Proof. Start with a special Aronszajn tree T which is normal and infinitely splitting. We build a normal $\omega _1$ -tree U satisfying that $U \upharpoonright (\text {Lim}(\omega _1) \cup \{ 0 \}) = T$ . For any $x \in T$ , insert $\omega $ -many new levels in between x and its immediate successors in T which is a copy of the tree $({}^{< \omega } n \setminus \{ \emptyset \},\subset )$ . Now pick distinct cofinal branches $\langle b_k : k < \omega \rangle $ of this copy of ${}^{< \omega } n \setminus \{ \emptyset \}$ so that every element of the copy belongs to some branch. Enumerating the immediate successors of x in T as $\langle y_k : k < \omega \rangle $ , specify $y_k$ to be the unique upper bound of $b_k$ for all $k < \omega $ . It is routine to check that the tree U obtained in this manner is normal and n-splitting. Since U restricted to the club set of countable limit ordinals is special, by Lemma 1.1(c) so is U.

Proof. By Lemma 2.1, we can fix a special Aronszajn tree T which is normal and $2$ -splitting. Let $\pi : T \to \omega $ be a specializing function. We construct by induction a sequence of normal $\omega _1$ -trees $\langle U^\alpha : \alpha < \omega _1 \rangle $ satisfying that $U^0 = T$ and for all ${\alpha < \beta < \omega _1}$ , $U^\alpha $ is a downwards closed subtree of $U^\beta $ and $U^\beta \upharpoonright (\alpha +1) = U^\alpha \upharpoonright (\alpha +1)$ . Note that since $U^\alpha $ is downwards closed in $U^\beta $ , $\mathrm {ht}_{U^\beta } \upharpoonright U^\alpha = \mathrm {ht}_{U^\alpha }$ .

Begin by letting $U^0 = T$ . Suppose that $U^\alpha $ is defined for some fixed $\alpha < \omega _1$ . Define $U^{\alpha +1}$ which extends $U^\alpha $ by adding above each $x \in U^\alpha _\alpha $ infinitely many pairwise disjoint trees, which we denote by $U(\alpha ,x,n)$ for $n < \omega $ , each of them isomorphic to T minus the root of T. Since T is special, we can fix a specializing function $\pi _{\alpha ,x,n} : U(\alpha ,x,n) \to \omega $ for each such x and n. This completes the constriction of $U^{\alpha +1}$ , which easily satisfies the required properties. Now assume that $\delta < \omega _1$ is a limit ordinal and $U^\alpha $ is defined for all $\alpha < \delta $ . Let $U^\delta = \bigcup _{\alpha < \delta } U^\alpha $ . Again, it is straightforward to verify that $U^\delta $ is as required. This completes the construction.

Now let $U = \bigcup _{\alpha < \omega _1} U^\alpha $ . It is routine to check that U is a normal $\omega _1$ -tree, and for all $\alpha < \omega _1$ , $U^\alpha $ is a downwards closed subtree of U and $U \upharpoonright (\alpha +1) = U^\alpha \upharpoonright (\alpha +1)$ . For each $\alpha $ , since $U^\alpha $ is downwards closed in U, $\mathrm {ht}_U \upharpoonright U^\alpha = \mathrm {ht}_{U^\alpha }$ . For each $z \in U$ , let $\alpha _z < \omega _1$ be the least ordinal such that $z \in U^{\alpha _z}$ . Then either $\alpha _z = 0$ or $\alpha _z$ is a successor ordinal. In the latter case, let $\alpha _z = \beta _z+1$ . Note that if $y <_U z$ , then $\alpha _y \le \alpha _z$ since $U^{\alpha _z}$ is downwards closed in U. For each $z \in U^0 = T$ , let $x_z$ denote the root of T. For each $z \in U \setminus U^0$ , let $x_z$ be the unique member of $U^{\beta _z}_{\beta _z}$ such that $x_z <_U z$ and let $n_z < \omega $ be such that $x_z \in U(\beta _z,x_z,n_z)$ .

To show that U is infinitely splitting, consider $z \in U$ . Let $\gamma = \mathrm {ht}_{U}(z)$ . If $\alpha _z = 0$ , then at stage $\gamma $ we have that $\mathrm {ht}_{U^\gamma }(z) = \gamma $ , and we added $\omega $ -many immediate successors of z in forming $U^{\gamma +1}$ . Suppose that $\alpha _z> 0$ . Every element of $U^{\alpha _z} \setminus U^{\beta _z}$ has height at least $\alpha _z$ in $U^{\alpha _z}$ . Therefore, $\gamma> \beta _z$ . At stage $\gamma $ we have that $\mathrm {ht}_{U^\gamma }(z) = \gamma $ and we added $\omega $ -many immediate successors of z in forming $U^{\gamma +1}$ . This completes the proof that U is infinitely splitting.

Claim. Suppose that $y <_U z$ . If $\alpha _y = \alpha _z$ then $x_y = x_z$ , and if also $\alpha _y> 0$ then $n_y = n_z$ . If $\alpha _y < \alpha _z$ then $x_y <_U x_z <_U z$ .

Proof. First, assume that $\alpha _y = \alpha _z$ . If $\alpha _y = 0$ then $x_y = x_z$ is the root of T. If $\alpha _y> 0$ , then easily $x_y = x_z$ and $n_y = n_z$ , for otherwise y and z are incomparable. Secondly, assume that $\alpha _y < \alpha _z$ . If $\alpha _y = 0$ , then clearly $x_y <_U x_z$ since $x_y$ is the root of T. Suppose that $\alpha _y> 0$ . Since $y <_U z$ , $x_y <_U y$ , and $x_z <_U z$ , it follows that $x_y$ and $x_z$ are comparable in U. But $x_y$ has height $\beta _y$ , which is less than the height of $x_z$ which is $\beta _z$ . So $x_y <_U x_z$ . $\dashv _{\ \text {Claim}}$

To show that U is special, we define a map $\pi _U : U \to {}^{<\omega } \omega \setminus \{ \emptyset \}$ by induction as follows. For any $z \in U^0 = T$ , let $\pi _U(z) = \langle \pi (z) \rangle $ . Now consider $z \in U \setminus U^0$ and assume that $\pi _U(y)$ is defined for all $y \in U$ such that $\alpha _y < \alpha _z$ . In particular, we have defined $\pi _U(x_z)$ . Let $\pi _U(z)$ be the concatenation $\pi _U(x_z)^\frown \langle \pi _{\beta _z,x_z,n_z}(z) \rangle $ .

To show that $\pi _U$ is a specializing map, suppose that $y <_U z$ . Assume by induction that for all $c <_U d$ with heights less than $\mathrm {ht}_U(z)$ , $\pi _U(c) \ne \pi _U(d)$ .

Case 1. $\alpha _y = \alpha _z$ . Then by the claim, $x_y = x_z$ , and $n_y = n_z$ in the case that $\alpha _y> 0$ . If $\alpha _y = 0$ , then $\pi _U(y) = \langle \pi (y) \rangle \ne \langle \pi (z) \rangle = \pi _U(z)$ . Suppose that $\alpha _y> 0$ . As $\pi _{\beta _y,x_y,n_y}$ is a specializing map on $U(\beta _y,x_y,n_y)$ , $\pi _{\beta _y,x_y,n_y}(y) \ne \pi _{\beta _z,x_z,n_z}(z)$ . So

$$ \begin{align*}\pi_U(y) = \pi_U(x_y)^\frown \langle \pi_{\beta_y,x_y,n_y}(y) \rangle \ne \pi_U(x_z)^\frown \langle \pi_{\beta_z,x_z,n_z}(z) \rangle = \pi_U(z). \end{align*} $$

Case 2. $\alpha _y < \alpha _z$ . Then by the claim, $x_y <_U x_z <_U z$ . By the inductive hypothesis, $\pi _U(x_y) \ne \pi _U(x_z)$ . If $\alpha _y = 0$ , then $\pi _U(y) = \langle \pi (y) \rangle $ , which has length $1$ . But $\pi _U(z) = \pi _U(x_z) \ \widehat {} \ \pi _{\beta _z,x_z,n_z}(z)$ , which has length at least $2$ and hence is different form $\pi _U(y)$ . If $\alpha _y> 0$ , then

$$ \begin{align*}\pi_U(y) = \pi_U(x_y)^\frown \langle \pi_{\beta_y,x_y,n_y}(y) \rangle \ne \pi_U(x_z)^\frown \langle \pi_{\beta_z,x_z,n_z}(z) \rangle = \pi_U(z). \end{align*} $$

So indeed U is special, and T is an uncountable downwards closed subtree of U which is normal and $2$ -splitting. Now for each $x \in U$ with some height $\gamma $ , $U_x$ contains $U(\gamma ,x,0) \cup \{ x \}$ , which is an uncountable downwards closed subtree of $U_x$ which is normal and $2$ -splitting.

Proof sketch.

Let W be some uncountable downwards closed $(\le \! m)$ -splitting subtree of T. In the case that T is normal, by thinning out further if necessary using Lemma 1.1(a) we may assume that W is normal. Similar to the proof of Proposition 2.2, we build a sequence $\langle U^\alpha : \alpha < \omega _1 \rangle $ of subtrees of T roughly as follows. Let $U^0 = W$ , and take unions at limit stages. Assuming that $U^\alpha $ is defined for a fixed $\alpha < \omega _1$ , pick for each $x \in U^{\alpha }_\alpha $ , $(m-m_x)$ -many immediate successors of x in $T \setminus U^\alpha $ , where $m_x \le m$ is the number of immediate successors of x in $U^\alpha $ . Now construct $U^{\alpha +1}$ by adding above each such immediate successor y some uncountable downwards closed $(\le \! m)$ -splitting subtree of $T_y$ , which we may take to be normal assuming that T is normal. We leave the details to the interested reader.

Proof. By Corollary 2.4 and Theorem 2.7.

Question 2.9. Is the topological square of a Lindelöf tree always Lindelöf?

Question 2.10. Is it consistent with CH that every $\omega _1$ -tree is non-Lindelöf?

Question 2.11. Is it consistent that there exists a Suslin tree, and within the class of infinitely splitting $\omega _1$ -trees, being Suslin is equivalent to being Lindelöf?

Proof. It is well-known that if there exists a Suslin tree, then there exists a normal Suslin tree. So by Lemma 1.1(b), if there exists a Suslin tree then there exists a normal infinitely splitting Suslin tree, which we denote by S.

For each limit ordinal $\alpha < \omega _1$ fix a cofinal branch $b_\alpha $ of the tree $S \upharpoonright \alpha $ which has no upper bound in $S_\alpha $ . This is possible since S has countable levels yet $S \upharpoonright \alpha $ has $2^\omega $ -many cofinal branches. Now define a tree T extending S by placing above each $b_\alpha $ a copy of S, which we denote by $S^\alpha $ . It is routine to check that T is a normal $\omega _1$ -tree which is infinitely splitting. If b is a cofinal branch of T, then since S has no cofinal branches, there exists some limit ordinal $\alpha < \omega _1$ and some $x \in b \cap S^\alpha $ . But $S^\alpha $ is upwards closed in T, so $S^\alpha $ contains the uncountable chain $x {\uparrow }$ , contradicting that $S^\alpha $ is Suslin. So T is Aronszajn. For each limit ordinal $\alpha < \omega _1$ let $x_\alpha $ be the root of $S^\alpha $ , that is, $x_\alpha $ is the unique upper bound of $b_\alpha $ on level $\alpha $ of T. Note that $\{ x_\alpha : \alpha \in \text {Lim}(\omega _1) \}$ is an uncountable antichain of T, and hence T is not Suslin.

We claim that T is Lindelöf. By Theorem 2.7, it suffices to show that T does not contain an uncountable downwards closed finitely splitting subtree. Suppose for a contradiction that U is such a subtree. By finding a subset of U if necessary using Lemma 1.1(a), we may assume without loss of generality that U is normal. Case 1. $U \subseteq S$ . Then S is not Lindelöf, which contradicts that S is Suslin. Case 2. $U \not \subseteq S$ . Then there exists some limit ordinal $\alpha < \omega _1$ such that $U \cap S^\alpha \ne \emptyset $ . Since U is normal and $S^\alpha $ is upwards closed in T, it follows that $U \cap S^\alpha $ is uncountable. But then $U \cap S^\alpha $ witnesses that $S^\alpha $ is not Lindelöf, which contradicts that $S^\alpha $ is Suslin.

Proof of Theorem 3.1.

We prove the statement by induction on $\beta $ . Fix T, n, and $\langle \mathbb {P}_i, \dot {\mathbb {Q}}_j : i \le \beta , \ j < \beta \rangle $ as in the theorem. Without loss of generality we may assume that for all $\gamma \le \beta $ , $\mathbb {P}_\gamma $ is separative, for otherwise there exists an equivalent forcing iteration which has this property (this step is not necessary, but it eliminates the need for an additional lemma).

The cases in which $\beta $ is equal to $0$ or is a successor ordinal are immediate. So assume that $\beta $ is a limit ordinal. By the inductive hypothesis, for all $\gamma < \beta $ , $\mathbb {P}_\gamma $ is proper and forces that T does not contain an uncountable downwards closed $(< \! n)$ -splitting subtree.

Suppose that $p \in \mathbb {P}_\beta $ and p forces that $\dot U$ is a downwards closed $(< \! n)$ -splitting subtree of T. We find an extension of p which forces that $\dot U$ is countable.

Claim. Suppose that D is a dense open subset of $\mathbb {P}_\beta $ , $p^* \le _\beta p$ , and $\alpha < \beta $ . Then $\mathbb {P}_\alpha $ forces that, if $p^* \upharpoonright \alpha \in \dot G_\alpha $ , then there exists some $\gamma < \omega _1$ so that for all $x \in T_\gamma $ , there exists $q_{x} \le _\beta p^*$ in D such that $q_{x} \upharpoonright \alpha \in \dot G_\alpha $ and $q_{x} \Vdash ^V_\beta x \notin \dot U$ .

Proof. We prove that any $r \in \mathbb {P}_\alpha $ has an extension in $\mathbb {P}_\alpha $ which forces the conclusion of the claim. So let $r \in \mathbb {P}_\alpha $ be given. Extending r further if necessary, we may assume that r decides whether or not $p^* \upharpoonright \alpha \in \dot G_\alpha $ . If it decides not, then we are done. Otherwise, by separativity $r \le _\alpha p^* \upharpoonright \alpha $ . Let $r^* = r \cup p^* \upharpoonright [\alpha ,\beta )$ , which is in $\mathbb {P}_\beta $ and extends $p^*$ . Since D is dense in $\mathbb {P}_\beta $ , we can fix some $u \le _\beta r^*$ in D. Then $u \upharpoonright \alpha \le _\alpha r$ .

We argue that $u \upharpoonright \alpha $ is as required. So let $G_\alpha $ be a generic filter on $\mathbb {P}_\alpha $ which contains $u \upharpoonright \alpha $ . In $V[G_\alpha ]$ , let $\mathbb {P}_\beta / G_\alpha $ be the suborder of $\mathbb {P}_\alpha $ consisting of those $q \in \mathbb {P}_\beta $ such that $q \upharpoonright \alpha \in G_\alpha $ . So $u \in \mathbb {P}_\beta / G_\alpha $ . Recall that in V, $\mathbb {P}_\beta $ is forcing equivalent to $\mathbb {P}_\alpha * (\mathbb {P}_\beta / \dot G_\alpha )$ . So it makes sense to define in the model $V[G_\alpha ]$

$$ \begin{align*}W = \{ x \in T : u \Vdash_{\mathbb{P}_\beta / G_\alpha} x \in \dot U \}. \end{align*} $$

Clearly, W is a downwards closed subtree of T and $u \Vdash _{\mathbb {P}_\beta / G_\alpha } W \subseteq \dot U$ . Since $u \le _\beta p$ and p forces that $\dot U$ is $(< \! n)$ -splitting, W must be $(< \! n)$ -splitting.

By the inductive hypothesis, T does not contain an uncountable downwards closed $(< \! n)$ -splitting subtree in $V[G_\alpha ]$ . Therefore, W must be countable. Fix some $\gamma < \omega _1$ such that $W \subseteq T \upharpoonright \gamma $ . Now consider any $x \in T_\gamma $ . Then $x \notin W$ , so by the definition of W there exists some $p_x \le _\beta u$ in $\mathbb {P}_\beta / G_\alpha $ such that $p_x \Vdash _{\mathbb {P}_\beta / G_\alpha } x \notin \dot U$ . Since $p_x \in \mathbb {P}_\beta / G_\alpha $ , $p_x \upharpoonright \alpha \in G_\alpha $ . Find $v_x \in G_\alpha $ which forces in $\mathbb {P}_\alpha $ over V that $p_x \in \mathbb {P}_\beta / \dot G_\alpha $ and $p_x \Vdash _{\mathbb {P}_\beta / \dot G_\alpha }^{V[\dot G_\alpha ]} x \notin \dot U$ . By separativity, $v_x \le _\alpha p_x \upharpoonright \alpha $ . Now let ${q_x = v_x \cup p_x \upharpoonright [\alpha ,\beta )}$ . Then $q_x \le _\beta p_x \le _\beta u$ . It is easy to check that $q_x \Vdash ^V_\beta x \notin \dot U$ . As $u \in D$ and D is open, $q_x \in D$ . $\dashv _{\ \text {Claim}}$

We now work towards finding a condition $q \le _\beta p$ which forces that $\dot U$ is countable. Let $\lambda $ be a large enough regular cardinal and let $N \prec H(\lambda )$ be countable such that N contains as members the objects $\langle \mathbb {P}_i, \dot {\mathbb {Q}}_j : i \le \beta , \ j < \beta \rangle $ , p, T, and $\dot U$ . Fix the following objects:

• • an enumeration $\langle D_m : m < \omega \rangle $ of all dense open subsets of $\mathbb {P}_\beta $ which lie in N;

• • an enumeration $\langle x_m : m < \omega \rangle $ of $T_{N \cap \omega _1}$ ;

• • an increasing sequence $\langle \alpha _m : m < \omega \rangle $ of ordinals in $N \cap \beta $ which is cofinal in $\sup (N \cap \beta )$ , where $\alpha _0 = 0$ .

We define by induction sequences $\langle \dot p_m : m < \omega \rangle $ and ${\langle q_m : m < \omega \rangle }$ satisfying that for all $m < \omega $ :

1. (1) $\dot p_m$ is a $\mathbb {P}_{\alpha _m}$ -name for a condition in $\mathbb {P}_\beta $ , and $\dot p_0$ is a $\mathbb {P}_0$ -name for p;

2. (2) $q_m$ is an $(N,\mathbb {P}_{\alpha _m})$ -generic condition and $q_m \Vdash _{\alpha _m} (\dot p_m \in N \ \land \ \dot p_m \upharpoonright \alpha _m \in \dot G_{\alpha _m})$ ;

3. (3) $q_{m+1} \upharpoonright \alpha _m = q_m$ , and $q_{m+1}$ forces in $\mathbb {P}_{\alpha _{m+1}}$ that:

   1. (a) $\dot p_{m+1} \le _\beta \dot p_m$ ;

   2. (b) $\dot p_{m+1} \in D_m$ ;

   3. (c) $\dot p_{m+1} \Vdash ^V_\beta x_m \notin \dot U$ .

Note that the statements above imply that for all $m < \omega $ , $q_m \Vdash _{\alpha _m} \dot p_m \le _\beta p$ .

Begin by letting $q_0$ be the empty condition in $\mathbb {P}_0$ and letting $\dot p_0$ be a $\mathbb {P}_0$ -name for p. Now let $m < \omega $ and assume that $\dot p_m$ and $q_m$ are defined as described. Note that properties (1) and (2) imply that the assumptions of Lemma 3.2 hold, where we let $\delta = \alpha _{m+1}$ , $\alpha = \alpha _m$ , $\dot p = \dot p_m \upharpoonright \alpha _{m+1}$ , and $q = q_m$ . So fix $q_{m+1} \in \mathbb {P}_{\alpha _{m+1}}$ such that $q_{m+1} \upharpoonright \alpha _m = q_{m}$ , $q_{m+1}$ is $(N,\mathbb {P}_{\alpha _{m+1}})$ -generic, and $q_{m+1} \Vdash _{\alpha _{m+1}} \dot p_m \upharpoonright \alpha _{m+1} \in \dot G_{\alpha _{m+1}}$ .

To define the $\mathbb {P}_{\alpha _{m+1}}$ -name $\dot p_{m+1}$ , consider a generic filter $G_{\alpha _{m+1}}$ for $\mathbb {P}_{\alpha _{m+1}}$ which contains $q_{m+1}$ . Since $q_{m+1}$ is $(N,\mathbb {P}_{\alpha _{m+1}})$ -generic, $N[G_{\alpha _{m+1}}] \cap V = N$ . Let $G_{\alpha _m} = G_{\alpha _{m+1}} \upharpoonright \alpha _m$ . Since $q_{m+1} \upharpoonright \alpha _{m} = q_m$ , $q_m \in G_{\alpha _m}$ . Let $p_m = \dot p_m^{G_{\alpha _m}}$ . Then $p_m \in N$ by (2). By the choice of $q_{m+1}$ , $p_m \upharpoonright \alpha _{m+1} \in G_{\alpha _{m+1}}$ . Also, $p_m \le _\beta p$ . Applying the claim for $D = D_m$ , $p^* = p_m$ , and $\alpha = \alpha _{m+1}$ , in $V[G_{\alpha _{m+1}}]$ there exists some $\gamma < \omega _1$ so that for all $x \in T_\gamma $ , there exists $q_x \le _\beta p_m$ in $D_m$ such that $q_x \upharpoonright \alpha _{m+1} \in G_{\alpha _{m+1}}$ and $q_x \Vdash ^V_\beta x \notin \dot U$ . Since the relevant parameters are in $N[G_{\alpha _{m+1}}]$ , we can assume that $\gamma \in N[G_{\alpha _{m+1}}] \cap \omega _1 = N \cap \omega _1$ .

Letting $x = x_m \upharpoonright \gamma $ , we can find a condition $p_{m+1} \le _\beta p_m$ in $D_m$ such that $p_{m+1} \upharpoonright \alpha _{m+1} \in G_{\alpha _{m+1}}$ and $p_{m+1} \Vdash ^V_\beta x_m \upharpoonright \gamma \notin \dot U$ . By elementarity, we may assume that $p_{m+1} \in N[G_{\alpha _{m+1}}]$ . But $N[G_{\alpha _{m+1}}] \cap V = N$ , so $p_{m+1} \in N$ . In summary, $p_{m+1} \le _\beta p_m$ , $p_{m+1} \in N \cap D_m$ , $p_{m+1} \upharpoonright \alpha _{m+1} \in G_{\alpha _{m+1}}$ , and $p_{m+1} \Vdash ^V_\beta x_m \upharpoonright \gamma \notin \dot U$ . Since $\dot U$ is forced to be downwards closed, $p_{m+1} \Vdash ^V_\beta x_m \notin \dot U$ . Now let $\dot p_{m+1}$ be a $\mathbb {P}_{\alpha _{m+1}}$ -name for a member of $\mathbb {P}_\beta $ which is forced to satisfy the above properties in the case that $q_{m+1} \in \dot G_{\alpha _{m+1}}$ .

This completes the construction. Let $q = \bigcup _m q_m$ , which is in $\mathbb {P}_{\beta }$ . We argue that $q \le _\beta p$ , q is $(N,\mathbb {P}_\beta )$ -generic, and q forces that $\dot U$ is countable. Fix a generic filter $G_\beta $ for $\mathbb {P}_\beta $ which contains q and let $U = \dot U^{G_\beta }$ . For each $m < \omega $ let $G_{\alpha _m} = G_\beta \upharpoonright \alpha _m$ and $p_m = \dot p_m^{G_{\alpha _m}}$ .

We claim that for all $m < \omega $ , $q \le _\beta p_m$ , and in particular, $q \le _\beta p_0 = p$ . Since $q \upharpoonright \alpha _m = q_m$ , $q_m \Vdash _{\alpha _m} \dot p_m \upharpoonright \alpha _m \in \dot G_{\alpha _m}$ , and $\mathbb {P}_{\alpha _m}$ is separative, it follows that ${q \upharpoonright \alpha _m \le _{\alpha _m} p_m \upharpoonright \alpha _m}$ . Hence, for all $m < k < \omega $ , since $p_k \le _\beta p_m$ we have that $q \upharpoonright \alpha _k \le _{\alpha _k} p_k \upharpoonright \alpha _k \le _{\alpha _k} p_m \upharpoonright \alpha _k$ . Now $\mathrm {dom}(q) \subseteq \bigcup _k \alpha _k = \sup (N \cap \beta )$ , and since $p_m \in N$ , also $\mathrm {dom}(p_m) \subseteq \sup (N \cap \beta )$ . So in fact for all $\delta < \beta $ , $q \upharpoonright \delta \le _\delta p_m \upharpoonright \delta $ , which implies that $q \le _\beta p_m$ .

Since $G_\beta $ was an arbitrary generic filter which contains q, q forces in $\mathbb {P}_\beta $ that for all $m < \omega $ , $q \le _\beta \dot p_m$ , and hence that $\dot p_m \in \dot G_\beta \cap N$ . So by (3), q forces in $\mathbb {P}_\beta $ that for all $m < \omega $ , $\dot p_{m+1} \in D_m \cap N \cap \dot G_\beta $ . This proves that q is $(N,\mathbb {P}_\beta )$ -generic. Also by (3), q forces in $\mathbb {P}_\beta $ that for all $m < \omega $ , $p_{m+1}$ is a condition in $\dot G_\beta $ which forces over V that $x_m \notin \dot U$ , and therefore in fact $x_m \notin \dot U$ . Since $\dot U$ is forced to be downwards closed, q forces that $\dot U \subseteq T \upharpoonright (N \cap \omega _1)$ and hence $\dot U$ is countable.

Proof. By Lemma 1.1(a).

Proof. Let $\vec x$ be the root of U. We define a sequence $\langle \delta _n : n < \omega \rangle $ and tuples $\vec b_n^0$ and $\vec b_n^1$ in $U \cap T_{\delta _n}^m$ for all $n < \omega $ by induction. We maintain inductively that for all $k < n < \omega $ and $j < 2$ , $\vec b_n^j \upharpoonright \delta _k = \vec b_k^0$ .

For the base case, apply Proposition 4.3 to find some countable ordinal $\delta _0> \beta $ such that there exist infinitely many pairwise disjoint members of $U \cap T_{\delta _0}^m$ above $\vec x$ . Pick two among these disjoint tuples and call them $\vec b_0^0$ and $\vec b_0^1$ . Now let $n < \omega $ and assume that $\delta _n$ , $\vec b_n^0$ , and $\vec b_n^1$ are defined. Apply Proposition 4.3 to find some countable ordinal $\delta _{n+1}> \delta _n$ such that there exist infinitely many pairwise disjoint members of $U \cap T_{\delta _{n+1}}^m$ above $\vec b_n^0$ . Take two such disjoint tuples and call them $\vec b_{n+1}^0$ and $\vec b_{n+1}^1$ . Note that the inductive hypothesis is maintained. This completes the induction. Let $\delta = \sup _n \delta _n$ .

For every $n < \omega $ pick some tuple $\vec b_n$ in $U \cap T_{\delta }^m$ above $\vec b_n^1$ . Consider $k < n < \omega $ . Then $\vec b_k \upharpoonright \delta _k = \vec b_k^1$ , which is disjoint from $\vec b_k^0$ . On the other hand, by our inductive hypothesis

$$ \begin{align*}\vec b_n \upharpoonright \delta_k = (\vec b_n \upharpoonright \delta_n) \upharpoonright \delta_k = \vec b_n^1 \upharpoonright \delta_k = \vec b_k^0. \end{align*} $$

So $\vec b_k \upharpoonright \delta _k$ and $\vec b_n \upharpoonright \delta _k$ are disjoint. It follows that $\vec b_k$ and $\vec b_n$ are disjoint as well.

Now fix $0 < k < \omega $ . Consider $n_0 < n_1 \le k$ . Then by the last paragraph, $\vec b_{n_0} \upharpoonright \delta _{n_0}$ and $\vec b_{n_1} \upharpoonright \delta _{n_0}$ are disjoint. Since $\delta _{k-1} \ge \delta _{n_0}$ , $\vec b_{n_0} \upharpoonright \delta _{k-1}$ and $\vec b_{n_1} \upharpoonright \delta _{k-1}$ are also disjoint. For any $n \le k$ , $\vec b_n \upharpoonright \beta = \vec x$ , which is injective, so $X_n$ has unique drop-downs to $\beta $ and hence also to $\delta _{k-1}$ . It easily follows from these observations that $X_0 \cup \cdots \cup X_{k}$ has unique drop-downs to $\delta _{k-1}$ .

By the above, $X_0 \cup X_1$ has unique drop-drowns to $\delta _0$ , and since $(X_0 \cup X_1) \upharpoonright \delta _0$ consists of the elements of $\vec b_0^0$ and $\vec b_0^1$ , which are disjoint and above $\vec x$ , $(X_0 \cup X_1) \upharpoonright \delta _0$ has $(\le \! 2)$ -to-one drop-downs to $\beta $ . By Lemma 4.5, $X_0 \cup X_1$ has $(\le \! 2)$ -to-one drop-downs to $\beta $ . Now let $1 < k < \omega $ and we show that $X_0 \cup \cdots \cup X_k$ has $(\le \! 2)$ -to-one drop-downs to $\delta _{k-2}$ . As proved above, $X_0 \cup \cdots \cup X_{k-1}$ has unique drop-downs to $\delta _{k-2}$ . Now $\vec b_k \upharpoonright \delta _k = \vec b_k^1$ and $\vec b_{k-1} \upharpoonright \delta _{k-1} = \vec b_{k-1}^1$ . By our inductive hypothesis, it follows that $\vec b_k \upharpoonright \delta _{k-2} = (\vec b_k \upharpoonright \delta _k) \upharpoonright \delta _{k-2} = \vec b_{k}^1 \upharpoonright \delta _{k-2} = \vec b_{k-2}^0$ and $\vec b_{k-1} \upharpoonright \delta _{k-2} = (\vec b_{k-1} \upharpoonright \delta _{k-1}) \upharpoonright \delta _{k-2} = \vec b_{k-1}^1 \upharpoonright \delta _{k-2} = \vec b_{k-2}^0$ . It easily follows that $X_{k-1} \cup X_k$ has $(\le \! 2)$ -to-one drop-downs to $\delta _{k-2}$ and so $X_0 \cup \cdots \cup X_{k}$ has $(\le \! 2)$ -to-one drop-downs to $\delta _{k-2}$ .

Finally, consider $0 < k < \omega $ . For any $k < n < \omega $ , we have that $\vec b_n \upharpoonright \delta _{k-1} = \vec b_{k-1}^0 = \vec b_k \upharpoonright \delta _{k-1}$ . It follows that $(\bigcup _n X_n) \upharpoonright \delta _{k-1} = (X_0 \cup \ldots \cup X_k) \upharpoonright \delta _{k-1}$ .

Proof. By Proposition 4.3, we can fix some $\gamma < \omega _1$ greater than $\zeta $ for which there exists a pairwise disjoint sequence $\langle \vec b_n : n < \omega \rangle $ of tuples in $U \cap T_\gamma ^m$ above the root of U. Then for all finite $t \subseteq T_\gamma $ , there is some n such that $\vec b_n$ is disjoint from t. For all $\gamma \le \alpha < \beta < \omega _1$ , the equality $(U \cap T_\beta ^m) \upharpoonright \alpha = U \cap T_\alpha ^m$ follows from the fact that U is downwards closed and every element of U has uncountably many elements above it.

Proof. The proof is by induction on $\gamma $ , where the base case $\gamma = \beta $ is immediate.

Successor case: Assume that the statement is true for some countable ordinal $\gamma \ge \beta $ , and we prove that it is true for $\gamma +1$ . So let $X \subseteq T_{\gamma +1}$ be finite with $(\le \! s)$ -to-one drop-downs to $\beta $ such that for all $x \in X$ , $F(x \upharpoonright \beta ) = 1$ . By Lemma 4.5, X has $(\le \! s)$ -to-one drop-downs to $\gamma $ and $X \upharpoonright \gamma $ has $(\le \! s)$ -to-one drop-downs to $\beta $ . Obviously, for all $y \in X \upharpoonright \gamma $ , $F(y \upharpoonright \beta ) = 1$ . By the inductive hypothesis, we can fix an $\mathcal S$ -splitting subtree function G with top level $\gamma $ such that $G(x \upharpoonright \gamma ) = 1$ for all $x \in X$ and G fulfills every generalized promise in $\Gamma $ .

Next we extend G to $G^+$ with top level $\gamma +1$ . To help with the construction of $G^+$ , fix the following objects:

• • an enumeration $\langle y_n : n < \omega \rangle $ of $T_\gamma $ ;

• • an enumeration $\langle \mathcal U_n : n < \omega \rangle $ of $\Gamma $ ;

• • for each $n < \omega $ , an enumeration $\langle \mathcal Y^n_k : k < \omega \rangle $ of $\mathcal U_n(\gamma +1)$ (possibly with repetitions);

• • an enumeration $\langle t_n : n < \omega \rangle $ of all finite subsets of $T_{\gamma +1}$ ;

• • a bijection $h : \omega \to 2 \times \omega ^3$ .

We define by induction an increasing sequence $\langle Z_n : n < \omega \rangle $ of finite subsets of $T_{\gamma +1}$ . Our inductive hypothesis is that for all $n < \omega $ and $y \in T_{\gamma }$ , y has at most s-many immediate successors in $Z_n$ .

To begin, let $Z_0 = X$ . Since X has $(\le \! s)$ -to-one drop-downs to $\gamma $ , the inductive hypothesis holds. Now let $n < \omega $ and assume that $Z_n$ is defined and satisfies the inductive hypothesis. Let $h(n) = (i,k_0,k_1,k_2)$ .

Case 1. $i = 0$ . Consider $y_{k_0}$ . If $G(y_{k_0}) = 0$ , then let $Z_{n+1} = Z_n$ . Assume that $G(y_{k_0}) = 1$ . Since $y_{k_0}$ has at least s-many immediate successors in T and $y_{k_0}$ has at most s-many immediate successors in $Z_n$ , we can find a finite set $Z_{n+1}$ containing $Z_n$ such that $y_{k_0}$ has exactly s-many immediate successors in $Z_{n+1}$ and every member of $Z_{n+1} \setminus Z_n$ is an immediate successor of $y_{k_0}$ . Clearly, the inductive hypothesis holds.

Case 2. $i = 1$ . Consider $\mathcal U_{k_0}$ , $\mathcal Y^{k_0}_{k_1}$ , and $t_{k_2}$ . Let l be the dimension of $\mathcal U_{k_0}$ . Applying the fact that G fulfills $\mathcal U_{k_0}$ , we can find some $\vec a = (a_0,\ldots ,a_{l-1}) \in \mathcal Y^{k_0}_{k_1} \upharpoonright \gamma $ which is disjoint from $(Z_n \cup t_{k_2}) \upharpoonright \gamma $ and satisfies that for all $i < l$ , $G(a_i) = 1$ . Fix ${\vec a^+ = (a_0^+,\ldots ,a_{l-1}^+) \in \mathcal Y^{k_0}_{k_1}}$ above $\vec a$ . Since $\vec a$ is injective, $\vec a^+$ has unique drop-downs to $\gamma $ . Note that $\vec a^+$ is disjoint from $Z_n \cup t_{k_2}$ . Define $Z_{n+1} = Z_n \cup \{ a_0^+,\ldots ,a_{l-1}^+ \}$ . By the inductive hypothesis for n and the fact that $\{ a_0,\ldots ,a_{l-1} \}$ is disjoint from $Z_n \upharpoonright \gamma $ , the inductive hypothesis is maintained.

This completes the construction. Now define $G^+$ extending G so that for all $x \in T_{\gamma +1}$ , $G^+(x) = 1$ if $x \in \bigcup _n Z_n$ , and $G^+(x) = 0$ otherwise. It is routine to check that $G^+$ is as required.

Limit case: Assume that $\delta> \beta $ is a countable limit ordinal and for all $\gamma $ with $\beta \le \gamma < \delta $ , the result holds for $\gamma $ . Let $X \subseteq T_\delta $ be a finite set which has $(\le \! s)$ -to-one drop-downs to $\beta $ such that for all $x \in X$ , $F(x \upharpoonright \beta ) = 1$ . To help with the construction, fix the following objects:

• • an increasing sequence of ordinals $\langle \delta _n : n < \omega \rangle $ cofinal in $\delta $ with $\delta _0 = \beta $ ;

• • an enumeration $\langle y_n : n < \omega \rangle $ of $T \upharpoonright \delta $ ;

• • an enumeration $\langle \mathcal U_n : n < \omega \rangle $ of $\Gamma $ ;

• • for each $n < \omega $ , an enumeration $\langle \mathcal Y^n_k : k < \omega \rangle $ of $\mathcal U_n(\delta )$ (possibly with repetitions);

• • an enumeration $\langle t_n : n < \omega \rangle $ of all finite subsets of $T_{\delta }$ ;

• • a surjection $h : \omega \to 2 \times \omega ^3$ such that every element of the codomain has an infinite preimage.

We define by induction an increasing sequence $\langle Z_n : n < \omega \rangle $ of finite subsets of $T_{\delta }$ together with a sequence $\langle F_n : n < \omega \rangle $ satisfying that for all $n < \omega $ :

1. (1) $Z_n$ has $(\le \! s)$ -to-one drop-downs to $\delta _n$ ;

2. (2) $F_n$ is an $\mathcal S$ -splitting subtree function with top level $\delta _n$ , $F_n(z \upharpoonright \delta _n) = 1$ for all $z \in Z_n$ , and $F_n \subseteq F_{n+1}$ .

To begin, let $F_0 = F$ and $Z_0 = X$ . Now let $n < \omega $ and assume that $Z_n$ and $F_n$ are defined and satisfy (1) and (2). By Lemma 4.5, $Z_n \upharpoonright \delta _{m+1}$ has $(\le \! s)$ -to-one drop-downs to $\delta _m$ . Let $h(n) = (i,k_0,k_1,k_2)$ . As a first step, we define $Z_{n+1}$ which has $(\le \! s)$ -to-one drop-downs to $\delta _{n}$ such that for all $x \in Z_{n+1}$ , $F_n(x \upharpoonright \delta _n) = 1$ .

Case 1. $i = 0$ . Consider $y_{k_0}$ . If either (a) $\mathrm {ht}_T(y_{k_0})> \delta _n$ , (b) $\mathrm {ht}_T(y_{k_0}) \le \delta _n$ and $F_n(y_{k_0}) = 0$ , or (c) $\mathrm {ht}_T(y_{k_0}) \le \delta _n$ and $y_{k_0} \in (Z_n \upharpoonright \mathrm {ht}_T(y_{k_0}))$ , then let $Z_{n+1} = Z_n$ (roughly speaking, (a) says $y_{k_0}$ is not yet ready to be handled, (b) says $y_{k_0}$ is not relevant, and (c) says $y_{k_0}$ has already been handled). Now suppose that ${\mathrm {ht}_T(y_{k_0}) \le \delta _n}$ , $F_n(y_{k_0}) = 1$ , and $y_{k_0} \notin (Z_n \upharpoonright \mathrm {ht}_T(y_{k_0}))$ . Since $F_n$ is a subtree function, we can fix $y' \ge _T y_{k_0}$ with height $\delta _n$ such that $F_n(y') = 1$ . Note that $y' \notin Z_n \upharpoonright \delta _n$ . As T is normal, fix some $z \in T_\delta $ above $y'$ . Let $Z_{n+1} = Z_n \cup \{ z \}$ . By the inductive hypothesis and the fact that $z \upharpoonright \delta _n = y'$ is not in $Z_n \upharpoonright \delta _n$ , it easily follows that $Z_{n+1}$ has $(\le \! s)$ -to-one drop-downs to $\delta _n$ .

Case 2. $i = 1$ . Consider $\mathcal U_{k_0}$ , $\mathcal Y^{k_0}_{k_1}$ , and $t_{k_2}$ . Let l be the dimension of $\mathcal U_{k_0}$ . Applying the fact that $F_n$ fulfills $\mathcal U_{k_0}$ , we can find $\vec a = (a_0,\ldots ,a_{l-1}) \in \mathcal Y^{k_0}_{k_1} \upharpoonright \delta _n$ which is disjoint from $(Z_n \cup t_{k_2}) \upharpoonright \delta _n$ . Fix $\vec a^+ = (a_0^+,\ldots ,a_{l-1}^+) \in \mathcal Y^{k_0}_{k_1}$ above $\vec a$ . Since $\vec a$ is injective, $\vec a^+$ has unique drop-downs to $\delta _n$ . Note that $\vec a^+$ is disjoint from $t_{k_2}$ . Define $Z_{n+1} = Z_n \cup \{ a_0^+,\ldots ,a_{l-1}^+ \}$ . By the inductive hypothesis and the fact that $\{ a_0^+,\ldots ,a_{l-1}^+ \} \upharpoonright \delta _n$ is disjoint from $Z_n \upharpoonright \delta _n$ , it easily follows that $Z_{n+1}$ has $(\le \! s)$ -to-one drop-downs to $\delta _{n+1}$ .

As a second step, we define $F_{n+1}$ . Note that by Lemma 4.5, $Z_{n+1} \upharpoonright \delta _{n+1}$ has $(\le \! s)$ -to-one drop-downs to $\delta _n$ , and for all $y \in Z_{n+1} \upharpoonright \delta _{n+1}$ , $F_n(y \upharpoonright \delta _n) = 1$ . By the inductive hypothesis for $\delta _{n+1}$ , we can find an $\mathcal S$ -splitting subtree function $F_{n+1}$ extending $F_n$ with top level $\delta _{n+1}$ such that for all $y \in Z_{n+1} \upharpoonright \delta _{n+1}$ , $F_{n+1}(y) = 1$ .

This completes the construction. Define $F^+$ extending $\bigcup _n F_n$ so that for all $x \in T_{\delta }$ , $F^+(x) = 1$ if $x \in \bigcup _n Z_n$ , and $F^+(x) = 0$ otherwise. It is routine to check that $F^+$ is as required.

Proof. Apply Lemma 4.6 to U and then argue as in Lemma 4.8.

Proof. Let $\mathcal U(\delta ) = \{ \mathcal Y_0 \}$ . Fix sequences $\langle \delta _n : n < \omega \rangle $ , $\langle \vec b_n : n < \omega \rangle $ , and ${\langle X_n : n < \omega \rangle }$ as in Definition 5.5. To help with the construction of $F^+$ , we fix the following objects:

• • an enumeration $\langle y_n : n < \omega \rangle $ of $T \upharpoonright \delta $ ;

• • an enumeration $\langle \mathcal U_n : n < \omega \rangle $ of $\Gamma $ ;

• • for each $n < \omega $ , an enumeration $\langle \mathcal Y^n_k : k < \omega \rangle $ of $\mathcal U_n(\delta )$ (possibly with repetitions);

• • an enumeration $\langle t_n : n < \omega \rangle $ of all finite subsets of $T_{\delta }$ ;

• • a surjection $h : \omega \to 2 \times \omega ^3$ such that every element of the codomain has an infinite preimage.

We define by induction an increasing sequence $\langle Z_n : n < \omega \rangle $ of finite subsets of $T_{\delta }$ together with a sequence $\langle F_n : n < \omega \rangle $ satisfying that for all $n < \omega $ :

1. (1) $Z_n$ has unique drop-downs to $\delta _n$ ;

2. (2) $Z_n \upharpoonright \delta _n$ and $(\bigcup _m X_m) \upharpoonright \delta _n$ are disjoint;

3. (3) $F_n$ is an $\mathcal S$ -splitting subtree function extending F with top level $\delta _n$ such that $F_n(z \upharpoonright \delta _n) = 1$ for all $z \in Z_n \cup \bigcup _m X_m$ ;

4. (4) $F_n \subseteq F_{n+1}$ .

To begin, let $Z_0 = \emptyset $ . Since $X_0 \cup X_1$ has $(\le \! 2)$ -to-one drop-downs to $\beta $ , by Lemma 4.5 also $(X_0 \cup X_1) \upharpoonright \delta _0$ has $(\le \! 2)$ -to-one drop-downs to $\beta $ . We also know that for all $x \in X_0 \cup X_1$ , $F(x \upharpoonright \beta ) = 1$ . By Proposition 5.4 (Extension), we can fix an $\mathcal S$ -splitting subtree function $F_0$ extending F with top level $\delta _0$ which fulfills every member of $\Gamma $ and satisfies that for all $x \in X_0 \cup X_1$ , $F_0(x \upharpoonright \delta _0) = 1$ . Inductive hypotheses (1) and (2) are immediate, and (3) follows from the fact that $(\bigcup _m X_m) \upharpoonright \delta _0 = (X_0 \cup X_1) \upharpoonright \delta _0$ .

Now let $n < \omega $ be given and assume that $Z_n$ and $F_n$ are defined as required. Let $h(n) = (i,k_0,k_1,k_2)$ . We begin by defining $Z_{n+1}$ satisfying the following properties:

• • $Z_{n+1}$ has unique drop-downs to $\delta _n$ ;

• • $Z_{n+1} \upharpoonright \delta _n$ and $(\bigcup _m X_m) \upharpoonright \delta _n$ are disjoint;

• • for all $x \in Z_{n+1}$ , $F_n(x \upharpoonright \delta _n) = 1$ .

Note that these properties imply inductive hypotheses (1) and (2) for $n+1$ .

Case 1. $i = 0$ . Consider $y_{k_0}$ . If either (a) $\mathrm {ht}_T(y_{k_0})> \delta _n$ , (b) $\mathrm {ht}_T(y_{k_0}) \le \delta _n$ and $F_n(y_{k_0}) = 0$ , or (c) $\mathrm {ht}_T(y_{k_0}) \le \delta _n$ and $y_{k_0} \in (Z_n \cup \bigcup _m X_m) \upharpoonright \mathrm {ht}_T(y_{k_0})$ , then let $Z_{n+1} = Z_n$ . Now assume that $\mathrm {ht}_T(y_{k_0}) \le \delta _n$ , $F_n(y_{k_0}) = 1$ , and $y_{k_0}$ has nothing above it in $Z_n \cup (\bigcup _m X_m)$ . Since $F_n$ is a subtree function, we can fix some $y' \ge _T y_{k_0}$ with height $\delta _n$ such that $F_n(y') = 1$ . As T is normal, let $z \in T_\delta $ be above $y'$ . Define $Z_{n+1} = Z_n \cup \{ z \}$ . By the inductive hypothesis it easily follows that $Z_{n+1}$ has unique drop-downs to $\delta _n$ . Since $Z_{n} \upharpoonright \delta _n$ and $(\bigcup _m X_m) \upharpoonright \delta _n$ are disjoint and $y'$ is not in $(\bigcup _m X_m) \upharpoonright \delta _{n}$ , it follows that $Z_{n+1} \upharpoonright \delta _{n}$ and $(\bigcup _m X_m) \upharpoonright \delta _{n}$ are disjoint.

Case 2. $i = 1$ . Consider $\mathcal U_{k_0}$ , $\mathcal Y^{k_0}_{k_1}$ , and $t_{k_2}$ . Let l be the dimension of $\mathcal U_{k_0}$ . Applying the fact that $F_n$ fulfills $\mathcal U_{k_0}$ , we can find $\vec a = (a_0,\ldots ,a_{l-1}) \in \mathcal Y^{k_0}_{k_1} \upharpoonright \delta _n$ which is disjoint from

$$ \begin{align*}(Z_n \cup t_{k_2} \cup X_0 \cup \cdots \cup X_{n+1}) \upharpoonright \delta_n, \end{align*} $$

such that for all $i < l$ , $F_n(a_i) = 1$ . Fix $\vec a^+ = (a_0^+,\ldots ,a_{l-1}^+) \in \mathcal Y^{k_0}_{k_1}$ above $\vec a$ . Since $\vec a$ is injective, $\vec a^+$ has unique drop-downs to $\delta _n$ . Note that $\vec a^+$ is disjoint from $t_{k_2}$ . Define $Z_{n+1} = Z_n \cup \{ a_0^+,\ldots ,a_{l-1}^+ \}$ . By the inductive hypothesis and the choice of $\vec a$ , it easily follows that $Z_{n+1}$ has unique drop-downs to $\delta _n$ . Also by the inductive hypothesis and the choice of $\vec a$ , $Z_{n+1} \upharpoonright \delta _n$ and $(X_0 \cup \cdots \cup X_{n+1}) \upharpoonright \delta _n$ are disjoint. But $(X_0 \cup \cdots \cup X_{n+1}) \upharpoonright \delta _n = (\bigcup _m X_m) \upharpoonright \delta _n$ , so $Z_{n+1} \upharpoonright \delta _n$ and $(\bigcup _m X_m) \upharpoonright \delta _n$ are disjoint.

This completes the definition of $Z_{n+1}$ in both cases. We know that for all ${x \in Z_{n+1} \cup X_0 \cup \cdots \cup X_{n+2}}$ , $F_n(x \upharpoonright \delta _n) = 1$ . Also, $(\bigcup _m X_m) \upharpoonright \delta _{n+1} = (X_0 \cup \cdots \cup X_{n+2}) \upharpoonright \delta _{n+1}$ , which by hypothesis and Lemma 4.5 has $(\le \! 2)$ -to-one drop-downs to $\delta _{n}$ . Since $Z_{n+1}$ has unique drop-downs to $\delta _n$ and $Z_{n+1} \upharpoonright \delta _n$ and $(\bigcup _m X_m) \upharpoonright \delta _n$ are disjoint, it easily follows that $(Z_{n+1} \cup X_0 \cup \cdots \cup X_{n+2}) \upharpoonright \delta _n$ has $(\le \! 2)$ -to-one drop-downs to $\delta _n$ . By Proposition 5.4 (Extension), fix an $\mathcal S$ -splitting subtree function $F_{n+1}$ extending $F_n$ with top level $\delta _{n+1}$ which fulfills every member of $\Gamma $ and satisfies that $F_{n+1}(x \upharpoonright \delta _{n+1}) = 1$ for all $x \in Z_{n+1} \cup X_0 \cup \cdots \cup X_{n+2}$ . Since $(X_0 \cup \cdots \cup X_{n+2}) \upharpoonright \delta _{n+1} = (\bigcup _m X_m) \upharpoonright \delta _{n+1}$ , for all $x \in \bigcup _m X_m$ , $F_{n+1}(x \upharpoonright \delta _{n+1}) = 1$ .

This completes the construction. Now define $F^+$ extending $\bigcup _m F_m$ so that for all $x \in T_{\delta }$ , $F^+(x) = 1$ if $x \in \bigcup _m Z_m \cup \bigcup _m X_m$ , and $F^+(x) = 0$ otherwise. It is routine to check that $F^+$ is as required.

Proof. Immediate from Proposition 5.4 (Extension).

Proof. Immediate from Proposition 5.7 (Adding generalized promises).

Proof. Since T is normal, we can fix $\xi < N \cap \omega _1$ greater than or equal to $\beta $ such that X has unique drop-downs to $\xi $ . By Lemma 4.5, $X \upharpoonright \xi $ has $(\le \! s)$ -to-one drop-downs to $\beta $ . Applying Proposition 6.2 (Extension) in N, find $F'$ in N such that $(F',\Gamma ) \le (F,\Gamma )$ , $(F',\Gamma )$ has top level $\xi $ , and for all $x \in X$ , $F'(x \upharpoonright \xi ) = 1$ . Now to simplify notation, let us just assume that $\beta = \xi $ and $F = F'$ .

Fix some injective tuple $\vec a = (a_0,\ldots ,a_{m-1})$ which lists the elements of X. Suppose for a contradiction that for every $(G,\Sigma ) \le (F,\Gamma )$ in $N \cap D$ with some top level $\gamma $ , there exists $j < m$ such that $G(a_j \upharpoonright \gamma ) = 0$ . Let $x_i := a_i \upharpoonright \beta $ for all $i < m$ and let $\vec x = (x_0,\ldots ,x_{m-1})$ .

Define W as the set of all tuples $\vec b = (b_0,\ldots ,b_{m-1})$ in the derived tree $T_{\vec x}$ satisfying that whenever $(G,\Sigma ) \le (F,\Gamma )$ is in D and has some top level $\gamma $ , where the height of the elements of $\vec b$ is greater than or equal to $\gamma $ , then for some $j < m$ , $G(b_j \upharpoonright \gamma ) = 0$ . By elementarity, $W \in N$ . Note that W is downwards closed in $T_{\vec x}$ .

We claim that W is uncountable. If not, then by elementarity there exists some $\zeta \in N \cap \omega _1$ greater than $\beta $ such that every member of W is in $(T \upharpoonright \zeta )^m$ . We claim that $\vec a \upharpoonright \zeta $ is in W, which is a contradiction. Otherwise, there exists some $(G,\Sigma ) \le (F,\Gamma )$ in D with some top level $\gamma $ , where $\gamma \le \zeta $ , such that for all $j < m$ , $G(a_j \upharpoonright \gamma ) = 1$ . Since $(a_0 \upharpoonright \zeta ,\ldots ,a_{m-1} \upharpoonright \zeta ) \in N$ , by elementarity we may assume that $(G,\Sigma )$ is in N. So $(G,\Sigma ) \le (F,\Gamma )$ is in $N \cap D$ and for all $j < m$ , $G(a_j \upharpoonright \gamma ) = 1$ , contradicting our assumptions.

Apply Lemma 4.2 to fix a promise $U \subseteq W$ with the same root as W, which by elementarity we may assume is in N. The root of U is $\vec x$ , so $\beta $ is both the base level of U and also the top level of $(F,\Gamma )$ . Also, $F(x_i) = 1$ for all $i < m$ . Applying Lemma 5.6, fix a countable limit ordinal $\delta> \beta $ such that the function $\mathcal U$ with domain $[\delta ,\omega _1)$ defined by $\mathcal U(\gamma ) = \{ U \cap T_\gamma ^m \}$ is a generalized promise which is suitable for F. By elementarity, we may assume that $\delta $ and $\mathcal U$ are in N. Apply Proposition 6.3 (Adding generalized promises) in N to find some G in N such that $(G,\Gamma \cup \{ \mathcal U \}) \in \mathbb {P}$ , $(G,\Gamma \cup \{ \mathcal U \}) \le (F,\Gamma )$ , and G has top level $\delta $ .

Now fix $(H,\Lambda ) \le (G,\Gamma \cup \{ \mathcal U \})$ in $N \cap D$ . Let $\gamma $ be the top level of H. Since H fulfills $\mathcal U$ and $\mathcal U(\gamma ) = \{ U \cap T_\gamma ^m \}$ , there exists a tuple $\vec b = (b_0,\ldots ,b_{m-1}) \in U \cap T_\gamma ^m$ such that for all $i < m$ , $H(b_i) = 1$ . Since $U \subseteq W$ , $\vec b \in W$ . But the condition $(H,\Lambda )$ witnesses that $\vec b \notin W$ , and we have a contradiction.