Proof (Sketch)

The result is immediate for iterations of length a successor ordinal, so let $\delta $ be a limit ordinal and suppose that we have a finite support forcing iteration $\langle \mathbb {P}_{i}, \dot {\mathbb {Q}}_{j} : i \le \delta , \ j < \delta \rangle $ of c.c.c. forcings such that for all $i < \delta $ , $\mathbb {P}_i$ forces that S is Suslin. Then $\mathbb {P}_{\delta }$ is c.c.c. Let $p \in \mathbb {P}_{\delta }$ and assume for a contradiction that

$$ \begin{align*} p \Vdash_{\delta} “\{ \dot x_{\alpha} : \alpha < \omega_1 \} \text{ is an uncountable antichain of }S.” \end{align*} $$

Then for each $\alpha $ we can choose $p_{\alpha } \le p$ in $\mathbb {P}_{\delta }$ and $y_{\alpha } \in S$ such that $p_{\alpha } \Vdash _{\mathbb {P}_{\delta }} “\dot x_{\alpha } = y_{\alpha }.”$ By a standard $\Delta $ -system argument on the domains of the $p_{\alpha }$ ’s, find $\beta < \delta $ and an uncountable set $X \subseteq \omega _1$ such that for all $i < j$ in X, if $p_i \upharpoonright \beta $ and $p_j \upharpoonright \beta $ are compatible in $\mathbb {P}_{\beta }$ , then $p_i$ and $p_j$ are compatible in $\mathbb {P}_{\delta }$ .

For all $i \in X$ , $p_i \upharpoonright \beta \le p \upharpoonright \beta $ in $\mathbb {P}_{\beta }$ . By a standard fact about c.c.c. forcings, there exists $u \le p \upharpoonright \beta $ in $\mathbb {P}_{\beta }$ such that

$$ \begin{align*}u \Vdash_{\beta} “\{ i \in X : p_i \upharpoonright \beta \in \dot G_{\mathbb{P}_{\beta}} \} \text{ is uncountable.”} \end{align*} $$

Let G be a generic filter on $\mathbb {P}_{\beta }$ which contains u. Let $Y := \{ i \in X : p_i \upharpoonright \beta \in G \}$ , which is uncountable. Then for all $i < j$ in Y, $p_i \upharpoonright \beta $ and $p_j \upharpoonright \beta $ are in G and hence are compatible in $\mathbb {P}_{\beta }$ . As $Y \subseteq X$ , $p_i$ and $p_j$ are compatible in $\mathbb {P}_{\delta }$ , which in turn easily implies that $y_i$ and $y_j$ are incomparable in S. Hence, in $V[G]$ the set $\{ y_i : i \in Y \}$ is an uncountable antichain of S, which contradicts our assumption that S is Suslin in $V^{\mathbb {P}_{\beta }}$ .

Proof Suppose that U is not Aronszajn and let B be a cofinal branch of U. For each $\alpha < \omega _1$ , write $B(\alpha ) = (B_0(\alpha ),\ldots ,B_{n-1}(\alpha ))$ . Then for each $i < n$ , $B_i$ is a cofinal branch of $T_{a_i}$ , so $T_{a_i}$ is not Aronszajn. Conversely, suppose that for each $i < n$ , $B_i$ is a cofinal branch of $T_{a_i}$ . Then B defined by $B(\alpha ) := (B_0(\alpha ),\ldots ,B_{n-1}(\alpha ))$ is a cofinal branch of U.

Proof Let $f : S \otimes T \to \omega $ be a specializing function. In $V^S$ , let B be a cofinal branch of S, and we define a specializing function $g : T \to \omega $ . For any $x \in T$ , define $g(x) := f(B(\mathrm {ht}_T(x)),x)$ . If $x <_T y$ , then $(B(\mathrm {ht}_T(x)),x) <_{S \otimes T} (B(\mathrm {ht}_T(y)),y)$ , so $g(x) = f(B(\mathrm {ht}_T(x)),x)) \ne f(B(\mathrm {ht}_T(y)),y) = g(y)$ .

Proof Suppose that $\{ (d_{i,0},\ldots ,d_{i,n-1}) : i < \omega _1 \}$ is an uncountable antichain of $S_{b_0} \otimes \cdots \otimes S_{b_{n-1}}$ . Since level $\xi $ of S is countable, we can find an uncountable set $X \subseteq \omega _1$ and some $c_0,\ldots ,c_{n-1}$ of height $\xi $ such that for all $i \in X$ , $d_{i,0} \upharpoonright \xi = c_0, \ldots , d_{i,n-1} \upharpoonright \xi = c_{n-1}$ . Then $\{ (d_{i,0},\ldots ,d_{i,n-1}) : i \in X \}$ is an uncountable antichain of $S_{c_0} \otimes \cdots \otimes S_{c_{n-1}}$ .

Proof Let $f : W \to \omega $ be a specializing function. Define $g : U \to \omega $ by letting $g(c_0,\ldots ,c_{n-1}) := f(c_{i_0},\ldots ,c_{i_{m-1}})$ . Then easily g is a specializing function for U, so U is special.

Proof See Chapter 4 of [Reference Baumgartner3] for the forward direction. Conversely, suppose that B is a cofinal branch of T. For each $\alpha < \omega _1$ define $p_{\alpha } := \{ (B(\alpha ),0) \}$ . Then $\{ p_{\alpha } : \alpha < \omega _1 \}$ is an uncountable antichain of $\mathbb {Q}(T)$ .

Proof We use the fact that for c.c.c. forcings $\mathbb {P}$ and $\mathbb {Q}$ , $\mathbb {P} \times \mathbb {Q}$ is c.c.c. iff $\Vdash _{\mathbb {P}} \text {“}\mathbb {Q} \text { is c.c.c.”}$ Recall that the tree S is Suslin iff the forcing poset S (with the reversed order) is c.c.c. So both forcings S and $\mathbb {Q}(T)$ are c.c.c. Therefore,

$$ \begin{gather*} \Vdash_{\mathbb{Q}(T)} “S \text{ is Suslin”} \ \Longleftrightarrow \ \Vdash_{\mathbb{Q}(T)} “S \text{ is c.c.c.”} \ \Longleftrightarrow \\ \mathbb{Q}(T) \times S \ \text{is c.c.c.} \ \Longleftrightarrow \ S \times \mathbb{Q}(T) \ \text{is c.c.c.} \ \Longleftrightarrow \ \Vdash_S “\mathbb{Q}(T) \text{ is c.c.c.”} \end{gather*} $$

Now since $\mathbb {Q}(T)$ is defined by finite conditions, by absoluteness $\mathbb {Q}(T)^V = \mathbb {Q}(T)^{V^S}$ . Thus, $\Vdash _S “\mathbb {Q}(T) \text { is c.c.c.”}$ iff $\Vdash _S “\mathbb {Q}(T)^{V^S} \text { is c.c.c.,”}$ which by Theorem 3.3 is equivalent to $\Vdash _S “T \text { is Aronszajn.”}$

Proof Using (5), it is easy to check that the pair $(x \cup y,f \cup g)$ satisfies all of the requirements of being a condition except that the domain of the function $f \cup g$ is not necessarily downwards closed in $T \upharpoonright (x \cup y)$ . So we extend $f \cup g$ to a function h whose domain is downwards closed in $T \upharpoonright (x \cup y)$ and then verify that $(x \cup y,h)$ is a condition. Such an extension h is obtained by adding to the domain of $f \cup g$ all elements of T of the form $a \upharpoonright \gamma $ , where $a \in \mathrm {dom}(g) \cap T_{\beta }$ and $\gamma \in x \setminus \alpha $ , and defining $h(a \upharpoonright \gamma ) := g(a) \upharpoonright \gamma $ . Note that by (4), for any new element c of height $\gamma $ there exists a unique element $a \in \mathrm {dom}(g) \cap T_{\beta }$ such that $c = a \upharpoonright \gamma $ .

Obviously h is height preserving and its domain is downwards closed in $T \upharpoonright (x \cup y)$ . If h is not injective, then there are distinct c and d of the same height $\gamma \in x \setminus \alpha $ , at least one of which is new, such that $h(c) = h(d)$ . Suppose that c and d are both new. Then there are distinct a and b in $\mathrm {dom}(g) \cap T_{\beta }$ such that $c = a \upharpoonright \gamma $ and $d = b \upharpoonright \gamma $ . But then $h(c) = h(d) \le _U g(a),g(b)$ , which contradicts that $\Delta _U(g(a),g(b)) < \alpha $ by (4). If just one of them is new, then we may assume $c = a \upharpoonright \gamma $ where $a \in \mathrm {dom}(g) \cap T_{\beta }$ and $d \in \mathrm {dom}(f)$ . Then $f(d) = h(d) = h(c) = g(a) \upharpoonright \gamma <_U g(a)$ , contradicting (5).

It remains to prove that for all $c, d \in \mathrm {dom}(h)$ , $c <_T d$ implies $h(c) <_U h(d)$ . It suffices to verify this in the case where at least one of c or d is new.

Case 1: $\mathrm {ht}_T(c) < \alpha $ and $d = a \upharpoonright \gamma $ is new. Note $c \in \mathrm {dom}(g)$ . If $c <_T d$ then $c <_T a$ , hence $g(c) <_U g(a)$ . Therefore, $h(c) = g(c) <_U g(a) \upharpoonright \gamma = h(a \upharpoonright \gamma ) = h(d)$ .

Case 2: $c \in \mathrm {dom}(f) \setminus (T \upharpoonright \alpha )$ and $d = a \upharpoonright \gamma $ is new. Then c cannot be below d in T, for otherwise $c <_T a$ which contradicts (5).

Case 3: $c = a \upharpoonright \gamma $ is new and $d \in \mathrm {dom}(f) \setminus \alpha $ . If $c <_T d$ , then $a \upharpoonright \alpha = c \upharpoonright \alpha <_T d$ , and hence $a \upharpoonright \alpha = d \upharpoonright \alpha $ . But $\mathrm {dom}(f)$ is downwards closed in $T \upharpoonright x$ and $\alpha \in x$ , so $d \upharpoonright \alpha $ is in $\mathrm {dom}(f)$ . then $d \upharpoonright \alpha = a \upharpoonright \alpha <_T a$ , which contradicts (5).

Case 4: $c = a \upharpoonright \gamma $ and $d = b \upharpoonright \xi $ are both new. Assume $c <_T d$ . Then $c <_T a, b$ . If $a \ne b$ , then $\Delta _T(a,b) < \alpha \le \gamma = \mathrm {ht}_T(c)$ , which contradicts that $c <_T a,b$ . So $a = b$ . Therefore, $h(c) = g(a) \upharpoonright \gamma <_U g(a) \upharpoonright \xi = h(d)$ .

Case 5: $c = a \upharpoonright \gamma $ is new and $d \in \mathrm {dom}(g) \setminus \beta $ . Assume $c <_T d$ . Then $a \le _T d$ , for otherwise by (4) $\Delta _T(a,d) = \Delta _T(a,d \upharpoonright \beta ) < \alpha \le \mathrm {ht}_T(c)$ , which contradicts that $c <_T a,d$ . Hence, $h(c) = g(a) \upharpoonright \gamma <_U g(a) \le _U g(d) = h(d)$ .

Proof By a straightforward pressing down argument, we can find a stationary set $Y_0 \subseteq Y$ , a function f, a set x, and an ordinal $\gamma < \omega _1$ less than $\min (Y_0)$ such that for all $\alpha \in Y_0$ :

1. (1) $x_{\alpha } \cap \alpha = x$ .

2. (2) $f_{\alpha } \upharpoonright (T \upharpoonright \alpha ) = f$ .

3. (3) For all distinct a and b in $\mathrm {dom}(f_{\alpha }) \cap T_{\alpha }$ , both of the ordinals $\Delta _T(a,b)$ and $\Delta _U(f_{\alpha }(a),f_{\alpha }(b))$ are less than $\gamma $ .

4. (4) For all $\beta \in Y_0$ larger than $\alpha $ , $x_{\alpha } \subseteq \beta $ .

Now applying Theorem 2.1 to the disjoint union of the trees T and U, we can find $\alpha < \beta $ in $Y_0$ such that every member of $\mathrm {dom}(f_{\alpha }) \setminus (T \upharpoonright \alpha )$ is incomparable in T with every member of $\mathrm {dom}(f_{\beta }) \setminus (T \upharpoonright \beta )$ , and every member of $\mathrm {ran}(f_{\alpha }) \setminus (U \upharpoonright \alpha )$ is incomparable in U with every member of $\mathrm {ran}(f_{\beta }) \setminus (U \upharpoonright \beta )$ . By Lemma 3.7, the conditions $(x_{\alpha },f_{\alpha })$ and $(x_{\beta },f_{\beta })$ are compatible.

Proof Suppose not. Then for all $a \in S$ we can fix $d_a>_S a$ and a club $C_a \subseteq \omega _1$ such that $C_a \cap Z_{d_a} = \emptyset $ . Now the set $\{ d_a : a \in S \}$ is obviously dense, so its upwards closure is dense open. Since S is Suslin, we can fix some $\gamma < \omega _1$ such that for all $y \in S_{\gamma }$ , there exists some $a \in S$ such that $d_a \le _S y$ .

Let $D := \bigcap \{ C_a : a \in S \upharpoonright \gamma \}$ , which is a club since $S \upharpoonright \gamma $ is countable. As Z is stationary, $D \cap Z$ is stationary. So we can fix some $\alpha \in D \cap Z$ such that $b_{\alpha }$ has height greater than or equal to $\gamma $ . Let $y := b_{\alpha } \upharpoonright \gamma $ . By the choice of $\gamma $ , there exists some $a \in S$ such that $d_a \le _S y$ . Then $a \in S \upharpoonright \gamma $ , so $D \subseteq C_a$ , and therefore $\alpha \in C_a$ . But $d_a \le _S y \le _S b_{\alpha }$ , which means that $\alpha \in Z_{d_a}$ . So $\alpha \in C_a \cap Z_{d_a} = \emptyset $ , which is a contradiction.

Proof We prove the contrapositive. Assume that there is a condition $p \in \mathbb {Q}(T,U)$ such that

$$ \begin{align*} p \Vdash_{\mathbb{Q}(T,U)} “\dot A = \{ \dot a_{\alpha} : \alpha < \omega_1 \} \text{ is an uncountable antichain of } S.” \end{align*} $$

We will find some $a \in S$ which forces in S that either T or U is not Aronszajn.

Write $p = (x,f)$ and let Z be the set of limit ordinals in $\omega _1$ above $\max (x)$ . For each $\alpha \in Z$ , let $p_{\alpha } = (x \cup \{ \alpha \},f)$ , which is clearly a condition below p. Extend each $p_{\alpha }$ to some $q_{\alpha } = (x_{\alpha },f_{\alpha })$ which forces, for some $b_{\alpha } \in S$ , that $\dot a_{\alpha }$ equals $b_{\alpha }$ . This gives us a family of conditions $\{ (x_{\alpha },f_{\alpha }) : \alpha \in Z \}$ satisfying that for all $\alpha \in Z$ , $\alpha \in x_{\alpha }$ .

Consider any $\alpha < \beta $ in Z and suppose that $q_{\alpha }$ and $q_{\beta }$ are compatible in $\mathbb {Q}(T,U)$ . Fix $r \le q_{\alpha }, q_{\beta }$ . Then r forces that $b_{\alpha }$ and $b_{\beta }$ are both in the antichain $\dot A$ and hence are incomparable in S. But then $b_{\alpha }$ and $b_{\beta }$ really are incomparable in S.

Applying Lemma 3.9 to the collection $\{ b_{\alpha } : \alpha \in Z \}$ , we can find some $a \in S$ such that for all $d>_S a$ , the set

$$ \begin{align*} Z_d := \{ \alpha \in Z : d \le_S b_{\alpha} \} \end{align*} $$

is stationary. We claim that

$$ \begin{align*} a \Vdash_S “\{ \alpha \in Z : b_{\alpha} \in \dot G_S \} \text{ is stationary.”} \end{align*} $$

If not, then there exists some $d>_S a$ and an S-name $\dot D$ for a club subset of $\omega _1$ such that

$$ \begin{align*} d \Vdash_S \text{“for all } \alpha \in \dot D \cap Z, b_{\alpha} \notin \dot G_S\text{.”} \end{align*} $$

Since S is c.c.c., we can find a club $E \subseteq \omega _1$ such that d forces that $E \subseteq \dot D$ . Now $Z_d$ is stationary, so we can fix some $\alpha \in Z_d \cap E$ . Then d forces that $\alpha \in \dot D$ and hence that $b_{\alpha } \notin \dot G_S$ . On the other hand, $\alpha \in Z_d$ so $d \le _S b_{\alpha }$ . But then $b_{\alpha }$ extends d in the forcing S, so $b_{\alpha }$ forces that $b_{\alpha } \notin \dot G_S$ , which is impossible. This completes the proof of the claim.

Let G be a generic filter on S such that $a \in G$ . We will prove that in $V[G]$ , either T or U is not Aronszajn. Suppose for a contradiction that both T and U are Aronszajn in $V[G]$ . Note that the definition of $\mathbb {Q}(T,U)$ is absolute between V and $V[G]$ due to the finiteness of the conditions. That is, $\mathbb {Q}(T,U)^V = \mathbb {Q}(T,U)^{V[G]}$ . In $V[G]$ , define $Y := \{ \alpha \in Z : b_{\alpha } \in G \}$ . By the claim, Y is stationary. Moreover, the collection $\{ (x_{\alpha },f_{\alpha }) : \alpha \in Y \}$ is a subset of $\mathbb {Q}(T,U)^{V[G]}$ which satisfies that for all $\alpha \in Y$ , $\alpha \in x_{\alpha }$ .

Since T and U are normal Aronszajn trees in $V[G]$ , we can apply Lemma 3.8 in $V[G]$ to find some $\alpha < \beta $ in Y such that $q_{\alpha } = (x_{\alpha },f_{\alpha })$ and $q_{\beta } = (x_{\beta },f_{\beta })$ are compatible in $\mathbb {Q}(T,U)^{V[G]}$ . By absoluteness, $q_{\alpha }$ and $q_{\beta }$ are compatible in $\mathbb {Q}(T,U)$ in V. As observed above, the compatibility of $q_{\alpha }$ and $q_{\beta }$ in $\mathbb {Q}(T,U)$ implies that $b_{\alpha }$ and $b_{\beta }$ are incomparable in S. But $b_{\alpha }$ and $b_{\beta }$ are both in G, so they are comparable in S. This contradiction completes the proof that either T or U is not an Aronszajn tree in $V[G]$ .

Proof Define by recursion a countable support forcing iteration

$$ \begin{align*}\langle \mathbb{P}_{\alpha}, \dot{\mathbb{Q}}_{\beta} : \alpha \le \omega_2, \ \beta < \omega_2 \rangle \end{align*} $$

of proper forcings which preserve S. After defining $\mathbb {P}_{\alpha }$ , we consider by some bookkeeping a pair of Aronszajn trees $T_{\alpha }$ and $U_{\alpha }$ in $V^{\mathbb {P}_{\alpha }}$ and ask whether or not S forces over $V^{\mathbb {P}_{\alpha }}$ that $T_{\alpha }$ and $U_{\alpha }$ remain Aronszajn. If so, then by Theorem 3.10 forcing with $\mathbb {Q}(T_{\alpha },U_{\alpha })$ over $V^{\mathbb {P}_{\alpha }}$ preserves S being Suslin. In this case, define $\dot {\mathbb {Q}}_{\alpha }$ as a $\mathbb {P}_{\alpha }$ -name for $\mathbb {Q}(T_{\alpha },U_{\alpha })$ , and otherwise let $\dot {\mathbb {Q}}_{\alpha }$ be a $\mathbb {P}_{\alpha }$ -name for the trivial forcing.

Consider on the other hand the case that there exists $x \in S$ and $W \in \{ T_{\alpha }, U_{\alpha } \}$ such that in $V^{\mathbb {P}_{\alpha }}$ , $x \Vdash _S “W \text {has a cofinal branch.”}$ By Proposition 3.1, there exists a club $C \subseteq \omega _1$ and a strictly increasing and height preserving function $f : S_x \upharpoonright C \to W \upharpoonright C$ in $V^{\mathbb {P}_{\alpha }}$ . By upwards absoluteness, f has the same property in $V^{\mathbb {P}_{\omega _2}}$ . Since S, and hence $S_x$ , is Suslin in $V^{\mathbb {P}_{\omega _2}}$ , it follows that in $V^{\mathbb {P}_{\omega _2}}$ the tree W contains a Suslin subtree, namely $f[S_x \upharpoonright C]$ .

By standard proper forcing iteration theorems and our cardinal arithmetic assumptions, for all $\beta < \omega _2$ , $\mathbb {P}_{\beta }$ is $\omega _2$ -c.c. and has cardinality at most $\omega _2$ (see Chapter VIII of [Reference Shelah8]). Thus, by a standard bookkeeping argument we can arrange that all pairs of Aronszajn trees in the final model have been handled at some stage less than $\omega _2$ .

Proof Assume $\textsf {PFA}(S)$ and consider normal Aronszajn trees T and U which have no Suslin subtree. By Proposition 3.1 and the comments which follow it, S forces that T and U are Aronszajn. By Theorem 3.10, $\mathbb {Q}(T,U)$ preserves S being Suslin. Also $\mathbb {Q}(T,U)$ is proper. By choosing a filter meeting an appropriate collection of dense subsets of $\mathbb {Q}(T,U)$ , it is easy to show that T and U are club isomorphic.

Proof Fix a positive integer n. We define by recursion a finite support forcing iteration

$$ \begin{align*} \langle \mathbb{P}_{\alpha}, \dot{\mathbb{Q}}_{\beta} : \alpha \le \omega_1, \ \beta < \omega_1 \rangle \end{align*} $$

of c.c.c. forcings. We will arrange that for each $\beta < \omega _1$ , there exists an $(n+1)$ -tuple $\vec a_{\beta } = (a_{\beta ,0},\ldots ,a_{\beta ,n})$ of distinct elements of S of the same height such that

$$ \begin{align*} \Vdash_{\beta} “S_{\vec a_{\beta}} \text{ is Suslin and } \dot{\mathbb{Q}}_{\beta} = \mathbb{Q}(S_{\vec a_{\beta}})\text{.”} \end{align*} $$

By Theorem 3.3, each $\dot {\mathbb {Q}}_{\beta }$ is forced to be c.c.c. We bookkeep our forcings in such a way that for all $\gamma < \beta < \omega _1$ , the height of the elements of $\vec a_{\gamma }$ are less than or equal to the height of the elements of $\vec a_{\beta }$ , which is possible since the levels of S are countable.

Let us say that an $(n+1)$ -tuple $\vec a$ of distinct elements of S of the same height has been handled by stage $\delta $ if $\Vdash _{\delta } \text {“}S_{\vec a} \text { is special.”}$ A given $(n+1)$ -tuple $\vec a$ can be handled either explicitly by forcing with $\mathbb {Q}(S_{\vec a})$ , or incidentally as a consequence of forcing other trees to be special. Our bookkeeping ensures that all $(n+1)$ -tuples of elements of one level of S are handled before we move on and handle the $(n+1)$ -tuples of the next level. We need to maintain at each step that every derived tree of S of dimension n remains Suslin. In order to make sure we can handle all $(n+1)$ -tuples, we also maintain that any derived tree of dimension $n+1$ which has not been handled by a given stage $\delta $ is still Suslin in $V^{\mathbb {P}_{\delta }}$ . The following inductive hypothesis achieves these goals.

Inductive Hypothesis on $\delta < \omega _1$ : Let $\vec b = (b_0,\ldots ,b_{m-1})$ be a tuple of distinct elements of S satisfying that for all $\gamma < \delta $ :

1. (1) the height of the elements of $\vec b$ are greater than or equal to the height of the elements of $\vec a_{\gamma }$ ;

2. (2) for all $\gamma < \delta $ there exists $i \le n$ such that for all $j < m$ , $a_{\gamma ,i} \not \le _S b_j$ .

Then $\Vdash _{\delta } \text {“}S_{\vec b} \text { is Suslin.”}$

Assume for now that the inductive hypothesis is true for all $\delta < \omega _1$ , and we describe how it can be used to prove the theorem. To begin, we show that we can arrange all derived trees of dimension $n+1$ to be special. So consider $\delta < \omega _1$ and assume that a particular $(n+1)$ -tuple $\vec a = (a_0,\ldots ,a_{n})$ has not been handled by stage $\delta $ . Since we are specializing derived trees one level of S at a time, it follows that for all $\gamma < \delta $ , the height of the elements of $\vec a_{\gamma }$ is less than or equal to the height of the elements of $\vec a$ .

We claim that $\Vdash _{\delta } \text {“}S_{\vec a} \text { is Suslin.”}$ By the inductive hypothesis, it suffices to show that for all $\gamma < \delta $ there exists $i \le n$ such that for all $j \le n$ , $a_{\gamma ,i} \not \le _S a_j$ . Suppose for a contradiction that $\gamma < \delta $ and for all $i \le n$ there is some $j_i \le n$ such that $a_{\gamma ,i} \le _S a_{j_i}$ . Since the elements of $\vec a_{\gamma }$ are distinct and S is a tree, it follows that the map $i \mapsto j_i$ from $n+1$ to $n+1$ is an injection, and hence a bijection. Consequently, $S_{a_{j_0}} \otimes \cdots \otimes S_{a_{j_n}}$ is a subtree of $S_{\vec a_{\gamma }}$ . But $S_{\vec a_{\gamma }}$ is special in $V^{\mathbb {P}_{\gamma +1}}$ , and hence in $V^{\mathbb {P}_{\delta }}$ . So $S_{a_{j_0}} \otimes \cdots \otimes S_{a_{j_n}}$ is also special in $V^{\mathbb {P}_{\delta }}$ , since any subtree of a special tree is special. As $S_{\vec a}$ and $S_{a_{j_0}} \otimes \cdots \otimes S_{a_{j_n}}$ are isomorphic, $S_{\vec a}$ is special in $V^{\mathbb {P}_{\delta }}$ as well, which contradicts our assumption that $\vec a$ has not been handled by stage $\delta $ .

In summary, any derived tree of dimension $n+1$ which has not been handled by stage $\delta < \omega _1$ is still Suslin, and hence Aronszajn, in $V^{\mathbb {P}_{\delta }}$ . Thus, we can easily arrange by bookkeeping that the forcing iteration $\mathbb {P}_{\omega _1}$ eventually handles all derived trees of S of dimension $n+1$ . Therefore, $\mathbb {P}_{\omega _1}$ forces that all derived trees of S of dimension $n+1$ are special. By Lemma 2.7, it follows that $\mathbb {P}_{\omega _1}$ forces that all derived trees of S of dimension greater than n are special.

Next let us see that the inductive hypothesis implies that all derived trees of S of dimension n are Suslin in $V^{\mathbb {P}_{\omega _1}}$ . So let $\vec b = (b_0,\ldots ,b_{n-1})$ be an n-tuple of distinct elements of S of the same height. To show that $S_{\vec b}$ is Suslin in $V^{\mathbb {P}_{\omega _1}}$ , it suffices to show that for all $\delta < \omega _1$ , $S_{\vec b}$ is Suslin in $V^{\mathbb {P}_{\delta }}$ . As $\delta $ is countable, we can fix some $\xi < \omega _1$ greater than the height of the elements of $\vec b$ such that for all $\gamma < \delta $ , $\xi $ is greater than the height of the elements of $\vec a_{\gamma }$ .

By Lemma 2.6, in order to prove that $S_{\vec b}$ is Suslin in $V^{\mathbb {P}_{\delta }}$ , it suffices to show that for all $\vec c$ in $S_{\vec b}$ whose elements have height $\xi $ , $S_{\vec c}$ is Suslin in $V^{\mathbb {P}_{\delta }}$ . So let such $\vec c = (c_0,\ldots ,c_{n-1})$ be given. By the inductive hypothesis, it suffices to show that for all $\gamma < \delta $ there exists some $i \le n$ such that for all $j < n$ , $a_{\gamma ,i} \not \le _S c_j$ . Let $\gamma < \delta $ . Let $\alpha $ be the height of the elements of $\vec a_{\gamma }$ , which is less than $\xi $ by the choice of $\xi $ . The set $\{ c_0 \upharpoonright \alpha , \ldots , c_{n-1} \upharpoonright \alpha \}$ has size at most n, whereas $\{ a_{\gamma ,0}, \ldots , a_{\gamma ,n} \}$ has size $n+1$ . So we can choose $i \le n$ such that $a_{\gamma ,i}$ is not an element of $\{ c_0 \upharpoonright \alpha , \ldots , c_{n-1} \upharpoonright \alpha \}$ . Then for all $j < n$ , $a_{\gamma ,i} \not \le _S c_j$ and we are done.

It remains to prove the inductive hypothesis. Let $\delta < \omega _1$ and assume that the inductive hypothesis holds for all $\beta < \delta $ . Let $\vec b = (b_0,\ldots ,b_{m-1})$ be a tuple of distinct elements of S of the same height satisfying that for all $\gamma < \delta $ : (a) the height of the elements of $\vec b$ are greater than or equal to the height of the elements of $\vec a_{\gamma }$ , and (b) there exists $i \le n$ such that for all $j < m$ , $a_{\gamma ,i} \not \le _S b_j$ . We will prove that $\Vdash _{\delta } \text {“}S_{\vec b} \text { is Suslin.”}$

Note that for all $\delta _0 < \delta $ , $\vec b$ satisfies properties (a) and (b) for all $\gamma < \delta _0$ . By the inductive hypothesis, for all $\delta _0 < \delta $ , $\Vdash _{\delta _0} \text {“}S_{\vec b} \text { is Suslin.”}$ If $\delta $ is a limit ordinal, then by Theorem 2.3 it follows that $\Vdash _{\delta } \text {“}S_{\vec b} \text { is Suslin”}$ and we are done.

Suppose that $\delta = \beta +1$ is a successor ordinal. Then as just observed, $\Vdash _{\beta } \text {“}_{\vec b} \text { is Suslin.”}$ So it suffices to prove that in $V^{\mathbb {P}_{\beta }}$ , $\Vdash _{\mathbb {Q}_{\beta }} \text {“}S_{\vec b} \text { is Suslin.”}$ Recall that $\mathbb {Q}_{\beta }$ is equal to $\mathbb {Q}(S_{\vec a_{\beta }})$ . By our assumptions on $\vec b$ , we can fix $i^* \le n$ such that for all $j < m$ , $a_{\beta ,i^*} \not \le _S b_j$ . Let $\alpha $ be the height of the elements of $\vec b$ .

We work in $V^{\mathbb {P}_{\beta }}$ . In order to show that $\Vdash _{\mathbb {Q}(S_{\vec a_{\beta }})} \text {“}S_{\vec b} \text { is Suslin,”}$ by Theorem 3.4 it suffices to show that $\Vdash _{S_{\vec b}} \text {“}S_{\vec a_{\beta }} \text { is Aronszajn.”}$ By Lemma 2.4, it is enough to show that $\Vdash _{S_{\vec b}} \text {“}S_{a_{\beta ,i^*}} \text { is Aronszajn.”}$ By a simple argument, it suffices to show that whenever $a \in S_{\alpha }$ is above $a_{\beta ,i^*}$ , then $\Vdash _{S_{\vec b}} \text {“}S_a \text { is Aronszajn.”}$ So let such an a be given. By the choice of $i^*$ , a is not equal to any of $b_0,\ldots ,b_{m-1}$ .

There are two possibilities to consider. First, assume that for all $\gamma < \beta $ , there exists some $i \le n$ such that $a_{\gamma ,i}$ is not less than or equal to any of $b_0,\ldots ,b_{m-1},a$ . Applying the inductive hypothesis for $\beta $ we get that $S_{b_0} \otimes \cdots \otimes S_{b_{m-1}} \otimes S_a$ is Suslin in $V^{\mathbb {P}_{\beta }}$ . As discussed in Section 2, it follows that $S_{\vec b}$ forces that $S_a$ is Suslin, and hence Aronszajn, and we are done.

Secondly, assume that there exists some $\gamma < \beta $ such that for all $i \le n$ , $a_{\gamma ,i}$ is less than or equal to one of $b_0,\ldots ,b_{m-1},a$ . By assumption (b) about $\vec b$ , there exists $i \le n$ such that for all $j < m$ , $a_{\gamma ,i} \not \le _S b_j$ , and therefore $a_{\gamma ,i} \le _S a$ . For any $l \le n$ different from i, $a_{\gamma ,i}$ and $a_{\gamma ,l}$ are different, so they cannot both be below a. Hence, we can pick $j_l < m$ such that $a_{\gamma ,l} \le _S b_{j_l}$ . Then the map $l \mapsto j_l$ is injective. Define $d_i := a$ and for $l \le n$ different from i, $d_l := b_{j_l}$ . Then $S_{d_0} \otimes \cdots \otimes S_{d_n}$ is a subtree of $S_{\vec a_{\gamma }}$ . Now in $V^{\mathbb {P}_{\gamma +1}}$ , and hence in $V^{\mathbb {P}_{\beta }}$ , $S_{\vec a_{\gamma }}$ is special. Since $S_{d_0} \otimes \cdots \otimes S_{d_n}$ is a subtree of $S_{\vec a_{\gamma }}$ , it is special as well. By Lemma 2.7, it follows that $S_{b_0} \otimes \cdots \otimes S_{b_{m-1}} \otimes S_a$ is special. By Lemma 2.5, $S_{\vec b}$ forces that $S_a$ is special and hence Aronszajn.