Introduction

Recall that an \(\omega _1\)-tree is said to be Souslin if it has no uncountable chain or antichain. In [2, 3], Fuchs and Hamkins considered various notions of rigidity of Souslin trees and studied the following question: How many generic branches can Souslin trees introduce, when they satisfy certain rigidity requirements? In [2], Fuchs asks a few questions which motivate the following theorem.

Theorem 1.1

It is consistent with \(\mathrm{GCH}\) that there is a Souslin tree S such that \(\Vdash _S ``S\) is Kurepa and \(S \upharpoonright C\) is rigid for every club \(C \subset \omega _1\)”.

Theorem 1.1 answers all questions in [2]. We refer the reader to [2, 3] for motivation and history.

In [1], Baumgartner proves that under \(\diamondsuit ^+\) there is a lexicographically ordered Souslin tree which is minimal as a tree and as an uncountable linear order. At the end of his construction he asks the following question: Does there exist a minimal Aronszajn line if \(\diamondsuit \) holds? This question is not settled here but motivates the following proposition.

Proposition 1.2

It is consistent with \(\diamondsuit \) that if S is a Souslin tree then there is a dense \(X \subset S\) which does not contain a copy of S.

Proposition 1.2 shows it is impossible to follow the same strategy as Baumgartner’s in [1], in order to show \(\diamondsuit \) implies that there is a minimal Aronszajn line. More precisely, it is impossible to find a lexicographically ordered Souslin tree which is minimal as a tree and as an uncountable linear order.

This paper is organized as follows. In the next section we prove Proposition 1.2. In the third section we introduce a Souslin tree which makes itself a Kurepa tree. This tree is used in the last section, where we prove Theorem 1.1.

Let’s fix some definitions, notations and conventions. Assume TS are trees and \(f : T \longrightarrow S\) is injective. Then f is said to be an embedding when \(t<_T s \Longleftrightarrow f(t) <_S f(s)\). T is called an \(\omega _1\)-tree if its levels are countable and \(\mathrm{ht}(T)= \omega _1\). T is said to be pruned if for all \(t \in T\) and \(\alpha \in \omega _1 \setminus \mathrm{ht}(t)\) there is \(s \ge t\) such that \(\mathrm{ht}(s) = \alpha \). If \(t \in T\) and \(\alpha \le \mathrm{ht}(t)\), \(t \upharpoonright \alpha \) refers to the \(\le _T\) predecessor of t in level \(\alpha \). \(C \subset T\) is called a chain if it consists of pairwise comparable elements. A chain \(b \subset T\) is called a branch if it intersects all levels of T. An \(\omega _1\)-tree U is called minimal if for every uncountable \(X \subset U\), U embeds into X. If T is a tree and \(\alpha \) is an ordinal, \(T(\alpha )= \{ t \in T: \mathrm{ht}(t) = \alpha \}\) and \(T(< \alpha ) = \{ t \in T: \mathrm{ht}(t) < \alpha \}\). If A is a set of ordinals, \(T \upharpoonright A = \{t \in T: \mathrm{ht}(t) \in A \}\). If \(t \in T\) and \(U \subset T\) then \(U_t = \{ u \in U: t \le _T u \}\). Assume Q is a poset and \(\theta \) is a regular cardinal. We say \(M \prec H_\theta \) is suitable for Q if Q and the power set of the transitive closure of Q are in M.

Minimality of Souslin trees and \(\diamondsuit \)

This section is devoted to the proof of Proposition 1.2. We will use the following terminology and notation in this section. By N we mean the set of all countable infinite successor ordinals, and \(\mathbb {P}\) refers to the countable support iteration \(\langle {P_i, \dot{Q}_j : i \le \omega _2, j < \omega _2} \rangle \), where \(Q_j=2^{<\omega _1}\) for each \(j \in \omega _2\).

Lemma 2.1

Assume \(U= (\omega _1, <)\) is a Souslin tree, \(p \in \mathbb {P}\), \(\dot{X}\) is the canonical \({\mathbb {P}}_1\)-name for the generic subset of \(\omega _1\), \(p \Vdash ``\dot{f}\) is an embedding from U to \(\dot{X}\)” and for every \(t \in U\) define \(\varphi (p,t) = \{ s \in U : \exists \bar{p} \le p \ \bar{p} \Vdash \dot{f}(t) = s \}.\) Then there is an \(\alpha \in \omega _1\) such that for all \(t \in U \setminus U(< \alpha ) \), \(\varphi (p,t)\) is not a chain.

Proof

Let \(Y_p = \{ y \in U : \varphi (p,y) \text { is a chain} \}.\) \(Y_p\) is downward closed and if it is countable we are done. Fix \(p \in \mathbb {P}\) and assume for a contradiction that \(Y_p\) is uncountable. Let \(A_p = \{ t \in U : p \Vdash t \in \dot{X} \text { or } p\Vdash t \notin \dot{X} \}.\) \(A_p\) is countable. Fix \(\alpha > \sup \{ \mathrm{ht}(a): a \in A_p \}\) and \(y \in Y_p \setminus U(\le \alpha )\). Since U is an Aronszajn tree and \(\varphi (p,y) \) is a chain, we can choose \(\beta \in \omega _1 \setminus \sup \{ \mathrm{ht}(s): s \in \varphi (p,y) \}\). For all \(s \in \varphi (p,y)\), \(\alpha< \mathrm{ht}(s) < \beta \) since \(\emptyset \Vdash \mathrm{ht}(y) \le \mathrm{ht}(\dot{f}(y))\). Then we can extend p to q such that \(q \Vdash \dot{X} \cap (U(\le \beta ) \setminus U(< \alpha )) = \emptyset \), which contradicts \(p \Vdash \dot{f}(y) \in \varphi (p,y)\). \(\square \)

Lemma 2.2

Assume \(U \in \textsc {V}\) is a pruned Souslin tree and \(G \subset \mathbb {P}\) is \(\textsc {V}\)-generic. Then in \(\textsc {V}{[G]}\), there is a dense \(X \subset U\) which does not have a copy of U.

Proof

Let \(\dot{X}\) be as in Lemma 2.1. Since U is pruned, \(1_\mathbb {P}\Vdash \dot{X} \subset U\) is dense. We will show \(1_\mathbb {P}\Vdash \dot{X}\) has no copy of U. Assume for a contradiction that \(p \Vdash _\mathbb {P}\dot{f}\) is an embedding from U to \(\dot{X}\). Fix a regular cardinal \(\theta \) and a countable \(M \prec H_\theta \) which contains \(U, p, \dot{f}, 2^\mathbb {P}\). Also let \(\langle {D_n : n \in \omega } \rangle \) be an enumeration of all dense open subsets of \(\mathbb {P}\) in M, \(\delta = M \cap \omega _1\) and \(t \in U(\delta )\). For each \(\sigma \in 2^{< \omega }\), find \(p_\sigma \in D_{|\sigma |} \cap M\), \(s_\sigma \) and \(t_{|\sigma |}< t\), such that:

  1. (1)

    if \(\sigma \sqsubset \tau \) then \(p_\tau \le p_\sigma \) and \(s_\sigma \le s_\tau \),

  2. (2)

    if \(\sigma \perp \tau \) then \(s_\sigma \perp s_\tau \),

  3. (3)

    \(p_\sigma \Vdash \dot{f}(t_{|\sigma |}) = s_\sigma \).

In order to see how these sequences are constructed, let \(t_0 < t\) be arbitrary and \(p_{\emptyset }, s_\emptyset \) be such that \(p_\emptyset \Vdash \text {``} \dot{f}(t_0)= s_\emptyset \text {''}\) and \(p_\emptyset \in D_0 \cap M\). Assuming these sequences are given for all \(\sigma \in 2^n\), use Lemma 2.1 to find \( t_{n+1} < t\) such that \(\varphi (p_\sigma , t_{n+1})\) is not a chain, for all \(\sigma \in 2^n\). Let \(s_{\sigma \frown 0}, s_{\sigma \frown 1}\) be in \(\varphi (p_\sigma , t_{n+1}) \cap M\) such that \(s_{\sigma \frown 0} \perp s_{\sigma \frown 1}\). Now find \(p_{\sigma \frown 0} , p_{\sigma \frown 1}\) in \(M \cap D_{n+1}\) which are extensions of \(p_\sigma \) such that \(p_{\sigma \frown i} \Vdash \text {``}\dot{f}(t_{n + 1}) = s_{\sigma \frown i}\text {''}, \text { for } i=0,1.\)

For each \(r \in 2^\omega \), let \(p_r\) be a lower bound for \(\{p_\sigma : \sigma \sqsubset r \}\) and let \(b_r \subset U \cap M\) be a downward closed chain such that \(p_r \Vdash \dot{f}[\{ s \in U : s < t \}] \subset b_r\). Note that \(b_r\) intersects all the levels of U below \(\delta \). It is obvious that \(p_r\) is an \((M,\mathbb {P})\)-generic condition below p. Moreover, if \(r,r'\) are two distinct real numbers then \(b_r \ne b_{r'}\). Let \(r \in 2^\omega \) such that U has no element on top of \(b_r\). Then \(p_r\) forces that \(\dot{f}(t)\) is not defined, which is a contradiction. \(\square \)

Now we are ready for the proof of Proposition 1.2. Let \(\textsc {V}\) be a model of \(\mathrm{ZFC}+ \mathrm{GCH}\) and \(G \subset \mathbb {P}\) be \(\textsc {V}\)-generic. Since \(\mathbb {P}\) is a countable support iteration of \(\sigma \)-closed posets of size \(\aleph _1\), it preserves all cardinals. The same argument as in Theorem 8.3 in [4] shows that \(\diamondsuit \) holds in \(\textsc {V}[G]\).

Let U be a Souslin tree in \(\textsc {V}[G]\). For some \(\alpha \in \omega _2\), \(U \in \textsc {V}[G \cap P_\alpha ]\) since \(|U| = \aleph _1\). Let \(\dot{R}\) be the canonical \(P_\alpha \)-name such that \(\mathbb {P}= P_\alpha * \dot{R}\). Then \(1_{P_\alpha } \Vdash \dot{R}\) is isomorphic to \(\mathbb {P}\). By Lemma 2.2, there is a dense \(X \subset U\) in \(\textsc {V}\)[G] which has no copy of U, as desired.

A Souslin tree with many generic branches

Definition 3.1

The poset Q is the set of all \(p=(T^p, \Pi _p)\) such that:

  1. (1)

    \(\Delta _p \in \omega _1\) and \(T^p = (\Delta _p, \le _p)\) is a countable binary tree of height \(\alpha _p\) such that for all \(t \in T^p\) and for all \(\beta \in \alpha _p \setminus \mathrm{ht}_{T^p}(t)\) there is \(s \in T^p(\beta )\) with \(t<_{T^p} s\).

  2. (2)

    \(\Pi _p = \langle {\pi _\xi ^p: \xi \in D_p} \rangle \) where \(D_p \subset \omega _2\) is countable and for each \(\xi \in D_p\) there are xy of the same height in \(T^p\) such that \(\pi _\xi ^p : (T^p)_x \longrightarrow (T^p)_y\) is a tree isomorphism.

We let \(q \le p \) if \(T^q\) end-extends \(T^p\), \(D_p \subset D_q\) and for all \(\xi \in D_p\), \(\pi _\xi ^q \upharpoonright T^p =\pi _\xi ^p \).

Lemma 3.2

Q is \(\sigma \)-closed. Moreover if \(\mathrm{CH}\) holds, Q has the \(\aleph _2\)-cc.

Proof

The first part of the lemma is obvious. Assume \(A \in Q^{\aleph _2}\). By thinning A out, we can assume that for all pq in A, \(T^p =T^q\), \(\{ D_p : p \in A \}\) is a \(\Delta \)-system with root R and \(|\{ \langle {\pi _\xi ^p : \xi \in R} \rangle : p \in A \}|=1\). Now all pq in A are compatible. \(\square \)

Lemma 3.3

If \(T = \bigcup _{p \in G}T^p\) for a generic \(G \subset Q\), then T is Souslin.

Proof

Obviously T is an \(\omega _1\)-tree. Let \(\tau \) be a Q-name and \(p \Vdash _Q ``\tau \subset T\) is a maximal antichain”. We show \(p \Vdash \tau \) is countable. Let \(M \prec H_\theta \) be countable, \(\theta \) regular and \(2^Q, \tau \) be in M. Let \(\langle {p_n = (T_n , \Pi _n) :n \in \omega } \rangle \), be a descending (MQ)-generic sequence with \(p_0 = p\). Let \(\pi _\xi ^{p_n}= \pi _\xi ^n\), \(\delta = M \cap \omega _1\), and \(R = \bigcup _{n \in \omega } T_n\). So \(\mathrm{ht}(R) = \delta \) and \(M \cap \omega _2 = \bigcup _{n \in \omega }D_{p_n}\). Let \({\mathcal {F}}\) be the set of all finite compositions of functions of the form \(\bigcup _{n \in \omega } \pi _\xi ^n\) with \(\xi \in M \cap \omega _2 \). Let \(\langle {f_n : n \in \omega } \rangle \) be an enumeration of \({\mathcal {F}}\) with infinite repetition and \(A= \{ t \in R: \exists n \in \omega \) \((p_n \Vdash t \in \tau ) \}\). Observe that for all \(t \in R\) there is \(a \in A\) such that at are comparable.

Let \(\langle {\alpha _m: m \in \omega } \rangle \) be an increasing cofinal sequence in \(\delta \). For each \(t \in R\) we build an increasing sequence \(\bar{t} = \langle {t_m : m \in \omega } \rangle \) as follows. Let \(t_0 =t\). Assume \(t_m\) is given. If \(R_{t_m} \cap \mathrm{dom}(f_m) = \emptyset \), choose \(t_{m+1} > t_m\) with \(\mathrm{ht}(t_{m +1}) > \alpha _m\). If \(R_{t_m} \cap \mathrm{dom}(f_m) \ne \emptyset \), let \(s \in \mathrm{dom}(f_m) \cap R_{t_m}\). Let \(a \in A\) such that \(a, f_m(s)\) are comparable. Let \(x = \max \{f_m(s),a \}\) and \(t_{m+1}> f^{-1}_m(x)\) with \(\mathrm{ht}(t_{m+1})> \alpha _m\). Let \(b_t\) be the downward closure of \(\bar{t}\).

Let \(B = \{ f_n[b_t] : t \in R \text { and } n \in \omega \}\). Let q be the lower bound for \(\langle {p_n: n \in \omega } \rangle \) described as follows. \(T^q = R \cup T^q(\delta )\) and for each cofinal branch \(c \subset R\) there is a unique \(y \in T^q(\delta )\) above c if and only if \( c \in B \). For each \(\xi \in M \cap \omega _2\), let \(\pi _\xi ^q \upharpoonright R = \bigcup _{n \in \omega } \pi _\xi ^n\). Note that this determines \(\pi _\xi ^q\) on \(T^q(\delta )\) as well and \(\pi _\xi ^q (y)\) is defined for all \(y \in T^q(\delta )\).

The condition q forces that for each \(y \in T(\delta ) = T^q(\delta )\) there is \(a \in A\) with \(a < y\). In other words q forces that \(\tau = A\). Since p was arbitrary, \(1_Q\) forces that every maximal antichain has to be countable. \(\square \)

From now on T is the same tree as in Lemma 3.3. For each \(\xi \in \omega _2\) let \(\pi _\xi = \bigcup _{p \in G} \pi ^p_\xi \), where \(G \subset Q\) is generic. Observe that if \(x \in \mathrm{dom}(\pi _\xi ) \cap \mathrm{dom}(\pi _\eta )\) and \(\xi \ne \eta \) are ordinals then there is \(\alpha > \mathrm{ht}(x)\) such that for all \(y \in T(\alpha ) \cap T_x\), \(\pi _\xi (y) \ne \pi _\eta (y)\). So forcing with T makes T Kurepa.

Highly rigid dense subsets of T

In this section we show the tree T, in the forcing extensions by \(P = (2^{<\omega _1}, \supset )\), has dense subsets which are witnesses for Theorem 1.1.

Lemma 4.1

Let \(U=(\omega _1, <)\) be a pruned Souslin tree and \(S \subset \omega _1\) be generic for P. Then in \(\textsc {V}[S]\) the following hold.

  1. (1)

    S is a Souslin tree when it is considered with the inherited order from U.

  2. (2)

    \(S \subset U\) is dense.

  3. (3)

    For all clubs \(C \subset \omega _1\), \(S \upharpoonright C\) is rigid.

Proof

In order to see that S is Souslin, note that \(\sigma \)-closed posets do not add uncountable antichains to Souslin trees. Moreover by standard density arguments \(S \subset U\) is dense.

Assume for a contradiction \(p \Vdash _P ``\dot{f}: \dot{S} \upharpoonright \dot{C} \longrightarrow \dot{S} \upharpoonright \dot{C}\) is a nontrivial tree embedding.” Let \(\langle {M_\xi : \xi \in \omega +1} \rangle \) be a continuous \(\in \)-chain of countable elementary submodels of \(H_\theta \) where \(\theta \) is regular and \(p, \dot{f}, 2^U \) are in \(M_0\). For each \(\xi \le \omega \), let \(\delta _\xi = M_\xi \cap \omega _1\) and \(t \in U({\delta _\omega })\). Let \(t_n = t \upharpoonright \delta _n\). For each \(\sigma \in 2^{< \omega }\) we find \(q_\sigma \in M_{|\sigma |+1} \cap P\), \(s_\sigma \) such that:

  1. (1)

    \(q_0 \le p\), and if \(\sigma \subset \tau \) then \(q_\tau \le q_\sigma \),

  2. (2)

    \(q_\sigma \) is \((M_{|\sigma |}, P)\)-generic and \(q_\sigma \subset M_{|\sigma |}\),

  3. (3)

    \(q_\sigma \) forces that \(\dot{f}(t_{|\sigma |-1}) = s_\sigma \),

  4. (4)

    if \(\sigma \perp \tau \) then \(s_\sigma \perp s_\tau \),

  5. (5)

    if \(\sigma \subset \tau \) then \(q_\tau \) forces that \(t_{|\sigma |} \in \dot{S} \upharpoonright \dot{C}\).

Assuming \(q_\sigma \) and \(s_\sigma \) are given for all \(\sigma \in 2^{n}\), we find \(q_{\sigma \frown 0}, q_{\sigma \frown 1},s_{\sigma \frown 0},\) and \( s_{\sigma \frown 1}\). Let \(\bar{q}_\sigma = q_\sigma \cup \{( t_n,1) \}\). Obviously, \(\bar{q}_\sigma \Vdash t_n \in \dot{S} \upharpoonright \dot{C}\) and for all \(\sigma \in 2^n\), \(\{ s \in U : \exists r \le \bar{q}_\sigma \) \(r \Vdash \dot{f}(t_n)=s\}\) is uncountable. In \(M_{n+1}\), find \(r_0,r_1\) below \(\bar{q}_\sigma \) and \(s_{\sigma \frown 0}, s_{\sigma \frown 1}\) such that \(s_{\sigma \frown 0} \perp s_{\sigma \frown 1}\) and \(r_i \Vdash ``\dot{f}(t_n) = s_{\sigma \frown i}\).” Let \(q_{\sigma \frown i} < r_i\) be \((M_{n+1}, P)\)-generic with \(q_{\sigma \frown i} \subset M_{n +1}\), and \(q_{\sigma \frown i} \in M_{n +2}\).

Let \(r \in 2^\omega \) such that \(\{ s_\sigma : \sigma \subset r \}\) does not have an upper bound in U. Let \(p_r\) be a lower bound for \(\{p_\sigma : \sigma \subset r \}\). Then \(p_r\) forces that \(\dot{f}(t)\) is not defined which is a contradiction. \(\square \)

Lemma 4.2

Suppose M is suitable for Q and \(\delta = M \cap \omega _1\). Let \(\langle {q_n : n \in \omega } \rangle \) be a decreasing (MQ)-generic sequence. Define a condition \(q \in Q\) by setting \(T^q = \bigcup _{n \in \omega } T^{q_n}\), \(D_q = \bigcup _{n \in \omega }D_{q_n}\) and for each \(\xi \in D_q\) let \(\pi _\xi ^q = \bigcup _{n \in \omega }\pi _\xi ^{q_n}\). Also let \(\Pi _q = \langle {\pi _\xi ^q: \xi \in D_q} \rangle \). Let \({\mathcal {F}}\) be the set of all finite compositions of functions of the form \(\pi ^q_\xi \) with \(\xi \in D_q\). Assume \(m \in \omega \) and \(\langle {b_i : i \in m} \rangle \) are branches through \(T^q\). Then there is an extension \(q' \le q\) such that \(\alpha _{q'} \ge \delta +1\) and for all branches \(c \subset T^q\), c has an upper bound iff for some \(f \in {\mathcal {F}}\text { and } i \in m,\) \( f(b_i)\) is cofinal in c.

Proof

Note that \(D_q= M \cap \omega _2\) and \(\alpha _q = \delta \). Let \(T^{q'} \upharpoonright \delta = T^q\). Let \(B= \{ f(b_i) : i \in m \text { and } f \in {\mathcal {F}} \}\). Obviously B is countable and we can fix an enumeration of B with \(n \in \omega \). Let \(T^{q'}(\delta +1) = [ \delta , \delta + \omega )\) and put \(\delta + n\) on top of the n’th element in B. It is obvious how we should extend \(\Pi _q\) to \(\Pi _{q'}\) with \(D_q = D_{q'}\). \(\square \)

Lemma 4.3

Let \(G \subset Q\) be \(\textsc {V}\)-generic, \(p \in P\) and \(\dot{S}\) be the canonical P-name for the generic subset of \(\omega _1\). Let \(\dot{f}, \dot{C}\) be \(P * T\)-names in \(\textsc {V}[G]\) and txy be pairwise incompatible in T. Suppose (pt) forces \(\dot{f}\) is an embedding from \(\dot{S}_x \upharpoonright \dot{C}\) to \(\dot{S}_y \upharpoonright \dot{C}\). For every \(u \in T_x\) define \(\psi (p,t,u) = \{s \in T : \exists t' > t \ \exists \bar{p} \le p \ (\bar{p}, t') \Vdash [u \in \dot{S}_x \upharpoonright \dot{C} \wedge \dot{f}(u) = s] \}.\) Then for any \(u \in T_x\) there is \(u' > u\) such that \(\psi (p,t, u')\) is not a chain.

Proof

Fix ptu as above and assume for a contradiction that for all \(u' > u\) in T, \(\psi (p,t,u')\) is a chain. Since T is ccc, without loss of generality we can assume that for all \(q \in P\) and \( \alpha \in \omega _1\), there is \(\bar{q} \le q\) such that \((\bar{q},1_T)\) decides the statement \(\alpha \in \dot{C}\). For each \(q \in P,r \in T, v \in T\) let \(\alpha _{q,r,v} = \sup \{ \mathrm{ht}_T(s) : s \in \psi (q,r, v) \}\). Note that if \(\bar{q} \le q\) and \(\bar{r} \ge r\) then \(\psi (\bar{q},\bar{r},v) \subseteq \psi (q,r,v)\) and \(\alpha _{\bar{q},\bar{r},v} \le \alpha _{q,r,v}\).

Let \(M_0, M_1\) be countable elementary submodels of \(H_\theta \), \(\theta \) be a regular cardinal and \(\{p,t,u,x,y, \dot{f}, \dot{C}\} \in M_0 \in M_1\). Suppose \(\langle {p_n: n \in \omega } \rangle \) is an \((M_0,P)\)-generic sequence which is in \(M_1\) and \(p_0 \le p\). Let \(p' = \bigcup _{n \in \omega } p_n\) and \(\delta _i = M_i \cap \omega _1\), for \(i \in 2\). Note that \(p' \Vdash \delta _0 \in \dot{C}\).

Let \(\bar{p} < p'\) such that:

  1. (1)

    \(\bar{p} \Vdash \forall v \in T_x \cap (M_1 \setminus M_0) \ [v \in \dot{S}]\)

  2. (2)

    \(\bar{p} \Vdash \forall v \in T_y \cap (M_1 \setminus M_0) \ [v \notin \dot{S}]\).

Let \(u_0 > u\) be in \(T(\delta _0)\). Since \(\bar{p}\) is \((M_0,Q)\)-generic, it forces that \(\delta _0 \in \dot{C} \wedge u_0 \in \dot{S} \wedge \mathrm{ht}_{\dot{S}}(u_0) = \delta _0\). In particular, by elementarity of \(M_0\) and basic facts on ordinal arithmetic, \(\bar{p} \Vdash u_0 \in \dot{S}_x \upharpoonright \dot{C}\).

Suppose \(q< \bar{p}, r > t\) such that (qr) decides \(\dot{f}(u_0)\). Then the condition (qr) forces that \(\mathrm{ht}(\dot{f}(u_0)) \ge \delta _1\). So, \(\delta _1 \le \alpha _{\bar{p},t,u_0 } \le \alpha _{p', t, u_0} \in M_1.\) But this is a contradiction. \(\square \)

In the next lemma we use the following standard fact: If U is a Souslin tree and \(X \subset U\) is uncountable and downward closed, then there is \(x \in U\) such that \(U_x \subset X\). In order to see this assume for all \(v \in U,\) \(U_v\) is not contained in X. Let A be the set of all minimal a outside of X. Observe that A is an uncountable antichain, contradicting the fact that U was Souslin. Lemma 4.4 finishes the proof of Theorem 1.1.

Lemma 4.4

Assume \(G*S*b\) is \(\textsc {V}\)-generic for \(Q*P*\dot{T}\). Let xy be incomparable in T. Then in \(\textsc {V}[G*S*b]\) for all clubs \(C \subset \omega _1\), \(S_x \upharpoonright C\) does not embed into \(S_y \upharpoonright C\).

Proof

Assume for a contradiction that \((q_0,p_0,t_0)\) is a condition in \(Q*P *\dot{T}\) which forces \(\dot{f}: \dot{S}_x \upharpoonright \dot{C} \longrightarrow \dot{S}_y \upharpoonright \dot{C}\) is a tree embedding and xy are incompatible in T. Note that \(\dot{f}(\dot{S}_x)\) is an uncountable subset of \(\dot{T}_y\) and \(\dot{T}\) is a Souslin tree in \(\textsc {V}[G][S]\). So the downward closure of \(\dot{f}(\dot{S}_x)\) contains \(\dot{T}_z\) for some \(z > y\). Therefore, by extending \(x,y,(q_0,p_0,t_0)\) if necessary, we can assume that \(\dot{f}(\dot{S}_x)\) is dense in \(\dot{S}_y\).

Again by extending \(x,y,(q_0,p_0,t_0)\) we may assume \((q_0, p_0,t_0)\) \(\Vdash \) [xy are in \(\dot{S} \upharpoonright \dot{C}\) and \(\dot{f}(x) = y]\). Furthermore, by extending \(t_0\) if necessary we can assume that \(\mathrm{ht}(t_0) > \mathrm{ht}(y)\) and \(x,y,t_0\) are pairwise incomparable. Since T is a ccc poset we can assume that for all \(\alpha \in \omega _1\), for all uv in T and for all \((a,b) \in P*Q\) we have \((a,b,u) \Vdash \alpha \in \dot{C}\) \(\longleftrightarrow \) \((a,b,v) \Vdash \alpha \in \dot{C}\).

Let M be a countable elementary submodel of \(H_\theta \) such that \(\theta \) is regular and \((q_0,p_0,t_0), \dot{f}\) are in M. Let \(\langle {q_n : n \in \omega } \rangle \) be a decreasing (MQ)-generic sequence. Define \(q \in Q\) as in Lemma 4.2. Let \({\mathcal {F}}\) be the set of all finite compositions of functions of the form \(\pi ^q_\xi \) with \(\xi \in M \cap \omega _2\). Let \(\Pi _q=\langle {\pi _\xi ^q: \xi \in M \cap \omega _2} \rangle \). Obviously, q is an (MQ)-generic condition. Let \(\langle {g_n : n \in \omega } \rangle \) be an enumeration of \({\mathcal {F}}\) with infinite repetition. Let \(\langle {\gamma _n : n \in \omega } \rangle \) be an increasing cofinal sequence in \(\delta = M \cap \omega _1\) with \(\gamma _0 = 0\).

We find a decreasing sequence \(\langle {p_n \in P \cap M: n \in \omega } \rangle \) and increasing sequences \(\langle {\delta _n \in \delta : n \in \omega } \rangle \), \(\langle {t_n \in T^q: n \in \omega } \rangle \), \(\langle {u_n \in T^q: n \in \omega } \rangle \) \(\langle {s_n \in T^q : n \in \omega } \rangle \) such that:

  1. (1)

    \(\delta _n \ge \gamma _n\) for all \(n \in \omega ,\)

  2. (2)

    \((q,p_n.t_n) \Vdash \min \{ \mathrm{ht}_{\dot{S}}(s_n), \mathrm{ht}_{\dot{S}}(u_n), \mathrm{dom}(p_n) \} \ge \delta _n\),

  3. (3)

    \(\mathrm{ht}_{T^q}(t_n) \ge \mathrm{ht}_{T^q}(s_n) +1\),

  4. (4)

    \((q,p_n,1_{T^q}) \Vdash \delta _n \in \dot{C}\),

  5. (5)

    \((q, p_n,t_n)\) \(\Vdash \) \(\dot{f}(u_n) = s_n\),

  6. (6)

    if \(n \in \omega \setminus 1\) and \(t_{n -1} \in \mathrm{dom}(g_n)\) then \(g_{n} (t_n) \perp s_{n}\),

  7. (7)

    if \(n \in \omega \setminus 1 \) and \(u_{n-1} \in \mathrm{dom}(g_n)\) then \(g_n(u_n) \perp s_n\).

We let \(u_0 =x, s_0=y, \delta _0 \in \omega _1\) such that \((q,p_0, t_0)\) forces that \(\min \{\mathrm{ht}_{\dot{S}}(x), \mathrm{ht}_{\dot{S}}(y), \alpha _{p_n} \} = \delta _0\). It is easy to see that this choice together with \(p_0,t_0\) will satisfy the corresponding conditions. For given \(p_n, t_n, s_n, u_n, \delta _n\) we introduce \(p_{n+1}, t_{n+1}, s_{n+1}, u_{n + 1}, \delta _{n+1}\).

If \(t_n \notin \mathrm{dom}(g_{n + 1}) \) let \(v = s_n\). If \(t_n \in \mathrm{dom}(g_{n + 1}) \), let \(v\ge s_{n}\) such that \(v \perp g_{n+1}(t_n)\). Such a v exists because \(\mathrm{ht}(t_n) > \mathrm{ht}(s_n)\), \(g_{n +1}\) is level preserving and the tree \(T^q\) is binary.

Claim 4.5

There are \(t_n' > t_n\), \(p_n' < p_n ,\) \(u_n' > u_n\) such that if \(u_n \in \mathrm{dom}(g_{n+1})\) then \((q,p_n', t_n')\) forces \([u_n' \in \mathrm{dom}(\dot{f})\) \(\wedge \) \(v < \dot{f}(u_n')\) \(\wedge \) \(\dot{f}(u_n') \perp g_{n+1}(u_n')]\).

Proof of Claim

Assume \(u_n \in \mathrm{dom}(g_{n+1})\). Recall that \(\dot{f}(\dot{S}_x)\) is forced to be dense in \(\dot{S}_y\). Let \(\bar{p}_n \le p_n , \bar{t}_n \ge t_n , a_0> u_n, v' > v\) such that \((q, \bar{p}_n, \bar{t}_n) \Vdash \dot{f}(a_0) = v'\). This is possible because q is (MQ)-generic. Let \(a > a_0\), \(t_n^0, t_n^1\) be extensions of \(\bar{t}_n\), and \(p_n^0,p_n^1\) be extensions of \(\bar{p}_n\) such that \((q,p_n^i, t_n^i) \Vdash \dot{f}(a) = s_n^i\) where \(i \in 2\) and \(s_n^0 \perp s_n^1\). Again, this is possible because of Lemma 4.3 and the fact that q is (MQ)-generic. Let \(a' > a\) such that \(\mathrm{ht}(a') > \max \{ \mathrm{ht}(s_n^0), \mathrm{ht}(s_n^1) \}\). Fix \(i \in 2\) such that \(g_{n+1}(a') \perp s_n^i\). Then for all \(e > a'\), \((q,p_n^i, t_n^i)\) forces that if \(e \in \mathrm{dom}(\dot{f})\) then \(\dot{f}(e) > s_n^i\). Moreover it forces that \(g_{n+1}(e) \perp s_\sigma ^{i}\). Therefore, \((q,p_n^i, t_n^i)\Vdash [\forall e>a' \ e \in \mathrm{dom}(\dot{f}) \longrightarrow g_{n+1}(e) \perp \dot{f}(e)]\). Let \(u_n' > a'\), \(p_n' <p_n^i\) and \(t_n' > t_n^i\) such that \((q,p_n',t_n') \Vdash [u_n' \in \mathrm{dom}(\dot{f})]\). Then this condition will also force \(\dot{f}(u_n') \perp g_{n+1}(u_n')\) and \( v < \dot{f}(u_n')\). \(\square \)

Fix \(p_n', t_n', u_n'\) as in the claim above. By extending \(p_n'\) if necessary, we can assume that \((q,p_n',1_{T^q})\) decides the \(\gamma _{n+1}\)’st element of \(\dot{C} \setminus \delta _n\) and we let \(\delta _{n+1}\) be this ordinal. Let \(u_{n+1} > u_n'\) such that for some \(p_{n+1}< p_n'\) with \(\mathrm{dom}(p_{n+1}) \ge \delta _{n+1}\), the condition \((q,p_{n+1},1_{T^q})\) forces that \(u_{n+1} \in \dot{S} \upharpoonright \dot{C}\) and \( \mathrm{ht}_{\dot{S}}(u_{n+1}) \ge \delta _{n+1}\). Let \(r > t_n'\). By extending \((q,p_{n+1},r)\) if necessary, we can assume this condition decides \(\dot{f}(u_{n+1})\). Let \(s_{n+1} \in T^q\) such that \((q,p_{n+1}, r) \Vdash \dot{f}(u_{n+1}) = s_{n+1}\). Let \(t_{n+1} \ge r\) such that \(\mathrm{ht}(t_{n+1}) > \mathrm{ht}(s_{n+1})\). We leave it to the reader to verify that all of the conditions above hold.

Let \(b_0,b_1\) be the downward closure of \(\{u_n: n \in \omega \}\) and \(\{t_{n} : n \in \omega \}\) respectively. By Lemma 4.2 there is \(q' < q\) such that \(\alpha _{q'} \ge \delta +1\) and for all branches \(c \subset T^{q}\), c has an upper bound in \(T^{q'}\) if and only if \(g_n(b_i)\) is cofinal in c for some \(n \in \omega \) and \( i \in 2\). Fix such a \(q'\) for the rest of the argument.

We claim that \(\langle {s_n : n \in \omega } \rangle \) does not have an upper bound in \(T^{q'}\). Suppose for a contradiction that it has an upper bound. Then for some \(m \in \omega \), either

  1. (1)

    \(\{ g_m (t_n) :n \in \omega \wedge t_n \in \mathrm{dom}(g_m)\}\) is cofinal in the downward closure of \(\{s_{n} : n \in \omega \}\) or

  2. (2)

    \(\{ g_m (u_n) : n \in \omega \wedge u_n \in \mathrm{dom}(g_m)\}\) is cofinal in the downward closure of \(\{s_{n} : n \in \omega \}\).

Due to similarity of the arguments, let’s assume that the first alternative happens. Since we enumerated the elements of \({\mathcal {F}}\) with infinite repetition, by increasing m if necessary, we can assume that \(t_{m} \in \mathrm{dom}(g_m)\). But then \(g_m(t_{m}) \perp s_{m}\), meaning that the first alternative cannot happen, which is a contradiction. Hence \(\{s_{n} : n \in \omega \}\) does not have an upper bound in \(T^{q'}\).

Let t be the upper bound of \(\langle {t_{n} : n \in \omega } \rangle \) in \(T^{q'}\), and u be the upper bound for \(\langle {u_n : n \in \omega } \rangle \) which has the lowest height \(\delta \). Let p be a lower bound for \(\langle {p_{n} : n \in \omega } \rangle \) which forces that \(u \in \dot{S}\). It is easy to see that \((q',p,t) \Vdash [\delta \in \dot{C} \wedge u \in \dot{S} \wedge \mathrm{ht}_{\dot{S}}(u) = \delta ].\) Also by (5), \((q', p,t)\) forces \(\dot{f}(u_n) = s_n\) for all \(n \in \omega \). Hence \((q',p,t) \) forces that \(\dot{f}(u)\) is an upper bound for \(\langle {s_{n} : n \in \omega } \rangle \) which is a contradiction. \(\square \)