On the rigidity of Souslin trees and their generic branches

Abstract

We show it is consistent that there is a Souslin tree S such that after forcing with S, S is Kurepa and for all clubs $C \subset \omega _1$, $S\upharpoonright C$ is rigid. This answers the questions in Fuchs (Arch Math Logic 52(1–2):47–66, 2013). Moreover, we show it is consistent with $\diamondsuit$ that for every Souslin tree T there is a dense $X \subseteq T$ which does not contain a copy of T. This is related to a question due to Baumgartner in Baumgartner (Ordered sets (Banff, Alta., 1981), volume 83 of NATO Adv. Study Inst. Ser. C: Math. Phys. Sci., Reidel, Dordrecht-Boston, pp 239–277, 1982).

1 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 T, S 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.

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

• (2) if $\sigma \perp \tau$ then $s_\sigma \perp s_\tau$,

• (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.

3 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) $\Delta _p \in \omega _1$ 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) $\Pi _p = \langle {\pi _\xi ^p: \xi \in D_p} \rangle$ where $D_p \subset \omega _2$ is countable and for each $\xi \in D_p$ there are x, y 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 p, q 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 p, q 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 (M, Q)-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 a, t 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.

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

• (2) $S \subset U$ is dense.

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

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

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

• (4) if $\sigma \perp \tau$ then $s_\sigma \perp s_\tau$,

• (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 (M, Q)-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 t, x, y be pairwise incompatible in T. Suppose (p, t) 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 p, t, u 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) $\bar{p} \Vdash \forall v \in T_x \cap (M_1 \setminus M_0) \ [v \in \dot{S}]$

• (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 (q, r) decides $\dot{f}(u_0)$. Then the condition (q, r) 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 x, y 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 x, y 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 $(q_0, p_0,t_0)$ $\Vdash$ [x, y 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 u, v 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 (M, Q)-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 (M, Q)-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) $\delta _n \ge \gamma _n$ for all $n \in \omega ,$

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

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

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

• (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) 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 (M, Q)-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 (M, Q)-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) $\{ g_m (t_n) :n \in \omega \wedge t_n \in \mathrm{dom}(g_m)\}$ is cofinal in the downward closure of $\{s_{n} : n \in \omega \}$ or

• (2) $\{ g_m (u_n) : n \in \omega \wedge u_n \in \mathrm{dom}(g_m)\}$ 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. $\square$