HIGHER SOUSLIN TREES AND THE GCH, REVISITED

ASSAF RINOT

Abstract. It is proved that for every uncountable cardinal 𝜆, GCH+�(𝜆+) entails the existenceof a cf(𝜆)-complete 𝜆+-Souslin tree. In particular, if GCH holds and there are no ℵ2-Souslin trees,then ℵ2 is weakly compact in Godel’s constructible universe, improving Gregory’s 1976 lower bound.Furthermore, it follows that if GCH holds and there are no ℵ2 and ℵ3 Souslin trees, then the Axiomof Determinacy holds in 𝐿(R).

1. Introduction

1.1. Background. A tree is a partially ordered set (𝑇,<𝑇 ) with the property that for every 𝑥 ∈ 𝑇 ,the downward cone 𝑥↓ := {𝑦 ∈ 𝑇 | 𝑦 <𝑇 𝑥} is well-ordered by <𝑇 . The order type of (𝑥↓, <𝑇 ) is

denoted by ht(𝑥), and the 𝛼𝑡ℎ-level of the tree is the set 𝑇𝛼 := {𝑥 ∈ 𝑇 | ht(𝑥) = 𝛼}.Let 𝜅 denote a regular uncountable cardinal. A 𝜅-Aronszajn tree is a tree of size 𝜅 having no

chains or levels of size 𝜅. A 𝜅-Souslin tree is a tree of size 𝜅 having no chains or antichains of size𝜅. As tree-levels are antichains, any 𝜅-Souslin tree is a 𝜅-Aronszajn tree.

The above concepts stemmed from a 1920 question of Mikhail Souslin [Sou20], asking whetherevery ccc dense complete linear ordering with no endpoints is isomorphic to the real line.1 Kureparealized [Kur35] that a negative answer is equivalent to the existence of (what we nowadays call)an ℵ1-Souslin tree. However, all of Kurepa’s attempts to construct such a tree were unsuccessful.At one point, Kurepa told Aronszajn about his goal, and in response, Aronszajn came up with aconstruction of a poor man’s version of a Souslin tree; Indeed, Aronszajn constructed (what wenowadays call) an ℵ1-Aronszajn tree.

It was only three decades after [Kur35], in [Ten68], [Jec67], [Jen68], and [ST71], when it wasdiscovered that — unlike ℵ1-Aronszajn trees — the existence of an ℵ1-Souslin tree is independentof the usual axioms of set theory (ZFC).

As these objects proven incredibly useful and important, a systematic study of their consistencyand interrelation was carried out. Jensen proved [Jen72] that in Godel’s constructible universe, 𝐿,for every regular uncountable cardinal 𝜅, the following are equivalent:

∙ There exists a 𝜅-Souslin tree;∙ There exists a 𝜅-Aronszajn tree;∙ 𝜅 is not a weakly compact cardinal.

In another work, Jensen proved (see the monograph [DJ74]) that the existence of an ℵ1-Souslintree is independent of ZFC+GCH.2 As for Aronszajn trees, Specker proved [Spe49] that GCHentails the existence of a 𝜆+-Aronszajn tree for every regular cardinal 𝜆, and Mitchell and Silver

Date: February 21, 2017.2010 Mathematics Subject Classification. Primary 03E05; Secondary 03E35.Key words and phrases. Souslin tree, microscopic approach, weakly compact cardinal, square.

1Here, ccc is a consequence of separability, asserting that every pairwise-disjoint family of open intervals is countable.2GCH is an abbreviation for the Generalized Continuum Hypothesis, asserting that 2ℵ𝛼 = ℵ𝛼+1 for every ordinal 𝛼.

1

2 ASSAF RINOT

proved [Mit73] that the nonexistence of an ℵ2-Aronszajn tree is equiconsistent with the existenceof a weakly compact cardinal.

Altogether, the four theorems crystallized the following question:

Question (folklore, see [KM78]). Does GCH entail the existence of an ℵ2-Souslin tree?If not, is the consistency strength of a negative answer a weakly compact cardinal?

Supporting the thesis of the question, Laver and Shelah [LS81] managed to construct a modelof ZFC + CH in which there are no ℵ2-Souslin trees, indeed assuming the consistency of a weaklycompact cardinal. However, GCH fails in their model.

More work connecting higher trees and weakly compact cardinals was then carried out in [Tod81],[SS82a], [SS82b], [SS88], [Tod89], yet the best known result on the original question remained thefollowing:

Theorem (Gregory, [Gre76]). If GCH holds and there are no ℵ2-Souslin trees, then ℵ2 is a Mahlocardinal in L.

In this paper, Gregory’s 1976 lower bound is increased to the anticipated value:

Theorem A. If GCH holds and there are no ℵ2-Souslin trees, then ℵ2 is weakly compact in 𝐿.

In [Tod81], Todorcevic proved that after Levy-collapsing a weakly compact cardinal to ℵ2 overa model of GCH: GCH holds, and every ℵ2-Aronszajn tree contains an ℵ1-Aronszajn subtree. Astrengthening of Theorem A, then, provides the following optimal result:

Theorem A’. If GCH holds, and ℵ2 is not weakly compact in 𝐿, then there exists an ℵ2-Souslintree with no ℵ1-Aronszajn subtrees.

We remind the reader that a regular uncountable cardinal 𝜅 is said to be Mahlo if the set ofregular cardinals below 𝜅 is stationary in 𝜅. A regular uncountable cardinal 𝜅 is said to be weaklycompact if it satisfies the generalized Ramsey partition relation: 𝜅 → (𝜅)22. By a theorem of Hanf[Han64], every weakly compact cardinal must have stationarily many Mahlo cardinals below it.

1.2. Some details. Recall that a coherent 𝐶-sequence (over a regular uncountable cardinal 𝜅) isa sequence ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ such that:

(1) for all 𝛼 < 𝜅, 𝐶𝛼 ⊆ 𝛼;(2) for all limit 𝛼 < 𝜅, 𝐶𝛼 is a club in 𝛼;(3) for all 𝛼 < 𝜅, if �� ∈ acc(𝐶𝛼), then 𝐶�� = 𝐶𝛼 ∩ ��.3

The easiest way to obtain such a sequence is to fix some club 𝐷 in 𝜅, and let 𝐶𝛼 := 𝐷 ∩ 𝛼 for all𝛼 ∈ acc(𝐷), and 𝐶𝛼 := 𝛼 ∖ sup(𝐷 ∩ 𝛼) for all other 𝛼. Of more interest is the following concept:

Definition 1.1 (Jensen, [Jen72]). �𝜆(𝐸) asserts the existence of a coherent 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩,such that otp(𝐶𝛼) ≤ 𝜆 and acc(𝐶𝛼) ∩ 𝐸 = ∅ for all 𝛼 < 𝜆+.

Write �𝜆 for �𝜆(∅).

Jensen proved [Jen72] that if there exists a stationary subset 𝐸 ⊆ 𝜆+ for which ♢(𝐸) + �𝜆(𝐸)holds,4 then there exists a 𝜆+-Souslin tree, and Solovay noticed that the existence of such an 𝐸follows from ♢(𝜆+) +�𝜆. Gregory proved [Gre76] that GCH implies ♢(𝜆+) for every cardinal 𝜆 ofuncountable cofinality, and Shelah [She79],[She81] improved this to every uncountable cardinal 𝜆.Altogether:

3Here, acc(𝐶) stands for the set of accumulation points of 𝐶, that is, acc(𝐶) := {𝛽 ∈ 𝐶 | sup(𝐶 ∩ 𝛽) = 𝛽 > 0}.4The definition of the Diamond principle may be found at the beginning of Section 3 below.

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 3

Fact 1.2 (1970’s). For every uncountable 𝜆, GCH +�𝜆 entails the existence of a 𝜆+-Souslin tree.

By results of Jensen and Solovay, the failure of �ℵ1 is equiconsistent with the existence of aMahlo cardinal, hence, in view of the goal of deriving a weakly compact cardinal, one should lookat weaker hypothesis than �ℵ1 .

One consequence of �𝜆 is that every stationary subset of 𝜆+ contains a nonreflecting stationarysubset. Baumgartner proved [Bau76b] that after forcing to Levy-collapse a weakly compact cardinal

to ℵ2, every stationary subset of 𝐸ℵ2=ℵ1

reflects, and in [Gre76], Gregory indeed managed to reduce

the hypothesis �𝜆 into that of the existence of nonreflecting stationary subset of 𝐸𝜆+

=𝜆, providedthat 𝜆 is regular.

However, ten years later, Harrington and Shelah [HS85] proved that the reflection of every

stationary subset of 𝐸ℵ2=ℵ1

is equiconsistent with the existence of a Mahlo cardinal, sending some

people back to the drawing table.

So what should we do?At the end of [KS93], Kojman and Shelah suggested to try proving that GCH entails the existence

of an ℵ2-Souslin tree, under the hypothesis that there exist two stationary subsets of 𝐸ℵ2=ℵ1

which do

not reflect simultaneously. The point here is that Magidor proved [Mag82] that the failure of suchsimultaneous reflection is equiconsistent with the existence of a weakly compact cardinal. However,as of now, this idea appears to be infertile.

Lastly, during our visit to the Erwin Schroedinger Institute in 2009, B. Konig suggested to usto try deriving an ℵ2-Souslin tree from the combination of GCH and Todorcevic’s principle �(ℵ2).Here, �(𝜅) asserts the existence of a coherent 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩, of the weakest possiblenontrivial form. That is, for every club 𝐷 in 𝜅, there exists some 𝛼 ∈ acc(𝐷) for which 𝐶𝛼 = 𝐷∩𝛼.This time, the key point is that by [Tod87], if 𝜅 is a regular uncountable cardinal and �(𝜅) fails,then 𝜅 is a weakly compact cardinal in 𝐿.

And indeed, the main result of this paper is the following.

Theorem B. If 𝜆 an uncountable cardinal, and GCH +�(𝜆+) holds, then:

(1) There exists a 𝜆+-Souslin tree which is cf(𝜆)-complete;5

(2) There exists a 𝜆+-Souslin tree which is club-regressive.6

So, Theorems A,A’ follow from Theorem B. As explained in Section 4, it also follows that thenonexistence of higher Souslin trees at two successive cardinals has considerably stronger strength:

Theorem C. If GCH holds and there are no 𝜅-Souslin trees for 𝜅 ∈ {ℵ2,ℵ3}, then the Axiom ofDeterminacy holds in 𝐿(R).

By Fact 1.2 and the main result of [Ste05], GCH and the nonexistence of an ℵ𝜔+1-Souslin treeentails that the Axiom of Determinacy holds in 𝐿(R). So the surprise here is the move down intothe realm of the ℵ𝑛 for finite 𝑛 (and away from the successors of singular cardinals).

1.3. Organization of this paper. In Section 2, we introduce a new ideal, which we denote by𝐽 [𝜅], and study its extent. It is established that if GCH holds, then for every uncountable cardinal𝜆, 𝐽 [𝜆+] contains a stationary set, yet, it is consistent that 𝐽 [ℵ2] is the nonstationary ideal over ℵ2.

In Section 3, we deal with the Diamond principle and a weak consequence of it — club hitting.

5A tree (𝑇,<𝑇 ) is said to be 𝜃-complete if any <𝑇 -increasing sequence of elements from 𝑇 , and of length < 𝜃, has anupper bound in 𝑇 .6The definition of a club-regressive tree may be found in [BR15]. For our purpose, it suffices to mention that aclub-regressive 𝜅-tree contains no 𝜈-Aronszajn subtrees nor 𝜈-Cantor subtrees for every regular cardinal 𝜈 < 𝜅.

4 ASSAF RINOT

In Section 4, we recall the principle �−(𝑆) from [BR15], and prove that for every regular un-countable cardinal 𝜅 and every stationary 𝑆 ∈ 𝐽 [𝜅], �(𝜅) + ♢(𝜅) entails �−(𝑆). This suffices toobtain an asymptotic version of Theorem B(2) that requires only a local instance of the full GCH:

Theorem D. For every cardinal 𝜆 ≥ i𝜔, CH𝜆 +�(𝜆+) entails the existence of a 𝜆+-Souslin treewhich is club-regressive.

Then, we introduce the principle �′(𝑆), and prove that �−(𝜅) + ♢(𝜅) entails �′(𝑆) for everystationary subset 𝑆 ⊆ 𝜅, from which Theorem B(1) follow. Finally, Theorems A,A’,C are derivedas corollaries.

1.4. Notations and conventions. Throughout, by a cardinal, we mean an infinite cardinal. WriteCH𝜆 to assert that 2𝜆 = 𝜆+. Denote 𝐸𝜅

𝜃 := {𝛼 < 𝜅 | cf(𝛼) = 𝜃}, and define 𝐸𝜅=𝜃, 𝐸

𝜅>𝜃 and 𝐸𝜅

≥𝜃 in a

similar fashion. Write [𝑋]𝜃 for the collection of all subsets of 𝑋 of cardinality 𝜃. A dense subfamilyof [𝑋]𝜃 is a collection ℱ ⊆ [𝑋]𝜃 such that for all 𝑌 ∈ [𝑋]𝜃, there exists some 𝑍 ∈ ℱ with 𝑍 ⊆ 𝑌 .The least size of such a dense subfamily is denoted by 𝒟(𝑋, 𝜃).

For sets of ordinals 𝐶,𝐷, write 𝐷 ⊑ 𝐶 iff there exists some ordinal 𝛽 such that 𝐷 = 𝐶 ∩ 𝛽, thatis, 𝐶 end-extends 𝐷. Denote nacc(𝐶) := 𝐶 ∖ acc(𝐶).

2. The regressive functions ideal

Recall that a function 𝑓 : 𝛼→ 𝛼 is said to be regressive iff 𝑓(𝛽) < 𝛽 for all nonzero 𝛽 < 𝛼.

Definition 2.1. For a regular uncountable cardinal 𝜅, define a collection 𝐽 [𝜅], as follows.A subset 𝑆 ⊆ 𝜅 is in 𝐽 [𝜅] iff there exists a club 𝐶 ⊆ 𝜅 and a sequence of functions ⟨𝑓𝑖 : 𝜅→ 𝜅 | 𝑖 < 𝜅⟩

satisfying the following. For every 𝛼 ∈ 𝑆∩𝐶, every regressive function 𝑓 : 𝛼→ 𝛼, and every cofinalsubset 𝐵 ⊆ 𝛼, there exists some 𝑖 < 𝛼 such that sup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝛼.

It is easy to see that 𝐽 [𝜅] is a 𝜅-complete normal ideal. We now turn to study its extent.

2.1. Positive results.

Proposition 2.2. Suppose that 𝜃 < cf(𝜆) ≤ 𝜆 are cardinals, and 𝒟(𝜆, 𝜃) = 𝜆.

Then 𝐸𝜆+

𝜃 ∈ 𝐽 [𝜆+].

Proof. If 𝜃 is singular, then 𝐸𝜆+

𝜃 = ∅ ∈ 𝐽 [𝜆+], so suppose that 𝜃 is regular. Fix a family {𝐴𝛼 |𝛼 < 𝜆+} ⊆ [𝜆]cf(𝜆) such that |𝐴𝛼 ∩ 𝐴𝛽| < cf(𝜆) for all 𝛼 < 𝛽 < 𝜆+.7 Let {𝑔𝑖 | 𝑖 < 𝜆} be a dense

subfamily of [𝜆 × 𝜆]𝜃. For all 𝛽 < 𝜆+, fix a bijection 𝜓𝛽 : 𝜆 ↔ max{𝜆, 𝛽}. Finally, for all 𝑖 < 𝜆,define a function 𝑓𝑖 : 𝜆+ → 𝜆+ as follows. For all 𝛽 < 𝜆+, if there exists a unique (𝛾, 𝛿) ∈ 𝑔𝑖 suchthat 𝛾 ∈ 𝐴𝛽, then let 𝑓𝑖(𝛽) := 𝜓𝛽(𝛿). Otherwise, let 𝑓𝑖(𝛽) := 0.

We claim that the club 𝐶 := 𝜆+ ∖ 𝜆 and the sequence ⟨𝑓𝑖 : 𝜆+ → 𝜆+ | 𝑖 < 𝜆⟩ together witness

that 𝐸𝜆+

𝜃 ∈ 𝐽 [𝜆+].8 To see this, fix an arbitrary 𝛼 ∈ 𝐸𝜆+

𝜃 ∩ 𝐶 along with a regressive function𝑓 : 𝛼 → 𝛼, and a cofinal subset 𝐵 of 𝛼. Let 𝐵′ be a cofinal subset of 𝐵 ∖ {0} of order-type 𝜃.Since 𝜃 < cf(𝜆), for all 𝛽 ∈ 𝐵′, we may pick 𝛾𝛽 ∈ 𝐴𝛽 ∖

⋃{𝐴𝜂 | 𝜂 ∈ 𝐵′, 𝜂 = 𝛽}. Also, for all

𝛽 ∈ 𝐵′, since 𝑓(𝛽) < 𝛽 ⊆ Im(𝜓𝛽), we may let 𝛿𝛽 := 𝜓−1𝛽 (𝑓(𝛽)). Put 𝑔 := {(𝛾𝛽, 𝛿𝛽) | 𝛽 ∈ 𝐵′}.

As 𝛽 ↦→ 𝛾𝛽 is one-to-one over 𝐵′, we get that 𝑔 ∈ [𝜆 × 𝜆]𝜃, and hence we may find some 𝑖 < 𝜆such that 𝑔𝑖 ⊆ 𝑔. Let 𝐵′′ := {𝛽 ∈ 𝐵′ | (𝛾𝛽, 𝛿𝛽) ∈ 𝑔𝑖}. Since 𝐵′′ ⊆ 𝐵′ and |𝐵′′| = otp(𝐵′),we get that sup(𝐵′′) = sup(𝐵′) = 𝛼. Thus, it suffices to verify that 𝑓 � 𝐵′′ = 𝑓𝑖 � 𝐵′′. Let

7See, e.g., [Bau76a, Theorem 2.3].8This is not a typing error; we simply settle for a sequence of length 𝜆.

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 5

𝛽 ∈ 𝐵′′ be arbitrary. Since (𝛾𝛽, 𝛿𝛽) is the unique pair in 𝑔𝑖 to satisfy 𝛾𝛽 ∈ 𝐴𝛽, we conclude that

𝑓𝑖(𝛽) = 𝜓𝛽(𝛿𝛽) = 𝜓𝛽(𝜓−1𝛽 (𝑓(𝛽))) = 𝑓(𝛽), as sought. �

In particular, GCH entails that 𝐸𝜆+

<𝜆 ∈ 𝐽 [𝜆+] for every regular cardinal 𝜆. To deal with the caseof 𝜆 singular, let us recall the definition of Shelah’s approachability ideal :

Definition 2.3 (Shelah, [She93]). For a regular uncountable cardinal 𝜅, define a collection 𝐼[𝜅],as follows. A subset 𝑆 ⊆ 𝜅 is in 𝐼[𝜅] iff there exists a club 𝐶 ⊆ 𝜅 and a sequence ⟨𝑎𝑖 | 𝑖 < 𝜅⟩ ofbounded subsets of 𝜅 satisfying the following. For every 𝛼 ∈ 𝑆 ∩ 𝐶, there exists a cofinal subset𝐵 ⊆ 𝛼 of order-type cf(𝛼) < 𝛼 such that {𝐵 ∩ 𝑖 | 𝑖 < 𝛼} ⊆ {𝑎𝑖 | 𝑖 < 𝛼}.

Proposition 2.4. Suppose that cf(𝜆) < 𝜃 < 𝜆 are cardinals, and 𝒟(𝜆, 𝜃) = 𝜆.

For every 𝑆 ⊆ 𝐸𝜆+

𝜃 , if 𝑆 ∈ 𝐼[𝜆+], then 𝑆 ∈ 𝐽 [𝜆+].

Proof. By Shelah’s celebrated theorem from [She94a, Theorem 1.5], let us fix a scale ℎ = ⟨ℎ𝛼 | 𝛼 < 𝜆+⟩for 𝜆. This means, in particular, that for all 𝛼 < 𝜆+, ℎ𝛼 is a function from cf(𝜆) to 𝜆. Fix a bijection𝜋 : cf(𝜆) × 𝜆↔ 𝜆, and let 𝐴𝛼 := 𝜋[ℎ𝛼] for all 𝛼 < 𝜆+.

Suppose that we are given 𝑆 ⊆ 𝐸𝜆+

𝜃 in 𝐼[𝜆+]. To avoid trivialities, assume that 𝑆 is stationary.

In particular, 𝜃 is regular. By 𝑆 ∈ 𝐼[𝜆+] and [CFM04, Corollary 2.15],9 there exists a club 𝐶 ⊆ 𝜆+

such that every 𝛼 ∈ 𝑆 ∩ 𝐶 is good for the scale ℎ. By [Eis10, Theorem 3.50], this means that forevery 𝛼 ∈ 𝑆 ∩ 𝐶 and every cofinal 𝐵 ⊆ 𝛼, there exists some cofinal 𝐵′ ⊆ 𝐵 of order-type 𝜃 and𝑗′ < cf(𝜆) such that ℎ𝛽(𝑗) < ℎ𝛾(𝑗) for all 𝛽 < 𝛾 both from 𝐵′ and all 𝑗 ∈ (𝑗′, cf(𝜆)).

Claim 2.4.1. Suppose that 𝛼 ∈ 𝑆 ∩ 𝐶, and 𝐵 is a cofinal subset of 𝛼.Then there exists some cofinal 𝐵′ ⊆ 𝐵 of order-type 𝜃 for which the following is nonempty:∏

𝛽∈𝐵′

𝐴𝛽 ∖⋃

{𝐴𝜂 | 𝜂 ∈ 𝐵′, 𝜂 = 𝛽}.

Proof. By 𝛼 ∈ 𝑆 ∩ 𝐶, 𝛼 is good, so let us pick some cofinal 𝐵′ ⊆ 𝐵 of order-type 𝜃 and 𝑗 < cf(𝜆)such that ℎ𝛽(𝑗) < ℎ𝛾(𝑗) for all 𝛽 < 𝛾 both from 𝐵′. Then ⟨𝜋(𝑗, ℎ𝛽(𝑗)) | 𝛽 ∈ 𝐵′⟩ is an element of∏

𝛽∈𝐵′ 𝐴𝛽 ∖⋃{𝐴𝜂 | 𝜂 ∈ 𝐵′, 𝜂 = 𝛽}. �

Now, continue as in the proof of Proposition 2.2, but using the sequence ⟨𝐴𝛼 | 𝛼 < 𝜆+⟩ we justconstructed. �

Corollary 2.5. Suppose that 𝜃 is a regular cardinal.For every cardinal 𝜆 ≥ 𝒟(𝜃, 𝜃), the following are equivalent:

(1) 𝒟(𝜆, 𝜃) = 𝜆;

(2) there exists some stationary subset 𝑆 ⊆ 𝐸𝜆+

𝜃 with 𝑆 ∈ 𝐽 [𝜆+].

Proof. (1) =⇒ (2): Suppose that 𝜆 is a cardinal, satisfying 𝒟(𝜆, 𝜃) = 𝜆. In particular, 𝜃 = cf(𝜆).

I If 𝜃 < cf(𝜆), then 𝐸𝜆+

𝜃 ∈ 𝐽 [𝜆+] by Proposition 2.2.

I If 𝜃 > cf(𝜆), then 𝜆 is singular, and by [She93, S1], there exists some stationary 𝑆 ⊆ 𝐸𝜆+

𝜃 suchthat 𝑆 ∈ 𝐼[𝜆+]. Now, appeal to Proposition 2.4.

(2) =⇒ (1): Suppose that 𝜆 ≥ 𝒟(𝜃, 𝜃) is some cardinal, and 𝐽 [𝜆+] contains a stationary subset

of 𝐸𝜆+

𝜃 . In particular, there exists a sequence of functions ⟨𝑓𝑖 : 𝜆+ → 𝜆+ | 𝑖 < 𝜆+⟩ and an ordinal

𝛼 ∈ 𝐸𝜆+

𝜃 with 𝛼 > 𝜆 such that for every regressive function 𝑓 : 𝛼 → 𝛼, and every cofinal subset𝐵 ⊆ 𝛼, there exists some 𝑖 < 𝛼 such that sup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝛼.

9See also [FM97, Claim 4.4].

6 ASSAF RINOT

Fix a cofinal subset 𝐵 ⊆ 𝛼 of order-type 𝜃, with min(𝐵) > 𝜆. For all 𝑖 < 𝛼, let 𝑌𝑖 := 𝑓𝑖[𝐵] ∪ 𝜃.As |𝑌𝑖| = 𝜃 and 𝒟(𝜃, 𝜃) ≤ 𝜆, pick ℱ𝑖 of size ≤ 𝜆 which is a dense subfamily of [𝑌𝑖]

𝜃. Then,ℱ := [𝜆]𝜃 ∩

⋃{ℱ𝑖 | 𝑖 < 𝛼} has size ≤ 𝜆. To see that ℱ is a dense subfamily of [𝜆]𝜃, let 𝑌 be an

arbitrary element of [𝜆]𝜃. Pick a regressive function 𝑓 : 𝛼 → 𝛼 such that 𝑓 � 𝐵 is a bijection from𝐵 to 𝑌 . Fix 𝑖 < 𝛼 for which 𝐵𝑖 := {𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} is cofinal in 𝛼. As otp(𝐵) = cf(𝛼),we have |𝐵𝑖| = 𝜃, so that |𝑌𝑖 ∩ 𝑌 | = 𝜃. By 𝑌𝑖 ∩ 𝑌 ∈ [𝑌𝑖]

𝜃 and the fact that ℱ𝑖 ⊆ ℱ , we infer theexistence of 𝑍 ∈ ℱ such that 𝑍 ⊆ 𝑌 . �

Definition 2.6 (Cummings-Foreman-Magidor, [CFM01]). ADS𝜆 asserts the existence of a sequence⟨𝐴𝛼 | 𝛼 < 𝜆+⟩ such that:

(1) 𝐴𝛼 is a cofinal subset of 𝜆 of order-type cf(𝜆);(2) For all 𝛽 < 𝜆+, there exists some 𝑔 : 𝛽 → 𝜆 such that the sequence ⟨𝐴𝛼 ∖ 𝑔(𝛼) | 𝛼 < 𝛽⟩

consists of pairwise disjoint sets.

Proposition 2.7. Suppose that ADS𝜆 holds for a given singular strong limit cardinal 𝜆.

Then 𝐸𝜆+

=cf(𝜆) ∈ 𝐽 [𝜆+].

Proof sketch. Since 𝐽 [𝜆+] is 𝜆+-complete, it suffices to prove that 𝐸𝜆+

𝜃 ∈ 𝐽 [𝜆+] for every regularcardinal 𝜃 < 𝜆 with 𝜃 = cf(𝜆). Fix such a cardinal 𝜃. Since 𝜆 is a strong limit and 𝜃 = cf(𝜆), wehave 𝒟(𝜆, 𝜃) = 𝜆 (cf. the proof of Claim 4.5.1). Then, continue as in the proof of Proposition 2.2,but using the sequence ⟨𝐴𝛼 | 𝛼 < 𝜆+⟩ witnessing ADS𝜆, instead. �

It follows that the model of [GS08] admits a singular cardinal 𝜆 for which 𝐽 [𝜆+] ⊆ 𝐼[𝜆+].

2.2. Negative results.

Proposition 2.8. Suppose that 𝜆 is an uncountable cardinal.

For every 𝑆 ∈ 𝐽 [𝜆+], we have that 𝑆 ∩ 𝐸𝜆+

cf(𝜆) is nonstationary.

Proof. Suppose not. In particular, there exists a sequence of regressive functions ⟨𝑓𝑖 : 𝜆+ → 𝜆+ | 𝑖 < 𝜆+⟩and an ordinal 𝛼 ∈ 𝐸𝜆+

cf(𝜆) with 𝛼 > 𝜆 such that for every regressive function 𝑓 : 𝛼 → 𝛼, and every

cofinal subset 𝐵 ⊆ 𝛼, there exists some 𝑖 < 𝛼 such that sup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝛼.For all 𝛽 < 𝜆+, fix a bijection 𝜓𝛽 : 𝜆↔ max{𝜆, 𝛽}. For all 𝑖 < 𝛼, define 𝑔𝑖 : 𝛼→ 𝜆 by stipulating:

𝑔𝑖(𝛽) := 𝜓−1𝛽 (𝑓𝑖(𝛽)).

Let ⟨𝜆𝑗 | 𝑗 < cf(𝜆)⟩ be a strictly increasing sequence of ordinals converging to 𝜆. Let ⟨𝛽𝑗 | 𝑗 < cf(𝜆)⟩be a strictly increasing sequence of ordinals converging to 𝛼, with 𝛽0 > 𝜆. Now, pick some function𝑔 : 𝛼→ 𝜆 that satisfies for all 𝑗 < cf(𝜆) :

𝑔(𝛽𝑗) := min(𝜆 ∖ {𝑔𝑖(𝛽𝑗) | 𝑖 ∈ 𝜓𝛼[𝜆𝑗 ]}).

Define a regressive function 𝑓 : 𝛼→ 𝛼 by stipulating:

𝑓(𝛽) :=

{0, if 𝛽 < 𝜆;

𝜓𝛽(𝑔(𝛽)), otherwise.

Now, by the choice of ⟨𝑓𝑖 | 𝑖 < 𝜆+⟩, we may find some 𝑖 < 𝛼 such that

sup{𝛽𝑗 | 𝑗 < cf(𝜆), 𝑓𝑖(𝛽𝑗) = 𝑓(𝛽𝑗)} = 𝛼.

In particular, sup{𝑗 < cf(𝜆) | 𝑔𝑖(𝛽𝑗) = 𝑔(𝛽𝑗)} = cf(𝜆). By 𝑖 ∈ 𝜓𝛼[𝜆], let us fix a large enough𝑗* < cf(𝜆) such that 𝑖 ∈ 𝜓𝛼[𝜆𝑗* ]. By definition of 𝑔, then, 𝑔(𝛽𝑗) = 𝑔𝑖(𝛽𝑗) for all 𝑗 ∈ [𝑗*, cf(𝜆)). Thisis a contradiction. �

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 7

Corollary 2.9. If non(ℳ) > ℵ1, then 𝐽 [ℵ2] is the nonstationary ideal over ℵ2.10

Proof. Towards a contradiction, suppose that 𝐽 [ℵ2] is not the nonstationary ideal over ℵ2. By

Proposition 2.8, there exists a sequence of functions ⟨𝑓𝑖 : ℵ2 → ℵ2 | 𝑖 < ℵ2⟩ and an ordinal 𝛼 ∈ 𝐸ℵ2ℵ0

with 𝛼 > 𝜔1 such that for every regressive function 𝑓 : 𝛼 → 𝛼, and every cofinal subset 𝐵 ⊆ 𝛼,there exists some 𝑖 < 𝛼 such that sup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝛼.

Let ⟨𝛽𝑛 | 𝑛 < 𝜔⟩ be a strictly increasing sequence of ordinals converging to 𝛼, with 𝛽0 > 𝜔. Forall 𝑖 < 𝛼, define a real 𝑟𝑖 : 𝜔 → 𝜔 by stipulating:

𝑟𝑖(𝑛) :=

{𝑓𝑖(𝛽𝑛), if 𝑓𝑖(𝛽𝑛) < 𝜔;

0, otherwise.

By non(ℳ) > ℵ1 = |𝛼| and [Bar87, S2], we may pick a real 𝑟 : 𝜔 → 𝜔 such that for all 𝑖 < 𝛼,𝑟(𝑛) = 𝑟𝑖(𝑛) for all but finitely many 𝑛 < 𝜔. Pick a regressive function 𝑓 : 𝛼 → 𝛼 such that𝑓(𝛽𝑛) = 𝑟(𝑛) for all 𝑛 < 𝜔. Put 𝐵 := {𝛽𝑛 | 𝑛 < 𝜔}. Then, there exists no 𝑖 < 𝛼 such thatsup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝛼. �

Proposition 2.10. For every inaccessible cardinal 𝜅, 𝐽 [𝜅] is the nonstationary ideal over 𝜅.

Proof. Suppose not. In particular, we may fix an inaccessible cardinal 𝜅, a sequence of functions⟨𝑓𝑖 : 𝜅→ 𝜅 | 𝑖 < 𝜅⟩, and an uncountable limit cardinal 𝜇 satisfying the following. For every re-gressive function 𝑓 : 𝜇 → 𝜇, and every cofinal subset 𝐵 ⊆ 𝜇, there exists some 𝑖 < 𝜇 such thatsup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} = 𝜇.

Let 𝐴 denote the set of cardinals below 𝜇. Pick a regressive function 𝑓 : 𝜇→ 𝜇 such that for all𝜆 ∈ 𝐴:

𝑓(𝜆+) := min(𝜆+ ∖ {𝑓𝑖(𝜆+) | 𝑖 ≤ 𝜆}).

Put 𝐵 := {𝜆+ | 𝜆 ∈ 𝐴}. Then 𝐵 is a cofinal subset of 𝜇. For all 𝑖 < 𝜇, we have

sup{𝛽 ∈ 𝐵 | 𝑓𝑖(𝛽) = 𝑓(𝛽)} ≤ |𝑖|+ < 𝜇,

contradicting the choice of ⟨𝑓𝑖 | 𝑖 < 𝜅⟩. �

3. Diamond and club hitting

Definition 3.1 (Jensen, [Jen72]). For a stationary subset 𝑇 of a regular uncountable cardinal 𝜅,♢(𝑇 ) asserts the existence of a sequence ⟨𝐴𝛾 | 𝛾 < 𝜅⟩ such that for every 𝐴 ⊆ 𝜅, the intersection𝐺(𝐴) ∩ 𝑇 is stationary, where 𝐺(𝐴) := {𝛾 < 𝜅 | 𝐴 ∩ 𝛾 = 𝐴𝛾}.

Proposition 3.2 (folklore). Suppose that 𝑇 is a stationary subset of a regular uncountable cardinal𝜅, and ♢(𝑇 ) holds. Then there exists a matrix

⟨𝐴𝑖

𝛾 | 𝑖 < 𝜅, 𝛾 < 𝜅⟩, such that for every sequence

�� =⟨𝐴𝑖 | 𝑖 < 𝜅

⟩of cofinal subsets of 𝜅, the intersection 𝐺(��) ∩ 𝑇 is stationary, where

𝐺(��) := {𝛾 < 𝜅 | ∀𝑖 < 𝛾(sup(𝐴𝑖 ∩ 𝛾) = 𝛾 & 𝐴𝑖 ∩ 𝛾 = 𝐴𝑖𝛾)}.

Proof. Let ⟨𝐴𝛾 | 𝛾 < 𝜅⟩ be a witness to ♢(𝑇 ). Fix a bijection 𝜋 : 𝜅 × 𝜅 ↔ 𝜅. For all 𝑖 < 𝜅 and

𝛾 < 𝜅, let 𝐴𝑖𝛾 := {𝛿 < 𝛾 | 𝜋(𝑖, 𝛿) ∈ 𝐴𝛾}. To see that

⟨𝐴𝑖

𝛾 | 𝑖 < 𝜅, 𝛾 < 𝜅⟩

works, let �� =⟨𝐴𝑖 | 𝑖 < 𝜅

⟩be an arbitrary sequence of cofinal subsets of 𝜅. For all 𝑖 < 𝜅, 𝐷𝑖 := {𝛾 < 𝜅 | sup(𝐴𝑖 ∩ 𝛾) = 𝛾} isa club, and hence 𝐷 := {𝛾 ∈ △𝑖<𝜅𝐷𝑖 | 𝜋[𝛾 × 𝛾] = 𝛾} is a club. Put 𝐴 := {𝜋(𝑖, 𝛿) | 𝑖 < 𝜅, 𝛿 ∈ 𝐴𝑖}.

Then 𝐺 := 𝐺(𝐴) ∩ 𝑇 ∩𝐷 is stationary, and 𝐺 ⊆ 𝐺(��) ∩ 𝑇 , so we are done. �

Fact 3.3 (Shelah, [She10]). For every uncountable cardinal 𝜆, CH𝜆 entails ♢(𝜆+).

10Here, “non(ℳ) > ℵ1” stands for the assertion that every ℵ1-sized set of reals is meager. This assertion is consistentwith ZFC + ¬CH, as witnessed by the model of [ST71].

8 ASSAF RINOT

Definition 3.4. Suppose that 𝑆 is a stationary subset of a regular uncountable cardinal 𝜅.A sequence ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is said to hit clubs at 𝑆 iff for every club 𝐷 ⊆ 𝜅, there exists some 𝛼 ∈ 𝑆

such that sup(nacc(𝐶𝛼) ∩𝐷) = 𝛼.

Proposition 3.5. Suppose that 𝜅 is a regular cardinal ≥ ℵ2.Then �(𝜅) holds iff for every stationary 𝑆 ⊆ 𝜅, there exists a coherent 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩,

that hits clubs at 𝑆.

Proof. (⇐=) Suppose that ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a coherent 𝐶-sequence that hits clubs at 𝑆 := acc(𝜅).Let 𝐶 be an arbitrary club in 𝜅, and we shall find some 𝛼 ∈ acc(𝐶) such that 𝐶𝛼 = 𝐶 ∩ 𝛼.

Consider the club 𝐷 := acc(𝐶). Pick 𝛼 ∈ 𝑆 such that sup(nacc(𝐶𝛼) ∩ 𝐷) = 𝛼. Then 𝛼 ∈acc(𝐷) ⊆ acc(𝐶) and nacc(𝐶𝛼) ∩ acc(𝐶) = ∅, so that 𝐶𝛼 = 𝐶 ∩ 𝛼.

(=⇒) Suppose that �(𝜅) holds, as witnessed by ⟨𝐶𝛼 | 𝛼 < 𝜅⟩. Let 𝑆 be an arbitrary stationarysubset of 𝜅. The proof is a simple combination of ideas from [She94b, S2] and [Rin14a, S3].

For every club 𝐸 ⊆ 𝜅 and 𝛼 < 𝜅, denote

𝐶𝛼[𝐸] := {sup(𝐸 ∩ 𝜂) | 𝜂 ∈ (𝐶𝛼 ∖ (min(𝐸) + 1))}.Notice that if 𝛼 ∈ acc(𝐸), then 𝐶𝛼[𝐸] is a club in 𝛼, and acc(𝐶𝛼[𝐸]) = acc(𝐶𝛼) ∩ acc(𝐸).

Claim 3.5.1. There exists a club 𝐸 ⊆ 𝜅 satisfying the following.For every club 𝐷 ⊆ 𝜅, there exists some 𝛿 ∈ 𝑆 such that sup(nacc(𝐶𝛿[𝐸]) ∩𝐷) = 𝛿.

Proof. Suppose not. Then, we can recursively construct a ⊆-decreasing sequence of clubs in 𝜅,⟨𝐷𝑖 | 𝑖 < 𝜔1⟩, such that 𝐷0 = 𝜅, 𝐷𝑖 =

⋂𝑗<𝑖𝐷𝑗 for all nonzero limit 𝑖 < 𝜔1, and for all 𝑖 < 𝜔1 and

𝛿 ∈ 𝑆,

sup(nacc(𝐶𝛿[𝐷𝑖]) ∩𝐷𝑖+1) < 𝛿.

As 𝜅 = cf(𝜅) ≥ ℵ2, 𝐷 :=⋂

𝑖<𝜔1𝐷𝑖 is a club. We claim that 𝐵 := {𝛿 ∈ 𝑆 | sup((𝐷∩ 𝛿) ∖𝐶𝛿) = 𝛿}

is nonempty.Suppose not. Then there exists some 𝜀 < 𝜅 for which

𝐻 := {𝛿 ∈ 𝑆 | sup((𝐷 ∩ 𝛿) ∖ 𝐶𝛿) = 𝜀} ∩ acc(𝐷 ∖ 𝜀)is stationary in 𝜅. Consequently, for every 𝛼 < 𝛿 both in 𝐻, we have 𝛼 ∈ acc(𝐶𝛿) and hence 𝐶𝛼 ⊑𝐶𝛿. So {𝐶𝛿 | 𝛿 ∈ 𝐻} is an ⊑-chain, converging to the club 𝐶 :=

⋃{𝐶𝛿 | 𝛿 ∈ 𝐻}. Let 𝛼 ∈ acc(𝐶) be

arbitrary. Then 𝛼 ∈ acc(𝐶𝛿) for some 𝛿 ∈ 𝐻, and then 𝐶 ∩ 𝛼 = (𝐶 ∩ 𝛿) ∩ 𝛼 = 𝐶𝛿 ∩ 𝛼 = 𝐶𝛼. Thatis, 𝐶 contradicts the fact that ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a �(𝜅)-sequence.

Thus, we have established that 𝐵 is nonempty. Pick 𝛿 ∈ 𝐵. For all 𝑖 < 𝜔1, by the choice of𝐷𝑖+1, we have that 𝛿𝑖 := sup(nacc(𝐶𝛿[𝐷𝑖]) ∩𝐷𝑖+1) is < 𝛿. We consider two cases:I If cf(𝛿) > ℵ0, then by 𝛿 ∈ 𝐵, let us pick 𝛽 ∈ (𝐷∩𝛿)∖𝐶𝛿 above sup𝑛<𝜔 𝛿𝑛. Put 𝛾 := min(𝐶𝛿∖𝛽).

Then 𝛿 > 𝛾 > 𝛽, and for all 𝑛 < 𝜔, since 𝛽 ∈ 𝐷𝑛, we infer that sup(𝐷𝑛 ∩ 𝛾) ≥ 𝛽. So min(𝐶𝛿[𝐷𝑛] ∖𝛽) = sup(𝐷𝑛 ∩ 𝛾) for all 𝑛 < 𝜔. Since {𝐷𝑛 | 𝑛 < 𝜔} is a decreasing chain, there exists some 𝑛 < 𝜔such that sup(𝐷𝑛∩𝛾) = sup(𝐷𝑛+1∩𝛾). Fix such an 𝑛. Then min(𝐶𝛿[𝐷𝑛]∖𝛽) = min(𝐶𝛿[𝐷𝑛+1]∖𝛽),and in particular, 𝛽* := min(𝐶𝛿[𝐷𝑛] ∖ 𝛽) is in 𝐷𝑛+1.

∙ If 𝛽* ∈ nacc(𝐶𝛿[𝐷𝑛]), then we get a contradiction to the fact that 𝛽* ≥ 𝛽 > 𝛿𝑛.∙ If 𝛽* ∈ acc(𝐶𝛿[𝐷𝑛]), then 𝛽* = 𝛽 and 𝛽* ∈ acc(𝐶𝛿), contradicting the fact that 𝛽 ∈ 𝐶𝛿.

I If cf(𝛿) = ℵ0, then we may pick an uncountable 𝐼 ⊆ 𝜔1, for which sup𝑖∈𝐼 𝛿𝑖 < 𝛿.11 By𝛿 ∈ 𝐵, pick 𝛽 ∈ (𝐷 ∩ 𝛿) ∖ 𝐶𝛿 above sup𝑖∈𝐼 𝛿𝑖, and put 𝛾 := min(𝐶𝛿 ∖ 𝛽). Then 𝛿 > 𝛾 > 𝛽,and min(𝐶𝛿[𝐷𝑖] ∖ 𝛽) = sup(𝐷𝑖 ∩ 𝛾) for all 𝑖 < 𝜔1. Since {𝐷𝑖 | 𝑖 < 𝜔1} is a decreasing chain

11See Fact 1.3 of [Rin10].

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 9

and 𝐼 is cofinal in 𝜔1, we may fix an 𝑖 ∈ 𝐼 such that sup(𝐷𝑖 ∩ 𝛾) = sup(𝐷𝑖+1 ∩ 𝛾). Thenmin(𝐶𝛿[𝐷𝑖] ∖ 𝛽) = min(𝐶𝛿[𝐷𝑖+1] ∖ 𝛽), and in particular, 𝛽* := min(𝐶𝛿[𝐷𝑖] ∖ 𝛽) is in 𝐷𝑖+1.

∙ If 𝛽* ∈ nacc(𝐶𝛿[𝐷𝑖]), then we get a contradiction to the fact that 𝛽* ≥ 𝛽 > 𝛿𝑖.∙ If 𝛽* ∈ acc(𝐶𝛿[𝐷𝑖]), then 𝛽* = 𝛽 and 𝛽* ∈ acc(𝐶𝛿), contradicting the fact that 𝛽 ∈ 𝐶𝛿.

�

Let 𝐸 be given by Claim 3.5.1. For all 𝛼 < 𝜅, define

𝐷𝛼 :=

{𝐶𝛼[𝐸], if 𝛼 ∈ acc(𝐸);

𝛼 ∖ sup(𝐸 ∩ 𝛼), if 𝛼 ∈ acc(𝐸).

Claim 3.5.2. ⟨𝐷𝛼 | 𝛼 < 𝜅⟩ is a coherent 𝐶-sequence.

Proof. Let 𝛼 < 𝜅 be an arbitrary limit ordinal. It is easy to see that 𝐷𝛼 is a club in 𝛼. Supposethat we are given �� ∈ acc(𝐷𝛼). We shall prove that 𝐷�� = 𝐷𝛼 ∩ ��.

Put 𝜏 := sup(𝐸 ∩ 𝛼), and consider two cases:I If 𝜏 < 𝛼, then 𝐷𝛼 is the interval [𝜏, 𝛼), and (𝜏, 𝛼)∩𝐸 = ∅. So, �� ∈ (𝜏, 𝛼) and sup(𝐸 ∩ ��) = 𝜏 .

Consequently, 𝐷𝛼 ∩ �� = 𝐷��.I If 𝜏 = 𝛼, then 𝐷𝛼 is the club 𝐶𝛼[𝐸], and by �� ∈ acc(𝐷𝛼), we have �� ∈ acc(𝐶𝛼) ∩ acc(𝐸),

meaning that 𝐶�� = 𝐶𝛼 ∩ �� and 𝐷�� = 𝐶��[𝐸]. Consequently, 𝐷𝛼 ∩ �� = 𝐷��. �

Finally, let 𝐷 be an arbitrary club in 𝜅. By the choice of 𝐸, let us pick 𝛼 ∈ 𝑆 such thatsup(nacc(𝐶𝛼[𝐸])∩𝐷) = 𝛼. In particular, 𝐷𝛼 = 𝐶𝛼[𝐸], and sup(nacc(𝐷𝛼)∩𝐷) = 𝛼, as sought. �

4. Main Results

An examination of all ♢-based constructions of 𝜅-Souslin trees in the literature (e.g., [Jen72],[Gre76], [Dev83], [She84], [BDS86], [Dav90], [KS93] ,[Cum97], [Sch05], [KLY07], and [Rin11b])reveals that they all involve an ingredient of sealing antichains at some nonreflecting stationary setof levels of the 𝜅-tree. As here, we are interested in deriving more than a Mahlo cardinal, we mustdevise a method of deriving a 𝜅-Souslin tree from principles that are compatible with reflection. Ayear ago, in a joint work with Brodsky, we came up with the following candidate:

Definition 4.1 (Brodsky-Rinot, [BR15]). For a regular uncountable cardinal 𝜅, and a stationarysubset 𝑆 ⊆ 𝜅, �−(𝑆) asserts the existence of a coherent 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩, such that forevery cofinal subset 𝐴 ⊆ 𝜅, there exists some 𝛼 ∈ 𝑆 for which sup(nacc(𝐶𝛼) ∩𝐴) = 𝛼.

Fact 4.2 (Brodsky-Rinot, [BR15]). Suppose 𝜅 is a regular uncountable cardinal, and ♢(𝜅) holds.

(a) If �−(𝜅) holds, then there exists a club-regressive 𝜅-Souslin tree;(b) If �−(𝐸𝜅

≥𝜃) holds, and 𝜆<𝜃 < 𝜅 for all 𝜆 < 𝜅, then there exists a 𝜃-complete 𝜅-Souslin tree.

By [Rin14b, Theorem A], if 𝜆 is an uncountable cardinal and CH𝜆 holds, then �𝜆 entails �−(𝑆)

for every stationary 𝑆 ⊆ 𝐸𝜆+

=cf(𝜆). One goal of the current section is to relax �𝜆 to �(𝜆+).12

Theorem 4.3. Suppose that 𝜅 is a regular uncountable cardinal, and �(𝜅) + ♢(𝜅) holds.Then �−(𝑆) holds for every stationary 𝑆 ∈ 𝐽 [𝜅].

Proof. Suppose that 𝑆 ∈ 𝐽 [𝜅] is a given stationary set. Fix a club 𝐶 ⊆ 𝜅 and a sequence of regressivefunctions ⟨𝑓𝑖 : 𝜅→ 𝜅 | 𝑖 < 𝜅⟩ as in Definition 2.1. By ♢(𝜅), let us fix a matrix

⟨𝐴𝑖

𝛾 | 𝑖 < 𝜅, 𝛾 < 𝜅⟩

as in Proposition 3.2. For all 𝑖, 𝛽 < 𝜅, put

𝑋𝑖𝛽 := {𝛽} ∪𝐴𝑖

𝑓𝑖(𝛽).

12In light of Proposition 2.7, we mention that by [She82, p. 440],�𝜆 entails ADS𝜆.

10 ASSAF RINOT

Since 𝑆 is stationary, we infer from Proposition 2.8 that 𝜅 ≥ ℵ2. So, by �(𝜅), appeal toProposition 3.5 to obtain a coherent 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩, that hits clubs at 𝑆 ∩ 𝐶. Finally,for all 𝑖, 𝛼 < 𝜅, let

𝐶𝑖𝛼 := 𝐶𝛼 ∪ {min(𝑋𝑖

𝛽 ∖ (sup(𝐶𝛼 ∩ 𝛽) + 1)) | 𝛽 ∈ nacc(𝐶𝛼) & 𝛽 > 0}.

Claim 4.3.1. Let 𝑖 < 𝜅 be arbitrary. Then⟨𝐶𝑖𝛼 | 𝛼 < 𝜅

⟩is a coherent 𝐶-sequence.

Proof. Let 𝛼 < 𝜅 be an arbitrary limit ordinal. By 𝐶𝛼 ⊆ 𝐶𝑖𝛼, we have sup(𝐶𝑖

𝛼) = 𝛼. For allsuccessive points 𝛽− < 𝛽 from 𝐶𝛼, the relative interval (𝛽−, 𝛽)∩𝐶𝑖

𝛼 contains at most one element,thus, as 𝐶𝛼 is closed below 𝛼, so does 𝐶𝑖

𝛼. Now, suppose that �� ∈ acc(𝐶𝑖𝛼). Then �� ∈ acc(𝐶𝛼), and

hence 𝐶�� = 𝐶𝛼∩��. Then the local nature of the definition of 𝐶𝑖𝛼 makes it clear that 𝐶𝑖

𝛼∩�� = 𝐶𝑖��. �

Claim 4.3.2. There exists an 𝑖 < 𝜅 for which⟨𝐶𝑖𝛼 | 𝛼 < 𝜅

⟩is a �−(𝑆)-sequence.

Proof. Suppose not. It then follows from Claim 4.3.1 that there exists a sequence of cofinal subsets

of 𝜅, �� =⟨𝐴𝑖 | 𝑖 < 𝜅

⟩, such that for all 𝑖 < 𝜅 and 𝛼 ∈ 𝑆, we have sup(nacc(𝐶𝑖

𝛼) ∩ 𝐴𝑖) < 𝛼.

Let 𝐺 be 𝐺(��) as in the statement of Proposition 3.2. Then 𝐺 is a stationary subset of 𝜅, and𝐷 := {𝛽 < 𝜅 | sup(𝐺 ∩ 𝛽) = 𝛽 > 0} is a club in 𝜅. As ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ hits clubs at 𝑆 ∩ 𝐶, let uspick 𝛼 ∈ 𝑆 ∩ 𝐶 such that sup(nacc(𝐶𝛼) ∩ 𝐷) = 𝛼. Put 𝐵 := nacc(𝐶𝛼) ∩ 𝐷. For all 𝛽 ∈ 𝐵, by𝛽 ∈ 𝐷, we know that the relative interval 𝐺∩ (sup(𝐶𝛼 ∩𝛽), 𝛽) is nonempty. Consequently, we mayfind some regressive function 𝑓 : 𝛼 → 𝛼 such that 𝑓(𝛽) ∈ 𝐺 ∩ (sup(𝐶𝛼 ∩ 𝛽), 𝛽) for all 𝛽 ∈ 𝐵. Pick𝑖 < 𝛼 and a cofinal subset 𝐵′ ⊆ 𝐵 such that 𝑓𝑖 � 𝐵′ = 𝑓 � 𝐵′. Fix a large enough 𝜂 ∈ 𝐶𝛼 such thatsup(𝐶𝛼 ∩ 𝜂) ≥ 𝑖. By omitting an initial segment, we may assume that 𝐵′ ∩ 𝜂 = ∅.

Let 𝛽 ∈ 𝐵′ be arbitrary. Write 𝛾 := 𝑓𝑖(𝛽). Then 𝛾 ∈ 𝐺∩ (sup(𝐶𝛼 ∩𝛽), 𝛽) ⊆ (𝑖, 𝛽). In particular,sup(𝐴𝑖∩𝛾) = 𝛾 and 𝐴𝑖∩𝛾 = 𝐴𝑖

𝛾 , so that 𝑋𝑖𝛽 = {𝛽}∪(𝐴𝑖∩𝛾), and min(𝑋𝑖

𝛽∖(sup(𝐶𝛼∩𝛽)+1)) ∈ 𝐴𝑖∩𝛾.

Thus we have shown that for all 𝛽 ∈ 𝐵′, 𝐶𝑖𝛼 ∩ 𝐴𝑖 ∩ (sup(𝐶𝛼 ∩ 𝛽), 𝛽) is a singleton, contradicting

the fact that sup(nacc(𝐶𝑖𝛼) ∩𝐴𝑖) < 𝛼 = sup(𝐵′). �

This completes the proof. �

Remark. Notice that since 𝐽 [𝜅] is an ideal, the conclusion of the preceding theorem may be strength-ened to: if 𝑆 ∈ 𝐽 [𝜅], then �−(𝑇 ) holds for every 𝑇 ∈ 𝒫(𝜅) such that 𝑆 ∩ 𝑇 is stationary. Comparethis with Kunen’s theorem [Kun80] that for 𝑆 ⊆ 𝜅, ♢*(𝑆) entails that ♢(𝑇 ) holds for every 𝑇 ∈ 𝒫(𝜅)such that 𝑆 ∩ 𝑇 is stationary.

Corollary 4.4. Suppose that 𝜃 < cf(𝜆) ≤ 𝜆 are cardinals, 𝒟(𝜆, 𝜃) = 𝜆 and 2𝜆 = 𝜆+.

For every stationary 𝑆 ⊆ 𝐸𝜆+

𝜃 , �(𝜆+) entails �−(𝑆).

Proof. Suppose that 𝑆 ⊆ 𝐸𝜆+

𝜃 is a given stationary set. By Proposition 2.2, 𝐸𝜆+

𝜃 ∈ 𝐽 [𝜆+]. Sincethe latter is an ideal, we get that 𝑆 ∈ 𝐽 [𝜆+]. By CH𝜆, 𝜆 ≥ ℵ1 and Fact 3.3, ♢(𝜆+) holds. So, byTheorem 4.3, �(𝜆+) entails �−(𝑆). �

Corollary 4.5. Suppose that GCH holds, and 𝜆 is an uncountable cardinal.

Then �(𝜆+) entails that �−(𝐸𝜆+

𝜃 ) holds for every regular cardinal 𝜃 < 𝜆.

Proof. By Corollary 4.4, �(𝜆+)+GCH entails �−(𝑆) for every stationary 𝑆 ⊆ 𝐸𝜆+

<cf(𝜆). Thus, from

now on, suppose that 𝜆 is a singular cardinal, and 𝜃 is a regular cardinal in [cf(𝜆), 𝜆). If 𝜃 = cf(𝜆),then let 𝜃′ := 𝜃+; otherwise, let 𝜃′ := 𝜃.

Claim 4.5.1. There exists a stationary 𝑆 ⊆ 𝐸𝜆+

𝜃′ such that 𝑆 ∈ 𝐽 [𝜆+].

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 11

Proof. By Corollary 2.5, it suffices to prove that 𝒟(𝜆, 𝜃′) = 𝜆. Let Σ : cf(𝜆) → 𝜆 be an increasing

function whose image is cofinal in 𝜆. Put ℱ :=⋃{[𝛼]𝜃

′ | 𝜃′ ≤ 𝛼 < 𝜆}. By GCH, 𝜆 is a strong

limit, and hence |ℱ| = 𝜆. Let 𝐵 ∈ [𝜆]𝜃′

be arbitrary. If there exists an ordinal 𝛼 < 𝜆 such that|𝐵 ∩ 𝛼| = 𝜃′, then 𝐵 ∩ 𝛼 is a subset of 𝐵 that belongs to ℱ , as sought.

Towards a contradiction, suppose that this is not case, and define ℎ : cf(𝜆) → 𝜃′ by stipulatingℎ(𝑗) := |𝐵 ∩Σ(𝑗)|. Since cf(𝜆) < cf(𝜃′), there exists some large enough 𝜀 < 𝜃′ such that Im(ℎ) ⊆ 𝜀.But then |𝐵| ≤ max{cf(𝜆), |𝜀|} < 𝜃′. This is a contradiction. �

Let 𝑆 be given by the preceding claim. Suppose that �(𝜆+) holds. By CH𝜆, 𝜆 ≥ ℵ1 and Fact 3.3,

♢(𝜆+) holds. So, by Theorem 4.3, �−(𝑆) holds, let alone �−(𝐸𝜆+

𝜃′ ). In particular, if 𝜃′ = 𝜃, thenwe are done.

Suppose now that 𝜃′ > 𝜃. Let ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ be a witness to �−(𝐸𝜆+

𝜃′ ). We claim that the same

sequence witnesses �−(𝐸𝜆+

𝜃 ). To see this, let 𝐴 ⊆ 𝜆+ be an arbitrary cofinal subset of 𝜆+. Fix𝛼 ∈ 𝑆 such that sup(nacc(𝐶𝛼)∩𝐴) = 𝛼. As cf(𝛼) > 𝜃, we may pick some �� ∈ acc(𝐶𝛼) of cofinality𝜃 such sup(nacc(𝐶𝛼) ∩𝐴 ∩ ��) = ��. Then 𝐶�� = 𝐶𝛼 ∩ �� and sup(nacc(𝐶��) ∩𝐴) = ��, as sought. �

Corollary 4.6. Suppose that 𝜆 is a successor of a regular cardinal 𝜃.

If every stationary subset of 𝐸𝜆+

𝜃 reflects, then �−(𝐸𝜆+

𝜃 ) entails �−(𝐸𝜆+

𝜆 ).

Proof. Suppose that every stationary subset of 𝐸𝜆+

𝜃 reflects, and that �� = ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ is a

witness to �−(𝐸𝜆+

𝜃 ). We claim that �� is also a witness to �−(𝐸𝜆+

𝜆 ).Let 𝐴 be an arbitrary cofinal subset of 𝜆+. First, let us point out that

𝑆 := {�� ∈ 𝐸𝜆+

𝜃 | sup(nacc(𝐶��) ∩𝐴) = ��}is stationary in 𝜆+. To see this, notice that given any club 𝐷 in 𝜆+, we may find some sparseenough cofinal subset 𝐴′ ⊆ 𝐴 such that for all 𝛾 < 𝛿 both from 𝐴′, the relative interval (𝛾, 𝛿) ∩𝐷is nonempty; then, by the choice of ��, we pick �� ∈ 𝐸𝜆+

𝜃 such that sup(nacc(𝐶��) ∩ 𝐴′) = ��, andhence �� ∈ acc(𝐷), so that �� ∈ 𝑆 ∩𝐷.

As every stationary subset of 𝐸𝜆+

𝜃 reflects, let us pick 𝛼 ∈ 𝐸𝜆+

𝜆 such that 𝑆 ∩ 𝛼 is stationary.Put 𝐵 := 𝑆 ∩ acc(𝐶𝛼). Then 𝐵 is stationary in 𝛼, and for all �� ∈ 𝐵, we have sup(nacc(𝐶𝛼)∩𝐴) ≥sup(nacc(𝐶��) ∩𝐴) = ��. Consequently, sup(nacc(𝐶𝛼) ∩𝐴) = sup(𝐵) = 𝛼, as sought. �

Corollary 4.7. For every cardinal 𝜆 ≥ i𝜔, CH𝜆 +�(𝜆+) entails �−(𝜆+), and hence the existenceof a club-regressive 𝜆+-Souslin tree.

Proof. Suppose that we are given a cardinal 𝜆 ≥ i𝜔. By the main result of [She00] (see also [Koj15,S2]), there exists some regular cardinal 𝜃 < i𝜔 such that 𝒟(𝜆, 𝜃) = 𝜆. So, by Corollary 2.5, let us

fix a stationary 𝑆 ⊆ 𝐸𝜆+

𝜃 which is in 𝐽 [𝜆+]. Assume that CH𝜆 + �(𝜆+) holds. By CH𝜆, 𝜆 ≥ ℵ1

and Fact 3.3, ♢(𝜆+) holds. Then, by Theorem 4.3, �−(𝑆) holds, let alone �−(𝜆+).By Fact 4.2(a), then, there exists a club-regressive 𝜆+-Souslin tree. �

Corollary 4.8. Suppose that GCH holds, and 𝜆 is an uncountable cardinal.

(1) If �(𝜆+) holds, then there exists a club-regressive 𝜆+-Souslin tree;(2) If there are no club-regressive 𝜆+-Souslin trees, then 𝜆+ is weakly compact in 𝐿;(3) If there are no ℵ2-Souslin trees and no ℵ3-Souslin trees, then the Axiom of Determinacy

holds in 𝐿(R).

Proof. (1) By Corollary 4.5, GCH+�(𝜆+) entails �−(𝐸𝜆+

𝜔 ). In particular, �−(𝜆+) holds. By CH𝜆,𝜆 ≥ ℵ1 and Fact 3.3, ♢(𝜆+) holds. Finally, by Fact 4.2(a), we infer the existence of a club-regressive𝜆+-Souslin tree.

12 ASSAF RINOT

(2) By [Tod87], if 𝜅 is a regular uncountable cardinal and �(𝜅) fails, then 𝜅 is weakly compactin 𝐿. Now, appeal to Clause (1).

(3) By [SS14], ¬�(𝜔2)+¬�𝜔2+2ℵ1 = ℵ2 implies that 𝐿(R) |= AD. Now, appeal to Clause (1). �

The last goal of this section is derive Souslin trees which are complete to the maximal possibleextent. For this, we shall be considering the following finer concept:

Definition 4.9. A coherent* 𝐶-sequence (over a regular uncountable cardinal 𝜅) is a sequence⟨𝐶𝛼 | 𝛼 < 𝜅⟩ such that:

(1) for all 𝛼 < 𝜅, 𝐶𝛼 ⊆ 𝛼;(2) for all limit 𝛼 < 𝜅, 𝐶𝛼 is a club in 𝛼;(3) for all 𝛼 < 𝜅, if �� ∈ acc(𝐶𝛼), then sup((𝐶��∆𝐶𝛼) ∩ ��) < ��.

Definition 4.10. For a regular uncountable cardinal 𝜅, and a stationary subset 𝑆 ⊆ 𝜅, �′(𝑆)asserts the existence of a coherent* 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩, such that for every cofinal subset𝐴 ⊆ 𝜅, there exists some 𝛼 ∈ 𝑆 for which sup(nacc(𝐶𝛼) ∩𝐴) = 𝛼.

The very same construction from the proof of Fact 4.2(b) demonstrates (cf. [BR16, S2]):

Proposition 4.11. Suppose that 𝜅 is a regular uncountable cardinal, and ♢(𝜅) +�′(𝐸𝜅≥𝜃) holds.

If 𝜆<𝜃 < 𝜅 for all 𝜆 < 𝜅, then there exists a 𝜃-complete 𝜅-Souslin tree.

Lemma 4.12. Suppose that 𝜅 is a regular uncountable cardinal, and ♢(𝜅) +�′(𝜅) holds.Then there exists a partition 𝜅 = 𝑇0 ⊎ 𝑇1 such that ♢(𝑇0) +�′(𝑇1) holds.

Proof. By ♢(𝜅) and Devlin’s lemma [Dev78], let us fix a partition 𝜅 = 𝑇0 ⊎ 𝑇1 such that ♢(𝑇0) +♢(𝑇1) holds. Of course, it suffices to show that there exists some 𝑖 < 2 such that �′(𝑇𝑖) holds.

Towards a contradiction, suppose that this is not the case. Let ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ be a �′(𝜅)-sequence.By ♢(𝜅), let us fix a matrix A =

⟨𝐴𝑖

𝛾 | 𝑖 < 𝜅, 𝛾 < 𝜅⟩

as in Proposition 3.2.Let 𝑖 < 2 be arbitrary. Put

∙ 𝑋𝑖𝛾 := {𝛾} ∪𝐴𝑖

𝛾 for all 𝛾 < 𝜅, and then

∙ 𝐶𝑖𝛼 := 𝐶𝛼 ∪ {min(𝑋𝑖

𝛾 ∖ (sup(𝐶𝛼 ∩ 𝛾) + 1)) | 𝛾 ∈ nacc(𝐶𝛼) & 𝛾 > 0} for all 𝛼 < 𝜅.

The proof of Claim 4.3.1 makes it clear that⟨𝐶𝑖𝛼 | 𝛼 < 𝜅

⟩is a coherent* 𝐶-sequence. So, since

�′(𝑇𝑖) fails, we may fix a cofinal subset 𝐴𝑖 ⊆ 𝜅, such that sup(nacc(𝐶𝑖𝛼) ∩ 𝐴𝑖) < 𝛼 for all 𝛼 ∈ 𝑇𝑖.

Put 𝐺𝑖 := {𝛾 < 𝜅 | sup(𝐴𝑖 ∩ 𝛾) = 𝛾 & 𝐴𝑖 ∩ 𝛾 = 𝐴𝑖𝛾 & 𝛾 > 0}.

Now, by the choice of A, 𝐺 := 𝐺0 ∩𝐺1 is stationary. As ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a �′(𝜅)-sequence, let usfix 𝛼 < 𝜅 such that sup(nacc(𝐶𝛼) ∩𝐺) = 𝛼. Let 𝑖 < 2 be the unique integer such that 𝛼 ∈ 𝑇𝑖. Let𝛾 be an arbitrary element of Γ := nacc(𝐶𝛼)∩𝐺. Then 𝐴𝑖

𝛾 = 𝐴𝑖 ∩ 𝛾 is a cofinal subset of 𝛾, so that

min(𝑋𝑖𝛾∖(sup(𝐶𝛼∩𝛾)+1)) ∈ 𝐴𝑖∩𝛾. Thus we have shown that for all 𝛾 ∈ Γ, 𝐶𝑖

𝛼∩𝐴𝑖∩(sup(𝐶𝛼∩𝛾), 𝛾)

is a singleton, contradicting the fact that sup(nacc(𝐶𝑖𝛼) ∩𝐴𝑖) < sup(Γ) = 𝛼 ∈ 𝑇𝑖. �

Theorem 4.13. Suppose that 𝜅 is a regular cardinal ≥ ℵ2, and ♢(𝜅) +�′(𝜅) holds.Then �′(𝑆) holds for every stationary subset 𝑆 ⊆ 𝜅.

Proof. By Lemma 4.12, let us fix a partition 𝜅 = 𝑇0⊎𝑇1 such that ♢(𝑇0)+�′(𝑇1) holds. By ♢(𝑇0),let us fix a matrix M =

⟨𝐴𝑖

𝛾 | 𝑖 < 𝜅, 𝛾 < 𝜅⟩

as in Proposition 3.2. Let 𝑆 be an arbitrary stationarysubset of 𝜅. Towards showing that �′(𝑆) holds, we prove the following.

Claim 4.13.1. There exist some 𝑖 < 𝜔1 and a coherent* 𝐶-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜅⟩, such that forevery subset 𝐴 ⊆ 𝜅 and every club 𝐷 ⊆ 𝜅, there exists some 𝛿 ∈ 𝑆 such that

sup{𝛾 ∈ nacc(𝐶𝛿) ∩𝐷 ∩ 𝑇0 | 𝐴𝑖𝛾 = 𝐴 ∩ 𝛾} = 𝛿.

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 13

Proof. Suppose not. Building on ideas from [KS93], we shall recursively construct a sequence of

triples ⟨(𝐴𝑖, 𝐷𝑖,−→𝐶𝑖) | 𝑖 < 𝜔1⟩, such that for all 𝑖 < 𝜔1:

(1) 𝐴𝑖 is a cofinal subset of 𝜅;(2) 𝐷𝑖 is a club in 𝜅;

(3)−→𝐶𝑖 =

⟨𝐶𝑖𝛿 | 𝛿 < 𝜅

⟩is a coherent* 𝐶-sequence. In addition, for all 𝛿 < 𝜅:

(a) 𝐶𝑖𝛿+1 = {𝛿} and 𝐶𝑖

𝛿 ⊆ 𝐶𝑖+1𝛿 ;

(b) if 𝛼 ∈ acc(𝐶𝑖𝛿) and 𝜂 ∈ 𝐶𝑖

𝛼∩𝐶𝑖𝛿 satisfies sup((𝐶𝑖

𝛼∆𝐶𝑖𝛿)∩𝛼) ≤ 𝜂, then sup((𝐶𝑖+1

𝛼 ∆𝐶𝑖+1𝛿 )∩

𝛼) ≤ 𝜂;

(c) if 𝑖 is a limit nonzero ordinal, then 𝐶𝑖𝛼 =

⋃𝑗<𝑖𝐶

𝑗𝛼 and acc(𝐶𝑖

𝛿) =⋃

𝑗<𝑖 acc(𝐶𝑗𝛿 ).

Whenever 𝐴𝑖 and 𝐷𝑖 will be defined, we shall also derive the set 𝐺𝑖, by letting:

𝐺𝑖 := {𝛾 ∈ 𝐷𝑖 ∩ 𝑇0 | 𝐴𝑖𝛾 = 𝐴𝑖 ∩ 𝛾}.

Here comes the recursion:

I Put 𝐴0 := 𝜅 and 𝐷0 := 𝜅. Let−→𝐶0 =

⟨𝐶0𝛿 | 𝛿 < 𝜅

⟩be some witness to �′(𝑇1). Of course, we

may also assume that 𝐶0𝛿+1 = {𝛿} for all 𝛿 < 𝜅.

I Suppose that 𝑖 < 𝜔1, and ⟨(𝐴𝑗 , 𝐷𝑗 ,−→𝐶𝑗) | 𝑗 ≤ 𝑖⟩ has already been defined. By the indirect

hypothesis, let us fix some cofinal subset 𝐴𝑖+1 ⊆ 𝜅 and some club 𝐷𝑖+1 ⊆ 𝜅 such that for all 𝛿 ∈ 𝑆,we have:

sup(nacc(𝐶𝑖𝛿) ∩𝐺𝑖+1) < 𝛿.

Next, define−−→𝐶𝑖+1 =

⟨𝐶𝑖+1𝛿 | 𝛿 < 𝜅

⟩by recursion on 𝛿 < 𝜅, as follows. Let 𝐶𝑖+1

0 := ∅, and

𝐶𝑖+1𝛿+1 := {𝛿} for all 𝛿 < 𝜅. Now, if 𝛿 < 𝜅 is a nonzero limit ordinal and

⟨𝐶𝑖+1𝛾 | 𝛾 < 𝛿

⟩has already

been defined, let:

(⋆)𝑖 𝐶𝑖+1𝛿 := 𝐶𝑖

𝛿 ∪⋃

{𝐶𝑖+1𝛾 ∖ sup(𝐶𝑖

𝛿 ∩ 𝛾) | 𝛾 ∈ nacc(𝐶𝑖𝛿) ∖𝐺𝑖+1}.

It is easy to see that Clauses (3)(a) and (3)(b) hold for 𝑖. So, as−→𝐶𝑖 is a coherent* 𝐶-sequence,

we get that−−→𝐶𝑖+1 is a coherent* 𝐶-sequence.

I Suppose that 𝑖 < 𝜔1 is a nonzero limit ordinal, and ⟨(𝐴𝑗 , 𝐷𝑗 ,−→𝐶𝑗) | 𝑗 < 𝑖⟩ has already been

defined. Put 𝐴𝑖 := 𝜅 and 𝐷𝑖 := 𝜅. For all 𝛿 < 𝜅, let 𝐶𝑖𝛿 :=

⋃𝑗<𝑖𝐶

𝑗𝛿 . As Clause (3) holds for

all 𝑗 < 𝑖, to see that−→𝐶𝑖 =

⟨𝐶𝑖𝛿 | 𝛿 < 𝜅

⟩is a coherent* 𝐶-sequence, it suffice to show that for all

𝛼 < 𝛿 < 𝜅 such that sup(𝐶𝑖𝛿 ∩ 𝛼) = 𝛼 > 0, there exists some 𝑗 < 𝑖 such that 𝛼 ∈ acc(𝐶𝑗

𝛿 ).

Suppose that 𝛼 < 𝛿 are as above. By Clause (3)(a), ⟨min(𝐶𝑗𝛿 ∖ 𝛼) | 𝑗 < 𝑖⟩ is a weakly decreasing

sequence of ordinals, and hence must stabilize at some 𝑗* < 𝑖. Denote 𝛾 := min(𝐶𝑗*

𝛿 ∖ 𝛼) and

𝛾− := sup(𝐶𝑗*

𝛿 ∩ 𝛾). If 𝛾 ∈⋂{nacc(𝐶𝑗

𝛿 ) | 𝑗* ≤ 𝑗 < 𝑖}, then we get by induction on 𝑗 ∈ [𝑗*, 𝑖) —

using (⋆)𝑗 at successor stages and Clause (3)(c) at limit stages — that 𝐶𝑗𝛿 ∩ (𝛾−, 𝛾] = {𝛾} for all

𝑗 ∈ [𝑗*, 𝑖), contradicting the fact that 𝛾− < sup(𝐶𝑖𝛿 ∩𝛼) = 𝛼 ≤ 𝛾.13 Thus, pick 𝑗 ∈ [𝑗*, 𝑖) such that

𝛾 ∈ acc(𝐶𝑗𝛿 ). As 𝛾 = min(𝐶𝑗

𝛿 ∖ 𝛼), this means that 𝛼 = 𝛾, and hence 𝛼 ∈ acc(𝐶𝑗𝛿 ), as sought.

At the end of the above process, we have obtained ⟨(𝐴𝑖, 𝐷𝑖,−→𝐶𝑖) | 𝑖 < 𝜔1⟩. Put 𝐺 :=

⋂𝑖<𝜔1

𝐺𝑖. By

the choice of the matrix M, we know that 𝐺 is a stationary subset of 𝑇0. As−→𝐶0 is a �′(𝑇1)-sequence,

13Note that 𝛾 ∈ nacc(𝐶𝑗*

𝛿 ) =⇒ 𝛾− < 𝛼.

14 ASSAF RINOT

for every 𝛽 < 𝜅, there exists some 𝛼 ∈ 𝑇1 such that sup(nacc(𝐶0𝛼) ∩ (𝐺 ∖ 𝛽)) = 𝛼. Consequently,

the following set is cofinal in 𝜅:

𝐴 := {𝛼 ∈ 𝑇1 | sup(nacc(𝐶0𝛼) ∩𝐺) = 𝛼}.

Let 𝛼 ∈ 𝐴 be arbitrary. Trivially, the set Γ𝛼 := nacc(𝐶0𝛼) ∩ 𝐺 is cofinal in 𝛼. Suppose that

𝑖 < 𝜔1, and Γ𝛼 ⊆ nacc(𝐶𝑖𝛼). Then, by Γ𝛼 ⊆ 𝐺 ⊆ 𝐺𝑖+1 and (⋆)𝑖, we also have Γ𝛼 ⊆ nacc(𝐶𝑖+1

𝛼 ).Recalling Clauses (3)(a) and (3)(c), we altogether get that for all 𝑖 < 𝜔:

(⋆⋆)𝑖 Γ𝛼 ⊆ nacc(𝐶𝑖𝛼) ∩𝐺𝑖+1.

As 𝑆 is stationary in 𝜅 = sup(𝐴), we now pick 𝛿 ∈ 𝑆 such that sup(𝐴 ∩ 𝛿) = 𝛿. For all 𝑖 < 𝜔1,by the choice of the pair (𝐴𝑖+1, 𝐷𝑖+1), the following ordinal is strictly smaller than 𝛿:

𝛿𝑖 := sup(nacc(𝐶𝑖𝛿) ∩ 𝐺𝑖+1).

At this stage, the analysis splits into two main cases and a few subcases:

Case 1. cf(𝛿) > ℵ0: In this case, 𝛿* := sup𝑖<𝜔 𝛿𝑖 is below 𝛿, so let us fix some 𝛼 ∈ 𝐴∩ (𝛿*, 𝛿).

Denote 𝛾𝑖 := min(𝐶𝑖𝛿 ∖ 𝛼) for all 𝑖 < 𝜔. By Clause (3)(a), ⟨𝛾𝑖 | 𝑖 < 𝜔⟩ is weakly decreasing,

so let us fix some 𝑖 < 𝜔 such that 𝛾𝑖 = 𝛾𝑖+1. Now, there are three cases to consider, each ofwhich yielding a contradiction:Case 1.1. 𝛾𝑖 > 𝛼: By 𝛾𝑖 = min(𝐶𝑖

𝛿 ∖𝛼) and 𝛾𝑖 > 𝛼, we have 𝛾𝑖 ∈ nacc(𝐶𝑖𝛿). By 𝛾𝑖 > 𝛼 >

𝛿* ≥ 𝛿𝑖, we moreover have 𝛾𝑖 ∈ nacc(𝐶𝑖𝛿) ∖𝐺𝑖+1. Denote 𝛾−𝑖 := sup(𝐶𝑖

𝛿 ∩ 𝛾𝑖). By (⋆)𝑖,we have

𝐶𝑖+1𝛿 ∩ [𝛾−𝑖 , 𝛾𝑖) = 𝐶𝑖+1

𝛾𝑖 ∩ [𝛾−𝑖 , 𝛾𝑖).

By 𝛾𝑖 = min(𝐶𝑖𝛿 ∖ 𝛼) and 𝛾𝑖 > 𝛼, we have 𝛾−𝑖 < 𝛼, and hence

𝐶𝑖+1𝛿 ∩ [𝛼, 𝛾𝑖) = 𝐶𝑖+1

𝛾𝑖 ∩ [𝛼, 𝛾𝑖).

So 𝛾𝑖+1 = min(𝐶𝑖+1𝛿 ∖ 𝛼) = min(𝐶𝑖+1

𝛾𝑖 ∖ 𝛼) < 𝛾𝑖, contradicting the choice of 𝑖.

Case 1.2. 𝛾𝑖 = 𝛼 ∈ acc(𝐶𝑖𝛿): In this case, sup((𝐶𝑖

𝛿∆𝐶𝑖𝛼)∩𝛼) < 𝛼 and hence by (⋆⋆)𝑖, we

get that sup(nacc(𝐶𝑖𝛿) ∩𝐺𝑖+1) ≥ 𝛼, contradicting the fact that 𝛼 > 𝛿* ≥ 𝛿𝑖.

Case 1.3. 𝛾𝑖 = 𝛼 ∈ nacc(𝐶𝑖𝛿): Write 𝛼− := sup(𝐶𝑖

𝛿 ∩𝛼). As 𝛼 ∈ 𝐴 ⊆ 𝑇1 and 𝐺𝑖+1 ⊆ 𝑇0,

we get from (⋆)𝑖 that 𝐶𝑖+1𝛿 ∩[𝛼−, 𝛼) = 𝐶𝑖+1

𝛼 ∩[𝛼−, 𝛼). In particular, by (⋆⋆)𝑖+1, we have

Γ𝛼∩ [𝛼−, 𝛼) ⊆ nacc(𝐶𝑖+1𝛿 )∩𝐺𝑖+2. But then sup(nacc(𝐶𝑖+1

𝛿 )∩𝐺𝑖+2) ≥ 𝛼, contradictingthe fact that 𝛼 > 𝛿* ≥ 𝛿𝑖+1.

Case 2. cf(𝛿) = ℵ0: Pick an uncountable 𝐼 ⊆ 𝜔1 and some 𝛿* < 𝛿 for which sup𝑖∈𝐼 max{𝛿𝑖, 𝛿𝑖+1} ≤𝛿*. Pick 𝛼 ∈ 𝐴 ∩ (𝛿*, 𝛿), and denote 𝛾𝑖 := min(𝐶𝑖

𝛿 ∖ 𝛼) for all 𝑖 < 𝜔1. By Clause (3)(a) andsince 𝐼 is cofinal in 𝜔1, let us fix some 𝑖 ∈ 𝐼 such that 𝛾𝑖 = 𝛾𝑖+1. As before, there are threecases to consider:Case 2.1. 𝛾𝑖 > 𝛼: By 𝛾𝑖 > 𝛼 > 𝛿* ≥ 𝛿𝑖, we have 𝛾𝑖 ∈ nacc(𝐶𝑖

𝛿) ∖ 𝐺𝑖+1. Then, by (⋆)𝑖,

we have 𝐶𝑖+1𝛿 ∩ [𝛼, 𝛾𝑖) = 𝐶𝑖+1

𝛾𝑖 ∩ [𝛼, 𝛾𝑖). So 𝛾𝑖+1 = min(𝐶𝑖+1𝛿 ∖𝛼) = min(𝐶𝑖+1

𝛾𝑖 ∖𝛼) < 𝛾𝑖,contradicting the choice of 𝑖.

Case 2.2. 𝛾𝑖 = 𝛼 ∈ acc(𝐶𝑖𝛿): In this case, sup((𝐶𝑖

𝛿∆𝐶𝑖𝛼)∩𝛼) < 𝛼 and hence by (⋆⋆)𝑖, we

get that sup(nacc(𝐶𝑖𝛿) ∩𝐺𝑖+1) ≥ 𝛼, contradicting the fact that 𝛼 > 𝛿* ≥ 𝛿𝑖.

Case 2.3. 𝛾𝑖 = 𝛼 ∈ nacc(𝐶𝑖𝛿): Write 𝛼− := sup(𝐶𝑖

𝛿 ∩ 𝛼). As 𝛼 ∈ 𝐺𝑖+1, we get from (⋆)𝑖

that 𝐶𝑖+1𝛿 ∩[𝛼−, 𝛼) = 𝐶𝑖+1

𝛼 ∩[𝛼−, 𝛼). In particular, by (⋆⋆)𝑖+1, we have sup(nacc(𝐶𝑖+1𝛿 )∩

𝐺𝑖+2) ≥ 𝛼, contradicting the fact that 𝛼 > 𝛿* ≥ 𝛿𝑖+1. �

Fix a coherent* 𝐶-sequence ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ and some 𝑖 < 𝜔1 as in Claim 4.13.1. Put:

HIGHER SOUSLIN TREES AND THE GCH, REVISITED 15

∙ 𝑌𝛾 := {𝛾} ∪𝐴𝑖𝛾 for all 𝛾 < 𝜅, and then

∙ 𝐷𝛼 := 𝐶𝛼 ∪ {min(𝑌𝛾 ∖ (sup(𝐶𝛼 ∩ 𝛾) + 1)) | 𝛾 ∈ nacc(𝐶𝛼) & 𝛾 > 0} for all 𝛼 < 𝜅.

The proof of Claim 4.3.1 makes it clear that ⟨𝐷𝛼 | 𝛼 < 𝜅⟩ is a coherent* 𝐶-sequence. Finally, let 𝐴be an arbitrary cofinal subset of 𝜅. Put 𝐷 := {𝛾 < 𝜅 | sup(𝐴 ∩ 𝛾) = 𝛾 > 0}. Pick 𝛿 ∈ 𝑆 such that

Γ := {𝛾 ∈ nacc(𝐶𝛿) ∩𝐷 ∩ 𝑇0 | 𝐴𝑖𝛾 = 𝐴 ∩ 𝛾}

is cofinal in 𝛿.Let 𝛾 be an arbitrary element of Γ. Then 𝐴𝑖

𝛾 = 𝐴 ∩ 𝛾 is a cofinal subset of 𝛾, so that min(𝑌𝛾 ∖(sup(𝐶𝛿 ∩ 𝛾) + 1)) ∈ 𝐴 ∩ 𝛾. Thus we have shown that for all 𝛾 ∈ Γ, 𝐷𝛿 ∩ 𝐴 ∩ (sup(𝐶𝛿 ∩ 𝛾), 𝛾) is asingleton, and hence sup(nacc(𝐷𝛿) ∩𝐴) = 𝛿, as sought. �

Corollary 4.14. For every uncountable cardinal 𝜆, GCH +�(𝜆+) entails the existence of a cf(𝜆)-complete 𝜆+-Souslin tree.

Proof. By Corollary 4.5, GCH + �(𝜆+) entails �−(𝐸𝜆+

𝜔 ). In particular, �′(𝜆+) holds, and then,

by Theorem 4.13, �′(𝐸𝜆+

cf(𝜆)) holds. By CH𝜆, 𝜆 ≥ ℵ1 and Fact 3.3, ♢(𝜆+) holds. Finally, by GCH

and Proposition 4.11, we infer the existence of a cf(𝜆)-complete 𝜆+-Souslin tree. �

Remark. The preceding provides an affirmative answer to Question 9 from the survey paper[Rin11a].

Acknowledgements

I thank A. M. Brodsky, M. Kojman, M. Gitik, C. Lambie-Hanson and D. Raghavan for theirinput that improved the exposition of this paper. Special thanks go to Y. Hayut for proofreading apreliminary version of this paper, and the anonymous referee for a thorough reading of this paper.

The results of this paper were presented at the Mini-Symposia of the 7𝑡ℎ European Congressof Mathematics, Berlin, July 2016, and the Set Theory and its Applications in Topology meeting,Oaxaca, September 2016. I thank the organizers for the invitations, and the participants for theirfeedback.

This work was partially supported by the Israel Science Foundation (grant #1630/14).

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Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel.URL: http://www.assafrinot.com