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).
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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:
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if \(\sigma \sqsubset \tau \) then \(p_\tau \le p_\sigma \) and \(s_\sigma \le s_\tau \),
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(2)
if \(\sigma \perp \tau \) then \(s_\sigma \perp s_\tau \),
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\(p_\sigma \Vdash \dot{f}(t_{|\sigma |}) = s_\sigma \).
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:
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\(\Delta _p \in \omega _1\) 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 \(t<_{T^p} s\).
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\(\Pi _p = \langle {\pi _\xi ^p: \xi \in D_p} \rangle \) 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.
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S is a Souslin tree when it is considered with the inherited order from U.
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(2)
\(S \subset U\) is dense.
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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:
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\(q_0 \le p\), and if \(\sigma \subset \tau \) then \(q_\tau \le q_\sigma \),
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\(q_\sigma \) is \((M_{|\sigma |}, P)\)-generic and \(q_\sigma \subset M_{|\sigma |}\),
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\(q_\sigma \) forces that \(\dot{f}(t_{|\sigma |-1}) = s_\sigma \),
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if \(\sigma \perp \tau \) then \(s_\sigma \perp s_\tau \),
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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:
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\(\bar{p} \Vdash \forall v \in T_x \cap (M_1 \setminus M_0) \ [v \in \dot{S}]\)
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\(\bar{p} \Vdash \forall v \in T_y \cap (M_1 \setminus M_0) \ [v \notin \dot{S}]\).
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 \(n \in \omega ,\)
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(2)
\((q,p_n.t_n) \Vdash \min \{ \mathrm{ht}_{\dot{S}}(s_n), \mathrm{ht}_{\dot{S}}(u_n), \mathrm{dom}(p_n) \} \ge \delta _n\),
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(3)
\(\mathrm{ht}_{T^q}(t_n) \ge \mathrm{ht}_{T^q}(s_n) +1\),
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(4)
\((q,p_n,1_{T^q}) \Vdash \delta _n \in \dot{C}\),
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(5)
\((q, p_n,t_n)\) \(\Vdash \) \(\dot{f}(u_n) = s_n\),
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(6)
if \(n \in \omega \setminus 1\) and \(t_{n -1} \in \mathrm{dom}(g_n)\) then \(g_{n} (t_n) \perp s_{n}\),
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(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
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(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
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(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. \(\square \)
References
Baumgartner, J.E.: Order types of real numbers and other uncountable orderings. In: Ordered sets (Banff, Alta., 1981), volume 83 of NATO Adv. Study Inst. Ser. C: Math. Phys. Sci., pp 239–277. Reidel, Dordrecht-Boston, Mass (1982)
Fuchs, G.: Club degrees of rigidity and almost Kurepa trees. Arch. Math. Logic 52(1–2), 47–66 (2013)
Fuchs, G., Hamkins, J.D.: Degrees of rigidity for Souslin trees. J. Symb. Logic 74(2), 423–454 (2009)
Kunen, K.: An Introduction to Independence Proofs, volume 102 of Studies in Logic and the Foundations of Mathematics. North-Holland, Amsterdam (1983)
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Lamei Ramandi, H. On the rigidity of Souslin trees and their generic branches. Arch. Math. Logic 62, 419–426 (2023). https://doi.org/10.1007/s00153-022-00843-5
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DOI: https://doi.org/10.1007/s00153-022-00843-5