Proof. The proof is a minor modification of the proof given in [Reference Feng, Magidor and Woodin2], which uses Choice for some steps. Let $\operatorname {RO}(X)$ denote the collection of all regular open subsets of X, which is a complete boolean algebra with negation given by the complement of the closure, and suprema given by the interior of the closure of the union. Considering $\mathrm {RO}(X)$ as a poset (under inclusion, with the empty set excluded), let S and T be trees witnessing that A is $\mathrm {RO}(X)$ -Baire, with $\mathrm {p}[S] = A$ . For each pair $n,m \in \omega $ , let $X_{n,m}$ be the set of $x \in X$ with $f(x)(n) = m$ . Then each $X_{n,m}$ is in $\mathrm {RO}(X)$ . Let $\dot {g}$ be the $\operatorname {RO}(X)$ -name for an element of $\omega ^{\omega }$ consisting of the pairs $(X_{n,m}, (n,m))$ for each $n, m \in \omega $ , so that each $X_{n,m}$ forces that $\dot {g}(n) = m$ .

Densely many conditions in $\mathrm {RO}(X)$ decide whether the realization of $\dot {g}$ is in the projection of S or T. Let U be the union of the conditions forcing $\dot {g}$ into $\mathrm {p}[S]$ , and let V be the corresponding set for T. Since intersections of regular open sets are regular open, U and V are disjoint. Since X has a regular open base, $U \cup V$ is dense. It suffices to see that $f^{-1}[A] \bigtriangleup U$ is meager in X.

For each $\sigma \in \omega ^{<\omega }$ , let $X_{\sigma }$ be $\bigcap _{n < |\sigma |}X_{n,\sigma (n)}$ (i.e., the set of $x \in X$ for which $\sigma $ is an initial segment of $f(x)$ ). For each $n \in \omega $ , let $\mathcal {D}_{n}$ be the set of conditions Y of $\mathrm {RO}(X)$ which are contained in either U or V and also in some set of the form $X_{\sigma }$ , for some $\sigma \in \omega ^{n}$ , and which decide the first n elements of the leftmost branch of whichever of $S_{\dot {g}}$ or $T_{\dot {g}}$ is ill-founded (i.e., $S_{\dot {g}}$ for sets contained in U and $T_{\dot {g}}$ for sets contained in V). Let $\tau _{n}(Y)$ denote the length-n initial segment of this leftmost branch as decided by Y. Note then that if $Y \in \mathcal {D}_{n}$ is contained in some such $X_{\sigma }$ , then the pair $(\sigma ,\tau _{n}(Y))$ is in the corresponding tree.

For each $n \in \omega $ , let $D_{n}$ be the union of all the members of $\mathcal {D}_{n}$ . Then each $D_{n}$ is dense open. It suffices now to see that if x is in $U \cap \bigcap _{n \in \omega }D_{n}$ , then $f(x) \in A$ , and if $x \in V \cap \bigcap _{n \in \omega }D_{n}$ , then $f(x) \not \in A$ .

To see the former claim, fix $x \in U$ and regular open sets $Y_{n} \in \mathcal {D}_{n} (n \in \omega )$ with $x \in \bigcap _{n \in \omega } Y_{n}$ . Each $Y_{n}$ is contained in some $X_{\sigma _{n}}$ , in such a way that $f(x) = \bigcup _{n\in \omega }\sigma _{n}$ . Furthermore, since the conditions $Y_{n}$ are compatible, the values $\tau _{n}(Y_{n})$ are compatible, and the pair $(f(x), \bigcup _{n \in \omega }\tau _{n}(Y_{n}))$ gives a branch through S, showing that $f(x) \in A$ . The proof for the latter claim is the same.

Proof. If $\mathrm {p}[T] \subseteq \mathrm {p}[U]$ , then $\mathrm {p}[T] \cap \mathrm {p}[\tilde {U}] = \emptyset $ . This implies that

$$\begin{align*}1_{\mathbb{P}} \mathbin{\Vdash} \mathrm{p}[T] \cap \mathrm{p}[\tilde{U}] = \emptyset,\end{align*}$$

by the absoluteness of wellfoundedness of the tree

$$\begin{align*}\{(s,t,u) : (s,t) \in T \wedge (s,u) \in \tilde{U}\}\end{align*}$$

on $\omega \times \mathrm {Ord} \times \mathrm {Ord}$ . Because $(U, \tilde {U})$ is $\mathbb {P}$ -absolutely complementing, it follows that $1_{\mathbb {P}} \mathbin {\Vdash } \mathrm {p}[T] \subseteq \mathrm {p}[U]$ .

Proof. Let $\eta = 2^{\mathord {<}\kappa }$ and let $\langle \mathbb {P}_\alpha : \alpha < \eta \rangle $ enumerate the set of all posets on cardinals less than $\kappa $ . For each $\alpha < \eta $ , choose a $\mathbb {P}_\alpha $ -absolutely complementing pair of trees $(T_\alpha , \tilde {T}_\alpha )$ witnessing that A is $\mathbb {P}_\alpha $ -Baire. Consider the tree T on $\omega \times \mathrm {Ord}$ defined by

$$\begin{align*}T = \{ (s, (\alpha^\frown t) \mathord{\restriction} \left|s\right|) : \alpha < \eta \text{ and } (s,t) \in T_\alpha \}.\end{align*}$$

This construction immediately implies that $\mathrm {p}[T] = \bigcup _{\alpha < \eta } \mathrm {p}[T_\alpha ]$ in every outer model of V. Similarly, letting

$$\begin{align*}\tilde{T} = \{ (s, (\alpha^\frown t) \mathord{\restriction} \left|s\right|) : \alpha < \eta \text{ and } (s,t) \in \tilde{T}_\alpha \},\end{align*}$$

$\mathrm {p}[\tilde {T}] = \bigcup _{\alpha < \eta } \mathrm {p}[\tilde {T}_\alpha ]$ in every outer model of V. In V, we have $\mathrm {p}[T] = A$ and $\mathrm {p}[\tilde {T}] = \omega ^{\omega } \setminus A$ . It follows immediately from the construction of T and $\tilde {T}$ that the union of their projections is $\omega ^{\omega }$ in any forcing extension by a poset of cardinality less than $\kappa $ . The absoluteness of the wellfoundedness of the tree

$$\begin{align*}\{ (s,t,u) : (s,t) \in T \wedge (s,u) \in \tilde{T}\}\end{align*}$$

implies that the projections of T and $\tilde {T}$ are still disjoint in any such extension. That is, the pair $(T, \tilde {T})$ is $\kappa $ -absolutely complementing.

Proof. Clearly, $A \subseteq \mathrm {p}[T_{\vec {\varphi }}]$ as before. Conversely, let $y \in \mathrm {p}[T_{\vec {\varphi }}]$ . Fix $f \in \mathrm {Ord}^\omega $ such that $(y,f) \in [T_{\vec {\varphi }}]$ . Take a tree $T_A$ on $\omega \times \mathrm {Ord}$ witnessing that A is Suslin, and take a tree $T_E$ on $\omega \times \omega \times \omega \times \mathrm {Ord}$ witnessing that E is Suslin.

For $N < \omega $ , define a full N-witness to be an object of the form

$$\begin{align*}( \langle x_i : i < N\rangle, \langle g_i : i < N\rangle, \langle h_{n,i,j} : n \le i \le j < N\rangle)\end{align*}$$

such that

• • for all $i < N, \ x_i \in \omega ^\omega $ witnesses that $(y \mathord {\restriction } i,f \mathord {\restriction } i) \in T_{\vec {\varphi }}$ , meaning that $x_i \in A,\ y \mathord {\restriction } i = x_i \mathord {\restriction } i$ and $f \mathord {\restriction } i = \langle \varphi _0(x_i),\ldots ,\varphi _{i-1}(x_i)\rangle $ ;

• • for all $i < N,\ g_i \in \mathrm {Ord}^\omega $ witnesses that $x_i \in A$ in the sense that $(x_i,g_i) \in [T_A]$ ;

• • for all $n \le i \le j < N, \ h_{n,i,j} \in \mathrm {Ord}^\omega $ witnesses $\varphi _n(x_i) = \varphi _n(x_j)$ in the sense that $(\bar {n}, x_i, x_j, h_{n,i,j}) \in [T_E]$ .

We define a partial N-witness to be an object that can be obtained by restricting some full N-witness as above to N to produce the following:

$$\begin{align*}( \langle x_i \mathord{\restriction} N : i < N\rangle, \langle g_i \mathord{\restriction} N: i < N\rangle, \langle h_{n,i,j} \mathord{\restriction} N: n \le i \le j < N\rangle).\end{align*}$$

Note that every full N-witness can be extended to a full $(N+1)$ -witness by choosing any $x_N \in \omega ^\omega $ witnessing $(y \mathord {\restriction } N,f \mathord {\restriction } N) \in T_{\vec {\varphi }}$ and then choosing $g_N$ and $h_{n,i,N}$ appropriately for all n and i such that $n \le i \le N$ .

It follows that every partial N-witness can be extended to a partial $(N+1)$ -witness: first extend it to a full N-witness, extend that to a full $(N+1)$ -witness, and then restrict that to a partial $(N+1)$ -witness. By extending a partial N-witness to a partial $(N+1)$ -witness, we mean that its sequences of length N are extended to sequences of length $N+1$ , and new sequences of length $N+1$ of the form $x_N \mathord {\restriction } (N+1),\ g_N \mathord {\restriction } (N+1)$ and $h_{n,i,N} \mathord {\restriction } (N+1)$ are added.

The advantage of partial witnesses (as compared with full witnesses) is that they are essentially finite sequences of ordinals, so the Axiom of Choice is not required to choose among them. We may therefore define a sequence $\langle \sigma _N : N < \omega \rangle $ where for each $N < \omega ,\ \sigma _N$ is an N-witness and $\sigma _{N+1}$ extends $\sigma _N$ . Taking $\bigcup _{N < \omega } \sigma _N$ , by which we really mean taking the union in each coordinate separately, we obtain an object

$$\begin{align*}( \langle x_i : i < \omega\rangle, \langle g_i : i < \omega\rangle, \langle h_{n,i,j} : n \le i \le j < \omega\rangle). \end{align*}$$

Then for all $i < N$ , we have $x_i \in A$ as witnessed by $g_i \in \mathrm {Ord}^\omega $ . Furthermore, the sequence $\langle x_i : i < \omega \rangle $ converges to y since $y \mathord {\restriction } i = x_i \mathord {\restriction } i$ for all $i < \omega $ . We might not have $\varphi _n(x_i) = f(n)$ for all n and i such that $n \le i < \omega $ because this property of full witnesses might be lost by passing to partial witnesses and taking unions. However, we do have $\varphi _n(x_i) = \varphi _n(x_j)$ for all $n,\ i$ and j such that $n \le i \le j < \omega $ because this is witnessed by $h_{n,i,j} \in \mathrm {Ord}^\omega $ . Therefore, the sequence $\langle x_i : i < \omega \rangle $ converges to y via $\vec {\varphi }$ . Since $\vec {\varphi }$ is a semiscale on A, it follows that $y \in A$ , as desired.

Proof. Let $(T_A, T_{\neg A})$ be a pair of trees on $\omega \times \mathrm {Ord}$ witnessing that A is $\mathbb {P}$ -Baire, and let $(T_R, T_{\neg R})$ be a pair of trees on $\omega \times \omega \times \omega \times \mathrm {Ord}$ witnessing that R is $\mathbb {P}$ -Baire. The desired conclusion will follow from the fact that the wellfoundedness of various trees defined from these four trees is absolute from V to $V[G]$ .

First, we show that $R^{V[G]}_n$ is a prewellordering of $A^{V[G]}$ for each $n \in \omega $ . Let $n \in \omega $ . We have $R_n \subseteq A \times A$ , so the trees

$$\begin{align*}\{ (s, t, u, v) : (\bar{n}, s, t, u) \in T_{R} \text{ and } (s, v) \in T_{\neg A} \}\end{align*}$$

and

$$\begin{align*}\{ (s, t, u, v) : (\bar{n}, s, t, u) \in T_{R} \text{ and } (t, v) \in T_{\neg A} \}\end{align*}$$

on $\omega \times \omega \times \mathrm {Ord} \times \mathrm {Ord}$ are wellfounded, and therefore, $R^{V[G]}_n \subseteq A^{V[G]} \times A^{V[G]}$ (here, $\bar {n}$ refers to the constant sequence of length $|s|$ with value n).

Since $R_n$ is transitive, the tree on $\omega \times \omega \times \omega \times \mathrm {Ord} \times \mathrm {Ord} \times \mathrm {Ord}$ consisting of all tuples $(s, t, u, v, w, x)$ such that

• • $(\bar {n}, s, t, v) \in T_{R}$ ,

• • $(\bar {n}, t, u, w) \in T_{R}$ and

• • $(\bar {n}, s, u, x) \in T_{\neg R}$

is wellfounded. It follows that $R^{V[G]}_n$ is transitive.

Since $R_n$ is total on A, the tree on $\omega \times \omega \times \mathrm {Ord} \times \mathrm {Ord} \times \mathrm {Ord} \times \mathrm {Ord}$ consisting of all tuples $(s, t, u, v, w, x)$ such that

• • $(s, u) \in T_A$

• • $(t, v) \in T_A$

• • $(\bar {n}, s, t, w) \in T_{\neg R}$ , and

• • $(\bar {n}, t, s, x) \in T_{\neg R}$

is wellfounded. It follows that $R^{V[G]}_n$ is total on $A^{V[G]}$ .

For the remaining parts, which are the wellfoundedness of each $R^{V[G]}_n$ and the semiscale condition, it will help to introduce the following notation. Fix a recursive bijection $\omega \times \omega \to \omega $ denoted by $(i, j) \mapsto \langle i, j \rangle $ . Given $s \in \omega ^{\mathord {<}\omega }$ and integers $i,j$ , define the finite sequence $s_{i,j}$ to be $\langle s(\langle i, 0\rangle ), \ldots , s(\langle i, j-1\rangle )\rangle $ if all necessary values of s are defined, and undefined otherwise.

The strict part of $R_n$ is wellfounded, so $R_n$ has no infinite strictly decreasing sequences. Then the tree on $\omega \times \mathrm {Ord} \times \mathrm {Ord}$ consisting of all tuples $(s, t, u)$ such that for all $i,j \in \omega $ such that $s_{i,j}$ (and hence also $t_{i,j}$ and $u_{i,j}$ ) and $s_{i+1,j}$ are defined,

• • $(\bar {n}, s_{i+1, j}, s_{i,j}, t_{i,j}) \in T_R$ and

• • $(\bar {n}, s_{i,j}, s_{i+1,j}, u_{i,j}) \in T_{\neg R}$

is wellfounded, so there is no strictly $R^{V[K]}_n$ -decreasing infinite sequence in any generic extension $V[K]$ (we apply this fact with $K = (G,H)$ in the next paragraph).

Because we are not assuming the Axiom of Choice, the previous paragraph does not immediately show that $R^{V[G]}_n$ is wellfounded. To show that it is, let S be a nonempty subset of $A^{V[G]}$ in $V[G]$ . Let $\mathbb {Q}$ be the poset of strictly $R^{V[G]}_n$ -decreasing finite sequences of reals in S, ordered by reverse inclusion, and take a $V[G]$ generic filter $H \subseteq \mathbb {Q}$ . Because the tree on $\omega \times \mathrm {Ord} \times \mathrm {Ord}$ defined above is still wellfounded in $V[G][H]$ , the sequence $\bigcup H$ cannot be infinite, so it has a last element, which must be an $R^{V[G]}_n$ -minimal element of S.

Lastly, we prove the semiscale condition. Because R codes a semiscale on A, the tree on $\omega \times \omega \times \mathrm {Ord} \times \mathrm {Ord} \times \mathrm {Ord}$ consisting of all tuples $(c, s, t, u, v)$ such that for all $n,i,j \in \omega $ such that $c(n) \le \min (i,j)$ and $s_{i,j}$ (and hence also $t_{i,j}$ and $u_{i,j}$ ) and $s_{i+i,j}$ are defined,

• • $(\bar {n}, s_{i,j}, s_{i+1,j}, t_{i,j}) \in T_R$ ,

• • $(\bar {n}, s_{i+1,j}, s_{i,j}, u_{i,j}) \in T_R$ ,

• • $s_{i,c(n)} = s_{i+1,c(n)}$ and

• • $(s_{i,n}, v \mathord {\restriction } n) \in T_{\neg A}$

is wellfounded (here, $c(n)$ indicates a point such that the reals $\bigcup _{j}s_{i,j}$ for $i> c(n)$ agree both in the n-th norm and also for their first n digits; we could have done without c by using Remark 3.2). This, along with the fact that $R^{V[G]}_n$ is a prewellordering of $A^{V[G]}$ for each $n \in \omega $ , implies that $R^{V[G]}$ is a semiscale on $A^{V[G]}$ in $V[G]$ .

Proof. Let $\mathcal {M}$ denote $L^{F \mathord {\restriction } \Gamma }(\mathbb {R}, \Gamma )$ , and fix $A \in \Gamma $ . Let $\vec {\varphi }_+$ and $\vec {\varphi }_-$ be semiscales on A and $\omega ^\omega \setminus A$ , respectively, such that the corresponding codes

$$ \begin{align*} R_+ &= \{ (\bar{n}, x, y) \in \omega^{\omega} \times A \times A : \varphi_{+,n}(x) \le \varphi_{+,n}(y) \}\\ \text{and}\\ R_- &= \{ (\bar{n}, x, y) \in \omega^{\omega} \times (\omega^\omega \setminus A) \times (\omega^\omega \setminus A) : \varphi_{-,n}(x) \le \varphi_{-,n}(y) \} \end{align*} $$

are in $\Gamma $ .

Consider an arbitrary set $Z \in \mathcal {M}$ . By Remark 2.3, it suffices to find a $\mathrm {Col}(\omega ,Z)$ -absolutely complementing pair of trees for A in $\mathcal {M}$ . To do this, we will fix a V-generic filter $H \subseteq \mathrm {Col}(\omega ,Z)$ , find a complementing pair of trees for A in $\mathcal {M}[H]$ , and then show that these trees exist in $\mathcal {M}$ and are as desired.

Fixing H, we have by Lemma 3.5 that the expanded relations $R_+^{V[H]}$ and $R_-^{V[H]}$ code semiscales on the expanded sets $A^{V[H]}$ and $(\omega ^\omega \setminus A)^{V[H]}$ , respectively. Letting $\rho _+$ and $\rho _-$ be the standard $\mathrm {Col}(\omega ,Z)$ -names for the canonical expansions of the relations $R_+$ and $R_-$ , respectively, we see that the restrictions $\rho _+ \cap \mathcal {M}$ and $\rho _- \cap \mathcal {M}$ are in $\mathcal {M}$ , as they are coded into the predicate $F \mathord {\restriction } \Gamma $ . Therefore, the restricted relations $R_+^{V[H]} \cap {\mathcal {M}[H]}$ and $R_-^{V[H]} \cap {\mathcal {M}[H]}$ are in the model $\mathcal {M}[H]$ . These restricted relations code semiscales $\vec {\psi }_+,\vec {\psi }_- \in \mathcal {M}[H]$ on the sets $A^{V[H]} \cap {\mathcal {M}[H]}$ and $(\omega ^\omega \setminus A)^{V[H]} \cap {\mathcal {M}[H]}$ , respectively, by Lemma 3.4. In $\mathcal {M}[H]$ , let $S_+$ and $S_-$ be respectively the trees of the semiscales $\vec {\psi }_{+}$ and $\vec {\psi }_{-}$ . It suffices to see that these trees are in $\mathcal {M}$ and that they project to complements in any forcing extension of $\mathcal {M}$ by $\mathrm {Col}(\omega , Z)$ .

These two claims are proved by the same symmetry argument, showing that $S_{+}$ and $S_{-}$ do not depend on the generic filter H. More precisely, suppose toward a contradiction that there exist conditions $p, p^{\prime } \in \mathrm {Col}(\omega , Z)$ , with $p \in H$ , which force contradictory statements about the membership of some sequence $\sigma $ (from $\mathcal {M}$ ) in either of the trees $S_{+}$ and $S_{-}$ , noting that these trees are definable in the $\mathrm {Col}(\omega , Z)$ -extension of $\mathcal {M}$ from the realizations of $\rho _{+} \cap \mathcal {M}$ and $\rho _{-} \cap \mathcal {M}$ . By the symmetry of $\mathrm {Col}(\omega , Z)$ , there exists a V-generic filter $H^{\prime } \subseteq \mathrm {Col}(\omega , Z)$ , with $p^{\prime } \in H^{\prime }$ , such that $\bigcup H$ and $\bigcup H^{\prime }$ differ by only finitely many coordinates, so that $V[H] = V[H^{\prime }]$ and $\mathcal {M}[H] = \mathcal {M}[H^{\prime }]$ . We have then (by Remark 2.8) that $A^{V[H]} = A^{V[H^{\prime }]}, \ (\omega ^{\omega } \setminus A)^{V[H]} = (\omega ^{\omega } \setminus A)^{V[H^{\prime }]}, \ R_{+}^{V[H]} = R_{+}^{V[H^{\prime }]}$ and $R_{-}^{V[H]} = R_{-}^{V[H^{\prime }]}$ , from which it follows that the corresponding versions of $S_{+}$ and $S_{-}$ in $\mathcal {M}[H]$ and $\mathcal {M}[H^{\prime }]$ are the same, giving a contradiction. The same argument shows that no condition $p^{\prime }$ can force that $S_{+}$ and $S_{-}$ fail to project to complements in a $\mathrm {Col}(\omega , Z)$ -extension of $\mathcal {M}$ , since they do project to complements in $\mathcal {M}[H]$ .

Proof. To show that $(F \mathord {\restriction } \Gamma )^{\mathcal {M}} = (F \mathord {\restriction } \Gamma ) \cap \mathcal {M}$ , fix $A \in \Gamma $ and let $\mathbb {P}$ be a poset in $\mathcal {M}$ . Let $p \in \mathbb {P}$ , and let $\dot {x} \in \mathcal {M}$ be a $\mathbb {P}$ -name for a real. We claim that

$$\begin{align*}\mathcal{M} \models (A, \mathbb{P}, p, \dot{x}) \in F \mathord{\restriction} \Gamma \iff (A, \mathbb{P}, p, \dot{x}) \in F \mathord{\restriction} \Gamma.\end{align*}$$

Take a pair of trees $(T, \tilde {T}) \in \mathcal {M}$ witnessing that A is $\mathbb {P}$ -Baire in $\mathcal {M}$ and take a pair of trees $(U,\tilde {U})$ witnessing that A is $\mathbb {P}$ -Baire in V. Because $\mathrm {p}[T] = \mathrm {p}[U] = A$ and $\mathrm {p}[\tilde {T}] = \mathrm {p}[\tilde {U}] = \omega ^\omega \setminus A$ in V and the pair $(U,\tilde {U})$ is $\mathbb {P}$ -absolutely complementing, Lemma 2.5 gives

$$\begin{align*}1_{\mathbb{P}} \mathbin{\Vdash} (\mathrm{p}[T] \subseteq \mathrm{p}[U] \text{ and } \mathrm{p}[\tilde{T}] \subseteq \mathrm{p}[\tilde{U}]).\end{align*}$$

We want to show that the following statements are equivalent for a condition $p \in \mathbb {P}$ and a $\mathbb {P}$ -name $\dot {x}$ in $\mathcal {M}$ :

1. 1. $\mathcal {M} \models p \mathbin {\Vdash } \dot {x} \in \mathrm {p}[T]$ ;

2. 2. $p \mathbin {\Vdash } \dot {x} \in \mathrm {p}[U]$ .

To see this, assume first that (1) is true and take a V-generic filter $G \subseteq \mathbb {P}$ containing p. Because G is also $\mathcal {M}$ -generic, we have

$$\begin{align*}\dot{x}_G \in \mathrm{p}[T]^{M[G]} \subseteq \mathrm{p}[T]^{V[G]} \subseteq \mathrm{p}[U]^{V[G]}.\end{align*}$$

This shows that (2) is true.

Now assume that (1) is false, and take a condition $q \le p$ such that

$$\begin{align*}\mathcal{M} \models q \mathbin{\Vdash} \dot{x} \notin \mathrm{p}[T].\end{align*}$$

Take a filter $G \subseteq \mathbb {P}$ containing q that is V-generic, and hence also M-generic. Then we have

$$ \begin{align*} \dot{x}_G &\in (\omega^\omega \setminus \mathrm{p}[T])^{\mathcal{M}[G]}\\ &= \mathrm{p}[\tilde{T}]^{\mathcal{M}[G]}\\ &\subseteq \mathrm{p}[\tilde{T}]^{V[G]}\\ &\subseteq \mathrm{p}[\tilde{U}]^{V[G]}\\ &= (\omega^\omega \setminus \mathrm{p}[U])^{V[G]}. \end{align*} $$

This shows that (2) is false.

To see that $L^{F \mathord {\restriction } \Gamma }(\Gamma , \mathbb {R}) \subseteq \mathcal {M}$ , consider the model $(L^{F \mathord {\restriction } \Gamma }(\Gamma , \mathbb {R}))^{\mathcal {M}} = L^{(F \mathord {\restriction } \Gamma )^{\mathcal {M}}}(\Gamma , \mathbb {R})$ . It is contained in $\mathcal {M}$ , and its construction agrees with that of the model $L^{F \mathord {\restriction } \Gamma }(\Gamma , \mathbb {R})$ at every stage because $(F \mathord {\restriction } \Gamma )^{\mathcal {M}} = (F \mathord {\restriction } \Gamma ) \cap \mathcal {M}$ .

Proof. Fix Z, let $\eta = \max (\lambda , \left |Z\right |{}^{V[G]})$ , and let $(S,T)$ be an $\eta ^{+}$ -absolutely complementing pair of trees in $V(\mathbb {R}^*_G)$ . Then, as above, because S and T are subsets of the ground model, a standard homogeneity argument shows that $(S,T) \in V[G \mathord {\restriction } \xi ]$ for some ordinal $\xi < \lambda $ . The pair $(S,T)$ is $\eta ^{+}$ -absolutely complementing in the model $V[G \mathord {\restriction } \xi ]$ also because this property is downward absolute (since $\lambda \leq \eta $ the cardinal successor of $\eta $ is the same in all models under consideration).

Since $\eta \ge \lambda $ , every generic extension of $V[G]$ by the poset $\mathrm {Col}(\omega ,\eta )$ is also a generic extension of $V[G \mathord {\restriction } \xi ]$ by the poset $\mathrm {Col}(\omega ,\eta )$ . It follows that the pair $(S,T)$ is $\eta ^{+}$ -absolutely complementing in $V[G]$ . Since $\eta \ge \left |Z\right |{}^{V[G]}$ , this implies that the pair $(S,T)$ is $\mathrm {Col}(\omega , Z)$ -absolutely complementing in $V[G]$ . This property is downward absolute, so the pair $(S,T)$ is $\mathrm {Col}(\omega , Z)$ -absolutely complementing in $V(\mathbb {R}^*_G)$ , as desired.

Proof. By the wellfoundedness of the Wadge hierarchy on homogeneously Suslin sets (see Theorem 3.3.5 of [Reference Larson10] and the discussion before it), there is minimal $\kappa < \lambda $ such that every $\kappa $ -homogeneously Suslin set of reals is $\mathord {<}\lambda $ -homogeneously Suslin. By the elementarity of $\pi $ , we have $\kappa \in \operatorname {\mathrm {ran}}(\pi _0)$ and $(S,T) \in \operatorname {\mathrm {ran}}(\pi _0)$ for some $\lambda $ -absolutely complementing pair of trees $(S,T)$ for C. Define $(\bar {\lambda },\bar {\kappa },\bar {S},\bar {T}) = \pi _0^{-1}(\lambda ,\kappa ,S,T)$ . Let $\delta _0$ denote the least Woodin cardinal of $P_0$ above $\bar {\kappa }$ .

Let W denote the set of all (reals coding) ordered pairs $(\mathcal {T},b)$ such that

• • $\mathcal {T}$ is a $2^\omega $ -closed iteration tree on $P_0$ of length $\omega $ (meaning that each extender produces a $2^{\omega }$ -closed ultrapower when applied to the model from which it is chosen),

• • $\mathcal {T}$ is above $\bar {\kappa }$ and based on $\delta _0$ (meaning that each extender chosen has critical point above $\bar {\kappa }$ and von Neumann rank below the corresponding image of $\delta _{0}$ ),

• • b is a cofinal branch of $\mathcal {T}$ , and

• • the model $\mathcal {M}_b^{\pi _0\mathcal {T}}$ is wellfounded, where $\pi _0\mathcal {T}$ is the lifted tree on $V_{\gamma }$ .

As shown by Martin and Steel [Reference Martin and Steel13, p. 27], the condition that the model $\mathcal {M}_b^{\pi _0\mathcal {T}}$ is wellfounded implies that the branch b is $\pi _0$ -realizable (meaning that the branch embedding $i^{\mathcal {T}}_b$ factors into $\pi _0$ ), which in turn implies that the model $\mathcal {M}_b^{\mathcal {T}}$ is wellfounded.

By a result of K. Windszus (see Steel [Reference Steel, Gao, Jackson and Zhang22, Lemma 1.1]) the set W is $\kappa $ -homogeneously Suslin and therefore $\mathord {<}\lambda $ -homogeneously Suslin by our choice of $\kappa $ . (We added the condition that $\mathcal {T}$ is based on $\delta _0$ , which is harmless and convenient.) Note that essentially all of the complexity of W comes from condition that $\mathcal {M}_{b}^{\pi _0\mathcal {T}}$ is wellfounded; the other conditions are arithmetic.

Next, we show (following Steel [Reference Steel, Gao, Jackson and Zhang22, Claim 2 on p. 7]) that the following statements are equivalent, for all $x \in \mathbb {R}$ :

1. 1. $x \in A$ .

2. 2. There exist a pair $(\mathcal {T},b) \in W$ and an $\mathcal {M}^{\mathcal {T}}_b$ -generic filter

   $$\begin{align*}g \subseteq \mathrm{Col}(\omega,i^{\mathcal{T}}_b(\delta_0))\end{align*}$$
   such that $x \in \mathcal {M}^{\mathcal {T}}_b[g]$ and $\psi \big [\mathrm {p}[i^{\mathcal {T}}_b(\bar {S})],x\big ]$ holds in the symmetric extension of $\mathcal {M}^{\mathcal {T}}_b[g]$ at its limit $i^{\mathcal {T}}_b(\bar {\lambda })$ of Woodin cardinals. (It does not matter which symmetric extension, by the homogeneity of the Levy collapse forcing.)

The two directions of the equivalence of (1) and (2) can be proved by constructing a suitable $\pi _{0}$ -realizable $\mathbb {R}$ -genericity iteration of $P_{0}$ at $\lambda $ . We prove the forward direction first and then note the changes needed for the reverse direction.

Fix $x \in \mathbb {R}$ (at the end of the proof, we will assume in addition that x is in A), and note that because $\delta _0$ is a Woodin cardinal of $P_0$ , by Theorem 7.16 of Neeman [Reference Neeman, Foreman and Kanamori18], there is a $2^\omega $ -closed iteration tree $\mathcal {T}$ on $P_0$ that is above $\bar {\kappa }$ , based on $\delta _0$ , of length $\omega $ , such that for every cofinal wellfounded branch b of $\mathcal {T}$ , there is a $\mathcal {M}^{\mathcal {T}}_b$ -generic filter $g \subseteq \mathrm {Col}(\omega ,i^{\mathcal {T}}_b(\delta _0))$ such that $x \in \mathcal {M}^{\mathcal {T}}_b[g]$ . By Martin and Steel [Reference Martin and Steel13, Corollary 5.7], the $2^\omega $ -closed iteration tree $\pi _0\mathcal {T}$ on $V_{\gamma }$ has a cofinal wellfounded branch b. For such a branch b, the pair $(\mathcal {T},b)$ is in W, and there is a $\mathcal {M}^{\mathcal {T}}_b$ -generic filter $g \subseteq \mathrm {Col}(\omega ,i^{\mathcal {T}}_b(\delta _0))$ such that $x \in \mathcal {M}^{\mathcal {T}}_b[g]$ .

Now let $x_0 = x$ , and let $\mathcal {T}_0,\ b_0$ and $g_0$ be such that $(\mathcal {T}_0,b_0) \in W, \ g_0 \subseteq \mathrm {Col}(\omega ,i^{\mathcal {T}}_b(\delta _0))$ is an $\mathcal {M}^{\mathcal {T}_0}_{b_0}$ -generic filter, and $x \in \mathcal {M}^{\mathcal {T}_0}_{b_0}[g_0]$ . (The previous paragraph shows that such $\mathcal {T}_0,\ b_0$ and $g_0$ exist.) Let $P_1 = \mathcal {M}^{\mathcal {T}_0}_{b_0}$ , and let $i_{0,1} = i^{\mathcal {T}_{0}}_{b_{0}}$ be the branch embedding from $P_{0}$ to $P_{1}$ . As noted above, the condition that $\mathcal {M}_{b_0}^{\pi _0\mathcal {T}_0}$ is wellfounded implies that the branch $b_0$ is $\pi _0$ -realizable, meaning that there is an elementary embedding $\pi _1 : P_1 \to V_{\gamma }$ such that $\pi _1 \circ i_{0,1} = \pi _0$ .

Now with the model $P_1$ in place of the model $P_0$ , the map $\pi _1$ in place of the map $\pi _0$ , a Woodin cardinal $\delta _1$ of $P_1$ such that $i_{0,1}(\delta _0) < \delta _1 < i_{0,1}(\bar {\lambda })$ in place of $\delta _0$ , and an arbitrary real $x_1$ , we can repeat the process above, applying the Neeman and Martin–Steel theorems again to get an iteration tree $\mathcal {T}_1$ on $P_1$ of length $\omega $ , above $i_{0,1}(\delta _0)$ and based on $\delta _1$ , a $\pi _1$ -realizable branch $b_1$ of $\mathcal {T}_1$ , and an $\mathcal {M}^{\mathcal {T}_1}_{b_1}$ -generic filter $g_1 \subseteq \mathrm {Col}(\omega , i^{\mathcal {T}_1}_{b_1}(\delta _1))$ such that $x_1 \in \mathcal {M}^{\mathcal {T}_1}_{b_1}[g_1]$ .

We can repeat this process for $\omega $ many stages. (Note that, for the purposes of showing the equivalence of (1) and (2), at a stage $j>0$ , it suffices to use a $\pi _j$ -realizable branch of $\mathcal {T}_j$ given by Martin and Steel [Reference Martin and Steel13, Theorem 3.12]; the requirement that $\mathcal {M}_{b_j}^{\pi _j\mathcal {T}_j}$ is wellfounded was only necessary for stage $j=0$ to establish that $(\mathcal {T}_{0}, b_{0})\in W$ .) Furthermore, given a generic enumeration $\{x_j : j < \omega \}$ of $\mathbb {R}^V$ in $V^{\mathrm {Col}(\omega ,\mathbb {R})}$ , we do this in such a way to produce a $\pi _0$ -realizable $\mathbb {R}$ -genericity iteration $P_0 \to P_1 \to \cdots P_\omega $ . With suitable bookkeeping, one can do this in such a way that the Woodin cardinals $i_{j,\omega }(\delta _j)$ for $j<\omega $ are cofinal in $i_{0,\omega }(\bar {\lambda })$ , and the generic filters $g_j$ for $j<\omega $ fit together into a $P_\omega $ -generic filter $g \subseteq \mathrm {Col}(\omega ,\mathord {<}i_{0,\omega }(\bar {\lambda }))$ (as in Remark 6.2) such that $\mathbb {R}^V = \bigcup \{\mathbb {R} \cap P_{\omega }[g \mathord {\restriction } \xi ] : \xi < i_{0\omega }(\bar {\lambda })\}$ .

To see that (1) implies (2), it remains to observe that the condition $x \in A$ is equivalent to the condition $P_\omega (\mathbb {R}^V) \models \psi \big [i_{0,\omega }(\bar {S}),x\big ]$ by the hypothesis of the lemma and the fact that $C = \mathrm {p}[i_{0,\omega }(\bar {S})]^{P_\omega (\mathbb {R}^V)}$ , as in Remark 6.4. This in turn is equivalent to the condition that $\psi \big [\mathrm {p}[i_{0,1}(\bar {S})],x\big ]$ holds in the symmetric extension of $P_1[g_0]$ at its limit $i_{0,1}(\bar {\lambda })$ of Woodin cardinals (i.e., that the first step of our iteration witnessed (2)) by the fact that the iteration map $i_{1,\omega } : P_1 \to P_\omega $ extends to an elementary embedding $i^*_{1,\omega } : P_1[g_0] \to P_\omega [g_0]$ .

The proof that (2) implies (1) is the same, except that the first step of the genericity iteration is given by assuming (2), and the two equivalences in the previous paragraph are applied in reverse.

This characterization of the set A given by the equivalence of (1) and (2) shows that A is a projection of a $\mathord {<}\lambda $ -homogeneously Suslin set (essentially all of whose complexity comes from W) – that is, that A is $\mathord {<}\lambda $ -weakly homogeneously Suslin. It follows that $\lambda $ is $\mathord {<}\lambda $ -homogeneously Suslin by the main theorem from the proof of projective determinacy (Martin and Steel [Reference Martin and Steel12, Theorem 5.11]; see also Theorem 3.3.13 of [Reference Larson10]).

Proof. Let $A = \textbf {S}_{\vec {\mu }}$ where $\vec {\mu }$ is a tree of $\bar {\delta }^+$ -complete measures in $P[g]$ . By the Levy–Solovay theorem, each measure $\mu _s$ is induced by the $\bar {\delta }^+$ -complete measure $\mu _s \cap P \in P$ , which we will also denote by $\mu _s$ . In P, there is a set Y consisting of $\bar {\delta }^+$ -complete measures such that $|Y|^{P} = \bar {\alpha }$ every member of $\vec {\mu }$ is induced by a member of Y.

Given a cardinal $\bar {\kappa }$ of P with $\bar {\alpha } < \bar {\kappa } < \bar {\delta }$ (as in the statement of the lemma), let $f \in P$ be a tower-flipping function that continuously associates to every tower of measures from Y a tower of $\bar {\kappa }$ -complete measures. In $P[g]$ , define the tree of $\bar {\kappa }$ -complete measures $\vec {\nu } = \langle \nu _s : s \in \omega ^{\mathord {<}\omega }\rangle $ that is continuously associated to $\vec {\mu }$ by f in the sense that $\langle \nu _{s \mathord {\restriction } n} : n \le \left |s\right |\rangle = f(\langle \mu _{s \mathord {\restriction } n} :n \le \left |s\right |\rangle )$ for every finite sequence $s \in \omega ^{\mathord {<}\omega }$ . So in $P[g]$ , we have $\textbf {S}_{\vec {\nu }} = \omega ^\omega \setminus \textbf {S}_{\vec {\mu }}$ by the tower-flipping property of f. Moreover, $\pi (f)$ is a tower-flipping function in $V_{\gamma }$ by the elementarity of $\pi $ , so we have $\textbf {S}_{\pi "\vec {\nu }} = \omega ^\omega \setminus \textbf {S}_{\pi "\vec {\mu }}$ in $V_{\gamma }$ .

In $P[g]$ , define the Martin–Solovay trees $\bar {S} = \operatorname {\mathrm {ms}}(\vec {\mu })$ and $\bar {T} = \operatorname {\mathrm {ms}}(\vec {\nu })$ , and in $V_{\gamma }$ , define the Martin–Solovay trees $S = \operatorname {\mathrm {ms}}(\pi "\vec {\mu })$ and $T = \operatorname {\mathrm {ms}}(\pi "\vec {\nu })$ . Because for each $n \in \omega $ the first n levels of the Martin–Solovay tree are determined by finitely many measures (in the ground model, by the absoluteness of the construction of the Martin-Solovay tree between P and $P[g]$ for $\bar {\delta }^{+}$ -complete measures), we have $(\bar {S} \mathord {\restriction } n,\bar {T} \mathord {\restriction } n) \in P$ for all $n<\omega $ . For the same reason, we have $S = \bigcup _{n<\omega } \pi (\bar {S}\mathord {\restriction } n)$ and $T = \bigcup _{n<\omega } \pi (\bar {T}\mathord {\restriction } n)$ .