SOUSLIN TREES AND DEGREES OF CONSTRUCTIBILITY

A Thesis

Submitted to the Faculty

in partial fulfillment of the requirements for the

degree of

Doctor of Philosophy

in

Mathematics

by

Francois G. Dorais

DARTMOUTH COLLEGE

Hanover, New Hampshire

June 18, 2007

Examining Committee:

Marcia Groszek, Chair

Rebecca Weber

Peter Winkler

Charles BarloweDean of Graduate Studies

Akihiro Kanamori

Copyright byFrancois G. Dorais

2007

Abstract

This thesis investigates possible initial segments of the degrees of con-

structibility. Specifically, we completely characterize the structure of de-

grees in generic extensions of the constructible universe L via forcing

with Souslin trees. Then we use this characterization to realize any con-

structible dual algebraic lattice as a possible initial segment of the degrees

of constructibility.

In a seminal paper [20], Gerald E. Sacks introduced perfect set forcing

(also known as Sacks forcing) to show that the two-element lattice can

be realized as an initial segment of the degrees of constructibility; i.e., if

S is a perfect set generic over L, then L[S] has precisely two degrees of

constructibility. Refining Sacks’s method, Marcia J. Groszek and Richard

A. Shore have shown that for every countable complete algebraic lattice

L ∈ L, there is a notion of forcing such that the degrees in the generic

extension form a lattice dual isomorphic to L.

Using a completely different method — namely forcing with Souslin

trees — we extend this result by removing the size restriction.

Theorem. (1.1.7) Assume V = L. Let κ be an infinite regular cardinal

and let L be a complete algebraic lattice with at most κ compact elements.

There is a Souslin tree T of height κ+ such that if G is a generic branch

through T , then the degrees of constructibility in L[G] form a lattice dual

isomorphic to L.

In view of this result, a natural question is: if the c-degrees in L[A]

form a complete lattice, must the lattice be algebraic? We show that this

is not the case.

ii

Theorem. (1.1.8) Assume V = L. There is a Souslin tree T of height

ω1 such that if G is any generic branch through T , then the degrees of

constructibility with representatives in L[G] form a lattice isomorphic to

the unit interval [0, 1].

The unit interval is not an algebraic lattice, though it is a continuous

lattice. The methods used to prove these results could be extended to

realize many other continuous lattices as initial segments of degrees of

constructibility in this way.

iii

Acknowledgements

First and foremost, I would like to thank my advisor, Marcia Groszek, for her infinite

support, uncountable advice, and imcommensurable wisdom. I would also like to

thank all members and staff of the department of mathematics for their invaluable

support. Warm thanks to my fellow logic students Brooke Andersen, John Bourke,

Jared Corduan, and Rachel Esselstein, as well as MIT/Harvard logic students Nate

Ackerman, Andrew Brooke-Taylor, Alice Chan, Cameron Freer, and Christina God-

dard. A special thank you to Stefan Bilaniuk for first introducing me to Souslin trees

while I was a student at Trent University. Also thanks to professors Richard Shore,

Akihiro Kanamori, and Gerald Sacks for their support, inspiration, and encourage-

ment. Very special thanks to my family and friends from home for their moral and

emotional support. Finally, thank you to Dartmouth College for its financial support

through a graduate fellowship.

Hanover, New Hampshire Francois G. Dorais

iv

Contents

Abstract ii

Acknowledgements iv

Contents v

Figures vii

1 Introduction 1

1.1 The Main Results in Context . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Souslin Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Degrees of Constructibility . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Degrees in Souslin Tree Extensions 18

2.1 Congruence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2 The Representation Theorem . . . . . . . . . . . . . . . . . . . . . . 22

2.3 The Comparison Theorem . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Embeddings into the Degrees via Souslin Trees 29

3.1 Lattices of Congruence Relations . . . . . . . . . . . . . . . . . . . . 30

3.2 Continuous Distances on Trees . . . . . . . . . . . . . . . . . . . . . . 36

v

3.3 Embedding Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Construction of Souslin Trees 49

4.1 First Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.2 Second Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.3 Third Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

A Some Set Theory 83

A.1 Forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

A.2 Combinatorial Principles . . . . . . . . . . . . . . . . . . . . . . . . . 88

B Lattices and Semi-Lattices 92

B.1 Lattices and Semi-Lattices . . . . . . . . . . . . . . . . . . . . . . . . 92

B.2 Complete Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

B.3 Algebraic and Continuous Lattices . . . . . . . . . . . . . . . . . . . 97

C Semi-Lattice Valued Distances 100

C.1 Distances and Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . 100

C.2 Pointed Spaces and Amalgamation . . . . . . . . . . . . . . . . . . . 103

C.3 Amalgamation of n-Pointed Spaces . . . . . . . . . . . . . . . . . . . 111

Bibliography 122

Index 124

vi

Figures

4.1 Illustration of the three distances d, d′, and τ . . . . . . . . . . . . . . 55

4.2 Illustration of the three distances d, d′, and τ . . . . . . . . . . . . . . 66

B.1 The lattices N5 and M3. . . . . . . . . . . . . . . . . . . . . . . . . . 94

C.1 A useful amalgamable 2-pointed premetric space. . . . . . . . . . . . 108

C.2 Two non-amalgamable N5-valued 2-pointed premetric spaces. . . . . . 112

C.3 Two non-amalgamable M3-valued 3-pointed premetric spaces. . . . . 113

C.4 Three 2-pointed premetric spaces. . . . . . . . . . . . . . . . . . . . . 120

C.5 Three amalgams of 2-pointed premetric spaces. . . . . . . . . . . . . 121

vii

Chapter 1

Introduction

We consider the problem of describing the structure of the join semi-lattice of degrees

of constructibility. This problem is impossible to answer in general because of the

lack of absoluteness of the degrees of constructibility. A more reasonable question is:

which constructible (semi-)lattices can be realized as initial segments of the degrees of

constructibility? In this thesis, we will show that every constructible algebraic lattice

(and some other continuous lattices) can be realized as initial segments of the degrees

of constructibility.

This chapter contains this brief introduction and the main cast of chacracters. The

main results of the thesis are stated in the first section together with some historical

context. The main protagonists, Souslin trees and degrees of constructibility, are

introduced in the next two sections of this chapter.

The plan for the remaining chapters of this thesis is that of a single long-winded

proof of the main results. In broad terms, the second chapter presents some basic

results to build intuition for the problem, the third chapter presents some technical

machinery to attack the problem, and finally the fourth chapter uses that machinery

to solve the problem.

1

The second chapter introduces congruence relations on Souslin trees and inves-

tigates how these are related to degrees of constructibility represented by sets of

ordinals in a generic extension of the universe via a Souslin tree.

The third chapter introduces continuous distances on Souslin trees. The role of

these is to control the congruence relations on a Souslin tree. Using results of the

second chapter, we will see how some of these continuous distances lead to embeddings

of continuous lattices into the degrees of constructibility.

The fourth chapter presents three constructions of Souslin trees together with

continuous distances. The first and third of these constructions will show how to

realize any constructible algebraic lattice as an initial segment of the degrees of con-

structibility. The second construction will show how to realize the constructible real

unit interval as an initial segment of the degrees of constructibility.

1.1 The Main Results in Context

In a seminal paper [20], Gerald E. Sacks introduced perfect set forcing (also known

as Sacks forcing) and he showed that this notion of forcing adds a minimal degree of

constructibility.

1.1.1 Theorem. (Sacks [20]) Assume V = L. There is a notion of forcing P such

that every P -generic set G represents a minimal degree of constructibility.

Part of Sacks’ inspiration for this result was Clifford Spector’s [24] construction of a

minimal Turing degree. Spector’s construction was improved successively by Hugill,

Lachlan, Lerman, Lebeuf, Abraham and Shore, among others, to achieve a complete

characterization of the initial segments of the Turing degrees up to size ω1.

1.1.2 Theorem. (Abraham–Shore [2]) Every locally countable join semi-lattice with

2

zero of size at most ω1 is isomorphic to an initial segment of the Turing degrees.

Groszek and Slaman [11] show that this is best possible in ZFC. One might hope that

similar techniques would yield a similar result for degrees of constructibility, but this

is not possible as shown by Robert S. Lubarsky in [18].

Perfect sets are omnipresent in investigations of Turing degrees and the results

alluded to above rely on some complex machinery to relate join semi-lattices with

zero to partial orderings of perfect sets. (See Manuel Lerman’s monograph [17] for a

nice treatment.) Zofia Smolska-Adamowicz observed that this perfect set machinery

could be used to modify Sacks’ original notion of forcing to produce similar results

in the context of degrees of constructibility. After proving in [21] and [23] that every

finite lattice can be realized as the lattice of degrees of constructibility in a forcing

extension of L, Smolska-Adamowicz finally proved the following very general result.

1.1.3 Theorem. (Smolska-Adamowicz [22]) Assume V = L. Suppose L is a lo-

cally countable,1 wellfounded join semi-lattice with zero. Then there is a symmetric

extension N of L such that the join semi-lattice of degrees of constructibility with

representatives in N is isomorphic to L.

The Turing degrees naturally form a locally countable join semi-lattice, but there is

no reason for the join semi-lattice of degrees of constructibility to be locally countable,

nor is there reason for it to be wellfounded.

The same basic perfect set machinery was further refined by Marcia J. Groszek

and Richard A. Shore to eliminate wellfoundedness to a certain extent.

1.1.4 Theorem. (Groszek–Shore [10]) Assume V = L. Let L be a complete count-

able algebraic lattice. There is a notion of forcing P such that if G is P -generic, then

the degrees of constructibility in L[G] form a lattice dual-isomorphic to L.

1L is locally countable if every element of L has at most countably many predecessors in L.

3

Note that the dual of a wellfounded lattice with unit is an algebraic lattice, so this

last result includes a significant part of Smolska-Adamowicz’s earlier result.

At the end of their paper, Groszek and Shore argue that perfect set techniques as

they used are limited to producing dual continuous lattices as initial segments of the

degrees of constructibility. Since every countable continuous lattice is algebraic, their

result is best possible for countable lattices using this approach. In another paper [8],

Groszek showed that 1 + ω∗1 can be realized as an initial segment of the degrees of

constructibility using perfect set techniques, thus showing that local countability is

not an essential restriction.

Perfect set forcing is not the only way to add a minimal degree. In [19], Kenneth

McAloon remarked that Ronald B. Jensen’s [14] construction of a Souslin tree in the

constructible universe also creates a minimal degree of constructibility.

1.1.5 Theorem. (Jensen) Assume V = L. There is a Souslin tree T of height ω1

such that any generic branch G through T represents a minimal degree of constructibil-

ity.

Using ideas of Groszek and Jech [9] on generalized iterations of forcing, C. Patrick

(Paddy) Farrington used Souslin trees to realize lattices of large size as initial segments

of the degrees of constructibility.

1.1.6 Theorem. (Farrington [4]) Assume V = L. Let κ be an infinite regular cardi-

nal and let L be a wellfounded lattice with unit of size at most κ. There is a Souslin

tree T of height κ+ such that if G is a generic branch through T , then the degrees of

constructibility with representatives in L[G] form a lattice isomorphic to L.

At the end of his paper, Farrington indicates that his methods could be used to realize

some non-wellfounded lattices as initial segments of the degrees of constructibility,

but no details are given.

4

The main result of this thesis improves on both the result of Groszek and Shore

(Theorem 1.1.4) by removing the size restriction, and on Farrington’s result (Theo-

rem 1.1.6) by weakening the wellfoundedness requirement.

1.1.7 Theorem. Assume V = L. Let κ be an infinite regular cardinal and let L

be a complete algebraic lattice with at most κ compact elements. There is a Souslin

tree T of height κ+ such that if G is a generic branch through T , then the degrees of

constructibility in L[G] form a lattice dual isomorphic to L.

Since the dual of a wellfounded lattice with unit is an algebraic lattice, our result

completely includes that of Farrington.

It is natural to ask whether the algebraicity requirement is essential. Our second

main result shows that it is not.

1.1.8 Theorem. Assume V = L. There is a Souslin tree T of height ω1 such that if

G is any generic branch through T , then the degrees of constructibility with represen-

tatives in L[G] form a lattice isomorphic to the unit interval [0, 1].

Note that [0, 1]L[G] = [0, 1]L since forcing with a Souslin tree does not add reals.

The unit interval [0, 1] is the prototypical example of a non-algebraic continuous

lattice. Many of the tools that we use to prove our main result on algebraic lattices

apply to the broader class of continuous lattices. We hope that our methods can one

day be extended to all continuous lattices. We feel confident that the following more

modest conjecture will be established soon.

1.1.9 Conjecture. Assume V = L. If L is a continuous separable lattice, then there

is a Souslin tree of height ω1 such that if G is any generic branch through T then the

degrees of constructibility with representatives in L[G] form a lattice dual isomorphic

to L.

5

The main obstructions are in adapting the methods from Chapter 4 to construct

Souslin trees continuous with suitable continuous distances rather than algebraic dis-

tances. The construction used to prove Theorem 1.1.8 suggessts that this is possible,

but it relies heavily on algebraic properties of [0, 1] that are not shared by all contin-

uous lattices.

1.2 Souslin Trees

A tree is a partial order T = (T, /T ) such that the initial segment (−, t)T =

{t′ ∈ T : t′ /T t} is wellordered (by /T ) for every t ∈ T . Elements of T are called

nodes. The height of a node, denoted htT (t), is the ordinal type of (−, t)T . The

α-th level of T is the set Tα = {t ∈ T : htT (t) = α} of all nodes of T with height α.

The height of the tree T , denoted ht(T ), is the smallest ordinal α such that Tα is

empty; equivalently, ht(T ) = {htT (t) : t ∈ T}. We will occasionally have to deal with

proper class trees which have nodes of every ordinal height; the height of such a tree

is then defined as Ord, which is in accord with the second definition of ht(T ).

If A ⊆ ht(T ), then the restriction T �A is the tree with base set⋃

α∈A Tα =

{T ∈ t : htT (t) ∈ A} with the ordering induced from T . For ordinals α, we often

write T<α instead of T �α. We say that B ⊆ T is bounded (in T ) if B ⊆ T<α for

some α < ht(T ). We say that the tree T is regular if every node t has extensions of

all heights up to ht(T ), i.e. the successors of t are unbounded in T .

Since the initial segment (−, t]T is wellordered, it contains a unique node of

height α for every α ≤ htT (t) — that node is denoted t�α. Also, the intersection

(−, t]T ∩ (−, t′]T is wellordered for all t, t′ ∈ T ; the ordinal type of this intersection

is the splitting height of t and t′, and it is denoted 4T (t, t′). A tree T is called

Hausdorff if nodes of limit height are uniquely determined by their sets of predeces-

6

sors, i.e. (−, t)T = (−, t′)T =⇒ t = t′ when htT (t) = htT (t′) is a limit ordinal. (This

is equivalent to saying that T is a Hausdorff space when endowed with the order

topology.) If T is Hausdorff, then (−, t]T ∩ (−, t′]T always has a maximal element, the

meet of t and t′, which is denoted t∧ t′. Note that 4T (t, t′) = htT (t∧ t′) + 1, and so

4T takes only successor ordinal values when T is Hausdorff.

A morphism of trees is an order- and level-preserving map f : T → U , i.e.

t /T t′ =⇒ f(t) /U f(t′) and htT (t) = htU(f(t)) for all t, t′ ∈ T . This notion makes

the class of trees into a category. This category has a terminal element, namely

the proper class tree Ord (with the usual ordering). For every tree T , the unique

morphism from T to Ord is the height map htT . The category of trees also has an

initial element, namely the empty tree ∅.

We say that S is a subtree of T if the inclusion map is a morphism from S

to T . In other words, S ⊆ T has the induced ordering and t ∈ S implies that

(−, t)T ⊆ S for every t ∈ T . An important class of subtrees of T are the derived trees

T t = {u ∈ T : u E t ∨ u D t} for t ∈ T . Note that the restriction T �A is a subtree of

T only when A ∩ ht(T ) ∈ Ord.

The sum (or coproduct) T ⊕U in this category is the side-by-side disjoint union

of T and U . More precisely, the base set of T ⊕ U is ({T} × T ) ∪ ({U} × U), and

(R, r) /T⊕U (S, s) ⇐⇒ R = S & r /R s. The morphisms iT : T → T ⊕ U and

iU : U → T ⊕ U are given by iT (t) = (T, t) and iU(u) = (U, u) respectively. It is easy

to verify that T ⊕U has the required universal property: if f : T → S and g : U → S

are morphisms then the map

h(R, r) =

f(r) when R = T

g(r) when R = U

7

is a morphism from T ⊕ U to S such that f = h ◦ iT and g = h ◦ iU .

The product T ⊗U in this category is a more interesting operation. Indeed, the

Cartesian product of T and U , which is the natural product in the category of partial

orderings, is not usually a tree. Nevertheless, products still exist in the category of

trees. The base set of T ⊗ U is

{(t, u) ∈ T × U : htT (t) = htU(u)} ,

and the ordering is (t, u) /T⊗U (t′, u′) ⇐⇒ t /T t′ & u /U u′. The canonical morphisms

πT : T ⊗ U → T and πU : T ⊗ U are simply the left and right coordinate projections

respectively, i.e. πT (t, u) = t and πU(t, u) = u. To see that T ⊗ U has the required

universal property, suppose that f : S → T and g : S → U are morphisms. Then the

map h(s) = (f(s), g(s)) is a morphism from S to T ⊗ U , and f = πT ◦ h, g = πU ◦ h

as required by the universal property of the product.

It is easy to extend the definitions of finite sums and products above to sums and

products of arbitrary families of trees. One can also show more generally that limits

and colimits always exist in the category of trees. Thus the category of trees is a

complete category.

A set C of nodes of a tree T is said to be a chain if any two elements of C are

comparable. We say that C is a maximal chain of T if C is not properly contained

in another chain of T . Equivalently, C is a maximal chain if it is a chain and every

node of T is incomparable with some element of C. Note that a maximal chain is

downward closed, i.e. if t ∈ C then (−, t] ⊆ C. A branch of T is an unbounded

maximal chain.

The concept dual to that of a chain is the concept of antichain. A set A of nodes

of a tree T is said to be an antichain if any two elements of A are incomparable. We

8

say that A is a maximal antichain of T if A is not properly contained in another

antichain of T . Equivalently, A is a maximal antichain if it is an antichain and every

node of T is comparable with some element of A. In a regular tree T , every level Tα

is a maximal antichain.

Let T be a regular tree. We say that

• T is an Aronszajn tree if every chain of T is bounded2; and

• T is a Souslin tree if every antichain of T is bounded.

Although the concepts of Aronszajn and Souslin trees are dual of each other, they

are not incompatible. We say that T is a splitting tree if every node of T has at

least two incomparable extensions.

1.2.1 Proposition. Every splitting Souslin tree is Aronszajn.

proof: Let T be a splitting Souslin tree of height η and suppose, for the sake of

contradiction, that B is a branch through T . For every α < η, let bα be the unique

node of B with height α. Since T is splitting, each bα has two incomparable extensions

aα and cα. Only one of these two nodes can be in B, say aα /∈ B. Define the sequence

〈αi〉i≤τ by recursion via the rule αi = supj<i ht(aαi) until ατ = η. Note that this

sequence is strictly increasing. Also note that bαi/ aαj

when i < j < τ , and bαi⊥ aαj

when j ≤ i < τ . It follows that aαi⊥ aαj

when i < j < τ and so A = {aαi}i<τ is

an antichain of T . Moreover, supi<τ ht(aαi) ≥ supi<τ αi = η which shows that A is

unbounded — which contradicts the fact that T is Souslin. qed

Non-splitting Souslin trees are easy to come by. For example, (η,<) is a Souslin tree.

More generally, any regular tree T of height η which is the union of fewer than cf(η)

2Many authors also require that |Tα| < ht(T ) for every α.

9

chains is a Souslin tree. These are trivial Souslin trees. If T is a nontrivial Souslin

tree, we can trim away the non-splitting nodes of T to obtain a splitting Souslin

subtree of the same height.

Other than their purely combinatorial interest, Souslin trees are important as no-

tions of forcing and this will be their central role in this thesis. (Refer to Appendix A

for a review of forcing.) Perhaps the most remarkable fact about Souslin trees is the

following, which sheds some light on why non-trivial Souslin trees are Aronszajn.

1.2.2 Proposition. Suppose that V |= ‘T is a Souslin tree.’ Then every branch

through T is a V -generic branch, and conversely.

proof: Suppose that G ⊆ T is a V -generic filter. Another way to say that G

is a filter is to say that G is a downward closed chain in T . So we only have to

show that G has nodes at every level α < ht(T ). Since T is a regular tree, the set

Dα = {t ∈ T : ht(t) ≥ α} is an open dense subset of T for every α < ht(T ). Clearly,

each Dα is definable in V and so G ∩Dα 6= ∅ for every α < ht(t). Therefore, G has

an element of height α as claimed.

Conversely, suppose that G is a branch through T . Suppose that D ∈ V is an

open dense subset of T . We must show that G ∩ D 6= ∅. The set A of minimal

elements of D is a maximal antichain in V . Since, V |= ‘T is a Souslin tree,’ it follows

that A is bounded by some height α < ht(T ). Let t be the unique element of G

with height α. Then t is comparable with some element of A since A is a maximal

antichain in T . Since t cannot lie below some element of A, it must lie above some

element of A and so t ∈ D. qed

Souslin trees are also remarkably tame as forcing notions. In fact, Souslin trees are

surgical tools compared to other notions of forcing.

10

1.2.3 Proposition. If T is a Souslin tree of regular height κ, then T satisfies the

κ-chain condition and has κ-Baire property. Therefore, forcing with T adds no new

sequences of ordinals with length less than κ, and preserves all cardinals and cofinal-

ities.

proof: To see that T satisfies the κ-chain condition, suppose, for the sake of con-

tradiction, that {tα}α<κ is an antichain in T . Since T is regular, we can pick uα D tα

such that ht(uα) ≥ α for every α. Then {uα}α<κ is an unbounded antichain in T —

which contradicts the fact that T is Souslin.

To see that T has the κ-Baire property, let {Di}i∈I be a family of open dense

subsets of T with |I| < κ. For each i ∈ I, the set Ai of all minimal elements of Di is

a maximal antichain in T . Therefore, Ai is bounded by some ordinal αi < κ. Note

that Di therefore contains every node of T with height at least αi. Since κ is regular

and |I| < κ, α = supi∈I αi < κ. Now the intersection⋂

i∈I Di contains every node of

T with height at least α. Thus⋂

i∈I Di is dense in T as required. qed

It turns out that the effect of forcing with a Souslin tree T is closely tied to

the position that T has in the category of trees. We can ask which trees have new

branches when forcing with T . This question has a relatively simple answer.

1.2.4 Lemma. If T is a Souslin tree of height η, U is a Hausdorff tree of height η

and B is a T -name such that T ‘B is a branch through U ’, then there is a closed

unbounded set C ⊆ η and a morphism f : T �C → U�C such that for any generic

branch through T , f [G�C] ⊆ BG.

proof: If t ∈ T then the set Bt = {u ∈ U : t T u ∈ B} is a downward closed chain

in U . Let βt be the order type of Bt. Note that since U is Hausdorff, βt is a successor

ordinal whenever βt < η. To wit, if u ∈ U is a limit and t ‘(−, u)U ⊆ B’ then there

11

cannot be a t′ D t such that t′ ‘u /∈ B’ for such a t′ would force that B is bounded

in U . So it must be that t ‘u ∈ B’.

Since B is forced to be unbounded in U , for every α < η the set Aα of all minimal

t ∈ T such that βt ≥ α is a maximal antichain in T . Since T is Souslin, Aα is bounded

in T . Let C be the closed unbounded subset of η consisting of all limit γ < η such

that α < γ =⇒ Aα ⊆ T<η.

Note that βt > ht(t) whenever t ∈ T �C, since βt ≥ ht(t) and ht(t) < η is a limit

ordinal. For each t ∈ T �C let f(t) be the unique element of Bt with height ht(t).

Clearly, ht(f(t)) = ht(t) for every t ∈ T �C so it suffices to check that f is order-

preserving. If t′ / t, then Bt′ is an initial segment of Bt and so f(t′), f(t) ∈ Bt. Since

Bt is a branch and ht(f(t′)) = ht(t′) < ht(t) = ht(f(t)) it follows that f(t′) / f(t) as

required. qed

1.3 Degrees of Constructibility

In 1940, Kurt Godel [6] introduced the constructible universe L, the smallest collection

of sets that contains all ordinals and satisfies ZF. Godel used this to show that the

axiom of choice and the generalized continuum hypothesis (which both hold in L) are

compatible with the basic ZF axioms. The constructible universe has been widely

studied by set theorists. Its very rich structure has spawned a great deal of research

in all branches of set theory, from combinatorial set theory to inner model theory and

large cardinals.

An interesting feature of Godel’s construction is that it relativizes: given a set

X of ordinals, we can define L[X] to be the smallest collection of sets that contains

all ordinals and the set X, and satisfies the axioms of ZF. We say that a set Y of

ordinals is constructible from X, and we write Y ≤c X, if Y ∈ L[X]. This is a

12

transitive, reflexive relation on sets of ordinals, analogous to Turing reducibility in

computability theory. This relation induces an equivalence relation (X ≡c Y ⇐⇒

X ≤c Y & Y ≤c X) and a partial ordering on the resulting equivalence classes.3 These

equivalence classes are called degrees of constructibility, or simply c-degrees.

Just like Turing degrees, the c-degrees form an join semi-lattice with zero — the

degree of all constructible sets of ordinals.

The classes L and L[A] are examples of inner models: a transitive class W is

an inner model if it contains all ordinals and satisfies the axioms of ZF. So L is the

smallest inner model, and for every set A ⊆ Ord, L[A] is the smallest inner model

that contains the set A as an element. It is a remarkable fact that L and L[A] are

definable classes.

In the sequel, we will often work with V -degrees rather than c-degrees. While this

is a concept well accepted by set theorists, it is only vaguely defined and there is no

widely accepted formal definition. Intuitively, we want to mimic the above situation

with a fixed ground model V (of ZFC) instead of L. Thus, for sets X, Y ⊆ Ord

we would like to say that X ≤V Y if X ∈ V [Y ] where V [Y ] is the smallest inner

model that contains Y and every element of V as an element. The problem with this

definition is that there is no reason for such an inner model V [Y ] to exist. Another

approach is to say that X ≤V Y if there is an effective set theoretic procedure to define

the set X in terms of Y and parameters from V . This is a more robust approach

than the first, provided that we have a suitable definition of ‘effective set theoretic

procedure.’

Following Jensen (see [3]), we define the family of rudimentary functions as

the smallest collection of functions that contains the basic functions (the projection

3In order to avoid dealing with proper classes, we follow Scott in only using the representativesof minimal rank to represent each degree of constructibility.

13

functions, the null function, the successor function, and the choice function) and is

closed under the composition and union schemes.

Projection Functions: For m < n < ω, the m-th n-ary projection function is the

n-ary function Pnm defined by

Pnm(x0, . . . , xn−1) = xm.

Null Function: The null function is the unary function defined by

N1(x) = ∅.

Successor Function: The successor function is the binary function defined by

S2(x, y) = x ∪ {y} .

Choice Function: The choice function is the 4-ary function defined by

Q4(x0, x1, y, z) =

x1 if y ∈ z,

x0 if y /∈ z.

Composition Scheme: If g is an m-ary function and f0, . . . , fm−1 are n-ary func-

tion, then the composition of these is the n-ary function Cnm[g, f0, . . . , fm−1]

defined by

Cnm[g, f0, . . . , fm−1](x0, . . . , xn−1) = g(f0(x0, . . . , xn−1), . . . , fm−1(x0, . . . , xn−1)).

14

Union Scheme: If f is an (n+1)-ary function, then the union of f is the (n+1)-ary

function Un+1[f ] defined by

Un+1[f ](x, z0, . . . , zn−1) =⋃y∈x

f(y, z0, . . . , zn−1).

It is easy to see that all of the above definitions are ∆0 and thus absolute for transitive

classes. The rudimentary functions contain many basic set theoretic functions.

1.3.1 Proposition. The following functions are rudimentary:

x, y 7→ x ∪ y, x, y 7→ x ∩ y, x0, . . . , xn−1 7→ {x0, . . . , xn−1} ,

x, y 7→ x× y, x, y 7→ x \ y, x0, . . . , xn−1 7→ (x0, . . . , xn−1),

as well as the constant functions x 7→ n for n < ω. In fact, if φ is any bounded

formula in the language of set theory, then the function

y, z 7→ {(x0, . . . , xn−1) ∈ y0 × · · · × yn−1 : φ(x, y, z)}

is rudimentary.

Although rudimentary functions suffice for most purposes, we will prefer the slightly

broader family primitive recursive functions as in [15]. This is the smallest col-

lection of functions that contains the basic functions and is closed under the union,

composition, and recursion schemes.

Recursion Scheme: If f is an (n + 2)-ary function, then Rn+1[f ] is the unique

(n+ 1)-ary function that satisfies the ∈-recursive equation

Rn+1[f ](y, z0, . . . , zn−1) = f(⋃

x∈y Rn+1[f ](x, z0, . . . , zn−1), y, z0, . . . , zn−1)

15

Examples of primitive recursive functions that are not rudimentary are the rank

function and the transitive closure function.

If X is a set of ordinals, let V [X] denote the smallest transitive class that is closed

under primitive recursive functions and that contains as an element X and every

element of V . We can now give a robust definition for V -degrees. If X and Y are

sets of ordinals, then X ≤V Y if and only if X ∈ V [Y ]. It is easy to check that

this is a reflexive and transitive relation on sets of ordinals. This relation induces

an equivalence relation (X ≡V Y ⇐⇒ X ≤V Y & Y ≤V X) and a partial ordering

on the resulting equivalence classes. These equivalence classes are called V -degrees.

The V -degrees form a join semi-lattice with zero — the V -degree of all sets of ordinals

in V . The L-degrees coincide with c-degrees defined above.

This class V [X] is well-defined, but it is not necessarily an inner model. However,

it is a reasonable candidate for the smallest inner model that contains X and every

element of V as an element.

1.3.2 Proposition. (Hajnal [12]) A transitive class W is an inner model if and only

if Ord ⊆ W , W is closed under rudimentary functions, and every set X ⊆ W is

contained in some Y ∈ W .

A class W such that every set X ⊆ W is contained in some Y ∈ W is called almost

universal by Jech [13]. This is a difficult condition to verify in general, so the above

result will not be used in the sequel.

However, it turns out that all instances of V [X] that we will consider in the sequel

are indeed inner models. So the reader who prefers the intuitive notion of V -degree

may use it. Moreover, we will use a very relaxed approach to V -degrees. When show-

ing that X ≤V Y , we will usually just give an explicit method to effectively compute

X from Y using parameters in V . It will usually be clear that the computation

16

method is primitive recursive, but we will not explicitly mention or prove that fact.

17

Chapter 2

Degrees in Souslin Tree Extensions

The object of this chapter is to characterize (within the ground model) the different

degrees of sets of ordinals in a generic extension obtained by forcing with a Souslin

tree T .

In the first section, we will introduce the concept of congruence relations on trees

and see how they are related to morphisms in the category of trees. At the end of the

section, the Factoring Theorem (2.1.3) will reveal how different congruence relations

on a Souslin tree T (in the ground model) can be used to represent different degrees

in the generic extension obtained by forcing with T .

The second and third section each prove one main theorem. The Representation

Theorem (2.2.1) shows that the degree of every set of ordinals in the generic extension

via forcing with a Souslin tree T can be represented by a closed congruence relation

on T . The Comparison Theorem (2.3.1) gives combinatorial criteria to compare the

degrees represented by closed congruence relations on a Souslin tree T .

Combining the main results of these three sections, we see that the structure of

congruence relations on a Souslin tree T completely characterizes the structure of

degrees in a generic extension obtained by forcing with T .

18

2.1 Congruence Relations

A congruence relation on a tree T is an equivalence relation E on the nodes of T

whose graph is a (downward closed) subtree of T ⊗ T . In other words, E is a binary

relation on T such that:

• E is an equivalence relation on each level Tα, i.e.

t E t

t E u⇐⇒ u E t

t E u & u E v =⇒ t E v

for all t, u, v ∈ Tα; and

• E is level and downward closed, i.e. if t E u then ht(t) = ht(u) and t�α E u�α

for all α < ht(t) = ht(u).

Moreover, if E doesn’t split at limit levels, i.e. if α = ht(t) = ht(u) is a limit ordinal

and t�β E u�β for all β < α then t E u, then we say that E is a closed congruence

relation on T . Finally, we say that E is a regular congruence relation if for all

t, u, u′ ∈ T such that u′ . u and t E u there is a t′ ∈ T such that t′ . t and t′ E u′.

If f : T → U is a morphism of trees, then the relation Ef defined by t Ef t′ ⇐⇒

f(t) = f(t′) is a congruence relation on T . Conversely, if E is a congruence relation

on T , then the quotient set T/E has a unique tree structure such that the quotient

map t 7→ [t]E is a morphism of trees. The tree ordering on the quotient T/E is given

by

[t]E / [u]E ⇐⇒ (∃t′)(t′ E t & t′ / u) ⇐⇒ ht(u) > ht(t) & t E u� ht(t)

19

for all t, u ∈ T . If E is regular, then we also have the equivalence

[t]E / [u]E ⇐⇒ (∃u′)(t / u′ & u′ E u).

The following proposition sheds some light on the role of closure for congruence

relations.

2.1.1 Proposition. Suppose that E is a congruence relation on the tree T . Then E

is closed if and only if the quotient T/E is Hausdorff.

proof: Suppose that E is closed and let t and u be nodes of T with the same limit

height α. If [t�β]E = [u�β]E for all β < α then t�β E u�β for all β < α and hence

t E u since E is closed. Thus [t]E = [u]E. Since any two nodes of the same limit

height in T/E are of the form [t]E and [u]E for t and u as above, this shows that the

quotient T/E is Hausdorff.

Conversely, suppose that T/E is Hausdorff and let t and u be nodes of T with

the same limit height α. If t�β E u�β for all β < α, then [t]E�β = [t�β]E = [u�β]E =

[u]E�β for all β < α. Since T/E is Hausdorff, it follows that [t]E = [u]E and hence

t E u. qed

Regularity is more subtle than closure for congruence relations, but it is equally

important in the sequel. If U is a subtree (often a branch) of T and E is a congruence

relation on T then the E-expansion of U is the subtree of T defined by

UE =⋃u∈U

[u]E = {t ∈ T : (∃u ∈ U) (t E u)} .

The following proposition reveals part of the role of regularity for congruence relations.

20

2.1.2 Proposition. Suppose that E is a congruence relation on the regular tree T .

Then E is regular if and only if UE is a regular subtree of T for every regular subtree

U of T .

proof: Suppose that E is regular and that U is a regular subtree of T . Let t be

a node of UE and let α < ht(U). We need to find t′ ∈ UE such that t′ D t and

ht(t′) ≥ α. Pick u ∈ U such that t E u. Since U is regular, there is a u′ D u such

that ht(u′) ≥ α. Since E is regular there is a t′ D t such that t′ E u′. Then t′ ∈ UE,

t′ D t, and ht(t′) = ht(u′) ≥ α as required.

Conversely, suppose that E is not regular. Pick nodes t, u, and u′ such that t E u,

u′ . u, and there is no t′ . t such that t′ E u′. Let U be the regular subtree of T

consisting of all nodes comparable with u′. Then t ∈ UE since t E u and u ∈ U .

However, t has no extension in UE with height ht(u′). Indeed, such a node t′ would

satisfy t′ . t and t′ E u′ since u′ is the only node of U with height ht(u′). qed

The following important theorem gives a relationship between regular congruence

relations on a Souslin tree T and degrees in a generic extension via forcing with T .

2.1.3 Factoring Theorem. Suppose that V |= ‘T is a Souslin tree of height η and

E is a regular congruence relation on T .’ If G is a V -generic branch through T , then:

• V |= ‘T/E is a (possibly trivial) Souslin tree of height η’ and GE/E = {[t]E : t ∈ G}

is a V -generic branch through T/E.

• V [GE] |= ‘GE is a (possibly trivial) Souslin tree of height η’ and G is a V [GE]-

generic branch through GE.

proof: Let π : T → T/E be the canonical projection.

21

First we show that T/E is Souslin in V . Note that if A is an antichain in T/E,

then

π−1[A] = {t ∈ T : [t]E ∈ A}

is an antichain in T since distinct elements of the same E-equivalence class are incom-

parable, and elements of incomparable E-equivalence classes are also incomparable.

Since π preserves height, A is bounded in T/E if and only if π−1[A] is bounded in T .

It follows that T/E is Souslin since T is.

Now we show that GE is Souslin in V [GE]. Let A be a T/E-name for an antichain

in GE. Let B be the set of all minimal v ∈ T such that [v]E (∃u E v)(u ∈ A).

Note that B is an antichain of T in the ground model V . Since T is Souslin, B is

bounded by some height β < ht(T ). We claim that A is also bounded by β. Suppose

that [t]E u ∈ A. We may assume that ht(t) ≥ ht(u). Since [t]E u ∈ GE,

it follows that [u]E E [t]E. Since E is regular, there is a v D u such that v E t.

Then [v]E u E v ∧ u ∈ A. It follows that v extends some element v0 of B. Now

[v]E (∃u0 E v0)(u0 ∈ A). Since [v]E ‘A is an antichain’, we cannot have u . v0.

Therefore, u E v0 and hence ht(u) ≤ ht(v0) < β. qed

Note that the regularity of E was only required to prove the second part of the

theorem. Thus the quotient T/E of a Souslin tree T is a (possibly trivial) Souslin

tree for every congruence relation E on T .

2.2 The Representation Theorem

The Factoring Theorem tells us that every congruence relation E on a Souslin tree T

represents an intermediate model V [GE] between the ground model V and the generic

extension V [G]. Which intermediate models arise in this way? The Representation

22

Theorem answers this question by showing that every intermediate model generated

by a set of ordinals corresponds to a congruence relation on T in this way.

2.2.1 The Representation Theorem. Suppose that V |= ‘T is a Souslin tree of

height η and X is a T -name for a set of ordinals.’ Then there is a closed congruence

relation E on T in the ground model such that V [XG] = V [GE] whenever G is a

generic branch through T .

A first guess for E would be the equivalence relation

(∀ξ ∈ Ord) (t ξ ∈ X ⇐⇒ u ξ ∈ X).

However, this relation almost never has the requisite properties even when restricted

to nodes of equal level. To correct this, for each α < η let Yα be the (proper) class

consisting of all ordinals ζ such that every t ∈ Tα decides ζ ∈ X, i.e.

Yα = {ζ ∈ Ord : (∀t ∈ Tα) (t ζ ∈ X ∨ t ζ /∈ X)} .

Then the relation

ht(t) = ht(u) = α & (∀ζ ∈ Yα) (t ζ ∈ X ⇐⇒ u ζ ∈ X)

is almost as required, except that it is not necessarily closed. To ensure closure, let

Zα =⋃

β<α Yβ and let

t E u⇐⇒ ht(t) = ht(u) = α & (∀ζ ∈ Zα) (t ζ ∈ X ⇐⇒ u ζ ∈ X).

We claim that this relation is as required.

23

proof of Theorem 2.2.1: It is clear that E is a level equivalence relation on the

nodes of T . To see that the graph of E is downward closed in T ⊗ T , pick (t0, u0) /

(t1, u1) in T ⊗T . Let Z = Zht(t0) = Zht(u0) and note that t0, t1, u0, u1 all decide ζ ∈ X

for each ζ ∈ Z. Since t0 / t1, t0 and t1 must decide ζ ∈ X in the same way for each

ζ ∈ Z. Similarly for u0 and u1. So if t1 E u1 then

t0 ζ ∈ X ⇐⇒ t1 ζ ∈ X ⇐⇒ u1 ζ ∈ X ⇐⇒ u0 ζ ∈ X

for all ζ ∈ Z which shows that t0 E u0 as required.

To see that E is closed, suppose that t1 6E u1 where t1 and u1 have the same limit

height α. We show that t0 6E u0 for some (t0, u0) / (t1, u1). By definition of E there is

a ζ ∈ Zα such that t1 ζ ∈ X ⇐⇒ u1 ζ /∈ X; say, by symmetry, that t1 ζ ∈ X

and u1 ζ /∈ X. Since α is a limit, there is a β < α such that ζ ∈ Yβ ⊆ Zβ+1. Then

t0 = t1�(β + 1) and u0 = u1�(β + 1) are such that t0 ζ ∈ X and u0 ζ /∈ X. It

follows that t0 6E u0 and (t0, u0) / (t1, u1) since β + 1 < α.

We now show that this congruence relation E satisfies V [XG] = V [GE] for any

generic branch G through T . On the one hand, we will show that

t ∈ GE ⇐⇒ (∀ζ ∈ Zht(t)) (t ζ ∈ X ⇐⇒ ζ ∈ XG) (1)

which implies that GE ∈ V [XG]. On the other hand, we will show

ξ ∈ XG ⇐⇒ (∃t ∈ GE) (ξ ∈ Zht(t) ∧ t ξ ∈ X) (2)

which implies that XG ∈ V [GE].

To verify (1), let t ∈ Tα and let g be the element of G with height α. Observe

that ζ ∈ XG ⇐⇒ g ζ ∈ X for all ζ ∈ Zα, since g decides ζ ∈ X for every ζ ∈ Zα.

24

So

t ∈ GE ⇐⇒ t E g ⇐⇒ (∀ζ ∈ Zα) (t ζ ∈ X ⇐⇒ g ζ ∈ X)

⇐⇒ (∀ζ ∈ Zα) (t ζ ∈ X ⇐⇒ ζ ∈ XG)

as required.

To verify (2), first observe that Ord =⋃

α<η Zα. Indeed, for every ordinal ξ there

is an open dense set of t ∈ T which decide ξ ∈ X; since T is Souslin, it follows that

there is a level Tα every element of which decides whether ξ ∈ X, i.e. ξ ∈ Zα. That

said, we have

ξ ∈ XG ⇐⇒ (∃g ∈ G) (g ξ ∈ X)

⇐⇒ (∃g ∈ G) (ξ ∈ Zht(g) ∧ g ξ ∈ X)

⇐⇒ (∃t ∈ GE) (ξ ∈ Zht(t) ∧ t ξ ∈ X)

as required. qed

2.3 The Comparison Theorem

We now know that closed congruence relations can be used to represent all of the in-

termediate models of a generic extension via a Souslin tree. These closed congruence

relations are tangible combinatorial objects in the ground model whereas the interme-

diate models are much more difficult to grasp. How can we use the structure of closed

congruence relations to compare the intermediate models that they represent? When

do two congruence relations represent the same model? The Comparison Theorem

answers this question by giving combinatorial criteria for when a closed congruence

25

relation represents an intermediate submodel of another.

2.3.1 The Comparison Theorem. Suppose that V |= ‘T is a Souslin tree of height

η and E1, E2 are closed congruence relations on T .’ If G is a V -generic branch through

T then V [GE1 ] ⊆ V [GE2 ] if and only if there are a (closed) unbounded set C ⊆ η in

V and a g ∈ G such that t E2 u =⇒ t E1 u for all t, u ∈ T g�C.

As an immediate corollary we obtain combinatorial criteria for the equality of the

intermediate models associated to two congruence relations.

2.3.2 Corollary. Suppose that V |= ‘T is a Souslin tree of height η and E1, E2

are closed congruence relations on T .’ If G is a V -generic branch through T , then

V [GE1 ] = V [GE2 ] if and only if there is a (closed) unbounded set C ⊆ η and a g ∈ G

such that t E1 u⇐⇒ t E2 u for all t, u ∈ T g�C.

The ‘if’ direction of the Comparison Theorem will be shown to be an immediate

consequence of the following general fact.

2.3.3 Lemma. Let T ∈ V be a tree and let E be a congruence relation on T in V . If

G is any branch through T and X is any unbounded subset of GE then V [GE] ⊆ V [X].

proof: Let U be the downward closure of X in T . Then G ⊆ U ⊆ GE and so

UE = GE which shows that GE is definable in V [X] with parameters E, T and X.

qed

Now to show the ‘if’ direction of the Comparison Theorem, suppose that C ⊆ η is an

unbounded set and g ∈ G is such that t E2 t′ =⇒ t E1 t

′ for all t, t′ ∈ T g�C. Then the

set X = GgE2

�C is an unbounded subset of GE1 in V [GE2 ], and so V [GE1 ] ⊆ V [GE2 ]

by Lemma 2.3.3.

26

For the ‘only if’ direction of the Comparison Theorem, first define the closed

congruence relation E0 by t E0 t′ ⇐⇒ t E1 t′ & t E2 t′ for all t, t′ ∈ T . Note

that E0 ∈ V , that GE0 = GE1 ∩ GE2 , and that GEi= (GE0)Ei

for i = 1, 2. Thus

GE1 ∈ V [GE2 ] if and only if GE0 ∈ V [GE2 ].

Now let H be the V -generic branch through T/E0 induced by G as in Theo-

rem 2.1.3. Let E be the closed congruence relation induced by E2 on T/E0, i.e.

[t]E0 E [t′]E0 ⇐⇒ t E2 t′ for all t, t′ ∈ T . (This makes sense since t E0 t

′ =⇒ t E2 t′

for all t, t′ ∈ T .) Note that we then have V [H] = V [GE0 ] and V [HE] = V [GE2 ] and

so

GE1 ∈ V [GE2 ] ⇐⇒ GE0 ∈ V [GE2 ] ⇐⇒ H ∈ V [HE].

We have thus reduced the question GE1 ∈ V [GE2 ] to the question H ∈ V [HE]. The

next proposition answers this question.

2.3.4 Proposition. Suppose that V |= ‘U is a Souslin tree of height η and E is a

closed congruence relation on U .’ If H is a V -generic branch through U and H ∈

V [HE] then there are a closed unbounded set C ⊆ η in V and an h ∈ H such that E

reduces to equality on Uh�C.

In our context, this proposition states that if H ∈ V [HE] (or equivalently GE2 ∈

V [GE1 ]) then there is a closed unbounded set C ⊆ η and an h ∈ H such that

u E u′ =⇒ u = u′ for all u, u′ ∈ T/E0�C with u, u′ D h. If g ∈ G is such that

[g]E0 = h, then we also have

t E2 t′ =⇒ [t]E0 E [t′]E0 =⇒ [t]E0 = [t′]E0 =⇒ t E0 t

′ =⇒ t E1 t′

for all t, t′ ∈ T g�C — which is the conclusion of the Comparison Theorem. So we

have reduced the proof of the Comparison Theorem to that of Proposition 2.3.4.

27

proof of Proposition 2.3.4: Let G be the V -generic branch through U/E in-

duced by H as in the Factoring Theorem (2.1.3). Note that V [HE] = V [G] and so

H ∈ V [G]. It follows that there is a U/E-name B such that H = BG. Furthermore,

there is a g ∈ G such that g U/E ‘B is an unbounded branch through U ’. Then by

Lemma 1.2.4 there is a closed unbounded set C and a morphism f : (U/E)g�C → U�C

such that f [G�C] ⊆ H. Since f and C are in the ground model V , there is an h ∈ H

with g E [h]E such that h U f [G�C] ⊆ H where G and H are the canonical U -names

for G and H respectively.

We claim that h and C are as required. Suppose that u1, u2 D h are such that

ht(u1) = ht(u2) = α ∈ C, u1 E u2. Then [u1]E = [u2]E and so f([u1]E) = f([u2]E).

On the other hand, for i ∈ {1, 2} we have that ui U f([ui]E) ∈ H ∩ Uα and so

f([ui]E) = ui since ui G ∩ Uα = {ui}. Therefore, u1 = f([u1]E) = f([u2]E) = u2.

qed

28

Chapter 3

Embeddings into the Degrees via

Souslin Trees

In the last chapter, we saw that that congruence relations on a Souslin tree T com-

pletely determine the structure of degrees in the generic extension obtained by forcing

with T . In this chapter, we will investigate the structure of congruence relations fur-

ther with the ultimate goal of realizing a broad class of lattices in the degrees of a

generic extension via a Souslin tree.

In the first section, we will investigate the general structure of congruence relations

on a tree T . More precisely, we will see that the set ET of all congruence relations on

a tree T forms a complete algebraic lattice. We will then define three more complete

lattices of congruence relations. The last of these will be the set E∗T of all normal

congruence relations on T which will play a central role in the remainder of the

chapter.

In the second section, we will introduce an new tool which will allow us to control

the structure of congruence relations on a Souslin tree. For every continuous lattice

L, we will define the notion of a continuous L-valued distance function on a tree T .

29

We will then show that a regular continuous distance naturally leads to an embedding

of L into the lattice E∗T of normal congruence relations on T .

In the third and last section of this chapter, we will show how a faithful regular

continuous L-valued distance function on a Souslin tree T leads to a dual embedding

of L into the degrees in the generic extension via T . Finally, we will introduce a

combinatorial property (#) which will ensure that this dual embedding is surjective.

The results of this chapter lend strong support for the conjecture that every dual

continuous lattice can be realized as an initial segment of the degrees of constructibil-

ity. However, this chapter does not address the question of constructing Souslin trees

and distances as above. The construction of such trees will be the subject of the next

chapter.

This chapter makes heavy use of the material in Appendix B. The reader who

wishes for a review of lattice theory may want to read this appendix first.

3.1 Lattices of Congruence Relations

Let ET denote the set of all congruence relations on the tree T . It is easy to check

that ET is a complete lattice with respect to inclusion. Indeed, if {Ei}i∈I is a family

of congruence relations on T , then the meet∧

i∈I Ei is the set-theoretic intersection

of the Ei’s, and the join∨

i∈I Ei is the transitive closure of the set-theoretic union of

the Ei’s. Note that if the family {Ei}i∈I is directed, then the join∨

i∈I Ei is simply

the set-theoretic union of the Ei’s.

For t, u ∈ T with ht(t) = ht(u), let Et,u be the smallest congruence relation on T

such that t Et,u u. It is not difficult to see that v1 Et,u v2 if and only if either v1 = v2,

or {v1, v2} = {t�α, u�α} for some α ≤ ht(t) = ht(u). It is also easy to check that each

Et,u is compact in the lattice ET by the above characterization of directed joins. This

30

characterization also implies that ET is an algebraic lattice.

3.1.1 Proposition. The lattice ET is a complete algebraic lattice and the compact

elements of ET are finite joins of relations of the form Et,u for (t, u) ∈ T ⊗ T .

So ET has a nice structure, but it is often too large for our purposes. Indeed, the

results of the previous chapter indicate that closed congruence relations and regular

congruence relations are the most interesting.

Recall that a congruence relation E on T is closed if it does not split at limit

levels, i.e. if α is a limit ordinal and t, u ∈ Tα then t E u ⇐⇒ t�β E u�β for every

β < α. Let EcloT be the set of all closed congruence relations on T .

3.1.2 Proposition. The set EcloT is closed under arbitrary meets. In fact, there is a

surjective closure operator clo : ET → EcloT which preserves all meets, i.e.

E ⊆ clo(E) = clo(clo(E))

for every E ∈ ET , and

clo(∧

i∈I Ei) =∧

i∈I clo(Ei)

for every family {Ei}i∈I ⊆ ET .

proof: Let E denote the topological closure of E ∈ ET with respect to the order

topology on T ⊗ T , i.e.

t E u⇐⇒ (∀(t′, u′) / (t, u))[E ∩ ((t′, u′), (t, u)]T⊗T 6= ∅].

In other words, t E u if and only if either t E u or ht(t) = ht(u) is a limit ordinal

and t�α E u�α for every α < ht(t) = ht(u). The key is to note that E is a closed

congruence relation on T .

31

It is clear that E is level, downward closed, reflexive, and symmetric. To see that

it is transitive, suppose that t E u and u E v. We may suppose that α = ht(t) =

ht(u) = ht(v) ∈ Lim for otherwise t E u E v which implies t E v. Thus t�β E u�β

and u�β E v�β for every β < α. It follows that t�β E v�β for every β < α and hence

t E v.

To see that E is closed, suppose that α is a limit ordinal, t, u ∈ Tα, and t�β E u�β

for every β < α. It follows that t�(β + 1) E u�(β + 1) for every β < α since E and E

agree on successor levels. Since E is downward closed, it follows that t�β E u�β for

every β < α and hence t E u as required.

It follows immediately that E = clo(E), i.e. that E is the smallest closed congru-

ence relation on T which contains E. Therefore, EcloT is closed under intersections.

Since meets in EcloT are set-theoretic intersections, we also have that clo preserves all

meets as this is a property of topological closure operators. qed

It follows from the above that EcloT is a complete lattice; joins and meets in Eclo

T will

usually be denoted∨clo and

∧clo respectively. Meets in EcloT agree with meets in the

larger lattice ET , i.e. if {Ei}i∈I is a family of closed congruence relations on T , then

∧cloi∈I Ei =

∧i∈I Ei =

⋂i∈I Ei.

Joins are more difficult, but one has the usual equation

∨cloi∈I Ei = clo

(∨i∈I Ei

).

Recall that a congruence relation E on T is regular if for all t, u, u′ ∈ T such that

t E u and u E u′ there is a t′ ∈ T such that t E t′ and t′ E u′. Let EregT be the set of

all regular congruence relations on T .

32

3.1.3 Proposition. The set EregT is closed under arbitrary joins. Consequently, there

is a surjective kernel operator reg : ET → EregT , i.e.

reg(reg(E)) = reg(E) ⊆ E

for every E ∈ ET .

proof: Let {Ei}i∈I be a family of regular congruence relations on T and let E =∨i∈I Ei. Let us show that E is regular. Let t, u, u′ ∈ T be such that t E t′ and t E u.

We need to find u′ D u such that t′ E u′. By the above characterization of joins in

ET , we can find v1, . . . , vn ∈ T and i0, . . . , in ∈ I such that

t Ei0 v1 Ei1 · · · Ein−1 vn Ein u.

Since each Ei is regular, we can successively find v′1 D v1,. . . ,v′n D vn, and u′ D u such

that

t′ Ei0 v′1 Ei1 · · · Ein−1 v

′n Ein u

′.

It follows that t′ E u′ and so u′ D u is as required.

Now the surjective kernel operator reg : ET → EregT can be defined by

reg(E0) =⋃{E ∈ Ereg

T : E ⊆ E0}

for every E0 ∈ ET . qed

It follows from the above that EregT is a complete lattice; joins and meets in Ereg

T will

usually be denoted∨reg and

∧reg respectively. Joins in EregT agree with joins in the

33

larger lattice ET , i.e. if {Ei}i∈I is a family of regular congruence relations on T , then

∨regi∈I Ei =

∨i∈I Ei = transitive closure of

⋃i∈I Ei.

In particular, joins of directed families are simply set-theoretic unions. Meets are

more difficult, but one has the usual equation

∧regi∈I Ei = reg

(∧i∈I Ei

).

The above definition of the kernel operator reg is not well suited to computations.

Fortunately, there is another way to define this operator. For E ∈ ET , the regular

derivative is the relation E ′ defined by

t E ′ u⇐⇒ ht(t) = ht(u) & (∀t′ D t)(∃u′ D u)(t′ E u′)

& (∀u′ D u)(∃t′ D t)(t′ E u′).

3.1.4 Lemma. For every E ∈ ET , the regular derivative E ′ is a congruence relation

on T such that E ′ ⊆ E. Moreover, E ′ = E if and only if E is regular.

proof: It is clear that E ′ is level, reflexive, and symmetric. To see that it is tran-

sitive, suppose that t E ′ u and u E ′ v. For every t′ D t there is a u′ D u such that

t′ E u′. In turn, there is a v′ D v such that u′ E v′. Therefore, for every t′ D t there

is a v′ D v such that t′ E v′. Similarly, for every v′ D v there is a t′ D t such that

t′ E v′. Since ht(t) = ht(u) = ht(v), it follows that t E ′ v.

The fact that E ′ ⊆ E is clear from the definition of the regular derivative. It is

also clear that the statement E ′ = E is equivalent to regularity. qed

34

Now it is not always the case that E ′ is regular, but we can iterate the derivative

until we get a regular congruence relation. More precisely, for every E ∈ ET define

E(0) = E and for every ordinal α > 0, let

E(α) =

(E(β))′ if α = β + 1,∧

β<αE(β) if α ∈ Lim.

Then reg(E) =∧

α∈Ord E(α). In fact, it is easy to see that there is a α < |T |+ such

that reg(E) = E(α).

It is almost never the case that a congruence relation is both regular and closed.

To remedy this, we introduce yet another type of congruence relation which combines

the advantages of both. We say that E is a normal congruence relation if E =

clo(reg(E)). Let E∗T be the set of all normal congruence relations on T .

3.1.5 Proposition. The set E∗T is closed under arbitrary joins in the complete lattice

Eclo. In fact, the map E 7→ E∗ = clo(reg(E)) is a surjective kernel operator from EcloT

onto E∗T . Moreover, the maps clo : EregT → E∗T and reg : E∗T → Ereg

T are mutually

inverse complete lattice isomorphisms.

The following lemma is useful for computations and is helpful in understanding the

above.

3.1.6 Lemma. If E ∈ EregT then reg(clo(E)) = E =

(E

)′.

proof: We clearly have that E ⊆ reg(clo(E)) ⊆(E

)′, so it suffices to show that(

E)′ ⊆ E. Suppose that t

(E

)′u and pick t′ . t. Then there is a u′ . u such that

t′ E u′. Since (t, u) / (t′, u′) it follows that t E u. qed

35

proof of Proposition 3.1.5: It follows immediately from Lemma 3.1.6 that

E∗∗ = clo(reg(clo(reg(E)))) = clo(reg(E)) = E∗ (3.1)

for every E ∈ EcloT . Since we clearly have E∗ ⊆ E for every E ∈ Eclo

T , it follows that

E 7→ E∗ is indeed a kernel operator in EcloT . Thus the set E∗T is closed under joins in

EcloT as per Proposition B.2.2.

It also follows from Lemma 3.1.6 that reg(clo(E)) = E for every E ∈ EregT . Since

clo(reg(E) = E for every E ∈ E∗T by (3.1), it follows that clo : EregT → E∗T and

reg : E∗T → EregT are inverses of each other. Since clo and reg are order-preserving, it

follows that E∗T and EregT are isomorphic as complete lattices. qed

Again, it follows that E∗T is a complete lattice under inclusion. Meets and joins in E∗T

will usually be denoted by∧∗ and

∨∗ respectively. Meets in E∗T are given by

∧∗i∈I Ei =

(∧i∈I Ei

)∗=

(⋂i∈I Ei

)∗.

The easiest way to compute joins in E∗T is via the following equation

∨∗i∈I Ei = clo

(∨i∈I Ei

′)which follows from the isomorphism between E∗T and Ereg

T together with Lemma 3.1.6.

3.2 Continuous Distances on Trees

Let L be a continuous lattice. A continuous (L-valued) predistance on T is a

map δ : T ⊗ T → L such that:

36

• δ is an L-valued predistance on each level Tα, i.e.

δ(t, t) = 0

δ(t, u) = δ(u, t)

δ(t, v) ≤ δ(t, u) ∨ δ(u, v)

for all t, u, v ∈ Tα;

• (monotonicity) if (t′, u′) / (t, u) then δ(t′, u′) � δ(t, u); and

• (continuity) if (t, u) has limit height and ξ � δ(t, u) then ξ � δ(t′, u′) for some

(t′, u′) / (t, u).

We say that δ is a regular continuous predistance if in addition to the above we

have that:

• (regularity) if δ(t, u) � ξ and t′ D t then there is a u′ D u such that δ(t′, u′) � ξ.

Since the lattice L is continuous, another way to state the continuity condition is that

δ(t, u) =∨

(t′,u′)/(t,u)

δ(t′, u′)

for all nodes (t, u) ∈ T⊗T with limit height. We say that δ is a (regular) continuous

distance on T if δ is an L-valued distance on each level of T , i.e. δ(t, u) = 0 ⇐⇒ t = u

for all t, u ∈ T .

Continuous predistances lead to order-preserving maps of L into EcloT in the fol-

lowing manner.

3.2.1 Proposition. Let δ be a continuous L-valued predistance on the tree T . For

each ξ ∈ L, the relation t Ecloξ t′ ⇐⇒ δ(t, t′) ≤ ξ is a closed congruence relation on T .

37

Moreover, the map ξ 7→ Ecloξ is an order preserving map from L to Eclo

T that preserves

all meets.

proof: Since δ is a predistance on each level of T , it is clear that Ecloξ is a level

equivalence relation on T . So it suffices to check that Ecloξ is downward closed and

topologically closed in T ⊗ T . Downward closure is an immediate from the mono-

tonicity of δ. Topological closure is more delicate.

Suppose that (t, u) ∈ T ⊗ T has limit height and that t�α Ecloξ u�α for all α <

ht(t, u). By continuity of δ, if ζ � δ(t, u) then ζ � δ(t�α, u�α) for some α < ht(t, u).

But δ(t�α, u�α) ≤ ξ by hypothesis. So ζ � δ(t, u) =⇒ ζ ≤ ξ which implies that

δ(t, u) ≤ ξ since L is a continuous lattice. Therefore, t Ecloξ u and it follows that Eclo

ξ

is a closed congruence relation on T .

The map ξ 7→ Ecloξ is clearly order-preserving. To see that it preserves meets, let

{ξi}i∈I be an arbitrary family of elements of L and let ξ =∧

i∈I ξi. Then

(t, u) ∈ Ecloξ ⇐⇒ δ(t, u) ≤ ξ ⇐⇒ (∀i ∈ I)(δ(t, u) ≤ ξi) ⇐⇒ (t, u) ∈

⋂i∈I E

cloξi

which shows that Ecloξ is indeed the meet of the family

{Eclo

ξi

}i∈I

in EcloT . qed

For regular continuous predistances, we have a similar order preserving map from

L into EregT .

3.2.2 Proposition. Let δ be a regular continuous predistance on the tree T . For

every ξ ∈ L, the relation t Eregξ u ⇐⇒ δ(t, u) � ξ is a regular congruence relation

on T . Moreover, the map ξ 7→ Eregξ is an order preserving map from ξ to Ereg

T that

preserves directed joins.

proof: The fact that Eregξ is a regular congruence relation on T follows immediately

from the properties of δ.

38

It is easy to see that the map ξ 7→ Eregξ is order preserving. To see that it

preserves directed joins, let {ξi}i∈I be a directed family of elements of L and let

ξ =∨

i∈I ξi. Note that{Ereg

ξi

}i∈I

is then a directed family of elements of EregT . Since

L is a continuous lattice, we have

(t, u) ∈ Eregξ ⇐⇒ δ(t, u) � ξ ⇐⇒ (∃i ∈ I)(δ(t, u) � ξi) ⇐⇒ (t, u) ∈

⋃i∈I E

regξi

and so Eregξ is the join of the family

{Ereg

ξi

}i∈I

in EregT . qed

If δ is a regular continuous predistance on T , then we have a meet preserving map

from L into EcloT and a directed join preserving map from L into Ereg

T . We will now see

how to combine these maps to form an order-preserving map from L into E∗T which

preserves both meets and directed joins.

3.2.3 Proposition. Let δ be a regular continuous predistance on the tree T . Then

Eregξ = reg(Eclo

ξ ) for every ξ ∈ L. The map ξ 7→ Eξ = clo(Eregξ ) = clo(reg(Eclo

ξ )) is an

order-preserving map from L to E∗T that preserves arbitrary meets and directed joins.

The following useful lemma shows that Eregξ and Eclo

ξ are closer than one might think.

3.2.4 Lemma. For every ξ ∈ L, the closed congruence relations clo(Eregξ ) and Eclo

ξ

agree on all limit levels of T .

proof: Suppose that α is a limit ordinal and that t, u ∈ Tα. Then (t, u) ∈ clo(Eregξ )

if and only if δ(t′, u′) � ξ for all (t′, u′) / (t, u). By continuity, this is equivalent to

δ(t, u) ≤ ξ. Therefore (t, u) ∈ clo(Eregξ ) ⇐⇒ (t, u) ∈ Eclo

ξ . qed

proof of Proposition 3.2.3: Since the map clo is an isomorphism from EregT onto

E∗T , it follows immediately from Proposition 3.2.2 that the map ξ 7→ Eξ preserves

directed joins.

39

To see that meets are preserved, let {ξi}i∈I be an arbitrary family of elements

of L and let ξ =∧

i∈I ξ. It is clear that Eξ ⊆∧∗

i∈I Eξi. On the other hand, since

Eξi⊆ Eclo

ξifor every i ∈ I, we have

∧∗i∈I Eξi

⊆∧clo

i∈I Ecloξi

= Ecloξ

by Proposition 3.2.1. Since Eξ = (Ecloξ )∗, it follows that

∧∗i∈I Eξi

⊆ Eξ. Therefore the

map ξ 7→ Eξ preserves meets. qed

What about finite joins? We say that the distance δ has interpolants if for all

ξ, ζ ∈ L and all t, t′ ∈ T with successor height if t Eξ∨ζ t′ then there are u0, . . . , u2k ∈ T

such that

t = u0 Eξ u1 Eζ · · · Eξ u2k−1 Eζ u2k = t′.

3.2.5 Lemma. If δ has interpolants, then the maps ξ 7→ Eregξ , ξ 7→ Eclo

ξ , and ξ 7→ Eξ

all preserve finite joins.

proof: See Proposition C.1.1. qed

The above is somewhat complicated and there is a simpler approach that works for

algebraic lattices. Let K be a join semi-lattice with minimal element 0. An algebraic

(K-valued) predistance on T is a map δ : T ⊗ T → K such that:

• δ is a K-valued predistance on each level Tα, i.e.

δ(t, t) = 0

δ(t, u) = δ(u, t)

δ(t, v) ≤ δ(t, u) ∨ δ(u, v)

for all t, u, v ∈ Tα;

40

• (monotonicity) if (t′, u′) E (t, u) then δ(t′, u′) ≤ δ(t, u); and

• (algebraicity) if (t, u) has limit height, then δ(t, u) = δ(t′, u′) for some (t′, u′) /

(t, u).

We say that δ is a regular algebraic predistance if in addition to the above

• (regularity) if t′ D t and ht(u) = ht(t), then there is a u′ D u such that

δ(t′, u′) = δ(t, u).

Naturally, we say that δ is a (regular) algebraic distance on T if δ is a K-valued

distance on each level of T , i.e. δ(t, u) = 0 ⇐⇒ t = u for all t, u ∈ T .

3.2.6 Proposition. If δ is a (regular) algebraic K-valued predistance on the tree

T , then the map (t, u) 7→ (δ(t, u)) = {ξ ∈ K : ξ ≤ δ(t, u)} is a (regular) continuous

Id(K)-valued predistance on T which takes only compact values.

Note that not every continuous Id(K)-valued predistance comes from an algebraic

K-valued predistance as above, but algebraic predistances form a large class which is

sufficient for most purposes.

3.3 Embedding Theorems

In the ground model V , let L be a continuous lattice and let T be a Souslin tree of

height η with a regular continuous predistance δ : T ⊗ T → L. If G is a V -generic

branch through T , then for every ξ let Gξ be the Eξ-expansion of G, i.e.

Gξ = GEξ= {t ∈ T : (∃g ∈ G)(t Eξ g)} .

By the Comparison Theorem (2.3.1), we know that V [Gζ ] ⊆ V [Gξ] if and only if

there is a closed unbounded set C and a g ∈ G such that t Eξ u =⇒ t Eζ u for all

41

t, u ∈ T g�C. Clearly, a sufficient condition for this is that ξ ≤ ζ. We would like for

this condition to be necessary, but this is not the case in general.

We say that δ is faithful if for all ξ, ζ ∈ L such that ξ � ζ and all t ∈ T there are

u, v D t such that δ(u, v) � ξ and δ(u, v) � ζ.

3.3.1 Proposition. Suppose that V |= ‘δ is a faithful regular continuous predistance

on the Souslin tree T .’ If G is a V -generic branch through T then V [Gζ ] ⊆ V [Gξ] if

and only if ξ ≤ ζ for all ξ, ζ ∈ L.

proof: We have already observed that if ξ ≤ ζ then V [Gζ ] ⊆ V [Gξ].

For the converse, suppose for the sake of contradiction that ξ � ζ and V [Gζ ] ⊆

V [Gξ]. By the Comparison Theorem (2.3.1) there is a closed unbounded set C and a

g ∈ G such that t Eξ u =⇒ t Eζ u for all t, u ∈ T g�C. But ξ � ζ, so by faithfulness

there are t, u D g such that δ(t, u) � ξ and δ(t, u) � ζ. By regularity, we can find

(t′, u′) D (t, u) such that ht(t′) = ht(u′) ∈ C and δ(t′, u′) � ξ. Note that since

δ(t′, u′) ≥ δ(t, u), we also have δ(t′, u′) � ζ. Thus t′, u′ ∈ T g�C and t′ Eξ u′ but

t′ 6Eζ u′ — a contradiction. qed

Therefore, if δ is faithful then the map ξ 7→ Gξ induces an order reversing embed-

ding of L into the join semi-lattice of V -degrees with representatives in the generic

extension V [G]. It is natural to wonder what happens with the lattice structure; are

joins/meets in L mapped to meets/joins in the V -degrees?

We first deal with the case of finite joins in L. We say that the distance δ has

enough interpolants if for all ξ, ζ ∈ L and all t, t′ . t0 with successor height if

t Eξ∨ζ t′ then there are u0, . . . , u2k . t0 such that

t = u0 Eξ u1 Eζ · · · Eξ u2k−1 Eζ u2k = t′.

42

In other words, the restriction of δ to T t0 has interpolants for every t0 ∈ T .

3.3.2 Proposition. Suppose that V |= ‘δ is a faithful continuous predistance on T

that has enough interpolants.’ If G is a V -generic branch through T , then for all

ξ, ζ ∈ L, the V -degree of Gξ∨ζ is the meet of the V -degrees of Gξ and Gζ.

proof: Suppose that X is a set of ordinals in V [Gξ]∩V [Gζ ]. By the Representation

Theorem (2.2.1) there is a closed congruence relation E on T such that V [X] = V [GE].

By the Comparison Theorem (2.3.1) there are a closed unbounded set C and a g ∈ G

such that t Eξ u =⇒ t E u and t Eζ t =⇒ t E u for all t, u ∈ T g�C. (Technically,

the Comparison Theorem gives two closed unbounded sets Cξ and Cζ and two nodes

gξ, gζ ∈ G, but then C = Cξ ∩ Cζ and g = max(gξ, gζ) are as required.) Since δ has

enough interpolants, if t, u ∈ T g�C and t Eξ∨ζ u then there are v1, . . . , v2k ∈ T g�C

such that

t Eξ v1 Eζ · · · Eζ v2k Eξ u.

It follows that

t E v1 E · · · E v2k E u

and hence t E u. Thus, t Eξ∨ζ u =⇒ t E u for all t, u ∈ T g�C which implies that

X ∈ V [Gξ∨ζ ]. qed

It is not difficult to prove directly that finite meets in L map to finite joins in the

V -degrees, but we can prove more.

3.3.3 The Embedding Theorem. Suppose that V |= ‘L is a continuous lattice,

that T is a Souslin tree of regular height η, and that δ : T ⊗ T → L is a faithful

regular continuous distance on T which has enough interpolants.’ If G is a V -generic

branch through T , then the map ξ ∈ L 7→ Gξ induces an η-complete dual lattice

embedding of L into the V -degrees with representatives in V [G].

43

To say that the map is η-complete is somewhat ambiguous. However, (L<η)V =

(L<η)V [G] since forcing with a Souslin tree of regular height η does not add new

sequences of length less than η. Thus ‘η-complete’ has the same meaning in V or

V [G].

proof: By Proposition 3.3.1, we know that the map ξ 7→ Gξ is an order reversing

embedding. Furthermore, by Proposition 3.3.2, we know that finite joins in L map

to finite meets in the V -degrees. So it suffices to show that meets of size less than η

in L map to joins in the V -degrees, and that directed joins of size less than η in L

map to meets in the V -degrees.

Let {ξi}i∈I be a family of elements of L with |I| < η and ξ =∧

i∈I ξi. Suppose

that X is a set of ordinals in V [G] such that Gξi∈ V [X] for every i ∈ I. By the

Representation Theorem (2.2.1) there is a closed congruence relation E on T in V such

that V [X] = V [GE]. By the Comparison Theorem (2.3.1) there are closed unbounded

sets Ci and gi ∈ G such that t E u =⇒ t Eξiu for all t, u ∈ T gi�Ci. Since |I| < η, we

may assume that there are a single closed unbounded set C ⊆ Lim and a single g ∈ G

such that Ci = C and gi = g for all i ∈ I. (Otherwise, replace Ci by⋂

i∈I Ci ∩ Lim

and gi by any g ∈ G with ht(g) ≥ supi∈I ht(gi).) Then for all t, u ∈ T g�C we have

t E u =⇒ (∀i ∈ I)(t Eξiu) =⇒ (t, u) ∈

∧cloi∈I Eξi

=⇒ t Ecloξ u.

Since Ecloξ andEξ agree on limit levels by Lemma 3.2.4, we have t E u =⇒ t Eξ u for all

t, u ∈ T g�C. It follows from the Comparison Theorem that V [Gξ] ⊆ V [X] = V [GE].

Therefore, the V -degree of Gξ is the least upper bound of the V -degrees of the sets

Gξifor i ∈ I.

Let {ξi}i∈I be a directed family of elements of L with |I| < η and ξ =∨

i∈I ξi.

Suppose that X is a set of ordinals in V [G] such that X ∈ V [Gξi] for every i ∈ I.

44

By the Representation Theorem (2.2.1) there is a closed congruence relation E on T

such that V [X] = V [GE]. As above, there are a closed unbounded set C and a g ∈ G

such that t Eξiu =⇒ t E u for all t, u ∈ T g�C. It follows that

(t, u) ∈∨reg

i∈I reg(Eξi) =⇒ (∃i ∈ I)(t Eξi

u) =⇒ t E u

for all t, u ∈ T g�C. Since E is closed and Eξ = clo(∨reg

i∈I reg(Eξi)), it follows that

t Eξ u =⇒ t E u for all t, u ∈ T g�C ′. It follows from the Comparison Theorem that

V [X] = V [GE] ⊆ V [Gξ]. Therefore, the V -degree of Gξ is the greatest lower bound

of the V -degrees of the sets Gξifor i ∈ I. qed

We say that δ has property (#) if for every closed congruence relation E on T

there is an unbounded set S ⊆ η such that if ht(u, v) ∈ S and u E v, then there are

(u0, v0) / (u, v) such that u′ E v′ for all (u′, v′) D (u0, v0) such that δ(u′, v′) � δ(u, v).

3.3.4 The Initial Segment Theorem. Suppose that V |= ‘L is a continuous lattice

with a dense set of size less than η, that T is a Souslin tree of height η, and that

δ : T ⊗ T → L is a faithful regular continuous distance on T which has enough

interpolants and has property (#).’ If G is a V -generic branch through T , then the

map ξ 7→ Gξ induces a dual isomorphism from the lattice L onto the V -degrees with

representatives in V [G].

The remainder of this section is devoted to proving this theorem. Suppose that

L, T and δ satisfy the hypotheses of the Initial Segment Theorem. We know from

the Embedding Theorem that the map ξ 7→ Gξ induces an η-complete dual lattice

embedding of L into the V -degrees with representatives in V [G], it suffices to show

that the map is surjective. By the Representation and Comparison Theorems, our

45

task reduces to showing that if E is a closed congruence relation on T , then there are

a ξ ∈ L, a g ∈ G, and a closed unbounded set C ⊆ η such that t E u⇐⇒ t Eξ u for

all t, u ∈ T g�C.

So fix a closed congruence relation E on T . For ξ ∈ L, consider the following open

subsets of T :

• Let Lξ be the set of all t ∈ T for which there is a closed unbounded set C such

that u Eξ v =⇒ u E v for all u, v ∈ T t�C.

• Let Uξ be the set of all t ∈ T for which there is a closed unbounded set C such

that u E v =⇒ u Eξ v for all u, v ∈ T t�C.

3.3.5 Lemma. Suppose that ξ ∈ L and t ∈ T .

(1) If t ∈ Lξ then u Eξ v =⇒ u E v for all u, v D t.

(2) If {ξi}i∈I is any family of elements of L and ξ =∨

i∈I ξi, then Lξ =⋂

i∈I Lξi.

(3) If t /∈ Uξ then there are u D t and ζ � ξ such that u ∈ Lζ.

proof:

(1) If t ∈ Lξ, then there is a closed unbounded set C such that u Eξ v =⇒ u E v

for all u, v ∈ T t�C. By regularity of δ, if δ(u, v) � ξ and u, v D t then we can find

(u′, v′) D (u, v) such that ht(u′) = ht(v′) ∈ C and δ(u′, v′) � ξ. It follows that u′ E v′

and hence that u E v. We have just shown that u Eregξ v =⇒ u E v for all u, v D t.

Since E is closed and Eξ = clo(Eregξ ), it follows that u Eξ v =⇒ u E v for all u, v D t

as claimed.

(2) Clearly, Lξ ⊆⋂

i∈I Lξi. So suppose that t ∈

⋂i∈I Lξi

and let us show that t ∈ Lξ.

By (1), we have that u Eξiv =⇒ u E v for all u, v D t. Since δ has enough interpolants

46

and L is continuous, if u, v E t and δ(u, v) � ξ then there are w1, . . . , wn ∈ T t and

i0, . . . , in ∈ I such that

u Eξi0w1 Eξi1

· · · Eξin−1wn Eξin

v.

It follows that u Eregξ v =⇒ u E v for all u, v D t. Since E is closed and Eξ = clo(Ereg

ξ )

it follows that u Eξ v =⇒ u E v for all u, v D t.

(3) Let S be as in property (#) for E. Since t /∈ Uξ there are u, v ∈ T t�S such

that u E v and u 6Eξ v. If ζ = δ(u, v) then ζ � ξ and, by property (#), there are

(u0, v0) / (u, v) such that u′ E v′ for all (u′, v′) D (u0, v0) such that δ(u′, v′) � ζ. We

may assume that u0 6Eξ v0 and that u0, v0 . t. By regularity of δ, for all u1, u2 D u0

we have

u1 Eregζ u2 ⇐⇒ δ(u1, u2) � ζ ⇐⇒ (∃v′ D v0)(δ(u1, v

′), δ(u2, v′) � ζ)

It follows that if u1, u2 D u0

u1 Eregζ u2 =⇒ (∃v′ D v) (u1 E v′ E u2) =⇒ u1 E u2.

Since E is closed and Eζ = clo(Eregζ ), it follows that u1 Eξ u2 =⇒ u1 E u2 for all

u1, u2 D u0. Hence u0 ∈ Lζ and we have found u0 D t and ζ � ξ such that u0 ∈ Lζ

as required. qed

Note that (3) implies that the set Dξ = Uξ ∪⋃

ζ�ξ Lξ is open dense in T for every

ξ ∈ L.

proof of Theorem 3.3.4: Let D be a dense subset of L with size less than η. We

may assume thatD is a join semi-lattice with 0. Consider the set I = {ξ ∈ L : G ∩ Lξ 6= ∅}.

47

Note that I is an ideal in L since Lξ∨ζ = Lξ ∩ Lζ for all ξ, ζ ∈ L. Since |D| < η,

D∩ I ∈ V and so ξ =∨

(D∩ I) exists in V . It follows from 3.3.5(2) and the fact that

|D∩ I| < η that G∩Lξ 6= ∅. We also know that G∩Dξ 6= ∅ since Dξ = Uξ ∪⋃

ζ�ξ Lξ

is open dense in T . We cannot have G ∩ Lζ 6= ∅ when ζ � ξ by definition of ξ, so

by the above we must have G ∩ Uξ 6= ∅. Let g ∈ G ∩ Lξ ∩ Uξ, then there is a closed

unbounded set C such that t E u⇐⇒ t Eξ u for all t, u ∈ T g�C. It follows from the

Comparison Theorem that V [Gξ] = V [GE] as required. qed

48

Chapter 4

Construction of Souslin Trees

In this final chapter we will give three constructions of Souslin trees with continuous

distances which satisfy the hypotheses of the Initial Segment Theorem (3.3.4) from

the previous chapter. These constructions will finally prove the main theorems of this

thesis. The first and third constructions will complete the proof Theorem 1.1.7.

Theorem 1.1.7 (restated). Assume V = L. Let κ be an infinite regular cardinal

and let L be a complete algebraic lattice with at most κ compact elements. There is

a Souslin tree T of height κ+ such that if G is a generic branch through T , then the

degrees of constructibility in L[G] form a lattice dual isomorphic to L.

The second construction will complete the proof of Theorem 1.1.8.

Theorem 1.1.8 (restated). Assume V = L. There is a Souslin tree T of height ω1

such that if G is any generic branch through T , then the degrees of constructibility

with representatives in L[G] form a lattice isomorphic to the unit interval [0, 1].

The first and second constructions are close relatives of each other, but the third

construction is very different from the other two.

49

The constructions make heavy use of Appendix A and Appendix C. The reader

may find it beneficial to be familiar with the contents of these appendices before

reading this chapter.

4.1 First Construction

The construction of this section will establish the following theorem.

4.1.1 Theorem. Assume ♦ω1. Let K be a countable join semi-lattice with zero.

There is a Souslin tree T of height ω1 with a faithful regular algebraic distance

δ : T ⊗ T → K which has enough interpolants and has property (#).

Since ♦ω1 holds in L by Proposition A.2.2, this theorem together with the Initial

Segment Theorem (3.3.4) and Proposition 3.2.6 proves Theorem 1.1.7 for κ = ω.

(The third construction covers the case κ > ω.)

Before we begin, we need to set up some background framework. First, fix a ♦ω1-

sequence 〈Xα : α < ω1〉. Let <H be a wellordering of Hω1 and let 〈Mα : α < ω1〉 be

a sequence of countable transitive elementary submodels of Hω1 as in Lemma A.2.1

such that Xα ∈ Mα for each α < ω1. The tree T will be a subtree of ω<ω1 ordered

by end-extension. For each α < ω1, we will make sure that the initial segments T<α

and δ�T 2<α of our tree and distance belong to Mα. We will achieve this by using the

wellordering <H to make canonical choices where necessary in such a way that T<α

and δ�T 2<α can always be defined in Mα. We will also assume that the base set of K

is ω and that the join operation ∨ : ω × ω → ω belongs to M0. This way, we can

always use K as a parameter throughout the construction.

We will assume that the following is true for every level α > 0 of the construction:

50

4.1.2 Induction Hypothesis. If τ : n2 → K is a distance and u ∈ T n<α is such

that δ(ui, uj) ≤ τ(i, j) for i, j < n, then there is a v ∈ T nα such that ui / vi and

δ(vi, vj) = τ(i, j) for i, j < n.

At the end of the construction, we will see that a stronger statement (Lemma 4.1.5)

is in fact true.

The base level is easy to deal with.

Base Level: Let T0 = {∅} and let δ(∅,∅) = 0.

For all higher levels, suppose that we have constructed T<α and δ�T 2<α according

to our specifications. Let Pα be the partial order defined as follows.

• The elements of Pα are triples (n, t, d) where n < ω, t ∈ T n<α, d : n2 → K is a

distance such that δ(ti, tj) ≤ d(i, j) for i, j < n.

• The ordering of Pα is given by (n, t, d) ≤ (n′, t′, d′) if and only if n ≤ n′, ti E t′i

for i < n, and d ⊆ d′.

Note that Pα is definable from the parameters T<α, K, and δ�T 2<α. All of these

parameters are in Mα, and so Pα ∈Mα.

4.1.3 Lemma. For all n0 < ω, α0 < α, and u ∈ T<α, the sets

D1α(n0) = {(n, t, d) ∈ Pα : n ≥ n0} ,

D2α(α0) = {(n, t, d) ∈ Pα : ht(t) ≥ α0} ,

D3α(u) = {(n, t, d) ∈ Pα : (∃i < n)(u E ti)} ,

are all dense in Pα.

51

proof: The fact that D1α is dense follows trivially from the arguments below.

To see that D2α(α0) is dense, suppose that (n, t, d) ∈ Pα and ht(t) < α0. By the

Induction Hypothesis 4.1.2, we can pick u ∈ T nα0

such that ti / ui and δ(ti, tj) =

δ(ui, uj) for i, j < n. Then (n, u, d) is an extension of (n, t, d) in D2α(α0).

To see that D3α(u) is dense, suppose that (n, t, d) ∈ Pα. By the previous argument,

we may assume that ht(t) ≥ ht(u). Pick u′ D u such that ht(t) = ht(u′). Let

d′ : (n + 1)2 → K be an end-extension of d such that δ(ti, u′) ≤ d′(i, n) for i < n.

(For example, let d′(i, n) = δ(ti, u) ∨ d(0, i) for i < n.) Then (n + 1, t_u′, d′) is an

extension of (n, t, d) in D3α(u). qed

Since all of these dense sets are definable in Mα, we see immediately that if G is a

Mα-generic filter over Pα, then

d∗ =⋃{d : (n, t, d) ∈ G} ,

defines a distance on ω,

t∗i =⋃{ti : (n, t, d) ∈ G ∧ n > i}

is a branch through T<α for each i < ω, and every t ∈ T<α lies on at least one (in

fact, infinitely many) of these branches.

We can now define the successor levels of the tree T .

Successor Levels: Suppose that α = α0+1, let Gα be the <H-first Mα-generic filter

over Pα, and let d∗ and 〈t∗i : i < ω〉 be as above with G = Gα. Note that each

t∗i is simply an element of Tα0 and that δ(t∗i , t∗j) ≤ d∗(i, j) for all i, j < ω. Let

Tα = {t∗i _i : i < ω} and set δ(t∗i_i, t∗j

_j) = d∗(i, j) for all i, j < ω.

Note that Gα is definable in Mα+1, so Tα and δ�T 2α can be defined in Mα+1.

52

Before we can define the limit levels, we need one more lemma.

4.1.4 Lemma. If α is a limit, then the set

Qα = {(n, t, d) ∈ Pα : (∀i, j < n)(δ(ti, tj) = d(i, j))}

is dense in Pα.

proof: Suppose that (n, t, d) ∈ Pα and ht(t) = β. By the Induction Hypothesis 4.1.2,

there is a u ∈ T nβ+1 such that ti / ui and δ(ui, uj) = d(i, j) for i, j < n. Now (n, u, d)

is an extension of (n, t, d) in Qα. qed

Note that if (n, t, d) ∈ Qα, then ti 6= tj for all i < j < n since δ(ti, tj) = d(i, j) 6= 0.

It follows that if G is an Mα-generic filter over Pα and i < j < ω, then the branches

t∗i and t∗j (as defined above) are distinct and δ(t∗i �β, t∗j�β) = d∗(i, j) for all sufficiently

large β < α. So we can define the limit levels in the following manner.

Limit Levels: Suppose that α is a limit, let Gα be the first Mα-generic filter over

Pα, and let d∗ and 〈t∗i : i < ω〉 be as above with G = Gα. Let Tα = {t∗0, t∗1, . . . }

and set δ(t∗i , t∗j) = d∗(i, j).

Again, note that Gα is definable in Mα+1, so Tα and δ�T 2α can be defined in Mα+1.

This completes the definition of the tree T and the algebraic distance δ : T ⊗T →

K. We need to verify that the construction respects the Induction Hypothesis 4.1.2.

In fact, we will establish a stronger result in the following lemma.

4.1.5 Lemma. Suppose that the distance τ : n2 → K is defines an amalgamable m-

pointed metric space on n (with distinguished points 0, . . . ,m−1), that v0, . . . , vm−1 ∈

Tα are such that δ(vi, vj) = τ(i, j) for i, j < m, and that u0, . . . , un−1 ∈ Tβ are

such that ui / vi for i < m and δ(ui, uj) ≤ τ(i, j) for i, j < n. Then there are

vm, . . . , vn−1 ∈ Tα such that ui / vi for i < n and δ(vi, vj) = τ(i, j) for i, j < n.

53

proof: We use the notation from the construction. Suppose, for simplicity, that

vi =

t∗i

_i if α is a successor,

t∗i if α is a limit

for i < m. Note that d∗(i, j) = δ(vi, vj) = τ(i, j) for i, j < m. Pick (n0, t0, d0) ∈ Gα

such that n0 ≥ m and ht(t0) ≥ ht(u). Note that we necessarily have t0,i D ui and

δ(t0,i, t0,j) ≤ d0(i, j) = τ(i, j) for i < m.

Consider the set Xα of all (k + n, t, d) ∈ Pα such that

ui E

ti when i < m,

tk+i when m ≤ i < n;

and

τ(i, j) =

d(i, j) when i, j < m,

d(i, k + j) when i < m ≤ j < n,

d(k + i, j) when j < m ≤ i < n,

d(k + i, k + j) when m ≤ i, j < n.

We claim that Xα is dense above (n0, t0, d0). Before we prove this, let us show that

this fact is sufficient to establish the lemma.

Assuming that Xα is dense above (n0, t0, d0) ∈ Gα, we can pick (k + n, t, d) ∈

Gα ∩ Xα extending (n0, t, d0). Let

vi =

t∗k+i

_(k + i) if α is a successor,

t∗k+i if α is a limit

54

for m ≤ i < n. Then ui E tk+i / vi for m ≤ i < n and

δ(vi, vj) =

d(i, j) when i, j < m

d(i, k + j) when i < m ≤ j < n

d(k + i, j) when j < m ≤ i < n

d(k + i, k + j) when m ≤ i, j < n

= τ(i, j).

So vm, . . . , vn−1 satisfy the requirements of the lemma.

Let us return to show that Xα is dense above (n0, t0, d0). Suppose that (k+m, t′, d′)

is an extension of (n0, t0, d0). Note that ui E t0,i E ti and τ(i, j) = d0(i, j) = d′(i, j)

for i, j < m. First, let u′i = t′i for i < m, and (using the lemma inductively if

necessary) pick u′i D ui for m ≤ i < n such that ht(u′) = ht(t′) and δ(u′i, u′j) ≤ τ(i, j)

for i, j < n. The distance d′ : (k + m)2 → K defines a m-pointed space on k + m

(with distinguished points 0, . . . ,m− 1) such that d′(i, j) = τ(i, j) for i, j < m. Since

τ : •0

•m− 1

•m

•n− 1

d : •0

•m− 1

•m

•m+ k − 1

•m+ k

•n+ k − 1

d′ : •0

•m− 1

•m

•m+ k − 1

Figure 4.1: Illustration of the three distances d, d′, and τ .

τ is amalgamable, we can find a distance d : (k + n)2 → K such that d′(i, j) = d(i, j)

55

for i, j < k +m, and that

τ(i, j) =

d(i, j) when i, j < m,

d(i, k + j) when i < m ≤ j < n,

d(k + i, j) when j < m ≤ i < n,

d(k + i, k + j) when m ≤ i, j < n.

Let ti = t′i for i < k+m, and let tk+i = u′i for m ≤ i < n. If necessary, replace d(i, j)

by d(i, j) ∨ δ(i, j) to make sure that (k + n, t, d) ∈ Pα. (This minor adjustment will

not change the fact that d is an amalgam of d′ and τ as above.) Then (k + n, t, d) is

an extension of (k +m, t′, d′) in Xα as required. qed

It is an immediate consequence of the previous lemma that δ is a regular distance on

T .

verification that δ is regular: Suppose that u0, u1 ∈ Tβ and v0 ∈ Tα is such

that u0 / v0. Let τ : {0, 1} → K be the distance with τ(0, 1) = δ(u0, u1). The

1-pointed space {0, 1} with distance τ and distinguished point 0 is amalgamable by

Proposition C.2.1. Therefore, by Lemma 4.1.5, there is a v1 ∈ Tα such that v1 . u1

and δ(v0, v1) = τ(0, 1) = δ(u0, u1) as required for regularity. qed

Another judicious choice of τ will show that δ has enough interpolants.

verification that δ has enough interpolants: Let t ∈ T<α and ξ, ζ ∈ K.

Suppose that u, v ∈ T tα are such that u Eξ∨ζ v. We will show that there are

w0, w1, w2, w3, w4 ∈ T tα such that

u = w1 Eξ w2 Eζ w3 Eξ w4 Eζ w0 = v.

56

Let τ : {0, 1, . . . , 4}2 → K be the distance of the 2-pointed metric space of Corol-

lary C.2.3 with ρ = ξ ∨ ζ. We know that this 2-pointed space is amalgamable, and

so, by Lemma 4.1.5, given w1 = u and w0 = v we can find w2, w3, w4 . t such

that δ(wi, wj) = τ(i, j) for i, j < 5. In particular, δ(w1, w2) = δ(w3, w4) = ξ and

δ(w2, w3) = δ(w4, w0) = ζ. So w0, w1, w2, w3, w4 are as required. qed

The following lemma is key to the verification that T has property (#).

4.1.6 Lemma. Suppose that α is a limit ordinal and that E is a closed congruence

relation on T<α. For all i < j < ω, the set Eα = Eα(E, i, j) is dense in Qα, where Eα

consists of all (n, t, d) ∈ Qα such that i, j < n, and either ti 6E tj or else u E v for all

(u, v) D (ti, tj) with δ(u, v) = δ(ti, tj).

proof: For simplicity, assume that i = 0 and j = 1, and let i and j be unreserved

variables for the remainder of this proof. Pick (n, t, d) ∈ Qα (with n ≥ 2). If there

is an extension (n′, t′, d′) of (n, t, d) such that t′0 6E t′1, then we are done. So let us

assume that t′0 E t′1 for every such (n′, t′, d′).

We will show that v0 E v1 for all (v0, v1) D (t0, t1) with δ(v0, v1) = δ(t0, t1).

Suppose, for the sake of contradiction, that (v0, v1) D (t0, t1) is such that v0 6E v1. The

distance d : n2 → K defines a 2-pointed space on n (with distinguished points 0 and 1).

Let τ : (3n−2)2 → K be the distance defined by the composition (n, d)∗ (n, d)∗(n, d).

For k < n, let k(1), k(2), and k(3) denote the element of 3n−2 corresponding to k in the

first, second, and third copy of (n, d) in (3n − 2, τ) respectively. Note that 0 = 0(1),

1(1) = 1(2), 0(2) = 0(3), and 1(3) = 1. Let u ∈ T 3n−2<α be defined by uk(1) = uk(2) =

uk(3) = tk for k < n. Then u0 / v0, u1 / v1, δ(v0, v1) = τ(0, 1), δ(ui, uj) ≤ τ(i, j)

for i, j < 3n − 2, and τ is amalgamable 2-pointed space by Proposition C.2.5. So,

by Lemma 4.1.5, there are v2, . . . , v3n−3 such that ui / vi and δ(vi, vj) = τ(i, j) for

57

i, j < 3n − 2. Now let t(1)k = vk(1) , t

(2)k = vk(2) , and t

(3)k = vk(3) for k < n. Then

(n, t(1), d), (n, t(2), d), and (n, t(3), d) are all extensions of (n, t, d) in Qα and so

t(1)0 E t

(1)1 = v1(1) = v1(2) = t

(2)1 E t

(2)0 = v0(2) = v0(3) = t

(3)0 E t

(3)1 .

However, t(1)0 = v0(1) = v0 and t

(3)1 = v1(3) = v1 — which contradicts the hypothesis

that v0 6E v1. qed

verification of property (#): Let E be a closed congruence relation on T . Let

〈Nα : α < ω1〉 be a continuous increasing sequence of countable elementary submodels

of Hλ (for some large enough λ) such that E, T, δ,K ∈ N0. There is a stationary set

S ⊆ ω1 such that α = Nα ∩ ω1 and E ∩ T 2<α ∈Mα for every α ∈ S. We claim that S

is as required by property (#).

Suppose that α ∈ S, t∗i , t∗j ∈ Tα (using the notation from the construction), and

that t∗i E t∗j . Since E ∩ T 2<α ∈ Mα, Gα meets the dense set E(E ∩ T 2

<α, i, j) from

Lemma 4.1.6. It follows that there are ti / t∗i , tj / t

∗j such that δ(ti, tj) = δ(t∗i , t

∗j) and

u E v for all (u, v) ∈ T 2<α such that (u, v) D (ti, tj) and δ(u, v) = δ(ti, tj) (the other

alternative of Lemma 4.1.6 is ruled out by the fact that t∗i E t∗j). Then Nα |= ‘u E v

for all (u, v) D (ti, tj) with δ(u, v) = δ(ti, tj)’ and, since Nα ≺ Hλ, we really have

u E v for all (u, v) D (ti, tj) with δ(u, v) = δ(ti, tj). qed

Finally, we need to verify that T is a Souslin tree.

4.1.7 Lemma. Suppose that α is a limit ordinal and that A is an arbitrary subset of

T<α. For every i < ω, the set Dα = Dα(A, i) is dense in Qα, where Dα consists of all

(n, t, d) ∈ Qα such that i < n and either ti extends an element of A or else ti has no

extension in A.

58

proof: For simplicity, assume that i = 0 and let i be an unreserved variable for the

remainder of this proof. Pick (n, t, d) ∈ Qα (with n ≥ 1). If there is an extension

(n′, t′, d′) of (n, t, d) such that t′0 extends an element of A, then we are done. So let

us assume that for every such (n′, t′, d′), t′0 doesn’t extend an element of A.

We will show that t0 has no extension in A. Suppose, for the sake of contra-

diction, that v0 is an extension of t0 in A. The distance d : n2 → K defines a

1-pointed space (with distinguished point 0). Every 1-pointed space is amalgamable

by Proposition C.2.1, so by Lemma 4.1.5 there are v1, . . . , vn−1 such that ti / vi for

and δ(vi, vj) = d(i, j) for i, j < n. Now (n, v, d) is an extension of (n, t, d) such that

v0 ∈ A — which contradicts our hypothesis that no such extension exists. qed

verification that T is Souslin: Suppose that A is a maximal antichain in T .

Let 〈Nα : α < ω1〉 be a continuous increasing sequence of countable elementary sub-

models of Hλ (for some large enough λ) such that A, T, δ,K ∈ N0. There is a α < ω1

such that α = Nα ∩ ω1 and A ∩ T<α = A ∩ Nα ∈ Mα. We claim that A ⊆ T<α or,

equivalently, that every element of Tα extends an element of A.

Suppose, for the sake of contradiction, that t∗i ∈ Tα (using the notation of the

construction) does not extend an element of A. Since A ∩ T<α ∈ Mα, Gα meets the

dense set D(A ∩ T<α, i) from Lemma 4.1.7. It follows that there is a ti / t∗i such that

ti has no extension in A ∩ T<α (and that ti does not already extend an element of

A). Since A ∩ T<α = A ∩Nα, it follows that Nα |= ‘ti has no extension in A’. Since

Nα ≺ Hλ the same is true in Hλ and hence ti really has no extension in A — this

contradicts the fact that A is a maximal antichain in T . qed

4.2 Second Construction

The construction of this section will establish the following theorem.

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4.2.1 Theorem. Assume ♦ω1. There is a Souslin tree T of height ω1 with a faithful

regular continuous distance δ : T ⊗ T → [0,∞] which has property (#).

Since ♦ω1 holds in L by Proposition A.2.2, this theorem together with the Initial

Segment Theorem (3.3.4) (and the fact that [0, 1] is dual isomorphic to [0,∞]) proves

Theorem 1.1.8. (Note that since [0,∞] is a linear ordering, we don’t need to worry

about interpolants.) The construction given here has many parts in common with

that of the previous section, but there are sufficiently many small differences that

every detail will be repeated exept for the verification that T is Souslin. Note that

since [0,∞] is a linear ordering, we will not have to worry about interpolants.

The background framework is the same as in the previous section. First, we fix a

♦ω1 sequence 〈Xα : α < ω1〉. Let <H be a wellordering of Hω1 and let 〈Mα : α < ω1〉

be a sequence of countable transitive elementary submodels of Hω1 as in Lemma A.2.1

such that Xα ∈ Mα for each α < ω1. The tree T will be a subtree of ω<ω1 ordered

by end-extension. For each α < ω1, we will make sure that the initial segments T<α

and δ�T 2<α of our tree and distance belong to Mα. We will achieve this by using the

wellordering <H to make canonical choices where necessary in such a way that T<α

and δ�T 2<α can always be defined in Mα. Note that the rationals are definable in M0.

We will assume that the following is true for every level α > 0 of the construction:

4.2.2 Induction Hypothesis. If τ : n2 → [0,∞)Q is a distance and u ∈ T n<α is

such that δ(ui, uj) < τ(i, j) for i, j < n, then there is a t ∈ T nα such that ui / ti and

δ(ti, tj) = τ(i, j) for i, j < n.

At the end of the construction, we will see that a stronger statement (Lemma 4.2.5)

is in fact true at every level α > 0.

The base level is easy to deal with.

60

Base Level: Let T0 = {∅} and let δ(∅,∅) = 0.

For all later levels, suppose that we have constructed T<α and δ�T 2<α according to

our specifications. Let Pα be the partial order such that:

• The elements of Pα are triples (n, t, d) where n < ω, t ∈ T n<α, and d : n2 →

[0,∞)Q is a distance such that δ(ti, tj) < d(i, j) for i < j < n.

• The ordering of Pα is given by (n, t, d) ≤ (n′, t′, d′) if and only if n ≤ n′, ti E t′i

for i < n, and d ⊆ d′.

Note that Pα is definable from the parameters T<α, [0,∞)Q, and δ�T 2<α. All of these

parameters are in Mα, and so Pα ∈Mα.

4.2.3 Lemma. For all n0 < ω, α0 < α, and u ∈ T<α, the sets

D1α(n0) = {(n, t, d) ∈ Pα : n ≥ n0} ,

D2α(α0) = {(n, t, d) ∈ Pα : ht(t) ≥ α0} ,

D3α(u) = {(n, t, d) ∈ Pα : (∃i < n)(u E ti)} ,

are all dense in Pα.

proof: The fact that D1α is dense is trivial.

To see that D2α(α0) is dense, suppose that (n, t, d) ∈ Pα and ht(t) < α0. Pick

a distance d : n2 → [0,∞)Q such that δ(ti, tj) < d(i, j) < d(i, j) for i, j < n. (For

example, d(i, j) = d(i, j))− ε for some small enough ε > 0.) By Hypothesis 4.2.2, we

can find u ∈ T nα0

such that ti / ui and δ(ui, uj) = d(i, j) for i, j < n. Then (n, u, d) is

an extension of (n, t, d) in D2α(α0).

To see that D3α(u) is dense, suppose that (n, t, d) ∈ Pα. By the previous argument,

we may assume that ht(t) ≥ ht(u). Pick u′ D u such that ht(t) = ht(u′). Let

61

d′ : (n + 1)2 → L be an end-extension of d such that δ(ti, u′) < d′(i, n) for i < n.

(For example, let d′(i, n) = M for i < n where M is a large enough rational number.)

Then (n+ 1, t_u′, d′) is an extension of (n, t, d) in D3α(u). qed

Since all of these dense sets are definable in Mα, we see immediately that if G is an

Mα-generic filter over Pα, then

d∗ =⋃{d : (n, t, d) ∈ G} ,

defines a [0,∞)Q-valued metric on ω,

t∗i =⋃{ti : (n, t, d) ∈ G ∧ n > i}

is a branch through T<α for each i < ω, and every t ∈ T<α lies on at least one (in

fact, infinitely many) of these branches.

We can now define the successor levels of the tree T :

Successor Levels: Suppose that α = α0 + 1, let Gα be the first Mα-generic filter

over Pα, and let d∗ and 〈t∗i : i < ω〉 be as above with G = Gα. Note that each

t∗i is simply an element of Tα0 and that δ(t∗i , t∗j) < d∗(i, j) for all i < j < ω. Let

Tα = {t∗i _i : i < ω} and set δ(t∗i_i, t∗j

_j) = d(i, j) for all i, j < ω.

Before we can define the limit levels, we need one more lemma.

4.2.4 Lemma. If α is a limit, then for every ε > 0, the set

Qα(ε) = {(n, t, d) ∈ Pα : (∀i, j < n)(δ(ti, tj) ≥ d(i, j)− ε)}

is dense in Pα.

62

proof: Pick (n, t, d) ∈ Pα. We may assume that ε is small enough that δ(ti, tj) <

d(i, j)− ε for all i < j < n. Then the map d : n2 → [0,∞)Q defined by

d(i, j) =

0 when i = j,

d(i, j)− ε when i 6= j.

is a distance on n such that δ(ti, tj) < d(i, j) < d(i, j) for all i < j < n. By the

Induction Hypothesis 4.2.2, there is a u ∈ T n<α such that ti / ui and δ(ui, uj) = d(i, j)

for i, j < n. Now (n, u, d) is an extension of (n, t, d) in Qα(ε) as required. qed

Suppose that G is an Mα-generic filter over Pα, and let t∗i and d∗ be as defined above.

By Lemma 4.2.4, for every ε > 0 there is a β < α such that δ(t∗i �β, t∗j�β) ≥ d∗(i, j)−ε.

Therefore the branches t∗i and t∗j are distinct and

d∗(i, j) =∨

β<α δ(t∗i �β, t

∗j�β).

So we can define the limit levels in the following manner.

Limit Levels: Suppose that α is a limit, let Gα be the first Mα-generic filter over

Pα, and let d∗ and 〈t∗i : i < ω〉 be as above with G = Gα. Let Tα = {t∗0, t∗1, . . . }

and set δ(t∗i , t∗j) = d∗(i, j).

This completes the definition of the tree T and the continuous distance δ : T ⊗ T →

[0,∞)Q.

We need to verify that the construction respects Hypothesis 4.2.2. In fact, we will

establish a stronger property in the following lemma.

4.2.5 Lemma. Let τ : n2 → [0,∞)Q be a distance which defines an m-pointed space

on n (with distinguished points 0, . . . ,m−1). Suppose that v0, . . . , vm−1 ∈ Tα are such

63

that δ(vi, vj) = τ(i, j) for i, j < m, and that u0, . . . , un−1 ∈ Tβ are such that ui / vi

for i < m and δ(ui, uj) < τ(i, j) for i, j < n. Then there are vm, . . . , vn−1 ∈ Tα such

that ui / vi for i < n and δ(vi, vj) = τ(i, j) for i, j < n.

proof: We use the notation from the construction. Suppose, for simplicity, that

vi =

t∗i

_i if α is a successor,

t∗i if α is a limit

for i < m. Note that d∗(i, j) = δ(vi, vj) = τ(i, j) for i, j < m. Pick (n0, t0, d0) ∈ Gα

such that n0 ≥ m and ht(t0) ≥ ht(u). Note that we necessarily have t0,i D ui and

δ(t0,i, t0,j) � d0(i, j) = τ(i, j) for i, j < m.

Consider the set Xα of all (k + n, t, d) ∈ Pα such that

ui E

ti when i < m,

tk+i when m ≤ i < n;

and

τ(i, j) =

d(i, j) when i, j < m,

d(i, k + j) when i < m ≤ j < n,

d(k + i, j) when j < m ≤ i < n,

d(k + i, k + j) when m ≤ i, j < n.

We claim that Xα is dense above (n0, t0, d0). Before we establish this, let us show

that this fact is sufficient to prove the lemma.

Assuming that Xα is dense above (n0, t0, d0) ∈ Gα, we can pick (k + n, t, d) ∈

64

Gα ∩ Xα. Let

vi =

t∗k+i

_(k + i) if α is a successor,

t∗k+i if α is a limit

for m ≤ i < n. Then ui E tk+i / vi for m ≤ i < n and

δ(vi, vj) =

d(i, j) when i, j < m

d(i, k + j) when i < m ≤ j < n

d(k + i, j) when j < m ≤ i < n

d(k + i, k + j) when m ≤ i, j < n

= τ(i, j).

So vm, . . . , vn−1 satisfy the requirements of the lemma.

It remains to show that Xα is indeed dense above (n0, t0, d0). Suppose that (k +

m, t′, d′) is an extension of (n0, t0, d0). Note that ui E t0,i E ti and τ(i, j) = d0(i, j) =

d′(i, j) for i, j < m. First, let u′i = t′i for i < m, and (using the lemma inductively if

necessary) pick u′i D ui for m ≤ i < n such that ht(u′) = ht(t′) and δ(ui, uj) < τ(i, j)

for i < j < n. The distance d′ : (k + m)2 → L defines a m-pointed space on k + m

(with distinguished elements 0, . . . ,m − 1) and d′(i, j) = τ(i, j) for i, j < m. Let

d : (k + n)2 → [0,∞)Q be defined by d′(i, j) = d(i, j) when i, j < k +m,

τ(i, j) =

d(i, j) when i, j < m,

d(i, k + j) when i < m ≤ j < n,

d(k + i, j) when j < m ≤ i < n,

d(k + i, k + j) when m ≤ i, j < n,

and

d(k + i,m+ j) = d(m+ j, k + i) = min`<m

τ(i, `) ∨ d′(`,m+ j)

65

when m ≤ i < n and j < k. Note that d is isometric to the canonical amalgam of d′

τ : •0

•m− 1

•m

•n− 1

d : •0

•m− 1

•m

•m+ k − 1

•m+ k

•n+ k − 1

d′ : •0

•m− 1

•m

•m+ k − 1

Figure 4.2: Illustration of the three distances d, d′, and τ .

and τ . Since [0,∞)Q is a distributive lattice, it follows from Proposition C.3.3 that

d is a distance on k + n. Let ti = t′i for i < k +m, and let tk+i = u′i for m ≤ i < n.

Note that if m ≤ i < n and j < k, then

δ(tk+i, tm+j) = δ(t′i, u′m+j) ≤ δ(t′i, t`) ∨ δ(u′`, u′m+j) < τ(i, `) ∨ d′(`,m+ j)

for every ` < m and so δ(tk+i, tm+j) < d(k + i,m + j). It follows that (k + n, t, d) is

an extension of (k +m, t′, d′) in Xα as required. qed

It is an immediate consequence of the previous lemma that δ is a regular faithful

continuous distance on T .

It remains to verify that T has property (#). The following lemma is key to that

proof.

4.2.6 Lemma. Suppose that α is a limit ordinal and that E is a closed congruence

relation on T<α. For all i < j < ω, the set Eα = Eα(E, i, j) is dense in Pα, where Eα

consists of all (n, t, d) ∈ Pα such that i, j < n, and either ti 6E tj or else u E v for all

u, v ∈ T<α such that (u, v) D (ti, tj) and δ(u, v) < d(i, j).

proof: For simplicity, assume that i = 0 and j = 1, and let i and j be unreserved

variables for the remainder of this proof. Pick (n, t, d) ∈ Pα (with n ≥ 2). If there

66

is an extension (n′, t′, d′) of (n, t, d) such that t′0 6E t′1, then we are done. So let us

assume that t′0 E t′1 for every such (n′, t′, d′).

We will show that v0 E v1 for all v0, v1 ∈ T<α such that (v0, v1) D (t0, t1) and

δ(v0, v1) < d(0, 1). Suppose, for the sake of contradiction, that v0, v1 ∈ T<α are such

that (v0, v1) D (t0, t1), δ(v0, v1) < d(0, 1), and v0 6E v1. The distance d : n2 → [0,∞)Q

defines a 2-pointed space (with distinguished elements 0 and 1). By Lemma 4.2.5,

there are v2, . . . , vn−1 such that ti / vi and δ(vi, vj) = d(i, j) for i, j < n. Then (n, v, d)

is an extension of (n, t, d) in Pα and so it follows that v0 E v1 — which contradicts

the hypothesis that v0 6E v1. qed

verification of property (#): Let E be a closed congruence relation on T . Let

〈Nα : α < ω1〉 be a continuous increasing sequence of countable elementary submodels

of Hλ (λ large enough) such that E, T, δ,K, · · · ∈ N0. There is a stationary set S ⊆ ω1

such that α = Nα ∩ ω1 and E ∩ T 2<α ∈Mα for all α ∈ S.

Suppose that α ∈ S, t∗i , t∗j ∈ Tα (using the notation from the construction), and

that t∗i E t∗j . Since E ∩ T 2<α ∈ Mα, Gα meets the dense set E(E ∩ T 2

<α, i, j) from

Lemma 4.2.6. Pick (n, t, d) ∈ E(E ∩ T 2<α, i, j). It follows that (ti, tj) / (t∗i , t

∗j) are

such that u E v for all u, v ∈ T<α such that (u, v) D (ti, tj) and δ(u, v) < d(i, j) =

d∗(i, j) = δ(t∗i , t∗j) (the other alternative of Lemma 4.2.6 is ruled out by the fact that

t∗i E t∗j). Then Nα |= ‘u E v for all (u, v) D (tj, ti) with δ(u, v) < d(i, j)’ and, since

Nα ≺ Hλ, we really have u E v for all (u, v) D (ti, tj) such that δ(u, v) < δ(t∗i , t∗j) as

required by property (#). qed

4.3 Third Construction

The construction of this section will establish the following theorem.

67

4.3.1 Theorem. Assume �� κ+ where κ is a regular uncountable cardinal. Let K be a

join semi-lattice with zero and size at most κ. There is a Souslin tree T of height κ+

with a regular faithful algebraic distance δ : T ⊗T → K which has enough interpolants

and has property (#).

Since �� κ+ holds in L by Proposition A.2.4, this theorem together with the Initial

Segment Theorem (3.3.4) and Proposition 3.2.6 proves Theorem 1.1.7 for κ > ω.

(The first construction covers the case κ = ω.)

Before we begin, we need to set up some background framework. First, we will

need a �κ+-sequence 〈Cα : α ∈ κ+ ∩ Lim〉 together with a built-in diamond sequence

〈Dα : α ∈ κ+ ∩ Lim〉. Next, let <H be a wellordering of Hκ+ and let 〈Mα : α < κ+〉

be a sequence of transitive elementary submodels of Hκ+ as in Lemma A.2.1 such

that (Cα, Dα) ∈ Mα for every α < κ+. The tree T will be a subtree of κ<κ+ordered

by end-extension. For each α < κ+, we will make sure that the initial segments T<α

and δ�T 2<α of our tree and distance belong to Mα. We will achieve this by using the

wellordering <H to make canonical choices where necessary in such a way that T<α

and δ�T 2<α can be defined in Mα. We will also assume that the base set of K is κ and

that the join operation ∨ : κ× κ→ κ belongs to M0. This way, we can always use K

as a parameter throughout the construction.

Suppose that β < α < κ+. A map e : Tβ → Tα is said to be an extender map if

e(t) . t and δ(e(t), e(u)) = δ(t, u) for all t, u ∈ Tβ. Our construction will rely on the

existence of many extender maps from Tβ to Tα for all β < α < κ+. More precisely,

we will ensure that T satisfies the following conditions.

4.3.2 Homogeneity Conditions. Suppose that β < α < κ+.

(H1) If t ∈ Tβ and u ∈ Tα are such that t / u, then there is an extender map

e : Tβ → Tα such that e(t) = u.

68

(H2) If t, u ∈ Tβ and t′, u′ ∈ Tα are such that t / t′, u / u′, and δ(t, u) = δ(t′, u′),

then there are extender maps e0, . . . , e2r : Tβ → Tα such that

t′ = e0(t), e0(u) = e1(u),

e1(t) = e2(t), e2(u) = e3(u),

......

e2r−1(t) = e2r(t), e2r(u) = u′

for some r < ω.

Note that (H1) already guarantees that δ is a regular distance on T . To see this, if

t, u ∈ Tβ and t′ ∈ Tα is such that t′ . t then there is an extender map e : Tβ → Tα

such that e(t) = t′. We then have that u′ = e(u) . u and δ(t′, u′) = δ(t, u) as required

for regularity.

In order to guide our construction at limit levels, we will pick a specific extender

map eα,β : Tβ → Tα for all β < α < κ+. Ideally, this would be a completely

coherent system, i.e. eα,γ = eα,β ◦ eβ,γ for all γ < β < α < κ+. Unfortunately, this

is incompatible with T being Souslin. (For example, {e0,α(∅) : α < κ} would be a

branch through T .) So we settle for some weaker form of coherence.

4.3.3 Coherence Conditions. Suppose that β < α < κ+.

(C1) If α = γ + 1 then eα,β = eα,γ ◦ eγ,β.

(C2) If α is a limit and γ ∈ C ′α \ β then eα,β = eα,γ ◦ eγ,β.

Finally, we also require that Tα =⋃

β<α eα,β[Tβ] when α is a limit ordinal.

The base level is easy to deal with.

69

Base Level: Let T0 = {∅} and let δ(∅,∅) = 0.

There is nothing to verify at this level.

We first deal with successor levels. Suppose that we have defined T≤α, δ�T 2≤α, and

〈eγ,β : β < γ ≤ α〉 according to our specifications. Let Pα be the partial order defined

as follows.

• The elements of Pα are pairs (ν, d) where ν < κ, and d : (Tα × ν)2 → K is a

predistance such that

d((t, i), (u, j)) ≥ d((t, i), (u, i)) = δ(t, u)

for all t, u ∈ Tα and i, j < ν.

• The ordering of Pα is given by (ν, d) ≤ (ν ′, d′) if and only if ν ≤ ν ′ and d ⊆ d′.

Note that Pα is definable from the parameters Tα, K, and δ�T 2α. All of these param-

eters are in Mα, and so Pα ∈Mα. The reader will have no problem proving following

basic properties of Pα.

4.3.4 Lemma. The partial order Pα is κ-closed and for every ν0 < κ the set

Dα(ν0) = {(ν, d) ∈ Pα : ν ≥ ν0} ,

is dense in Pα.

Since M<κα ⊆Mα, this guarantees that every condition is contained in an Mα-generic

filter over Pα. If G is an Mα-generic filter over Pα, then

d∗ =⋃{d : (ν, d) ∈ G}

70

defines a predistance on Tα × κ such that

d∗((t, i), (u, j)) ≥ d∗((t, i), (u, i)) = δ(t, u)

for all t, u ∈ Tα and i, j < κ.

We can now define the successor levels of the tree T .

Successor Levels: Let Gα be the <H-first Mα-generic filter over Pα, and let d∗ be

as above with G = Gα. Then define

Tα+1 = {t_i : i < κ ∧ (∀j < i)(d∗((t, i), (t, j)) 6= 0)} ;

δ(t_i, u_j) = d∗((t, i), (u, j)) for all t_i, u_j ∈ Tα+1; and eα+1,α(t) = t_0 for

every t ∈ Tα.

Note that the definition of Tα+1 above ensures that δ is a distance on Tα+1. To wit,

if δ(t_i, u_j) = 0 then we must have t = u since δ(t_i, u_j) ≥ δ(t, u). If j < i then

t_i /∈ Tα+1 by definition and similarly u_j /∈ Tα+1 when i < j. So we must have

i = j.

Let us verify the homogeneity conditions for this level. For every i < κ define the

map e(i) = e(i)α+1,α : Tα → Tα+1 by

e(i)(t) = t_ min {j ≤ i : d∗((t, i), (t, j)) = 0} .

for every t ∈ Tα. It is easy to see that e(i) is an extender map since

δ(e(i)(t), e(i)(u)) = d∗((t, i), (u, i)) = δ(t, u)

71

for all t, u ∈ Tα. Using these maps, it is easy to verify the first homogeneity condition

(H1). Suppose u ∈ Tβ, t_i ∈ Tα+1, and u E t. By the induction hypothesis, there is

an extender map e : Tβ → Tα such that e(u) = t. Then the composition e(i) ◦ e is an

extender map such that e(i)(e(u)) = e(i)(t) = t_i.

To verify the second homogeneity condition (H2), we need a lemma.

4.3.5 Lemma. For all t, u ∈ Tα and i, j < κ, the set Hα(t, i, u, j) is dense in Pα

where Hα(t, i, u, j) consists of all (ν, d) ∈ Pα such that i, j < ν and if d((t, i), (u, j)) =

δ(t, u) then there are k, `,m < ν such that

d((t, i), (t, k)) = 0, d((u, k), (u, `)) = 0,

d((u, j), (u,m)) = 0, d((t,m), (t, `)) = 0.

proof: Let (ν, d) ∈ Pα be given. By Lemma 4.3.4 we may assume that i, j < ν. We

may also suppose that d((t, i), (u, j)) = δ(t, u) or else there is nothing to prove. View

Tα as a 2-pointed metric space with distance δ and distinguished points 0 = t and 1 =

u, and view Tα×ν as a 2-pointed premetric space with predistance d and distinguished

points 0 = (t, i) and 1 = (u, j). Define the predistance d′ : (Tα × (ν + 3))2 → K as

the amalgam of Tα × ν with the composition Tα ∗ Tα ∗ Tα wherein the three copies of

Tα are identified with Tα × {k}, Tα × {`}, and Tα × {m} respectively where k = ν,

` = ν + 1, m = ν + 2. Then (ν + 3, d′) ∈ Pα is an extension of (ν, d) and k, `, m

witness that (ν + 3, d′) belongs to Hα(t, i, u, j). qed

Note that the sets Hα(t, i, u, j) are definable in Mα and so each one is met by the

Mα-generic filter Gα.

To verify (H2), suppose that t0, u0 ∈ Tβ and t1_i, u1

_j ∈ Tα+1 are such that

t0 E t1, u0 E u1 and δ(t1_i, u1

_j) = δ(t1, u1). By the induction hypothesis, there are

72

extender maps e0, . . . , e2r : Tβ → Tα such that

t1 = e0(t0), e0(u0) = e1(u0),

e1(t0) = e2(t0), e2(u0) = e3(u0),

......

e2r−1(t0) = e2r(t0), e2r(u0) = u1.

Set t = e2r(t0), u = u1, and note that t_i = e(i)(t). By Lemma 4.3.5, there are

k, `,m < κ such that

d∗((t, i), (t, k)) = 0, d∗((u, k), (u, `)) = 0,

d∗((u, j), (u,m)) = 0, d∗((t,m), (t, `)) = 0.

In terms of extender maps, this means that

t_i = e(i)(t) = e(k)(t), e(k)(u) = e(`)(u),

u_j = e(j)(u) = e(m)(u), e(m)(t) = e(`)(t).

73

Let e′s = e(i) ◦ es for s = 0, . . . , 2r− 1, let e′2r = e(k) ◦ e2r, let e′2r+1 = e(`) ◦ e2r, and let

e′2r+2 = e(m) ◦ e2r. Then it is easy to check that

t1_i = e′0(t0), e′0(u0) = e′1(u0),

e′1(t0) = e′2(t0), e′2(u0) = e′3(u0),

......

e′2r−1(t0) = e′2r(t0), e′2r(u0) = e′2r+1(u0),

e′2r+1(t0) = e′2r+2(t0), e′2r+2(u0) = u1_j

as required by (H2).

Limit levels are divided into two cases according to whether or not supC ′α = α.

The case supC ′α = α is easy to deal with since the coherence conditions completely

determine Tα. Indeed, if β < α and t ∈ Tβ, the coherence conditions force that

eα,β(t) = eα,γ(eγ,β(t)) ≥ eγ,β(t) for every γ ∈ C ′α \ β. Since supC ′

α = α, there is only

one choice for eα,β(t), namely

eα,β(t) =⋃

γ∈C′α\β

eγ,β(t). (4.1)

Limit levels α = supC ′α: Let Tα =

⋃β<α eα,β[Tβ] where eα,β is defined by (4.1). If

t, u ∈ Tα then let δ(t, u) = δ(t�β, u�β) for any β < α such that t = eα,β(t�β)

and u = eα,β(u�β).

Note that the homogeneity conditions (H1) and (H2) are automatically satisfied at

limit levels.

The case when α > α0 = supC ′α is the heart of the construction. Note that

cf(α) = ω since tp(Cα\α0) = ω and sup(Cα\α0) = α. The plan is to define inductively

74

an increasing sequence α1, α2, . . . of ordinals cofinal in α and a commuting system of

extender maps fn,m : Tαm → Tαn for m < n < ω. We will then have a sequence of

extender maps fm : Tαm → Tα defined by

fm(t) =∞⋃

n=m+1

fn,m(t)

for every t ∈ Tαn . Once this is done, we will define the α-th level as follows.

Limit levels α > supC ′α: Let Tα =

⋃n<ω fn[Tαn ]. For t, u ∈ Tα let δ(t, u) = δ(t�αn, u�αn)

for any n < ω such that t = fn(t�αn) and u = fn(u�αn). Finally, let eα,β =

fn ◦ eαn,β where n is least such that β ≤ αn.

We will use some important parameters in our definitions. Let 〈−,−〉 : Ord2 →

Ord be a standard ordinal pairing function. We assume that ξ, ζ < 〈ξ, ζ〉 for all

ξ, ζ ∈ Ord. The first important parameter is the set D†α ⊆ T<α⊗T<α which is defined

as follows. Let Dα be the α-th set from the diamond part of the �� κ+-sequence. This

set codes a subset D†α of T<α ⊗ T<α in the following way: if t and u are the τ -th

and υ-th elements of Tγ respectively, then (t, u) ∈ D†α if and only if 〈γ, 〈τ, υ〉〉 ∈ Dα.

(Since Dα ⊆ α this is always well defined.)

The next important parameters are the nodes t†, u† ∈ T<α which are defined as

follows. Write tp(C ′α) = 〈β, 〈τ, υ〉〉. Let t† and u† be the τ -th and υ-th elements of Tγ

where γ is the β-th element of Cα. (Note that β < tp(C ′α) ≤ tp(Cα), so this always

makes sense.)

Finally, let t0 = eα0,γ(t†) and u0 = eα0,γ(u

†).

The key step in the inductive definition is the choice of α1 and f1,0. If possible,

let α1 be the least element of Cα greater than α0 such that there is an extender map

f : Tα0 → Tα1 with the property that (f(t0), f(u0)) extends an element of D†α. Then

also let f1,0 be the <H-first such extender map in Mα. If such a choice is impossible,

75

let α1 be the first element of Cα after α0 and let f1,0 = eα1,α0 . For n ≥ 2, simply

let αn be the first element of Cα after αn−1 and let fn,m = eαn,αn−1 ◦ fn−1,m for every

m < n.

Since tp(Cα \ α0) = ω, we necessarily have α = supn<ω αn. It is easy to show

by induction that the system of extender maps 〈fn,m : m < n < ω〉 is commuting, i.e.

that fn,m = fn,` ◦ f`,m for all m < ` < n < ω.

This completes our construction of T and δ. It is clear that δ is a regular algebraic

distance on T . We need to verify that δ has enough interpolants and has property

(#), and finally that T is a Souslin tree.

4.3.6 Lemma. For every t ∈ Tα, i, j < κ, and ξ, ζ ∈ K, the set Iα(t, i, j, ξ, ζ) is

dense in Pα, where Iα(t, i, , j, ξ, ζ) consists of all (ν, d) ∈ Pα such that i, j < ν and if

d((t, i), (t, j)) ≤ ξ ∨ ζ then there are k0, k1, k2, k3, k4 < ν such that

d((t, i), (t, k0)) = 0 = d((t, j), (t, k1)),

d((t, k1), (t, k2)) = ξ = d((t, k3), (t, k4)),

d((t, k2), (t, k3)) = ζ = d((t, k4), (t, k0)).

proof: Suppose that (ν0, d0) ∈ Pα and t ∈ Tα, i, j < ν0 are such that δ((t, i), (t, j)) ≤

ξ∨ζ. (If δ((t, i), (t, j)) � ξ∨ζ then there is nothing to do.) Set k0 = ν0, . . . , k4 = ν0+4.

By Corollary C.2.3, we can extend d0 to a predistance d′ on (Tα × ν0) ∪ ({t} ×

{k0, . . . , k4}) in such a way that

d′((t, i), (t, k0)) = 0 = d′((t, j), (t, k1)),

d′((t, k1), (t, k2)) = ξ = d′((t, k3), (t, k4)),

d′((t, k2), (t, k3)) = ζ = d′((t, k4), (t, k0)).

76

Now extend d′ to a distance d on Tα × (ν0 + 5) by setting

d((u, `), (v, ka)) = d′((u, `), (t, ka)) ∨ δ(t, v) when ` < ν0,

d((u, ka), (v, kb)) = δ(u, t) ∨ δ(t, v) when a 6= b,

d((u, ka), (v, ka)) = δ(u, v) otherwise.

Note that this is the same as successively amalgamating the 1-pointed spaces Tα with

distinguished point t and Tα× (ν0 + a)∪ {t}× {ka, . . . , k4} with distinguished points

(t, ka) for a = 0, . . . , 4. Then (ν0+5, d) ∈ Pα is an extension of (ν0, d0) in I(t, i, j, ξ, ζ)

as required. qed

verification that δ has enough interpolants: We will show by induction on

α that if u, v ∈ Tα, u, v D t, and δ(u, v) ≤ ξ ∨ ζ, then there are interpolants

w0, . . . , w2n D t such that

u = w0 Eξ w1 Eζ · · · Eξ w2k−1 Eζ w2k = v.

The base case α = 0 is trivial.

First, let us consider the case when α is a limit ordinal. Then there is a β < α such

that δ(u′, v′) = δ(u, v) where u′ = u�β and v′ = v�β. We may assume that β ≥ ht(t)

so that u′, v′ D t. By the induction hypothesis there are interpolants w′0, . . . , w′2n D t

such that

u′ = w′0 Eξ w′1 Eζ · · · Eξ w

′2n−1 Eζ w

′2n = v′.

By homogeneity condition (H2) there are extender maps e0, . . . , e2r : Tβ → Tα such

77

that

u = e0(u′), e0(v

′) = e1(v′),

e1(u′) = e2(u

′), e2(v′) = e3(v

′),

......

e2r−1(u′) = e2r(u

′) e2r(v′) = v.

Now let wi,j = ei(wj) for i < 2r and j < 2n, then

u = w0,0 Eξ · · · Eζ w0,2n = w1,0

w0,2n = w1,0 Eξ · · · Eζ w1,2n = w2,0

......

......

w2r−1,2n = w2r,0 Eξ · · · Eζ w2r,2n = v.

These are the required interpolants for u, v D t.

Now suppose that α = β + 1. We first consider the case when t = u�β = v�β.

Then u = e(i)(t) = t_i and v = e(j)(t) = t_j for some i, j < κ. By Lemma 4.3.6, we

can find k0, k1, k2, k3, k4 such that

u = e(k1)(t) Eξ e(k2)(t) Eζ e

(k3)(t) Eξ e(k4)(t) Eζ e

(k0)(t) = v

which immediately gives the required interpolants.

If u�β = u′ 6= v′ = v�β then let w′0, . . . , w′2n D t be interpolants for u′ and v′. Let

e : Tβ → Tα be an extender map such that e(u′) = u and set w0 = e(w′0), . . . , w2n =

e(w′2n). Then

u = w0 Eξ w1 Eζ · · · Eξ w2n−1 Eζ w2n = e(v′).

78

If e(v′) = v, then we are done. Otherwise, note that e(v′)�β = v′ = v�β and

δ(e(v′), v) ≤ ξ ∨ ζ. So (by extending t as necessary) we find ourselves in the pre-

vious case and we can extend the sequence with w2n+1, w2n+2, w2n+3, w2n+4 in such

that

w2n Eξ w2n+1 Eζ w2n+2 Eξ w2n+3 Eζ w2n+4 = v.

qed

For the remaining verifications, the reader may find it useful to review the defini-

tions of the paramatersD†α, t†, u†, t0, u0 from the definition of limit levels α > sup(C ′

α).

The following lemma is key to the verification of property (#).

4.3.7 Lemma. Suppose that

D†α = {(t, u) ∈ T<α ⊗ T<α : t 6E u}

where E is a closed congruence relation on T . Then either

eα,α0(t0) = eα,γ(t†) 6E eα,γ(u

†) = eα,α0(u0)

or else t E u for all t, u ∈ T<α such that (t, u) D (t0, u0) and δ(t, u) = δ(t0, u0).

proof: Suppose that there are (t, u) ∈ T<α⊗T<α such that (t, u) D (t0, u0), δ(t, u) =

δ(t0, u0), and t 6E u. Without loss of generality we may assume that η = ht(t, u) ∈ Cα.

79

By (H2) there are extender maps e0, . . . , e2r : Tα0 → Tη such that

t = e0(t0), e0(u0) = e1(u0),

e1(t0) = e2(t0), e2(u0) = e3(u0),

......

e2r−1(t0) = e2r(t0), e2r(u0) = u.

Since t 6E u and E is transitive there must be a s ∈ {0, . . . , 2r} such that es(t0) 6E

es(u0). It follows that (es(t0), es(u0)) ∈ D†α and, by our construction, this implies that

α1 and f1,0 have been chosen such a way that (f1,0(t0), f1,0(u0)) ∈ D†α or, equivalently.

f1,0(u0) 6E f1,0(u0). Since f1,0(t0) / eα,α0(t0) and f1,0(u0) / eα,α0(u0) it follows that

eα,α0(t0) 6E eα,α0(u0) as required. qed

verification of property (#): Let E be a closed congruence relation on T . Let

〈Nα : α < κ+〉 be a continuous increasing sequence of elementary submodels of Hλ (for

some large enough λ) such that {E, T, δ,K}∪κ ⊆ N0 and |Nα| = κ for every α < κ+.

Let C be the closed unbounded set consisting of all α < κ+ such that Nα ∩ κ+ = α.

By �� κ+ there is an unbounded set S ⊆ Sκ = {α < κ+ : cf(α) = κ} such that if α ∈ S

then C ′α ⊆ C and

D†α = {(t, u) ∈ T<α ⊗ T<α : t 6E u}

for every α ∈ C ′α. We claim that S is as required for property (#).

Suppose that α ∈ S, t∗, u∗ ∈ Tα, and t∗ E u∗. We must show that there are

(t0, u0) / (t∗, u∗) such that δ(t0, u0) = δ(t∗, u∗) and t′ E u′ for all (t′, u′) D (t0, u0)

with δ(t′, u′) = δ(t0, u0). Let γ ∈ Cα and t†, u† ∈ Tγ be such that eα,γ(t†) = t∗ and

eα,γ(u†) = u∗. Say that γ is the β-th element of Cα and t†, u† are the τ -th and υ-th

elements of Tγ respectively. Let α0 and α be the 〈β, τ, υ〉-th and (〈β, τ, υ〉 + 1)-th

80

elements of C ′α. Note that t† and u† are the parameters used for the construction of

level α.

By Lemma 4.3.7, the construction of level α ensured that either eα,γ(t†) 6E eα,γ(u

†)

or else that t E u for all t, u ∈ T<α such that (t, u) D (t0, u0) and δ(t, u) = δ(t0, u0)

where t0 = eα0,γ(t†) and u0 = eα0,γ(u

†). Since α ∈ C ′α it follows from (C2) that

eα,γ(t†) / eα,γ(t

†) = t∗ and eα,γ(u†) / eα,γ(u

†) = u∗. Since t∗ E u∗ we must have that

eα,γ(t†) E eα,γ(u

†). Therefore, it must be the case that t E u for all t, u ∈ T<α such

that (t, u) D (t0, u0) and δ(t, u) = δ(t0, u0). Now t0, u0 ∈ Nα and T<α = T ∩ Nα,

so Nα |= ‘t E u for all (t, u) D (t0, u0) with δ(t, u) = δ(t0, u0)’. Since Nα ≺ Hλ the

same is true in Hλ and hence we really have that t E u for all (t, u) D (t0, u0) with

δ(t, u) = δ(t0, u0). Thus (t0, u0) / (t∗, u∗) are as required by property (#). qed

Finally, we verify that T is a Souslin tree.

4.3.8 Lemma. Suppose that

D†α = {(t, u) ∈ T<α ⊗ T<α : (∃t′ E t)(t′ ∈ A)}

where A ⊆ T is arbitrary. Then either eα,γ(t†) = eα,α0(t0) extends an element of A

or else t0 has no extension in A ∩ T<α.

proof: Indeed, suppose that t0 has an extension in A∩T<α. Then there is a γ ∈ Cα

and t ∈ Tγ such that t extends an element of A. By (H1) there is an extender map

f : Tα0 → Tγ such that f(t0) = t. It follows that (f(t0), f(u0)) ∈ D†α. So by the

construction α1 and f1,0 were chosen such that (f1,0(t0), f1,0(u0)) ∈ D†α. This implies

that f1,0(t0) / eα,α0(t0) extends an element of A as required. qed

verification that T is Souslin: LetA be a maximal antichain in T . Let 〈Nα : α < κ+〉

be a continuous increasing sequence of elementary submodels of Hλ (for some large

81

enough λ) such that {A, T, δ,K} ∪ κ ⊆ N0 and |Nα| = κ for every α < κ+. Let C be

the closed unbounded set consisting of all α < κ+ such that Nα ∩ κ+ = α. By �� κ+ ,

there is a α < κ+ such that cf(α) = κ, C ′α ⊆ C, and

D†α = {(t, u) ∈ T<α ⊗ T<α : (∃t′ E t)(t′ ∈ A)}

for every α ∈ C ′α. We claim that A ⊆ T<α or, equivalently, that every element of Tα

extends an element of A.

Suppose for the sake of contradiction that t∗ ∈ Tα doesn’t extend an element

of A. Let γ ∈ Cα and t ∈ Tγ be such that eα,γ(t) = t∗. Say that γ is the β-th

element of Cα and t is the τ -th elements of Tγ. Let α0 and α be the 〈β, 〈τ, 0〉〉-th and

(〈β, 〈τ, 0〉〉+ 1)-th elements of C ′α. Note that t = t† for the construction of level α.

By Lemma 4.3.8, the construction of Tα ensured that either eα,γ(t) extends an

element of A or else that t0 = eα0,γ(t) has no extension in A ∩ T<α. Since α ∈ C ′α it

follows from (C2) that eα,α0(t0) = eα,γ(t) / eα,γ(t) = t∗. So eα,γ(t) cannot extend an

element of A. it follows that t0 has no extension in T<α ∩ A (and that t0 does not

already extend an element of A). Now t0 ∈ Nα and T<α ∩ A = A ∩Nα, so Nα |= ‘t0

has no extension in A’. Since Nα ≺ Hλ the same is true in Hλ and hence we really

have that t0 has no extension in A, but this contradicts the fact that A is a maximal

antichain in T . qed

82

Appendix A

Some Set Theory

In this section, we briefly review necessary set theory background. All of the material

here can be found in the general references [13] and [16]; the more specialized reference

[3] is also useful.

A.1 Forcing

In this section, we will briefly outline the basics of forcing. The reader who knows

forcing can safely skip this section. Note, however, that our ordering convention are

opposite that of [13] and [16] in that p ≤ q here means that q is a stronger condition

than p.

Let P be a partially ordered set in the ground model V . In the context of forcing, P

will often be called a notion of forcing and elements of P will be called conditions.

Let X ⊆ P . We say that:

• X is open iff p ≤ q and p ∈ X always entails q ∈ X.

• X is dense iff for every p ∈ P there is a q ∈ X such that p ≤ q.

83

• X is dense above p0 iff for every p ≥ p0 there is a q ∈ X such that p ≤ q.

• X is a filter1 iff X satisfies the following three conditions

X 6= ∅,

p ∈ X & q ≤ p =⇒ q ∈ X,

p, q ∈ X =⇒ (∃r ∈ X)(p, q ≤ r).

With these basic definitions in mind, we are ready to define the first fundamental

object of forcing.

We say that a set G ⊆ P is a V -generic filter over P is G is a filter over P and

G∩D 6= ∅ for every open dense set D ⊆ P in the ground model V . Note that if P has

no maximal elements, then G cannot be in the ground model since the complement

of any filter in P is an open dense set in P . (If P has a maximal element p, then

{q ∈ p : q ≤ p} is a V -generic filter.)

We will now see that given a V -generic filter over P , there is an simple way to

construct an extension of V which containsG and satisfies ZFC. The class of P -names

V P is defined by induction on rank as follows: a set τ is a P -name iff every element of

τ has the form (p, σ) where p ∈ P and σ is a P -name. For example, ∅ is a P -name,

and if p ∈ P then so are {(p,∅)} and {(q, {(p,∅)}) : (∀r ∈ P )(r � q ∨ r � p)}.

If G is a V -generic filter over P , then we define the evaluation of the P -name τ

at G by induction on rank using the rule

τ [G] = {σ[G : ]} (p, σ) ∈ τ ∧ p ∈ G.1We really should say ‘ideal’ instead of ‘filter’ given our ordering convention, but we prefer not

to deviate too much from the standard forcing terminology.

84

For example, ∅[G] = ∅, {(p,∅)} [G] = {∅} if p ∈ G and {(p,∅)} [G] = ∅ if p /∈ G.

It is a fun exercise to evaluate {(q, {(p,∅)}) : (∀r ∈ P )(r � q ∨ r � p)}. (Note that if

(∀r ∈ P )(r � p ∨ r � q) then p and q cannot both be in G.)

We write V [G] for the class of all evaluations τ [G] of P -names τ . The following

proposition shows that V [G] agrees with our previous usage of the notation V [G] in

Section 1.3.

A.1.1 Proposition. Suppose that G is a V -generic filter over P .

(1) V ⊆ V [G] and G ∈ V [G].

(2) V [G] is transitive and OrdV [G] = OrdV .

(3) If V ∪ {G} ⊆ W , W is transitive and closed under primitive recursive functions,

then V [G] ⊆ W .

We will see later that V [G] is in fact a model of ZFC, but first we will introduce the

second fundamental object of forcing.

For a partial order P , the forcing relation p φ(τ1, . . . , τn) is defined as follows.

Atomic Relations: We define p σ R τ for R ∈ {∈, /∈,=, 6=} by simultaneous

induction on the rank of σ and τ :

• p σ ∈ τ iff the set

{q ∈ P : (∃(π, r) ∈ τ)(q ≥ r ∧ q σ = π)}

is dense above p.

• p σ /∈ τ iff there is no q ≥ p such that q σ ∈ τ .

• p σ = τ iff there is no q ≥ p such that q σ 6= τ .

85

• p σ 6= τ iff the set

{q ∈ P : (∃(π, r) ∈ σ ∪ τ)(q ≥ r ∧ (q π /∈ σ ∨ q π /∈ τ)))}

is dense above p.

Note that the second and third clauses agree with the definition of negation

below.

Logical Connectives: Logical connectives, ∧ and ∧ are handled as follows:

• p ¬φ(τ1, . . . , τn) iff there is no q ≥ p such that q φ(τ1, . . . , τn).

• p (φ ∧ ψ)(τ1, . . . , τn) iff p φ(τ1, . . . , τn) and p ψ(τ1, . . . , τn).

Other connectives (∨, →, ↔) can be defined in the in terms of the above two

in the usual way.

Quantifiers: Existential and universal quantifiers are handled as follows:

• p (∀v)φ(v, τ1, . . . , τn) iff p φ(σ, τ1, . . . , τn) for every P -name σ.

• p (∃v)φ(v, τ1, . . . , τn) iff the set

{q ∈ P : (∃σ)(q φ(σ, τ1, . . . , τn))}

is dense above p.

It is an instructive exercise to verify that p (∀v)¬φ(v) ⇐⇒ p ¬(∃v)φ(v).

Although the above definition of the forcing relation is quite complex, it is easy to

see that that p φ has the same complexity as φ in the Levy hierarchy. (Note that

(∀p ∈ P ) and (∃p ∈ P ) are bounded quantifiers since P is a set.)

86

The Truth Lemma is the fundamental result of forcing.

A.1.2 The Truth Lemma. If G is a V -generic filter over P , then

(V [G] |= φ(τ1[G], . . . , τn[G])) ⇐⇒ (∃p ∈ G)(p φ(τ1, . . . , τn))

for any formula φ and P -names τ1, . . . , τn ∈ V .

Note that the relation p φ(τ1, . . . , τn) is definable in V . So we have an effective way

of predicting within V which sentences may be true or false in the generic extension

V [G]. With some tedious computations, it is possible to show that if φ is an axiom

of ZFC, then p φ for every p ∈ P . Therefore, V [G] is always a model of ZFC.

Certain combinatorial properties of the partial order P can help us predict certain

features of the generic extension V [G]. We say that P has the κ-Baire property if

the intersection of any family of fewer than κ open dense subsets of P is again open

dense in P .

A.1.3 Proposition. Suppose that V |= ‘P has the κ-Baire property’ and that G is

a V -generic filter over P . Then any sequence of ordinals with length less than κ in

V [G] was already in V . In particular, every V -cardinal up to and including κ remains

a cardinal in V [G] and every limit ordinal with cofinality less than κ in V has the

same cofinality in V [G].

We say that p, q ∈ P are incompatible if there is no r ∈ P such that r ≥ p, q. A

set A ⊆ P is an antichain if any two elements of A are incompatible. Note that in a

tree, incompatibility and incomparability agree, so this definition of antichain agrees

with the one given in Section 1.2. We say that P satisfies the κ-chain condition2 if

every antichain of P has size less than κ.

2We should really say the ‘κ-antichain condition,’ but this terminology (which comes from Booleanalgebras) is now standard.

87

A.1.4 Proposition. Suppose that V |= ‘P satisfies the κ-chain condition’ and that

G is a V -generic filter over P . Then every V -cardinal from κ on remains a cardinal

in V [G] and any limit ordinal of cofinality κ or higher in V has the same cofinality

in V [G].

Note that a Souslin tree of height κ has the κ-Baire property and satisfies the κ-chain

condition. So all cardinals and cofinalities are preserved in the generic extension after

forcing with a Souslin tree.

A.2 Combinatorial Principles

In this section, we will review some combinatorial principles that we need for the

constructions of Chapter 4.

If κ is a regular cardinal, let Hκ denote the collection of all sets with transitive

closure of size less than κ. The following elementary lemma is very useful for inductive

constructions such as those of Chapter 4.

A.2.1 Lemma. Assume that κ<κ = κ and 2κ = κ+, and let 〈Xα : α < κ+〉 be a

sequence of elements of Hκ+. There is a wellordering <H of Hκ+ with order type

κ+ and a sequence 〈Mα : α < κ+〉 such that (Mα,∈, <H) is a transitive elementary

submodel of (Hκ+ ,∈, <H) with |Mα| = κ, M<κα ⊆ Mα, and Xα, 〈Mβ : β < α〉 ∈ Mα

for every α < κ+.

proof: First observe that there is a wellordering <H of Hκ+ with order type κ+ such

that x ∈ y =⇒ x <H y for all x, y ∈ Hκ+ . To see this, let f : Hκ+ → κ+ be any

injection. Define the function f : Hκ+ → κ+ by ∈-recursion using the rule

f(x) = sup({f(x)} ∪ {f(y) + 1: y ∈ x}).

88

Note that f(x) ≤ f(x) and x ∈ y =⇒ f(x) < f(y) for all x, y ∈ Hκ+ . Define

x <H y ⇐⇒ f(x) < f(y) or else f(x) = f(y) & f(x) < f(y).

It is clear that <H is a wellordering of Hκ+ . Moreover, tp(Hκ+ , <H) = κ+ since

x <H y =⇒ f(x) < f(y) for all x, y ∈ Hκ+ .

We consider the structure (Hκ+ ,∈, <H). For each formula φ(v1, . . . , vn) of the

first-order language L(∈, <H) (with all free variables shown), let hφ : Hnκ+ → Hκ+ be

the Skolem function defined by

hφ(x) = y ⇐⇒ (Hκ+ ,∈, <H) |= φ(x, y) ∧ (∀z)(z <H y → ¬φ(x, z)).

For each α < κ+, let Mα be the smallest initial segment of (Hκ+ , <H) such that

cf(Mα, <H) = κ, Xα, 〈Mβ : β < α〉 ∈ Mα, and hφ[Mnα ] ⊆ Mα for every Skolem func-

tion hφ as above. Then (Mα,∈, <H , P ) is an elementary submodel of (Hκ+ ,∈, <H , P )

with |Mα| = κ. Mα is transitive since (Mα, <H) is an initial segment of (Hκ+ , <H) and

x ∈ y =⇒ x <H y. Note that Mα is closed under the map x 7→ x<κ by elementarity,

so M<κα ⊆Mα since cf(Mα, <H) = κ. qed

Let S ⊆ κ. A diamond sequence for S is a sequence 〈Xα : α ∈ S〉 such that

if X ⊆ κ and C ⊆ κ is closed unbounded, then there is a α ∈ C ∩ S such that

X ∩ α = Xα. The assertion that there is a diamond sequence for S is denoted

♦κ(S). The statement ♦κ(κ) is abbreviated ♦κ. Note that ♦κ implies that κ<κ = κ

since [κ]<κ = {Xα : α < κ}. In particular, it follows that κ is a regular uncountable

cardinal.

A.2.2 Proposition. (Jensen; see [3]) Assume V = L. Then ♦κ holds for every

uncountable regular cardinal κ.

89

A square sequence for S ⊆ η is a sequence 〈Cα : α ∈ η ∩ Lim〉 such that

• Cα is a closed unbounded subset of α, and

• if β ∈ C ′α then β /∈ S and Cβ = Cα ∩ β

for every limit α < η. The assertion that there is a square sequence for S ⊆ η is

denoted �η(S). If �η(S) holds then S ⊆ η is nonreflecting since C ′α ∩ S = ∅ for

every limit α < η. The prototypical example of a stationary nonreflecting set is the

set Sκ = {α < κ+ : cf(α) = κ} where κ is an infinite regular cardinal. The statement

�κ+(Sκ) is abbreviated �κ+ .3 Note that 〈Cα : α ∈ κ+ ∩ Lim〉 is a �κ+-sequence if and

only if

• Cα is a closed unbounded subset of α with tp(Cα) ≤ κ, and

• if β ∈ C ′α then Cβ = Cα ∩ β

for every limit α < κ+. Thus �κ+ is the traditional square principle as formulated by

Jensen.

A.2.3 Proposition. (Jensen; see [3]) Assume V = L. Then �κ+ holds for every

regular infinite cardinal κ.

A built-in diamond sequence for the �η(S)-sequence 〈Cα : α ∈ η ∩ Lim〉 is a

sequence 〈Xα : α ∈ η ∩ Lim〉 such that

• if β < α are limit ordinals and β ∈ C ′α then Xβ = Xα ∩ β, and

• if X ⊆ η and C ⊆ η is closed unbounded, then there is a α ∈ C ∩ S such that

C ′α ⊆ C and Xα = X ∩ α.

3Note that many authors, e.g. [3], use �κ for �κ+ and �κ(S) for what we would call �κ+(S∪Sκ).

90

The assertion that there is a square sequence with built-in diamond sequence for

S ⊆ η is denoted �� η(S). The assertion �� κ+(Sκ) is abbreviated �� κ+ .

A.2.4 Proposition. (Abraham–Shelah–Solovay [1]) Assume V = L. Then �� κ+

holds for every infinite regular cardinal κ.

91

Appendix B

Lattices and Semi-Lattices

This appendix reviews some lattice theory, including complete, continuous, and alge-

braic lattices. All of the material here and much more can be found in [7] and [5], for

example.

B.1 Lattices and Semi-Lattices

A lattice is a partial order L in which every two elements ξ and ζ have a least upper

bound, which is denoted ξ ∨ ζ and is called the join of ξ and ζ, and a greatest lower

bound, which is denoted ξ ∧ ∨ and is called the meet of ξ and ζ. Alternatively, we

could define a lattice as an algebraic structure with two binary operations ∧ and ∨

which satisfy the following axioms

ξ ∧ ξ = ξ ξ = ξ ∨ ξ

ξ ∧ ζ = ζ ∧ ξ ξ ∨ ζ = ζ ∨ ξ

ξ ∨ (ζ ∨ ρ) = (ξ ∨ ζ) ∨ ρ ξ ∧ (ζ ∧ ρ) = (ξ ∧ ζ) ∧ ρ

ξ ∧ (ξ ∨ ζ) = ξ ξ = (ζ ∧ ξ) ∨ ξ

92

for all ξ, ζ, ρ ∈ L. It is an easy exercise to verify that these axioms imply that

ξ ≤ ζ ⇐⇒ ξ ∨ ζ = ζ ⇐⇒ ξ = ξ ∧ ζ is a partial ordering and that ξ ∨ ζ and ξ ∧ ζ are

the least upper bound greatest lower bound of ξ and ζ respectively with respect to

this ordering.

A lattice L can have a least element, which is denoted 0L or simply 0 and is called

the zero of L, and it can have a greatest element, which is denoted 1L or simply 1

and is called the unit of L. These elements, when they exist, are characterized by

the following identities:

ξ ∨ 0 = ξ, ξ ∧ 0 = 0,

ξ ∧ 1 = ξ, ξ ∨ 1 = 1.

A join semi-lattice (resp. meet semi-lattice) is a partial ordering K where

every two elements ξ and ζ have a least upper bound ξ ∨ ζ (resp. greatest lower

bound ξ ∧ ζ). Algebraically, any idempotent, commutative, and associative binary

operation ◦ on a set defines a semi-lattice; this semi-lattice can either be a join or a

meet semi-lattice depending on whether one defines ξ ≤ ζ as ξ = ξ ◦ ζ or ξ ◦ ζ = ζ

respectively. If a semi-lattice K has an identity element, then it is denoted 0 or 1

and it is called zero or unit, depending on whether K is a join or meet semi-lattice

respectively.

A distributive lattice is a lattice L which satisfies the two distributive laws:

ξ ∧ (ζ ∨ ρ) = (ξ ∧ ζ) ∨ (ξ ∧ ρ), ξ ∨ (ζ ∧ ρ) = (ξ ∨ ζ) ∧ (ξ ∨ ρ).

It is a fun exercise to show that the two distributive laws above are each equivalent

93

to the single identity

(ξ ∨ ζ) ∧ (ζ ∨ ρ) ∧ (ρ ∨ ξ) = (ξ ∧ ζ) ∨ (ζ ∧ ρ) ∨ (ρ ∧ ξ).

An easy consequence of either distributive law is the modular law:

ξ ≤ ζ =⇒ ξ ∨ (ρ ∧ ζ) = (ξ ∨ ρ) ∧ ζ.

A lattice in which the modular law holds is called a modular lattice. So every

distributive lattice is modular. The lattices N5 and M3 illustrated in Figure B.1 show

that not every lattice is modular and that not every modular lattice is distributive

respectively. In fact, these lattices completely characterize the classes of modular and

distributive lattices.

B.1.1 Proposition. ([7], II.1.2)

(1) A lattice is modular if and only if it contains no sublattice isomorphic to the

lattice N5.

(2) A modular lattice is distributive if and only if it contains no lattice isomorphic

to the lattice M3.

N5 •1

•0

•α

•β•γ

wwwww??

????

?

GGGGG�������

M3 •1

•0

•α •β • γ

JJJJJJJJJ

ttttttttt

ttttttttt

JJJJJJJJJ

Figure B.1: The lattices N5 and M3.

94

B.2 Complete Lattices

A complete lattice is a partial order L in which every set S has a least upper bound,

denoted∨S and called the join of S, and a greatest lower bound, denoted

∧S and

is called the meet of S. Clearly, a complete lattice is a lattice where ξ ∨ ζ =∨{ξ, ζ}

and ξ ∧ ζ =∧{ξ, ζ} for all ξ, ζ ∈ L. A complete lattice always has a zero and unit

since 0 =∧

L =∨

∅ and 1 =∨

L =∧

∅. Also note that any partial order L in

which every set S has a greatest lower bound (resp. least upper bound) is a complete

lattice since the least upper bound of S is the greatest lower bound of the set of all

upper bounds of S (resp. least upper bound of the set of lower bounds of S).

A complete lattice L is said to be meet-continuous if

ξ ∧∨

J =∨

(ξ ∧ J)

for every ideal J of L where ξ ∧ J = {ξ ∧ ζ : ζ ∈ J}. Dually, we say that L is join-

continuous if

ξ ∨∧

F =∧

(ξ ∨ F )

for every filter F of L where ξ ∨ F = {ξ ∨ ζ : ζ ∈ F}.

B.2.1 Proposition. ([5], O-4.3) Let L be a complete lattice.

(1) The infinite distributive law

ξ ∨∧i∈I

ζi =∧i∈I

(ξ ∨ ζi)

holds if and only if L is distributive and meet-continuous.

95

(2) The infinite distributive law

ξ ∧∨i∈I

ζi =∨i∈I

(ξ ∧ ζi)

holds if and only if L is distributive and join-continuous.

A common way of defining new complete lattices from old is through closure and

kernel operators. A closure operator is an order-preserving map c : L → L such

that

ξ ≤ c(ξ) = c(c(ξ))

for all ξ ∈ L. Dually, a kernel operator is an order-preserving map k : L → L such

that

ξ ≥ k(ξ) = k(k(ξ))

for all ξ ∈ L.

B.2.2 Proposition. ([5], O-3.12)

(1) If c : L → L is a closure operator, then the range Lc of c is closed under arbitrary

meets in L and thus forms a complete lattice with the induced ordering. Meets in

Lc agree with meets in L, but joins are given by the equation∨cX = c(

∨X) for

every set X ⊆ Lc.

(2) If k : L → L is a kernel operator, then the range Lk of k is closed under arbitrary

joins in L and thus forms a complete lattice with the induced ordering. Joins in

Lk agree with joins in L, but meets are given by the equation∧k X = c(

∧X) for

every set X ⊆ Lk.

96

B.3 Algebraic and Continuous Lattices

Let K be a join semi-lattice with zero. A set J ⊆ K is an ideal if:

• 0 ∈ J ,

• ξ ≤ ζ & ζ ∈ J =⇒ ξ ∈ J , and

• ξ, ζ ∈ J =⇒ ξ ∨ ζ ∈ J .

For example, for every ξ ∈ K, the principal ideal (ξ) = {ζ ∈ K : ζ ≤ ξ} is an ideal.

The set of all ideals of K is denoted Id(K). This set is a complete lattice under

inclusion. Indeed, it is easy to check that the set-theoretic intersection of any family

of ideals of K is again an ideal of K. The join of a directed family of ideals of is simply

the set-theoretic union of the family. Finally, the join of two ideals J and K of K is

given by

J ∨K = {ξ ∨ ζ : ξ ∈ J, ζ ∈ K} .

An element ξ of a complete lattice L is said to be compact if for every set S

we have∨S ≥ ξ if and only if

∨F ≥ ξ for some finite set F ⊆ S. For example,

0 is compact in every complete lattice. Also, it is easy to check that the compact

elements of Id(K) are precisely the principal ideals. A complete lattice L is algebraic

if ζ =∨{ξ ≤ ζ : ξ compact} for every ζ ∈ L.

It is a trivial exercise to check that Id(K) is an algebraic lattice for every join

semi-lattice K with zero. In fact, every algebraic lattice is of this form.

B.3.1 Proposition. ([5], I-4.10) If L is an algebraic lattice, then the compact ele-

ments K(L) of L form a join semi-lattice with 0 and the join map∨

: Id(K(L)) → L

is a complete lattice isomorphism.

97

Algebraic lattices are special cases of continuous lattices. Let L be a complete

lattice and let ξ, ζ ∈ L. We say that ξ is way-below ζ, which is denoted ξ � ζ, if

for every set S ⊆ L, if∨S ≥ ζ then

∨F ≥ ξ for some finite set F ⊆ S.

B.3.2 Proposition. ([5], I-1.2) For all ξ, ζ, ρ ∈ L, the following are true:

(1) 0 � ξ

(2) ξ � ρ & ζ � ρ =⇒ ξ ∨ ζ � ρ

(3) ξ ≤ ζ & ζ � ρ =⇒ ξ � ρ

(4) ξ � ζ & ζ ≤ ρ =⇒ ξ � ρ

The first three items of the previous proposition say that the set 〈ξ〉 = {ζ ∈ L : ζ � ξ}

is an ideal of L. In fact, it is easy to see that

〈ξ〉 =⋂{J ∈ Id(L) :

∨J ≥ ξ} .

A continuous lattice is a complete lattice L which satisfies the axiom of approx-

imation:

ξ =∨ζ�ξ

ζ =∨〈ξ〉

In other words, 〈ξ〉 is the smallest ideal J ⊆ L such that∨J = ξ.

The reader is advised not to confuse the concepts of continuity, meet-continuity,

and join-continuity. However, we note the following useful fact.

B.3.3 Proposition. ([5], I-1.8) A continuous lattice is meet-continuous.

proof: Let L be a continuous lattice. To show that L is meet-continuous we need

to show that ξ∧∨J ≤

∨(ξ∧J) for every ideal J of L. We may assume that ξ ≤

∨J

98

since ξ ∧J = ξ′∧J where ξ′ = ξ ∧∨J . Now ξ ≤

∨J implies that 〈ξ〉 ⊆ J . It follows

that 〈ξ〉 = ξ ∧ 〈ξ〉 ⊆ ξ ∧ J and hence that ξ ≤∨

(ξ ∧ J). qed

It is a fun exercise to show that if L is an algebraic lattice then ζ1 � ζ2 if and

only if there is a compact element ξ of L such that ζ1 ≤ ξ ≤ ζ2. It is then easy to see

that every algebraic lattice satisfies the axiom of approximation. The prototypical

example of a continuous lattice which is not algebraic is the real unit interval [0, 1].

Indeed, it is easy to see that the way-below relation on [0, 1] is ξ � ζ if and only if

ξ = 0 or ξ < ζ. The following proposition shows how to produce more examples of

continuous lattices.

B.3.4 Proposition. ([5] I-2.11)

(1) If {Li}i∈I is a family of continuous lattices, then the product∏

i∈I Li is also a

continuous lattice and the way-below relation is given by ξ � ζ if and only if

ξi � ζi for all i ∈ I and ξi = 0 for all but finitely many i ∈ I.

(2) If L is a continuous lattice and L′ ⊆ L is closed under arbitrary meets and directed

joins, then L′ is also a continuous lattice (but not necessarily a sublattice) with

the induced ordering.

(3) If L is a continuous lattice, L′ is a partial ordering, and h : L → L′ is a surjective

homomorphism which preserves arbitrary meets and directed joins, then L′ is also

a continuous lattice.

99

Appendix C

Semi-Lattice Valued Distances

C.1 Distances and Metric Spaces

Let K be a join semi-lattice with zero. A (K-valued) predistance on the set X is

a map d : X ×X → K such that

d(x, x) = 0,

d(x, y) = d(y, x),

d(x, z) ≤ d(x, y) ∨ d(y, z)

for all x, y, z ∈ X. The last of the three conditions above is called the triangle

inequality; by permuting x, y, z cyclically we see that this condition is equivalent to

the triangle identity:

d(x, y) ∨ d(y, z) = d(y, z) ∨ d(z, x) = d(z, x) ∨ d(x, y).

100

We say that d is a (K-valued) distance if the first condition can be strengthened

to d(x, y) = 0 ↔ x = y. A K-valued (pre)metric space is a set X endowed with

a K-valued (pre)distance dX : X ×X → K.

If X and Y are K-valued premetric spaces, an isometry from X to Y is a map

e : X → Y such that dY (e(x), e(y)) = dX(x, y) for all x, y ∈ X. Note that if X is a

K-valued metric space, then any isometry with domain X is injective.

Let X be a premetric space. If J is an ideal on K, then the relation

x EJ y ⇐⇒ dX(x, y) ∈ J

is an equivalence relation on X; x EJ x since dX(x, x) = 0, x EJ y ⇐⇒ y EI x since

dX(x, y) = dX(y, x), and x EJ y & y EJ z =⇒ x EJ z since dX(x, y), dX(y, z) ∈ J

implies that dX(x, y)∨dY (y, z) ∈ J and hence dX(x, z) ∈ J by the triangle inequality.

When I = (ξ) we write simply Eξ instead of the cumbersome E(ξ).

We say that the premetric space X has interpolants if for all ξ, ζ ∈ K and all

x, y ∈ X, if x Eξ∨ζ y then there are z1, . . . , z2k+1 ∈ X such that

x Eξ z1 Eζ · · · Eξ z2k+1 Eζ y

(where ξ’s and ζ’s alternate) for some k < ω. If there is a fixed k < ω such that

the above holds for every choice of ξ, ζ ∈ K and x, y ∈ X, then we say that X has

(2k + 1)-interpolants; we say that X has 2k-interpolants if in addition we can

always choose z1, . . . , z2k+1 such that either x = z1 or y = z2k+1. To say that X has

interpolants is equivalent to saying that Eξ∨ζ is Eξ ∨ Eζ (the transitive closure of

Eξ ∪ Eζ), i.e. that the map ξ 7→ Eξ is a morphism of join semi-lattices. Note that

this morphism additionally preserves zero if and only if X is a metric space.

101

C.1.1 Proposition. If X is a K-valued premetric space, then the map Id(K) → PX

defined by J 7→ EJ is a morphism of continuous lattices, i.e. it preserves all meets

and all directed joins. The map additionally preserves finite joins if and only if X

has interpolants and then the map is a morphism of complete lattices, i.e. it preserves

all joins and all meets.

proof: If 〈Ji : i ∈ I〉 is an arbitrary family of ideals on K and J =⋂

i∈I Ji, then

x EJ y ⇐⇒ dX(x, y) ∈ J

⇐⇒ (∀i ∈ I)(dX(x, y) ∈ Ji)

⇐⇒ (∀i ∈ I)(x EJiy).

Similarly, if 〈Ji : i ∈ I〉 is a directed family of ideals of K and J =⋃

i∈I Ji, then

x EJ y ⇐⇒ dX(x, y) ∈ J

⇐⇒ (∃i ∈ I)(dX(x, y) ∈ Ji)

⇐⇒ (∃i ∈ I)(x EJiy).

So the map J 7→ EJ preserves arbitrary meets and directed joins.

We have already observed that X has interpolants if and only if Eξ∨ζ = Eξ ∨ Eζ

for all ξ, ζ ∈ K. Since (ξ ∨ ζ) is the join of the principal ideals (ξ) and (ζ) in Id(K),

it follows that if J 7→ EJ preserves finite joins then X has interpolants. For the

converse, suppose that X has interpolants. Let J,K be two ideals of K and recall

that

J ∨K = {θ ∈ K : (∃ξ ∈ J, ζ ∈ K)(θ ≤ ξ ∨ ζ)} .

Since we clearly have EJ ∨EK ⊆ EJ∨K , we only have to show that EJ ∨EK ⊇ EJ∨K .

102

Clearly, x EJ∨K y implies that x Eξ∨ζ y for some ξ ∈ J and ζ ∈ K. Therefore, there

are z1, . . . , z2k+1 ∈ X such that

x Eξ z1 Eζ · · · Eξ z2k+1 Eζ y

for some k < ω. It follows that

x EJ z1 EK · · · EJ z2k+1 EK y

which shows that x EJ ∨ EK y. qed

C.2 Pointed Spaces and Amalgamation

An n-pointed (pre)metric space is a (pre)metric space X with n distinguished

(but not necessarily distinct) points labelled 0X , . . . , (n − 1)X ∈ X. If X and Y are

n-pointed premetric spaces, an (n-pointed) isometry from X to Y is an isometry

f : X → Y such that f(iX) = iY for all i < n.

If X and Y are n-pointed (pre)metric spaces, an amalgam of X and Y is a

n-pointed (pre)metric space Z together with n-pointed isometries f : X → Z and

g : Y → Z such that Z = f [X] ∪ g[Y ]. Since f(iX) = iZ = g(iY ) for all i < n, a

necessary condition for X and Y to have an amalgam is that dX(iX , jX) = dY (iY , jY )

for all i, j < n. This condition is not sufficient in general; we say that a n-pointed

(pre)metric space X is amalgamable if X and Y have an amalgam for every n-

pointed premetric space Y such that dY (iY , jY ) = dX(iX , jX) for all i, j < n.

When dealing with amalgams of n-pointed premetric spaces X and Y , it is of-

ten convenient to assume that iX = i = iY for every i < n and that X ∩ Y =

{0, . . . , n− 1}. Then every amalgam of X and Y is isometric to a unique n-pointed

103

premetric space Z where Z = X ∪ Y and dZ�X2 = dX , dZ�Y 2 = dY . Observe that if

d1, d2 : Z × Z → K define two amalgams of X and Y , then the map d : Z × Z → K

defined by d(x, y) = d1(x, y) ∨ d2(x, y) for all x, y ∈ Z also defines an amalgam of X

and Y . It follows that the (isometry classes) of amalgams of X and Y naturally form

a join semi-lattice. If this lattice has a greatest element, it is called the maximal

amalgam of X and Y . When K is a lattice, we have the natural upper bound

dZ(x, y) ≤∧i<n

(dX(x, iX) ∨ dY (iY , y))

whenever x ∈ X and y ∈ Y . If Z achieves this upper bound for all x ∈ X and y ∈ Y ,

then we say that Z is the canonical amalgam of X and Y . Clearly, the canonical

amalgam, when it exists, is also the maximal amalgam of X and Y .

C.2.1 Proposition. Any two K-valued 1-pointed premetric spaces are canonically

amalgamable.

proof: Let X and Y be 1-pointed premetric spaces. We may assume as above

that 0X = 0 = 0Y and X ∩ Y = {0}. Write Z = X ∪ Y , let 0Z = 0, and define

dZ : Z × Z → K as follows

dZ(z1, z2) =

dX(z1, z2) when z1, z2 ∈ X,

dY (z1, z2) when z1, z2 ∈ Y ,

dX(z1, 0) ∨ dY (0, z2) when z1 ∈ X and z2 ∈ Y ,

dY (z1, 0) ∨ dX(0, z2) when z1 ∈ Y and z2 ∈ X.

Some of the above cases overlap when z1 = 0 or z2 = 0, but it is easy to check that

no conflict arises from this overlap. It is also easy to check that dZ(z1, z1) = 0 and

dZ(z1, z2) = dZ(z2, z1) for all z1, z2 ∈ Z.

104

To check that dZ is a predistance, we only have to verify that dZ satisfies the

triangle identity

dZ(z1, z2) ∨ dZ(z2, z3) = dZ(z2, z3) ∨ dZ(z3, z1) = dZ(z3, z1) ∨ dZ(z1, z2)

for all z1, z2, z3 ∈ Z. Now either two of the three points are in X or two of the three

points are in Y . By symmetry, we may suppose that z1, z3 ∈ X so that dZ(z1, z3) =

dX(z1, z3). If we also have z2 ∈ X, then the result follows from the fact that the

triangle identity holds in X. Suppose that z2 ∈ Y , so dZ(z1, z2) = dX(z1, 0)∨dY (0, z2)

and dZ(z2, z3) = dY (z2, 0) ∨ dX(0, z3). Remembering that

dX(z1, 0) ∨ dX(0, z3) = dX(0, z3) ∨ dX(z3, z1) = dX(z3, z1) ∨ dX(z1, 0)

we have

dZ(z1, z2) ∨ dZ(z2, z3) = [dX(z1, 0) ∨ dY (0, z2)] ∨ [dY (z2, 0) ∨ dX(0, z3)]

= dY (z2, 0) ∨ dX(z1, 0) ∨ dX(0, z3) ∨ dX(z3, z1),

dZ(z2, z3) ∨ dZ(z3, z1) = [dY (z2, 0) ∨ dX(0, z3)] ∨ dX(z3, z1)

= dY (z2, 0) ∨ dX(0, z3) ∨ dX(z3, z1) ∨ dX(z1, 0),

dZ(z3, z1) ∨ dZ(z1, z2) = dX(z3, z1) ∨ [dX(z1, 0) ∨ dY (0, z2)]

= dY (z2, 0) ∨ dX(z3, z1) ∨ dX(z1, 0) ∨ dX(0, z3).

Combining the above we obtain the triangle identity for z1, z2, z3. qed

A n-pointed premetric spaceX is said to be split if there is a partitionX0, . . . , Xn−1

of X such that iX ∈ Xi and if x ∈ Xi then dX(x, jX) = dX(iX , jX) ∨ dX(x, iX) for

j < n. (Note that the condition iX ∈ Xi is necessary if dX(iX , jX) 6= 0 when

105

i < j < n.)

C.2.2 Proposition. Every split K-valued n-pointed space is canonically amalgam-

able.

proof: Let X be a split n-pointed space and let X0, . . . , Xn−1 be a splitting of X.

Let Y be any n-pointed space such that dX(iX , jX) = dY (iY , jY ) for all i, j < n. We

may assume that iX = i = iY for i < n and that X ∩ Y = {0, . . . , n− 1}. Define the

n-pointed space Z where Z = X ∪ Y , iZ = i for i < n, and

dZ(z1, z2) =

dX(z1, z2) when z1, z2 ∈ X,

dY (z1, z2) when z1, z2 ∈ Y ,

dX(z1, i) ∨ dY (i, z2) when z1 ∈ Xi and z2 ∈ Y ,

dY (z1, i) ∨ dX(i, z2) when z1 ∈ Y and z2 ∈ Xi.

Observe that when restricted to Xi ∪ Y , dZ is just the canonical amalgam of the

1-pointed spaces Xi and Y with distinguished point i. So, by the proof of C.2.1,

we know that the triangle inequality holds when z1, z2, z3 ∈ Xi ∪ Y for some i < n.

We also know that the triangle inequality holds when z1, z2, z3 ∈ X. It remains to

check that the triangle inequality holds when two of the points z1, z2, z3 belong to two

different sets of the splitting of X and the other belongs to Y . Suppose that z1 ∈ Xi,

z2 ∈ Y , and z3 ∈ Xj where i < j < n. Note that dZ(z1, z3) = dX(z1, z3) and

dZ(z1, z2) = dX(z1, i) ∨ dY (i, z2), dZ(z3, z2) = dX(z3, j) ∨ dY (j, z2).

106

Thus by the triangle inequality in X and Y , and the splitting of X, we find that

dZ(z3, z1) ∨ dZ(z1, z2) = dX(z3, z1) ∨ dX(z1, i) ∨ dY (i, z2)

≥ dX(z3, i) ∨ dY (i, z2)

= dX(z3, j) ∨ dX(j, i) ∨ dY (i, z2)

≥ dX(z3, j) ∨ dY (j, z2) = dZ(z3, z2).

Similarly, we find that

dZ(z1, z3) ∨ dZ(z3, z2) ≥ dZ(z1, z2).

Finally, using just the triangle inequality in X and Y , we find that

dZ(z1, z2) ∨ dZ(z2, z3) = dX(z1, i) ∨ dY (i, z2) ∨ dY (z2, j) ∨ dX(j, z3)

= dX(z1, i) ∨ dY (i, z2) ∨ dY (i, j) ∨ dY (z2, j) ∨ dX(j, z3)

≥ dX(z1, i) ∨ dX(i, j) ∨ dX(j, z3)

≥ dX(z1, z3) = dZ(z1, z3).

This completes the verification of the triangle inequality in Z. qed

C.2.3 Corollary. Suppose that ρ, ξ, ζ ∈ K are such that ρ ∨ ξ = ρ ∨ ζ = ξ ∨ ζ; then

the 2-pointed space (with distinguished points 0 and 1) illustrated in Figure C.1, where

all missing distances are ξ ∨ ζ, is amalgamable.

We will now introduce two practical operations on the class of 2-pointed premetric

spaces. The involution of a 2-pointed premetric space X is the 2-pointed space X

with the same base set and predistance but with 0 and 1 interchanged, i.e. 0 eX = 1X

107

•0

•1

•2

•3

• 4

ρ

ξ

$$$$$$$

ζ zzzzzzzz

ξ

DDDD

DDDD

ζ

�������

Figure C.1: A useful amalgamable 2-pointed premetric space.

and 1 eX = 0X . The composition of the 2-pointed premetric spaces X and Y is the

2-pointed space X ∗Y that is the canonical 1-amalgam of X and Y (seen as 1-pointed

spaces) with distinguished points 0X∗Y = 0X and 1X∗Y = 1Y . In other words, suppose

that 1X = p = 0Y and that X ∩ Y = {p}. Then X ∗ Y is the premetric space with

base set X ∪ Y , distinguished points 0X∗Y = 0X and 1X∗Y = 1Y , and predistance

dX∗Y (z1, z2) =

dX(z1, z2) when z1, z2 ∈ X,

dY (z1, z2) when z1, z2 ∈ Y ,

dX(z1, 1X) ∨ dY (0Y , z2) when z1 ∈ X and z2 ∈ Y ,

dY (z1, 0Y ) ∨ dX(1X , z2) when z1 ∈ Y and z2 ∈ X.

C.2.4 Proposition. The class of 2-pointed K-valued premetric spaces forms a monoid

under the composition operation where the identity is the trivial 2-pointed space with

only one point.

proof: It is clear that the trivial 2-pointed space is a two-sided identity for com-

position, so it suffices to show that composition is associative. So let X, Y , and Z

be 2-pointed K-valued premetric spaces. Without loss of generality 1X = p = 0Y ,

1Y = q = 0Z , X ∩ Y = {p}, Y ∩ Z = {q}, and X ∩ Z = ∅ so that we can compute

X ∗ (Y ∗ Z) and (X ∗ Y ) ∗ Z as above.

108

If x, x′ ∈ X, y, y′ ∈ Y , and z, z′ ∈ Z, then:

dX∗(Y ∗Z)(x, x′) = dX(x, x′),

d(X∗Y )∗Z(x, x′) = dX∗Y (x, x′) = dX(x, x′);

dX∗(Y ∗Z)(y, y′) = dY ∗Z(y, y′) = dY (y, y′),

d(X∗Y )∗Z(y, y′) = dX∗Y (y, y′) = dY (y, y′);

dX∗(Y ∗Z)(z, z′) = dY ∗Z(z, z′) = dZ(z, z′),

d(X∗Y )∗Z(z, z′) = dZ(z, z′).

If x ∈ X, y ∈ Y , and z ∈ Z, then:

dX∗(Y ∗Z)(x, y) = dX(x, 1X) ∨ dY ∗Z(0Y ∗Z , y) = dX(x, 1X) ∨ dY (0Y , y),

d(X∗Y )∗Z(x, y) = dX∗Y (x, y) = dX(x, 1X) ∨ dY (0Y , y);

dX∗(Y ∗Z)(y, z) = dY ∗Z(y, z) = dY (y, 1Y ) ∨ dZ(0Z , z),

d(X∗Y )∗Z(y, z) = dX∗Y (y, 1X∗Y ) ∨ dZ(0Z , z) = dY (1Y , q) ∨ dZ(0Z , z);

dX∗(Y ∗Z)(x, z) = dX(x, 1X) ∨ dY ∗Z(0Y ∗Z , z)

= dX(x, 1X) ∨ dY (0Y , 1Y ) ∨ dZ(0Z , z),

d(X∗Y )∗Z(x, z) = dX∗Y (x, 1X∗Y ) ∨ dZ(0Z , z)

= dX(x, 1X) ∨ dY (0Y , 1Y ) ∨ dZ(0Z , z).

Thus d(X∗Y )∗Z = dX∗(Y ∗Z). qed

C.2.5 Proposition. If X and Y are 2-pointed premetric spaces with dX(0X , 1X) =

dY (0Y , 1Y ), then X ∗ Y is an amalgamable 2-pointed space.

proof: Let X, Y, Z be 2-pointed spaces with dX(0X , 1X) = dY (0Y , 1Y ) = dZ(0Z , 1Z).

109

We will show how to amalgamate X ∗ Y and Z. For convenience, suppose that

1X = p = 0Y , 1Y = q = 0Z , 1Z = r = 0X , and X ∩ Y = {p}, Y ∩ Z = {q},

Z ∩X = {r}. Let W = X ∪ Y ∪ Z and define the map dW : W 2 → K by

dW (w1, w2) =

dX∗Y (w1, w2) when w1, w2 ∈ X ∪ Y ,

dY ∗Z(w1, w2) when w1, w2 ∈ Y ∪ Z,

dZ∗X(w1, w2) when w1, w2 ∈ Z ∪X.

This is a sound definition since no conflict arises from overlapping cases. For example,

if w1, w2 ∈ X then the first and third cases apply, but dX∗Y (w1, w2) = dX(w1, w2) =

dZ∗X(w1, w2).

It is clear that dW (w1, w1) = 0 and that dW (w1, w2) = dW (w2, w1) for all w1, w2 ∈

W . So to check that dW is a predistance, we only have to verify that dW satisfies

the triangle identity for all w1, w2, w3 ∈ W . If w1, w2, w3 ∈ X ∪ Y , then the triangle

identity for w1, w2, w3 follows from the fact that dX∗Y satisfies the triangle identity.

Similarly, we see that the triangle identity holds for all w1, w2, w3 ∈ Y ∪ Z and all

w1, w2, w3 ∈ Z ∪X.

The only remaining case is when we have one element from each of X, Y, Z. So

pick x ∈ X, y ∈ Y , and z ∈ Z. By definition of dX∗Y and dY ∗Z we have

dW (x, y) ∨ dW (y, z) = [dX(x, 1X) ∨ dY (0Y , y)] ∨ [dY (y, 1Y ) ∨ dZ(0Z , z)].

110

Remembering that dX(0X , 1X) = dY (0Y , 1Y ) = dZ(0Z , 1Z) and that

dX(x, 1X) ∨ dX(1X , 0X) = dX(x, 1X) ∨ dX(0X , x),

dY (0Y , y) ∨ dY (y, 1Y ) = dY (0Y , y) ∨ dY (y, 1Y ) ∨ dY (1Y , 0Y ),

dZ(1Z , 0Z) ∨ dZ(0Z , z) = dZ(z, 1Z) ∨ dZ(0Z , z),

we find that

dW (x, y) ∨ dW (y, z) = [dX(x, 1X) ∨ dX(0X , x)]∨

[dY (y, 1Y ) ∨ dY (0Y , y)] ∨ [dZ(z, 1Z) ∨ dZ(0Z , z)].

Similarly, we find that

dW (y, z) ∨ dW (z, x) =

[dX(x, 1X) ∨ dX(0X , x)] ∨ [dY (y, 1Y ) ∨ dY (0Y , y)] ∨ [dZ(z, 1Z) ∨ dZ(0Z , z)]

= dW (y, z) ∨ dW (z, x)

which establishes the triangle identity for x, y, z. qed

C.3 Amalgamation of n-Pointed Spaces

C.3.1 Proposition. If K is a lattice and every 2-pointed K-valued premetric space

is amalgamable, then K is modular.

proof: Recall that a lattice is modular if and only if it contains no copy of the

pentagonal lattice N5 (see Figure B.1). So it suffices to show that there is a pair of

non-amalgamable N5-valued premetric spaces. Consider the 2-pointed spaces X =

111

{0, 1, x} and Y = {0, 1, y} with distances as illustrated in Figure C.2. Suppose that d

X

•x

• 0

•1

αssssssssss

γ KKKKKKKKKK 1

Y

• y

•0

•1

βKKKKKKKKKK

γssssssssss1

Figure C.2: Two non-amalgamable N5-valued 2-pointed premetric spaces.

is a predistance on the set Z = {0, 1, x, y} which is an amalgam of X and Y . Observe

that

d(x, y) ≤ [d(x, 0) ∨ d(0, y)] ∧ [d(x, 1) ∨ d(1, y)] = [α ∨ β] ∧ [γ ∨ γ] = 0.

We then obtain the contradiction

α = d(0, x) ≤ d(0, y) ∨ d(y, x) ≤ β ∨ 0 = β.

We conclude that there is no such predistance exists and hence that the 2-pointed

premetric spaces X and Y are not amalgamable. qed

C.3.2 Proposition. If K is a lattice and every 3-pointed K-valued premetric space

is amalgamable, then K is a distributive lattice.

proof: Suppose that K is a lattice and every 3-pointed K-valued premetric space is

amalgamable. It follows that every 2-pointed K-valued premetric space is amalgam-

able and hence K is a modular lattice by Proposition C.3.1. Recall that a modular

lattice is distributive if and only if it contains no copy of the lattice M3 (see Fig-

ure B.1). So it suffices to show that there is a pair of non-amalgamable M3-valued

112

metric spaces. Consider the 3-pointed spaces X = {0, 1, 2, x} and Y = {0, 1, 2, y}

with distances as illustrated in Figure C.3. Suppose that d is a predistance on the

X

•x

•0

•1

•2

α

β

qqqqqqqqqqqγ

MMMMMMMMMMM

γ

����

����

����

����

���

α

β

2222222222222222222

Y

•y

•0

•1

•2

1

β

qqqqqqqqqqqγ

MMMMMMMMMMM

γ

����

����

����

����

���

α

β

2222222222222222222

Figure C.3: Two non-amalgamable M3-valued 3-pointed premetric spaces.

set Z = {0, 1, x, y} which is an amalgam of X and Y . Observe that

d(x, y) ≤ [d(x, 1) ∨ d(1, y)] ∧ [d(x, 2) ∨ d(2, y)] = β ∧ γ = 0.

We then obtain the contradiction

1 = d(0, y) ≤ d(0, x) ∨ d(x, y) ≤ α ∨ 0 = α.

We conclude that there is no such predistance exists and hence that the 3-pointed

premetric spaces X and Y are not amalgamable. qed

C.3.3 Proposition. If K is a distributive lattice, then every n-pointed premetric

space is canonically amalgamable.

proof: LetX and Y be n-pointed premetric spaces such that dX(iX , jX) = dY (iY , jY )

for all i, j < n. Without loss of generality, we may suppose that iX = i = iY for every

i < n and that that X ∩ Y = {0, . . . , n− 1}.

113

Let Z = X ∪Y , let iZ = i for i < n, and let dZ : Z ×Z → K be defined as follows

dZ(z1, z2) =

dX(z1, z2) when z1, z2 ∈ X,

dY (z1, z2) when z1, z2 ∈ Y ,∧i<n dX(z1, i) ∨ dY (i, z2) when z1 ∈ X and z2 ∈ Y ,∧i<n dY (z1, i) ∨ dX(i, z2) when z1 ∈ Y and z2 ∈ X.

(C.1)

Some of the above clauses overlap when z1 ∈ {0, . . . , n− 1} or z2 ∈ {0, . . . , n− 1},

but it is easy to check that no conflict arises from this. For example, say z1 ∈ X and

z2 = j so that the first and third clauses apply. First note that dY (i, j) = dX(i, j) for

all i < n, thus

dX(z1, j) = dX(z1, j) ∨ dY (j, j)

≥∧

i<n dX(z1, i) ∨ dY (i, j) =∧

i<n dX(z1, i) ∨ dX(i, j)

≥∧

i<n dX(z1, j) = dX(z1, j)

which shows that the definitions of clauses one and three agree.

It is also easy to check that dZ(z1, z1) = 0 and dZ(z1, z2) = dZ(z2, z1) for all

z1, z2 ∈ Z.

To check that dZ is a predistance, we only have to verify that dZ satisfies the

triangle identity

dZ(z1, z2) ∨ dZ(z2, z3) = dZ(z2, z3) ∨ dZ(z3, z1) = dZ(z3, z1) ∨ dZ(z1, z2)

for all z1, z2, z3 ∈ Z. Now either two of the three points are in X or two of the three

points are in Y . By symmetry, we may suppose that z1, z3 ∈ X so that dZ(z1, z3) =

114

dX(z1, z3). If we also have z2 ∈ X, then dZ(z1, z2) = dX(z1, z2), dZ(z2, z3) = dZ(z2, z3),

and the result follows from the triangle identity for dX . So suppose that z2 ∈ Y , then

dZ(z1, z2) =∧

i<n dX(z1, i) ∨ dY (i, z2),

dZ(z2, z3) =∧

i<n dY (z2, i) ∨ dX(i, z3).

Remembering that

dX(i, z3) ∨ dX(z3, z1) = dX(z3, z1) ∨ d(z1, i) = dX(i, z3) ∨ dX(z3, z1) ∨ d(z1, i)

for every i < n, we have

dZ(z2, z3) ∨ dZ(z3, z1) =

[∧i<n

dY (z2, i) ∨ dX(i, z3)

]∨ dX(z3, z1)

=∧i<n

dY (z2, i) ∨ dX(i, z3) ∨ dX(z3, z1)

=∧i<n

dY (z2, i) ∨ dX(i, z3) ∨ dX(z3, z1) ∨ dX(z1, i),

and similarly

dZ(z3, z1) ∨ dZ(z1, z2) = dX(z3, z1) ∨

[∧i<n

dX(z1, i) ∨ dY (i, z2)

]

=∧i<n

dX(z3, z1) ∨ dX(z1, i) ∨ dY (i, z2)

=∧i<n

dX(i, z3) ∨ dX(z3, z1) ∨ dX(z1, i) ∨ dY (i, z2).

115

Therefore

dZ(z2, z3) ∨ dZ(z3, z1) = dZ(z3, z1) ∨ dZ(z1, z2)

= dZ(z1, z2) ∨ dZ(z2, z3) ∨ dZ(z3, z1).

Also, remembering that

dY (i, z2) ∨ dY (z2, j) = dY (i, j) ∨ dY (i, z2) ∨ dY (z2, j)

and dX(i, j) = dY (i, j) for all i, j < n, we have

dZ(z1, z2) ∨ dZ(z2, z3)

=

[∧i<n

dX(z1, i) ∨ dY (i, z2)

]∨

[∧j<n

dY (z2, j) ∨ dX(j, z3)

]

=∧

i,j<n

dX(z1, i) ∨ dY (i, z2) ∨ dY (z2, j) ∨ dX(j, z3)

=∧

i,j<n

dX(z1, i) ∨ dX(i, j) ∨ dX(j, z3) ∨ dY (i, j) ∨ dY (i, z2) ∨ dY (z2, j)

≥ dX(z1, z3) = dZ(z1, z3)

which implies that

dZ(z1, z2) ∨ dZ(z2, z3) = dZ(z1, z2) ∨ dZ(z2, z3) ∨ dZ(z3, z1).

Combining the above, we obtain the triangle identity for z1, z2, z3. qed

Let us return to amalgamation of 2-pointed spaces. A modular lattice L is said

116

to be strongly Arguesian if for all π, ρ, ξ, ξ′, ξ′′, ζ, ζ ′, ζ ′′ ∈ L such that

π ∨ ξ = π ∨ ζ = ξ ∨ ζ,

π ∨ ξ′ = π ∨ ζ ′ = ξ′ ∨ ζ ′,

π ∨ ξ′′ = π ∨ ζ ′′ = ξ′′ ∨ ζ ′′,

(C.2)

and

ρ ∨ ξ′ = ρ ∨ ξ′′ = ξ′ ∨ ξ′′,

ρ ∨ ζ ′ = ρ ∨ ζ ′′ = ζ ′ ∨ ζ ′′,(C.3)

then

ρ ∨ ρ′ = ρ′ ∨ ρ′′ = ρ′′ ∨ ρ (C.4)

where

ρ′ = (ξ ∨ ξ′) ∧ (ζ ∨ ζ ′),

ρ′′ = (ξ ∨ ξ′′) ∧ (ζ ∨ ζ ′′).

Note that if any π ∈ L satisfies (C.2), then

π = (ξ ∨ ζ) ∧ (ξ′ ∨ ζ ′) ∧ (ξ′′ ∨ ζ ′′)

has that property too. Similarly, if any ρ satisfies (C.3), then

ρ = (ξ′ ∨ ξ′′) ∧ (ζ ′ ∨ ζ ′′)

has that property too. In fact, if L is modular and (C.2) holds, then the above ρ

117

always satisfies (C.3). Indeed, we have

ξ′ ∨ ρ = (ξ′ ∨ ξ′′) ∧ (ξ′ ∨ ζ ′ ∨ ζ ′′)

= (ξ′ ∨ ξ′′) ∧ (ζ ′ ∨ π ∨ ζ ′′)

= (ξ′ ∨ ξ′′) ∧ (ζ ′ ∨ ζ ′′ ∨ ξ′′) = ρ ∨ ξ′′,

and similarly ζ ′ ∨ ρ = ρ ∨ ζ ′′ which is easily seen to imply (C.3). A modular lattice

is said to be weakly Arguesian if (C.2) implies (C.4) for these choices of π and ρ

and any ξ, ξ′, ξ′′, ζ, ζ ′, ζ ′′ ∈ L.

C.3.4 Proposition. If L is a strongly Arguesian lattice, then every 2-pointed L-

valued premetric space is canonically amalgamable.

proof: LetX and Y be 2-pointed premetric spaces such that dX(0X , 1X) = dY (0Y , 1Y ).

Without loss of generality, we may suppose that 0X = 0 = 0Y and 1X = 1 = 1Y and

that that X ∩ Y = {0, 1}.

Let Z = X ∪ Y , 0Z = 0, 1Z = 1, and let dZ : Z × Z → K be defined exactly as in

equation (C.1) in the proof of Proposition C.3.3 for n = 2. As shown there, if dZ is

a predistance then Z is the canonical amalgam of X and Y .

To check that dZ is a predistance, we only have to verify that dZ satisfies the

triangle identity

dZ(z1, z2) ∨ dZ(z2, z3) = dZ(z2, z3) ∨ dZ(z3, z1) = dZ(z3, z1) ∨ dZ(z1, z2)

for all z, z′, z′′ ∈ Z such that either z ∈ X and z′, z′′ ∈ Y or z ∈ Y and z′, z′′ ∈ X.

118

The two cases being symmetric, we may assume the latter. Let

ξ = dY (z, 0), ζ = dY (z, 1),

ξ′ = dX(z′, 0), ζ ′ = dX(z′, 1),

ξ′′ = dX(z′′, 0), ζ ′′ = dX(z′′, 1),

and

π = dX(0, 1) = dY (0, 1), ρ = dX(z′, z′′) = dZ(z′, z′′).

The triangle identity in X and Y show that these choices satisfy the hypotheses (C.2)

and (C.3) of the strong Arguesian law. Therefore, the conclusion

ρ ∨ ρ′ = ρ′ ∨ ρ′′ = ρ′′ ∨ ρ

where

ρ′ = (ξ ∨ ξ′) ∧ (ζ ∨ ζ ′) = dZ(z, z′),

ρ′′ = (ξ ∨ ξ′′) ∧ (ζ ∨ ζ ′′) = dZ(z, z′′)

by definition of dZ . Therefore the triangle identity holds for z, z′, z′′. qed

It is not difficult to see that L is a strongly Arguesian lattice if and only if any two

2-pointed spaces are canonically amalgamable.

C.3.5 Proposition. If K is a lattice and every 2-pointed K-valued premetric space

is amalgamable, then K is weakly Arguesian.

119

proof: Suppose that

π ∨ ξ = π ∨ ζ = ξ ∨ ζ,

π ∨ ξ′ = π ∨ ζ ′ = ξ′ ∨ ζ ′,

π ∨ ξ′′ = π ∨ ζ ′′ = ξ′′ ∨ ζ ′′,

where

π = (ξ ∨ ζ) ∧ (ξ′ ∨ ζ ′) ∧ (ξ′′ ∨ ζ ′′).

These hypotheses above show that the spaces X,X ′, X ′′ illustrated in Figure C.4 are

2-pointed premetric spaces. Let

X

•0

•1

•a

π

ξ �������

ζ???????

X ′

•0

•1

•a′

π

ξ′ �������

ζ′???????

X ′′

•0

•1

•a′′

π

ξ′′ �������

ζ′′???????

Figure C.4: Three 2-pointed premetric spaces.

ρ = (ξ′ ∨ ξ′′) ∧ (ζ ′ ∨ ζ ′′),

ρ′ = (ξ ∨ ξ′) ∧ (ζ ∨ ζ ′),

ρ′′ = (ξ ∨ ξ′′) ∧ (ζ ∨ ζ ′′).

Since L is modular by Proposition C.3.1, we can show that

ρ ∨ ξ′ = ρ ∨ ξ′′ = ξ′ ∨ ξ′′, ρ ∨ ζ ′ = ρ ∨ ζ ′′ = ζ ′ ∨ ζ ′′;

ρ′ ∨ ξ = ρ′ ∨ ξ′ = ξ ∨ ξ′, ρ′ ∨ ζ = ρ′ ∨ ζ ′ = ζ ∨ ζ ′;

ρ′′ ∨ ξ = ρ′′ ∨ ξ′′ = ξ ∨ ξ′′, ρ′′ ∨ ζ = ρ′′ ∨ ζ ′′ = ζ ∨ ζ ′′,

120

as we did above for the first line. This shows that the spaces Y.Y ′, Y ′′ illustrated in

Figure C.5 are 2-pointed premetric spaces. Observe that Y, Y ′, Y ′′ are the canonical

Y •0

•1

•a′ •a′′π

ξ′

����

����

�ξ′′

????

????

?

ζ′ ζ′′

ρ

Y ′•0

•1

•a •a′′π

ξ

����

����

�ξ′′

????

????

?

ζ ζ′′

ρ′′

Y ′′•0

•1

•a •a′π

ξ

����

����

�ξ′

????

????

?

ζ ζ′

ρ′

Figure C.5: Three amalgams of 2-pointed premetric spaces.

amalgams of X ′ and X ′′, X and X ′′, X and X ′, respectively.

Let Z,Z ′, Z ′′ be any amalgams of X and Y , X ′ and Y ′, and X ′′ and Y ′′, respec-

tively. We may suppose that Z,Z ′, Z ′′ all have base space Z = 0, 1, a, a′, a′′. Define

d : Z × Z → L by

d(x, y) = dZ(x, y) ∨ dZ′(x, y) ∨ dZ′′(x, y)

for all x, y ∈ Z. We know that d is a distance on Z, and it is easy to check that

d(a′, a′′) = ρ, d(a, a′′) = ρ′, d(a, a′) = ρ′′.

So the triangle identity for a, a′, a′′ says that

ρ ∨ ρ′ = ρ′ ∨ ρ′′ = ρ′′ ∨ ρ,

which is the conclusion of the weak Arguesian law. qed

121

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124

Index

Abraham, Uri, 2

Ackerman, Nathanael, iv

almost universal, 16

amalgam, 103

canonical, 104

maximal, 104

amalgamable, 103

Andersen, Brooke M., iv

antichain, 8, 87

maximal, 9

Aronszajn, Nachman, 9, 10

axiom of approximation, 98

Baire property, 87

Baire, Rene, 11, 87

Bilaniuk, Stefan, iv

bounded

in a tree, 6

Bourke, John G., iv

branch, 8

Brooke-Taylor, Andrew, iv

c-degrees, 13

chain, 8

maximal, 8

chain condition, 87

Chan, Alice, iv

closure operator, 96

compact, 97

Comparison Theorem, 18, 25, 26, 26, 27,

41–45, 48

composition, 108

conditions, 83

congruence relation, 19

closed, 19

normal, 35

regular, 19

coproduct (of trees), 7

Corduan, Jared R., iv

degrees of constructibility, 13

dense, 83

above, 84

diamond sequence, 89

built-in, 90

125

distance, 101

algebraic

regular, 41

continuous, 37

regular, 37

distributive laws, 93

Embedding Theorem, 43, 45

Esselstein, Rachel M., iv

expansion

by a congruence relation, 20

extender map, 68

Factoring Theorem, 18, 22, 28

Farrington, C. Patrick, 4, 5

Farrington, C. Patrick (Paddy), 4, 5

filter, 84

generic, 84

forcing relation, 85

Freer, Cameron, iv

Goddard, Christina, iv

Groszek, Marcia J., iv, 3–5

Godel, Kurt, 12

Hajnal, Andras, 16

Hausdorff, Felix, 6, 7, 11, 20

height

of a node, 6

of a tree, 6

Hugill, D. F., 2

ideal, 97

principal, 97

incompatible, 87

Initial Segment Theorem, 45, 68

inner model, 13

interpolants

have, 40, 101

have 2k-, 101

have 2k + 1-, 101

have enough, 42

involution, 107

isometry, 101

pointed, 103

Jech, Thomas J., 4

Jensen, Ronald B., 4, 13, 89, 90

join, 92

arbitrary, 95

Kanamori, Akihiro, iv

kernel operator, 96

Lachlan, Alistair, 2

lattice, 92

126

algebraic, 97

Arguesian

strongly, 117

weakly, 118

complete, 95

continuous, 98

distributive, 93

join-continuous, 95

meet-continuous, 95

modular, 94

Lebeuf, Robert, 2

Lerman, Manuel, 2, 3

level

of a tree, 6

Lubarsky, Robert S., 3

Levy, Azriel, 86

McAloon, Kenneth, 4

meet, 7, 92

arbitrary, 95

metric space, 101

pointed, 103

split, 105

modular law, 94

morphism

of trees, 7

name, 84

evaluation, 84

nodes, 6

notion of forcing, 83

open, 83

predistance, 100

algebraic, 40

regular, 41

continuous, 36

faithful, 42

property (#), 45

regular, 37

primitive recursive, 85

primitive recursive function, 15

product (of trees), 8

regular derivative, 34

Representation Theorem, 18, 22, 23, 43–

45

restriction

of a tree, 6

rudimentary function, 13

Sacks, Gerald E., ii, iv, 2, 3

Scott, Dana S., 13

semi-lattice

127

join, 93

meet, 93

Shore, Richard A., iv, 2–5

Slaman, Theodore A., 3

Smolska-Adamowicz, Zofia, 3, 4

Souslin, Mikhail, iv, 4–6, 9–12, 18, 21–23,

25–27, 29, 30, 41–45, 49, 50, 58,

60, 68, 69, 76, 81

Spector, Clifford, 2

splitting height, 6

square sequence, 90

subtree, 7

sum (of trees), 7

tree, 6

Aronszajn, 9

Hausdorff, 6

regular, 6

Souslin, 9

trivial, 10

splitting, 9

triangle identity, 100

triangle inequality, 100

Truth Lemma, 87, 87

Turing, Alan, 2, 3, 13

unit, 93

V -degrees, 16

way-below, 98

zero, 93

128