Proof Let D be a subclub of $V^-(\mathbf T)$ (resp., $V(\mathbf T)$ ), and consider the tree $\mathbf T':=(T\mathbin \upharpoonright D,{<_T})$ . We claim that $V^-(\mathbf T')=\operatorname {\mathrm {acc}}(\kappa )$ (resp., $V(\mathbf T')=\operatorname {\mathrm {acc}}(\kappa )$ ). To this end, pick $\alpha \in \operatorname {\mathrm {acc}}(\kappa )$ . Let $\delta <\kappa $ be least ordinal to satisfy $\operatorname {\mathrm {otp}}(D\cap \delta )=\alpha $ . Then, $\delta \in \operatorname {\mathrm {acc}}(D)\subseteq D$ and hence $\delta \in V^-(\mathbf T)$ (resp., $\delta \in V(\mathbf T)$ ). As every vanishing $\delta $ -branch in $\mathbf T$ induces a vanishing $\alpha $ -branch in $\mathbf T'$ , we infer that $\alpha \in V^-(\mathbf T')$ (resp., $\alpha \in V(\mathbf T')$ ).

Proof (1) Suppose not, and fix a club $D\subseteq \kappa $ disjoint from $V^{-}(\mathbf T)$ . We shall construct a $<_T$ -increasing sequence $\langle t_\alpha \mathrel {|} \alpha \in D\rangle $ in such a way that $t_\alpha \in T_\alpha $ for all $\alpha \in D$ , contradicting the fact that $\mathbf T$ is $\kappa $ -Aronszajn. We start by letting $t_{\min (D)}$ be an arbitrary element of $T_{\min (D)}$ . Next, for every $\alpha \in D$ such that $t_\alpha $ has already been successfully defined, we set $\beta :=\min (D\setminus (\alpha +1))$ , and use the normality of $\mathbf T$ to pick $t_\beta $ in $T_\beta $ extending $t_\alpha $ . For every $\alpha \in \operatorname {\mathrm {acc}}(D)$ such that $\langle t_\epsilon \mathrel {|} \epsilon \in D\cap \alpha \rangle $ has already been defined, the latter clearly induces an $\alpha $ -branch, so the fact that $\alpha \notin V^-(\mathbf T)$ implies that there exists some $t_\alpha \in T_\alpha $ such that $t_\epsilon <_T t_\alpha $ for all $\epsilon \in D\cap \alpha $ . This completes the description of the recursion.

(2) Suppose that $\mathbf T$ is homogeneous. Let $\alpha \in V^-(\mathbf T)$ , and fix a vanishing $\alpha $ -branch b. Now, given a node x of $\mathbf T$ of height less than $\alpha $ , let y be the unique element of b to have the same height as x. Since $\mathbf T$ is homogeneous, there exists an automorphism $\pi $ of $\mathbf T$ sending y to x, and it is clearly the case that $\pi [b]$ is a vanishing $\alpha $ -branch through x.

Proof By [Reference KönigKön03, Theorem 3.9], $\square (\kappa )$ yields a sequence of functions $\langle f_\beta :\beta \rightarrow \beta \mathrel {|} \beta \in \operatorname {\mathrm {acc}}(\kappa )\rangle $ such that:

• • for every $(\beta ,\gamma )\in [\operatorname {\mathrm {acc}}(\kappa )]^2$ , $\{\alpha <\beta \mathrel {|} f_\beta (\alpha )\neq f_\gamma (\alpha )\}$ is finite;

• • there is no cofinal $B\subseteq \operatorname {\mathrm {acc}}(\kappa )$ such that $\{ f_\beta \mathrel {|} \beta \in B\}$ is linearly ordered by $\subseteq $ .

Set $T:=\{ f\in {}^\alpha \alpha \mathrel {|} \alpha <\kappa , f\text { disagrees with }f_\alpha \text { on a finite set}\}$ . Then, $\mathbf T=(T,{\subsetneq })$ is a uniformly coherent $\kappa $ -Aronszajn tree. By [Reference Rinot and ShalevRS23, Remark 2.20], then, $V(\mathbf T)=E^\kappa _\omega $ .

Proof Write $\mathbf T=(T,{<_T})$ . Toward a contradiction, suppose that $\alpha \in S$ is a counterexample. As $\alpha \notin V(\mathbf T)$ , we may fix $x\in T$ with $\operatorname {\mathrm {ht}}(x)<\alpha $ such that every $\alpha $ -branch B with $x\in B$ has an upper bound in $\mathbf T$ . Since either $\operatorname {\mathrm {cf}}(\alpha )\le \omega $ or $V^-(\mathbf T)\cap \alpha $ is nonstationary in $\alpha $ , we may fix a club C in $\alpha $ of order-type $\chi $ such that $\min (C)=\operatorname {\mathrm {ht}}(x)$ and such that $\operatorname {\mathrm {acc}}(C)\cap V^-(\mathbf T)=\emptyset $ .

Let $\langle \alpha _i\mathrel {|} i<\chi \rangle $ denote the increasing enumeration of C. We shall recursively construct an array of nodes $\langle t_s\mathrel {|} s\in {}^{<\chi }\varsigma \rangle $ in such a way that $t_s\in T_{\alpha _{\operatorname {\mathrm {dom}}(s)}}$ . Set $t_\emptyset :=x$ . For every $i<\chi $ and every $s:i\rightarrow \varsigma $ such that $t_s$ has already been defined, since T is normal and $\varsigma $ -splitting, we may find an injective sequence $\langle t_{s{}^\smallfrown \langle j\rangle }\mathrel {|} j<\varsigma \rangle $ of nodes of $T_{\alpha _{i+1}}$ all extending $t_s$ . For every $i\in \operatorname {\mathrm {acc}}(\chi )$ such that $\langle t_s\mathrel {|} s\in {}^{<i}\varsigma \rangle $ has already been defined, for every $s:i\rightarrow \varsigma $ , since $\{ t_{s\mathbin \upharpoonright \iota }\mathrel {|} \iota <i\}$ induces an $\alpha _i$ -branch, the fact that $\alpha _i\notin V^-(\mathbf T)$ implies that we may find $t_s\in T_{\alpha _i}$ that is a limit of that $\alpha _i$ -branch. This completes the recursive construction of our array.

For every $s\in {}^{\chi }\varsigma $ , $B_s:=\{ t\in T\mathrel {|} \exists i<\chi \,(t<_T t_{s\mathbin \upharpoonright i})\}$ is an $\alpha $ -branch containing x, and hence there must be some $b_s\in T_\alpha $ extending all elements of $B_s$ . Our construction also ensures that $B_s\neq B_{s'}$ whenever $s\neq s'$ . We now consider a few options:

1. (1) Suppose that $\varsigma ^\chi \ge \kappa $ . Then, $|T_\alpha |\ge |\{ b_s\mathrel {|} s\in {}^\chi \varsigma \}|=\varsigma ^\chi \ge \kappa $ . This is a contradiction.

2. (2) Suppose that $\mathbf T$ is S-regressive, as witnessed by $\rho :T\mathbin \upharpoonright S\rightarrow T$ . For every $s\in {}^{\chi }\varsigma $ , $\rho (b_s)$ belongs to $B_s$ , but by Remark 2.9, we may assume that $\rho (b_s)= t_{s\mathbin \upharpoonright i}$ for some $i<\chi $ .

   • ▸ If $\varsigma ^{<\chi }<\varsigma ^\chi $ , then we may now find $s\neq s'$ in ${}^\chi \varsigma $ such that $\rho (b_s)=\rho (b_{s'})$ . Then, $\rho (b_{s'})<_T t_s$ and $\rho (b_s)<_T t_{s'}$ , contradicting the fact that $b_s\neq b_{s'}$ .

   • ▸ If $\chi =\varsigma $ and there exists a weak $\chi $ -Kurepa tree, then this may be witnessed by a tree of the form $(K,{\subsetneq })$ for some $K\subseteq {}^{<\chi }\varsigma $ . Let $\langle s_\beta \mathrel {|} \beta <\chi ^+\rangle $ be an injective enumeration of branches through $(K,{\subsetneq })$ . Since $|K|\le \chi $ , there must exist $\beta \neq \beta '$ such that $\rho (b_{s_\beta })=\rho (b_{s_{\beta '}})$ , which yields a contradiction as in the previous case.

Proof Suppose that $\kappa $ is not a strong limit. It is not hard to see that there exists some infinite cardinal $\varsigma <\kappa $ for which there exists a regular cardinal $\chi <\kappa $ such that $\varsigma ^\chi \ge \kappa $ . Now, given a normal and splitting $\kappa $ -tree $\mathbf T=(T,<_T)$ , as shown in the proof of [Reference Rinot and ShalevRS23, Proposition 2.16], the club $D:=\{\alpha <\kappa \mathrel {|} \alpha =\varsigma ^\alpha \}$ satisfies that $\mathbf T'=(T\mathbin \upharpoonright D,{<_T})$ is normal and $\varsigma $ -splitting. By Lemma 2.10, $V^-(\mathbf T')$ is stationary. As D is a club in $\kappa $ , this means that $V^-(\mathbf T)$ is stationary, as well.

Proof Suppose that $\kappa $ and $\lambda $ are as above. Now, given a normal and splitting $\kappa $ -tree $\mathbf T=(T,<_T)$ , the club $D:=\{\alpha <\kappa \mathrel {|} \alpha =\lambda ^\alpha \}$ satisfies that $\mathbf T'=(T\mathbin \upharpoonright D,{<_T})$ is normal and $\lambda $ -splitting. By Lemma 2.10, $V(\mathbf T')\supseteq E^\kappa _\omega $ . As D is a club in $\kappa $ , this means that $E^\kappa _\omega \setminus V(\mathbf T)$ is nonstationary.

Proof Let D be an arbitrary club in $\kappa $ . Let $\pi :\kappa \leftrightarrow D$ be the unique order-preserving map. For every $\gamma <\kappa $ , for every $t\in T_{\pi (\gamma )}$ , define a corresponding $x_t:\gamma \rightarrow T$ via

$$ \begin{align*}x_t(\alpha):=t\mathbin\upharpoonright\pi(\alpha).\end{align*} $$

Consider $X:=\{ x_t\mathrel {|} t\in T\mathbin \upharpoonright D\},$ which is again a subset of ${}^{<\kappa }H_\kappa $ .

Proof Let $\beta <\gamma <\kappa $ and $x\in X\cap {}^\gamma H_\kappa $ . Pick $t\in T_{\pi (\gamma )}$ such that $x=x_t$ . As T is streamlined, $t\mathbin \upharpoonright \pi (\beta )$ is in T, so that $x_{t\mathbin \upharpoonright \pi (\beta )}$ is in X. In addition, for every $\alpha <\beta $ ,

$$ \begin{align*}x_{t\mathbin\upharpoonright\pi(\beta)}(\alpha)=(t\mathbin\upharpoonright\pi(\beta))\mathbin\upharpoonright\pi(\alpha)=t\mathbin\upharpoonright\pi(\alpha)=x_t(\alpha)=x(\alpha),\end{align*} $$

and hence $x\mathbin \upharpoonright \beta $ is in X.

It follows that for every $\gamma <\kappa $ , $X_\gamma =X\cap {}^\gamma H_\kappa =\{ x_t\mathrel {|} t\in T_{\pi (\gamma )}\}$ and in particular, $0<|X_\gamma |<\kappa $ . That is, X is a streamlined $\kappa $ -tree.

Next, consider the club of fixed-points $E:=\{\gamma \in \operatorname {\mathrm {acc}}(\kappa )\mathrel {|} \pi (\gamma )=\gamma \}$ .

A similar proof shows that if $D\subseteq V^-(T)$ , then $V^-(X)=\operatorname {\mathrm {acc}}(\kappa )$ .

Proof Let $s\neq s'$ be two nodes in $T_\beta $ for some $\beta <\kappa $ . For every $t\in T$ , consider

$$ \begin{align*}\begin{aligned} \alpha(t)&:=\min(\{\epsilon<\beta\mathrel{|} s(\epsilon)\neq t(\epsilon)\}\cup\{\beta,\operatorname{\mathrm{dom}}(t)\}),\text{ and}\\ \alpha'(t)&:=\min(\{\epsilon<\beta\mathrel{|} s'(\epsilon)\neq t(\epsilon)\}\cup\{\beta,\operatorname{\mathrm{dom}}(t)\}). \end{aligned}\end{align*} $$

For the next definition, we take the convention that for a function f, whenever we write $f(x)$ , then we implicitly express that x is in $\operatorname {\mathrm {dom}}(f)$ . Now, define a map $\pi :T\rightarrow T$ via:

$$ \begin{align*}\pi(t):=\begin{cases} (s'\mathbin\upharpoonright \alpha(t))*t,&\text{if }t\subseteq s\text{ or }s\subseteq t;\\ (s'\mathbin\upharpoonright \alpha(t))*t,&\text{if }\alpha'(t)<\alpha(t) \text{ and } t(\alpha(t))\neq s'(\alpha(t));\\ ((s'\mathbin\upharpoonright \alpha(t))*(s\mathbin\upharpoonright (\alpha(t)+1)))*t,&\text{if }\alpha'(t)<\alpha(t)\text{ and } t(\alpha(t))=s'(\alpha(t));\\ (s\mathbin\upharpoonright \alpha'(t))*t,&\text{if }t\subseteq s'\text{ or }s'\subseteq t;\\ (s\mathbin\upharpoonright \alpha'(t))*t,&\text{if }\alpha(t)<\alpha'(t) \text{ and } t(\alpha'(t))\neq s(\alpha'(t))\\ ((s\mathbin\upharpoonright \alpha'(t))*(s'\mathbin\upharpoonright (\alpha'(t)+1))*t,&\text{if }\alpha(t)<\alpha'(t)\text{ and } t(\alpha'(t))=s(\alpha'(t));\\ t,&\text{otherwise}. \end{cases}\end{align*} $$

It is not hard to see that $\pi $ is well-defined and satisfies $\pi (s)=s'$ .

Proof Write $\gamma :=\operatorname {\mathrm {dom}}(t)$ . Let $\delta <\beta $ be the least such that $s(\delta )\neq s'(\delta )$ . Note that if $\alpha (t)\neq \alpha '(t)$ , then $\min \{\alpha (t),\alpha '(t)\}=\delta $ . Now, by symmetry, it suffices to analyze the first three cases in the definition of $\pi $ . See Figure 1 for an illustration.

1. Case 1: If $t\subseteq s$ , then $\pi (t)=s'\mathbin \upharpoonright \gamma $ and hence $\pi (\pi (t))=s\mathbin \upharpoonright \gamma =t$ . Likewise, if $s\subseteq t$ , then $\pi (t)=s'*t$ and hence $\pi (\pi (t))=s*t=t$ .

2. Case 2: If $\delta =\alpha '(t)<\alpha (t)<\min \{\beta ,\gamma \}$ and $t(\alpha (t))\neq s'(\alpha (t))$ , then $\pi (t)=(s'\mathbin \upharpoonright \alpha (t))*t$ and hence $\alpha '(\pi (t))=\alpha (t)$ and $\alpha (\pi (t))=\delta =\alpha '(t)$ . So $\alpha (\pi (t))<\alpha '(\pi (t))$ . In addition, $\pi (t)(\alpha (t))=t(\alpha (t))$ , so that $\pi (t)(\alpha (t))\neq s(\alpha (t))$ . Altogether,

   $$ \begin{align*}\pi(\pi(t))=(s\mathbin\upharpoonright \alpha'(\pi(t)))*\pi(t)=(s\mathbin\upharpoonright\alpha(t))*t=t.\end{align*} $$

3. Case 3: If $\delta =\alpha '(t)<\alpha (t)<\min \{\beta ,\gamma \}$ and $t(\alpha (t))= s'(\alpha (t))$ , then $\pi (t)=((s'\mathbin \upharpoonright \alpha (t))*(s\mathbin \upharpoonright (\alpha (t)+1)))*t$ . So $\alpha (\pi (t))=\delta =\alpha '(t)$ and $\alpha '(\pi (t))\ge \alpha (t)$ . If $\alpha '(\pi (t))>\alpha (t)$ , then $t(\alpha (t))=s'(\alpha (t))=\pi (t)(\alpha (t))=s(\alpha (t))$ , contradicting the definition of $\alpha (t)$ . So $\alpha '(\pi (t))=\alpha (t)$ . Therefore, $\pi (t)(\alpha '(\pi (t)))=\pi (t)(\alpha (t))=s(\alpha (t))=s(\alpha '(\pi (t)))$ . Altogether,

   $$ \begin{align*}\begin{aligned}\pi(\pi(t)) =&\ (s\mathbin\upharpoonright \alpha'(\pi(t)))*(s'\mathbin\upharpoonright (\alpha'(\pi(t))+1))*\pi(t)\\ =&\ (s\mathbin\upharpoonright \alpha(t))*(s'\mathbin\upharpoonright (\alpha(t)+1))*t\\ =&\ (t\mathbin\upharpoonright \alpha(t))*(s'\mathbin\upharpoonright (\alpha(t)+1))*t\\ =&\ (t\mathbin\upharpoonright \alpha(t)){}^\smallfrown\langle t(\alpha(t)+1)\rangle*t=t. \end{aligned}\end{align*} $$

For the reader’s convenience, the above proof is illustrated in Figure 1.

Figure 1:

Two main cases from the proof of Claim 2.19.1.

At this point, to prove that $\pi $ is an automorphism, it suffices to show that it is order-preserving.

This completes the proof.

Proof $(1)\implies (2)$ : This is immediate.

$(2)\implies (3)$ : Suppose that D and $\langle A_\alpha \mathrel {|} \alpha \in S\cap D\rangle $ are as in (2). Let $\langle x_i\mathrel {|} i<\kappa \rangle $ be an injective enumeration of $\langle A_\alpha \cap \varepsilon \mathrel {|} \varepsilon <\alpha , \alpha \in S\cap D\rangle $ . For each $\alpha \in S\cap D$ , let $k_\alpha :\alpha \rightarrow \kappa $ be the unique function to satisfy for all $\varepsilon <\alpha $ :

$$ \begin{align*}A_\alpha\cap\varepsilon=x_{k_\alpha(\varepsilon)}.\end{align*} $$

Define first an auxiliary collection K by letting

$$ \begin{align*}K:=\{ k_\beta\mathbin\upharpoonright\alpha\mathrel{|} \alpha<\beta, \beta\in S\cap D\}.\end{align*} $$

Note that $\{ \operatorname {\mathrm {dom}}(y)\mathrel {|} y\in K\}=\kappa $ and that K is closed under taking initial segments. So K is a streamlined $\kappa $ -tree because otherwise there must exist some $\varepsilon <\kappa $ such that $\{ k_\beta \mathbin \upharpoonright \varepsilon \mathrel {|} \beta \in S\cap D\}$ has size $\kappa $ , contradicting the fact that $\langle A_\beta \mathrel {|} \beta \in S\cap D\rangle $ is thin. We shall use K to construct a uniformly homogeneous streamlined $\kappa $ -tree T by defining its levels $T_\alpha $ by recursion on $\alpha <\kappa $ .

Start by letting $T_0:=K_0$ . Clearly, $T_0=\{\emptyset \}$ , so that $|T_0|<\kappa $ . Next, for every nonzero $\alpha <\kappa $ such that $T\mathbin \upharpoonright \alpha $ has already been defined and has size less than $\kappa $ , let

$$ \begin{align*}T_{\alpha}:=\{ x*y\mathrel{|} x\in T\mathbin\upharpoonright\alpha,\, y\in K_\alpha\}\end{align*} $$

and note that $|T_\alpha |<\kappa $ . Altogether, T is a streamlined $\kappa $ -tree.

By the preceding claim together with Proposition 2.6, it now suffices to prove that $V^{-}(T)\supseteq S\cap D\cap \operatorname {\mathrm {acc}}(\kappa )$ . To this end, let $\alpha \in S\cap D\cap \operatorname {\mathrm {acc}}(\kappa )$ . Clearly, $b:=\{ k_\alpha \mathbin \upharpoonright \varepsilon \mathrel {|} \varepsilon <\alpha \}$ is an $\alpha $ -branch in K and hence in T. If b is not vanishing in T, then we may find $x\in T\mathbin \upharpoonright \alpha $ and $y\in K_\alpha $ such that $x*y=k_\alpha $ . Recalling the definition of $K_\alpha $ , we may pick $\beta \in S\cap D$ above $\alpha $ such that $y=k_\beta \mathbin \upharpoonright \alpha $ . As $\alpha <\beta $ , it is the case that $A_\alpha \neq A_{\beta }\cap \alpha $ , so we may pick $\delta \in A_\alpha \Delta (A_{\beta }\cap \alpha )$ . Then, $\varepsilon :=\max \{\delta ,\operatorname {\mathrm {dom}}(x)\}+1$ is smaller than $\alpha $ and satisfies $k_\alpha (\varepsilon )\neq k_{\beta }(\varepsilon )$ , contradicting the fact that $k_\alpha (\varepsilon )=(x*y)(\varepsilon )=y(\varepsilon )=k_\beta (\varepsilon )$ .

$(3)\implies (4)$ : This is immediate.

$(4)\implies (1)$ Every $\kappa $ -tree is order-isomorphic to an ordinal-based tree (see, e.g., [Reference Rinot and ShalevRS23, Proposition 2.16]), so we may assume that we are given a tree $\mathbf T$ of the form $(\kappa ,<_T)$ and a club $D\subseteq \kappa $ such that $V^-(\mathbf T)\supseteq S\cap D$ . By possibly shrinking D, we may also assume that $D\subseteq \operatorname {\mathrm {acc}}\{\beta <\kappa \mathrel {|} T\mathbin \upharpoonright \beta =\beta \}$ . It follows that for every $\alpha \in D$ , every $\alpha $ -branch is a cofinal subset of $\alpha $ . For every $\alpha \in S\cap D$ , let $A_\alpha $ be a vanishing $\alpha $ -branch. As $\mathbf T$ is a $\kappa $ -tree, the ladder system $\langle A_\alpha \mathrel {|} \alpha \in S\cap D\rangle $ is thin. In addition, for every $(\alpha ,\beta )\in [S\cap D]^2$ , if it were the case that $\sup (A_\beta \cap A_\alpha )=\alpha $ , then $\min (A_\beta \setminus A_\alpha )$ is a node extending all elements of $A_\alpha $ , contradicting the fact that $A_\alpha $ is vanishing. So, $\sup (A_\beta \cap A_\alpha )<\alpha $ .

Proof By Lemmas 2.20 and 2.15.

Proof Work in $\mathsf L$ , and fix a subtle cardinal $\kappa $ that is not weakly compact. Now, pass to a generic extension $\mathsf L[G]$ by Mitchell’s forcing of length $\kappa $ . By [Reference WeißWei10, Theorem 2.3.1], $\aleph _2$ - $\textsf {STP}$ holds, and hence $V(\mathbf T)$ cannot contain a club for every $\aleph _2$ -tree $\mathbf T$ . Since $\kappa $ is not weakly compact in $\mathsf {L}$ , $\square (\aleph _2)$ holds. In addition, this is a model in which $2^{\aleph _0}=2^{\aleph _1}=\aleph _2$ and hence Corollary 2.12 implies that $E^{\aleph _2}_{\aleph _0}\setminus V(\mathbf T)$ is nonstationary for every normal and splitting $\aleph _2$ -tree $\mathbf T$ . Therefore, $E^{\aleph _2}_{\aleph _1}\setminus V(\mathbf T)$ is stationary for every normal and splitting $\aleph _2$ -tree $\mathbf T$ .

Since the model $\mathsf L[G]$ is a forcing extension by a projection of $\operatorname {\mathrm {Add}}(\omega , \kappa )\times \mathbb R$ for some countably-closed forcing $\mathbb R$ , all of our reals live in $\mathsf L^{\operatorname {\mathrm {Add}}(\omega ,\kappa )}$ . Recalling that already $\operatorname {\mathrm {Add}}(\omega ,\omega _1)$ adds a Luzin set, we altogether infer that $\mathfrak b=\aleph _1$ . Finally, by [Reference RinotRin22, Theorem A], whenever $\mathfrak b=\aleph _1$ , $2^{\aleph _1}=\aleph _2$ and $\square (\aleph _2)$ all hold, there indeed exist an $\aleph _2$ -Souslin tree.

Proof By Lemma 2.20, it suffices to find a stationary $S^-\subseteq S$ that carries a thin almost disjoint C-sequence. We consider two cases:

$\blacktriangleright $ If $S\cap E^\kappa _\omega $ is stationary, then set $S^-:=S\cap E^\kappa _\omega $ , and let $\langle C_\alpha \mathrel {|} \alpha \in S^-\rangle $ be some $\omega $ -bounded C-sequence over $S^-$ .

$\blacktriangleright $ Otherwise, let $S^-:=S\setminus (E^\kappa _\omega \cup \operatorname {\mathrm {Tr}}(S))$ . Then, $S^-$ is stationary, and for every $\alpha \in S^-$ , we may pick a club $C_\alpha $ in $\alpha $ that is disjoint from S. Evidently, $\sup (C_\alpha \cap C_{\alpha '})<\alpha $ for every pair $\alpha <\alpha '$ of ordinals in $S^-$ .

Proof Just take a $\theta $ -bounded C-sequence over $E^\kappa _\theta $ .

Proof $(i)\implies (ii)$ Assuming that there exists a special $\kappa $ -Aronszajn tree, by [Reference KruegerKru13, Lemma 1.2 and Theorem 2.5], we may fix a C-sequence $\vec C=\langle C_\beta \mathrel {|} \beta <\kappa \rangle $ and a club $C\subseteq \operatorname {\mathrm {acc}}(\kappa )$ satisfying the following:

1. (1) for every $\beta \in C$ , $\min (C_\beta )>\operatorname {\mathrm {otp}}(C_\beta )$ ;

2. (2) for every $\beta \in \operatorname {\mathrm {acc}}(\kappa )\setminus C$ , $\min (C_\beta )>\sup (C\cap \beta )$ ;

3. (3) for every $\epsilon <\kappa $ , $|\{ C_\beta \cap \epsilon \mathrel {|} \beta <\kappa \}|<\kappa $ .

Consider the following additional requirement:

1. (4) $\min (C_\beta )=\operatorname {\mathrm {otp}}(C_\beta )+1$ for every $\beta \in C$ .

Now, let $\rho _0$ be the characteristic function from [Reference TodorcevicTod07, Section 6] obtained by walking along $\vec C$ satisfying (1)–(4), and consider the following streamlined $\kappa $ -tree:

$$ \begin{align*}T(\rho_0):=\{ \rho_{0\beta}\mathbin\upharpoonright\alpha\mathrel{|} \alpha\le\beta<\kappa\}.\end{align*} $$

Using (1)–(3), the proof of [Reference KruegerKru13, Theorem 4.4] provides a club $D\subseteq C$ and a function $g:T(\rho _0)\mathbin \upharpoonright D\rightarrow \kappa $ satisfying the following two:

• • $g(t)<\operatorname {\mathrm {dom}}(t)$ for all $t\in T(\rho _0)\mathbin \upharpoonright D$ ;

• • for every pair $s\subsetneq t$ of nodes from $T(\rho _0)\mathbin \upharpoonright D$ , $g(s)\neq g(t)$ .

Next, consider the following subfamily of $T(\rho _0)$ :

$$ \begin{align*}T:=\{ \rho_{0\beta}\mathbin\upharpoonright\alpha\mathrel{|} \alpha<\beta<\kappa\}.\end{align*} $$

Clearly, T is downward-closed and $\{\operatorname {\mathrm {dom}}(y)\mathrel {|} y\in T\}=\kappa $ , so that T is a streamlined $\kappa $ -Aronszajn subtree of $T(\rho _0)$ .

Proof The “in particular” part will follow from the fact that $\{ \rho _{0\alpha }\mathbin \upharpoonright \epsilon \mathrel {|} \epsilon <\alpha \}$ is an $\alpha $ -branch of T for every $\alpha <\kappa $ . Thus, let $\alpha \in C$ and we shall prove that $\rho _{0\alpha }\notin T$ . Suppose not, and pick some $\beta>\alpha $ such that $\rho _{0\alpha }=\rho _{0\beta }\mathbin \upharpoonright \alpha $ . Recall that for every $\gamma <\kappa $ ,

$$ \begin{align*}C_\gamma=\{ \xi<\gamma\mathrel{|} \rho_{0\gamma}(\xi)\text{ is a sequence of length }1\}.\end{align*} $$

In particular, $\min (C_\alpha )=\min (C_\beta )$ . As $\sup (C\cap \beta )\ge \alpha>\min (C_\alpha )$ , it follows from Clause (2) that $\beta \in C$ . So, by Clause (4), $\operatorname {\mathrm {otp}}(C_\alpha )=\operatorname {\mathrm {otp}}(C_\beta )$ . It follows that may fix some $\delta \in C_\alpha \setminus C_\beta $ . But then $\rho _{0\alpha }(\delta )$ is a sequence of length $1$ , whereas $\rho _{0\beta }(\delta )$ is a longer sequence. This is a contradiction.

For every $t\in T\mathbin \upharpoonright \operatorname {\mathrm {acc}}(\kappa )$ , define a function $k_t:\operatorname {\mathrm {dom}}(t)\rightarrow T$ via

$$ \begin{align*}k_t(\varepsilon):=t\mathbin\upharpoonright\varepsilon.\end{align*} $$

Let K be the following downward-closed subfamily of ${}^{<\kappa }H_\kappa $ :

$$ \begin{align*}K:=\{ k_t\mathbin\upharpoonright\alpha\mathrel{|} \alpha\le\operatorname{\mathrm{dom}}(t), t\in T\mathbin\upharpoonright\operatorname{\mathrm{acc}}(\kappa)\}.\end{align*} $$

Evidently, for all $x,y\in K$ and $\varepsilon \in \operatorname {\mathrm {dom}}(x)\cap \operatorname {\mathrm {dom}}(y)$ , if $x(\varepsilon )=y(\varepsilon )$ , then $x\mathbin \upharpoonright \varepsilon =y\mathbin \upharpoonright \varepsilon $ . In addition, $t\mapsto k_t$ constitutes an isomorphism between $(T\mathbin \upharpoonright \operatorname {\mathrm {acc}}(\kappa ),{\subsetneq })$ and $(K\mathbin \upharpoonright \operatorname {\mathrm {acc}}(\kappa ),{\subsetneq })$ , and hence K is a streamlined $\kappa $ -Aronszajn tree with $V^-(K)\supseteq D$ . The fact that the above map is an isomorphism also implies that a function $f:K\mathbin \upharpoonright D\rightarrow \kappa $ defined via $f(k_t):=g(t)$ satisfies that $f(x)<\operatorname {\mathrm {dom}}(x)$ for all $x\in K\mathbin \upharpoonright D$ , and that $f(x)\neq f(y)$ for every pair $x\subsetneq y$ of nodes from $K\mathbin \upharpoonright D$ .

$(ii)\implies (iii)$ : Suppose that K and $f:K\mathbin \upharpoonright D\rightarrow \kappa $ are as in Clause (ii). By possibly shrinking D, we may assume that for all $\beta \in D$ and $\alpha <\beta $ , it is the case that $\omega \cdot \alpha <\beta $ .

Using Remark 2.17, we may define a family T to be the collection of all elements of the form $x_0*\cdots *x_n,$ whereFootnote ¹⁴

1. (a) $n<\omega $ ,

2. (b) $x_i\in K$ for all $i\le n$ , and

3. (c) $\operatorname {\mathrm {dom}}(x_i)<\operatorname {\mathrm {dom}}(x_{i+1})$ for all $i<n$ .

It is clear that $t\mathbin \upharpoonright \alpha \in T$ for all $t\in T$ and $\alpha <\kappa $ . Thus, recalling the proof of Claim 2.20.1, to establish that T is a uniformly homogeneous streamlined $\kappa $ -tree, it suffices to prove the following claim.Footnote ¹⁵

For each node $t\in T$ , we define $n(t)$ and $x(t)$ by first letting $n(t)$ denote the least n for which there exists a sequence $(x_0,\ldots ,x_n)$ satisfying (a)–(c) for which $t=x_0*\cdots *x_n$ , and then letting $x(t)$ be such an $x_n$ . Note that $\operatorname {\mathrm {dom}}(x(t))=\operatorname {\mathrm {dom}}(t)$ , and that $K=\{ t\in T\mathrel {|} n(t)=0\}$ .

Define a function $g:T\mathbin \upharpoonright D\rightarrow \kappa $ via

$$ \begin{align*}g(t):=(\omega\cdot f(x(t)))+n(t).\end{align*} $$

It is easy to see that the two features of g together imply that T admits no $\kappa $ -branch. The beginning of the proof of [Reference KruegerKru13, Theorem 4.4] shows furthermore that T must be a special $\kappa $ -Aronszajn tree.

The implication $(iii)\implies (iv)$ follows from Lemma 2.15, and the implication $(iv)\implies (i)$ is trivial.

Proof Left to the reader.

Proof Left to the reader.

Proof Denote $\Theta :=\operatorname {\mathrm {Reg}}(\chi +1)$ . By Lemmas 2.26 and 2.20, for every $\theta \in \Theta $ , we may pick a $\kappa $ -tree $\mathbf T^\theta $ such that $E^\kappa _{\theta }\setminus V^-(\mathbf T^\theta )$ is nonstationary. In fact, the proof of $(2)\implies (3)$ of Lemma 2.20 shows that we may secure $V^-(\mathbf T^\theta )\supseteq E^\kappa _{\theta }$ . Let $\mathbf T:=\sum \{\mathbf T^\theta \mathrel {|} \theta \in \Theta \}$ be the disjoint sum of these trees. By Proposition 2.32, $V^-(\mathbf T)=\bigcup _{\theta \in \Theta }V^-(\mathbf T^\theta )\supseteq \bigcup _{\theta \in \Theta }E^\kappa _\theta = \operatorname {\mathrm {acc}}(\kappa )\cap E^\kappa _{\leq \chi }$ .

Question 2.36 Is it consistent that for some regular uncountable cardinal $\kappa $ , there are $\kappa $ -Souslin trees, but $V(\mathbf T)$ is nonstationary for every $\kappa $ -Souslin tree $\mathbf T$ ?

Proof (1) Suppose that $\square _\lambda \mathrel {+}\diamondsuit (\kappa )$ holds. By Lemma 2.5, it suffices to find a $\kappa $ -Souslin tree $\mathbf T$ for which $V(\mathbf T)$ covers a club in $\kappa $ .

• ▸ For $\lambda =\aleph _0$ , $\diamondsuit (\aleph _1)$ implies the existence of a normal and splitting $\aleph _1$ -Souslin tree $\mathbf T$ , and by Corollary 2.12, $V(\mathbf T)$ covers a club in $\aleph _1$ .

• ▸ For $\lambda \ge \aleph _1$ , by [Reference Brodsky and RinotBR17a, Corollary 3.9], $\square _\lambda +\operatorname {\mathrm {CH}}_\lambda $ is equivalent to $\operatorname {\mathrm {P}}_\lambda (\kappa ,2,{\sqsubseteq },1)$ . In addition, by a theorem of Jensen, $\square _\lambda $ gives rise to a special $\lambda ^+$ -Aronszajn tree. Thus, we infer from Theorem 2.27 the existence of a streamlined $\kappa $ -tree K for which $V(K)$ covers a club in $\kappa $ . It thus follows from Theorem 3.7(1) below that there exists a $\kappa $ -Souslin tree $\mathbf T$ for which $V(\mathbf T)$ is a club in $\kappa $ .

(2) By [Reference RinotRin17, Corollary 4.4], the hypothesis implies that $\operatorname {\mathrm {P}}^-(\kappa ,2,{\sqsubseteq },1)$ holds. In addition, by a theorem of Specker, $\lambda =\lambda ^{<\lambda }$ implies the existence of a special $\lambda ^+$ -Aronszajn tree. Now, continue as in the proof of Clause (1).

(3) Similar to the proof of Clause (1), using Theorem 3.7(2), instead.

Convention 3.4 We write $\operatorname {\mathrm {P}}_\xi (\kappa , \mu , \mathcal R, \theta , \mathcal S)$ to assert that $\operatorname {\mathrm {P}}^-_\xi (\kappa , \mu , \mathcal R, \theta , \mathcal S)$ and $\diamondsuit (\kappa )$ both hold.

Convention 3.5 If we omit $\xi $ , then we mean $\xi :=\kappa $ . If we omit $\mathcal S$ , then we mean $\mathcal S:=\{\kappa \}$ . In the case $\mu =2$ , we identify $\langle \mathcal {C}_\alpha \mathrel {|} \alpha <\kappa \rangle $ with the unique element $\langle C_\alpha \mathrel {|} \alpha < \kappa \rangle $ of $\prod _{\alpha <\kappa }\mathcal {C}_\alpha $ .

Fact 3.6 [Reference Brodsky and RinotBR17a, Lemma 2.2]

The following are equivalent:

1. (1) $\diamondsuit (\kappa )$ , i.e., there is a sequence $\langle f_\beta \mathrel {|} \beta <\kappa \rangle $ such that for every function $f:\kappa \rightarrow \kappa $ , the set $\{ \beta <\kappa \mathrel {|} f\mathbin \upharpoonright \beta =f_\beta \}$ is stationary in $\kappa $ .

2. (2) $\diamondsuit ^-(H_\kappa )$ , i.e., there is a sequence $\langle \Omega _\beta \mathrel {|} \beta < \kappa \rangle $ such that for all $p\in H_{\kappa ^{+}}$ and $\Omega \subseteq H_\kappa $ , there exists an elementary submodel $\mathcal M\prec H_{\kappa ^{+}}$ such that:

   • • $p\in \mathcal M$ ;

   • • $\mathcal M\cap \kappa \in \kappa $ ;

   • • $\mathcal M\cap \Omega =\Omega _{\mathcal M\cap \kappa }$ .

3. (3) $\diamondsuit (H_\kappa )$ , i.e., there are a partition $\langle R_i \mathrel {|} i < \kappa \rangle $ of $\kappa $ and a sequence $\langle \Omega _\beta \mathrel {|} \beta < \kappa \rangle $ such that for all $p\in H_{\kappa ^{+}}$ , $\Omega \subseteq H_\kappa $ , and $i<\kappa $ , there exists an elementary submodel $\mathcal M\prec H_{\kappa ^{+}}$ such that:

   • • $p\in \mathcal M$ ;

   • • $\mathcal M\cap \kappa \in R_i$ ;

   • • $\mathcal M\cap \Omega =\Omega _{\mathcal M\cap \kappa }$ .