Introduction - uni-muenster.de · (2)For every infinite cardinal , there exists a respecting +-Kurepa tree; Roughly speaking, this means that this +-Kurepa tree looks very much like

SQUARE WITH BUILT-IN DIAMOND-PLUS

ASSAF RINOT AND RALF SCHINDLER

Abstract. We formulate combinatorial principles that combine the square principle with variousstrong forms of the diamond principle, and prove that the strongest amongst them holds in 𝐿 forevery infinite cardinal.

As an application, we prove that the following two hold in 𝐿:(1) For every infinite regular cardinal 𝜆, there exists a special 𝜆+-Aronszajn tree whose projection

is almost Souslin;(2) For every infinite cardinal 𝜆, there exists a respecting 𝜆+-Kurepa tree; Roughly speaking,

this means that this 𝜆+-Kurepa tree looks very much like the 𝜆+-Souslin trees that Jensenconstructed in 𝐿.

1. Introduction

In his seminal paper [13], Jensen initiated the study of the fine structure of Godel’s constructibleuniverse, 𝐿, and proved that in this model, for every uncountable cardinal 𝜅 which is not weaklycompact, there exists a 𝜅-Souslin tree. These fine-structural-constructions of Souslin trees werethen factored through the combinatorial principles and (also due to Jensen), making theconstruction accessible to a wider audience of mathematicians.

In [9], Gray introduced a combinatorial principle that forms a strong combination of and ,which he denoted by . This principle turned out to be very fruitful, and, for instance, has recentlybeen used to answer an old question of Hajnal in infinite graph theory [17].

In this paper, we introduce principles that combine with stronger forms of , such as *

and +. We study the implication between these principles, and prove that the strongest amongst

them, †𝜆, holds in 𝐿 for every infinite cardinal 𝜆.

As and have countless applications in infinite combinatorics, we expect the principles of thispaper to prove fruitful, and allow deeper applications of the nature of 𝐿, outside of 𝐿.

In this paper, we demonstrate the utility of the new principles by presenting applications to thetheory of trees. It is proved:

(1) *𝜆 + 𝜆<𝜆 = 𝜆 entails the existence of a 𝜆-respecting special 𝜆+-Aronszajn tree whose

projection is almost Souslin (to be defined below);(2) +

𝜆 entails the existence of a 𝜆-respecting 𝜆+-Kurepa tree;

(3) †𝜆 entails the existence of a 𝜆-respecting 𝜆+-Kurepa tree with the additional feature of

having no 𝜆+-Aronszajn subtrees.

Before giving the definition of a respecting tree, we first motivate it by briefly recalling how Jensenconstructed a 𝜆+-Souslin tree from the hypothesis that 𝜆(𝐸) + (𝐸) holds for some stationarysubset 𝐸 ⊆ 𝜆+.1

Date: December 7, 2016.2010 Mathematics Subject Classification. Primary 03E45. Secondary 03E05.Key words and phrases. diamond principle, square principle, constructibility, walks on ordinals, Kurepa tree,

almost Souslin tree, parameterized proxy principle.1The full details may be found in, e.g., [6, Theorem IV.2.4] or [18, Lemma 11.68].

1

2 ASSAF RINOT AND RALF SCHINDLER

Let ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ and ⟨𝑆𝛼 | 𝛼 ∈ 𝐸⟩ witness the hypothesis. The construction of the tree 𝑇 is byrecursion, where at stage 𝛼 < 𝜆+, the 𝛼th-level, 𝑇𝛼, is constructed. We start with 𝑇0 a singleton,and for 𝑇𝛼+1, we simply make sure that any node of 𝑇𝛼 admits two incompatible extensions in𝑇𝛼+1. The heart of the matter is the definition of 𝑇𝛼 for 𝛼 limit nonzero, once 𝑇 𝛼 =

⋃𝛽<𝛼 𝑇𝛽

has already been constructed. Here, for every node 𝑥 ∈ 𝑇 𝛼, one identifies a canonical branch b𝛼𝑥

which is cofinal in 𝑇 𝛼 and goes through 𝑥. Of course, to be able to construct such a branch, weneed to make sure that the process of climbing up through the levels of 𝑇 𝛼 is always successful,i.e., that we never get stuck when trying to take a limit. For this, we advise with the -sequence,ensuring in advance that if ∈ acc(𝐶𝛼), then b

𝑥 would make an initial segment of b𝛼𝑥 .2

But we also need to seal antichains! For this, we advise with the -sequence, to decide whether𝑇𝛼 should be equal to b𝛼

𝑥 | 𝑥 ∈ 𝑇 𝛼, or only to some carefully-chosen subset of it.

Here is a possible abstraction of the above process.

Definition 1.1. Suppose that 𝑇 is a downward-closed family of functions from ordinals to somefixed set Ω, so that (𝑇,⊂) forms a 𝜆+-tree. Denote 𝑇 𝑋 = 𝑡 ∈ 𝑇 | dom(𝑡) ∈ 𝑋.

We say that 𝑇 is 𝜆-respecting if there exists a stationary subset 𝐸 ⊆ 𝜆+, and a sequence ofmappings

⟨b𝛼 : 𝑇 𝐶𝛼 → 𝛼Ω ∪ ∅ | 𝛼 < 𝜆+⟩such that:

(1) ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ is a 𝜆(𝐸)-sequence;(2) 𝑇𝛼 ⊆ Im(b𝛼) for every 𝛼 ∈ 𝐸;(3) if ∈ acc(𝐶𝛼) and 𝑥 ∈ 𝑇 𝐶, then b(𝑥) = b𝛼(𝑥) .

In particular, for every 𝛼 ∈ 𝐸, any node 𝑦 ∈ 𝑇𝛼 is essentially the limit of some canonical branchb𝛼(𝑥) for some 𝑥 ∈ 𝑇 𝐶𝛼.

While working on their paper, the authors of [2] were considering the problem of constructing,say, an ℵ3-Souslin tree whose reduced ℵ0-power is ℵ3-Aronszajn, and whose reduced ℵ1-power is ℵ3-Kurepa. They realized that such a tree may be constructed, provided that there exists a respectingℵ3-Kurepa tree.

Question. Can a 𝜆+-Kurepa tree be 𝜆-respecting?

At a first glance, this sounds unlikely, as 𝜆+-Kurepa trees are usually obtained in a top-downfashion (one outright identifies 𝜆++ many whole functions from 𝜆+ to 2, and then verifies that thenumber of traces on any 𝛼 < 𝜆+ is rather small), while respecting trees are described in a bottom-up langauge. However, in this paper, we shall demonstrate that +

𝜆 allows the construction of sucha 𝜆+-Kurepa tree. In fact, we shall construct a 𝜆+-Kurepa tree satisfying a considerably strongerform of 𝜆-respecting, that is, 𝜆(𝜆+)-respecting (see Definition 4.4 below).

In another front, we shall prove that *𝜆 entails the existence of a 𝜆-sequence for which the

𝜆+-Aronszajn trees derived from the process of walks on ordinals [23] along are respecting. Thisis also somewhat surprising, as the standard description of these trees is also top-down. However,the fact that these trees could be 𝜆-respecting is already hinted in [21, Equation (*) on p. 267],based on a concept affine to Definition 1.1 and implicit in the proof of [20, Theorem 4.1]. Of course,the derived trees obtained here will be moreover 𝜆(𝜆+)-respecting.

For 𝜆 regular uncountable, we shall also ensure that the derived tree 𝒯 (𝜌1) (which is a projectionof the special tree 𝒯 (𝜌0)) is almost Souslin.3 As for 𝜆 = ℵ0, in [22], it was proved that Cohen

2Here, acc(𝐴) = 𝛼 < sup(𝐴) | sup(𝐴 ∩ 𝛼) = 𝛼 > 0.3The relevant definitions may be found on Section 4 below.

SQUARE WITH BUILT-IN DIAMOND-PLUS 3

modification of a 𝐶-sequence on 𝜔1 makes its 𝒯 (𝜌1) almost Souslin. In [10], a similar result wasobtained from *(𝜔1). Altogether, we conclude that in 𝐿, for every infinite regular cardinal 𝜆,there exists a special 𝜆+-Aronszajn tree whose projection is almost Souslin.

1.1. Organization of this paper. In Section 2, we recall the definition of the principle 𝜆,

introduce the principles *𝜆,

+𝜆 ,

†𝜆, and discuss the interrelations between them.

In Section 3, we prove that if 𝑉 = 𝐿, then †𝜆 holds for every infinite cardinal 𝜆.

In Section 4, we use the new principles to derive new types of 𝜆+-trees.

2. Hybrid squares and diamonds

In [9], Gray introduced the principle 𝜆 for 𝜆 a regular uncountable cardinal. In [1, S2], thedefinition was generalized to cover the case of 𝜆 singular. Then, in [3], the principle was generalizedto cover the case 𝜆 = ℵ0, as well. The outcome is as follows:

Definition 2.1 ([9],[1],[3]). 𝜆 asserts the existence of ⟨(𝐶𝛼, 𝑆𝛼) | 𝛼 < 𝜆+⟩ such that:

(1) 𝐶𝛼 is a club in 𝛼 of order-type ≤ (𝜔 · 𝜆);(2) 𝑆𝛼 ⊆ 𝛼;(3) if ∈ acc(𝐶𝛼), then

(a) 𝐶 = 𝐶𝛼 ∩ ;(b) 𝑆 = 𝑆𝛼 ∩ ;

(4) for every𝑋 ⊆ 𝜆+ and every club𝐷 ⊆ 𝜆+, there exists a limit 𝛼 < 𝜆+ with otp(acc(𝐶𝛼)) = 𝜆,such that 𝑆𝛼 = 𝑋 ∩ 𝛼 and acc(𝐶𝛼) ⊆ 𝐷.

Note that ⟨𝑆𝛼 | 𝛼 < 𝜆+⟩ forms a (𝐸𝜆+

cf(𝜆))-sequence, and if 𝜆 is uncountable, then ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩forms a 𝜆-sequence. By [3], 𝜔 is equivalent to (𝜔1). By [19], 𝜆 + (𝜆+) does not imply 𝜆

for 𝜆 regular uncountable. By [16], 𝜆 + (𝜆+) is equivalent to 𝜆 for every singular cardinal 𝜆.

Definition 2.2. *𝜆 asserts the existence of ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ such that:

(1) 𝐶𝛼 is a club in 𝛼 of order-type ≤ 𝜆;(2) 𝑋𝛼 is a subset of 𝒫(𝛼), of size ≤ 𝜆;(3) 𝑓𝛼 : 𝐶𝛼 → 𝑋𝛼 is a function, and a surjection whenever otp(𝐶𝛼) = 𝜆;(4) if ∈ acc(𝐶𝛼), then

(a) 𝐶 = 𝐶𝛼 ∩ ;(b) 𝑓(𝛽) = 𝑓𝛼(𝛽) ∩ for all 𝛽 ∈ 𝐶;

(5) for every subset 𝑋 ⊆ 𝜆+ and a club 𝐶 ⊆ 𝜆+, the set𝛼 < 𝜆+ | 𝐶𝛼 ⊆* 𝐶 & 𝑋 ∩ 𝛼 ∈ 𝑋𝛼

contains a club;4

(6) 𝛼 < 𝜆+ | otp(𝐶𝛼) = 𝜆 is stationary in 𝜆+.

Of course, ⟨𝑋𝛼 | 𝛼 < 𝜆+⟩ forms a *(𝜆+)-sequence, and ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ forms a strong club-guessing sequence in the sense of [8]. In particular, by [12], *

𝜔 is actually stronger than *(𝜔1).

Definition 2.3. +𝜆 asserts the existence of ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ such that:

(1) 𝐶𝛼 is a club in 𝛼 of order-type ≤ 𝜆;(2) 𝑁𝛼 is a rud-closed transitive set, 𝜆 = |𝑁𝛼| ⊆ 𝑁𝛼, with 𝑓𝛽 | 𝛽 < 𝛼 ∪ 𝛼 ⊆ 𝑁𝛼;(3) 𝑓𝛼 : 𝐶𝛼 → 𝒫(𝛼) ∩𝑁𝛼 is a function;(4) if ∈ acc(𝐶𝛼), then

4Here, 𝐴 ⊆* 𝐵 stands for the assertion that sup(𝐴 ∖𝐵) < sup(𝐴).


(a) 𝐶 = 𝐶𝛼 ∩ ;(b) 𝑓(𝛽) = 𝑓𝛼(𝛽) ∩ for all 𝛽 ∈ 𝐶;

(5) for every subset 𝑋 ⊆ 𝜆+ and a club 𝐶 ⊆ 𝜆+, there exists a club 𝐷 ⊆ 𝜆+ such that for all𝛼 ∈ 𝐷:

∙ 𝐶𝛼 ⊆* 𝐶;∙ 𝑋 ∩ 𝛼,𝐷 ∩ 𝛼 ∈ 𝑁𝛼;∙ if otp(𝐶𝛼) = 𝜆, then 𝑋 ∩ 𝛼 ∈ Im(𝑓𝛼);

(6) 𝛼 < 𝜆+ | 𝑓𝛼 is surjective is stationary in 𝜆+;(7) 𝑁𝛼 | 𝛼 < 𝜆+ is an increasing ⊆-chain converging to 𝐻𝜆+ .

Note that ⟨𝑃 (𝛼) ∩𝑁𝛼 | 𝛼 < 𝜆+⟩ forms a +(𝜆+)-sequence. As before, the result of [12] entailsthat +

𝜔 is stronger than +(𝜔1).

Definition 2.4. †𝜆 asserts the existence of ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ such that Clauses (1)–(7) of

Definition 2.3 hold, with Clause (6) strengthened to

(6) for every 𝑛,𝑚 < 𝜔, end every Π𝑛𝑚-sentence 𝜂 valid in a structure

(𝜆+,∈, ⟨𝐴𝑖 | 𝑖 < 𝜔⟩), 5

there are stationarily many 𝛼 < 𝜆+ for which all of the following hold:∙ 𝑓𝛼 is surjective;∙ ⟨𝐴𝑖 𝛼 | 𝑖 < 𝜔⟩ ∈ 𝑁𝛼;∙ 𝑁𝛼 |= “𝜂 is valid in (𝛼,∈, ⟨𝐴𝑖 𝛼 | 𝑖 < 𝜔⟩)”.

Note that if ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ witnesses †𝜆, then ⟨𝑁𝛼 | 𝛼 < 𝜆+⟩ is not far from being a

witness to Devlin’s notion of a ♯(𝜆+)-sequence (see [5]). Specifically, there are two differences:

∙ In ♯(𝜆+), the models 𝑁𝛼 are required to be p.r.-closed, while here they are only rud-closed;∙ In ♯(𝜆+), the reflection property is restricted to Π1

2-sentences, while here there is norestriction on the complexity of the sentences.

Lemma 2.5. *𝜆 entails the existence of a *

𝜆-sequence ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ with the additionalproperty that for every club 𝐷 ⊆ 𝜆+, there exists some limit 𝛼 < 𝜆+ with otp(𝐶𝛼) = 𝜆 such that𝐶𝛼 ⊆ 𝐷.

Proof. Let ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ be a witness to *𝜆. Fix 𝜍 : 𝜆 → 𝜆 such that for all 𝑖 < 𝜆,

𝜍(𝑖) ≤ 𝑖 and 𝜍−1𝑖 is cofinal in 𝜆. Let 𝛼 < 𝜆+ be arbitrary. For all 𝑗 < 𝜆, denote

𝐶𝑗𝛼 = 𝛾 ∈ 𝐶𝛼 | otp(𝐶𝛼 ∩ 𝛾) ≥ 𝑗.

Let 𝜋𝛼 : otp(𝐶𝛼) → 𝐶𝛼 denote the monotone enumeration of 𝐶𝛼. Define 𝑓 ′𝛼 : 𝐶𝛼 → 𝑋𝛼 bystipulating

𝑓 ′𝛼(𝛽) = 𝑓𝛼(𝜋𝛼(𝜍(𝜋−1𝛼 (𝛽))).

It is easy to see that if ∈ acc(𝐶𝛼), then 𝜋 = 𝜋𝛼 otp(𝐶), and hence 𝑓 ′(𝛽) = 𝑓 ′𝛼(𝛽) ∩ forall 𝛽 ∈ 𝐶. In addition, if otp(𝐶𝛼) = 𝜆, then for all 𝑗 < 𝜆:

Im(𝑓 ′𝛼 𝐶𝑗𝛼) = Im(𝑓𝛼) = 𝑋𝛼.

Claim 2.5.1. There exists some 𝑗 < 𝜆 such that for every club 𝐷 ⊆ 𝜆+, there exists some limit

𝛼 < 𝜆+ with otp(𝐶𝛼) = 𝜆 and 𝐶𝑗𝛼 ⊆ 𝐷.

5“Validity in a structure” here has the obvious meaning, cf. [5, p. 891].


Proof. Suppose not. Then for all 𝑗 < 𝜆, we may pick a club counterexample 𝐷𝑗 ⊆ 𝜆+. Let𝐶 =

⋂𝑗<𝜆𝐷𝑗 . By Clauses (5) and (6) of Definition 2.3, then, there must exist some 𝛼 < 𝜆+ with

otp(𝐶𝛼) = 𝜆 and 𝐶𝛼 ⊆* 𝐶. Pick 𝑗 < 𝜆 such that 𝐶𝑗𝛼 ⊆ 𝐶. In particular, 𝐶𝑗

𝛼 ⊆ 𝐷𝑗 contradictingthe choice of 𝐷𝑗 .

Let 𝑗 < 𝜆 be given by the preceding claim. For all 𝛼 < 𝜆+, let

𝐶∙𝛼 =

𝐶𝑗𝛼, if otp(𝐶𝛼) > 𝑗;

𝐶𝛼, otherwise.

Put 𝑓∙𝛼 = 𝑓 ′𝛼 𝐶∙𝛼, and 𝑋∙

𝛼 = 𝑋𝛼. Then ⟨(𝐶∙𝛼, 𝑋

∙𝛼, 𝑓

∙𝛼) | 𝛼 < 𝜆+⟩ forms a *

𝜆-sequence with theadditional desired property.

Lemma 2.6. For every infinite cardinal 𝜆, †𝜆 =⇒ +

𝜆 =⇒ *𝜆 =⇒ 𝜆.

Proof. Let 𝜆 be an arbitrary infinite cardinal. The implication †𝜆 =⇒ +

𝜆 is trivial. To see that

+𝜆 =⇒ *

𝜆, let ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ be a +𝜆 -sequence. For all 𝛼 < 𝜆+, let

𝑋𝛼 =

𝒫(𝛼) ∩𝑁𝛼, if otp(𝐶𝛼) < 𝜆;

Im(𝑓𝛼), otherwise.

Then ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ forms a *𝜆-sequence.

Out next task is showing that *𝜆 =⇒ 𝜆. By [3], (𝜔1) implies 𝜔, hence we hereafter

assume that 𝜆 is uncountable. In particular, 𝜔 · 𝜆 = 𝜆.Let ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ be given by Lemma 2.5. Fix a bijection 𝜓 : 𝜆 × 𝜆+ → 𝜆+. Let

𝐸 = 𝛼 < 𝜆+ | 𝜓[𝜆 × 𝛼] = 𝛼. For every 𝛼 < 𝜆+, let 𝜋𝛼 : otp(𝐶𝛼) → 𝐶𝛼 denote the monotoneenumeration of 𝐶𝛼, and then for 𝑖 < 𝜆, write

𝑆𝑖𝛼 = 𝜏 < 𝛼 | 𝜓(𝑖, 𝜏) ∈ 𝑓𝛼(𝜋𝛼(𝑖)).

Claim 2.6.1. There exists an 𝑖 < 𝜆 such that for every club 𝐷 ⊆ 𝜆+ and every 𝑋 ⊆ 𝜆+, thereexists some limit 𝛼 < 𝜆+ with otp(acc(𝐶𝛼)) = 𝜆, such that 𝑆𝑖

𝛼 = 𝑋 ∩ 𝛼 and 𝐶𝛼 ⊆ 𝐷.

Proof. Suppose not, and so for all 𝑖 < 𝜆, pick a counterexample (𝐷𝑖, 𝑋𝑖). Let 𝑋 = 𝜓(𝑖, 𝜏) |𝑖 < 𝜆, 𝜏 ∈ 𝑋𝑖. By Clause (5) of Definition 2.2, we may find a club 𝐷 ⊆ 𝐸 ∩

⋂𝑖<𝜆𝐷𝑖 such that

𝑋 ∩ 𝛼 ∈ 𝑋𝛼 for all 𝛼 ∈ 𝐷. Now, fix a limit ordinal 𝛼 < 𝜆+ such that otp(𝐶𝛼) = 𝜆 and 𝐶𝛼 ⊆ 𝐷.In particular, 𝛼 ∈ 𝐸. Note that since 𝜆 is uncountable, moreover otp(acc(𝐶𝛼)) = 𝜆. Next, asotp(𝐶𝛼) = 𝜆, we infer from Clause (3) of Definition 2.2 the existence of some 𝑖 < 𝜆 such that𝑋 ∩ 𝛼 = 𝑓𝛼(𝜋𝛼(𝑖)). By definition of 𝑋 and since 𝜋[𝜆× 𝛼] = 𝛼, we conclude that

𝑆𝑖𝛼 = 𝜏 < 𝛼 | 𝜓(𝑖, 𝜏) ∈ 𝑋 ∩ 𝛼 = 𝑋𝑖 ∩ 𝛼.

By the choice of (𝐷𝑖, 𝑋𝑖), it must then be the case that 𝐶𝛼 * 𝐷𝑖. However, 𝐶𝛼 ⊆ 𝐷 ⊆ 𝐷𝑖. This isa contradiction.

Let 𝑖 < 𝜆 be given by the previous claim. For all 𝛼 ∈ 𝐸𝜆+

𝜔 , let 𝑐𝛼 be a cofinal subset of 𝛼 oforder-type 𝜔. Finally, for all 𝛼 < 𝜆+, let

𝐶∙𝛼 =

⎧⎪⎨⎪⎩𝐶𝛼 ∩ 𝐸, if sup(𝐸 ∩ 𝐶𝛼) = 𝛼;

𝐶𝛼 ∖ sup(𝐸 ∩ 𝛼), if sup(𝐸 ∩ 𝛼) < 𝛼;

𝑐𝛼, otherwise,


and

𝑆∙𝛼 =

𝑆𝑖𝛼, if sup(𝐸 ∩ 𝐶𝛼) = 𝛼;

∅, otherwise.

Claim 2.6.2. ⟨(𝐶∙𝛼, 𝑆

∙𝛼) | 𝛼 < 𝜆+⟩ is a 𝜆-sequence.

Proof. It is clear that for all limit 𝛼 < 𝜆+, 𝐶∙𝛼 is a club subset of 𝛼 of order-type ≤ 𝜆. Next,

suppose that 𝛼 < 𝜆+ and ∈ acc(𝐶𝛼).I If sup(𝐸 ∩ 𝐶𝛼) = 𝛼, then 𝐶∙

𝛼 = 𝐶𝛼 ∩ 𝐸 and hence ∈ acc(𝐶𝛼), and sup(𝐸 ∩ 𝐶) = sup(𝐸 ∩𝐶𝛼 ∩ ) = . Consequently, 𝐶∙

= 𝐶 ∩ 𝐸 = 𝐶∙𝛼 ∩ , 𝑓(𝜋(𝑖)) = 𝑓𝛼(𝜋𝛼(𝑖)) ∩ , and

𝑆∙𝛼 ∩ = 𝑆𝑖

𝛼 ∩ = 𝜏 < | 𝜓(𝑖, 𝜏) ∈ 𝑓𝛼(𝜋𝛼(𝑖)).As sup(𝐸 ∩ 𝐶), we get in particular that ∈ 𝐸, and hence the right hand side of the precedingis equal to

𝜏 < | 𝜓(𝑖, 𝜏) ∈ 𝑓𝛼(𝜋𝛼(𝑖)) ∩ = 𝑆𝑖 = 𝑆∙

.

I If sup(𝐸 ∩ 𝛼) < 𝛼, then 𝐶∙𝛼 = 𝐶𝛼 ∖ sup(𝐸 ∩ 𝛼), and ∈ acc(𝐶𝛼). Consequently,

𝐶∙ = 𝐶 ∖ sup(𝐸 ∩ ) = 𝐶𝛼 ∩ ∖ sup(𝐸 ∩ ) = 𝐶𝛼 ∩ ∖ sup(𝐸 ∩ 𝛼) = 𝐶∙

𝛼 ∩ ,and 𝑆∙

𝛼 ∩ = ∅ = 𝑆∙.

Finally, let 𝑋 ⊆ 𝜆+ and a club 𝐷 ⊆ 𝜆+ be arbitrary. By the choice of 𝑖, let us pick a limit ordinal𝛼 < 𝜆+ with otp(acc(𝐶𝛼)) = 𝜆 such that 𝑆𝑖

𝛼 = 𝑋 ∩ 𝛼 and 𝐶𝛼 ⊆ 𝐷 ∩ 𝐸. Then 𝑆∙𝛼 = 𝑆𝑖

𝛼 = 𝑋 ∩ 𝛼,as sought.

3. The strongest principle holds in 𝐿

Theorem 3.1. †𝜆 holds in 𝐿 for all infinite cardinals 𝜆.

Proof. We follow the proof of [18, Theorem 11.64], making use of arguments from [1], [11], and [14].We assume 𝑉 = 𝐿. Let 𝜆 denote an arbitrary infinite cardinal.

Consider the set 𝐶 = 𝛼 < 𝜆+ | 𝐽𝛼 ≺Σ𝜔 𝐽𝜆+, which is a club subset of 𝜆+ consisting of limit

ordinals above 𝜆. Note that 𝐶 ∩ 𝐸𝜆+

>𝜔 ⊆ acc(𝐶). Let 𝜙 : 𝜆+ → 𝐶 be the monotone enumeration of𝐶.

Let 𝛼 ∈ 𝐶. Obviously, 𝜆 is the largest cardinal of 𝐽𝛼. Note that as 𝐽𝛼 |= ZFC−, we have𝜌𝜔(𝐽𝛼) = 𝛼. We may therefore define 𝜈(𝛼) to be the largest 𝜈 > 𝛼 such that 𝛼 is a cardinal in theviewpoint of 𝐽𝜈 .6

Note also that if 𝛽 < 𝛼, then by 𝐽𝛼 |= |𝐽𝛽| = 𝜆, we have that 𝒫(𝜆) ∩ 𝐽𝛽 ( 𝒫(𝜆) ∩ 𝐽𝛼. So,𝜈(𝛽) < 𝛼 < 𝜈(𝛼) for all 𝛽 ∈ 𝐶 ∩ 𝛼.

As 𝜌𝜔(𝐽𝜈(𝛼)) = 𝜆, let 𝑛(𝛼) be the unique 𝑛 < 𝜔 such that 𝜆 = 𝜌𝑛+1(𝐽𝜈(𝛼)) < 𝛼 ≤ 𝜌𝑛(𝐽𝜈(𝛼)).

Let us write 𝑅 for the set of all 𝛼 < 𝜆+ such that 𝜈(𝜙(𝛼)) = 𝜈 +𝜔 for some 𝜈 such that 𝛼 is theonly cardinal of 𝐽𝜈 strictly above 𝜆.

For all 𝛼 < 𝜆+, set

𝑁𝛼 =

𝐽𝜈(𝜙(𝛼))+𝜔, if 𝛼 ∈ 𝑅;

𝐽𝜈(𝜙(𝛼)), otherwise.

As 𝜈 ∘ 𝜙 is an increasing function from 𝜆+ to 𝜆+, we have just established Clause (7) of Defini-tion 2.4.

6In particular, 𝐽𝜈+𝜔 can see that 𝛼 has cardinality 𝜆.


Let us now define a𝜆-sequence by what became the standard construction, cf. e.g. [18, pp. 270ff.],modulo various crucial adjustments. As we’ll have to refer to some details of this construction lateron in this proof, let us repeat this construction here for the convenience of the reader.

First, for 𝛼 ∈ 𝐶, we define 𝐷𝛼 as follows. We let 𝐷𝛼 consist of all ∈ 𝐶∩𝛼 such that 𝑛() = 𝑛(𝛼)and there is a weakly 𝑟Σ𝑛(𝛼)+1-elementary embedding

𝜎 : 𝐽𝜈()−→ 𝐽𝜈(𝛼)

such that 𝜎 = id, 𝜎(𝑝𝑛()+1(𝐽𝜈())) = 𝑝𝑛(𝛼)+1(𝐽𝜈(𝛼)), and if ∈ 𝐽𝜈(), then 𝛼 ∈ 𝐽𝜈(𝛼) and𝜎() = 𝛼. It is easy to see that if ∈ 𝐷𝛼, then there is exactly one map 𝜎 witnessing this, namelythe one which is given by

ℎ𝑛()+1,𝑝𝑛()+1(𝐽𝜈())

𝐽𝜈()(𝑖, ) ↦→ ℎ

𝑛(𝛼)+1,𝑝𝑛(𝛼)+1(𝐽𝜈(𝛼))

𝐽𝜈(𝛼)(𝑖, ),(1)

where 𝑖 < 𝜔 and ∈ [𝜆]<𝜔. We here make use of the notation for fine structural iterated Σ1 Skolemfunctions as presented e.g. in [18, Equation (11.29) on p. 252]. We shall denote the unique map asgiven by (1) by 𝜎,𝛼.

Notice that if 𝛼 ∈ 𝐶, then

𝐽𝜈(𝛼) = ℎ𝑛(𝛼)+1,𝑝𝑛(𝛼)+1(𝐽𝜈(𝛼))

𝐽𝜈(𝛼)“(𝜔 × [𝜆]<𝜔),

so that if ∈ 𝐷𝛼, then

Im(𝜎,𝛼) ⊆ ℎ𝑛(𝛼)+1,𝑝𝑛(𝛼)+1(𝐽𝜈(𝛼))

𝐽𝜈(𝛼)“(𝜔 × [𝜆]<𝜔),

which means that there must be 𝑖 < 𝜔 and ∈ [𝜆]<𝜔 such that the left hand side of (1) is undefined,whereas the right hand side of (1) is defined.

Also notice that the maps 𝜎,𝛼 commute, i.e., if ∈ 𝐷𝛼 and 𝛼 ∈ 𝐷𝛼′ , then ∈ 𝐷𝛼′ and

𝜎,𝛼′ = 𝜎𝛼,𝛼′ ∘ 𝜎,𝛼.Having constructed ⟨𝐷𝛼 | 𝛼 ∈ 𝐶⟩, we claim:

Claim 3.1.1. Let 𝛼 ∈ 𝐶. All of the following hold true:

(a) 𝐷𝛼 is closed;(b) If cf(𝛼) > 𝜔, then 𝐷𝛼 is unbounded in 𝛼;(c) If ∈ 𝐷𝛼 then 𝐷𝛼 ∩ = 𝐷.

Proof. This is Claim 11.65 of [18].

Notation. For 𝛼 ∈ 𝐶, 𝑖 < 𝜔 and ∈ [𝜆]<𝜔, we shall denote:

ℎ𝛼(𝑖, ) = ℎ𝑛(𝛼)+1,𝑝𝑛(𝛼)+1(𝐽𝜈(𝛼))

𝐽𝜈(𝛼)(𝑖, ).

Let 𝛼 ∈ 𝐶. If sup(𝐷𝛼) < 𝛼, let 𝜃(𝛼) = 0. Now, suppose sup(𝐷𝛼) = 𝛼. We shall obtain somelimit ordinal 𝜃(𝛼), and sequences ⟨𝜇𝛼𝑖 | 𝑖 ≤ 𝜃(𝛼)⟩ and ⟨𝜉𝛼𝑖 | 𝑖 < 𝜃(𝛼)⟩, by recursion, as follows.I Set 𝜇𝛼0 = min(𝐷𝛼).I Given 𝜇𝛼𝑖 with 𝜇𝛼𝑖 < 𝛼, we let 𝜉𝛼𝑖 be the least 𝜉 < 𝜆 such that

ℎ𝛼(𝑘, ) /∈ Im(𝜎𝜇𝛼𝑖 ,𝛼

)

for some 𝑘 < 𝜔 and some ∈ [𝜉]<𝜔. Given 𝜉𝛼𝑖 , we let 𝜇𝛼𝑖+1 be the least ∈ 𝐷𝛼 such that

ℎ𝛼(𝑘, ) ∈ Im(𝜎,𝛼)

for all 𝑘 < 𝜔 and ∈ [𝜉𝛼𝑖 ]<𝜔 such that ℎ𝛼(𝑘, ) exists.I Given ⟨𝜇𝛼𝑗 | 𝑗 < 𝑖⟩, where 𝑖 is a limit ordinal, we set 𝜇𝛼𝑖 = sup𝑗<𝑖 𝜇

𝛼𝑗 .


Naturally, 𝜃(𝛼) will be the least 𝑖 such that 𝜇𝛼𝑖 = 𝛼.

For any 𝛼 ∈ 𝐶, denote 𝐸𝛼 = 𝜇𝛼𝑖 | 𝑖 < 𝜃(𝛼).

Claim 3.1.2. Let 𝛼 ∈ 𝐶. All of the following hold true:

(a) ⟨𝜉𝛼𝑖 | 𝑖 < 𝜃(𝛼)⟩ is a strictly increasing sequence of ordinals below 𝜆;(b) otp(𝐸𝛼) = 𝜃(𝛼) ≤ 𝜆;(c) 𝜃(𝛼) > 0 iff 𝐸𝛼 is a club in 𝛼;(d) If ∈ acc(𝐸𝛼), then 𝐸𝛼 ∩ = 𝐸.

Proof. (a) is immediate, and it implies (b).(c) is trivial.(d) Let ∈ acc(𝐷𝛼). We have 𝐸𝛼 ⊆ 𝐷𝛼, and 𝐷 = 𝐷𝛼 ∩ by Claim 3.1.1(c). We now show

that ⟨𝜇𝑖 | 𝑖 < 𝜃()⟩ = ⟨𝜇𝛼𝑖 | 𝑖 < 𝜃()⟩ and ⟨𝜉𝑖 | 𝑖 < 𝜃()⟩ = ⟨𝜉𝛼𝑖 | 𝑖 < 𝜃()⟩ by induction:Say 𝜇𝑖 = 𝜇𝛼𝑖 , where 𝑖+ 1 ∈ 𝜃()∩ 𝜃(𝛼). Write 𝜇 = 𝜇𝑖 = 𝜇𝛼𝑖 . As 𝜎𝜇,𝛼 = 𝜎,𝛼 ∘ 𝜎𝜇,, for all 𝑘 < 𝜔

and ∈ [𝜆]<𝜔,

ℎ(𝑘, ) ∈ Im(𝜎𝜇,) =⇒ ℎ𝛼(𝑘, ) ∈ Im(𝜎𝜇,𝛼) = ∅.This gives 𝜇𝛼𝑖+1 ≤ 𝜇𝑖+1 On the other hand, Im(𝜎𝜇𝛼

𝑖+1,𝛼) contains the relevant witness so as to

guarantee conversely that 𝜇𝑖+1 ≤ 𝜇𝛼𝑖+1.

Recall that 𝜙 : 𝜆+ → 𝐶 denotes the monotone enumeration of 𝐶. For all 𝛼 ∈ acc(𝜆+), we have𝜙(𝛼) ∈ acc(𝐶), so we set

𝐶 ′𝛼 = 𝜙−1“𝐸𝜙(𝛼).

Since 𝐸𝜙(𝛼) ⊆ 𝐷𝜙(𝛼) ⊆ 𝐶 = Im(𝜙), we have otp(𝐶 ′𝛼) = 𝜃(𝜙(𝛼)) ≤ 𝜆, and 𝐶 ′

𝛼 is a club in 𝛼 iff𝜃(𝜙(𝛼)) > 0.

Because 𝐸𝜇 ⊆ 𝐷𝜇 ⊆ 𝐶 ∩ 𝜇 for every 𝜇 ∈ 𝐶, we have that if ∈ acc(𝐶 ′𝛼), then 𝐶 ′

= 𝐶 ′𝛼 ∩ .

Claim 3.1.3. Let 𝛼 ∈ 𝐸𝜆+

𝜔 be such that 𝜙(𝛼) = 𝛼.Then there exists a cofinal subset 𝑑 of 𝛼 of order-type 𝜔 satisfying the following:

(a) If 𝑐 ∈ 𝐽𝜈(𝛼) is a club in 𝛼, then 𝑑 ⊆* 𝑐;

(b) 𝑑 ∈ 𝑁𝛼′ whenever 𝛼 < 𝛼′ < 𝜆+.

Proof. The argument is a simplified version of the proof of [11, Theorem 4.15].

Fix 𝛼 ∈ 𝐸𝜆+

𝜔 such that 𝜙(𝛼) = 𝛼, so that 𝛼 ∈ 𝐶. Recall that 𝜆 = 𝜌𝑛(𝛼)+1(𝐽𝜈(𝛼)) < 𝛼 ≤𝜌𝑛(𝛼)(𝐽𝜈(𝛼)). Let us write 𝜈 = 𝜈(𝛼), 𝑛 = 𝑛(𝛼), and

𝜏 = sup(𝜉 < 𝜈 | 𝐽𝜈 |= |𝜉| ≤ 𝛼).

I.e., either 𝛼 is the largest cardinal of 𝐽𝜈 in which case 𝜏 = 𝜈, or else 𝜏 = 𝛼+𝐽𝜈 .

Case 1. cf(𝜏) = 𝜔.

Let (𝜏𝑚 | 𝑚 < 𝜔) be a sequence witnessing cf(𝜏) = 𝜔. For each 𝑚 < 𝜔, we may inside 𝐽𝜈 picksome club 𝑐𝑚 ⊆ 𝛼 such that 𝑐𝑚 ⊆* 𝑐 for every club 𝑐 ⊆ 𝛼, 𝑐 ∈ 𝐽𝜏𝑚 . E.g. we may let 𝑐𝑚 be thediagonal intersection of all clubs 𝑐 ⊆ 𝛼, 𝑐 ∈ 𝐽𝜏𝑚 , as being given by some enumeration in 𝐽𝜈 in ordertype 𝛼. Let then (𝜌𝑚 | 𝑚 < 𝜔) be a strictly increasing sequence which is cofinal in 𝛼 and such thatfor every 𝑚 < 𝜔, 𝜌𝑚 ∈

⋂𝑘≤𝑚 𝑐𝑚. Set 𝑑 = 𝜌𝑚 | 𝑚 < 𝜔. Then for every 𝑐 ∈ 𝐽𝜈 which is a club in

𝛼, we have 𝑑 ⊆* 𝑐.

Case 2. cf(𝜏) > 𝜔.


By [14, Lemma 1.2], we must then have that 𝜌𝑛(𝐽𝜈) = 𝛼, and if ℓ < 𝑛 is largest such that𝜌ℓ(𝐽𝜈) > 𝛼, then cf(𝜌ℓ(𝐽𝜈)) = cf(𝜏) > 𝜔. We then have ℓ+ 1 ≤ 𝑛 and

𝜌ℓ(𝐽𝜈) > 𝜌ℓ+1(𝐽𝜈) = 𝜌𝑛(𝐽𝜈) = 𝛼 > 𝜌𝑛+1(𝐽𝜈) = 𝜆.

Let (𝛼𝑚 | 𝑚 < 𝜔) be a sequence witnessing cf(𝛼) = 𝜔. By cf(𝜏) = 𝜔, there must be some 𝑚 < 𝜔such that

ℎ𝐽ℓ𝜈“(𝛼𝑚 ∪ 𝑝(𝐽 ℓ

𝜈)) ∩ 𝜏is cofinal in 𝜏 . Note that as 𝜔 = cf(𝛼) < cf(𝜌ℓ(𝐽𝜈)) and 𝛼 is a regular cardinal in 𝐽𝜈 ,

ℎ𝐽ℓ𝜈“(𝛼𝑚 ∪ 𝑝(𝐽 ℓ

𝜈)) ∩ 𝛼cannot be cofinal in 𝛼. Set

𝐻 = ℎ𝐽ℓ𝜈“(𝛼𝑚 ∪ 𝑝(𝐽 ℓ

𝜈)) and = sup(𝐻 ∩ 𝛼).

Let us write 𝜃 = cf(𝜌ℓ(𝐽𝜈)) = cf(𝜏), and let (𝜏 𝑖 : 𝑖 < 𝜃) be a strictly increasing sequence witnessingcf(𝜏) = 𝜃, where 𝜏 𝑖 ∈ 𝐻 for every 𝑖 < 𝜃. For each 𝑖 < 𝜃, in much the same way as in Case 1 wemay inside 𝐻 pick some club 𝑐𝑖 ⊆ 𝛼 such that 𝑐𝑖 ⊆* 𝑐 for every 𝑐 ∈ 𝐽𝜏 𝑖 which is a club in 𝛼. Noticethat we may arrange 𝑐𝑖 ∈ 𝐻 by 𝜏 𝑖 ∈ 𝐻. We may assume without loss of generality that 𝑐𝑖 ∈ 𝐽𝜏 𝑖+1

for every 𝑖 < 𝜃. We then get𝐻 |= 𝑐𝑖 ⊆* 𝑐𝑗

for all 𝑗 < 𝑖 < 𝜃, which by the choice of readily implies that

𝑐𝑖 ∖ ⊆ 𝑐𝑗 ∖ for all 𝑗 < 𝑖 < 𝜃. But as cf(𝛼) = 𝜃, this gives that

𝑐 =⋂𝑖<𝜃

(𝑐𝑖 ∖ )

is club in 𝛼.We may then let 𝑑 be a cofinal subset of 𝑐 of order type 𝜔. Then for every club 𝑐 ∈ 𝐽𝜈 which is

a club in 𝛼, we have 𝑑 ⊆* 𝑐.We thus in both cases found a set 𝑑 which satisfies (a) from Claim 3.1.3. The existence of some

such 𝑑 is a Σ1–statement in the parameter 𝐽𝜈(𝛼). Therefore, if we pick 𝑑 <𝐿–least with (a), thenby 𝑁𝛼 ∈ 𝐽𝜙(𝛼′) ≺ 𝐽𝜆+ for 𝛼′ > 𝛼 we will also get (b) from Claim 3.1.3.

Let us now for each 𝛼 ∈ 𝐸𝜆+

𝜔 define 𝑐𝛼 ⊆ 𝛼 as follows. If 𝜙(𝛼) = 𝛼, then we let 𝑐𝛼 be the <𝐿–least𝑑 which satisfies the conclusion of Claim 3.1.3. If 𝜙(𝛼) = 𝛼, then just let 𝑐𝛼 be the <𝐿-least cofinalsubset of 𝛼 of order-type 𝜔.

Finally, for all 𝛼 < 𝜆+, let

𝐶𝛼 =

⎧⎪⎪⎪⎨⎪⎪⎪⎩∅, if 𝛼 = 0;

𝛽, if 𝛼 = 𝛽 + 1;

𝐶 ′𝛼, if 𝛼 ∈ acc(𝜆+) & 𝜃(𝜙(𝛼)) > 0;

𝑐𝛼, otherwise.

Clearly, 𝐶𝛼 is a club in 𝛼 of order-type ≤ 𝜆. By Claim 3.1.1(c) and Claim 3.1.2(d), if ∈ acc(𝐶𝛼),then 𝐶 = 𝐶𝛼 ∩ . Altogether, ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ satisfies Clauses (1) and (4a) of Definition 2.3.

Let Γ: OR → 𝜔× [OR]<𝜔 be some simply-definable enumeration of 𝜔× [OR]<𝜔. Using Kleene’s“≃” notation, for 𝛼 < 𝜆+, let 𝑔𝛼 be the partial function with dom(𝑔𝛼) ⊆ 𝜆 such that

𝑔𝛼(𝜒) ≃ ℎ𝜙(𝛼)(Γ(𝜒)).


Claim 3.1.4. Let 𝛼 < 𝜆+. All of the the following hold:

(a) 𝐽𝜈(𝜙(𝛼)) = 𝑔𝛼(𝑖) | 𝑖 < 𝜆, 𝑔𝛼(𝑖) is defined;(b) If sup(acc(𝐶𝛼)) = 𝛼, 𝜒 < otp(𝐶𝛼), and 𝑔𝛼(𝜒) is defined, then there exists some ∈ acc(𝐶𝛼)

such that 𝑔(𝜒) is defined;

(c) If ∈ acc(𝐶𝛼), 𝜒 < otp(𝐶) and 𝑔(𝜒) ∈ 𝜙()2, then 𝑔𝛼(𝜒) ∈ 𝜙(𝛼)2 and 𝑔(𝜒) = 𝑔𝛼(𝜒) 𝜙().

Proof. (a) Since 𝐽𝜈(𝜙(𝛼)) = ℎ𝜙(𝛼)“(𝜔 × [𝜆]<𝜔), as pointed out earlier.(b) is clear from the construction, as 𝐽𝜈(𝜙(𝛼)) results from the direct limit of the system (𝐽𝜈(𝜙(𝛽)), 𝜎𝜙(𝛽),𝜙(𝛽′) |

𝛽 ≤ 𝛽′ ∈ 𝐶𝛼).

(c) Let 𝛼, , and 𝜒 be as in the hypothesis of (c). Then 𝑔𝛼(𝜒) ∈ 𝜙(𝛼)2 follows immediately from

the fact that 𝑔(𝜒) ∈ 𝜙()2, as 𝜎𝜙(),𝜙(𝛼) is Σ0–elementary and sends its critical point 𝜙() to 𝜙(𝛼).Moreover, if 𝜉 < 𝜙(), then 𝑔𝛼(𝜉) = 𝜎𝜙(),𝜙(𝛼)(𝑔)(𝜉) = 𝜎𝜙(),𝜙(𝛼)(𝑔(𝜉)) = 𝑔(𝜉).

We have already defined ⟨(𝐶𝛼, 𝑁𝛼) | 𝛼 < 𝜆+⟩, and our next goal is to define ⟨𝑓𝛼 | 𝛼 < 𝜆+⟩.I If 𝜆 = ℵ0, then for every 𝛼 ∈ acc(𝜆+), we have acc(𝐶𝛼) = ∅, and hence we simply let

𝑓𝛼 : 𝐶𝛼 → 𝒫(𝛼) ∩𝑁𝛼 be the <𝐿-least surjection. For 𝛼 ∈ 𝜆+ ∖ acc(𝜆+), we let 𝑓𝛼 : 𝐶𝛼 → 0 beconstant.I If 𝜆 > ℵ0, we do the following. Let 𝜓 : 𝜆 → 𝜆 be the <𝐿-least function such that 𝜓−1𝑖 is

cofinal in 𝜆 for all 𝑖 < 𝜆. Given 𝛼 ∈ acc(𝜆+), let 𝜑𝛼 : 𝐶𝛼 → otp(𝐶𝛼) denote the order-preservingisomorphism. Then, for all 𝛽 ∈ 𝐶𝛼, set

𝑍𝛽𝛼 = 𝛿 ∈ acc(𝐶𝛼) ∩ 𝛽 | 𝜓(𝜑𝛼(𝛽)) < 𝜑𝛼(𝛿) & 𝑔𝛿(𝜓(𝜑𝛼(𝛽))) ∈ 𝜙(𝛿)2,

and

𝑓𝛼(𝛽) =

𝜀 < 𝛼 | 𝑔𝛼(𝜓(𝜑𝛼(𝛽)))(𝜀) = 1, if 𝑍𝛽

𝛼 = ∅;

∅, otherwise.

For 𝛼 ∈ 𝜆+ ∖ acc(𝜆+), let 𝑓𝛼 : 𝐶𝛼 → 0 be constant.Having constructed ⟨𝑓𝛼 | 𝛼 < 𝜆+⟩, we claim:

Claim 3.1.5. Let 𝛼 < 𝜆+ be arbitrary. All of the following hold:

(a) 𝑓𝛼 is a (well-defined) function from 𝐶𝛼 to 𝒫(𝛼) ∩𝑁𝛼. Moreover, Im(𝑓𝛼) ⊆ 𝐽𝜈(𝜙(𝛼));(b) If ∈ acc(𝐶𝛼) and 𝛽 ∈ 𝐶, then 𝑓(𝛽) = 𝑓𝛼(𝛽) ∩ ;(c) If otp(𝐶𝛼) = 𝜆, then Im(𝑓𝛼) = 𝒫(𝛼) ∩ 𝐽𝜈(𝜙(𝛼));(d) 𝑓𝛼 is surjective iff 𝛼 ∈ 𝜆+ ∖𝑅<, where

𝑅< = 𝑅 ∪ 𝛼 < 𝜆+ | otp(𝐶𝛼) < 𝜆.

Proof. To avoid trivialities, assume 𝜆 > ℵ0 and 𝛼 ∈ acc(𝜆+).

(a) Let 𝛽 ∈ 𝐶𝛼 be arbitrary. If 𝑍𝛽𝛼 = ∅, then 𝑓𝛼(𝛽) = ∅ which is indeed an element of 𝒫(𝛼) ∩

𝐽𝜈(𝜙(𝛼)). Suppose that 𝑍𝛽𝛼 = ∅. Write 𝜒 = 𝜓(𝜑𝛼(𝛽)). Fix 𝛿 ∈ 𝑍𝛽

𝛼 . Then 𝛿 ∈ acc(𝐶𝛼) and

𝜒 < 𝜑𝛼(𝛿). Consequently, 𝜒 < otp(𝐶𝛿) and 𝑔𝛿(𝜒) ∈ 𝜙(𝛿)2, and then by Clause (c) of Claim 3.1.4,

𝑔𝛼(𝜒) ∈ 𝜙(𝛼)2. By Clause (a) of Claim 3.1.4, 𝑔𝛼(𝜒) ∈ 𝐽𝜈(𝜙(𝛼)). Since the latter is rud-closed, itfollows that 𝑓𝛼(𝛽) ∈ 𝒫(𝛼) ∩ 𝐽𝜈(𝜙(𝛼)). By definition of 𝑁𝛼 (cf. page 6), we have 𝐽𝜈(𝜙(𝛼)) ⊆ 𝑁𝛼.

(b) Suppose ∈ acc(𝐶𝛼) and 𝛽 ∈ 𝐶. Then 𝐶 = 𝐶𝛼 ∩ , 𝜑 = 𝜑𝛼 , and 𝜓(𝜑(𝛽)) =

𝜓(𝜑𝛼(𝛽)), say, it is 𝜒. As 𝑍𝛽𝛼 ⊆ 𝐶𝛼 ∩ 𝛽 and 𝑍𝛽

⊆ 𝐶 ∩ 𝛽, we altogether infer that 𝑍𝛽𝛼 = 𝑍𝛽

. Inparticular, if the latter is empty, then 𝑓(𝛽) = ∅ = 𝑓𝛼(𝛽) ∩ .

Next, suppose that 𝑍𝛽 is nonempty, and fix a witnessing element 𝛿. By 𝛿 ∈ 𝑍𝛽

, we know that𝜒 < otp(𝐶𝛿) and 𝑔𝛿(𝜒) ∈ 𝜙(𝛿)2, and then by Clause (c) of Claim 3.1.4, we know that 𝑔(𝜒) ∈ 𝜙()2.


By ∈ acc(𝐶𝛼) and 𝜒 < otp(𝐶𝛿) < otp(𝐶) and Clause (c) of Claim 3.1.4, we then know that

𝑔𝛼(𝜒) ∈ 𝜙(𝛼)2 and 𝑔(𝜒) = 𝑔𝛼(𝜒) . Consequently, 𝑓(𝛽) = 𝑓𝛼(𝛽) ∩ .(c) Suppose that otp(𝐶𝛼) = 𝜆, and let 𝑥 be an arbitrary element of 𝒫(𝛼) ∩ 𝐽𝜈(𝜙(𝛼)). By 𝑥 ⊆

𝛼 ⊆ 𝜙(𝛼), let 𝜚𝑥 : 𝜙(𝛼) → 2 denote the characteristic function of 𝑥. Since 𝜙(𝛼) ∈ 𝐽𝜈(𝜙(𝛼)) andsince the latter is rud-closed, we have 𝜚𝑥 ∈ 𝐽𝜈(𝜙(𝛼)). By Clause (a) of Claim 3.1.4, we may fixsome ordinal 𝑖 < 𝜆 such that 𝑔𝛼(𝑖) = 𝜚𝑥. Since otp(𝐶𝛼) = 𝜆 is an uncountable cardinal, we havesup(acc(𝐶𝛼)) = sup(𝐶𝛼) = 𝛼, and then by Clauses (b) and (c) of Claim 3.1.4, let us fix a largeenough ∈ acc(𝐶𝛼) such that 𝑖 < otp(𝐶) and 𝑔(𝑖) = 𝑔𝛼(𝑖) 𝜙(). Now, by the choice of 𝜓,there exists a large enough 𝑗 with 𝜑𝛼() < 𝑗 < 𝜆 such that 𝜓(𝑗) = 𝑖. As otp(𝐶𝛼) = 𝜆 > 𝑗, let𝛽 ∈ 𝐶𝛼 be the unique element to satisfy 𝜑𝛼(𝛽) = 𝑗. Then:

𝑍𝛽𝛼 = 𝛿 ∈ acc(𝐶𝛼) ∩ 𝛽 | 𝑖 < 𝜑𝛼(𝛿) & 𝑔𝛿(𝑖) ∈ 𝜙(𝛿)2.

By 𝑖 < otp(𝐶) = 𝜑𝛼() and 𝑔(𝑖) ∈ 𝜙()2, we have ∈ 𝑍𝛽𝛼 . So

𝑓𝛼(𝛽) = 𝜀 < 𝛼 | 𝑔𝛼(𝜓(𝜑𝛼(𝛽)))(𝜀) = 1= 𝜀 < 𝛼 | 𝑔𝛼(𝜓(𝑗))(𝜀) = 1= 𝜀 < 𝛼 | 𝑔𝛼(𝑖)(𝜀) = 1= 𝜀 < 𝛼 | 𝜚𝑥(𝜀) = 1= 𝑥,

as sought.(d) If 𝛼 ∈ 𝑅, then by Clause (a), Im(𝑓𝛼) ⊆ 𝐽𝜈(𝜙(𝛼)) 𝐽𝜈(𝜙(𝛼))+𝜔 = 𝑁𝛼. If otp(𝐶𝛼) < 𝜆, then

| Im(𝑓𝛼)| < 𝜆 = |𝑁𝛼|, so 𝛼 is not onto. Finally, if 𝛼 ∈ 𝑅 and otp(𝐶𝛼) = 𝜆, then by Clause (c) andthe definition of 𝑁𝛼 in this case, Im(𝑓𝛼) = 𝒫(𝛼) ∩ 𝐽𝜈(𝜙(𝛼)) = 𝒫(𝛼) ∩𝑁𝛼.

Thus, we have established Clauses (3) and (4b) of Definition 2.4.

Claim 3.1.6. Let 𝛼 < 𝜆+. All of the following hold:

(a) 𝐶𝛽 | 𝛽 < 𝛼 ⊆ 𝑁𝛼. Moreover:(b) 𝑓𝛽 | 𝛽 < 𝛼 ⊆ 𝑁𝛼.

Proof. Denote 𝛾 = 𝜙(𝛼) and 𝜈 = 𝜈(𝛾). Clearly, 𝛼 ≤ 𝛾 < 𝜈 and 𝑁𝛼 = 𝐽𝜈 or 𝑁𝛼 = 𝐽𝜈+𝜔. Let 𝛽 < 𝛼be arbitrary.

(a) If 𝐶𝛽 = 𝑐𝛽, then by 𝛾 ∈ 𝐶 ∖ (𝛽 + 1), we get that 𝐽𝛾 ≺Σ𝜔 𝐽𝜆+ and 𝐽𝛾 |= cf(𝛽) = 𝜔. Inparticular, if 𝑐𝛽 is the <𝐿-least cofinal subset of 𝛽 of order-type 𝜔, then 𝑐𝛽 ∈ 𝐽𝛾 ⊆ 𝐽𝜈 . If 𝑐𝛽 is notof this form, then it was obtained by Claim 3.1.3 which also ensures that 𝑐𝛽 ∈ 𝑁𝛼.

Next, suppose that 𝐶𝛽 = 𝐶 ′𝛽. Notice first that

𝜙(𝛽) < 𝜈(𝜙(𝛽)) < 𝜈(𝜙(𝛽)) + 𝜔 < 𝜙(𝛼) < 𝜈(𝜙(𝛼)) = 𝑁𝛼 ∩ OR.(2)

We have that 𝐶𝛽 = 𝜑−1“𝐸𝜙(𝛽), where 𝜑 = 𝜙 𝛽. By definition of 𝐶 and since 𝜙(𝛽) ∈ 𝐶, we knowthat 𝐶 ∩ 𝜙(𝛽) is Σ1-definable over 𝐽𝜙(𝛽)+𝜔. By Equation (2), 𝐽𝜙(𝛽)+𝜔 ∈ 𝑁𝛼 and 𝜑 ∈ 𝑁𝛼, and itsuffices to show that 𝐸𝜙(𝛽) ∈ 𝑁𝛼. But an inspection of the construction of 𝐸𝜙(𝛽) yields that 𝐸𝜙(𝛽)

is Σ1-definable over 𝐽𝜈(𝜙(𝛽))+𝜔. Hence by Equation (2) again, 𝐸𝜙(𝛽) ∈ 𝑁𝛼.(b) We have already shown that 𝐶𝛽 ∈ 𝑁𝛼. This immediately gives 𝜑𝛽 ∈ 𝑁𝛼. We have that

𝑔𝛽 is definable over 𝐽𝜈(𝜙(𝛽)), so that 𝑔𝛽 ∈ 𝑁𝛼 by 𝑁𝛽 ∈ 𝑁𝛼. Certainly, 𝜓 ∈ 𝑁𝛼. Taken together,𝑓𝛽 ∈ 𝑁𝛼.

Having Clause (c) of Claim 3.1.5 in mind, we now turn to verify Clause (5) of Definition 2.4.

Claim 3.1.7. Suppose that 𝐴 ⊆ 𝜆+ is some set, and 𝐵 ⊆ 𝜆+ is a club.Then there exists a club 𝐷 ⊆ 𝛼 ∈ 𝐶 | 𝜙(𝛼) = 𝛼 such that for all 𝛼 ∈ 𝐷:


(a) 𝐴 ∩ 𝛼,𝐵 ∩ 𝛼 ∈ 𝐽𝜈(𝛼);(b) 𝐷 ∩ 𝛼 ∈ 𝑁𝛼;(c) 𝐶𝛼 ⊆* 𝐵.

Proof. (a) Let ℎ𝜔𝐽𝜆++denote the closure under Σ𝑛 Skolem functions for 𝐽𝜆++ , for all 𝑛 < 𝜔, which

are uniformly definable over all 𝐽𝛾 , 𝛾 ≥ 𝜔 · 𝜔, in a canonical fashion.Let us recursively define ⟨𝑋𝑖 | 𝑖 < 𝜆+⟩ as follows.

𝑋0 = ℎ𝜔𝐽𝜆++(𝜔 × [𝜆 ∪ 𝐴,𝐵]<𝜔),

𝑋𝑖+1 = ℎ𝜔𝐽𝜆++(𝜔 × [𝜆 ∪ 𝐴,𝐵, ⟨𝑋𝑗 | 𝑗 ≤ 𝑖⟩]<𝜔), and

𝑋𝑖 =⋃

𝑋𝑗 | 𝑗 < 𝑖 for limit 𝑖 > 0.

For 𝑖 < 𝜆+, let 𝛾(𝑖) be such that

𝜋𝑖 : 𝐽𝛾(𝑖) ∼= 𝑋𝑖 ≺ 𝐽𝜆++ ,

and write 𝛼(𝑖) = 𝑋𝑖 ∩ 𝜆+ = 𝜆+𝐽𝛾(𝑖) . Notice that 𝐶,𝜙 ⊆ 𝑋0, and hence 𝛼(𝑖) is a closure pointunder 𝜙, and 𝜙−1(𝛼(𝑖)) = 𝛼(𝑖). Consider the set 𝐷 = 𝛼(𝑖) | 𝑖 < 𝜆+ which is a club subset of𝛼 ∈ 𝐶 | 𝜙(𝛼) = 𝛼.

For 𝑖 < 𝜆+,

𝐽𝛾(𝑖)+𝜔 |= 𝛼(𝑖) is a cardinal,

while the definition of 𝜈(𝛼(𝑖)) entails that 𝜈(𝛼(𝑖)) ≥ 𝛾(𝑖) + 𝜔, and hence

𝜈(𝛼(𝑖)) > 𝛾(𝑖).(3)

Of course, 𝐴 ∩ 𝛼(𝑖) = 𝜋−1𝑖 (𝐴) ∈ 𝐽𝛾(𝑖). Thus, 𝐴 ∩ 𝛼 ∈ 𝐽𝜈(𝛼) for every 𝛼 ∈ 𝐷, and likewise,

𝐵 ∩ 𝛼 ∈ 𝐽𝜈(𝛼) for every 𝛼 ∈ 𝐷.

(b) First, we point out that for all 𝑖 < 𝜆+, Equation (3) implies:

⟨𝜋−1𝑖 “𝑋𝑗 | 𝑗 < 𝑖⟩ ∈ 𝑁𝛼(𝑖).(4)

To see this, notice that the elementary embedding 𝜋𝑖 will respect the uniformly defined Σ𝑛 Skolemfunctions for 𝐽𝜆++ and 𝐽𝛾(𝑖), respectively, and hence

𝜋−1𝑖 “𝑋0 = ℎ𝜔𝐽𝛾(𝑖)(𝜔 × [𝜆 ∪ 𝐴 ∩ 𝛼(𝑖), 𝐵 ∩ 𝛼(𝑖)]<𝜔),

𝜋−1𝑖 “𝑋𝑖+1 = ℎ𝜔𝐽𝛾(𝑖)(𝜔 × [𝜆 ∪ 𝐴 ∩ 𝛼(𝑖), 𝐵 ∩ 𝛼(𝑖), ⟨𝜋−1

𝑖 “𝑋𝑗 | 𝑗 ≤ 𝑖⟩]<𝜔), and

𝑋𝑖 =⋃

𝑋𝑗 | 𝑗 < 𝑖 for limit 𝑖 > 0.

This gives that if 𝛼 = 𝛼(𝑖) ∈ 𝐷, then the sequence from (4) is ∆1–definable over 𝐽𝛾(𝑖)+𝜔. However,we obviously have that in this situation 𝛼 is the only cardinal of 𝐽𝛾(𝑖), so that if 𝛾(𝑖) + 𝜔 = 𝜈(𝛼),then 𝛼 ∈ 𝑅 and 𝑁𝛼 ∩ OR = 𝛾(𝑖) + 𝜔 · 2, hence Equation (4) holds true. If 𝜈(𝛼) > 𝛾(𝑖) + 𝜔, thenthe sequence from Equation (4) is in 𝐽𝜈(𝛼) ⊆ 𝑁𝛼.

Thus, we have verified that 𝐷 ∩ 𝛼 ∈ 𝑁𝛼 for every 𝛼 ∈ 𝐸.

(c) Let 𝛼 ∈ 𝐷 be arbitrary, and we shall show that 𝐶𝛼 ⊆* 𝐵. By 𝐵 ∩ 𝛼 ∈ 𝐽𝜈(𝛼), 𝜙(𝛼) = 𝛼 and

Claim 3.1.3, we may assume that 𝐶𝛼 = 𝑐𝛼. That is, 𝐶𝛼 = 𝐶 ′𝛼 = 𝜙−1“𝐸𝛼, and we must show that

𝐸𝛼 ⊆* 𝜙“𝐵.Let 𝑖 < 𝜆+ be such that 𝛼 = 𝛼(𝑖), and let 𝜇 ∈ 𝐸𝛼 be large enough such that

𝛾(𝑖), (𝜙“𝐵) ∩ 𝛼 ⊆ Im(𝜎𝜙(𝜇),𝛼).(5)


Write 𝛾 = 𝜎−1𝜙(𝜇),𝛼(𝛾(𝑖)). We have that 𝜎𝜙(𝜇),𝛼 𝐽𝛾 : 𝐽𝛾 → 𝐽𝛾(𝑖) is fully elementary, so that in fact

𝜙(𝜇) is a limit point of (𝜙“𝐵)∩𝛼, and hence 𝜙(𝜇) ∈ (𝜙“𝐵)∩𝛼, i.e., 𝜇 ∈ 𝐵. As Equation (5) holdstrue for a tail end of 𝜇 ∈ 𝐸𝛼, we have 𝐸𝛼 ⊆* 𝜙“𝐵.

We now verify Clause (6) of Definition 2.4, using an argument from [1, S2]. Assume (6) wereto fail. Recalling Clause (d) of Claim 3.1.5, there are 𝑛, 𝑚 < 𝜔, some Π𝑛

𝑚-sentence 𝜂 valid in astructure (𝜆+,∈, ⟨𝐴𝑖 | 𝑖 < 𝜔⟩), and some club 𝐷 ⊆ 𝜆+ such that for every 𝛼 ∈ 𝐷, 𝛼 ∈ 𝑅< or else

𝑁𝛼 |= “𝜂 is not valid in (𝛼,∈, ⟨𝐴𝑖 𝛼 | 𝑖 < 𝜔⟩).”Let (𝐷, ⟨𝐴𝑖 | 𝑖 < 𝜔⟩) be the <𝐿–least such pair. Notice that 𝐶, 𝜙, and 𝑅< are all definable over

𝐽𝜆++𝜔 by some formulas with no parameters. Also, 𝐷 and ⟨𝐴𝑖 | 𝑖 < 𝜔⟩ are both definable over𝐽𝜆+𝑛 by some formulas with no parameters. But as 𝜆+ and 𝜆+𝑛, and hence 𝐽𝜆++𝜔 and 𝐽𝜆+𝑛 , areboth Σ1–definable over 𝐽𝜆+𝑛+1 from the parameter 𝜆+𝑛, we get that 𝐷, ⟨𝐴𝑖 | 𝑖 < 𝜔⟩, 𝐶, and 𝜙, andalso 𝜆, 𝜆+, 𝜆+2, . . ., 𝜆+𝑛 are all Σ1–definable over 𝐽𝜆+𝑛+1 from the parameter 𝜆+𝑛.

Let us here and in what follows use the notation from [18, p. 194] which for 𝑋 ⊆ 𝐽𝛾 writesℎ𝐽𝛾 (𝑋) for ℎ𝐽𝛾“(𝜔 × [𝑋]<𝜔), where ℎ𝐽𝛾 is the canonical Σ1 Skolem function for 𝐽𝛾 .

We now have, setting 𝐷* = 𝜙“𝐷,

𝐷, ⟨𝐴𝑖 | 𝑖 < 𝜔⟩, 𝐶, 𝜙,𝐷*, 𝜆, 𝜆+, 𝜆+2, . . . , 𝜆+𝑛 ⊆ ℎ𝐽𝜆+𝑛+1 (𝜆+𝑛).

Let 𝜈 be such that

𝜎 : 𝐽𝜈 ∼= ℎ𝐽𝜆+𝑛+1 (𝜆 ∪ 𝜆+𝑛) ≺Σ1 𝐽𝜆+𝑛+1 ,(6)

and write 𝛼 = 𝜎−1(𝜆+) = crit(𝜎) and 𝛽 = 𝜎−1(𝜆+𝑛).Of course, 𝐽𝜈 = ℎ𝐽𝜈 (𝜆 ∪ 𝛽), so that 𝜌1(𝐽𝜈) = 𝜆 and 𝑝1(𝐽𝜈) ≤* 𝛽.7 However, if we had

𝑝1(𝐽𝜈) <* 𝛽, then 𝛽 ∈ ℎ𝐽𝜈 (𝛽), so that 𝜆+𝑛 ∈ ℎ𝐽𝜆+𝑛+1 (𝜆+𝑛); but it easily follows from the

Condensation Lemma that ℎ𝐽𝜆+𝑛+1 (𝜆+𝑛) ⊆ 𝐽𝜆+𝑛 . Therefore, 𝑝1(𝐽𝜈) = 𝛽.

Obviously, 𝜈(𝛼) = 𝜈. By 𝐷,𝜙,𝐷* ⊆ ℎ𝐽𝜆+𝑛+1 (𝜆+𝑛), we have that

𝐷 ∩ 𝛼,𝜙 ∩ (𝛼× 𝛼), 𝐷* ∩ 𝛼 = 𝜎−1(𝐷,𝜙,𝐷*) ∈ 𝐽𝜈 ,

so that 𝛼 ∈ 𝐷 ∩𝐷*, and we also get that 𝛼 is a closure point of 𝜙, and 𝛼 = 𝜙(𝛼).By 𝛼 = 𝜙(𝛼), we have 𝜈(𝜙(𝛼)) = 𝜈. As ⟨𝐴𝑖 | 𝑖 < 𝜔⟩ ∈ ℎ𝐽𝜆+𝑛+1 (𝜆+𝑛), 𝐴𝑖 𝛼 = 𝜎−1(𝐴𝑖) for

every 𝑖 < 𝜔. By elementarity then,

𝑁𝛼 = 𝐽𝜈 |= “𝜂 is valid in (𝛼,∈, ⟨𝐴𝑖 𝛼 | 𝑖 < 𝜔⟩)”.(7)

Claim 3.1.8. (a) There is no 𝜉 < 𝜆 such that ℎ𝐽𝜈 (𝜉 ∪ 𝛽) ∩ 𝛼 is cofinal in 𝛼, so that inparticular 𝜉𝛼𝑖 | 𝑖 < 𝜃(𝛼) is cofinal in 𝜆;

(b) For every 𝑖 < 𝜃(𝛼), |𝜉𝛼𝑖+1| = |𝜉𝛼𝑖 |;(c) 𝜃(𝛼) = 𝜆;(d) 𝛼 ∈ 𝑅<.

Proof. Write 𝑑 = (𝑛, ) ∈ 𝜔 × [𝜆]<𝜔 | ℎ𝐽𝜈 (𝑛, ∪ 𝛽) is defined. We now introduce the followingnotation. For 𝜉 < 𝜆, let us write 𝜁(𝜉) ≤ 𝜆+𝑛+1 for the least 𝜁 such that if (𝑛, ) ∈ 𝑑 ∩ (𝜔 × [𝜉]<𝜔),then ℎ𝐽𝜁 (𝑛, (, 𝜆+𝑛)) is defined. As 𝜆+𝑛+1 is regular, 𝜁(𝜉) < 𝜆+𝑛+1.

(a) Assume that 𝜉 < 𝜆 were such that ℎ𝐽𝜈 (𝜉 ∪ 𝛽) ∩ 𝛼 is cofinal in 𝛼. Using the map 𝜎 ofEquation (6), ℎ𝐽𝜆+𝑛+1 (𝜉 ∪ 𝜆+𝑛) ∩ 𝛼 is then cofinal in 𝛼. Let 𝜁 = 𝜁(𝜉) < 𝜆+𝑛+1. Trivially,

𝑑 ∩ (𝜔 × [𝜉]<𝜔) ∈ ℎ𝐽𝜆+𝑛+1 (𝜆), so that

𝜁 ∈ ℎ𝐽𝜆+𝑛+1 (𝑑 ∩ (𝜔 × [𝜉]<𝜔), 𝜆+𝑛) ⊆ ℎ𝐽𝜆+𝑛+1 (𝜆 ∪ 𝜆+𝑛) = Im(𝜎),

7Here, ≤* is the canonical well–ordering of finite sets of ordinals, cf. [18, Problem 5.19 and p. 254].


so that by using 𝜎 again,𝛼 = sup(ℎ𝐽𝜎−1(𝜁)

(𝜉 ∪ 𝛽) ∩ 𝛼).

However, ℎ𝐽𝜎−1(𝜁)(𝜉 ∪ 𝛽) ∈ 𝐽𝜈 and 𝛼 is regular in 𝐽𝜈 . This is a contradiction.

(b) This follows from the proof of (a). We have that

𝜁(𝜉𝛼𝑖 ) ∈ ℎ𝐽𝜆+𝑛+1 (𝑑 ∩ (𝜔 × [𝜉𝛼𝑖 ]<𝜔), 𝜆+𝑛),

and then

:= sup(ℎ𝐽𝜈 (𝜉𝛼𝑖 ∪ 𝛽) ∩ 𝛼) = sup(ℎ𝐽𝜎−1(𝜁(𝜉𝛼𝑖))

(𝜉𝛼𝑖 ∪ 𝛽) ∩ 𝛼)

∈ ℎ𝐽𝜈 (𝑑 ∩ (𝜔 × [𝜉𝛼𝑖 ]<𝜔), 𝛽).

However, by the Condensation Lemma, 𝑑 ∩ (𝜔 × [𝜉𝛼𝑖 ]<𝜔) ∈ ℎ𝐽𝜆+𝑛+1 ((𝜉𝛼𝑖 )+), so that

∈ ℎ𝐽𝜈 ((𝜉𝛼𝑖 )+ ∪ 𝛽).

But this means that there is (𝑛, ) ∈ 𝜔 × [(𝜉𝛼𝑖 )+]<𝜔 such that

= ℎ𝐽𝜈 (𝑛, (, 𝛽)).

But /∈ ℎ𝐽𝜈 (𝜉𝛼𝑖 ∪ 𝛽), which now readily gives 𝜉𝛼𝑖+1 < (𝜉𝛼𝑖 )+.(c) By Clause (b), for every 𝑖 < 𝜃(𝛼),

otp(𝑗 < 𝜃(𝛼) | 𝜉𝛼𝑗 < |𝜉𝛼𝑖 |+) = |𝜉𝛼𝑖 |+,which together with Clause (a) gives that 𝜃(𝛼) = 𝜆.

(d) As 𝜈 is certainly a limit of limit ordinals, 𝛼 /∈ 𝑅. But then Clause (c) gives 𝛼 /∈ 𝑅<.

Altogether 𝛼 ∈ 𝐷 ∖𝑅<, contradicting Equation (7).

4. Applications

4.1. Preliminaries. For a family of functions 𝑇 and a set of ordinals 𝐷, write 𝑇 𝐷 = 𝑓 ∈ 𝑇 |dom(𝑓) ∈ 𝐷, and succ𝜔(𝐷) = 𝛿 ∈ 𝐷 | 0 < otp(𝐷 ∩ 𝛿) < 𝜔.

Definition 4.1. We say that 𝑇 is a 𝜅-tree, whenever there exists a set Ω of size ≤ 𝜅, for which

∙ 𝑇 ⊆ <𝜅Ω;∙ dom(𝑓) | 𝑓 ∈ 𝑇 = 𝜅;∙ for every 𝑓 ∈ 𝑇 , we have 𝑓 𝛼 | 𝛼 < 𝜅 ⊆ 𝑇 ;∙ 𝑇𝛼 := 𝑓 ∈ 𝑇 | dom(𝑓) = 𝛼 has size < 𝜅 for all 𝛼 < 𝜅.

A 𝜅-Aronszajn tree is a 𝜅-tree with no cofinal branches. A 𝜅-Kurepa tree is a 𝜅-tree admittingat least 𝜅+ many cofinal branches. A 𝜆+-tree is special if it may be covered by 𝜆 many antichains.Following [2], which generalizes the case 𝜆 = ℵ0 from [7], we say that a 𝜆+-tree 𝑇 is almost Souslin

if for every antichain 𝐴 ⊆ 𝑇 , the set dom(𝑧) | 𝑧 ∈ 𝐴 ∩ 𝐸𝜆+

cf(𝜆) is nonstationary.

Of course, almost Souslin and special are contradictory concepts.

In [3],[4], the parameterized principle P−(𝜅, 𝜇,ℛ, 𝜃,𝒮, 𝜈, 𝜎, ℰ) was introduced and studied inrelation with 𝜅-Souslin tree constructions. Here, we shall only give the definition of the special case

(𝜅, 𝜇,ℛ, 𝜃,𝒮, 𝜈, 𝜎, ℰ) = (𝜅, 2,⊑, 𝜅, 𝑆, 2, 𝜔, (𝒫(𝜅))2),

which, for simplicity, is denoted by (𝑆).

Definition 4.2. For any regular uncountable cardinal 𝜅, and stationary 𝑆 ⊆ 𝜅, (𝑆) asserts theexistence of a sequence ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ such that:

∙ 𝐶𝛼 is a club subset of 𝛼 for every limit ordinal 𝛼 < 𝜅;


∙ 𝐶 = 𝐶𝛼 ∩ for every ordinal 𝛼 < 𝜅 and every ∈ acc(𝐶𝛼);∙ for every sequence ⟨𝐴𝑖 | 𝑖 < 𝜅⟩ of cofinal subsets of 𝜅, there exist stationarily many 𝛼 ∈ 𝑆

such that sup𝛽 < 𝛼 | succ𝜔(𝐶𝛼 ∖ 𝛽) ⊆ 𝐴𝑖 = 𝛼 for all 𝑖 < 𝛼.

Note that for 𝑆 ⊆ 𝑆′ ⊆ 𝜅, every (𝑆)-sequence is also a (𝑆′)-sequence. Clearly, every (𝜅)-sequence is in particular a (𝜅)-sequence.

Definition 4.3. The principle 𝜆(𝑆) asserts the existence of a (𝑆)-sequence ⟨𝐶𝛼 | 𝛼 < sup(𝑆)⟩with the additional property that otp(𝐶𝛼) ≤ 𝜆 for all 𝛼.

Note that every 𝜆(𝜆+)-sequence is in particular a 𝜆-sequence.

Definition 4.4 ([2]). Suppose that 𝑇 ⊆ <𝜅Ω is a 𝜅-tree, and 𝑆 is stationary in 𝜅.We say that 𝑇 is (𝑆)-respecting if there exist a subset S ⊆ 𝑆 and a sequence

⟨b𝛼 : 𝑇 𝐶𝛼 → 𝛼Ω ∪ ∅ | 𝛼 < 𝜅⟩such that:

(1) ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a (S)-sequence;(2) 𝑇𝛼 ⊆ Im(b𝛼) for every 𝛼 ∈ S;(3) if ∈ acc(𝐶𝛼) and 𝑥 ∈ 𝑇 𝐶, then b(𝑥) = b𝛼(𝑥) .

The notion of 𝜆(𝑆)-respecting is defined in a similar fashion.

4.2. Walks on ordinals. In this subsection, we address the trees obtained from walks on ordinals,

as introduced in [21] (see also [23]). Suppose that = ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a 𝐶-sequence over somefixed regular uncountable cardinal 𝜅. That is, 𝐶𝛼 is a club in 𝛼 for all limit 𝛼 < 𝜅, and 𝐶𝛼+1 = 𝛼for all 𝛼 < 𝜅. Recall few of the characteristic functions of walks on ordinals:

Definition 4.5 ([21],[23]). Define Tr : [𝜅]2 → 𝜔𝜅, 𝜌2 : [𝜅]2 → 𝜔, 𝜌1 : [𝜅]2 → 𝜅 and 𝜌0 : [𝜅]2 → <𝜔𝜅as follows.

For all 𝛼 < 𝛿 < 𝜅, let

∙ Tr(𝛼, 𝛿)(𝑛) :=

⎧⎪⎨⎪⎩𝛿, 𝑛 = 0;

min(𝐶Tr(𝛼,𝛿)(𝑛−1) ∖ 𝛼), 𝑛 > 0, & Tr(𝛼, 𝛿)(𝑛− 1) > 𝛼;

𝛼, otherwise;

∙ 𝜌2(𝛼, 𝛿) := min𝑛 < 𝜔 | Tr(𝛼, 𝛿)(𝑛) = 𝛼;∙ 𝜌1(𝛼, 𝛿) := max(𝜌0(𝛼, 𝛿)), where∙ 𝜌0(𝛼, 𝛿) := ⟨otp(𝐶Tr(𝛼,𝛿)(𝑖) ∩ 𝛼) | 𝑖 < 𝜌2(𝛼, 𝛿)⟩.

Definition 4.6 ([15]). Define 𝜙2 : [𝜅]2 → 2 by stipulating 𝜙2(𝛼, 𝛿) = 1 iff 𝛼 ∈ acc(𝐶Tr(𝛼,𝛿)(𝜌2(𝛼,𝛿)−1)).

Definition 4.7 ([21],[23]). For all 𝛿 < 𝜅, let 𝜌0𝛿 : 𝛿 → <𝜔𝛿, 𝜌1𝛿 : 𝛿 → 𝛿 and 𝜌2𝛿 : 𝛿 → 𝜔 denote thefiber maps 𝛼 ↦→ 𝜌0(𝛼, 𝛿), 𝛼 ↦→ 𝜌1(𝛼, 𝛿) and 𝛼 ↦→ 𝜌2(𝛼, 𝛿), respectively. Then, put

∙ 𝒯 (𝜌0) := 𝜌0𝛿 𝛽 | 𝛽 ≤ 𝛿 < 𝜅;∙ 𝒯 (𝜌1) := 𝜌1𝛿 𝛽 | 𝛽 ≤ 𝛿 < 𝜅;∙ 𝒯 (𝜌2) := 𝜌2𝛿 𝛽 | 𝛽 ≤ 𝛿 < 𝜅.

It is easy to see that if |𝐶𝛼 ∩ 𝛽 | 𝛼 < 𝜅| < 𝜅 for all 𝛽 < 𝜅, then 𝒯 (𝜌0), 𝒯 (𝜌1) and 𝒯 (𝜌2) are𝜅-trees.

Fact 4.8 ([21],[23]). Suppose that 𝒯 (𝜌0) is derived from walks along a 𝜆-sequence,8 then 𝒯 (𝜌0) is

a special 𝜆+-Aronszajn tree.

8This means that the 𝐶-sequence that was used to define 𝜌0 in Definition 4.5 is a 𝜆-sequence, ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩.


Fact 4.9 ([21],[23]). Suppose that 𝜆 is a regular cardinal, and 𝒯 (𝜌1) is derived from walks along a𝐶-sequence ⟨𝐶𝛼 | 𝛼 < 𝜆+⟩ for which otp(𝐶𝛼) ≤ 𝜆 for all 𝛼 < 𝜆+. Then:

(1) 𝒯 (𝜌1) ⊆ <𝜆+𝜆;

(2) for every 𝑧 ∈ 𝒯 (𝜌1) and 𝑖 < 𝜆, the set 𝑧−1𝑖 has size < 𝜆;(3) for every 𝛾 < 𝛿 < 𝜆+, the set 𝜉 ≤ 𝛾 | 𝜌1𝛾(𝜉) = 𝜌1𝛿(𝜉) has size < 𝜆.

Theorem 4.10. If is a (𝑆)-sequence, then the corresponding trees 𝒯 (𝜌0), 𝒯 (𝜌1), 𝒯 (𝜌2) are

(𝑆)-respecting, as witnessed by the very same .

Proof. Suppose that = ⟨𝐶𝛼 | 𝛼 < 𝜅⟩ is a (𝑆)-sequence. Fix bijections 𝜋 : 𝜅 ↔ 2 × 𝜅 and

𝜓 : 𝜅 ↔ <𝜔𝐶𝛼 ∩ 𝛽 | 𝛼, 𝛽 < 𝜅. By the coherence of , we have |𝐶𝛼 ∩ 𝛽 | 𝛼 < 𝜅| < 𝜅 for all𝛽 < 𝜅, and hence, the set

𝐸 := 𝛽 < 𝜅 | 𝜋[𝛽] = 2 × 𝛽 & <𝜔𝐶𝛼 ∩ 𝛽 | 𝛼 < 𝜅, sup(𝐶𝛼 ∩ 𝛽) < 𝛽 = 𝜓[𝛽]is a club in 𝜅. Fix a surjection 𝜙 : 𝜅 → 𝜅 such that the preimage of any singleton is cofinal in 𝜅.Put

S := 𝛼 ∈ 𝑆 ∩ 𝐸 | 𝛼 ⊆ 𝜙[𝐶𝛼].

Claim 4.10.1. is a (S)-sequence.

Proof. Given a sequence ⟨𝐴𝑖 | 𝑖 < 𝜅⟩ of cofinal subsets of 𝜅, write:

𝐴′𝑖 :=

𝐴𝑗 , if 𝜋(𝑖) = (0, 𝑗);

𝜙−1𝑗, if 𝜋(𝑖) = (1, 𝑗).

As is a (𝑆)-sequence, the following set is stationary:

𝑅 := 𝛼 ∈ 𝑆 ∩ 𝐸 | ∀𝑖 < 𝛼 sup𝛽 ∈ 𝐶𝛼 | succ𝜔(𝐶𝛼 ∖ 𝛽) ⊆ 𝐴′𝑖 = 𝛼.

Let 𝛼 ∈ 𝑅 be arbitrary.I By 𝜋[𝛼] ⊇ 0 × 𝛼, we have sup𝛽 ∈ 𝐶𝛼 | succ𝜔(𝐶𝛼 ∖ 𝛽) ⊆ 𝐴𝑖 = 𝛼 for all 𝑖 < 𝛼.I By 𝜋[𝛼] ⊇ 1 × 𝛼, we have sup𝛽 ∈ 𝐶𝛼 | succ𝜔(𝐶𝛼 ∖ 𝛽) ⊆ 𝜙−1𝑖 = 𝛼 for all 𝑖 < 𝛼. In

particular, 𝛼 ∈ S.

For all 𝜀 < 𝜅 and nonzero limit 𝛽 < 𝜅, let 𝜓𝛽𝜀 denote 𝜓(𝜀)𝐶𝛽. Then define Σ𝛽

𝜀 : 𝛽 → <𝜔𝜅, asfollows. Given 𝛼 < 𝛽, put

∙ 𝑗′ = min𝑗 ∈ dom(𝜓𝛽𝜀 ) | 𝜓𝛽

𝜀 (𝑗) ∖ 𝛼 = ∅;

∙ 𝛼+ = min(𝜓𝛽𝜀 (𝑗′) ∖ 𝛼), and

∙ Σ𝛽𝜀 (𝛼) = ⟨otp(𝜓𝛽

𝜀 (𝑗) ∩ 𝛼) | 𝑗 ≤ 𝑗′⟩a⟨otp(𝐶Tr(𝛼,𝛼+)(𝑖+1) ∩ 𝛼) | 𝑖+ 1 < 𝜌2(𝛼, 𝛼+)⟩.

Let 𝑚 : <𝜔𝜅 → 𝜅 denote a map that satisfies 𝜎 ↦→ max(𝜎) for all nonempty sequence 𝜎. Letℓ : <𝜔𝜅→ 𝜔 denote the map that satisfies 𝜎 ↦→ |𝜎| for all sequence 𝜎.

Denote Ω0 := <𝜔𝜅, Ω1 := 𝜅 and Ω2 := 𝜔. For all 𝑖 < 3, define−→b𝑖 = ⟨b𝛽

𝑖 : 𝒯 (𝜌𝑖) 𝐶𝛽 → 𝛽Ω𝑖 |𝛽 < 𝜅⟩ by stipulating:

b𝛽0 (𝑥) = Σ𝛽

𝜙(dom(𝑥)),

b𝛽1 (𝑥) = 𝑚 ∘ Σ𝛽

𝜙(dom(𝑥)),

b𝛽2 (𝑥) = ℓ ∘ Σ𝛽

𝜙(dom(𝑥)).

Claim 4.10.2. Suppose 𝑖 < 3, 𝛽 ∈ acc(𝐶𝛽) and 𝑥 ∈ 𝒯 (𝜌𝑖) 𝐶𝛽. Then b𝛽𝑖 (𝑥) = b𝛽

𝑖 (𝑥) 𝛽.


Proof. By 𝛽 ∈ acc(𝐶𝛽), we have 𝐶𝛽 = 𝐶𝛽 ∩ 𝛽, and it suffices to show that Σ𝛽𝜀 = Σ𝛽

𝜀 𝛽 for all𝜀 < 𝜅. But the latter is straight-forward to verify.

Claim 4.10.3. 𝒯 (𝜌𝑖)𝛽 ⊆ Im(b𝛽𝑖 ) for every 𝑖 < 3 and 𝛽 ∈ S.

Proof. We concentrate on the case 𝑖 = 0. Let 𝛽 ∈ S and 𝑧 ∈ 𝒯 (𝜌0)𝛽 be arbitrary. Pick 𝛿 ∈ [𝛽, 𝜅)such that 𝑧 = 𝜌0𝛿 𝛽. Let 𝑛 = 𝜌2(𝛽, 𝛿) − 𝜙2(𝛽, 𝛿). Define 𝜎 : 𝑛 → 𝒫(𝛽), by stipulating 𝜎(𝑗) :=𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽. By 𝛽 ∈ S ⊆ 𝐸 and the definition of 𝑛, there exists some 𝜀 < 𝛽 such that 𝜓(𝜀) = 𝜎.By 𝛽 ∈ S, there exists some 𝛾 ∈ 𝐶𝛽 such that 𝜙(𝛾) = 𝜀. Let 𝑥 = 𝑧 𝛾. By 𝑧 ∈ 𝒯 (𝜌0)𝛽, we have𝑥 ∈ 𝒯 (𝜌0)𝛾 , let alone 𝑥 ∈ 𝒯 (𝜌0) 𝐶𝛽.

We have 𝜙(dom(𝑥)) = 𝜀, and so, to show that b𝛽0 (𝑥) = 𝑧, it suffices to prove that Σ𝛽

𝜀 = 𝑧. First,we make the following observation.I If 𝜙2(𝛽, 𝛿) = 0, then

𝜓𝛽𝜀 = 𝜓(𝜀)𝐶𝛽 = 𝜎𝐶𝛽

= ⟨𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽 | 𝑗 < 𝜌2(𝛽, 𝛿)⟩𝐶𝛽

= ⟨𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽 | 𝑗 ≤ 𝑛⟩.

I If 𝜙2(𝛽, 𝛿) = 1, then 𝛽 ∈ acc(𝐶Tr(𝛽,𝛿)(𝜌2(𝛽,𝛿)−1)), and hence 𝐶𝛽 = 𝐶Tr(𝛽,𝛿)(𝜌2(𝛽,𝛿)−1) ∩ 𝛽.Consequently

𝜓𝛽𝜀 = 𝜓(𝜀)𝐶𝛽 = 𝜎𝐶𝛽

= ⟨𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽 | 𝑗 < 𝜌2(𝛽, 𝛿) − 1⟩𝐶𝛽

= ⟨𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽 | 𝑗 < 𝜌2(𝛽, 𝛿) − 1⟩(𝐶Tr(𝛽,𝛿)(𝜌2(𝛽,𝛿)−1) ∩ 𝛽)= ⟨𝐶Tr(𝛽,𝛿)(𝑗) ∩ 𝛽 | 𝑗 ≤ 𝑛⟩.

Now, let 𝛼 < 𝛽 be arbitrary. By 𝑧 = 𝜌0𝛿 𝛽 and definition of 𝜌0𝛿, we have:

𝑧(𝛼) = ⟨otp(𝐶Tr(𝛼,𝛿)(𝑗) ∩ 𝛼) | 𝑗 < 𝜌2(𝛼, 𝛿)⟩.

Let 𝑗′ = min𝑗 ∈ dom(𝜓𝛽𝜀 ) | 𝜓𝛽

𝜀 (𝑗)∖𝛼 = ∅. Then, for all 𝑗 < 𝑗′, we have (𝐶Tr(𝛽,𝛿)(𝑗)∩𝛽)∖𝛼 = ∅,and hence min(𝐶Tr(𝛽,𝛿)(𝑗) ∖ 𝛼) = min(𝐶Tr(𝛽,𝛿)(𝑗) ∖ 𝛽). As Tr(𝛼, 𝛿)(0) = 𝛿 = Tr(𝛽, 𝛿)(0), it followsthat

Tr(𝛼, 𝛿) 𝑗′ + 1 = Tr(𝛽, 𝛿) 𝑗′ + 1,

and hence

𝑧(𝛼) 𝑗′ + 1 = Σ𝛽𝜀 𝑗

′ + 1.

Let 𝛼+ := Tr(𝛼, 𝛿)(𝑗′). By definition

𝑧(𝛼) = (𝑧(𝛼) 𝑗′ + 1)⟨otp(𝐶Tr(𝛼,𝛼+)(𝑖+1) ∩ 𝛼) | 𝑖+ 1 < 𝜌2(𝛼, 𝛼+)⟩.

By Tr(𝛽, 𝛿)(𝑗′) = Tr(𝛼, 𝛿)(𝑗′) = 𝛼+, we have min(𝜓𝛽𝜀 (𝑗′) ∖ 𝛼) = 𝛼+. Altogether:

𝑧(𝛼) = Σ𝛽𝜀 (𝛼).

So for each 𝑖 < 3, S and−→b𝑖 witness that 𝒯 (𝜌𝑖) is (𝑆)-respecting.

Theorem 4.11. Suppose that *𝜆 holds for a given regular uncountable cardinal 𝜆. Then there

exists a 𝜆(𝐸𝜆+

𝜆 )-sequence ⟨𝐷𝛼 | 𝛼 < 𝜆+⟩ satisfying the following. For every stationary 𝑆 ⊆ 𝐸𝜆+

𝜆 ,there are 𝛿 ∈ 𝑆 and 𝛾 ∈ nacc(𝐷𝛿) ∩ 𝑆 such that 𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾.


Proof. Let ⟨(𝐶𝛼, 𝑋𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ witness *𝜆. Fix a surjection 𝜓 : 𝜆→ 𝜆 with the property that

for all 𝑗 < 𝜆, the set 𝑖 < 𝜆 | 𝜓“(𝑖, 𝑖+ 𝜔) = 𝑗 has size 𝜆. We may assume that 𝐶𝛼+1 = ∅ for all

𝛼 < 𝜅. Denote Λ = 𝛼 < 𝜆+ | otp(𝐶𝛼) = 𝜆. Clearly, Λ = 𝐸𝜆+

𝜆 . Write 𝜅 = 𝜆+.For all 𝛼 < 𝜅, let 𝜋𝛼 : otp(𝐶𝛼) → 𝐶𝛼 denote the monotone enumeration of 𝐶𝛼. We now define a

sequence of functions −→𝜎 = ⟨𝜎𝛼 : otp(𝐶𝛼) → 𝛼 | 𝛼 < 𝜅⟩ by recursion over 𝛼 < 𝜅. For this, suppose𝛼 < 𝜅, and −→𝜎 𝛼 has already been defined. The definition of 𝜎𝛼 : otp(𝐶𝛼) → 𝛼 is obtained byrecursion over 𝑖 < otp(𝐶𝛼). For this, suppose 𝑖 < otp(𝐶𝛼) and 𝜎𝛼 𝑖 has already been defined. Let

∙ 𝑋𝑖𝛼 = 𝜉 ∈ 𝑓𝛼(𝜋𝛼(𝜓(𝑖))) | 𝜋𝛼(𝑖) < 𝜉 ≤ 𝜋𝛼(𝑖+ 1), 𝜓(𝑖) ≤ 𝑖, and

∙ 𝑌 𝑖𝛼 = 𝜉 ∈ 𝑋𝑖

𝛼 | 𝜎𝛼 (𝑖+ 1) = 𝜎𝜉 (𝑖+ 1).

Then, let

𝜎𝛼(𝑖) =

⎧⎪⎨⎪⎩min(𝑌 𝑖′

𝛼 ), if 𝑖 = 𝑖′ + 1, 𝑌 𝑖′𝛼 = ∅;

min(𝑋𝑖′𝛼 ), if 𝑖 = 𝑖′ + 1, 𝑌 𝑖′

𝛼 = ∅, 𝑋𝑖′𝛼 = ∅;

𝜋𝛼(𝑖), otherwise.

Put 𝐷𝛼 = Im(𝜎𝛼).

Claim 4.11.1. 𝐷𝛼 is a club in 𝛼, otp(𝐷𝛼) = otp(𝐶𝛼), acc(𝐷𝛼) = acc(𝐶𝛼) and if ∈ acc(𝐷𝛼),then 𝐷 = 𝐷𝛼 ∩ .

Proof. For all 𝑖 < otp(𝐶𝛼), we have

(a) 𝜋𝛼(𝑖) < 𝜎𝛼(𝑖+ 1) ≤ 𝜋𝛼(𝑖+ 1);(b) 𝜎𝛼(𝑖) = 𝜋𝛼(𝑖) for all limit 𝑖 < otp(𝐶𝛼), including 𝑖 = 0.

So otp(𝐷𝛼∩𝛾) ≤ otp(𝐶𝛼∩𝛾)+1 for all 𝛾 < 𝛼, and acc(𝐷𝛼) = acc(𝐶𝛼). Towards a contradiction,suppose that ∈ acc(𝐷𝛼), while 𝐷𝛼∩ = 𝐷. Let 𝑖 < otp(𝐶) be the least such that 𝜎𝛼(𝑖) = 𝜎(𝑖).By ∈ acc(𝐷𝛼) = acc(𝐶𝛼), we have 𝐶 = 𝐶𝛼∩ , 𝜋 ⊆ 𝜋𝛼. So by Clause (b), 𝑖 must be a successorordinals, say 𝑖 = 𝑖′ + 1. We have:

𝑋𝑖′ = 𝜉 ∈ 𝑓(𝜋(𝜓(𝑖′))) | 𝜋(𝑖′) < 𝜉 ≤ 𝜋(𝑖+ 1), 𝜓(𝑖′) ≤ 𝑖′

= 𝜉 ∈ 𝑓𝛼(𝜋𝛼(𝜓(𝑖′))) ∩ | 𝜋𝛼(𝑖′) < 𝜉 ≤ 𝜋𝛼(𝑖+ 1), 𝜓(𝑖′) ≤ 𝑖 = 𝑋𝑖′𝛼 .

By minimality of 𝑖, we then also have 𝑌 𝑖′ = 𝑌 𝑖′

𝛼 . But then 𝜎𝛼(𝑖) = 𝜎(𝑖). This is a contradiction.

So := ⟨𝐷𝛼 | 𝛼 < 𝜅⟩ is a 𝜆-sequence. Next, we prove that is a 𝜆(S)-sequence for everystationary S ⊆ Λ.

Claim 4.11.2. For every stationary S ⊆ Λ, and every sequence ⟨𝐴𝛿 | 𝛿 < 𝜅⟩ of cofinal subsets of𝜅, there exist stationarily many 𝛼 ∈ S such that for every 𝛿 < 𝛼, we have

sup𝛽 ∈ 𝐷𝛼 | succ𝜔(𝐷𝛼 ∖ 𝛽) ⊆ 𝐴𝛿 = 𝛼.

Proof. Let S and ⟨𝐴𝛿 | 𝛿 < 𝜅⟩ be as in the hypothesis. By Clause (5) of Definition 2.2, for every𝛿 < 𝜅, fix a club 𝐸𝛿 ⊆ acc+(𝐴𝛿) such that 𝐴𝛿 ∩ 𝛼 ∈ 𝑁𝛼 for all 𝛼 ∈ 𝐸𝛿. Let 𝐶 := ∆𝛿<𝜅𝐸𝛿. ByClause (5) of Definition 2.2, let us fix 𝛼 ∈ S such that 𝐶𝛼 ⊆* 𝐶.

Let 𝜖, 𝛿 < 𝛼 be arbitrary. We shall find 𝛽 ∈ 𝐷𝛼 ∖ 𝜖 such that

succ𝜔(𝐷𝛼 ∖ 𝛽) ⊆ 𝐴𝛿.

Without loss of generality, 𝜖 is also large enough so that 𝐶𝛼 ∖ 𝜖 ⊆ 𝐶.By 𝛿 ∈ 𝛼 ∈ 𝐶, we have 𝛼 ∈ 𝐸𝛿, and hence 𝐴𝛿 ∩ 𝛼 ∈ 𝑋𝛼. Since 𝛼 ∈ S ⊆ Λ, we appeal

to Clause (3) of Definition 2.2 to obtain some 𝑗 < otp(𝐶𝛼) such that 𝑓𝛼(𝜋𝛼(𝑗)) = 𝐴𝛿 ∩ 𝛼. Fixa large enough 𝑖 < otp(𝐶𝛼) such that 𝜓“(𝑖, 𝑖 + 𝜔) = 𝑗, and 𝜋𝛼(𝑖) > max𝜖, 𝛿, 𝜋𝛼(𝑗). Write𝛽 := min(𝐷𝛼 ∖ 𝜋𝛼(𝑖) + 1). Then 𝛽 > 𝜖 and succ𝜔(𝐷𝛼 ∖ 𝛽) = 𝜎𝛼“(𝑖, 𝑖+ 𝜔).


Let 𝑖′ ∈ [𝑖, 𝑖+ 𝜔) be arbitrary. We shall show that 𝜎𝛼(𝑖′ + 1) ∈ 𝐴𝛿. We have

𝑋𝑖′𝛼 = 𝜉 ∈ 𝑓𝛼(𝜋𝛼(𝜓(𝑖′))) | 𝜋𝛼(𝑖′) < 𝜉 ≤ 𝜋𝛼(𝑖′ + 1), 𝜓(𝑖′) ≤ 𝑖′ =

= 𝜉 ∈ 𝑓𝛼(𝜋𝛼(𝑗)) | 𝜋𝛼(𝑖′) < 𝜉 ≤ 𝜋𝛼(𝑖′ + 1) == 𝜉 ∈ 𝐴𝛿 ∩ 𝛼 | 𝜋𝛼(𝑖′) < 𝜉 ≤ 𝜋𝛼(𝑖′ + 1).

By 𝜋𝛼(𝑖′+1) > max𝜖, 𝛿, we have 𝜋𝛼(𝑖′+1) ∈ 𝐸𝛿 ⊆ acc+(𝐴𝛿), and hence 𝑋𝑖′𝛼 = ∅. Consequently,

𝜎𝛼(𝑖′ + 1) ∈ 𝑋𝑖′𝛼 ⊆ 𝐴𝛿, as sought.

Claim 4.11.3. For every stationary 𝑆 ⊆ Λ, there are 𝛿 ∈ 𝑆 and 𝛾 ∈ nacc(𝐷𝛿) ∩ 𝑆 such that𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾.

Proof. Fix a surjection 𝜙 : 𝜅 → 𝐷𝛼 ∩ 𝛽 | 𝛼, 𝛽 < 𝜅. Note that by Claim 4.11.1, the following setis a club

𝐷 := 𝛿 < 𝜅 | 𝐷𝛼 ∩ 𝛽 | 𝛼, 𝛽 < 𝜅, sup(𝐷𝛼 ∩ 𝛽) < 𝛿 = 𝜙[𝛿].Let 𝑆 be an arbitrary stationary subset of Λ. Define a function 𝑔 : 𝜅×𝜅→ 𝜅 by letting for every

𝛼 < 𝜅:

𝑔(𝛼, 𝜁) :=

sup𝜉 ∈ 𝑆 | 𝜙(𝛼) ⊑ 𝐷𝜉 + 1, if sup𝜉 ∈ 𝑆 | 𝜙(𝛼) ⊑ 𝐷𝜉 < 𝜅;

min𝜉 ∈ 𝑆 | 𝜉 > 𝜁, 𝜙(𝛼) ⊑ 𝐷𝜉, otherwise.

Let 𝐸 := 𝛿 < 𝜅 | 𝑔[𝛿 × 𝛿] ⊆ 𝛿. Now, fix 𝛿 ∈ 𝑆 ∩ 𝐸 and 𝜖 < 𝛿 satisfying the following:

∙ 𝐶𝛿 ∖ 𝜖 ⊆ 𝐷 ∩ 𝐸;∙ 𝑆 ∩ 𝛿 ∈ 𝑋𝛿.

Since 𝛿 ∈ Λ and 𝑆 ∩ 𝛿 ∈ 𝑋𝛿, let us fix some 𝑗 < otp(𝐶𝛿) such that 𝑓𝛿(𝜋𝛿(𝑗)) = 𝑆 ∩ 𝛿. Fix a largeenough 𝑖 < otp(𝐶𝛿) such that 𝜓(𝑖) = 𝑗, and 𝜋𝛿(𝑖) > max𝜖, 𝜋𝛼(𝑗).

By 𝜋𝛿(𝑖+ 1) ∈ 𝐷, let us fix some 𝛼 < 𝜋𝛿(𝑖+ 1) such that 𝜙(𝛼) = 𝐷𝛿 ∩ 𝜋𝛿(𝑖+ 1). By 𝛼 ∈ 𝛿 ∈ 𝐸,we have 𝑔[𝛼 × 𝛿] ⊆ 𝛿. As 𝜙(𝛼) ⊑ 𝐷𝛿 and 𝛿 ∈ 𝑆, we thus infer that 𝑔(𝛼, 𝜁) > 𝜁 for all 𝜁 < 𝜅. Inparticular, by 𝜋𝛿(𝑖+ 1) ∈ 𝐸, we have 𝜋𝛿(𝑖) < 𝑔(𝛼, 𝜋𝛿(𝑖)) < 𝜋𝛿(𝑖+ 1). Recalling that

𝑋𝑖𝛿 = 𝜉 ∈ 𝑆 ∩ 𝛿 | 𝜋𝛿(𝑖) < 𝜉 ≤ 𝜋𝛿(𝑖+ 1),

and

𝑌 𝑖𝛿 = 𝜉 ∈ 𝑋𝑖

𝛿 | 𝜎𝛿 (𝑖+ 1) = 𝜎𝜉 (𝑖+ 1) = 𝜉 ∈ 𝑋𝑖𝛿 | 𝜙(𝛼) ⊑ 𝐷𝜉,

we see that 𝑔(𝛼, 𝜋𝛿(𝑖)) witnesses that 𝑌 𝑖𝛿 is nonempty. Write 𝛾 := 𝜎𝛿(𝑖 + 1). Then 𝛾 ∈ 𝑌 𝑖

𝛿 ∩nacc(𝐷𝛿) ∩ 𝑆, and hence 𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾 .

We now address the trees 𝒯 (𝜌0) and 𝒯 (𝜌1). Note that the latter is a projection of the former.9

Corollary 4.12. Suppose that *𝜆 holds for a given uncountable cardinal 𝜆 = 𝜆<𝜆. Then there

exists a 𝜆(𝐸𝜆+

𝜆 )-sequence which is respected by the corresponding trees 𝒯 (𝜌0) and 𝒯 (𝜌1). Moreover:

(1) 𝒯 (𝜌0) is a special 𝜆+-Aronszajn tree;(2) 𝒯 (𝜌1) is an almost Souslin, 𝜆+-Aronszajn tree;(3) 𝒯 (𝜌1) can be made special by means of a cofinality-preserving forcing.

9Indeed, let 𝑚 be some map satisfying 𝜎 ↦→ max(𝜎) for every nonempty sequence of ordinals, 𝜎. Then 𝒯 (𝜌1) isthe image of 𝒯 (𝜌0) under the map 𝑡 ↦→ 𝑚 ∘ 𝑡.


Proof. Let = ⟨𝐷𝛼 | 𝛼 < 𝜆+⟩ be given by Theorem 4.11. In particular, is a 𝜆(𝐸𝜆+

𝜆 )-sequence.

Let 𝒯 (𝜌0) and 𝒯 (𝜌1) denote the trees derived from walks along . By Theorem 4.10, 𝒯 (𝜌0) and

𝒯 (𝜌1) are 𝜆(𝐸𝜆+

𝜆 )-respecting as witnessed by our .

(1) Since is in particular a 𝜆-sequence, we get from Fact 4.8 that 𝒯 (𝜌0) is special.(2) It is easy to see that 𝒯 (𝜌1) is a 𝜆+-tree. By Clause (2) of Fact 4.9, 𝒯 (𝜌1) is moreover

Aronszajn. To see that 𝒯 (𝜌1) is almost Souslin, suppose that 𝐴 ⊆ 𝒯 (𝜌1), and that 𝑆 = dom(𝑧) |𝑧 ∈ 𝐴 ∩ 𝐸𝜆+

𝜆 is stationary. For all 𝛿 ∈ 𝑆, fix 𝑧𝛿 ∈ 𝐴 such that dom(𝑧𝛿) = 𝛿. We now run thearguments from [10]. For all 𝛿 ∈ 𝑆, put

𝑦𝛿 = 𝑧𝛿 𝜉 < 𝛿 | 𝑧𝛿(𝜉) = 𝜌1𝛿(𝜉).By 𝜆<𝜆 = 𝜆 and Fact 4.9, we may find a stationary subset 𝑆′ ⊆ 𝑆 such that 𝑦𝛿 | 𝛿 ∈ 𝑆′ is a

singleton. It follows that 𝑧𝛿 | 𝛿 ∈ 𝑆′ is an antichain iff 𝜌1𝛿 | 𝛿 ∈ 𝑆′ is an antichain. So, let us

show that the latter is not an antichain. By the choice of , let us fix 𝛿 ∈ 𝑆′ and 𝛾 ∈ nacc(𝐷𝛿)∩𝑆′

such that 𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾 . Let 𝛼 < 𝛾 be arbitrary, and we shall show that 𝜌1𝛾(𝛼) = 𝜌1𝛿(𝛼). Write𝛽 = sup(𝐷𝛿 ∩ 𝛾). Clearly, 𝛽 ∈ 𝐷𝛾 ∩𝐷𝛿 and 𝐷𝛾 ∩ 𝛽 = 𝐷𝛿 ∩ 𝛽.I If 𝛼 < 𝛽, then there exists some 𝛽′ ∈ (𝐷𝛾 ∩𝐷𝛿 ∩ (𝛽 + 1)) for which 𝜌0𝛿(𝛼) = maxotp(𝐷𝛿 ∩

𝛼), 𝜌0𝛿(𝛽′) and 𝜌0𝛾(𝛼) = maxotp(𝐷𝛾 ∩ 𝛼), 𝜌0𝛾(𝛽′). Since otp(𝐷𝛿 ∩ 𝛼) = otp(𝐷𝛿 ∩ 𝛽 ∩ 𝛼) =

otp(𝐷𝛾 ∩ 𝛽 ∩ 𝛼) = otp(𝐷𝛾 ∩ 𝛼), we infer that 𝜌1𝛾(𝛼) = 𝜌1𝛿(𝛼).I If 𝛽 ≤ 𝛼 < 𝛾, then min(𝐶𝛿 ∖ 𝛼) = 𝛾. Consequently 𝜌1𝛿(𝛼) = maxotp(𝐶𝛿 ∩ 𝛼), 𝜌1𝛾(𝛼). By

definition, we have 𝜌1𝛾(𝛼) ≥ otp(𝐶𝛾 ∩𝛼). As 𝐶𝛾 ⊒ 𝐶𝛿 ∩𝛾, we have otp(𝐶𝛾 ∩𝛼) ≥ otp(𝐶𝛿 ∩𝛾∩𝛼) =otp(𝐶𝛿 ∩ 𝛼), and hence 𝜌1𝛿(𝛼) = 𝜌1𝛾(𝛼).

(3) Let P denote the collection of all partial specializing functions of size < 𝜆. That is, 𝑝 ∈ P iffit is a function with dom(𝑝) ∈ [𝒯 (𝜌1)]

<𝜆, Im(𝑝) ⊆ 𝜆, such that 𝑝(𝑦) = 𝑝(𝑧) for all 𝑦 ( 𝑧 in dom(𝑝).Clearly, P is (< 𝜆)-closed. It remains to verify that P has the 𝜆+-cc.

Towards a contradiction, suppose that P admits an antichain of size 𝜆+. Then by 𝜆<𝜆 = 𝜆 anda standard ∆-system argument, one could find some cardinal 𝜇 < 𝜆 and a family ℱ ⊆ [𝒯 (𝜌1)]

𝜇

consisting of 𝜆+ many pairwise disjoint sets with the property that for every two distinct 𝑎, 𝑏 ∈ ℱ ,there exist 𝑥 ∈ 𝑎 and 𝑦 ∈ 𝑏 such that 𝑥 and 𝑦 are comparable. For all 𝑎 ∈ ℱ , let 𝑎(𝑖) | 𝑖 < 𝜇 be

some enumeration of 𝑎. For every 𝛿 ∈ 𝐸𝜆+

𝜆 , pick 𝑎𝛿 ∈ ℱ such that mindom(𝑥) | 𝑥 ∈ 𝑎 > 𝛿, and

define 𝑓𝛿 : 𝜇→ 𝛿𝜆 by stipulating 𝑓𝛿(𝑖) = 𝑎(𝑖) 𝛿. Then, for all 𝛾, 𝛿 ∈ 𝐸𝜆+

𝜆 there exist 𝑖, 𝑗 < 𝜇 such

that 𝑓𝛾(𝑖) and 𝑓𝛿(𝑗) are compatible. For all 𝛿 ∈ 𝐸𝜆+

𝜆 , let 𝐷𝛿 := 𝜉 < 𝛿 | ∃𝑖 < 𝜇[𝑓𝛿(𝑖)(𝜉) = 𝜌1𝛿(𝜉)].

By Clause (3) of Fact 4.9, |𝐷𝛿| < 𝜆. By 𝜆<𝜆 = 𝜆, we may find a stationary set 𝑆 ⊆ 𝐸𝜆+

𝜆 for which

(⟨𝑓𝛿(𝑖) 𝐷𝛿) | 𝑖 < 𝜇⟩, (𝜌1𝛿 𝐷𝛿)) | 𝛿 ∈ 𝑆is a singleton. Consequently, 𝜌1𝛿 | 𝛿 ∈ 𝑆 forms a chain in 𝒯 (𝜌1), contradicting the fact that𝒯 (𝜌1) is Aronszajn.

If one is willing to give away the respecting feature of the preceding, then it is possible to relax *

𝜆 to just *(𝜆+):

Corollary 4.13. Suppose that *(𝜆+) holds for a given infinite cardinal 𝜆 = 𝜆<𝜆. Then thereexists a 𝐶-sequence for which the corresponding trees 𝒯 (𝜌0) and 𝒯 (𝜌1) satisfy Clauses (1)–(3) ofCorollary 4.12.

Notice that the same ideas of this section provides a proof to the following, which unlike Corollary4.12, also apply to the case of 𝜆 singular:

Corollary 4.14. If 𝜆 holds for a given uncountable cardinal 𝜆, then there exists a 𝜆(𝐸𝜆+

cf(𝜆))-

sequence which is respected by the corresponding trees 𝒯 (𝜌0) and 𝒯 (𝜌1). Moreover:


(1) 𝒯 (𝜌0) is a special 𝜆+-Aronszajn tree;(2) 𝒯 (𝜌1) is a nonspecial 𝜆+-Aronszajn tree.

4.3. Kurepa.

Theorem 4.15. Suppose that 𝜆 is an uncountable cardinal, and ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ is a +

𝜆 -sequence. Denote Λ = 𝛼 < 𝜆+ | otp(𝐶𝛼) = 𝜆.Then, there exist a 𝜆(Λ)-sequence = ⟨𝐷𝛼 | 𝛼 < 𝜆+⟩ and 𝑇 ⊆ <𝜆+

2 such that:

∙ for every stationary 𝑆 ⊆ Λ, there are 𝛿 ∈ 𝑆 and 𝛾 ∈ nacc(𝐷𝛿) ∩ 𝑆 such that 𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾;

∙ 𝑇 is a 𝜆(Λ)-respecting 𝜆+-Kurepa tree, as witnessed by .

Proof. Let ⟨(𝐶𝛼, 𝑁𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ witness +𝜆 . Use ⟨(𝐶𝛼, 𝑓𝛼) | 𝛼 < 𝜆+⟩ to construct the sequence

= ⟨𝐷𝛼 | 𝛼 < 𝜆+⟩ exactly as in the proof of Theorem 4.11. Then:

(1) otp(𝐷𝛼) = otp(𝐶𝛼) and acc(𝐷𝛼) = acc(𝐶𝛼) for all 𝛼 < 𝜆+;

(2) is a 𝜆(S)-sequence for every stationary S ⊆ Λ;(3) for every stationary 𝑆 ⊆ Λ, there are 𝛿 ∈ 𝑆 and 𝛾 ∈ nacc(𝐷𝛿) ∩ 𝑆 such that 𝐷𝛿 ∩ 𝛾 ⊑ 𝐷𝛾 .

Let S = 𝛼 < 𝜆+ | 𝑓𝛼 is surjective. Then, S is a stationary subset of Λ. Denote 𝜅 = 𝜆+, and

∙ ℬ = 𝑏 ∈ 𝜅2 | ∀𝛼 < 𝜅[(𝑏 𝛼) ∈ 𝑁𝛼],∙ 𝑇 = 𝑏 𝛼 | 𝑏 ∈ ℬ, 𝛼 < 𝜅.

Then 𝑇 ⊆ <𝜅2 is downward closed, and |𝑇𝛼| ≤ |𝑁𝛼| ≤ 𝜆 < 𝜅 for all 𝛼 < 𝜅. For all 𝛼 < 𝜅, since𝑁𝛼 is rud-closed and 𝛼 ∈ 𝑁𝛼, we know that the constant function from 𝛼 to 0 belongs to 𝑇 , andso dom(𝑓) | 𝑓 ∈ 𝑇 = 𝜅.

We have shown that 𝑇 is a 𝜅-tree. Let us show it is Kurepa.

Claim 4.15.1. 𝑇 has at least 𝜅+ many cofinal branches.

Proof. For all 𝛼 < 𝜅, since 𝛼 ∈ 𝑁𝛼 and the latter is rud-closed, a subset of 𝛼 is 𝑁𝛼 iff its charac-teristic function is in 𝑁𝛼. Thus, it suffices to show that the following set has size > 𝜅:

𝒜 := 𝐴 ⊆ 𝜅 | ∀𝛼 < 𝜅[(𝐴 ∩ 𝛼) ∈ 𝑁𝛼].

Suppose not. Then we can find an enumeration (possibly with repetitions) 𝐴𝛽 | 𝛽 < 𝜅 of 𝒜.Consider the club 𝑍 = ∆𝛽<𝜅 acc+(𝐴𝛽). That is,

𝑍 = 𝜁 < 𝜅 | ∀𝛽 < 𝜁(sup(𝐴𝛽 ∩ 𝜁) = 𝜁).

Pick a club 𝐷 ⊆ 𝜅 such that for every 𝛼 ∈ 𝐷, we have 𝐷 ∩ 𝛼,𝑍 ∩ 𝛼 ⊆ 𝒫(𝛼) ∩ 𝑁𝛼. Then𝐸 := 𝑍 ∩ 𝐷 is a club. Let 𝜁𝛽 | 𝛽 < 𝜅 be the increasing enumeration of 𝐸. For all 𝛽 < 𝜅, wehave 𝛽 ≤ 𝜁𝛽 < 𝜁𝛽+1 and 𝜁𝛽+1 ∈ 𝑍, and hence sup(𝐴𝛽 ∩ 𝜁𝛽+1) = 𝜁𝛽+1 > 𝜁𝛽 = sup(𝐸 ∩ 𝜁𝛽+1).Consequently, 𝐴𝛽 = 𝐸 for all 𝛽 < 𝜅, and there must exist some 𝛼 < 𝜅 such that 𝐸 ∩ 𝛼 ∈ 𝑁𝛼. Fixsuch an 𝛼, and let 𝛼′ := sup(𝐸 ∩ 𝛼). As (𝐸 ∩ 𝛼) ∖ 𝛼′ is a finite subset of 𝛼, it is an element of𝑁𝛼. Since 𝑁𝛼 is closed under unions, the set 𝐸 ∩ 𝛼′ must be outside of 𝑁𝛼. In particular, it isnonempty, and 𝛼′ ∈ acc(𝐸) ⊆ 𝐷. But then, 𝐷 ∩ 𝛼′, 𝑍 ∩ 𝛼′ ⊆ 𝑁𝛼′ ⊆ 𝑁𝛼 and since 𝑁𝛼 is closedunder intersections, we get that 𝐸 ∩ 𝛼′ = (𝐷 ∩ 𝛼′) ∩ (𝑍 ∩ 𝛼′) is in 𝑁𝛼. This is a contradiction.

Next, we show that 𝑇 is respecting , by defining ⟨b𝛼 : 𝑇 𝐷𝛼 → 𝛼2 | 𝛼 < 𝜅⟩, as follows. Given𝛼 < 𝜅, let 𝜓𝛼 : 𝐷𝛼 ↔ 𝐶𝛼 denote the order-preserving bijection, and let 𝑔𝛼 : 𝐷𝛼 → 𝛼2 be such thatfor all 𝛽 ∈ 𝐷𝛼, 𝑔𝛼(𝛽) is the characteristic function of 𝑓𝛼(𝜓𝛼(𝛽)).

Now, for all 𝑥 ∈ 𝑇 𝐷𝛼, let

b𝛼(𝑥) = 𝑔𝛼(dom(𝑥)).


Suppose that ∈ acc(𝐷𝛼). Then ∈ acc(𝐶𝛼) and 𝜓 ⊆ 𝜓𝛼, so that for all 𝛽 ∈ 𝐷, 𝑓(𝜓(𝛽)) =𝑓𝛼(𝜓𝛼(𝛽)) ∩ and 𝑔(𝛽) = 𝑔𝛼(𝛽) . Consequently, b(𝑥) = b𝛼(𝑥) for all 𝑥 ∈ 𝑇 𝐷.

Finally, let 𝛼 ∈ S, and we shall show that 𝑇𝛼 ⊆ Im(b𝛼). Let 𝑦 ∈ 𝑇𝛼 be arbitrary. By definitionof 𝑇 , we have 𝑦 ∈ 𝑁𝛼. Since 𝑁𝛼 is rud-closed, the set 𝛾 < 𝛼 | 𝑦(𝛾) = 1 is in 𝑁𝛼, and so by𝛼 ∈ S, there exists some 𝛽 ∈ 𝐶𝛼 such that 𝑓𝛼(𝛽) = 𝛾 < 𝛼 | 𝑦(𝛾) = 1. Let 𝛽′ = 𝜓−1

𝛼 (𝛽). Then𝑔𝛼(𝛽′) = 𝑦. Now, put 𝑥 := 𝑦 𝛽′. Then 𝑥 ∈ 𝑇 𝐷𝛼, and by definition of b𝛼, we have b𝛼(𝑥) = 𝑦,as sought.

Corollary 4.16. Suppose that †𝜆 holds for a given uncountable cardinal 𝜆. Then there exists a

𝜆(𝐸𝜆+

cf(𝜆))-respecting 𝜆+-Kurepa tree that has no 𝜆+-Aronszajn subtrees.

Proof. The construction of all involved objects is identical to that of the proof of Theorem 4.15, but

this time we consult with a †𝜆-sequence rather than +

𝜆 . Consequently, the reflection argumentof [5, Theorem 2] shows that the 𝜆+-Kurepa tree will contain no 𝜆+-Aronszajn subtrees.

Acknowledgements

This work was engaged when the authors met at the MAMLS meeting at Carnegie MellonUniversity, May 2015. We are grateful to the organizers for the invitation.

The first author was partially supported by grant No. 1630/14 of the Israel Science Foundation.He would also like to acknowledge the German-Israeli Foundation for Scientific Research and De-velopment, grant No. I-2354-304.6/2014, for supporting his travel to the second author on July2015.

The second author partially supported by the SFB 878 of the Deutsche Forschungsgemeinschaft(DFG).

The authors thank the referee for his/her feedback.

References

[1] U. Abraham, S. Shelah, and R. M. Solovay. Squares with diamonds and Souslin trees with special squares. Fund.Math., 127(2):133–162, 1987.

[2] Ari M. Brodsky and Assaf Rinot. Reduced powers of Souslin trees. Forum Math. Sigma, 5(e2): 1–82, 2017.[3] Ari M. Brodsky and Assaf Rinot. A microscopic approach to Souslin-tree constructions. Part I. arXiv:1601.01821,

2016. http://www.assafrinot.com/paper/22[4] Ari M. Brodsky and Assaf Rinot. A microscopic approach to Souslin-tree constructions. Part II. In preparation,

2017, http://www.assafrinot.com/paper/23[5] Keith J. Devlin. The combinatorial principle ♯. J. Symbolic Logic, 47(4):888–899 (1983), 1982.[6] Keith J. Devlin. Constructibility. Perspectives in Mathematical Logic. Springer-Verlag, Berlin, 1984.[7] Keith J. Devlin and Saharon Shelah. Souslin properties and tree topologies. Proc. London Math. Soc. (3),

39(2):237–252, 1979.[8] Matthew Foreman and Peter Komjath. The club guessing ideal: commentary on a theorem of Gitik and Shelah.

J. Math. Log., 5(1):99–147, 2005.[9] Charles William Gray. Iterated forcing from the strategic point of view. ProQuest LLC, Ann Arbor, MI, 1980.

Thesis (Ph.D.)–University of California, Berkeley.[10] Michael Hrusak and Carlos Martınez Ranero. Some remarks on non-special coherent Aronszajn trees. Acta Univ.

Carolin. Math. Phys., 46(2):33–40, 2005.[11] Tetsuya Ishiu. Club guessing sequences and filters. J. Symbolic Logic, 70(4):1037–1071, 2005.[12] Tetsuya Ishiu and Paul B. Larson. Some results about (+) proved by iterated forcing. J. Symbolic Logic,

77(2):515–531, 2012.[13] R. Bjorn Jensen. The fine structure of the constructible hierarchy. Ann. Math. Logic, 4:229–308; erratum, ibid.

4 (1972), 443, 1972. With a section by Jack Silver.[14] Ronald Jensen, Ernest Schimmerling, Ralf Schindler, and John Steel. Stacking mice. J. Symbolic Logic, 74(1):315–

335, 2009.


[15] Assaf Rinot. Chain conditions of products, and weakly compact cardinals. Bull. Symb. Log., 20(3):293–314, 2014.[16] Assaf Rinot. Putting a diamond inside the square. Bull. Lond. Math. Soc., 47(3):436–442, 2015.[17] Assaf Rinot. Hedetniemi’s conjecture for uncountable graphs. J. Eur. Math. Soc., 19(1): 285–298, 2017.[18] Ralf Schindler. Set theory. Exploring independence and truth. Universitext. Springer, Cham, 2014.[19] Charles I. Steinhorn and James H. King. The uniformization property for ℵ2. Israel J. Math., 36(3-4):248–256,

1980.[20] Stevo Todorcevic. Trees, subtrees and order types. Ann. Math. Logic., 20(3):233–268, 1981.[21] Stevo Todorcevic. Partitioning pairs of countable ordinals. Acta Math., 159(3-4):261–294, 1987.[22] Stevo Todorcevic. Representing trees as relatively compact subsets of the first Baire class. Bull. Cl. Sci. Math.

Nat. Sci. Math., 30:29–45, 2005.[23] Stevo Todorcevic.Walks on ordinals and their characteristics, volume 263 of Progress in Mathematics. Birkhauser

Verlag, Basel, 2007.

Department of Mathematics, Bar-Ilan University, Ramat-Gan 5290002, Israel.URL: http://www.assafrinot.com

Institut fur mathematische Logik und Grundlagenforschung, Universitat Munster, Einsteinstr.62, 48149 Munster, Germany

URL: http://wwwmath.uni-muenster.de/logik/Personen/rds

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Introduction - uni-muenster.de · (2)For every infinite cardinal , there exists a respecting +-Kurepa tree; Roughly speaking, this means that this +-Kurepa tree looks very much like