting in L by the proof of Feng et al. [Reference Feng, Magidor and Woodin2, Theorem 3.3]. Also because ${\boldsymbol {\Delta }}^1_2 \subset \operatorname {\mathrm {uB}}$ , every ${\boldsymbol {\Sigma }}^1_2$ set of reals has the perfect set property by Feng et al. [Reference Feng, Magidor and Woodin2, Theorem 2.4]. Combining this with the assumption that $\operatorname {\mathrm {uB}} \subset {\boldsymbol {\Delta }}^1_2$ shows that every $\operatorname {\mathrm {uB}}$ set has the perfect set property, so $\omega _1^V$ is VSS in L by Lemmas 5.3 and 5.4.