introduction - assaf rinot
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HIGHER SOUSLIN TREES AND THE GCH, REVISITED
ASSAF RINOT
Abstract. It is proved that for every uncountable cardinal π, GCH+οΏ½(π+) entails the existenceof a cf(π)-complete π+-Souslin tree. In particular, if GCH holds and there are no β΅2-Souslin trees,then β΅2 is weakly compact in Godelβs constructible universe, improving Gregoryβs 1976 lower bound.Furthermore, it follows that if GCH holds and there are no β΅2 and β΅3 Souslin trees, then the Axiomof Determinacy holds in πΏ(R).
1. Introduction
1.1. Background. A tree is a partially ordered set (π,<π ) with the property that for every π₯ β π ,the downward cone π₯β := {π¦ β π | π¦ <π π₯} is well-ordered by <π . The order type of (π₯β, <π ) is
denoted by ht(π₯), and the πΌπ‘β-level of the tree is the set ππΌ := {π₯ β π | ht(π₯) = πΌ}.Let π denote a regular uncountable cardinal. A π -Aronszajn tree is a tree of size π having no
chains or levels of size π . A π -Souslin tree is a tree of size π having no chains or antichains of sizeπ . As tree-levels are antichains, any π -Souslin tree is a π -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 β΅1-Souslin tree. However, all of Kurepaβs 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 manβs version of a Souslin tree; Indeed, Aronszajn constructed (what wenowadays call) an β΅1-Aronszajn tree.
It was only three decades after [Kur35], in [Ten68], [Jec67], [Jen68], and [ST71], when it wasdiscovered that β unlike β΅1-Aronszajn trees β the existence of an β΅1-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 Godelβs constructible universe, πΏ,for every regular uncountable cardinal π , the following are equivalent:
β There exists a π -Souslin tree;β There exists a π -Aronszajn tree;β π is not a weakly compact cardinal.
In another work, Jensen proved (see the monograph [DJ74]) that the existence of an β΅1-Souslintree is independent of ZFC+GCH.2 As for Aronszajn trees, Specker proved [Spe49] that GCHentails the existence of a π+-Aronszajn tree for every regular cardinal π, 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 2β΅πΌ = β΅πΌ+1 for every ordinal πΌ.
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2 ASSAF RINOT
proved [Mit73] that the nonexistence of an β΅2-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 β΅2-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 β΅2-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 β΅2-Souslin trees, then β΅2 is a Mahlocardinal in L.
In this paper, Gregoryβs 1976 lower bound is increased to the anticipated value:
Theorem A. If GCH holds and there are no β΅2-Souslin trees, then β΅2 is weakly compact in πΏ.
In [Tod81], Todorcevic proved that after Levy-collapsing a weakly compact cardinal to β΅2 overa model of GCH: GCH holds, and every β΅2-Aronszajn tree contains an β΅1-Aronszajn subtree. Astrengthening of Theorem A, then, provides the following optimal result:
Theorem Aβ. If GCH holds, and β΅2 is not weakly compact in πΏ, then there exists an β΅2-Souslintree with no β΅1-Aronszajn subtrees.
We remind the reader that a regular uncountable cardinal π is said to be Mahlo if the set ofregular cardinals below π is stationary in π . A regular uncountable cardinal π is said to be weaklycompact if it satisfies the generalized Ramsey partition relation: π β (π )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 πΆ-sequence (over a regular uncountable cardinal π ) isa sequence β¨πΆπΌ | πΌ < π β© such that:
(1) for all πΌ < π , πΆπΌ β πΌ;(2) for all limit πΌ < π , πΆπΌ is a club in πΌ;(3) for all πΌ < π , if οΏ½οΏ½ β acc(πΆπΌ), then πΆοΏ½οΏ½ = πΆπΌ β© οΏ½οΏ½.3
The easiest way to obtain such a sequence is to fix some club π· in π , and let πΆπΌ := π· β© πΌ for allπΌ β acc(π·), and πΆπΌ := πΌ β sup(π· β© πΌ) for all other πΌ. Of more interest is the following concept:
Definition 1.1 (Jensen, [Jen72]). οΏ½π(πΈ) asserts the existence of a coherent πΆ-sequence, β¨πΆπΌ | πΌ < π+β©,such that otp(πΆπΌ) β€ π and acc(πΆπΌ) β© πΈ = β for all πΌ < π+.
Write οΏ½π for οΏ½π(β ).
Jensen proved [Jen72] that if there exists a stationary subset πΈ β π+ for which β’(πΈ) + οΏ½π(πΈ)holds,4 then there exists a π+-Souslin tree, and Solovay noticed that the existence of such an πΈfollows from β’(π+) +οΏ½π. Gregory proved [Gre76] that GCH implies β’(π+) for every cardinal π ofuncountable cofinality, and Shelah [She79],[She81] improved this to every uncountable cardinal π.Altogether:
3Here, acc(πΆ) stands for the set of accumulation points of πΆ, that is, acc(πΆ) := {π½ β πΆ | sup(πΆ β© π½) = π½ > 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 (1970βs). For every uncountable π, GCH +οΏ½π entails the existence of a π+-Souslin tree.
By results of Jensen and Solovay, the failure of οΏ½β΅1 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 οΏ½β΅1 .
One consequence of οΏ½π is that every stationary subset of π+ contains a nonreflecting stationarysubset. Baumgartner proved [Bau76b] that after forcing to Levy-collapse a weakly compact cardinal
to β΅2, every stationary subset of πΈβ΅2=β΅1
reflects, and in [Gre76], Gregory indeed managed to reduce
the hypothesis οΏ½π into that of the existence of nonreflecting stationary subset of πΈπ+
=π, providedthat π is regular.
However, ten years later, Harrington and Shelah [HS85] proved that the reflection of every
stationary subset of πΈβ΅2=β΅1
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 β΅2-Souslin tree, under the hypothesis that there exist two stationary subsets of πΈβ΅2=β΅1
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 β΅2-Souslin tree from the combination of GCH and Todorcevicβs principle οΏ½(β΅2).Here, οΏ½(π ) asserts the existence of a coherent πΆ-sequence, β¨πΆπΌ | πΌ < π β©, of the weakest possiblenontrivial form. That is, for every club π· in π , there exists some πΌ β acc(π·) for which πΆπΌ = π·β©πΌ.This time, the key point is that by [Tod87], if π is a regular uncountable cardinal and οΏ½(π ) fails,then π is a weakly compact cardinal in πΏ.
And indeed, the main result of this paper is the following.
Theorem B. If π an uncountable cardinal, and GCH +οΏ½(π+) holds, then:
(1) There exists a π+-Souslin tree which is cf(π)-complete;5
(2) There exists a π+-Souslin tree which is club-regressive.6
So, Theorems A,Aβ 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 π -Souslin trees for π β {β΅2,β΅3}, then the Axiom ofDeterminacy holds in πΏ(R).
By Fact 1.2 and the main result of [Ste05], GCH and the nonexistence of an β΅π+1-Souslin treeentails that the Axiom of Determinacy holds in πΏ(R). So the surprise here is the move down intothe realm of the β΅π for finite π (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 byπ½ [π ], and study its extent. It is established that if GCH holds, then for every uncountable cardinalπ, π½ [π+] contains a stationary set, yet, it is consistent that π½ [β΅2] is the nonstationary ideal over β΅2.
In Section 3, we deal with the Diamond principle and a weak consequence of it β club hitting.
5A tree (π,<π ) is said to be π-complete if any <π -increasing sequence of elements from π , and of length < π, has anupper bound in π .6The definition of a club-regressive tree may be found in [BR15]. For our purpose, it suffices to mention that aclub-regressive π -tree contains no π-Aronszajn subtrees nor π-Cantor subtrees for every regular cardinal π < π .
4 ASSAF RINOT
In Section 4, we recall the principle οΏ½β(π) from [BR15], and prove that for every regular un-countable cardinal π and every stationary π β π½ [π ], οΏ½(π ) + β’(π ) entails οΏ½β(π). 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 π β₯ iπ, CHπ +οΏ½(π+) entails the existence of a π+-Souslin treewhich is club-regressive.
Then, we introduce the principle οΏ½β²(π), and prove that οΏ½β(π ) + β’(π ) entails οΏ½β²(π) for everystationary subset π β π , from which Theorem B(1) follow. Finally, Theorems A,Aβ,C are derivedas corollaries.
1.4. Notations and conventions. Throughout, by a cardinal, we mean an infinite cardinal. WriteCHπ to assert that 2π = π+. Denote πΈπ
π := {πΌ < π | cf(πΌ) = π}, and define πΈπ =π, πΈ
π >π and πΈπ
β₯π in a
similar fashion. Write [π]π for the collection of all subsets of π of cardinality π. A dense subfamilyof [π]π is a collection β± β [π]π such that for all π β [π]π, there exists some π β β± with π β π .The least size of such a dense subfamily is denoted by π(π, π).
For sets of ordinals πΆ,π·, write π· β πΆ iff there exists some ordinal π½ such that π· = πΆ β© π½, thatis, πΆ end-extends π·. Denote nacc(πΆ) := πΆ β acc(πΆ).
2. The regressive functions ideal
Recall that a function π : πΌβ πΌ is said to be regressive iff π(π½) < π½ for all nonzero π½ < πΌ.
Definition 2.1. For a regular uncountable cardinal π , define a collection π½ [π ], as follows.A subset π β π is in π½ [π ] iff there exists a club πΆ β π and a sequence of functions β¨ππ : π β π | π < π β©
satisfying the following. For every πΌ β πβ©πΆ, every regressive function π : πΌβ πΌ, and every cofinalsubset π΅ β πΌ, there exists some π < πΌ such that sup{π½ β π΅ | ππ(π½) = π(π½)} = πΌ.
It is easy to see that π½ [π ] is a π -complete normal ideal. We now turn to study its extent.
2.1. Positive results.
Proposition 2.2. Suppose that π < cf(π) β€ π are cardinals, and π(π, π) = π.
Then πΈπ+
π β π½ [π+].
Proof. If π is singular, then πΈπ+
π = β β π½ [π+], so suppose that π is regular. Fix a family {π΄πΌ |πΌ < π+} β [π]cf(π) such that |π΄πΌ β© π΄π½| < cf(π) for all πΌ < π½ < π+.7 Let {ππ | π < π} be a dense
subfamily of [π Γ π]π. For all π½ < π+, fix a bijection ππ½ : π β max{π, π½}. Finally, for all π < π,define a function ππ : π+ β π+ as follows. For all π½ < π+, if there exists a unique (πΎ, πΏ) β ππ suchthat πΎ β π΄π½, then let ππ(π½) := ππ½(πΏ). Otherwise, let ππ(π½) := 0.
We claim that the club πΆ := π+ β π and the sequence β¨ππ : π+ β π+ | π < πβ© together witness
that πΈπ+
π β π½ [π+].8 To see this, fix an arbitrary πΌ β πΈπ+
π β© πΆ along with a regressive functionπ : πΌ β πΌ, and a cofinal subset π΅ of πΌ. Let π΅β² be a cofinal subset of π΅ β {0} of order-type π.Since π < cf(π), for all π½ β π΅β², we may pick πΎπ½ β π΄π½ β
β{π΄π | π β π΅β², π = π½}. Also, for all
π½ β π΅β², since π(π½) < π½ β Im(ππ½), we may let πΏπ½ := πβ1π½ (π(π½)). Put π := {(πΎπ½, πΏπ½) | π½ β π΅β²}.
As π½ β¦β πΎπ½ is one-to-one over π΅β², we get that π β [π Γ π]π, and hence we may find some π < πsuch that ππ β π. Let π΅β²β² := {π½ β π΅β² | (πΎπ½, πΏπ½) β ππ}. Since π΅β²β² β π΅β² and |π΅β²β²| = otp(π΅β²),we get that sup(π΅β²β²) = sup(π΅β²) = πΌ. Thus, it suffices to verify that π οΏ½ π΅β²β² = ππ οΏ½ π΅β²β². Let
7See, e.g., [Bau76a, Theorem 2.3].8This is not a typing error; we simply settle for a sequence of length π.
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 5
π½ β π΅β²β² be arbitrary. Since (πΎπ½, πΏπ½) is the unique pair in ππ to satisfy πΎπ½ β π΄π½, we conclude that
ππ(π½) = ππ½(πΏπ½) = ππ½(πβ1π½ (π(π½))) = π(π½), as sought. οΏ½
In particular, GCH entails that πΈπ+
<π β π½ [π+] for every regular cardinal π. To deal with the caseof π singular, let us recall the definition of Shelahβs approachability ideal :
Definition 2.3 (Shelah, [She93]). For a regular uncountable cardinal π , define a collection πΌ[π ],as follows. A subset π β π is in πΌ[π ] iff there exists a club πΆ β π and a sequence β¨ππ | π < π β© ofbounded subsets of π satisfying the following. For every πΌ β π β© πΆ, there exists a cofinal subsetπ΅ β πΌ of order-type cf(πΌ) < πΌ such that {π΅ β© π | π < πΌ} β {ππ | π < πΌ}.
Proposition 2.4. Suppose that cf(π) < π < π are cardinals, and π(π, π) = π.
For every π β πΈπ+
π , if π β πΌ[π+], then π β π½ [π+].
Proof. By Shelahβs celebrated theorem from [She94a, Theorem 1.5], let us fix a scale β = β¨βπΌ | πΌ < π+β©for π. This means, in particular, that for all πΌ < π+, βπΌ is a function from cf(π) to π. Fix a bijectionπ : cf(π) Γ πβ π, and let π΄πΌ := π[βπΌ] for all πΌ < π+.
Suppose that we are given π β πΈπ+
π in πΌ[π+]. To avoid trivialities, assume that π is stationary.
In particular, π is regular. By π β πΌ[π+] and [CFM04, Corollary 2.15],9 there exists a club πΆ β π+
such that every πΌ β π β© πΆ is good for the scale β. By [Eis10, Theorem 3.50], this means that forevery πΌ β π β© πΆ and every cofinal π΅ β πΌ, there exists some cofinal π΅β² β π΅ of order-type π andπβ² < cf(π) such that βπ½(π) < βπΎ(π) for all π½ < πΎ both from π΅β² and all π β (πβ², cf(π)).
Claim 2.4.1. Suppose that πΌ β π β© πΆ, and π΅ is a cofinal subset of πΌ.Then there exists some cofinal π΅β² β π΅ of order-type π for which the following is nonempty:β
π½βπ΅β²
π΄π½ ββ
{π΄π | π β π΅β², π = π½}.
Proof. By πΌ β π β© πΆ, πΌ is good, so let us pick some cofinal π΅β² β π΅ of order-type π and π < cf(π)such that βπ½(π) < βπΎ(π) for all π½ < πΎ both from π΅β². Then β¨π(π, βπ½(π)) | π½ β π΅β²β© is an element ofβ
π½βπ΅β² π΄π½ ββ{π΄π | π β π΅β², π = π½}. οΏ½
Now, continue as in the proof of Proposition 2.2, but using the sequence β¨π΄πΌ | πΌ < π+β© we justconstructed. οΏ½
Corollary 2.5. Suppose that π is a regular cardinal.For every cardinal π β₯ π(π, π), the following are equivalent:
(1) π(π, π) = π;
(2) there exists some stationary subset π β πΈπ+
π with π β π½ [π+].
Proof. (1) =β (2): Suppose that π is a cardinal, satisfying π(π, π) = π. In particular, π = cf(π).
I If π < cf(π), then πΈπ+
π β π½ [π+] by Proposition 2.2.
I If π > cf(π), then π is singular, and by [She93, S1], there exists some stationary π β πΈπ+
π suchthat π β πΌ[π+]. Now, appeal to Proposition 2.4.
(2) =β (1): Suppose that π β₯ π(π, π) is some cardinal, and π½ [π+] contains a stationary subset
of πΈπ+
π . In particular, there exists a sequence of functions β¨ππ : π+ β π+ | π < π+β© and an ordinal
πΌ β πΈπ+
π with πΌ > π such that for every regressive function π : πΌ β πΌ, and every cofinal subsetπ΅ β πΌ, there exists some π < πΌ such that sup{π½ β π΅ | ππ(π½) = π(π½)} = πΌ.
9See also [FM97, Claim 4.4].
6 ASSAF RINOT
Fix a cofinal subset π΅ β πΌ of order-type π, with min(π΅) > π. For all π < πΌ, let ππ := ππ[π΅] βͺ π.As |ππ| = π and π(π, π) β€ π, pick β±π of size β€ π which is a dense subfamily of [ππ]
π. Then,β± := [π]π β©
β{β±π | π < πΌ} has size β€ π. To see that β± is a dense subfamily of [π]π, let π be an
arbitrary element of [π]π. Pick a regressive function π : πΌ β πΌ such that π οΏ½ π΅ is a bijection fromπ΅ to π . Fix π < πΌ for which π΅π := {π½ β π΅ | ππ(π½) = π(π½)} is cofinal in πΌ. As otp(π΅) = cf(πΌ),we have |π΅π| = π, so that |ππ β© π | = π. By ππ β© π β [ππ]
π and the fact that β±π β β± , we infer theexistence of π β β± such that π β π . οΏ½
Definition 2.6 (Cummings-Foreman-Magidor, [CFM01]). ADSπ asserts the existence of a sequenceβ¨π΄πΌ | πΌ < π+β© such that:
(1) π΄πΌ is a cofinal subset of π of order-type cf(π);(2) For all π½ < π+, there exists some π : π½ β π such that the sequence β¨π΄πΌ β π(πΌ) | πΌ < π½β©
consists of pairwise disjoint sets.
Proposition 2.7. Suppose that ADSπ holds for a given singular strong limit cardinal π.
Then πΈπ+
=cf(π) β π½ [π+].
Proof sketch. Since π½ [π+] is π+-complete, it suffices to prove that πΈπ+
π β π½ [π+] for every regularcardinal π < π with π = cf(π). Fix such a cardinal π. Since π is a strong limit and π = cf(π), wehave π(π, π) = π (cf. the proof of Claim 4.5.1). Then, continue as in the proof of Proposition 2.2,but using the sequence β¨π΄πΌ | πΌ < π+β© witnessing ADSπ, instead. οΏ½
It follows that the model of [GS08] admits a singular cardinal π for which π½ [π+] β πΌ[π+].
2.2. Negative results.
Proposition 2.8. Suppose that π is an uncountable cardinal.
For every π β π½ [π+], we have that π β© πΈπ+
cf(π) is nonstationary.
Proof. Suppose not. In particular, there exists a sequence of regressive functions β¨ππ : π+ β π+ | π < π+β©and an ordinal πΌ β πΈπ+
cf(π) with πΌ > π such that for every regressive function π : πΌ β πΌ, and every
cofinal subset π΅ β πΌ, there exists some π < πΌ such that sup{π½ β π΅ | ππ(π½) = π(π½)} = πΌ.For all π½ < π+, fix a bijection ππ½ : πβ max{π, π½}. For all π < πΌ, define ππ : πΌβ π by stipulating:
ππ(π½) := πβ1π½ (ππ(π½)).
Let β¨ππ | π < cf(π)β© be a strictly increasing sequence of ordinals converging to π. Let β¨π½π | π < cf(π)β©be a strictly increasing sequence of ordinals converging to πΌ, with π½0 > π. Now, pick some functionπ : πΌβ π that satisfies for all π < cf(π) :
π(π½π) := min(π β {ππ(π½π) | π β ππΌ[ππ ]}).
Define a regressive function π : πΌβ πΌ by stipulating:
π(π½) :=
{0, if π½ < π;
ππ½(π(π½)), otherwise.
Now, by the choice of β¨ππ | π < π+β©, we may find some π < πΌ such that
sup{π½π | π < cf(π), ππ(π½π) = π(π½π)} = πΌ.
In particular, sup{π < cf(π) | ππ(π½π) = π(π½π)} = cf(π). By π β ππΌ[π], let us fix a large enoughπ* < cf(π) such that π β ππΌ[ππ* ]. By definition of π, then, π(π½π) = ππ(π½π) for all π β [π*, cf(π)). Thisis a contradiction. οΏ½
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 7
Corollary 2.9. If non(β³) > β΅1, then π½ [β΅2] is the nonstationary ideal over β΅2.10
Proof. Towards a contradiction, suppose that π½ [β΅2] is not the nonstationary ideal over β΅2. By
Proposition 2.8, there exists a sequence of functions β¨ππ : β΅2 β β΅2 | π < β΅2β© and an ordinal πΌ β πΈβ΅2β΅0
with πΌ > π1 such that for every regressive function π : πΌ β πΌ, and every cofinal subset π΅ β πΌ,there exists some π < πΌ such that sup{π½ β π΅ | ππ(π½) = π(π½)} = πΌ.
Let β¨π½π | π < πβ© be a strictly increasing sequence of ordinals converging to πΌ, with π½0 > π. Forall π < πΌ, define a real ππ : π β π by stipulating:
ππ(π) :=
{ππ(π½π), if ππ(π½π) < π;
0, otherwise.
By non(β³) > β΅1 = |πΌ| and [Bar87, S2], we may pick a real π : π β π such that for all π < πΌ,π(π) = ππ(π) for all but finitely many π < π. Pick a regressive function π : πΌ β πΌ such thatπ(π½π) = π(π) for all π < π. Put π΅ := {π½π | π < π}. Then, there exists no π < πΌ such thatsup{π½ β π΅ | ππ(π½) = π(π½)} = πΌ. οΏ½
Proposition 2.10. For every inaccessible cardinal π , π½ [π ] is the nonstationary ideal over π .
Proof. Suppose not. In particular, we may fix an inaccessible cardinal π , a sequence of functionsβ¨ππ : π β π | π < π β©, and an uncountable limit cardinal π satisfying the following. For every re-gressive function π : π β π, and every cofinal subset π΅ β π, there exists some π < π such thatsup{π½ β π΅ | ππ(π½) = π(π½)} = π.
Let π΄ denote the set of cardinals below π. Pick a regressive function π : πβ π such that for allπ β π΄:
π(π+) := min(π+ β {ππ(π+) | π β€ π}).
Put π΅ := {π+ | π β π΄}. Then π΅ is a cofinal subset of π. For all π < π, we have
sup{π½ β π΅ | ππ(π½) = π(π½)} β€ |π|+ < π,
contradicting the choice of β¨ππ | π < π β©. οΏ½
3. Diamond and club hitting
Definition 3.1 (Jensen, [Jen72]). For a stationary subset π of a regular uncountable cardinal π ,β’(π ) asserts the existence of a sequence β¨π΄πΎ | πΎ < π β© such that for every π΄ β π , the intersectionπΊ(π΄) β© π is stationary, where πΊ(π΄) := {πΎ < π | π΄ β© πΎ = π΄πΎ}.
Proposition 3.2 (folklore). Suppose that π is a stationary subset of a regular uncountable cardinalπ , and β’(π ) holds. Then there exists a matrix
β¨π΄π
πΎ | π < π , πΎ < π β©, such that for every sequence
οΏ½οΏ½ =β¨π΄π | π < π
β©of cofinal subsets of π , the intersection πΊ(οΏ½οΏ½) β© π is stationary, where
πΊ(οΏ½οΏ½) := {πΎ < π | βπ < πΎ(sup(π΄π β© πΎ) = πΎ & π΄π β© πΎ = π΄ππΎ)}.
Proof. Let β¨π΄πΎ | πΎ < π β© be a witness to β’(π ). Fix a bijection π : π Γ π β π . For all π < π and
πΎ < π , let π΄ππΎ := {πΏ < πΎ | π(π, πΏ) β π΄πΎ}. To see that
β¨π΄π
πΎ | π < π , πΎ < π β©
works, let οΏ½οΏ½ =β¨π΄π | π < π
β©be an arbitrary sequence of cofinal subsets of π . For all π < π , π·π := {πΎ < π | sup(π΄π β© πΎ) = πΎ} isa club, and hence π· := {πΎ β β³π<π π·π | π[πΎ Γ πΎ] = πΎ} is a club. Put π΄ := {π(π, πΏ) | π < π , πΏ β π΄π}.
Then πΊ := πΊ(π΄) β© π β©π· is stationary, and πΊ β πΊ(οΏ½οΏ½) β© π , so we are done. οΏ½
Fact 3.3 (Shelah, [She10]). For every uncountable cardinal π, CHπ entails β’(π+).
10Here, βnon(β³) > β΅1β stands for the assertion that every β΅1-sized set of reals is meager. This assertion is consistentwith ZFC + Β¬CH, as witnessed by the model of [ST71].
8 ASSAF RINOT
Definition 3.4. Suppose that π is a stationary subset of a regular uncountable cardinal π .A sequence β¨πΆπΌ | πΌ < π β© is said to hit clubs at π iff for every club π· β π , there exists some πΌ β π
such that sup(nacc(πΆπΌ) β©π·) = πΌ.
Proposition 3.5. Suppose that π is a regular cardinal β₯ β΅2.Then οΏ½(π ) holds iff for every stationary π β π , there exists a coherent πΆ-sequence, β¨πΆπΌ | πΌ < π β©,
that hits clubs at π.
Proof. (β=) Suppose that β¨πΆπΌ | πΌ < π β© is a coherent πΆ-sequence that hits clubs at π := acc(π ).Let πΆ be an arbitrary club in π , and we shall find some πΌ β acc(πΆ) such that πΆπΌ = πΆ β© πΌ.
Consider the club π· := acc(πΆ). Pick πΌ β π such that sup(nacc(πΆπΌ) β© π·) = πΌ. Then πΌ βacc(π·) β acc(πΆ) and nacc(πΆπΌ) β© acc(πΆ) = β , so that πΆπΌ = πΆ β© πΌ.
(=β) Suppose that οΏ½(π ) holds, as witnessed by β¨πΆπΌ | πΌ < π β©. Let π be an arbitrary stationarysubset of π . The proof is a simple combination of ideas from [She94b, S2] and [Rin14a, S3].
For every club πΈ β π and πΌ < π , denote
πΆπΌ[πΈ] := {sup(πΈ β© π) | π β (πΆπΌ β (min(πΈ) + 1))}.Notice that if πΌ β acc(πΈ), then πΆπΌ[πΈ] is a club in πΌ, and acc(πΆπΌ[πΈ]) = acc(πΆπΌ) β© acc(πΈ).
Claim 3.5.1. There exists a club πΈ β π satisfying the following.For every club π· β π , there exists some πΏ β π such that sup(nacc(πΆπΏ[πΈ]) β©π·) = πΏ.
Proof. Suppose not. Then, we can recursively construct a β-decreasing sequence of clubs in π ,β¨π·π | π < π1β©, such that π·0 = π , π·π =
βπ<ππ·π for all nonzero limit π < π1, and for all π < π1 and
πΏ β π,
sup(nacc(πΆπΏ[π·π]) β©π·π+1) < πΏ.
As π = cf(π ) β₯ β΅2, π· :=β
π<π1π·π is a club. We claim that π΅ := {πΏ β π | sup((π·β© πΏ) βπΆπΏ) = πΏ}
is nonempty.Suppose not. Then there exists some π < π for which
π» := {πΏ β π | sup((π· β© πΏ) β πΆπΏ) = π} β© acc(π· β π)is stationary in π . Consequently, for every πΌ < πΏ both in π», we have πΌ β acc(πΆπΏ) and hence πΆπΌ βπΆπΏ. So {πΆπΏ | πΏ β π»} is an β-chain, converging to the club πΆ :=
β{πΆπΏ | πΏ β π»}. Let πΌ β acc(πΆ) be
arbitrary. Then πΌ β acc(πΆπΏ) for some πΏ β π», and then πΆ β© πΌ = (πΆ β© πΏ) β© πΌ = πΆπΏ β© πΌ = πΆπΌ. Thatis, πΆ contradicts the fact that β¨πΆπΌ | πΌ < π β© is a οΏ½(π )-sequence.
Thus, we have established that π΅ is nonempty. Pick πΏ β π΅. For all π < π1, by the choice ofπ·π+1, we have that πΏπ := sup(nacc(πΆπΏ[π·π]) β©π·π+1) is < πΏ. We consider two cases:I If cf(πΏ) > β΅0, then by πΏ β π΅, let us pick π½ β (π·β©πΏ)βπΆπΏ above supπ<π πΏπ. Put πΎ := min(πΆπΏβπ½).
Then πΏ > πΎ > π½, and for all π < π, since π½ β π·π, we infer that sup(π·π β© πΎ) β₯ π½. So min(πΆπΏ[π·π] βπ½) = sup(π·π β© πΎ) for all π < π. Since {π·π | π < π} is a decreasing chain, there exists some π < πsuch that sup(π·πβ©πΎ) = sup(π·π+1β©πΎ). Fix such an π. Then min(πΆπΏ[π·π]βπ½) = min(πΆπΏ[π·π+1]βπ½),and in particular, π½* := min(πΆπΏ[π·π] β π½) is in π·π+1.
β If π½* β nacc(πΆπΏ[π·π]), then we get a contradiction to the fact that π½* β₯ π½ > πΏπ.β If π½* β acc(πΆπΏ[π·π]), then π½* = π½ and π½* β acc(πΆπΏ), contradicting the fact that π½ β πΆπΏ.
I If cf(πΏ) = β΅0, then we may pick an uncountable πΌ β π1, for which supπβπΌ πΏπ < πΏ.11 ByπΏ β π΅, pick π½ β (π· β© πΏ) β πΆπΏ above supπβπΌ πΏπ, and put πΎ := min(πΆπΏ β π½). Then πΏ > πΎ > π½,and min(πΆπΏ[π·π] β π½) = sup(π·π β© πΎ) for all π < π1. Since {π·π | π < π1} is a decreasing chain
11See Fact 1.3 of [Rin10].
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 9
and πΌ is cofinal in π1, we may fix an π β πΌ such that sup(π·π β© πΎ) = sup(π·π+1 β© πΎ). Thenmin(πΆπΏ[π·π] β π½) = min(πΆπΏ[π·π+1] β π½), and in particular, π½* := min(πΆπΏ[π·π] β π½) is in π·π+1.
β If π½* β nacc(πΆπΏ[π·π]), then we get a contradiction to the fact that π½* β₯ π½ > πΏπ.β If π½* β acc(πΆπΏ[π·π]), then π½* = π½ and π½* β acc(πΆπΏ), contradicting the fact that π½ β πΆπΏ.
οΏ½
Let πΈ be given by Claim 3.5.1. For all πΌ < π , define
π·πΌ :=
{πΆπΌ[πΈ], if πΌ β acc(πΈ);
πΌ β sup(πΈ β© πΌ), if πΌ β acc(πΈ).
Claim 3.5.2. β¨π·πΌ | πΌ < π β© is a coherent πΆ-sequence.
Proof. Let πΌ < π be an arbitrary limit ordinal. It is easy to see that π·πΌ is a club in πΌ. Supposethat we are given οΏ½οΏ½ β acc(π·πΌ). We shall prove that π·οΏ½οΏ½ = π·πΌ β© οΏ½οΏ½.
Put π := sup(πΈ β© πΌ), and consider two cases:I If π < πΌ, then π·πΌ is the interval [π, πΌ), and (π, πΌ)β©πΈ = β . So, οΏ½οΏ½ β (π, πΌ) and sup(πΈ β© οΏ½οΏ½) = π .
Consequently, π·πΌ β© οΏ½οΏ½ = π·οΏ½οΏ½.I If π = πΌ, then π·πΌ is the club πΆπΌ[πΈ], and by οΏ½οΏ½ β acc(π·πΌ), we have οΏ½οΏ½ β acc(πΆπΌ) β© acc(πΈ),
meaning that πΆοΏ½οΏ½ = πΆπΌ β© οΏ½οΏ½ and π·οΏ½οΏ½ = πΆοΏ½οΏ½[πΈ]. Consequently, π·πΌ β© οΏ½οΏ½ = π·οΏ½οΏ½. οΏ½
Finally, let π· be an arbitrary club in π . By the choice of πΈ, let us pick πΌ β π such thatsup(nacc(πΆπΌ[πΈ])β©π·) = πΌ. In particular, π·πΌ = πΆπΌ[πΈ], and sup(nacc(π·πΌ)β©π·) = πΌ, as sought. οΏ½
4. Main Results
An examination of all β’-based constructions of π -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 π -tree. As here, we are interested in deriving more than a Mahlo cardinal, we mustdevise a method of deriving a π -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 π , and a stationarysubset π β π , οΏ½β(π) asserts the existence of a coherent πΆ-sequence, β¨πΆπΌ | πΌ < π β©, such that forevery cofinal subset π΄ β π , there exists some πΌ β π for which sup(nacc(πΆπΌ) β©π΄) = πΌ.
Fact 4.2 (Brodsky-Rinot, [BR15]). Suppose π is a regular uncountable cardinal, and β’(π ) holds.
(a) If οΏ½β(π ) holds, then there exists a club-regressive π -Souslin tree;(b) If οΏ½β(πΈπ
β₯π) holds, and π<π < π for all π < π , then there exists a π-complete π -Souslin tree.
By [Rin14b, Theorem A], if π is an uncountable cardinal and CHπ holds, then οΏ½π entails οΏ½β(π)
for every stationary π β πΈπ+
=cf(π). One goal of the current section is to relax οΏ½π to οΏ½(π+).12
Theorem 4.3. Suppose that π is a regular uncountable cardinal, and οΏ½(π ) + β’(π ) holds.Then οΏ½β(π) holds for every stationary π β π½ [π ].
Proof. Suppose that π β π½ [π ] is a given stationary set. Fix a club πΆ β π and a sequence of regressivefunctions β¨ππ : π β π | π < π β© as in Definition 2.1. By β’(π ), let us fix a matrix
β¨π΄π
πΎ | π < π , πΎ < π β©
as in Proposition 3.2. For all π, π½ < π , put
πππ½ := {π½} βͺπ΄π
ππ(π½).
12In light of Proposition 2.7, we mention that by [She82, p. 440],οΏ½π entails ADSπ.
10 ASSAF RINOT
Since π is stationary, we infer from Proposition 2.8 that π β₯ β΅2. So, by οΏ½(π ), appeal toProposition 3.5 to obtain a coherent πΆ-sequence, β¨πΆπΌ | πΌ < π β©, that hits clubs at π β© πΆ. Finally,for all π, πΌ < π , let
πΆππΌ := πΆπΌ βͺ {min(ππ
π½ β (sup(πΆπΌ β© π½) + 1)) | π½ β nacc(πΆπΌ) & π½ > 0}.
Claim 4.3.1. Let π < π be arbitrary. Thenβ¨πΆππΌ | πΌ < π
β©is a coherent πΆ-sequence.
Proof. Let πΌ < π be an arbitrary limit ordinal. By πΆπΌ β πΆππΌ, we have sup(πΆπ
πΌ) = πΌ. For allsuccessive points π½β < π½ from πΆπΌ, the relative interval (π½β, π½)β©πΆπ
πΌ contains at most one element,thus, as πΆπΌ is closed below πΌ, so does πΆπ
πΌ. Now, suppose that οΏ½οΏ½ β acc(πΆππΌ). Then οΏ½οΏ½ β acc(πΆπΌ), and
hence πΆοΏ½οΏ½ = πΆπΌβ©οΏ½οΏ½. Then the local nature of the definition of πΆππΌ makes it clear that πΆπ
πΌβ©οΏ½οΏ½ = πΆποΏ½οΏ½. οΏ½
Claim 4.3.2. There exists an π < π for whichβ¨πΆππΌ | πΌ < π
β©is a οΏ½β(π)-sequence.
Proof. Suppose not. It then follows from Claim 4.3.1 that there exists a sequence of cofinal subsets
of π , οΏ½οΏ½ =β¨π΄π | π < π
β©, such that for all π < π and πΌ β π, we have sup(nacc(πΆπ
πΌ) β© π΄π) < πΌ.
Let πΊ be πΊ(οΏ½οΏ½) as in the statement of Proposition 3.2. Then πΊ is a stationary subset of π , andπ· := {π½ < π | sup(πΊ β© π½) = π½ > 0} is a club in π . As β¨πΆπΌ | πΌ < π β© hits clubs at π β© πΆ, let uspick πΌ β π β© πΆ such that sup(nacc(πΆπΌ) β© π·) = πΌ. Put π΅ := nacc(πΆπΌ) β© π·. For all π½ β π΅, byπ½ β π·, we know that the relative interval πΊβ© (sup(πΆπΌ β©π½), π½) is nonempty. Consequently, we mayfind some regressive function π : πΌ β πΌ such that π(π½) β πΊ β© (sup(πΆπΌ β© π½), π½) for all π½ β π΅. Pickπ < πΌ and a cofinal subset π΅β² β π΅ such that ππ οΏ½ π΅β² = π οΏ½ π΅β². Fix a large enough π β πΆπΌ such thatsup(πΆπΌ β© π) β₯ π. By omitting an initial segment, we may assume that π΅β² β© π = β .
Let π½ β π΅β² be arbitrary. Write πΎ := ππ(π½). Then πΎ β πΊβ© (sup(πΆπΌ β©π½), π½) β (π, π½). In particular,sup(π΄πβ©πΎ) = πΎ and π΄πβ©πΎ = π΄π
πΎ , so that πππ½ = {π½}βͺ(π΄πβ©πΎ), and min(ππ
π½β(sup(πΆπΌβ©π½)+1)) β π΄πβ©πΎ.
Thus we have shown that for all π½ β π΅β², πΆππΌ β© π΄π β© (sup(πΆπΌ β© π½), π½) is a singleton, contradicting
the fact that sup(nacc(πΆππΌ) β©π΄π) < πΌ = sup(π΅β²). οΏ½
This completes the proof. οΏ½
Remark. Notice that since π½ [π ] is an ideal, the conclusion of the preceding theorem may be strength-ened to: if π β π½ [π ], then οΏ½β(π ) holds for every π β π«(π ) such that π β© π is stationary. Comparethis with Kunenβs theorem [Kun80] that for π β π , β’*(π) entails that β’(π ) holds for every π β π«(π )such that π β© π is stationary.
Corollary 4.4. Suppose that π < cf(π) β€ π are cardinals, π(π, π) = π and 2π = π+.
For every stationary π β πΈπ+
π , οΏ½(π+) entails οΏ½β(π).
Proof. Suppose that π β πΈπ+
π is a given stationary set. By Proposition 2.2, πΈπ+
π β π½ [π+]. Sincethe latter is an ideal, we get that π β π½ [π+]. By CHπ, π β₯ β΅1 and Fact 3.3, β’(π+) holds. So, byTheorem 4.3, οΏ½(π+) entails οΏ½β(π). οΏ½
Corollary 4.5. Suppose that GCH holds, and π is an uncountable cardinal.
Then οΏ½(π+) entails that οΏ½β(πΈπ+
π ) holds for every regular cardinal π < π.
Proof. By Corollary 4.4, οΏ½(π+)+GCH entails οΏ½β(π) for every stationary π β πΈπ+
<cf(π). Thus, from
now on, suppose that π is a singular cardinal, and π is a regular cardinal in [cf(π), π). If π = cf(π),then let πβ² := π+; otherwise, let πβ² := π.
Claim 4.5.1. There exists a stationary π β πΈπ+
πβ² such that π β π½ [π+].
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 11
Proof. By Corollary 2.5, it suffices to prove that π(π, πβ²) = π. Let Ξ£ : cf(π) β π be an increasing
function whose image is cofinal in π. Put β± :=β{[πΌ]π
β² | πβ² β€ πΌ < π}. By GCH, π is a strong
limit, and hence |β±| = π. Let π΅ β [π]πβ²
be arbitrary. If there exists an ordinal πΌ < π such that|π΅ β© πΌ| = πβ², then π΅ β© πΌ is a subset of π΅ that belongs to β± , as sought.
Towards a contradiction, suppose that this is not case, and define β : cf(π) β πβ² by stipulatingβ(π) := |π΅ β©Ξ£(π)|. Since cf(π) < cf(πβ²), there exists some large enough π < πβ² such that Im(β) β π.But then |π΅| β€ max{cf(π), |π|} < πβ². This is a contradiction. οΏ½
Let π be given by the preceding claim. Suppose that οΏ½(π+) holds. By CHπ, π β₯ β΅1 and Fact 3.3,
β’(π+) holds. So, by Theorem 4.3, οΏ½β(π) holds, let alone οΏ½β(πΈπ+
πβ² ). In particular, if πβ² = π, thenwe are done.
Suppose now that πβ² > π. Let β¨πΆπΌ | πΌ < π+β© be a witness to οΏ½β(πΈπ+
πβ² ). We claim that the same
sequence witnesses οΏ½β(πΈπ+
π ). To see this, let π΄ β π+ be an arbitrary cofinal subset of π+. FixπΌ β π such that sup(nacc(πΆπΌ)β©π΄) = πΌ. As cf(πΌ) > π, we may pick some οΏ½οΏ½ β acc(πΆπΌ) of cofinalityπ such sup(nacc(πΆπΌ) β©π΄ β© οΏ½οΏ½) = οΏ½οΏ½. Then πΆοΏ½οΏ½ = πΆπΌ β© οΏ½οΏ½ and sup(nacc(πΆοΏ½οΏ½) β©π΄) = οΏ½οΏ½, as sought. οΏ½
Corollary 4.6. Suppose that π is a successor of a regular cardinal π.
If every stationary subset of πΈπ+
π reflects, then οΏ½β(πΈπ+
π ) entails οΏ½β(πΈπ+
π ).
Proof. Suppose that every stationary subset of πΈπ+
π reflects, and that οΏ½οΏ½ = β¨πΆπΌ | πΌ < π+β© is a
witness to οΏ½β(πΈπ+
π ). We claim that οΏ½οΏ½ is also a witness to οΏ½β(πΈπ+
π ).Let π΄ be an arbitrary cofinal subset of π+. First, let us point out that
π := {οΏ½οΏ½ β πΈπ+
π | sup(nacc(πΆοΏ½οΏ½) β©π΄) = οΏ½οΏ½}is stationary in π+. To see this, notice that given any club π· in π+, we may find some sparseenough cofinal subset π΄β² β π΄ such that for all πΎ < πΏ both from π΄β², the relative interval (πΎ, πΏ) β©π·is nonempty; then, by the choice of οΏ½οΏ½, we pick οΏ½οΏ½ β πΈπ+
π such that sup(nacc(πΆοΏ½οΏ½) β© π΄β²) = οΏ½οΏ½, andhence οΏ½οΏ½ β acc(π·), so that οΏ½οΏ½ β π β©π·.
As every stationary subset of πΈπ+
π reflects, let us pick πΌ β πΈπ+
π such that π β© πΌ is stationary.Put π΅ := π β© acc(πΆπΌ). Then π΅ is stationary in πΌ, and for all οΏ½οΏ½ β π΅, we have sup(nacc(πΆπΌ)β©π΄) β₯sup(nacc(πΆοΏ½οΏ½) β©π΄) = οΏ½οΏ½. Consequently, sup(nacc(πΆπΌ) β©π΄) = sup(π΅) = πΌ, as sought. οΏ½
Corollary 4.7. For every cardinal π β₯ iπ, CHπ +οΏ½(π+) entails οΏ½β(π+), and hence the existenceof a club-regressive π+-Souslin tree.
Proof. Suppose that we are given a cardinal π β₯ iπ. By the main result of [She00] (see also [Koj15,S2]), there exists some regular cardinal π < iπ such that π(π, π) = π. So, by Corollary 2.5, let us
fix a stationary π β πΈπ+
π which is in π½ [π+]. Assume that CHπ + οΏ½(π+) holds. By CHπ, π β₯ β΅1
and Fact 3.3, β’(π+) holds. Then, by Theorem 4.3, οΏ½β(π) holds, let alone οΏ½β(π+).By Fact 4.2(a), then, there exists a club-regressive π+-Souslin tree. οΏ½
Corollary 4.8. Suppose that GCH holds, and π is an uncountable cardinal.
(1) If οΏ½(π+) holds, then there exists a club-regressive π+-Souslin tree;(2) If there are no club-regressive π+-Souslin trees, then π+ is weakly compact in πΏ;(3) If there are no β΅2-Souslin trees and no β΅3-Souslin trees, then the Axiom of Determinacy
holds in πΏ(R).
Proof. (1) By Corollary 4.5, GCH+οΏ½(π+) entails οΏ½β(πΈπ+
π ). In particular, οΏ½β(π+) holds. By CHπ,π β₯ β΅1 and Fact 3.3, β’(π+) holds. Finally, by Fact 4.2(a), we infer the existence of a club-regressiveπ+-Souslin tree.
12 ASSAF RINOT
(2) By [Tod87], if π is a regular uncountable cardinal and οΏ½(π ) fails, then π is weakly compactin πΏ. Now, appeal to Clause (1).
(3) By [SS14], Β¬οΏ½(π2)+Β¬οΏ½π2+2β΅1 = β΅2 implies that πΏ(R) |= AD. Now, appeal to Clause (1). οΏ½
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* πΆ-sequence (over a regular uncountable cardinal π ) is a sequenceβ¨πΆπΌ | πΌ < π β© such that:
(1) for all πΌ < π , πΆπΌ β πΌ;(2) for all limit πΌ < π , πΆπΌ is a club in πΌ;(3) for all πΌ < π , if οΏ½οΏ½ β acc(πΆπΌ), then sup((πΆοΏ½οΏ½βπΆπΌ) β© οΏ½οΏ½) < οΏ½οΏ½.
Definition 4.10. For a regular uncountable cardinal π , and a stationary subset π β π , οΏ½β²(π)asserts the existence of a coherent* πΆ-sequence, β¨πΆπΌ | πΌ < π β©, such that for every cofinal subsetπ΄ β π , there exists some πΌ β π for which sup(nacc(πΆπΌ) β©π΄) = πΌ.
The very same construction from the proof of Fact 4.2(b) demonstrates (cf. [BR16, S2]):
Proposition 4.11. Suppose that π is a regular uncountable cardinal, and β’(π ) +οΏ½β²(πΈπ β₯π) holds.
If π<π < π for all π < π , then there exists a π-complete π -Souslin tree.
Lemma 4.12. Suppose that π is a regular uncountable cardinal, and β’(π ) +οΏ½β²(π ) holds.Then there exists a partition π = π0 β π1 such that β’(π0) +οΏ½β²(π1) holds.
Proof. By β’(π ) and Devlinβs lemma [Dev78], let us fix a partition π = π0 β π1 such that β’(π0) +β’(π1) holds. Of course, it suffices to show that there exists some π < 2 such that οΏ½β²(ππ) holds.
Towards a contradiction, suppose that this is not the case. Let β¨πΆπΌ | πΌ < π β© be a οΏ½β²(π )-sequence.By β’(π ), let us fix a matrix A =
β¨π΄π
πΎ | π < π , πΎ < π β©
as in Proposition 3.2.Let π < 2 be arbitrary. Put
β πππΎ := {πΎ} βͺπ΄π
πΎ for all πΎ < π , and then
β πΆππΌ := πΆπΌ βͺ {min(ππ
πΎ β (sup(πΆπΌ β© πΎ) + 1)) | πΎ β nacc(πΆπΌ) & πΎ > 0} for all πΌ < π .
The proof of Claim 4.3.1 makes it clear thatβ¨πΆππΌ | πΌ < π
β©is a coherent* πΆ-sequence. So, since
οΏ½β²(ππ) fails, we may fix a cofinal subset π΄π β π , such that sup(nacc(πΆππΌ) β© π΄π) < πΌ for all πΌ β ππ.
Put πΊπ := {πΎ < π | sup(π΄π β© πΎ) = πΎ & π΄π β© πΎ = π΄ππΎ & πΎ > 0}.
Now, by the choice of A, πΊ := πΊ0 β©πΊ1 is stationary. As β¨πΆπΌ | πΌ < π β© is a οΏ½β²(π )-sequence, let usfix πΌ < π such that sup(nacc(πΆπΌ) β©πΊ) = πΌ. Let π < 2 be the unique integer such that πΌ β ππ. LetπΎ be an arbitrary element of Ξ := nacc(πΆπΌ)β©πΊ. Then π΄π
πΎ = π΄π β© πΎ is a cofinal subset of πΎ, so that
min(πππΎβ(sup(πΆπΌβ©πΎ)+1)) β π΄πβ©πΎ. Thus we have shown that for all πΎ β Ξ, πΆπ
πΌβ©π΄πβ©(sup(πΆπΌβ©πΎ), πΎ)
is a singleton, contradicting the fact that sup(nacc(πΆππΌ) β©π΄π) < sup(Ξ) = πΌ β ππ. οΏ½
Theorem 4.13. Suppose that π is a regular cardinal β₯ β΅2, and β’(π ) +οΏ½β²(π ) holds.Then οΏ½β²(π) holds for every stationary subset π β π .
Proof. By Lemma 4.12, let us fix a partition π = π0βπ1 such that β’(π0)+οΏ½β²(π1) holds. By β’(π0),let us fix a matrix M =
β¨π΄π
πΎ | π < π , πΎ < π β©
as in Proposition 3.2. Let π be an arbitrary stationarysubset of π . Towards showing that οΏ½β²(π) holds, we prove the following.
Claim 4.13.1. There exist some π < π1 and a coherent* πΆ-sequence, β¨πΆπΌ | πΌ < π β©, such that forevery subset π΄ β π and every club π· β π , there exists some πΏ β π such that
sup{πΎ β nacc(πΆπΏ) β©π· β© π0 | π΄ππΎ = π΄ β© πΎ} = πΏ.
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 13
Proof. Suppose not. Building on ideas from [KS93], we shall recursively construct a sequence of
triples β¨(π΄π, π·π,ββπΆπ) | π < π1β©, such that for all π < π1:
(1) π΄π is a cofinal subset of π ;(2) π·π is a club in π ;
(3)ββπΆπ =
β¨πΆππΏ | πΏ < π
β©is a coherent* πΆ-sequence. In addition, for all πΏ < π :
(a) πΆππΏ+1 = {πΏ} and πΆπ
πΏ β πΆπ+1πΏ ;
(b) if πΌ β acc(πΆππΏ) and π β πΆπ
πΌβ©πΆππΏ satisfies sup((πΆπ
πΌβπΆππΏ)β©πΌ) β€ π, then sup((πΆπ+1
πΌ βπΆπ+1πΏ )β©
πΌ) β€ π;
(c) if π is a limit nonzero ordinal, then πΆππΌ =
βπ<ππΆ
ππΌ and acc(πΆπ
πΏ) =β
π<π acc(πΆππΏ ).
Whenever π΄π and π·π will be defined, we shall also derive the set πΊπ, by letting:
πΊπ := {πΎ β π·π β© π0 | π΄ππΎ = π΄π β© πΎ}.
Here comes the recursion:
I Put π΄0 := π and π·0 := π . LetββπΆ0 =
β¨πΆ0πΏ | πΏ < π
β©be some witness to οΏ½β²(π1). Of course, we
may also assume that πΆ0πΏ+1 = {πΏ} for all πΏ < π .
I Suppose that π < π1, and β¨(π΄π , π·π ,ββπΆπ) | π β€ πβ© has already been defined. By the indirect
hypothesis, let us fix some cofinal subset π΄π+1 β π and some club π·π+1 β π such that for all πΏ β π,we have:
sup(nacc(πΆππΏ) β©πΊπ+1) < πΏ.
Next, defineβββπΆπ+1 =
β¨πΆπ+1πΏ | πΏ < π
β©by recursion on πΏ < π , as follows. Let πΆπ+1
0 := β , and
πΆπ+1πΏ+1 := {πΏ} for all πΏ < π . Now, if πΏ < π is a nonzero limit ordinal and
β¨πΆπ+1πΎ | πΎ < πΏ
β©has already
been defined, let:
(β)π πΆπ+1πΏ := πΆπ
πΏ βͺβ
{πΆπ+1πΎ β sup(πΆπ
πΏ β© πΎ) | πΎ β nacc(πΆππΏ) βπΊπ+1}.
It is easy to see that Clauses (3)(a) and (3)(b) hold for π. So, asββπΆπ is a coherent* πΆ-sequence,
we get thatβββπΆπ+1 is a coherent* πΆ-sequence.
I Suppose that π < π1 is a nonzero limit ordinal, and β¨(π΄π , π·π ,ββπΆπ) | π < πβ© has already been
defined. Put π΄π := π and π·π := π . For all πΏ < π , let πΆππΏ :=
βπ<ππΆ
ππΏ . As Clause (3) holds for
all π < π, to see thatββπΆπ =
β¨πΆππΏ | πΏ < π
β©is a coherent* πΆ-sequence, it suffice to show that for all
πΌ < πΏ < π such that sup(πΆππΏ β© πΌ) = πΌ > 0, there exists some π < π such that πΌ β acc(πΆπ
πΏ ).
Suppose that πΌ < πΏ are as above. By Clause (3)(a), β¨min(πΆππΏ β πΌ) | π < πβ© is a weakly decreasing
sequence of ordinals, and hence must stabilize at some π* < π. Denote πΎ := min(πΆπ*
πΏ β πΌ) and
πΎβ := sup(πΆπ*
πΏ β© πΎ). If πΎ ββ{nacc(πΆπ
πΏ ) | π* β€ π < π}, then we get by induction on π β [π*, π) β
using (β)π at successor stages and Clause (3)(c) at limit stages β that πΆππΏ β© (πΎβ, πΎ] = {πΎ} for all
π β [π*, π), contradicting the fact that πΎβ < sup(πΆππΏ β©πΌ) = πΌ β€ πΎ.13 Thus, pick π β [π*, π) such that
πΎ β acc(πΆππΏ ). As πΎ = min(πΆπ
πΏ β πΌ), this means that πΌ = πΎ, and hence πΌ β acc(πΆππΏ ), as sought.
At the end of the above process, we have obtained β¨(π΄π, π·π,ββπΆπ) | π < π1β©. Put πΊ :=
βπ<π1
πΊπ. By
the choice of the matrix M, we know that πΊ is a stationary subset of π0. AsββπΆ0 is a οΏ½β²(π1)-sequence,
13Note that πΎ β nacc(πΆπ*
πΏ ) =β πΎβ < πΌ.
14 ASSAF RINOT
for every π½ < π , there exists some πΌ β π1 such that sup(nacc(πΆ0πΌ) β© (πΊ β π½)) = πΌ. Consequently,
the following set is cofinal in π :
π΄ := {πΌ β π1 | sup(nacc(πΆ0πΌ) β©πΊ) = πΌ}.
Let πΌ β π΄ be arbitrary. Trivially, the set ΞπΌ := nacc(πΆ0πΌ) β© πΊ is cofinal in πΌ. Suppose that
π < π1, and ΞπΌ β nacc(πΆππΌ). Then, by ΞπΌ β πΊ β πΊπ+1 and (β)π, we also have ΞπΌ β nacc(πΆπ+1
πΌ ).Recalling Clauses (3)(a) and (3)(c), we altogether get that for all π < π:
(ββ)π ΞπΌ β nacc(πΆππΌ) β©πΊπ+1.
As π is stationary in π = sup(π΄), we now pick πΏ β π such that sup(π΄ β© πΏ) = πΏ. For all π < π1,by the choice of the pair (π΄π+1, π·π+1), the following ordinal is strictly smaller than πΏ:
πΏπ := sup(nacc(πΆππΏ) β© πΊπ+1).
At this stage, the analysis splits into two main cases and a few subcases:
Case 1. cf(πΏ) > β΅0: In this case, πΏ* := supπ<π πΏπ is below πΏ, so let us fix some πΌ β π΄β© (πΏ*, πΏ).
Denote πΎπ := min(πΆππΏ β πΌ) for all π < π. By Clause (3)(a), β¨πΎπ | π < πβ© is weakly decreasing,
so let us fix some π < π such that πΎπ = πΎπ+1. Now, there are three cases to consider, each ofwhich yielding a contradiction:Case 1.1. πΎπ > πΌ: By πΎπ = min(πΆπ
πΏ βπΌ) and πΎπ > πΌ, we have πΎπ β nacc(πΆππΏ). By πΎπ > πΌ >
πΏ* β₯ πΏπ, we moreover have πΎπ β nacc(πΆππΏ) βπΊπ+1. Denote πΎβπ := sup(πΆπ
πΏ β© πΎπ). By (β)π,we have
πΆπ+1πΏ β© [πΎβπ , πΎπ) = πΆπ+1
πΎπ β© [πΎβπ , πΎπ).
By πΎπ = min(πΆππΏ β πΌ) and πΎπ > πΌ, we have πΎβπ < πΌ, and hence
πΆπ+1πΏ β© [πΌ, πΎπ) = πΆπ+1
πΎπ β© [πΌ, πΎπ).
So πΎπ+1 = min(πΆπ+1πΏ β πΌ) = min(πΆπ+1
πΎπ β πΌ) < πΎπ, contradicting the choice of π.
Case 1.2. πΎπ = πΌ β acc(πΆππΏ): In this case, sup((πΆπ
πΏβπΆππΌ)β©πΌ) < πΌ and hence by (ββ)π, we
get that sup(nacc(πΆππΏ) β©πΊπ+1) β₯ πΌ, contradicting the fact that πΌ > πΏ* β₯ πΏπ.
Case 1.3. πΎπ = πΌ β nacc(πΆππΏ): Write πΌβ := sup(πΆπ
πΏ β©πΌ). As πΌ β π΄ β π1 and πΊπ+1 β π0,
we get from (β)π that πΆπ+1πΏ β©[πΌβ, πΌ) = πΆπ+1
πΌ β©[πΌβ, πΌ). In particular, by (ββ)π+1, we have
ΞπΌβ© [πΌβ, πΌ) β nacc(πΆπ+1πΏ )β©πΊπ+2. But then sup(nacc(πΆπ+1
πΏ )β©πΊπ+2) β₯ πΌ, contradictingthe fact that πΌ > πΏ* β₯ πΏπ+1.
Case 2. cf(πΏ) = β΅0: Pick an uncountable πΌ β π1 and some πΏ* < πΏ for which supπβπΌ max{πΏπ, πΏπ+1} β€πΏ*. Pick πΌ β π΄ β© (πΏ*, πΏ), and denote πΎπ := min(πΆπ
πΏ β πΌ) for all π < π1. By Clause (3)(a) andsince πΌ is cofinal in π1, let us fix some π β πΌ such that πΎπ = πΎπ+1. As before, there are threecases to consider:Case 2.1. πΎπ > πΌ: By πΎπ > πΌ > πΏ* β₯ πΏπ, we have πΎπ β nacc(πΆπ
πΏ) β πΊπ+1. Then, by (β)π,
we have πΆπ+1πΏ β© [πΌ, πΎπ) = πΆπ+1
πΎπ β© [πΌ, πΎπ). So πΎπ+1 = min(πΆπ+1πΏ βπΌ) = min(πΆπ+1
πΎπ βπΌ) < πΎπ,contradicting the choice of π.
Case 2.2. πΎπ = πΌ β acc(πΆππΏ): In this case, sup((πΆπ
πΏβπΆππΌ)β©πΌ) < πΌ and hence by (ββ)π, we
get that sup(nacc(πΆππΏ) β©πΊπ+1) β₯ πΌ, contradicting the fact that πΌ > πΏ* β₯ πΏπ.
Case 2.3. πΎπ = πΌ β nacc(πΆππΏ): Write πΌβ := sup(πΆπ
πΏ β© πΌ). As πΌ β πΊπ+1, we get from (β)π
that πΆπ+1πΏ β©[πΌβ, πΌ) = πΆπ+1
πΌ β©[πΌβ, πΌ). In particular, by (ββ)π+1, we have sup(nacc(πΆπ+1πΏ )β©
πΊπ+2) β₯ πΌ, contradicting the fact that πΌ > πΏ* β₯ πΏπ+1. οΏ½
Fix a coherent* πΆ-sequence β¨πΆπΌ | πΌ < π β© and some π < π1 as in Claim 4.13.1. Put:
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 15
β ππΎ := {πΎ} βͺπ΄ππΎ for all πΎ < π , and then
β π·πΌ := πΆπΌ βͺ {min(ππΎ β (sup(πΆπΌ β© πΎ) + 1)) | πΎ β nacc(πΆπΌ) & πΎ > 0} for all πΌ < π .
The proof of Claim 4.3.1 makes it clear that β¨π·πΌ | πΌ < π β© is a coherent* πΆ-sequence. Finally, let π΄be an arbitrary cofinal subset of π . Put π· := {πΎ < π | sup(π΄ β© πΎ) = πΎ > 0}. Pick πΏ β π such that
Ξ := {πΎ β nacc(πΆπΏ) β©π· β© π0 | π΄ππΎ = π΄ β© πΎ}
is cofinal in πΏ.Let πΎ be an arbitrary element of Ξ. Then π΄π
πΎ = π΄ β© πΎ is a cofinal subset of πΎ, so that min(ππΎ β(sup(πΆπΏ β© πΎ) + 1)) β π΄ β© πΎ. Thus we have shown that for all πΎ β Ξ, π·πΏ β© π΄ β© (sup(πΆπΏ β© πΎ), πΎ) is asingleton, and hence sup(nacc(π·πΏ) β©π΄) = πΏ, as sought. οΏ½
Corollary 4.14. For every uncountable cardinal π, GCH +οΏ½(π+) entails the existence of a cf(π)-complete π+-Souslin tree.
Proof. By Corollary 4.5, GCH + οΏ½(π+) entails οΏ½β(πΈπ+
π ). In particular, οΏ½β²(π+) holds, and then,
by Theorem 4.13, οΏ½β²(πΈπ+
cf(π)) holds. By CHπ, π β₯ β΅1 and Fact 3.3, β’(π+) holds. Finally, by GCH
and Proposition 4.11, we infer the existence of a cf(π)-complete π+-Souslin tree. οΏ½
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 7π‘β 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).
References
[Bar87] Tomek Bartoszynski. Combinatorial aspects of measure and category. Fund. Math., 127(3):225β239, 1987.[Bau76a] James E. Baumgartner. Almost-disjoint sets, the dense set problem and the partition calculus. Ann. Math.
Logic, 9(4):401β439, 1976.[Bau76b] James E. Baumgartner. A new class of order types. Ann. Math. Logic, 9(3):187β222, 1976.[BDS86] Shai Ben-David and Saharon Shelah. Souslin trees and successors of singular cardinals. Ann. Pure Appl.
Logic, 30(3):207β217, 1986.[BR15] Ari Meir Brodsky and Assaf Rinot. A microscopic approach to Souslin-tree constructions. Part I. arXiv
preprint arXiv:1601.01821, 2015. submitted December 2015.[BR16] Ari Meir Brodsky and Assaf Rinot. More notions of forcing add a Souslin tree. arXiv preprint
arXiv:1607.07033, 2016. submitted July 2016.[CFM01] James Cummings, Matthew Foreman, and Menachem Magidor. Squares, scales and stationary reflection.
J. Math. Log., 1(1):35β98, 2001.[CFM04] James Cummings, Matthew Foreman, and Menachem Magidor. Canonical structure in the universe of set
theory. I. Ann. Pure Appl. Logic, 129(1-3):211β243, 2004.[Cum97] James Cummings. Souslin trees which are hard to specialise. Proc. Amer. Math. Soc., 125(8):2435β2441,
1997.[Dav90] R. David. Some results on higher Suslin trees. J. Symbolic Logic, 55(2):526β536, 1990.[Dev78] Keith J. Devlin. A note on the combinatorial principles β’(πΈ). Proc. Amer. Math. Soc., 72(1):163β165,
1978.
16 ASSAF RINOT
[Dev83] Keith J. Devlin. Reduced powers of β΅2-trees. Fund. Math., 118(2):129β134, 1983.[DJ74] Keith J. Devlin and Havard Johnsbraten. The Souslin problem. Lecture Notes in Mathematics, Vol. 405.
Springer-Verlag, Berlin, 1974.[Eis10] Todd Eisworth. Successors of singular cardinals. In Handbook of set theory. Vols. 1, 2, 3, pages 1229β1350.
Springer, Dordrecht, 2010.[FM97] Matthew Foreman and Menachem Magidor. A very weak square principle. J. Symbolic Logic, 62(1):175β196,
1997.[Gre76] John Gregory. Higher Souslin trees and the generalized continuum hypothesis. J. Symbolic Logic, 41(3):663β
671, 1976.[GS08] Moti Gitik and Assaf Sharon. On SCH and the approachability property. Proc. Amer. Math. Soc.,
136(1):311β320 (electronic), 2008.[Han64] W. Hanf. Incompactness in languages with infinitely long expressions. Fund. Math., 53:309β324, 1963/1964.[HS85] Leo Harrington and Saharon Shelah. Some exact equiconsistency results in set theory. Notre Dame J.
Formal Logic, 26(2):178β188, 1985.[Jec67] Tomas Jech. Non-provability of Souslinβs hypothesis. Comment. Math. Univ. Carolinae, 8:291β305, 1967.[Jen68] Ronald B Jensen. Souslinβs hypothesis is incompatible with V=L. Notices Amer. Math. Soc, 15(6), 1968.[Jen72] R. Bjorn Jensen. The fine structure of the constructible hierarchy. Ann. Math. Logic, 4:229β308; erratum,
ibid. 4 (1972), 443, 1972. With a section by Jack Silver.[KLY07] Bernhard Konig, Paul Larson, and Yasuo Yoshinobu. Guessing clubs in the generalized club filter. Fund.
Math., 195(2):177β191, 2007.[KM78] A. Kanamori and M. Magidor. The evolution of large cardinal axioms in set theory. In Higher set theory
(Proc. Conf., Math. Forschungsinst., Oberwolfach, 1977), volume 669 of Lecture Notes in Math., pages99β275. Springer, Berlin, 1978.
[Koj15] Menachem Kojman. Splitting families of sets in ZFC. Adv. Math., 269:707β725, 2015.[KS93] Menachem Kojman and Saharon Shelah. π-complete Souslin trees on π+. Arch. Math. Logic, 32(3):195β201,
1993.[Kun80] Kenneth Kunen. Set theory, volume 102 of Studies in Logic and the Foundations of Mathematics. North-
Holland Publishing Co., Amsterdam-New York, 1980. An introduction to independence proofs.[Kur35] Duro Kurepa. Ensembles ordonnes et ramifies. Publications de lβInstitut Mathematique Beograd, 4:1β138,
1935.[LS81] Richard Laver and Saharon Shelah. The β΅2-Souslin hypothesis. Trans. Amer. Math. Soc., 264(2):411β417,
1981.[Mag82] Menachem Magidor. Reflecting stationary sets. J. Symbolic Logic, 47(4):755β771 (1983), 1982.[Mit73] William Mitchell. Aronszajn trees and the independence of the transfer property. Ann. Math. Logic, 5:21β46,
1972/73.[Rin10] Assaf Rinot. A relative of the approachability ideal, diamond and non-saturation. J. Symbolic Logic,
75(3):1035β1065, 2010.[Rin11a] Assaf Rinot. Jensenβs diamond principle and its relative. In Set theory and its applications, volume 533 of
Contemp. Math., pages 125β156. Amer. Math. Soc., Providence, RI, 2011.[Rin11b] Assaf Rinot. On guessing generalized clubs at the successors of regulars. Ann. Pure Appl. Logic, 162(7):566β
577, 2011.[Rin14a] Assaf Rinot. Chain conditions of products, and weakly compact cardinals. Bull. Symb. Log., 20(3):293β314,
2014.[Rin14b] Assaf Rinot. The Ostaszewski square and homogeneous Souslin trees. Israel J. Math., 199(2):975β1012,
2014.[Sch05] Ernest Schimmerling. A question about Suslin trees and the weak square hierarchy. Notre Dame J. Formal
Logic, 46(3):373β374 (electronic), 2005.[She79] Saharon Shelah. On successors of singular cardinals. In Logic Colloquium β78 (Mons, 1978), volume 97 of
Stud. Logic Foundations Math., pages 357β380. North-Holland, Amsterdam-New York, 1979.[She81] Saharon Shelah. Models with second order properties. III. Omitting types for πΏ(π). Arch. Math. Logik
Grundlag., 21(1-2):1β11, 1981.[She82] Saharon Shelah. Proper forcing, volume 940 of Lecture Notes in Mathematics. Springer-Verlag, Berlin-New
York, 1982.[She84] Saharon Shelah. An β΅2 Souslin tree from a strange hypothesis. Abs. Amer. Math. Soc., 160:198, 1984.
HIGHER SOUSLIN TREES AND THE GCH, REVISITED 17
[She93] Saharon Shelah. Advances in cardinal arithmetic. In Finite and infinite combinatorics in sets and logic(Banff, AB, 1991), volume 411 of NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., pages 355β383. KluwerAcad. Publ., Dordrecht, 1993.
[She94a] Saharon Shelah. β΅π+1 has a Jonsson Algebra. In Cardinal Arithmetic, volume 29 of Oxford Logic Guides.Oxford University Press, 1994.
[She94b] Saharon Shelah. There are Jonsson algebras in many inaccessible cardinals. In Cardinal Arithmetic, vol-ume 29 of Oxford Logic Guides. Oxford University Press, 1994.
[She00] Saharon Shelah. The generalized continuum hypothesis revisited. Israel J. Math., 116:285β321, 2000.[She10] Saharon Shelah. Diamonds. Proc. Amer. Math. Soc., 138(6):2151β2161, 2010.[Sou20] Mikhail Yakovlevich Souslin. Probleme 3. Fundamenta Mathematicae, 1(1):223, 1920.[Spe49] E. Specker. Sur un probleme de Sikorski. Colloquium Math., 2:9β12, 1949.[SS82a] S. Shelah and L. Stanley. Generalized Martinβs axiom and Souslinβs hypothesis for higher cardinals. Israel
J. Math., 43(3):225β236, 1982.[SS82b] S. Shelah and L. Stanley. π -forcing. i. a βblack-boxβ theorem for morasses, with applications to super-souslin
trees. Israel J. Math., 43(3):185β224, 1982.[SS88] Saharon Shelah and Lee Stanley. Weakly compact cardinals and nonspecial Aronszajn trees. Proc. Amer.
Math. Soc., 104(3):887β897, 1988.[SS14] Ralf Schindler and John Steel. The core model induction. 2014. unpublished.[ST71] R. M. Solovay and S. Tennenbaum. Iterated Cohen extensions and Souslinβs problem. Ann. of Math. (2),
94:201β245, 1971.[Ste05] John R. Steel. PFA implies ADπΏ(R). J. Symbolic Logic, 70(4):1255β1296, 2005.[Ten68] S. Tennenbaum. Souslinβs problem. Proc. Nat. Acad. Sci. U.S.A., 59:60β63, 1968.[Tod81] Stevo B. Todorcevic. Trees, subtrees and order types. Ann. Math. Logic, 20(3):233β268, 1981.[Tod87] Stevo Todorcevic. Partitioning pairs of countable ordinals. Acta Math., 159(3-4):261β294, 1987.[Tod89] Stevo Todorcevic. Special square sequences. Proc. Amer. Math. Soc., 105(1):199β205, 1989.
Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel.URL: http://www.assafrinot.com