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The Hebrew University, Jerusalem

arXiv:math/9209204v1 [math.LO] 10 Sep 1992µ-Complete Souslin Trees on µ+by Menachem Kojman and Saharon Shelah†The Hebrew University, JerusalemABSTRACTWe prove that µ = µ<µ, 2µ = µ+ and “there is a non reflecting stationary subset of µ+composed of ordinals of cofinality < µ” imply that there is a µ-complete Souslin tree on µ+.Introduction The old problem of the existence of Souslin trees has attracted the attention ofmany (see [Je] for history). While the ℵ1 case is settled, the consistency of GCH + SH(ℵ2) is stillan open question.

Gregory showed in [G] that GCH + “there is a non reflecting stationary set ofω-cofinal elements of ω2” implies the existence of an ℵ2-Souslin tree. Gregory’s result showed thatthe consistency strength of GCH + SH(ℵ2) is at least that of the existence of a Mahlo cardinal.Without GCH, the consistency of CH + SH(ℵ2) is known from [LvSh 104].

In [ShSt 279] theequiconsistency of the existence of a weakly compact cardinal with “every ℵ2-Aronszajn tree isspecial” is shown. In [ShSt 154] it is shown that under CH, the consistency strength of “there areno ℵ1-complete ℵ2-Souslin trees” is at least that of an inaccessible cardinal.We show how a Souslin tree which is µ-complete (µ regular) can be constructed on a cardinalµ+ from a certain combinatorial principle (Theorem 2 below), and then show how this principlemay be gotten from GCH and a non reflecting stationary set of ordinals with cofinality < µ inµ+ (Theorem 3 below).

As a corollary (Corollary 5 below), GCH + “there is a non reflectingstationary set of ω-cofinal elements of ω2” implies the existence of an ℵ1-complete Souslin tree onℵ2.As mentioned in [G], 1.10(3), CH and the existence of a diamond sequence on {δ < ℵ2 : cfδ =ℵ1} imply the existence of a Souslin tree on ℵ2 which is ℵ1-complete. The construction of such atree is by induction on levels, where at a stage of countable cofinality all branches are realized (forℵ1-completeness), while at stages of cofinality ℵ1 the diamond is consulted to realize only a partof the cofinal branches in a way which kills all future big antichains.

The combinatorial principleused in Theorem 2 to construct a µ-complete Souslin tree on µ+ under GCH can be viewed as aweaker substitute for a diamond sequence on {δ < µ+ : cfδ = µ}: instead of using a guess at asingle guessat stage of cofinality µ, we use unboundedly many guesses, each at a level of cofinality< µ.Unlike a diamond sequence on the stationary set of critical cofinality, this principle makes sensealso in the case of an inaccessible cardinal (where there is no “critical cofinality”). This principleis closely related to club guessing (see [Sh-g] and [Sh-e]), which was discovered while the second† The second author thanks the Binational Science Foundation for supporting this research.Publication no.

4491souslin12.8.2021

author was trying to prove some results in Model Theory. This principle continues the principlethat appears in [AbShSo 221], in which Souslin trees on successors of singulars are treated.We learned from the referee that Gregory presented in the seventies in a seminar at Buffaloa construction of a countably complete Souslin tree on ℵ2 from GCH and a square, but that thiswas not written.1.

Notation: : (1) If C is a set of ordinals, then acc C is the set of accumulation points of Cand nacc C =df C \ acc C. By Tα we denote the α-th level of the tree T and by T(α) we denoteSb<α Tβ.2. Theorem:Suppose that(a) λ = µ+ = 2µ, µ = µ<µ;(b) S∗⊆{α ∈λ : cfα = µ} and C = ⟨cδ : δ ∈S∗}, δ = sup cδ, cδ is a closed set of limit ordinals.

(c) For every δ ∈S∗and α ∈nacc cδ, Pδ,α ⊆P(α), |Pδ,α| ≤cfα, and if α ∈S∗, then |Pδ,α| < µ;(d) For every set A ⊆λ and club E ⊆λ, there is a stationary SA,E ⊆S∗such that for everyδ ∈SA,E, δ = sup{α ∈nacc cδ : A ∩α ∈Pδ,α ∧α ∈E};(e) If δ, δ∗∈S∗, δ ∈acc cδ∗, then there is some α < δ such that ⟨Pδ,β : β ∈nacc cδ ∧β > α⟩=⟨Pδ∗,β : β ∈(nacccδ ∩δ) ∧β > α⟩. (f) for every γ < λ, |{⟨Pδ,α : α ∈Cδ ∩γ⟩: δ ∈S}| ≤µ.Then there is a µ-complete Souslin tree on λ.Discussion:Condition (d) is the prediction demand.

It says that for every club E and a set Athere is a stationary set of δ-s, such that for unboundedly many non-accumulation points α of cδtwo things happen: α ∈E it and A ∩α is guessed by Pδ,α.Proof:We assume, without loss of generality, that for every δ ∈S∗and α ∈cδ, α = µα. Byinduction on α < λ we construct a tree T(α) of height α such that:(i) The universe of T(α) is µ(α+1), the β-th level in T(α), Tβ, consists of the elements [µβ, µ(β +1)), and every x ∈Tβ for β < α has an extension in Tγ for every γ < α.

Every x ∈T(α) suchthat Lev(x) + 1 < α has at least two immediate successors in Tα. (ii) T(α) is µ-complete;(iii) For α < β, T(α) = T(β)↾|T(α)|Also, we define a partial function (which, intuitively speaking, chooses branches which helpus in preserving the maximality of small antichains that occur along the way):(iv) For every x ∈T(α) and a sequence t = ⟨Pδ,β : β ∈nacc cδ ∩α⟩such that sup(cδ ∩α) < α andLev(x) < max(cδ ∩α) and max(cδ ∩α) ∈nacc cδ, y(x, t) is defined, and is an element in thelevel max(cδ ∩α) which extends x and has the property that for every A ∈Pδ,max(cδ∩α) whichis a maximal antichain of T(max cδ ∩α), there is an element of A below y(x, t).

(v) If the sequence s extends the sequence t and y(x, t), y(x, s) exist, then T(α) |= y(x, t) < y(x, s).2souslin12.8.2021

(vi) For every increasing sequence ⟨ti : i < i∗⟩there is an upper bound (in the tree order) to⟨y(x, ti) : i < i∗⟩.The last demand is:(vii) If α = δ+1, δ ∈S∗then every y ∈Tδ satisfies that there is some δ∗≥δ ∈acc cδ∗and x ∈T(δ),such that y is the least upper bound (in the tree order) of ⟨y(x, tα) : α ∈nacc cδ∗∩δ ∧αx < α <δ⟩where αx is the least in nacc cδ∗such that αx > Lev(x), and tα = ⟨Pδ,β : β ∈nacc cd∗∧β ≤α⟩.We first show that this construction, once carried out, yields a µ-complete Souslin tree onλ. The completeness of T = ∪T(α) is clear from the regularity of λ.

Suppose that A ⊆λ is amaximal antichain of T of size λ. Let E be the club of points δ < λ such that T↾δ = T(δ) andA↾δ is a maximal antichain of T(δ).

Pick a point δ ∈S∗such that δ = sup{α ∈nacc cδ : α ∈E ∧A↾α ∈Pδ,α}. As |T(δ)| < λ there is an element a ∈A, Lev(a) > δ.

Let y be the unique suchthat Lev(y) = δ and y < a. Then by demand (vii), there is some δ∗≥δ and x ∈T(δ) such thaty is the least upper bound (in the tree order) of ⟨y(x, tα) : α ∈nacc cδ∗∩δ ∧α > Lev(x)⟩.

Thereis some α∗< δ such that ⟨Pδ,β : α < β < δ ∧β ∈nacc cδ⟩= ⟨Pδ∗,β : α∗< β ∈nacc cδ∗∩δ⟩.Pick some α ∈nacc cδ such that α > max{Lev(x), α∗}, α ∈E and A↾α ∈Pδ,αi. So α ∈nacc cδ∗and A ∩α ∈Pδ∗,α.

Then the unique x′ < y with Lev(x′) = α (which equals y(x, ⟨Pδ∗,γ : γ ∈(nacccδ∗∩(α + 1))⟩)) is above an element a′ ∈A↾α. But x′ < a — a contradiction to the factthat A is an antichain.Next let us show that we can carry out the construction by induction.

When α = β + 1 andβ is a successor or zero, add two immediate successors to every point in the β-th level. When βis limit, cfβ < µ, add an element above every infinite branch.

This addition amounts to the totalof µ<µ = µ points. If, in addition, β ∈nacc cδ for some δ ∈S∗, then for every x ∈T(β) definey(x, ⟨Pδ,γ : γ ∈nacc cδ ∧γ ≤β⟩) as follows: let γ0 = max(cδ ∩β).

When Lev(x) < γ0 set x0 asthe supremum (in T(α)) of ⟨y(x, ⟨Pδ,α : α ≤α∗)⟩) : α∗≤γ0 ∧α∗∈nacccδ⟩; else, x0 = x. As|Pδ,β| ≤cfβ, we can in cfβ steps choose a cofinal branch above x0 which has a point above anelement from A for every A ∈Pδ,β which is a maximal antichain of Tβ. Let the required y be thesupremum of this branch.If β is a limit and cfβ = µ, distinguish two cases: case (a): β = δ ∈S∗.

So we shouldsatisfy demand (vii), namely add bounds precisely to those branches which for some δ∗≥δ in S∗,δ ∈acc cδ∗, are of the form ⟨y(x, tγ) : γ ∈nacc cδ∗∩β⟩where tγ = ⟨Pδ∗,ζ : ζ ≤γ ∧ζ ∈nacc cδ∗⟩. By(f) this costs only the addition of µ new elements.

If, in addition, there is some δ′ ∈S∗such thatδ ∈nacc cδ′, we should define y(x, ⟨Pδ′,γ : γ ∈nacc cδ′ ∧γ ≤δ⟩) for all x ∈T(δ). This presents noproblem: as |Pδ′,δ| < µ, we attach to each x some x0 such that x0 = x or Tδ |= x0 > x and such thatx0 is above members from every maximal antichain in Pδ′,δ; now y(x, ⟨Pδ′,γ : γ ∈nacc cδ′ ∧γ ≤δ⟩)will be the point in level δ above x0 we obtained anyway to satisfy demand (vii).3souslin12.8.2021

Case (b): cfβ = µ and β /∈S∗. Then when there is some δ such that β ∈acc cδ we realizeenough limits to obtain completeness under increasing sequences of the form ⟨y(x, ti) : i < i∗⟩.

By(f), we add thus ≤µ elements. If there is no such δ, just make sure, by adding µ points to the treein level β, that above every x ∈T(β) there is a point in level β.

This takes care also of (i). If thereis some δ′ such that β ∈nacc cδ′, then for every x ∈T(β) define y(x, ⟨Pδ′,γ : γ ∈nacc cδ′ ∧γ ≤δ⟩)exactly as in the case of smaller cofinality.⊣2We will show now how to obtain from a non-reflecting stationary set a special case of theprediction principle we used in the previous theorem.

One should substitute Pδ,α in the previoustheorem by Bα from the next theorem to get the assumptions of the previous theorem.3. Theorem:Suppose λ = cfλ > ℵ1 , S ⊆λ is stationary, non-reflecting, and carries a diamondsequence ⟨Aα : α ∈S⟩, S∗is a given non reflecting stationary subset of λ, S∗∩S = ∅andδ ∈S∗⇒cfδ > ℵ0.

Then there are sequences C = ⟨cδ : δ ∈S∗⟩and B = ⟨Bα : α ∈S⟩such that:(i) Bα ⊆α;(ii) sup cδ = δ and cδ is a closed set of limit ordinals;(iii) if δ, δ∗∈S∗and δ ∈acc cδ∗, then there is some α < δ such that cδ∗∩(α, δ) = cδ ∩(α, δ);(iv) for every club E ⊆λ and set X ⊆λ there are stationarily many δ ∈S∗such that δ = sup{α ∈nacccδ : α ∈S ∩E ∧X ∩α = Aα}.Proof:We fix some 1-1 pairing function ⟨−, −⟩from λ×ω0 onto λ and let Aαn = {β < α : ⟨β, n⟩∈A}. We may assume that for every countable sequence X = ⟨Xn : n < ω⟩of subsets of λ thereare stationarily many α ∈S such that for every n, Xn ∩α = Aαn.

Denote by S(X), for a (finite orinfinite) sequence of subsets of λ the stationary set {α ∈S : Vn Xn ∩α = Aαn}.To every limit α < λ we attach a club of α, eα, satisfying eα ∩S = eα ∩S∗= ∅, otp eα = cfαand eα contains only limit ordinals whenever α ∈S∗. Let C0 = ⟨eδ : δ ∈S∗⟩.

Suppose thatCn = ⟨cnδ : δ ∈S⟩is a bad candidate for the job, namely that there are a club En and a set Xnsuch that for every δ ∈En∩S∗the set {α ∈nacc cnδ : α ∈S(Xn)∩En} is bounded below δ. (Surely,we may assume that En is as thin as we like — in particular that all its members are limits).

DefineCn+1 by induction on δ: For every γ ∈cnδ , we define cn+1δ∩(γ, min cnδ \(γ+1)) (where (γ, β) denotes,as usual, an open interval of ordinals), and we let cn+1δ= cnδ ∪S{cn+1δ∩(γ, min cnδ \(γ+1)) : γ ∈cnδ }.This is well defined, as every γ ∈cnδ has a successor in cnδ . So denote by β the ordinal min cnδ \(γ+1),and let(∗)cn+1δ∩(γ, β) =cn+1β∩(γ, β)if β ∈S∗∅if β ∈S(X0, · · · , Xn){α : γ < α < β ∧(∃ζ ∈eβ)(α = sup(ζ ∩En))}otherwiseNote that for the definition to be consistent, β ∈cnδ must always be limit (and this is indeedthe case).4souslin12.8.2021

3.1 Lemma:Suppose that Cn is defined for n ≤m. Then for For every n < m and δ ∈S∗:(0) If α ∈cnδ then β is a limit ordinal;(1) cnδ is closed.

(2) cnδ ⊆cn+1δ. (3) If α ∈S∗∩acc cnδ , then cnα and cn+1δ∩α have a common end segment.

(4) If α ∈cn+1δ∩S(X0, · · · , Xn), then α ∈nacc cn+1δ.Proof: : (2) is true by the definition of cn+1δfor every n and δ ∈S∗. (0), (1), (3) and (4) areproved by induction on n and δ.For n = 0 we know that eδ = c0δ is all limits and is closed, so (0) and (1) hold.

(3) is vacuouslytrue, because eδ ∩S∗= ∅, and (4) is vacuously true because eδ ∩S = ∅.For n + 1:(0): Suppose α ∈cn+1δ. If α ∈cnδ then it is a limit ordinal by (0) and the induction hypothesison n. If α /∈cnδ , let γ = sup cnδ ∩α.

Because of (1) and the induction hypothesis γ < α. Letβ = min cnδ \ (α + 1). If β ∈S∗then cn+1δ∩(γ, β) = cn+1β∩(γ, β).

So α ∈cn+1β, and by theinduction hypotheses on β, α is a limit ordinal. If β /∈S∗, the cn+1δ∩(γ, β) = {α : γ < α <β,(∃ζ ∈eβ)(α = sup ζ ∩En)}.

Therefore for some ζ ∈eβ our α is sup(ζ ∩En). Since En is aclub, α ∈En.

But En is a club of limits, so α is limit. (1) Suppose that α ∈acc cn+1δ, and we wish to show α ∈cn+1δ.

If α ∈acc cnδ , then becauseof (1) and the induction hypothesis on n α ∈cnδ and (by (2)) α ∈cn+1δ. Else, γ = sup α ∩cnδ andβ = min cnδ \ (a + 1), γ < α < β.

If β ∈S∗then α ∈acc cn+1β. By the induction hypothesis on βand (1), α ∈cn+1δ.

Otherwise, α is a limit of ⟨ai : i < i∗⟩such that αi = sup ζi ∩En ∈cn+1δ. Soclearly α ∈En.

Let ζ∗be the minimal in eβ above α. So α = sup ζ∗∩En.

Therefore α ∈cn+1δ.Before proving (3) we note:3.2 Fact:Suppose γ ∈cnδ and β = min cnδ \ (γ + 1). If β /∈S∗and α ∈cn+1δ∩(γ, β) is a limit ofcn+1δ, then α ∈eβ.Indeed, if α = sup{α(i) : i < i∗}, where α(i) = sup ζ(i) ∩En are elements in cn+1δ, α ∈En.Therefore every ζ(i) < α (or else sup ζ(i) ∩En ≥α > α(i)).

But ζ(i) ≥α(i), so α is a limit of eβ.As α < β and eβ is closed, α ∈eβ.⊣3.2(3): Let α ∈acc cn+1δ∩S∗, and we wish to show that cn+1δand cn+1αhave a common endsegment. If α ∈acc cnδ , then by the induction hypothesis on n and (3), we know that cnδ and cnαhave a common end segment; say they agree on the interval (α(0), α).

This means in particularthat for every γ ∈cnδ ∩(α(0), α), α ∈cnα and min cnδ \ (γ + 1) = min cnα \ (γ + 1) =: β. Thereforealso cn+1δ∩(γ, β) = cn+1α∩(γ, β), and consequently cn+1δ∩(α(0), α) = cn+1α∩(α(0), α). So assumethat α /∈acc cnδ .

The first possibility is that α /∈cnδ altogether. In this case let γ < α < β assumetheir traditional roles as the last ordinal of cnδ below α and the first above.

If β ∈S∗, then by5souslin12.8.2021

the induction hypothesis on β we know that cn+1βand cn+1αhave a common end segment; butcn+1δ∩(γ, β) = cn+1β∩(γ, β), so it follows that cn+1δand cn+1αhave a common end segment.If β /∈S∗, then by the Fact above, α ∈eβ — contradiction to eβ ∩S∗is empty. So this subcaseis non existent.The last case is: α /∈acc cnδ but α ∈cnδ , or in short α ∈nacccnδ .

Let γ be the last element ofcnδ ∩α. Then by (∗), cn+1δ∩(γ, α) = cn+1α∩(γ, α).

(4): Suppose that α ∈S(X0, ·, Xn) ∩cn+1δ. We should see that α ∈nacc cn+1δ.

Let m ≤n + 1be the minimal such that α ∈cmδ . It is enough to prove that α ∈nacc cmδ , because by (∗) it isclear that if α ∈S(X0, · · · , Xm) ∩nacc cmδ then α will remain a non-accumulation point in cm+1δ(because nothing will be added in the interval below it).

So without loss of generality we mayassume that α ∈cn+1δ\ cnδ . So denote by (γ, β), as usual, the unique minimal interval with endpoints in cnδ to which α belongs.

First case: β ∈S∗. So α ∈cn+1β; and by the induction hypothesison β, α ∈nacc cn+1β.

So this case is done. Otherwise, β /∈S∗.

So by the Fact above, if α were alimit of cn+1δ, it would be in eβ. But α ∈S, and therefore cannot be in eβ by the very choice of eβ.Therefore α ∈nacc cn+1δ.

(This is where the non reflection of S is used in an essential way).⊣3.13.3 Claim: : There is some n < ω for which Cn and ⟨Aαn : α ∈S⟩are as required.Proof: : Suppose not. Let Xω = ⟨Xn : n < ω⟩.

Let E = ∩nEn and E′ = acc (S(Xω) ∩E).So E′ is a club.Pick some δ ∈S∗∩E′.For every n there is a bound below δ of the set{α ∈nacc cnδ : α ∈S(Xn) ∩En}. As cfδ > ℵ0, let α∗< δ bound α(n) for all n. Let δ > β > α∗be in S(Xω) ∩E.

So for every n, Xn ∩β = Aβn and β ∈En. If β ∈cnδ for some n, then by (4)β ∈nacc cnδ — a contradiction to β > α(n).

So β /∈cnδ for all n. Therefore for every n we maydefine (γ(n), β(n)) as the minimal interval with ends in cnδ which contains β.3.4 Claim:β(n + 1) < β(n).Proof:By its definition, β(n) ∈nacc cnδ . In the case β(n) = δ∗∈S∗, there are clearly elements incn+1δ∗above β and below β(n), so the claim is obvious.

The case β(n) ∈S(X0, · · · , Xn) is impossiblebecause of (4). In the remaining case, cn+1δ∩(γ(n), β(n)) = {α : γ(n) < α < β(n), (∃ζ ∈eβ(n))(α =sup ζ ∩En)}.

Let ζ∗> β be in eβ(n). As β ∈E ⊆En, sup ζ∗∩En ≥β.

But the right hand sideof this inequality belongs to cn+1δ, while β does not; therefore sup ζ∗∩En > β. So we see thatthere are elements of cn+1δin (β, β(n)), therefore the least of them, namely β(n+1) is smaller thanβ(n).⊣3.4This is clearly a contradiction.

We conclude that after finitely many steps, Cn+1 cannot bedefined due to the lack of a counterexample. This means that Cn and ⟨Bα : α ∈S⟩where Bα = Anαsatisfy (i), (ii) and (iv).

By (3) above, they satisfy (iii) as well. ⊣3.3This shows that after finitely many corrections all the requirements are satisfied, and ourtheorem is proved.⊣34.

Theorem:If the eδ we pick in the proof of Theorem 3 satisfy the additional condition that6souslin12.8.2021

for every γ < λ the set {eδ ∩γ : α ∈λ is limit} has cardinality smaller then λ, then the resultinggood C = ⟨cδ : δ ∈S∗⟩satisfies that for every γ < λ, |{cδ ∩γ : δ ∈S∗}| < λProof:Let γ < λ be given. We must show that |{cnδ ∩γ : δ ∈S∗}| ≤µ.

Let N ≺H(χ, ∈) for somelarge enough χ, |N| < λ, γ ⊆N, γ ∈N, {eα ∩γ : α < λ is limit} ⊆N, ⟨eα : ν < λ is limit⟩∈N,and En, Xn ∈N for every n.We shall see that for every n and δ, cnδ ∩γ ∈N. Since |N| < λ, this is enough.First we notice that if δ < γ then cδ ∈N and by elementarity also cnδ ∈N for every n. Nowwe use induction on n to show that for every δ > γ, cnδ ∩γ ∈N.

For n = 0: if δ > γ thenc0δ ∩γ = eδ ∩γ ∈N by the assumptions on N. For n + 1 we use induction on δ. Suppose, then,that for all δ′ < δ, cn+1δ′∩γ ∈N.We need the easy4.1 Fact:If (α0, α1) is a minimal interval of cnδ ∩(γ + 1) then cn+1δ∩(α0, α1) ∈N.Proof: By (∗) above, the definition of cn+1δ∩(α0, α1) depends only on eα1, En and (if case thereis such) cn+1α1 .

All these objects are in N, so also cn+1δ∩(α0, α1) ∈N.⊣4.1Denote γ(δ) = sup cnδ ∩γ. So γ(δ) ≤γ.

If γ(δ) = γ, then cn+1δ∩γ = cnδ ∩γ ∪SI cn+1δ∩Iwhere I runs over all minimal intervals of cnδ ∩(γ + 1). So by the Fact above we are done.

Else,γ(δ) < γ. In this case define β(δ) = min cnδ \ γ.

If β(δ) = γ then again we are done by the Fact.The remaining case is γ(δ) < γ < β(δ). By the same Fact, cn+1δ∩γ(δ) ∈N.

If β(δ) ∈S∗, thencn+1δ∩(γ(δ), γ) = cn+1β(γ) ∩(γ(δ), γ). By the induction hypothesis, and since β(δ) < δ, the latterset is in N, and we are done.

If β(δ) /∈S∗, then either nothing is added into (γ(δ), β(δ)) (whenβ(δ) ∈S(X0, · · · , Xn)), or cn+1δ∩(γ(δ), β(δ)) = {α : γ(δ) < α < β(d) (∃ζ ∈eβ(δ))(α = sup En∩ζ)}.So in this definition N might not know who β(δ) is, but eβ(δ) ∩γ ∈N. Therefore, denoting by α∗the last member in En ∩γ, we can determine in N the set cn+1δ∩α∗.

As to whether α∗itself isin this set or not, we need knowledge which is not available in N, but who cares, as long as bothpossibilities are in N. ⊣45. Corollary:If there is a non-reflecting stationary set S ⊆{α < µ+ : cfα < µ}, and 2µ = µ+,µ<µ = µ, then there is a µ-complete Souslin tree on µ+.6.

Remark:This improves the result by Gregory in [G].Proof:It is known (see [G] 2.1) that if S ⊆{δ ∈µ+ : cfδ < µ} is stationary, then µ = µ<µimplies ♦(S). As S is non reflecting, we can, for every limit α < µ+, choose a closed set eα,α = sup eα and otp eα = cfα such that eα ∩S = ∅.

µ = µ<µ implies that for every γ < µ+ theset {eα ∩γ : α < λ, α is limit} is of cardinality at most µ. Use Theorem 3 and Theorem 4 toobtain the assumptions of Theorem 2, S being the given non reflecting stationary set and S∗being{δ < λ : cfδ = µ}.

By Theorem 2 there is an µ-complete Souslin tree on µ+.⊣57. Problem:(1) Can the existence of such a tree be proved in ZFC + GCH?

(2) Can a Souslin tree on ℵ2 be constructed from GCH and two stationary sets, each composed7souslin12.8.2021

of ordinals of countable cofinality, which do not reflect simultaneously? By [Mg] this would raisethe consistency strength of GCH +SH(ℵ2) to the consistency of the existence of a weakly compactcardinal.References[AbShSo] U. Abraham, S. Shelah and R. M. Solovay Squares with diamonds and Souslin treeswith special squares, Fundamenta Mathematicae vol.

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