Hrushovski and Ramsey Properties of Classes of Finite by Jakub … · 2012. 10. 31. · jfgj= kand x(i) = (fg)(i) for i

Hrushovski and Ramsey Properties of Classes of Finite

Inner Product Structures, Finite Euclidean Metric Spaces,

and Boron Trees

by

Jakub Jasi«ski

A thesis submitted in conformity with the requirements

for the degree of Doctor of Philosophy

Graduate Department of Mathematics

University of Toronto

Copyright c© 2011 by Jakub Jasi«ski

Abstract

Hrushovski and Ramsey Properties of Classes of Finite Inner Product Structures, Finite

Euclidean Metric Spaces, and Boron Trees

Jakub Jasi«ski

Doctor of Philosophy

Graduate Department of Mathematics

University of Toronto

2011

We investigate two combinatorial properties of classes of �nite structures, as well as re-

lated applications to topological dynamics. Using the Hrushovski property of classes

of �nite structures � a �nite extension property of homomorphisms � we can show the

existence of ample generics. For example, Solecki proved the existence of ample gener-

ics in the context of �nite metric spaces that do indeed possess this extension property.

Furthermore, Kechris, Pestov and Todorcevic have shown that the Ramsey property of

Fraïssé classes of �nite structures implies that the automorphism group of the corre-

sponding Fraïssé limit is extremely amenable, i.e., it possesses a very strong �xed point

property.

Gromov and Milman had shown that the unitary group of the in�nite-dimensional

separable Hilbert space is extremely amenable using non-combinatorial methods. This

result encourages a deeper look into structural Euclidean Ramsey theory, i.e., Euclidean

Ramsey theory in which we colour more than just points. In particular, we look at

complete �nite labeled graphs whose vertex sets are subsets of the Hilbert space and

whose labels correspond to the inner products. We prove �Ramsey-type� and �Hrushovski-

type� theorems for linearly ordered metric subspaces of �su�ciently� orthogonal sets.

In particular, the latter is used to show a �Hrushovski version� of the Ramsey-type

Matou²ek-Rödl theorem for simplices.

ii

It is known that the square root of the metric induced by the distance between vertices

in graphs produces a metric space embeddable in a Euclidean space if and only if the

graph is a metric subgraph of the Cartesian product of three types of graphs. These three

are the half-cube graphs, the so-called cocktail party graphs, and the Gosset graph. We

show that the class of metric spaces related to half-cube graphs � metric spaces on sets

with the symmetric di�erence metric � satis�es the Hrushovski property up to 3 points,

but not more. Moreover, the amalgamation in this class can be too restrictive to permit

the Ramsey Property.

Finally, following the work of Fouché, we compute the Ramsey degrees of structures

induced by the leaf sets of boron trees. Also, we brie�y show that this class does not

satisfy the full Hrushovski property. Fouché's trees are in fact related to ultrametric

spaces, as was observed by Lionel Nguyen van Thé. We augment Fouché's concept of

orientation so that it applies to these boron tree structures. The upper bound compu-

tation of the Ramsey degree in this case, turns out to be an �asymmetric� version of

the Graham-Rothschild theorem. Finally, we extend these structures to �oriented� ones,

yielding a Ramsey class and a corresponding Fraïssé limit whose automorphism group is

extremely amenable.

iii

Acknowledgements

I would like to thank Stevo Todorcevic for suggesting and supervising this project, as well

as his signi�cant feedback. I would also like to thank Lionel Nguyen Van Thé, Norbert

Sauer, and William Weiss for their valuable feedback.

iv

Contents

1 Background 1

1.1 Basic Notation and Terminology . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Ramsey Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Euclidean Ramsey Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Fraïssé Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Structural Ramsey Theory and the Hrushovski Property . . . . . . . . . 10

2 The Hilbert Space 16

2.1 Dense Fraïssé Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2 Inner Product Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 The Euclidean Metric 28

3.1 A Rigid Subset of `2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 Colouring using Linear Dependence and Orientation . . . . . . . . . . . . 29

3.3 Spreads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4 Graphs and Metrics 43

4.1 Hypermetric Distance Functions . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 The Symmetric Di�erence Metric . . . . . . . . . . . . . . . . . . . . . . 46

5 Boron Trees 53

5.1 Boron Tree Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

v

5.2 Ramsey Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.3 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Bibliography 81

vi

Chapter 1

Background

1.1 Basic Notation and Terminology

For any two sets A,B, de�neAB to be the set of functions from A with values in B. By

P(A) we denote the set of subsets of A. For any function f ∈AB and C ⊆ A, de�ne

f ′′C = {f(x) : x ∈ C}

and ran(f) = f ′′A, i.e., the range of f . If C ⊆ A and f : A → B is a function, then

f � C is the restriction of f to C, i.e., f � C = f ∩ (C ×B).

The symbol R denotes the set of real numbers, while ω = {0, 1, 2, . . .} and N = ω\{0}.

By [ω]<ω we denote the set of �nite subsets of ω. For A ⊆ n ∈ ω, de�ne [n] = {1, . . . , n}

and

Sym(A) = {ρ ∈AA : ρ is a permutation of A}.

If x ∈ Rk, for some k ∈ N, we will sometimes denote the i-th coordinate of x as xi. A

graph G is an ordered pair (V,E) such that

E ⊆ {{v0, v1} ∈ P(V ) : v0 6= v1} = [V ]2.

1

Chapter 1. Background 2

We will denote the vertex set of a graph G by V (G) and its edge set by E(G). A labeled

graph is a graph G with a function λG : E(G)→ I, for some �nite I. A homomorphism

f : V (G0)→ V (G1) is a map such that for all v0, v1 ∈ V (G0),

{v0, v1} ∈ E(G0)↔ (f(v0), f(v1)) ∈ E(G0)

and

λG0(v0) = λG1(f(v1))

in the case of labeled graphs. A weak homomorphism f : V (G0)→ V (G1) is a map such

that for all v0, v1 ∈ V (G0),

{v0, v1} ∈ E(G0)→ (f(v0), f(v1)) ∈ E(G0)

and if (f(v0), f(v1)) ∈ E(G1), then

λG0(v0, v1) = λG1(f(v0), f(v1))

in the case of labeled graphs. Suppose G0 and G1 are graphs such that V (G0) ⊆ V (G1)

and E(G0) is the set of edges {v0, v1} ∈ E(G1) with vertices v0, v1 ∈ V (G0). In case G0

and G1 are labeled graphs, suppose furthermore that for any edge{v0, v1} ∈ E(G0),

λG0(v0, v1) = λG1(v0, v1).

We call such G0 a subgraph of G1.


1.2 Ramsey Theory

For any n ∈ ω, n = {0, . . . , n − 1}. Let N, n, k, r ∈ N. De�ne(nk

)= {A ⊆ n : |A| = k}.

De�ne N → (n)kr to mean that for any �nite colouring (function) c :(Nk

)→ r there is

B ⊆ N of size n such that∣∣c′′(B

k

)∣∣ = 1 (i.e.,(Bk

)is monochromatic). In [R], Ramsey

proved the following:

Theorem 1.2.1 (Ramsey, 1930). For any r, n, k ∈ N, there is N ∈ N such that N →

(n)kr .

De�nition 1.2.2. Let Λ be a �nite set; we shall refer to it as the alphabet. Let n,m ∈ ω.

De�ne Λn = {f : n → Λ}. Let V = {v0, . . . , vm−1} be a �nite set such that Λ ∩ V = ∅;

we shall refer to its elements as the variables. Let the special symbol ∗ signify a �xed

element not in Λ. De�ne [Λ]∗(nm

)to be the set of functions f : n→ Λ ∪ {∗} ∪ V (called

words) that satisfy the following conditions:

1. f−1(vi) 6= ∅ for all i ∈ m.

2. min f−1(vi) < min f−1(vj) for all i < j ∈ m.

3. If f(i) = ∗ for some i ∈ n, then f(j) = ∗ for all j > i.

Let f ∈ [Λ]∗(nm

), de�ne |f | = min{j : f(j) = ∗} to be the length of f . De�ne

[Λ]

(n

m

)=

{f ∈ [Λ]∗

(n

m

): |f | = n

}.

Let f ∈ [Λ]∗(nm

)and g ∈ [Λ]∗

(mk

). The composite f · g ∈ [Λ]∗

(nk

)is de�ned recursively,

as follows:

1. (f · g)(0) = g(j) if f(0) = vj and (f · g) = f(0), otherwise.


2. Suppose (f · g)(j) has been de�ned for j < i. Then,

(f · g)(i) =

g(j) if f(i) = vj

∗ if (f · g)(j) = ∗ for some j < i

f(i) otherwise

X ⊆ [Λ](n0

)= Λn is called an m-parameter set corresponding to f ∈ [Λ]

(nm

)if and only if

X =

{f · g : g ∈ [Λ]

(m

0

)}

If α ∈ Λ and f ∈ [Λ]∗(nm

), de�ne F (f, α) ∈ [Λ]

(nm

)by setting

(F (f, α))(i) =

f(i) if f(i) 6= ∗

α if f(i) = ∗

Theorem 1.2.3 (Voigt, 1980 [V]). For every �nite Λ and every r,m ∈ N, there exists

N = N∗(r,m) ∈ N such that for every colouring c : [Λ]∗(N0

)→ r, there exists a variable

word f ∈ [Λ]∗(Nm

)such that the auxiliary colouring c(g) = c(f · g) assumes exactly one

colour on [A](m0

).

Suppose X ⊆ Λn is an m-parameter set corresponding to f ∈ [Λ](nm

). Let k < n.

The element x ∈ Λk is a partial point in X if and only if there is g ∈ [Λ]∗(m0

)such that

|f · g| = k and x(i) = (f · g)(i) for i < k. Given an n-parameter set X, with n ≥ p, de�ne(Xp

)to be the set of all p-parameter subsets of X.

Theorem 1.2.4 (Graham-Rothschild, 1971 [Ne]). Let n and t be positive integers. Let

p ∈ ω and Λ ∈ [ω]<ω . Then there exists N = GR(Λ, p, r, n) such that if X is an N-

parameter set and(Xp

)is coloured by r-colours then there exists an n-parameter set Y

such that(Yp

)is a monochromatic set.


The corollary 1.2.5 below is essential in proving Theorem 5.2.5 and is also used to

prove the related Theorem 1.2.4, in [Ne]. Note that Voigt used the one variable Hales-

Jewett theorem (Graham-Rothschild above with n = 1 and p = 0) to prove his truncated

version (see Lemma 2.6 in [V]). We can close this loop of implications by showing

that Voigt's truncated version of the Hales-Jewett theorem follows from the Graham-

Rothschild theorem: Let N = GR(Λ, 1, r,m + 1). Set X = ΛN . Let c : [Λ]∗(N0

)→ r

be a colouring. Set c′ :(X1

)→ r to be de�ned as follows: Let Z ∈

(X1

)and suppose

h ∈ [Λ](N1

)de�nes Z. Set c′(Z) = c(h · (∗)). Let Y be the (m + 1)-parameter set such

that(Y1

)is monochromatic with respect to c′. Suppose Y is de�ned using g ∈ [Λ]

(Nm+1

).

Set f = g · (v1, v2, . . . , vm−1, vm, ∗) ∈ [Λ]∗(Nm

).

Corollary 1.2.5 (Neset°il [Ne]). For every �nite Λ and every r,m ∈ N, there exists

N = HJ∗(r,m) ∈ N such that for every colouring c : Λ≤N → r, there exists an m-

parameter set X ⊆ ΛN such that every partial point in X is assigned the same colour by

c.

Proof. Pick any α0 ∈ Λ. Set N = N∗(r,m). De�ne a colouring c0 on [Λ]∗(N0

), as follows:

If g ∈ [Λ]∗(N0

), then there exists a unique x ∈ Λ|g| such that g(i) = x(i) for i < |g|. Set

c0(g) = c(x). By the Theorem 1.2.3, there exists f ∈ [Λ]∗(Nm

)such that

∣∣c′′0 ([A](m0

))∣∣ = 1,

with c0(g) = c0(f · g). Let x ∈ Λk, y ∈ Λl be partial points in the m-parameter set X

corresponding to F (f, α0). Note that neither x nor y are in X itself. Therefore, they

are in fact, partial points of the m-parameter set corresponding to f . Thus, there exists

g ∈ [Λ]∗(m0

)such that |f · g| = k and x(i) = (f · g)(i) for i < k. Likewise, there exists

h ∈ [Λ]∗(m0

)such that |f · h| = l and y(i) = (f · h)(i) for i < l. Therefore,

c(x) = c0(f · g) = c0(f · h) = c(y).


1.3 Euclidean Ramsey Theory

Call E the class of �nite sets A such that A ⊆ Rk, for some k. Let A,B ∈ E . A function

f : A → B is an isometry if and only if for all x, y ∈ A, ||x − y|| = ||f(x) − f(y)||.

Observe that f(x) = f(y) implies ||f(x)− f(y)|| = 0, i.e., ||x− y|| = 0, i.e., x = y. Thus,

every isometry is an injection. De�ne

(B

A

)||.||

= {A′ ⊆ B : there is an isometric bijection f : A′ → A}

De�ne(BA

)〈.,.〉 to be the set of subsets of B for which there is an inner product preserving

bijection to A. De�ne C →∗ (B)Ar (where ∗ is either ||.|| or 〈., .〉) to mean that for any

r-colouring of(CA

)∗ there is a B

′ ∈(CB

)∗ such that

(B′

A

)∗ is monochromatic.

Including the early papers of the rather extensive team of P. Erdös, R.L. Graham,

P. Montgomery, B.L. Rothschild, J. Spencer and E.G. Straus (see [EGMRS1, EGMRS2,

EGMRS3]), there has been some interest in Ramsey-type phenomena in Euclidean space.

Given B0 ⊆ Rm, B1 ⊆ Rn, for some k ∈ N, de�ne

B0 ×B1 = {(x0, . . . , xm, y0, . . . , yn) ∈ Rm+n : x ∈ B0 ∧ y ∈ B1}.

A fundamental result in this area is the following proposition.

Proposition 1.3.1. Suppose B0, B1 ∈ E are such that for any r ∈ N, there are C0, C1 ∈ E

with C0 →||.|| (B0){1}r and C1 →||.|| (B1)

{1}r . Then for any r ∈ N, there is C ∈ E such that

C →||.|| (B0 ×B1){1}r .

Moreover, in [EGMRS1, EGMRS2, EGMRS3], it has been shown that if B ∈ E is such

that for any r ∈ N, there is C ∈ E , C → (B){1}r , then B can be isometrically embedded

in a Sk(ρ) = {x ∈ Rk : ||x|| = ρ}, for some k ∈ N, ρ ∈ R. We list more results involving

colourings of singletons below.


Theorem 1.3.2 (K°iº, 1991 [Kr]). If B ⊆ Rn has a transitive solvable group of isome-

tries, then for any r, there is C ⊆ RN such that C →||.|| (B){1}r .

In particular, the platonic solids have transitive solvable groups of isometries. How-

ever, as we shall quickly see, colouring their faces � as opposed to their vertices � does

not yield positive results. In particular, the faces of a platonic solid (or, for that matter,

any subset of size larger than 2), span R3, if � for example � we place the barycenter of

the said platonic solid at the origin.

Recall that a set A ⊆ Rk is a�nely independent if and only if for any x ∈ A,

{y − x : y ∈ A\{x}} is linearly independent. A simplex is de�ned to be an a�nely

independent subset of Sk = {x ∈ Rk : ||x|| = 1}.

Theorem 1.3.3 (Frankl-Rödl, 1990 [FR]). For any simplex B ⊆ Rk and any r ∈ N,

there is N ∈ N and C ⊆ RN , such that C →||.|| (B){1}r .

In the above theorem, a key lemma showed that �almost-regular� simplexes satisfy

the theorem. By further imposing linear orderings, we shall prove that �su�ciently-

orthogonal� sets satisfy the theorem with the superscript �{1}� replaced by larger sets.

Theorem 1.3.4 (Matou²ek-Rödl, 1995 [MR]). For any simplex B, any r, and any ε > 0,

there exists N = N(B, r, ε) and a �nite subset C ⊆ SN(1 + ε) such that

C →||.|| (B){1}r

In the proof of the above theorem, the concept of �spread� was used to �rst approx-

imate the simplex with a set that satis�es the theorem. These spreads can be seen to

satisfy the theorem immediately, by referring to Ramsey's theorem. We will later look

at classes of ordered structures related to these spreads.

Although results pertaining to colourings of sets larger than singletons are scarce, in

[Can1, Can2], we �nd work pertaining to edge-colourings. Let E be a �nite con�guration


of line segments in Rn. Suppose, for any r ∈ N, there exists N ∈ N such that for any

colouring of the line segments in RN , there is a monochromatic �copy� of E. If this is the

case, then we shall call E edge-Ramsey (see [Gr]). In particular, if E is edge-Ramsey, then

all edges of E have the same length (see [EGMRS1, EGMRS2, EGMRS3]). The following

list of theorems can also be found in [Gr]. Refer to Cantwell's papers [Can1, Can2] for

the proofs.

Theorem 1.3.5. If E is edge-Ramsey, then the end-points of the edges of E are contained

in the surface of two spheres.

Theorem 1.3.6. If the end-points of E do not lie on a sphere and the graph formed by

E is not bipartite, then E is not edge-Ramsey.

Theorem 1.3.7 (Cantwell, 1996). The edge set of an n-cube is edge-Ramsey.

Theorem 1.3.8. The edge set of a regular n-gon is not E-Ramsey, if n 6= 6.

Corollary 1.3.9. The edge set of a regular hexagon is not edge-Ramsey.

1.4 Fraïssé Theory

Given a structure A, A will refer to its universe. Let A, B be two structures in the same

relational language with signature L = {Ri}i∈I , for some countable index set I. By the

notation A ⊆ B, we mean that A ⊆ B and RAi = RB

i ∩ An(i) for all i, where n(i) is

the arity of Ri. We say that A is a substructure of B, in this case. A homomorphism

f : A→ B is a function such that (a1, . . . , ak) ∈ RAi if and only if (f(a1), . . . , f(ak)) ∈ RB

i .

An isomorphism is a bijective homomorphism. A weak homomorphism f : A → B is a

function such that if (a1, . . . , ak) ∈ RAi then (f(a1), . . . , f(ak)) ∈ RB

i . We call an injective

homomorphism an embedding. We denote the existence of an embedding f : A ↪→ B by

A ≤ B. The class K of �nite structures for a �xed language is hereditary if and only if

it is closed under embeddings, i.e., for all B ∈ K, if A ≤ B, then A ∈ K. The class K


satis�es the joint embedding property (JEP) if and only if for any pair A, B ∈ K, one can

�nd C ∈ K such that A, B ≤ C. The class K satis�es the amalgamation property (AP)

if and only if given A, B1, B2 ∈ K, and embeddings fi : A ↪→ Bi, one can �nd C ∈ K and

embeddings gi : Bi ↪→ C such that g1f1 = g2f2. We call a countable (up to isomorphism)

class of arbitrarily large �nite structures which satis�es these three properties a Fraïssé

class. A structure A is ultrahomogeneous if and only if every �nite partial isomorphism

of A can be extended to an automorphism of A. The age of a structure A is the class of

all �nite structures which can be embedded in A. A substructure B ⊆ A is called �nitely

generated if and only if for some �nite set X ⊆ A, B is the smallest structure containing

X. Observe that in the context of relational languages, �nitely generated structures are

always �nite.

Theorem 1.4.1 (Fraïssé [Fr1, Fr2]). Given a Fraïssé class K, there is a unique (up

to isomorphism) countable ultrahomogeneous structure � called the Fraïssé limit of K �

whose �nitely generated substructures are �nite and whose age is K.

We will denote the Fraïssé limit of K by Flim(K). Given a class K of structures with

signature L, we will denote by OK, the class of structures with signature L∪{<} which

are identical to members of K except that <A is a linear ordering of A for every A ∈ OK.

Notable examples of Fraïssé classes include the class of �nite graphs with its limit being

the countable random graph and the class of �nite metric spaces with its limit being the

Urysohn space.

Let L = {Ri}i∈I be a relation language with I �nite and the arity of each Ri at least

2. We call a structure A in with this signature a hypergraph of type L if and only if every

RAi satis�es the following:

(irre�exivity) If (a1, . . . , ak) ∈ RAi , then ak 6= aj for k 6= j.

(symmetry) Given a permutation ρ ∈ Sym([k]), if

(a1, . . . , ak) ∈ RAi


then

(aρ(1), . . . , aρ(k)) ∈ RAi .

In other words, RAi ⊆ [A0]k.

Suppose G0 is a �nite labeled graph with labels in {0, . . . , n}. Then we can assign to

it a hypergraph of type L with I = {0, . . . , n} and arity of each Ri exactly 2, by setting

A = V (G0) and

RAi = {(v0, v1) : (v0, v1) ∈ E(G0) ∧ λG0(v0, v1) = i}.

Let L = {0, 1,−,∧,∨}, where 0, 1 have arity 0, − has arity 1 and ∧,∨ have arity 2.

A structure A with this signature corresponds to a Boolean algebra.

Suppose (X, d) is a metric space. De�ne ∆(X, d) = {δ : (∃x, y)d(x, y) = δ}. Let

∆0 ⊆ R be countable. Let L = {Rδ}δ∈∆0 . Suppose (X, d) is a metric space with

∆(X, d) ⊆ ∆0. We can de�ne a corresponding structure X as follows: (x, y) ∈ RXδ if

and only if d(x, y) < δ. Let M∆0 be the class of �nite structure with signature L that

correspond to a �nite metric space in this manner.

1.5 Structural Ramsey Theory and the Hrushovski Prop-

erty

In [KPT], signi�cant relationships between structural Ramsey theory and topological

dynamics had been explored. De�ne(BA

)to be the set of subsets of B isomorphic to A.

A class K satis�es the Ramsey property if and only if given members A, B and k ∈ N,

one can �nd C ∈ K, such that C ≥ B and given any k-colouring of the copies of A

in C, there is a copy B′ of B in C such that any copy of A in B′ assumes the same

colour. We denote the latter statement by C → (B)Ak . In [KPT], it had been shown

that a Fraïssé class satis�es the Ramsey property if and only if the automorphism group


of its Fraïssé limit is extremely amenable, i.e., any continuous action of this group G on

compact Hausdor� spaces has a �xed point. We refer to the aforementioned action by G

as a G-�ow.

The class of �nite graphs does not have the Ramsey property. However, the result in

[NR] of Ne²et°il and Rödl implies that the class of linearly ordered graphs does in fact

have the Ramsey property. Likewise, the class of linearly ordered labeled graphs has the

Ramsey property. To a class of <-linearly ordered structures K, let us assign the class

K0 whose members are reducts of members of K, i.e., if A ∈ K then A0 ∈ K0 is its reduct

when A0 is A with the relation <A removed. The class K satis�es the ordering property

if and only if given A0 ∈ K0, there is B0 ∈ K0 such that for every linear ordering ≺

on A0 and every linear ordering ≺′ on B0, if A and B are the extensions of A0 and B0

with ≺ and ≺′, respectively, then A ≤ B. Set F to be the Fraïssé limit of the class K,

denoted by Flim(K). Similarly, set F0 to be the Fraïssé limit of the class K0, denoted by

Flim(K0). A linear ordering on F0 is called K-admissible if and only if for any A0 ⊆ F0,

the restricted relation ≺� A0 extends A0 to A ∈ K. We call K reasonable if and only

if for every A0 ∈ K0, B0 ∈ K0, embedding π : A0 → B0, and linear ordering ≺ on A0

extending A0 to A ∈ K, there is a linear ordering ≺′ on B0 extending B0 to B ∈ K and

π preserves the ordering.

A G-�ow is a continuous action of a topological group G on a compact Hausdor�

space. A minimal G-�ow is one such that every orbit is dense. A universal minimal

�ow M(G) is a minimal G-�ow which can be homomorphically mapped onto any other

minimal G-�ow. In this context, a homomorphism is a map which is continuous and

preserves the action. Given a set X of n-ary relations on a countable structure F, we can

construct a G-�ow, where G = Aut(F) and the set acted upon corresponds to X. First,

encode n-tuples of elements of F as elements of ω. We can then encode any countable

n-ary relation as an element of 2ω, i.e., the compact Cantor space. The automorphism

group G is endowed with the pointwise convergence topology. Finally, the action of this


group on subsets of 2ω corresponds to the logic action, as follows: given g ∈ G and

r ∈ 2ω, we have that (g · r)(k) = 1 if and only if r(g−1(k)) = 1, for all k ∈ ω. Theorems

7.4, 7.5, and 10.8 in [KPT] provide a method of computing universal minimal �ows of

automorphism groups using the ordering property. We summarize 7.4 and 10.8 below.

Theorem 1.5.1. Suppose K is a reasonable Fraïssé class. Let XK be the set of linear

orderings on F which are K-admissible. Let also G0 = Aut(F0). Then the following are

equivalent

(i) XK is a minimal G0-�ow.

(ii) K satis�es the ordering property.

Moreover, the following are also equivalent,

(a) XK is the universal minimal �ow of G0.

(b) K has the Ramsey and ordering properties.

IfA0 ∈ K0, we de�ne t(A0,K0) to be the least t, if it exists, such that for anyA0 ≤ B0

in K0, k ≥ 2, there is C0 ∈ K0 such that C0 ≥ B and given any k-colouring of the copies

of A0 in C0, there is a copy B′0 of B0 in C0 such that any copy of A0 in B′0 assumes at

most t colours. If such a t does not exist, we de�ne t(A0,K0) =∞. We call t(A0,K0) the

Ramsey degree of A0 (in K0). Theorems 10.5 and 10.6 in [KPT] show the relationship

between the ordering property and Ramsey degrees.

Deuber in [De] and Milliken in [Mi] presented Ramsey-type theorems for trees. Below

a tree T will be a partially ordered set � a set with a re�exive, antisymmetric and transitive

relation ≤T � such that for each t ∈ T , the set {s ∈ T : s ≤T t ∧ s 6= t} is well-ordered

� every non-nonempty subset has a least element. A total (or linear) order has the

additional property that any two elements are comparable. A branch of T is a maximal

chain � a totally ordered set such that any extra element would make it no longer totally

ordered. De�ne Succ(t, T ) = {t′ ∈ T : t′ ≥ t} and IS(t, T ) to be the set of immediate

successors of t. A tree T is an α-tree if and only if each branch of T has cardinality α.

For α ≤ β, an α-tree S is embedded strongly in a β-tree T if and only if


1. S ⊆ T , while ≤S agrees with ≤T .

2. If s ∈ S is non-maximal in S and t ∈ IS(s, T ), then Succ(t, T ) ∩ IS(s, S) is a

singleton.

3. There is a strictly increasing function f : α→ β such that S(n) ⊆ T (f(n)).

We state a version of Milliken's theorem (1.3 in [Mi] and 6.5 in [T]) below.

Theorem 1.5.2. Suppose α and r are positive integers and that T is a tree of height ω

for which |IS(t, T )| < ω for all non-maximal t ∈ T . Let S be the set of all α-trees strongly

embedded in T . Let C1 ∪ . . . ∪ Cr be a partition of S. Then there is k and an ω-tree R

such that the set of α-trees strongly embedded in R is contained in Ck.

Of particular interest is the result pertaining to linearly ordered set systems (or lin-

early ordered hypergraphs), found in [NR]. Let ∆ ⊆ N be a �nite set. A set system of

type ∆ is a pair (X,M), where X is a �nite linearly ordered set, M = (Mδ : δ ∈ ∆),

andMδ ⊆ {Y ⊆ X : |Y | = nδ} for some nδ's with δ ∈ ∆. Moreover,Mδ ∩Mδ′ = ∅ for

δ 6= δ′.

A set Y ⊆ X is a cut of a set system B if there is a partition of X\Y into two

disjoint sets Y1, Y2 such that no pair {y1, y2}, y1 ∈ Y1, y2 ∈ Y2 is such that {y1, y2} ⊆ e

for some e ∈ Mδ, for some δ ∈ ∆. The system A is a weak subsystem of B if and

only if every vertex (edge, respectively) of A is a vertex (edge, respectively) of B. If

(X,M) and (Y,N ) are set systems of type ∆, then we call f : X → Y a (strong)

homomorphism of set systems if and only if for all δ ∈ ∆, {x1, . . . , xnδ} ∈ Mδ if and only

if {f(x1), . . . , f(xnδ)} ∈ Nδ. An isomorphism of set systems is a bijective homomorphism.

A system B is called homA-connected if and only if no cut Y of B has a homomorphism

into A.

For a class F of systems, de�ne Forb(F) to be the class of those systems A which do

not contain any system F ∈ F as a weak subsystem of B. As in the case of structures,


we de�ne C → (B)At to mean that for set systems A,B there exists a set system C such

that for any t-colouring of the copies of A in C, there is a copy of B in C such that all

of the copies of A within it assume the same colour.

Proposition 1.5.3 (Neset°il-Rödl, 1989). Let A be a set system. Let F be a set of

homA-connected systems. Then for every positive t and every B ∈ Forb(F) there exists

C ∈ Forb(F) such that C → (B)At .

In [Fo], Fouché computed the Ramsey degrees of trees, in which case the embeddings

were height preserving. This entailed Nguyen Van The's theorem stating that the class

of �nite convexly ordered ultrametric spaces � which can be represented using lexico-

graphically ordered trees with labeled levels � has the Ramsey Property. The boron tree

structures considered here are induced by leaf sets of boron graphs and the embeddings

here are in no sense height preserving.

Recall that a Hilbert space H is a real or complex inner product space that is also

a complete metric space with respect to the distance function induced by the inner

product. Note that any separable in�nite dimensional Hilbert space is isometric to `2 =

{(ai)∞i=1 : (∀i)ai ∈ R ∧ Σ∞i=1a2i <∞}. A unitary operator is de�ned to be a bounded linear

operator U : H → H on a Hilbert space H such that

U∗U = UU∗ = I

where U∗ is the adjoint of U and I : H → H is the identity operator. Moreover, we can

equivalently de�ne U to be unitary if and only if it satis�es the following

1. ran(U) is dense.

2. For all x, y ∈ H, 〈Ux, Uy〉 = 〈x, y〉 .

The strong operator topology is de�ned to be such that Tnx→ Tx (for all x ∈ H) if

and only if Tn → T . The strong topology on `2 is de�ned using the metric√

Σ∞i=1(ai − bi)2

for a, b ∈ `2. Solving a problem of Furstenberg, Gromov and Milman proved that the


unitary group U(`2) with the strong topology is extremely amenable, using the concept

of Lévy group (see [GM]). We construct a related extremely amenable group, in the

search for a combinatorial proof of the said theorem. Moreover, we will show that the

corresponding age satis�es the Hrushovski property.

We say that a class K of �nite structures has the Hrushovski property if and only if

given A ∈ K, we can �nd B ∈ K and an embedding ϕ : A ↪→ B such that any partial

isomorphism of the image ϕ′′A extends to an automorphism of B. In [Hr], Hrushovski

showed that the class of �nite graphs has this property. This property and related

theorems are also applicable in topological dynamics: Solecki proved in [So] that �nite

metric spaces with rational distances also have this property using a general result found

in [HL]. A Polish group is a topological group whose topology is separable and completely

metrizable. By a Polish group having ample generics we mean that for any n ∈ N, the

diagonal action of G on Gn has a generic element, i.e., its orbit with respect to this action

is a dense Gδ. Solecki used the Hrushovski property to prove that the automorphism

group of the rational Urysohn space has ample generics.

Chapter 2

The Hilbert Space

2.1 Dense Fraïssé Limit

Recall that Sk(ρ) ={x ∈ Rk : ||x|| = ρ

}and de�ne S∞ = {x ∈ `2 : ||x|| = 1}. In

a later chapter, we show that if the age of a metric subspace of S∞ has the Ramsey

Property, then the members of the age must be linearly independent. Moreover, to show

that the automorphisms of this subspace extend to unitary operators, we would like the

space to be dense and ultrahomogeneous. Towards this end, we construct below a dense

ultrahomogeneous subspace of S∞.

De�ne, for X ⊆ Rd, tr(X) = {i : (∃x ∈ X)xi 6= 0} . Suppose χ ⊆ R is countable

and closed under +,−,×,÷,√. Recall thatMχ is the class of structures corresponding

to �nite metric spaces with distances in χ. Set K ⊆ Mχ to be the class of structures

corresponding to �nite metric spaces which are isometric to some linearly independent

X ⊆ (χ ∪ {0})d inside the �nite dimensional unit sphere centered at ~0, for some d ≥ 1.

Note that in contrast to the Gram-Schmidt process, the following construction preserves

the Euclidean metric.

Proposition 2.1.1. K satis�es the amalgamation property.

Proof. We establish the property, by working with the corresponding metric spaces. De-

16

Chapter 2. The Hilbert Space 17

�ne Ri(θ) to be a rotation in the (i, i + 1)-plane, i.e., Ri(θ) is represented by the d × d

matrix

I

cos θ − sin θ

sin θ cos θ

I

De�ne for x ∈ Rd,

tr(x) = {i : xi 6= 0}

and for A ⊆ Rd

tr(A) =⋃x∈A

tr(x)

Say B,C ⊆ (χ∪ {0})d are linearly independent (and therefore �nite), while AB ⊆ B,

AC ⊆ C, and AB and AC are isometric.

Choose σB : [d] → [d] to be a permutation of [d] = {1, . . . , d} taking tr(AB) to

{1, . . . , |tr(AB)|} and choose σC : [d] → [d] to be a permutation of [d] taking tr(AC) to

{1, . . . , |tr(AC)|}. Given a permutation σ of [d], de�ne φσ(x1, . . . , xd) = (xσ(1), . . . , xσ(d)),

which is an isometry of Rd.

Set B′ = (φσB)′′B , C ′ = (φσC )′′C, AB′ = (φσB)′′AB and AC′ = (φσC )′′AC . Note that

we still have B′, C ′ ⊆ (χ ∪ {0})d.

Say

AB′ ={b1 = (b1

1, . . . , b1|tr(AB)|, 0, . . . , 0), . . . ,b|AB′ | = (b

|AB′ |1 , . . . , b

|AB′ ||tr(AB)|, 0, . . . , 0)

}

and

AC′ ={c1 = (b1

1, . . . , b1|tr(AB)|, 0, . . . , 0), . . . , c|AB′ | = (b

|AB′ |1 , . . . , b

|AB′ ||tr(AB)|, 0, . . . , 0)

}


while bj 7→ cj is an isometry.

Adjusting the rotation depending in which quadrant of the (|tr(AB)| − 1, |tr(AB)|)-

plane b1 lies, de�ne

f 11 = R|tr(AB)|−1

(arctan

(b1|tr(AB)|

b1|tr(AB)|−1

))

so that tr(f 11 (b1)) ⊆ {1, . . . , |tr(AB)| − 1}. Note that the above expression with arctan

can be replaced with an expression involving only +,−,×,÷,√, b1|tr(AB)| and b

1|tr(AB)|−1.

Continue until f 1|tr(AB)|−1 ◦ . . . ◦ f 1

1 (b1) = (1, 0, . . . , 0).

In the inductive step of our construction, suppose

tr((f i|tr(AB)−1|−1 ◦ . . . ◦ f i1) ◦ . . . ◦ (f 1|tr(AB)−1|−1 ◦ . . . ◦ f 1

1 )(bj)) = {1, . . . , j}

for all j ≤ i, for 1 < i ≤ |AB′|. Construct the composition of rotations f i+1|tr(AB)|−1◦. . .◦f

i+11

as in the base case but applying the construction to

(f i|tr(AB)−1|−1 ◦ . . . ◦ f i1) ◦ . . . ◦ (f 1|tr(AB)−1|−1 ◦ . . . ◦ f 1

1 )(bi+1)

instead of b1. Therefore,

tr((f i+1|tr(AB)−1|−1 ◦ . . . ◦ f

i+11 ) ◦ . . . ◦ (f 1

|tr(AB)−1|−1 ◦ . . . ◦ f 11 )(bj)) = {1, . . . , j}

for all j ≤ i+ 1.

We arrive at

AB′′ ={x1 = (1, 0 . . . , 0), . . . ,x|AB | = (x

|AB |1 , . . . , x

|AB ||AB |, 0 . . . , 0)

}⊆ (χ ∪ {0})d

isometric to AB and AC , and B′′ ⊇ AB′′ isometric to B.


Do this for AC′ and C′ and arrive at

AC′′ ={y1 = (1, 0 . . . , 0), . . . ,y|AB | = (y

|AB |1 , . . . , y

|AB ||AB |, 0 . . . , 0)

}⊆ (χ ∪ {0})d

isometric to AB and AC′ , and C′′ ⊇ AC′′ isometric to C.

Moreover, xj 7→ yj is an isometry. Finally, by induction, we can show that xji = ±yji ,

using the fact that the isometries of the sphere preserve the dot product: Recall that for

any d0 ∈ R, if f : Sd0 → Sd0 is an isometry, then for all x, y ∈ Sd0 ,

2− 2 〈x, y〉 = ||x||2 + ||y||2 − 2 〈x, y〉

= ||x− y||2

= ||f(x)− f(y)||2

= ||f(x)||2 + ||f(y)||2 − 2 〈f(x), f(y)〉

= 2− 2 〈f(x), f(y)〉 .

Therefore, without loss of generality, assume AB′′ = AC′′ . Note that |tr(AB′′)| = |AB′′ | =

|AB|.

Set e = max {|tr(B′′)|+ |tr(C ′′)| − |AB|, d}.

Choose σB′′ : [e]→ [e] to be a permutation taking {1, . . . , tr(B′′)} to

{|tr(C ′′)| − |AB|+ 1, . . . e} .

Choose σC′′ : [e]→ [e] to be a permutation taking {1, . . . , |AB|} to

{|tr(C ′′)| − |AB|+ 1, . . . , |tr(C ′′)|}

and {|AB|+ 1, . . . |tr(C ′′)|} to {1, . . . , |tr(C ′′)| − |AB|}.

Then A′′′ = (φσB′′ )′′AB′′ , B

′′′ = (φσB′′ )′′B′′ and C ′′′ = (φσC′′ )

′′C ′′ are such that A′′′ =


B′′′ ∩ C ′′′ and tr(A′′′) = tr(B′′′) ∩ tr(C ′′′)�this follows from linear independence of A′′′,

C ′′′ and B′′′. Set D = C ′′′ ∪B′′′.

Proposition 2.1.2. There is a countable dense ultrahomogeneous subset X of S∞ such

that every �nite subset of X is linearly independent (i.e., when embedded in a �nite

Euclidean space).

Proof. For X ⊆ Rd, de�ne tr(X) = {i : (∃x ∈ X)xi 6= 0} . Suppose χ ⊆ R is countable

and closed under +,−,×,÷,√. Set K to be the class of �nite metric spaces which are

isometric to some linearly independent χ ⊆ (χ ∪ {0})d inside the �nite dimensional unit

sphere centered at ~0, for some d ≥ 1. The class K is a Fraïssé class:

Heredity : A ≤ B and B ∈ K: Without loss of generality, we may assume that

B ⊆ Sd and B is linearly independent. The fact that A ⊆ B implies that A is linearly

independent.

Joint Embedding Property : A,B ∈ K. Without loss of generality, we may assume

that A,B ⊆ Sd and tr(A) ∩ tr(B) = ∅. Set C = A ∪ B. Then the coordinates of the

elements of C are still in χ ∪ {0}. Also, the fact that A and B are linearly independent,

while tr(A) ∩ tr(B) = ∅, implies that A ∪B is linearly independent.

Amalgamation Property : Refer to proposition 2.1.1.

Set X to be the Fraïssé limit of K. By Fraïssé's theorem, it is ultrahomogeneous.

Cover S∞ with countably many open balls, B1, B2, . . .. Enumerate σ : ω → X.

We will construct a X ′ ⊂ `2 and a bijective isometry to X, concurrently. Let en =

(0, . . . , 1, 0, . . .) be the vector with 1 in the n-th spot and 0's everywhere else.

We proceed by induction on l: Let l > 1. Say we have chosen a linearly independent

X ′n = {x′1, . . . , x′n} ⊆ S∞ and an isometric copyXn = {x1, . . . , xn} ⊆ X so that X ′n∩Bi 6=

∅ for all i < l; each nonzero coordinate of x′i is in χ; i = max(tr(x′i)). Note that the

linear span of �nitely many elements of `2 is nowhere dense in `2 (refer to [Ha]). Pick

x′ ∈ Bl so that it is of the form (w1, . . . , wd 6= 0, 0, . . .), with wi ∈ χ and X ′n∪{x′} linearly


independent, and d > n. Set x′n+i = en+i, i < d − n. Set x′d = x′. By ultrahomogeneity

of X, we can pick xn+i ∈ X so that Xn ∪ {xn+i}i≤d−n is isometric to X ′n ∪{x′n+i

}i≤d−n.

Let K ={j : σ(j) /∈ {xi}i≤d

}.

Enumerate K ∩ l: γ : |K ∩ l| → K ∩ l. We can embed

ψ : {x1, . . . , xd} ∪ {σ(γ(j)) : j < |K ∩ l|} → span{e1, . . . , ed+|K∩l|

}so that the image of xj is still x′j and d + j + 1 = max(tr(ψ(σ(γ(j))))) and all the

coordinates of ψ(σ(γ(j))) are in χ (refer to the proof of the amalgamation property

above). Set x′d+j = ψ(σ(γ(j)))) and set xd+j = σ(γ(j)).

X ′ is dense, if the Bi's are chosen to be an enumeration of B 1n(zi) for all n ∈ ω\{0},

and a �xed countable and dense {zi}∞i=1 ⊆ `2. Such an enumeration is possible, because

the union of countably many countable sets is still countable, i.e., {B 1n(zi)}n,i∈ω\(0) is

countable.

2.2 Inner Product Structures

Suppose A,B ⊆ Rn and <A, <B are linear orderings of A,B. De�ne(BA

)||.||,< to be the

set of subsets of B such that there is an order preserving isometry from A to B. De�ne(BA

)〈,〉,< to be the set of subsets of B such that there is an order preserving bijection from

A to B that also preserves the inner product. Supposing that the symbol ∗ is either

||.|| or 〈, 〉, de�ne C →∗,< (B)Ar to mean that for any r-colouring of(CA

)∗,< there is a

B′ ∈(CB

)∗,< such that

(B′

A

)∗,< is monochromatic. De�ne GA to be the complete linearly

ordered graph with V (GA) = A , labeled by λGA({x, y}) = 〈x, y〉 for {x, y} ∈ E(GA)

with the linear ordering on V (GA) = A being exactly the same as the ordering on A .

De�ne

ιX = {〈x, y〉 : x, y ∈ X}.


Let G′ be a �nite linearly ordered labeled graph such that G′ → (GB)GAr . Without loss

of generality we can assume G′ is complete by adding missing edges and labeling them

0. We can moreover assume that the labels in G′ are from among ιB ∪ {0}. Let δ > 0.

Set N =(|V (G′)|

2

)+ |V (G′)|. Let

f : [V (G′)]2 → {1, . . . ,(|V (G′)|

2

)}

and

g : V (G′)→ {1, . . . , |V (G′)|}

be bijections. Recall that tr(x) = {i : xi 6= 0}. We will de�ne {xi}1≤i≤|V (G′)| ⊆ RN .

In particular, we want |tr(xi) ∩ tr(xj)| = 1, for i 6= j. De�ne sgn : V (G′) × V (G′) →

{−1, 0, 1} to be

sgn(a, b) = sgn(λG′({a, b}))

if g(a) < g(b) and sgn(a, b) = 1, otherwise. Let a ∈ V (G′). For all b 6= a, set xg(a)f({a,b}) =

sgn(a, b) ·√|λG′({a, b})| and x

g(a)f({c,b}) = 0 if c 6= a. Let i be such that

(|V (G′)|2

)< i ≤(|V (G′)|

2

)+ |V (G′)| and i 6=

(|V (G′)|2

)+ g(a); set x

g(a)i = 0. Set

Rδ =√

maxi

Σ1≤j≤(|V (G′)|

2 )(xij)2 + δ.

Finally, set

xg(a)

(|V (G′)|2 )+g(a)

=

√R2δ − Σ

1≤j≤(|V (G′)|2 )(x

g(a)j )2.

It follows that if a, b ∈ V (G),⟨xg(a), xg(b)

⟩= λG′({a, b}). Set CRδ = {xi}1≤i≤|V (G′)|.

Note that CRδ ⊆ SN(Rδ) is linearly independent. De�ne a linear ordering on CRδ by

setting xi < xj if and only if g−1(i) < g−1(j). Therefore,

Proposition 2.2.1. CRδ →〈.,.〉,< (B)Ar .

Note that for x, y ∈ Sk, ||x− y||2 = 2− 2 〈x, y〉, i.e., on the unit sphere, the distance


between points depends only on the inner product. Set M to be the minimum possible

|V (G′)|. M depends on the graphs GA and GB (and r), but not on the actual values

of the labels. Let us de�ne an equivalence ∼ on the class of �nite linearly ordered

labeled graphs (with labels in R): G0 ∼ G1 if and only if there is an order preserving

bijection α : V (G0) → V (G1) and a bijection β : range(λG0) → range(λG1) such that

{α(a), α(b)} ∈ E(G1) if and only if {a, b} ∈ E(G0), while if a, b ∈ V (G0), then

λG1(α(a), α(b)) = β(λG0(a, b)).

Now, de�ne an equivalence relation on triples consisting of a pair of �nite linearly or-

dered labeled graphs and an embedding between them: Given f : V (G0) → V (G1) and

g : V (G2) → V (G3), de�ne (G0, G1, f) ≡ (G2, G3, g) to mean that G1 ∼ G3 with corre-

sponding bijections α and β (from above) for which g−1αf is also an order preserving

isomorphism of G0 and G2, while if a, b ∈ V (G0), then

λG2((g−1αf)(a), (g−1αf)(b)) = β(λG0(a, b)).

For A ⊆ B, de�ne [A,B] to be the set of triples (GA′ , GB′ , f) equivalent to (GA, GB, id)

where id the the identity embedding id : V (GA)→ V (GB). In other words, M depends

on the equivalence classes [A,B] (and r). Therefore, we arrive at the following:

Proposition 2.2.2. Say A ⊆ B ⊆ Sk are �nite and linearly ordered. Let M =

M([A,B], r) . Suppose(M2

)· max{|x| : x ∈ ιB} < 1. Then we can construct CRδ as

above, but with Rδ = 1 and |V (G′)| = M . Consequently, CRδ ⊆ S(M2 )+M and

CRδ →||.||,< (B)Ar .

Remark 2.2.3. Note that the condition(M2

)max{|x| : x ∈ ιB} < 1 implies that B gets

�closer� to being an orthogonal set, asM gets larger. Also, although max{|x| : x ∈ ιB} =


max{|x| : x ∈ ιCRδ}, we have that M([A,CRδ ], r) > M([A,B], r).

Corollary 2.2.4. (Su�ciently-orthogonal sets) Say ρ > 0 and A ⊆ B ⊆ Sk(ρ) are �nite

and linearly ordered. Let M = M([A,B], r). Suppose(M2

)·max{|x| : x ∈ ιB} < ρ2. Then

we can construct a linearly independent C ⊆ S(M2 )+M(ρ) such that

C →||.||,< (B)Ar .

For x ∈ Rk, y ∈ Rl, de�ne x × y to be the vector z in Rk+l such that zi = xi

for 1 ≤ i ≤ k and zk+i = yi for 1 ≤ i ≤ l. For A ⊆ Rk, B ⊆ Rl, de�ne A × B =

{x× y : x ∈ A, y ∈ B}. Let 0k be the origin in Rk. Note that if x0 ∈ Rk, we then have

{x0}×Sl(ρ) ⊆{x ∈ Rk+l : ||x− (x0 × 0l)|| = ρ

}. We can now say something about �nite

linearly ordered subsets contained in an intersections of two spheres:

Corollary 2.2.5. Let ρ1 > ρ2 > 0. Suppose A ⊆ B ⊆ Sk(ρ1) are �nite and linearly

ordered. Suppose there are x0 ∈ Rl, for 0 < l < k, and A′ ⊆ B′ ⊆ Sk−l(ρ2) such that

{x0} × A′ = A and {x0} × B′ = B, while(M2

)· max{|x| : x ∈ ιB′} < ρ2

2, where M =

M([A′, B′], r). Then we can construct C ′ ⊆ S(M2 )+M(ρ2) with {x0} × C ′ ⊆ Sl+(M2 )+M(ρ1)

and

{x0} × C ′ →||.||,< (B)Ar .

Moreover, {x0} × C ′ is linearly independent.

Let L be a �nite relational language with Ri (1 ≤ i ≤ n) as the relation symbols with

arity ni. A weak homomorphism f : A→ B between structures satis�es the following: if

(a1, . . . , ani) ∈ RAi , then (f(a1), . . . , f(ani)) ∈ RB

i . Let T be a �nite set of structures. A

structureA is T -free if and only if there is no structureB ∈ T and a weak homomorphism

f : B → A. Due to Theorem 3.2 in [HL], the class of T -free structures has the Hrushovski

property.


Suppose ni = 2 for 1 ≤ i ≤ n. Let G be the class of �nite labeled graphs with labels

in {1, . . . , n}. Let g ∈ {f : k → 2 : k ⊆ {1, . . . , n}, |k| = 2} = I. Set Tg = {0, 1}. For

i /∈ dom(g), set RTgi = ∅. For i ∈ dom(g), set R

Tgi = {(g(i)(mod 2), g(i) + 1(mod 2)}.

For 1 ≤ i ≤ n, set Si = {0} and RSij = ∅ for j 6= i, and RSi

i = {(0, 0)}. Set T = {Tg :

g ∈ I} ∪ {Si : 1 ≤ i ≤ n}.

Let A be T -free. Then there is a unique (up to isomorphism) labeled graph GA ∈ G

corresponding to A: Since A is T -free, for every {x, y} ∈ [A]2, if (x, y) ∈ RAi , then

(x, y) /∈ RAj and (y, x) /∈ RA

j , for i 6= j. Moreover, (x, x) /∈ RAj for all x ∈ A, and

1 ≤ j ≤ n. Let V (GA) = A and {x, y} ∈ E(GA) if and only if (x, y) ∈ RAi for some i.

For {x, y} ∈ E(GA), set λ({x, y}) = i, if (x, y) ∈ RAi or (y, x) ∈ RA

i . Moreover if G ∈ G,

then there is a corresponding T -free structure AG: Set AG = V (G) and (x, y) ∈ RAGi if

and only if λ({x, y}) = i . Therefore, by Theorem 3.2 in [HL],

Proposition 2.2.6. G has the Hrushovski property.

Corollary 2.2.7. If B ∈ E, then there is C ∈ E and an injection f : B ↪→ C such that

any partial bijection of f ′′B which preserves inner products can be extended to a bijection

of C which preserves inner products.

For unordered labeled graphs G0 and G1, de�ne the equivalence relation G0 ∼2 G1

to mean that there is a bijection α : V (G0) → V (G1) and a bijection β : range(λG0) →

range(λG1) such that {α(a), α(b)} ∈ E(G0) if and only if {a, b} ∈ E(G1), while if a, b ∈

V (G0), then

λG1(α(a), α(b)) = β(λG0(a, b)).

De�ne [B]2 = {GB′ : GB′ ∼2 GB}. As was in the case of the Ramsey property, if

B ⊆ Sk(ρ), there is M2 = M2([B]2) such that if(M2

2

)· max{|x| : x ∈ ιB} < ρ2, then

there are C ⊆ S(M22 )+M2

(ρ) and an isometric injection f : B → C such that any partial

isometry of f ′′B extends to an isometry of C. Corollary analogous to 2.2.5 also follows:


Corollary 2.2.8. Let ρ1 > ρ2 > 0. Suppose B ⊆ Sk(ρ1) are �nite. Suppose there are

x0 ∈ Rl, for 0 < l < k, and B′ ⊆ Sk−l(ρ2) such that {x0} × B′ = B, while(M2

)·

max{|x| : x ∈ ιB′} < ρ22, where M = M2([B′]). Then there are C ′ ⊆ S(M2 )+M(ρ2) with

{x0} × C ′ ⊆ Sl+(M2 )+M(ρ1) and an isometric injection f : B → {x0} × C ′ such that any

partial isometry of f ′′B extends to an isometry of {x0} × C ′.

Suppose χ ⊆ R is countable. Let the signature L consist of 2-ary relation symbols Rx

with x ∈ χ ∪ {0}. De�ne OK′ to be such that A ∈ OK′ if and only if A has signature

L, A ⊆finite Rk, for some k ∈ N, while (a, b) ∈ RAx if and only if a 6= b and 〈a, b〉 = x.

Essentially, OK′ is the class of �nite ordered labeled graphs with labels in χ. In case

we want to relate these structures to the ones from Proposition 2.1.1, we should further

assume that χ is closed under +,−,×,÷,√. Note that if A,B ∈ OK′, then C ∈ OK′,

where C = CRδ from the discussion leading up to Proposition 2.2.1. Therefore,

Corollary 2.2.9. OK′ has the Ramsey Property.

Corollary 2.2.10. Aut(Flim(OK′)) is extremely amenable.

Example 2.2.11. Set Flim(OK′) = K and K = {x1, x2, . . .}. We can recursively con-

struct a map ϕ : K → `2 such that (a, b) ∈ RKx if and only if a 6= b and 〈ϕ(a), ϕ(b)〉 = x:

De�ne λ(a, b) = x if and only if a 6= b and (a, b) ∈ RKx . Set ϕ(x1) = (1, 0, . . .). Suppose

ϕ(xl) has been de�ned for l < i. Now recursively de�ne ϕ(xi): Suppose ϕ(xi)k has been

de�ned for k < j < i . Set

ϕ(xi)j = λ(xi, xj)− Σk<jϕ(xi)kϕ(xj)k

Finally, set ϕ(xi)i = 1 and ϕ(xi)j = 0 for j > i.

Note that there is f ∈ Aut(K) such that ϕ ◦ f ◦ ϕ−1 does not extend to a unitary

operator on the Hilbert space `2: ||ϕ(x1)|| = 1 but ||ϕ(xj)|| > 1 for some j. Then f(x1) =

xj for some f ∈ Aut(K), by ultrahomogeneity of K. Thus, (ϕ ◦ f ◦ϕ−1)(ϕ(x1)) = ϕ(xj),

yet ||(ϕ ◦ f ◦ ϕ−1)(ϕ(x1))|| 6= ||ϕ(x1)||.


Fact 2.2.12. Say χ contains a positive nonzero element and is closed under addition (or

simply assume that χ is not bounded above). If ϕ : K → `2 is such that (a, b) ∈ RKx if

and only if a 6= b and 〈ϕ(a), ϕ(b)〉 = x. Then there is f ∈ Aut(K) such that ϕ ◦ f ◦ ϕ−1

does not extend to a unitary operator on the Hilbert space `2.

Proof. χ is closed under addition and since it was assumed it contains a positive nonzero

element, χ contains arbitrarily large numbers. Let x, y ∈ `2. The polarization identity

holds:

0 ≤ ||x− y||2 = ||x||2 + ||y||2 − 2 〈x, y〉 .

Thus,

2 〈x, y〉 ≤ ||x||2 + ||y||2.

Therefore, since χ contains arbitrarily large numbers, we have that for any r > 0, there

is x ∈ ϕ′′K such that ||x|| > r.

Let x, y ∈ K be such that ||ϕ(y)|| > ||ϕ(x)||. By ultrahomogeneity of K, there is

f ∈ Aut(K) such that f(x) = y. Thus, (ϕ ◦ f ◦ ϕ−1)(ϕ(x)) = ϕ(y), yet ||(ϕ ◦ f ◦

ϕ−1)(ϕ(x))|| 6= ||ϕ(x)||.

Chapter 3

The Euclidean Metric

3.1 A Rigid Subset of `2

We begin with a construction of a countable dense subset of `2 (or more generally any

dense subset of `2) whose corresponding age satis�es some Ramsey-type properties, triv-

ially � and such that every �nite subset is linearly independent. However, the set will

fail to be ultrahomogeneous for singletons. For any A, de�ne

∆A = {δ : (∃x, y ∈ A)d(x, y) = δ} .

The construction below was motivated by a list of �strange� subsets of Rn found in [Ci],

constructed using (trans�nite) induction.

Fact 3.1.1. Let D′ ⊆ `2 be dense, then there is a countable dense D ⊆ D′ such that for

all A ⊆ X with |A| ≥ 2 and X ⊇ B ⊇ A, we have(BA

)||.|| = {A} and so B →||.|| (B)A2 ,

trivially. Moreover, if X = `2, then D can be chosen to be such that every �nite subset

of D is linearly independent.

Proof. For any n we have balls Bn0 , . . . of radius

1nthat cover D′. Thus {Bn

i }n,i∈ω is

countable. Enumerate it: B0, B1, B2, . . . De�ne D inductively. Pick x0 ∈ B0 ∩ D′. Say

28

Chapter 3. The Euclidean Metric 29

we have chosen Dn−1 = {x0, . . . , xn−1}. Pick

xn ∈ (Bn ∩D′)\{x ∈ X : (∃y ∈ Dn−1)(∃δ ∈ ∆Dn−1)d(x, y) = δ

}Then D = {xn}n∈ω is such that no two distinct pairs of points have the same distance

assigned to them. We can choose xn so that every Dn is linearly independent, because

the linear span of a �nite number of vectors is nowhere dense (as before, see [Ha]).

3.2 Colouring using Linear Dependence and Orienta-

tion

In this section, we show the signi�cance of linear dependence and the geometric no-

tion of orientation in avoiding non-Ramsey classes of �nite linearly ordered structures

corresponding to �nite linearly ordered metric spaces. In particular, colouring faces of

platonic solids � as opposed to their vertices, as mentioned earlier � yields negative

results. Below, let ∗ be a symbolic placeholder for either ||.|| or ||.||, <. Recall that

Sn = {x ∈ Rn : ||x|| = 1}.

Proposition 3.2.1. Let n ≥ 2. Let B ⊆ Sn be linearly ordered. If there is a subset

A ⊆ B such that span(A) = Rn and∣∣(BA

)∗

∣∣ ≥ 2, then there is r such that for any N , and

any C ⊆ SN , it is not the case that C →∗ (B)Ar+1.

Proof. Let r = |Iso(A)| ·∣∣(BA

)∣∣2. Let C ⊆ SN be any set of points such that B ⊆ C.

Let G be a graph with V (G) =(CA

)∗ and (A′, A′′) ∈ E(G) if and only if A′, A′′ ∈

(B′

A

)for some B′ ∈

(CA

)∗. Consider a connected component G′. Let C ′ =

⋃A′∈V (G′) A

′. Let

A′ ∈ V (G′). Let B be the set of B′ ∈(C′

B

)∗ such that A′ ∈

(B′

A

)∗. Let B′, B′′ ∈ B. Let

f : B → B′ and g : B → B′′ be isometric bijections. Without loss of generality, we can

assume that B′ = B ⊆ Rn. However, because A ⊆ B′′ ⊆ SN and span(A) = Rn, we have

that B′′ ⊆ Rn ∩SN = Sn. Suppose we have an isometry f : B′ ∼= B′′ such that f ′′A = A.


Thus, f � A ∈ Iso(A). Therefore, there are at most |Iso(A)| such f . In other words,

G′ is a graph whose vertices are of valence at most r. Consequently, we may recursively

colour its vertex set using r+ 1 many colours, so that no edge has endpoints of the same

colour. Since∣∣(BA

)∗

∣∣ ≥ 2, the proposition follows.

In the case of boron tree structures, the concept of orientation will be used to compute

the respective Ramsey degree. Trees are planar, and orientation evokes the concept

of orientation as computed using, for example, the triple product in the case of R3

and the wedge product in higher dimensions. In fact, using linear independence and

the properties of the wedge product (i.e., the exterior product), we may also construct

counterexamples similar to the preceding one. Let {e1, e2, e3} be the standard basis of

R3. Let x1, x2, x3 ∈ R3. Recall that

x1 × x2 = det

e1 e2 e3

x11 x1

2 x13

x21 x2

2 x23

.

Since we are here considering only 3-dimensional Euclidean space, for the sake of brevity,

let us de�ne the symbolic ∧-product only for elements of R3, as follows:

x1 ∧ x2 ∧ x3 = x1 · (x2 × x3)(e1 ∧ e2 ∧ e3)

where x1 · (x2 × x3) represents a signed volume. Let T0 = {v1, v2, v3} ⊆ S3 be a scalene

triangle whose vectors are linearly independent. Consequently, since they are linearly

independent, v1 ∧ v2 ∧ v3 6= 0(e1 ∧ e2 ∧ e3). Then Let T1 = {u1, u2, u3} ⊆ S3 be a scalene


triangle congruent to T0, but such that T0 ∩ T1 = ∅ and

u1 ∧ u2 ∧ u3 = u1 · (u2 × u3)(e1 ∧ e2 ∧ e3)

= det

u1

u2

u3

(e1 ∧ e2 ∧ e3)

= − det

v1

v2

v3

(e1 ∧ e2 ∧ e3)

= −v1 ∧ v2 ∧ v3.

Set B0 = T0 ∪T ′0. Suppose further that B0 ⊆ Sn is linearly ordered so that T0 and T1 are

isometric via an order preserving map. This is possible because T0 ∩ T1 = ∅.

Proposition 3.2.2. For r = 2, for any N , and any C ⊆ SN , it is not the case that

C →∗ (B0)T0r .

Proof. We begin as in the previous proposition: Let C ⊆ SN be any set of points such

that B0 ⊆ C. Let G be a graph with V (G) =(CT0

)∗and (T ′, T ′′) ∈ E(G) if and only

if T ′, T ′′ ∈(B′

T

)for some B′ ∈

(CT0

)∗. Consider a connected component G′. Let C ′ =⋃

T ′∈V (G′) T′. Let T ′ ∈ V (G′). Let B be the set of B′ ∈

(C′

B0

)∗such that T ′ ∈

(B′

T0

)∗.

Let B′ 6= B′′ ∈ B. Without loss of generality, we can assume that B′ = B0 ⊆ R3.

However, because T0 ⊆ B′′ ⊆ SN and span(T0) = R3, we have that B′′ ⊆ R3 ∩ SN = S3.

Now, set T ′′ to be the unique member of(B′′

T0

)∗\{T0}. Let f : vi 7→ wi be an isometry

where {w1, w2, w3} = T ′′. Then u1 ∧ u2 ∧ u3 = −w1 ∧ w2 ∧ w3. We may also conclude

that⋃B′∈B B

′ ⊆ S3. Enumerate V (G′) = T ′1, . . . T′k. Recall that for any rotation (i.e.,

a direct isometry) f and any vectors z1, z2, z3 ∈ R3 , f(z1 ∧ z2 ∧ z3) = z1 ∧ z2 ∧ z3.

Enumerate the members of each T ′j so that T′j = {wj1, w

j2, w

j3} is isometric via a rotation

(i.e., a direct isometry) ρ to either T0 or T1 so that ρ(wjl ) = vjl or ρ(wjl ) = ujl . There


is a unique such rotation, because T0 and T1 are scalene. Now, colour V (G′) as follows:

c(T ′j) = sgn(wj1 · (wj2 × w

j3)).

If we disregard linear ordering, then we may also use scalene triangles to �nd a

counterexample, even in a more general setting: Let T0 and T1 be as before, but let them

share an edge E in B0. Let us also remove the assumption that B0 is in a Euclidean space.

Then, with G′ as in the above proof, set B to be every B′ = T ′∪T ′′ with T ′, T ′′ ∈ V (G′).

Then for any B′, B′′ ∈ B, B′ ∩B′′ is an edge E ′ = {x, y} congruent to E. Colour(C′

T0

)as

follows: c(T ′) = d(x0, x) where {x0} = T ′\E.

3.3 Spreads

For a ∈ Rd, d ≤ d′, d, d′ ∈ ω, J ⊆ [d′], de�ne Spread(a, J) to mean the set of b ∈ Rd′

such that bji = ai for some j1 < . . . < jd ∈ J , while bk = 0 for other k's. De�ne

tr(a) = {i : ai 6= 0} and if X ⊆ Rd, tr(X) =⋃

a∈X tr(a). De�ne Spread(a, J) ∈ M∆0

to be the corresponding structure, where ∆(Spread(a, J)) ⊆ ∆0. In particular, we are

interested in spreads because of the proposition below.

Proposition 3.3.1 (Matou²ek-Rödl, 1995 [MR]). For every δ > 0 and every m, there

exist N , n and an n-dimensional unit vector a ∈ Sn such that for some m-dimensional

linear subspace Z ⊆ RN and for every z ∈ SN ∩ Z there is a point y ∈ Spread(a, [N ]) at

distance at most δ from z.

If the ai's are �close enough� to each other, we only need to invoke Ramsey's theorem

to prove the conclusion of the following proposition.

Proposition 3.3.2. Say a ∈ Rd, r, e ∈ N, e ≥ d2. Also, we require that ai > 0 for all

1 ≤ i ≤ d and for some integer c > 0, 0 < aiaj − c < 1d, for all 1 ≤ i, j ≤ d. Then there

is an N such that

Spread(a, [N ])→ (Spread(a, [e]))Spread(a,[d2])r


Proof. By Ramsey's theorem, there is an N such that N → (e)d2

r . (*) Suppose

κ :

(Spread(a, [N ])

Spread(a, [d2])

)→ r

is a colouring. We will show that if X ∈(

Spread(a,[N ])Spread(a,[d2])

), then X = Spread(a, J) where

|J | = d2 and J ⊆ [N ]. The conclusion will follow from (*).

Say X ∈(

Spread(a,[N ])Spread(a,[d2])

). Because X is isometric to Spread(a, [d2]), there are d many

x1, . . . ,xd ∈ X such that 〈xi,xj〉 = 0 for i 6= j. (**)

Set

J =d⋃i=1

tr(xi).

Let x ∈ X be such that x 6= xi for all i. Set

Ii ={k : xk · xik 6= 0

}and

I =d⋃i=1

Ii

Claim 3.3.3. Ii ∩ Ij = ∅, for i 6= j.

Proof. In the statement of the proposition, we assumed c > 0 and 0 < akal − c < 1d, for

all k, l. Thus, akal > 0, for all k, l. Suppose there are i, j such that Ii ∩ Ij 6= ∅, while

i 6= j. Then there is k0 such that xik0 6= 0 and xjk0 6= 0, i.e.,

{xik0 , xjk0} ⊆ {a1, . . . , ad}.

Therefore, xik0xjk0> 0 and consequently 〈xi,xj〉 > 0, which contradicts (**).

Thus, |I| = Σdi=1|Ii|. Recall that given x ∈ R, the �oor function bxc is equal to the


largest integer smaller or equal to x. Observe that

⟨x,xi

⟩= xγ(0)x

iγ(0) + . . .+ xγ(|Ii|−1)x

iγ(|Ii|−1)︸︷︷︸

|Ii| many terms

.

for some bijection γ : |Ii| → Ii. Thus,

⟨x,xi

⟩− c · |Ii| = (xγ(0)x

iγ(0) − c) + . . .+ (xγ(|Ii|−1)x

iγ(|Ii|−1) − c).

Since a ∈ Rd, |Ii| ≤ d. Thus, by the assumption made on ai's,

0 < (xγ(0)xiγ(0) − c) + . . .+ (xγ(|Ii|−1)x

iγ(|Ii|−1) − c) < |Ii| ·

1

d≤ d · 1

d= 1

and consequently ⌊d∑i=1

⟨x,xi

⟩⌋= c · |I| .

Moreover, there are d many non-zero coordinates in x. Since a ∈ Rd,

d∑i=1

⟨x,xi

⟩= a1aσ(0) + . . .+ adaσ(d−1)︸︷︷︸

d many terms

for some function σ : d→ [d]. Observe that

d∑i=1

⟨x,xi

⟩− d · c = (a1aσ(0) − c) + . . .+ (adaσ(d−1) − c).

By the assumption made on ai's,

0 < (a1aσ(0) − c) + . . .+ (adaσ(d−1) − c) < d · 1

d= 1.


Hence,

d · c =

⌊d∑i=1

⟨x,xi

⟩⌋

Thus, |I| = d. Now, x ∈ X ⊆ Spread(a, [N ]) implies that |tr(x)| = d. Therefore,

tr(x) = I ⊆ J . Hence, x ∈ Spread(a, J). Therefore, X ⊆ Spread(a, J). Finally,

note that X ∼= Spread(a, [d2]) implies that |X| =(d2

d

)= |Spread(a, J)| and so X =

Spread(a, J).

�d2� is necessary for this particular proof: Here is an example for which an argument

similar to the one above would not work:

Spread((1, 1), [3]) = {(1, 1, 0), (1, 0, 1), (0, 1, 1)}

is congruent to

{(1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)}

and consequently we would not be able to �nd an appropriate J and show that X =

Spread(a, J).

Given a �xed a ∈ Rd, de�ne Sa to be the class of all metric spaces isometric to

Spread(a, [k]), for some k ≥ d. Given a ∈ Rd consider the following property: For all

k ∈ N, if X ∈ Sa and X is a metric subspace of Spread(a, [k]), then X = Spread(a, J)

for some J ⊆ [k]. For example, (1) satis�es this property, while (1, 1) does not, as we

have shown in the preceding example. If a does satisfy this property, then the class Sa

satis�es the Ramsey property, by Ramsey's theorem.

We will now investigate what happens when we augment our spreads by removing

the condition �j1 < . . . < jd ∈ J .� Let (a1, . . . , an) ∈ Rn. Let J = (ji)1≤i≤n be a

sequence of positive natural numbers from [N ]. De�ne spread(a, J) ∈ RN to be such that

(spread(a, J))k = 0 for k /∈ J and (spread(a, J))ji = ai for 1 ≤ i ≤ n. For I ⊆ N , recall


that

Spread(a, I) = {spread(a, (ji)1≤i≤n) : j1 < . . . < jn ∧ (∀i)ji ∈ I}.

Now, we also de�ne,

Spread∗(a, I) = {spread(a, (ji)1≤i≤n) : (∀i)ji ∈ I}.

Let e ≥ d2 and let a ∈ Rd. From the proof of Proposition 3.3.2, if Xi ∈(

Spread(a,[e])Spread(a,[d2])

),

then Xi = Spread(a, Ji) where |Ji| = d2 and Ji ⊆ [e]. Thus, an isometry f from Xi to Xj

can be de�ned using a monotone bijection between Ji and Jj. Suppose Iso(Spread(a, [d2]))

consists of only isometries de�ned using permutations of [d2]. Let g be some isometry

between Xj and Xi. Then f ◦ g is an automorphism of Xj. Thus, g can be de�ned using

a bijection between Ji and Jj. This extends to an automorphism of Spread∗(a, [e]).

Given a �xed a ∈ Rd, de�ne S∗a to be the class of all metric spaces isometric to

Spread∗(a, [k]), for some k ≥ d. Suppose a ∈ Rd has the following property: For all

k ∈ N, if X ∈ S∗a and X is a metric subspace of Spread∗(a, [k]), then X = Spread∗(a, J)

for some J ⊆ [k]. Therefore, if X1, X2 ∈ S∗a and X1, X2 are metric subspaces of Y =

Spread∗(a, [k]), then there is a permutation matrix taking X1 to X2 � which is also an

isometry of all of Y .

Example 3.3.4. Let a = (1, 2, 3), I = {1, 2, 3}. Then

Spread∗(a, I) = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}.

Note that

A = {(1, 2, 3), (1, 3, 2), (2, 1, 3)} ∼= {(3, 2, 1), (2, 3, 1), (3, 1, 2)}

and Spread∗(a, I) both span R3. Therefore, by Proposition 3.2.1 there is an r such that


there cannot be a metric space C embeddable in Rn for which C →||.|| (Spread∗(a, I))Ar .

We shall now combine spreads and su�ciently-orthogonal sets to show a Hrushovski-

type property for simplices, by closely following Matou²ek and Rödl's proof of Theorem

1.3.4. First, let us verify that an almost-regular set can be made su�ciently-orthogonal

at a crucial point in the proof.

Fact 3.3.5. Let β ≥ 0. We can �nd Y = {y1, . . . , yd} ⊆ Rd so that 〈yi, yj〉 = 0 and

||yi − yj|| = β for i 6= j, while ||yi|| = β√2.

Proof. Set (yi)i = β√2and (yi)j = 0 for i 6= j.

Then 〈yi, yj〉 = 0 for i 6= j and

||yi − yj||2 =β2

2+β2

2= β2

Fact 3.3.6. Let β > 0. Let X = {x1, . . . , xd}. If 0 < ||xi−xj||2−β2 < ε , for some small

ε > 0, then we can �nd {y1, . . . , yd} ⊆ Rd such that for all i, j, ||xi−xj|| = ||yi− yj|| and

|〈yi, yj〉| < δ(ε, β, d), for some small δ that depends on ε, β and d � which is small if ε is

small.

Proof. First, we show that X can be approximated with a regular simplex.

Let f : [d]2 →(d2

)be a bijection. De�ne wi (i < d) so that

(wi)f({i,j})+1 =1√2

√||xi − xj||2 − β2

(wi)f({k,j})+1 = 0

for k 6= i, j 6= i. Set R = 1√2

√dε and

(wi)(d2)+i+1 =

√R2 − 1

2Σdk=1(||xi − xk||2 − β2)


so that the set {w0, . . . , wd−1} is in the sphere with radius R = 1√2

√dε, centred at 0.

Finally, set

(wi)(d2)+j+1 = 0

for j 6= i.

Set y′i = xi ∗ wi. Then

||y′i − y′j||2 = ||xi − xj||2 + 2R2 − 21

2(||xi − xj||2 − β2)

= β2 + 2R2 = β2 + dε

Also

||y′i − xi||2 = ||wi||2 =dε

2< dε

and

∣∣||y′i − y′j||2 − ||xi − xj||2∣∣ = ||wi − wj||2 = 2R2 − 2 〈wi, wj〉 < dε

Find the isometry of Rd that maps the set {y′i}di=1 to the simplex {vi}di=1(from Fact

3.3.5) with 〈vi, vj〉 = 0 for i 6= j, as well as

||vi − vj||2 = β2 + dε

for i 6= j and ||vi|| = 1√2

√β2 + dε. Map xi's via the same isometry, yielding yi's such

that ||xi − xj|| = ||yi − yj|| and ||yi − vi||2 < dε.

Observe that

||yi|| = ||yi + vi − vi|| ≤ ||yi − vi||+ ||vi|| <√dε+ ||vi||


and

||vi|| ≤ ||yi − vi||+ ||yi|| <√dε+ ||yi||

Thus,

||vi|| −√dε < ||yi|| < ||vi||+

√dε

and, assuming ε is small enough, we can conclude that

||vi||2 − 2||vi||√dε+ dε < ||yi||2 < ||vi||2 + 2||vi||

√dε+ dε

Consequently,

∣∣||vi||2 − ||yi||2∣∣ < 2||vi||√dε+ dε = 2

(1√2

√β2 + dε

)√dε+ dε

Now, if i 6= j, then

2 |〈yi, yj〉| = 2 |〈yi, yj〉 − 〈vi, vj〉| ≤∣∣||yi||2 − ||vi||2∣∣

+∣∣||yj||2 − ||vj||2∣∣+

∣∣||yi − yj||2 − ||vi − vj||2∣∣< 2||vi||

√dε+ dε+ 2||vj||

√dε+ dε+ 2dε+ ε

= 4

(1√2

√β2 + dε

)√dε+ 4dε+ ε

which is small if ε is small.

Let us recall the simpler form of Corollary 2.2.8:

Lemma 3.3.7. (Su�ciently-orthogonal sets and the Hrushovski property) Say ρ > 0 and

B ⊆ Sk(ρ) is �nite. Let M = M([B]2). Suppose(M2

)· max {|〈x, y〉| : x, y ∈ B} < ρ2.

Then we can construct a linearly independent C ⊆ S(M2 )+M(ρ) such that B embeds in C

via an isometry f and any partial isometry of f ′′B extends to an isometry of C.


Given n, there are only �nitely many classes [B]2 with |B| = n. Thus, we can replace

M above with M(|B|). Therefore, by Fact 3.3.6, an almost-regular simplex can be made

su�ciently-orthogonal, in the proof of Theorem 3.3.9 below.

Observe that if x0, y0 ∈ Spread∗(a, J) are �xed, then the map of size 1 which takes

x0 to y0, extends to an isometry of Spread∗(a, J) (this is simply a permutation matrix).

If f is an isometric automorphism of X and g is an isometric automorphism of Y ,

then

x ∗ y 7→ f(x) ∗ g(y)

is an isometric automorphism of X ∗ Y .

Recall Schönberg's characterization of embaddability of �nite metric spaces in �nite

dimensional Euclidean spaces.

Theorem 3.3.8 (Schönberg, 1938 [Sch]). Suppose X = {x1, . . . xd} is a �nite metric

space. Set dij = d(xi, xj) for 1 ≤ i < j ≤ d. Then X is embeddable in Rd−1 if and only if

Σ1≤i<j≤dλiλjd2ij ≤ 0

for all choices of λi's such that Σλi = 0 and Σλ2i = 1. Moreover, if the inequality is

always strict, X can be embedded as an a�nely independent set.

Now, we modify the �nal steps of Matou²ek and Rödl's proof of Theorem 1.3.4 to

arrive at the following.

Theorem 3.3.9. If B ⊆ Sk is a simplex, then there is a �nite set C ⊆ SN(ρ) (for some

N, ρ) and an embedding f : B → C such that any partial map of f ′′B of size 1 extends

to an isometry of C.

Proof. Let B = {x1, . . . , xd} and mij = ||xi − xj||2. Since B is a�nely independent, by


Schönberg's theorem, there exists γ > 0 such that the l.h.s. of

Σ1≤i<j≤dλiλjmij ≤ 0

is always less than −γ, for all choices of λ1, . . . , λd with Σλi = 0 and Σλ2i = 1.

Let β < γd2

be su�ciently small. Then the distances d′ij = mij − β are of negative

type and consequently there is a set Z = {z1, . . . , zd} ⊆ Rd−1 such that

||zi − zj||2 = mij − β

Fix a small enough δ > 0 and �nd a

Y = {y1, . . . , yd} ⊆ Spread(a, J)

which approximates zi's, i.e., ||yi − zi|| < δ. Such a and J exist, by Proposition 3.3.1.

Therefore,

||zi − zj|| − 2δ < ||yi − yj|| < ||zi − zj||+ 2δ

and by squaring, we get

mij − β − 4δ√mij − β + 4δ2

< ||yi − yj||2 < mij − β + 4δ√mij − β + 4δ2

Let v1, . . . , vd be the points ||vi−vj||2 = mij−||yi−yj||2. Hence, by the above inequality,

β − 4δ√mij − β − 4δ2 < ||vi − vj||2 < β + 4δ

√mij − β − 4δ2

Becausemij ≤ 4 and δ is much smaller than β, we have that V = {v1, . . . , vd} is an almost

regular simplex. By the comments preceding this theorem, we can make it su�ciently-

orthogonal, by making δ small enough. Thus, there is C1 and an embedding g : V → C1


such that any partial isometry of f ′′V extends to an automorphism of C1.

We also have C2 = Spread∗(a, J) such that any partial isometry of Y extends to an

automorphism of C2. Finally, set C = C1 ∗C2. Note also that the map f : X → C1 ∗C2,

de�ned by f(xi) = g(vi) ∗ yi is an isometry.

Chapter 4

Graphs and Metrics

4.1 Hypermetric Distance Functions

We shall borrow terminology and notation from [DL]. Let n ≥ 2 and let λ1, . . . , λn be

integers. Suppose Σni=1λi = 1.

Σ1≤i<j≤nλiλjdij ≤ 0

is called a hypermetric inequality. A given metric space (X, d) is de�ned to be hypermetric

if and only if we have X = {xi}|X|i=1 and dij = d(xi, xj), for 1 ≤ i, j ≤ |X|, while

{dij}1≤i,j≤|X| satis�es all hypermetric inequalities. In particular, if (X, d) is hypermetric,

then (X, d′) is isometrically embeddable in `2 ([DL]), where d′(a, b) =√d(a, b), for all

a, b ∈ X. Given a connected graph G, de�ne the metric space (V (G), dG) by setting

dG(a, b) to be the length of the shortest path connecting a, b ∈ V (G). Given two graphs

G1, G2, de�ne the Cartesian product G1 ×G2 to be the graph with

V (G1 ×G2) = {{v1, v2} : v1 ∈ V1 ∧ v2 ∈ V2}

43

Chapter 4. Graphs and Metrics 44

such that {{u1, u2}, {v1, v2}} ∈ E(G1 × G2) if and only if either {h1, v1} ∈ V (G1) and

u2 = v2 or u1 = v1 and {u2, v2} ∈ E2. De�ne a graph G to be hypermetric if and only if

(V (G), dG) is hypermetric.

De�ne the half-cube graph G = 12H(n, 2), with vertex set

V (G) ={x ∈ {0, 1}n : Σn−1

i=0 xi is even}

and such that (x, y) ∈ E(G) if and only if |{i : xi 6= yi}| = 2. Thus, dG(x, y) =

12|{i : xi 6= yi}|. Moreover, observe that |{i : xi 6= yi}| = |{i : xi = 1}4{i : yi = 1}|.

The cocktail-party graph G = Kn×2 with V (G) = {v1, . . . , vn, vn+1, . . . , v2n} is such

that E(G) consists of all pairs except {vi, vn+i}. Note that the subscript is a Cartesian

product, not the usual product of integers. See [DL] for the full version of the following

theorem.

Theorem 4.1.1. Let G be a connected graph. Then G is hypermetric if and only if G is

an isometric subgraph (subgraph whose induced metric coincides with dG) of a Cartesian

product of half-cube graphs, cocktail-party graphs and copies of the Gosset graph G56.

Although we shall not look closely at the Gosset graph G56, let us de�ne it (see,

for example, [BHM]): Its vertex set V (G56) is the set of vectors in R8 with coordinates

consisting of either two 1's and six 0's, or six 12's and two −1

2's. If v1, v2 ∈ V (G56), then

{v1, v2} ∈ E(G56) if and only if 〈v1, v2〉 = 1.

It would be fortuitous to �nd a Ramsey (or Hrushovski) class of �nite structures that

corresponds simultaneously to �nite metric spaces and structures with a graph theoretic

formulation. For example, in [Ng], Nguyen van Thé observed the correspondence between

lexicographically ordered trees with levels labeled by reals and the convexly ordered

ultrametric spaces, which in turn are embeddable in `2. We can view the vertices of a

half-cube as characteristic functions of �nite sets. Notice that if the size of the symmetric

di�erence of two sets corresponding to two vertices is equal to one, then the values of the


characteristic functions must di�er in exactly two spots. Before investigate the symmetric

di�erence metric, let us �rst consider metric spaces induced by cocktail party graphs.

Fact 4.1.2. We can extend the map x 7→ u, y 7→ v for x, y, u, v ∈ Kn×2 � with x, y, u, v

distinct and {x, y}, {u, v} ∈ E(Kn×2) � to an automorphism of Kn×2.

Proof. Suppose G = Kn×2 with V (G) = {v1, . . . , vn, vn+1, . . . , v2n} is such that E(G)

consists of all pairs except {vi, vn+i}. Say x = vi, y = vj, u = vk, v = vl. Suppose i, k ≤ n,

k, l ≤ n. In this case, set f(x) = u, f(u) = v, f(u) = x, f(v) = u, f(vn+i) = vn+k,

f(vn+k) = vn+i, f(vn+j) = vn+l, and f(vn+l) = vn+j. For p /∈ {i, j, k, l, n + i, n + j, n +

k, n+ l}, set f(vp) = vp. The other cases are treated similarly, except with �n+� replaced

with �n−,� as needed. The function f is the required automorphism.

c

b

a

f

e

d

Set A1 = {a, b, e}, A2 = {b, d, e} and B1 = {a, b, d, e}. Note that A1 and A2 are

isomorphic as graphs and that the path metric induced by them coincides with the metric

induced by B1. Also, note that A1 and A2 are not cocktail-party graphs themselves. Let

C1 = Kn1×2 be a cocktail-party graph with V (C1) = {vi}2n1i=1 such that (vi, vj) ∈ E(C1)

unless j = n1 + i. Observe that if B′ is a copy of B1 in C1, then B′ = {vk, vn1+k, vl, vn1+l}

for some k, l. Now, colour every copy A′ of A1 in C, so that c(A′) = 0 if and only if

A′ = {vp, vq, vr} with p, q < n1 and c(A′) = 1, otherwise. With this colouring, there

cannot be a copy B′ with a monochromatic set of subsets isomorphic to A1.

Remark 4.1.3. In fact, for any cocktail-party graphs B = Ku×2, C = Kv×2 with u ≤ v,

the copies of B in C correspond to subsets of [v] of size u: When embedding B in C, �rst

consider the pairs of disconnected vertices. This means that, by Ramsey's theorem, for


any cocktail-party graphs A,B � with A being a subgraph of B � there is a cocktail-party

graph C such that for any colouring of the copies of A in C, there is a copy B′ of B in

C, such that the copies of A in B′ assume the same colour.

Now, let us investigate the Hrushovski property. Set A3 = {a, b, e} and A4 = {a, e, f}

and B2 = {a, b, c, d, e, f}. As before, A1 and A2 are isomorphic as graphs and the

corresponding path metrics coincide with the metric induced by B2. Without loss of

generality, we can assume that if C is a cocktail-party graph that extends B2, then

the copy of B2 in C is B2 itself. However, the graph isomorphism (and path-metric

isometry) a 7→ a, b 7→ f , e 7→ e cannot be extended to an automorphism: Such an

extending automorphism would necessarily map e to c.

4.2 The Symmetric Di�erence Metric

We shall now investigate the symmetric di�erence metric which is related to the path

metric on half-cube graphs. Fix nonzero C ∈ R. Set d to be the metric on [ω]<ω de�ned

by d(a, b) = C · |a4b|, for any a, b ∈ [ω]<ω. If X ⊆ [ω]<ω is a �nite subspace, then

the metric space (X, d′) is embeddable in `2, where d′(a, b) =

√|a4b|, for all a, b ∈ X:

Observe that the characteristic functions χa, χb (shifted by one spot) can be viewed as

vectors in `2. Thus, ||χa − χb||2 = |a4b|. Below, we assume C = 1.

Let k ∈ N. De�ne a class of �nite structures K to have the k-Hrushovski property if

and only if given B ∈ K , there is C ∈ K and an embedding j : B → C, such that if

1. A,A′ ⊆ j′′B,

2. |A| ≤ k, and

3. g : A→ A′ is an isomorphism,

then there is an automorphism f : C → C such that f � A = g.


Proposition 4.2.1. Let N ∈ ω. For any x1, x2, x3 ∈ P(N) and an isometry j with

j(xα) = yα for 1 ≤ α ≤ 3, there is a (bijective) isometry k : P(N) → P(N) such that

k(xα) = yα (1 ≤ α ≤ 3).

Proof. Set Y = P(N). Let xα, yα ∈ Y and let j be an isometry such that j(xα) = yα, for

1 ≤ α ≤ 3.

Set f : Y → Y to be such that f(a) = a∆x1, for all a ∈ Y . Then f(x1) = ∅. Note

that f is an isometry of Y . Set x2 = f(x2) and x′2 to be the set {0, . . . , |x14x2| − 1}. Pick

σ1 to be some permutation of N such that σ′′1 x2 = x′2, while �xing all other elements. Set

p1 : Y → Y to be such that p1(a) = σ′′1a. Then p1 is an isometry and p1(∅) = ∅ = f(x1).

Set x3 = p1(f(x3)). Let k = |x14x2|, l = |x14x3|, m = |x24x3|, n = |x′2 ∩ x3|,

o = |(N\x′2) ∩ x3)|. Then l = n + o, m = (k − n) + o. Therefore, we have a unique

solution for n and o. Now, pick σ2 to be some permutation of N such that

σ′′2(x′2 ∩ x3) = {0, . . . , |x′2 ∩ x3| − 1}

and

σ′′2((N\x′2) ∩ x3) = {|x′2| , . . . , |(N\x′2) ∩ x3| − 1} ,

while �xing all other elements. Set p2 : Y → Y to be such that p2(a) = σ′′2a. Note that

p2 is an isometry of Y , while p2(∅) = ∅ and p2(x′2) = x′2. Set x′3 = p2(p1(f(x3))) and

Z = {∅, x′2, x′3}.

Repeat the above process for {y1, y2, y3}, constructing g, r1 and r2, W = {∅, y′2, y′3}

corresponding to f, p1, p2 and Z, respectively. As above, we will have two linear equations

with two variables and a unique solution. Since {x1, x2, x3} and {y1, y2, y3} are isometric,

the solution will be the same and consequently x′2 = y′2 and x′3 = y′3. In other words,

g−1r−12 r−1

1 p2p1f(xα) = yα, while g−1r−1

2 r−11 p2p1f is an isometry of all of Y .

Corollary 4.2.2. Let X ⊆ [ω]<ω. There is a Y ⊆ [ω]<ω and an embedding i : X → Y

such that for any partial isometry j : {x1, x2, x3} ⊆ X → {x′1, x′2, x′3} ⊆ X, i ◦ j ◦ i−1


extends to an isometry of all of Y.

Proof. Embed X into P(|⋃X|) and apply the previous proposition.

Theorem 4.2.3. For any X ⊆ [ω]<ω, there is Y ⊆ [ω]<ω which satis�es the following:

there is an embedding i : X ↪→ Y such that for any partial isometry j : A → A′ with

A,A′ ⊆ X and |A| ≤ 3, we have that i ◦ j ◦ i−1 extends to an isometry of all of Y .

However, there exists X0 ∈ [ω]<ω and a partial isometry j : A → A′ with |A| = 4 such

that for any Y ⊆ [ω]<ω and any embedding i : X ↪→ Y , i ◦ j ◦ i−1 does not extend to an

isometry of all of Y .

We shall provide the example X0, below.

Proposition 4.2.4. Suppose X ⊆ [ω]<ω is isometric to P(N) for some N ∈ ω. Then

there exists a ∈ [ω]<ω such that

X ={b ∪ a : b ∈ P((

⋃X)\a)

}.

Proof. Pick a ∈ X so that a = min {|b| : b ∈ X}. Since X and P(N) are isometric,

there are precisely N many distinct points a0, . . . , aN−1 in X such that d(a, aα) = 1

and d(aα, aβ) = 2 for α 6= β. Let f : X → P(N) be any isometry. Then g : X →

P(N) de�ned by g(b) = f(a)4f(b) is also an isometry. In particular, g(a) = ∅, while

{g(aα)}α∈N = N .

|aα\a| = 1, for all α ∈ N : We know |a\aα| + |aα\a| = 1. However, aα cannot be

a subset of a, because a is of the smallest cardinality, by assumption. In other words,

|a\aα| = 0.

For all α ∈ N , set xα to be such that {xα} = aα\a.

Let b ∈ X\ {a, a0, . . . , aN−1}. Then there are distinct α0, . . . , α|g(b)|−1, such that

d(g(aαβ), g(b)) = d(g(a), g(b)) − 1. In particular, for β ∈ |g(b)|, aαβ4b = (a4b)\{xαβ}.

Recall that xαβ 's are distinct and note that there are |g(b)| many of them. Thus, |b\a| ≥

|g(b)|. (*)


Also,

|a4b| = d(g(a), g(b))

= |g(a)4g(b)|

= |∅4g(b)|

= |g(b)|

Thus, 0 ≤ |a\b| = |g(b)| − |b\a|. (**)

Combining (*) and (**), we get |a\b| = 0 and |b\a| = |g(b)|. Thus, b\a ⊆ {xα}α∈N

and consequently a is as required.

Proposition 4.2.5. For any Y ⊆ [ω]<ω and any isometric embedding i : P(7) → Y ,

there exists a partial isometry j : {x1, x2, x3, x4} → {y1, y2, y3, y4} of P(7) such that

i ◦ j ◦ i−1 does not extend to an isometry of all of Y .

Proof. Without loss of generality, assume Y = {0, . . . , |Y | − 1}. By the previous propo-

sition, there is a ∈ [ω]<ω, such that

i′′P(7) ={b ∪ a : b ∈ P((

⋃i′′P(7))\a)

}.

Without loss of generality, we can assume a = {7, . . . , 7 + |a| − 1} and

⋃i′′P(7) = {0, . . . , |a| − 1 + 7} .

Set x1 = a, x2 = {0, 1, 2, 3} ∪ a, x3 = {0, 1, 4, 5} ∪ a, x4 = {0, 2, 6} ∪ a, y1 = x1,

y2 = x2, y3 = x3, y4 = {2, 3, 5} ∪ a.

x1 :x2 :x3 :x4 :y4 : a

aaaa


Then j de�ned by j(xα) = yα is an isometry. Suppose j : Y → Y is an isometry that

extends j to all of Y . In particular, j(a) = a. Consider x = ({0} ∪ a) ∈ i′′P(7) ⊆ Y .

Consequently, d(x, x1) = d(j(x), y1) = 1. Therefore, |j(x)\a| = 1 or |a\j(x)| = 1.

The latter implies that d(j(x), y2) = 5, but d(x, x2) = 3, which contradict j being an

isometry. The unique element of j(x)\amust be in {0, . . . , 6}. However, this is impossible

by inspection and since j is an isometry: Let e be such that {e} = j(x)\a.

First, note that d(x, x1) = 1, d(x, x2) = 3, d(x, x3) = 3, d(x, x4) = 2. Moreover, recall

that j(xα) = yα.

Suppose e = 0: Then d(j(x), y1) = 1, d(j(x), y2) = 3, d(j(x), y3) = 3, d(j(x), y4) = 4.







Thus, in all of these cases, j is not an isometry.

Corollary 4.2.6. The class of metric spaces corresponding to the symmetric di�erence

metric (with C = 1) is k-Hrushovski if and only if k ∈ {1, 2, 3}.

We will now look at the amalgamation and Ramsey properties as they relate to the

symmetric di�erence metric. Let F be the class of (potentially ordered) �nite metric

spaces embeddable in ([ω]<ω, d) where d(a, b) = |a4b|.

Lemma 4.2.7. If A ⊆ [ω]<ω is isometric to X = {∅} ∪ {{i} : i ∈ N}, for some N ∈ ω,

N ≥ 1, then there exist a0, I ∈ [ω]<ω such that |I| = N , and

A = {a0} ∪ {{i}4a0 : i ∈ I}


Proof. Let α : X → A be an isometry. Set a0 = α(∅). De�ne a function f as follows:

f(a) = a4a0 for a ∈ A. Then f is an isometry. Now,

|f(α({i}))| = |α({i})4a0| = |α({i})4α(∅)| = |{i}4∅| = 1

Set I = {f(α({i})) : i ∈ N}. Note that for i ∈ N ,

f(α({i}))4a0 = (α({i})4a0)4a0 = α({i}).

Corollary 4.2.8. (Amalgamation for some spaces) Let N ∈ ω, N ≥ 1. If A = {∅} ∪

{{i} : i ∈ N}, B1, B2 ⊆�nite [ω]<ω, while fi : A ↪→ Bi is an embedding for i ∈ {1, 2},

then there exist a �nite C ⊆ [ω]<ω and embeddings gi : Bi ↪→ C (i ∈ {1, 2}) such that

g1 ◦ f1 = g2 ◦ f2 and (g1 ◦ f1)′′A = A.

Proof. By the previous fact, we can �nd a1, a2, I1, I2 ∈ [ω]<ω such that f ′′i A = {ai} ∪

{{j}4ai : j ∈ Ii} for i ∈ {1, 2}. De�ne for all a ∈ Bi, gi(a) = a4ai. Then C =

g′′1B1 ∪ g′′2B2, g1 and g2 are as required.

Lemma 4.2.9. If B ∼= P(N), for some N ∈ ω, N ≥ 1, while ∅ ∈ B and Y = {{i} : i ∈

J} ⊆ B with |J | = N , then B = P(J).

Proof. B ∼= P(N) implies that there is a ∈ [ω]<ω for which B = {b∪a : b ∈ P((⋃B)\a)}

and a has the smallest cardinality of all the elements of B. Because ∅ ∈ B, a = ∅.

Therefore, B contains exactly N many elements of the form {k}, with k ∈ I for some

I ∈ [ω]<ω. The fact that Y ⊆ B implies that I = J , which concludes the proof.

Lemma 4.2.10. Suppose A = {∅} ∪ {{i} : i ∈ N} for some N ∈ ω, N ≥ 1. Moreover,

suppose that B1, B2 ⊆ [ω]<ω are such that Bi∼= P(N) for some N ∈ ω, while there are

embeddings fi : A ↪→ Bi for i ∈ {1, 2} such that f ′′1A = f ′′2A. We can conclude that

B1 = B2.


Proof. By Lemma 4.2.7, f ′′i A = {a} ∪ {{i}4a : i ∈ I} for some a, I ∈ [ω]<ω. Set

B′i = {b4a : b ∈ Bi} for i ∈ {1, 2}. Then {∅} ∪ {{i} : i ∈ I} ⊆ B′i for i ∈ {1, 2}. By

Lemma 4.2.9, B′i = P(I) for i ∈ {1, 2}, which concludes the proof.

Corollary 4.2.11. For any r ∈ ω, there exist A,B ∈ F such that if B ≤ C, then there

exists a colouring χ :(CA

)→ r, such that for all B′ ∈

(CB

),∣∣∣χ′′(B′A)∣∣∣ = r.

Remark 4.2.12. On the other hand, for any r ∈ ω, if A = [K]1 and B = [L]1 with

K,L ∈ ω, K < L, then there exists N ∈ ω, such that [N ]1 → (B)Ar , by Ramsey's

theorem.

Remark 4.2.13. The following is an alternative proof of Cor 4.2.11, in case we disregard

orderings: Recall that for any A ⊆ [ω]<ω such that A ∼= P(K), where K ∈ ω, there

exist unique aA, IA ∈ [ω]<ω such that aA ∩ IA = ∅ and A = {b ∪ aA : b ∈ P(IA)}. Say

B ∼= P(L), while B ⊆ [ω]<ω, L ∈ ω, L > K. Let C ∈ F be such that B ≤ C ⊆ [ω]<ω.

Consider the colouring χ :(

CP(K)

)→ (L − K), where χ(A) = |aA|(mod (L − K)): Let

B′ ∈(CB

). Note that for any a′ ∈ P(IB) such that |a′| < L−K, there is A′ ∈

(B′

P(K)

)such

that aA′ = aB ∪ a′. Thus, |χ′′(

B′

P(K)

)| = L−K.

Fact 4.2.14. (Characterization of the Automorphisms of P(N)) Let N ∈ ω. If f :

P(N)→ P(N) is a (bijective) isometry then there exists a permutation σ : N → N such

that f(a) = (σ′′a)4f(∅) for all a ∈ P(N).

Proof. Set ai = f({i}), for i ∈ N . Note that |f({i})4f(∅)| = 1 while xi ∈ f({i})4f(∅)

(i ∈ N) are distinct. Then σ(i) = xi is as required: Note that xi ∈ f(a)4f(∅) if and

only if i ∈ a.

Chapter 5

Boron Trees

5.1 Boron Tree Structures

The class of structures considered in this chapter will have signature L = (R), where

R is a 4-ary symbol. Boron is a trivalent chemical element. A boron tree T is a graph

theoretic tree such that every vertex has valance either 1 or 3. Essentially, this is a binary

tree without an assigned root. De�ne (T ) to be the model with universe consisting of

the leaves of T and R(T ) de�ned as follows: If a, b, c, d are four distinct leaves of T then

(a, b, c, d) ∈ R(T ) if and only if there is a path p1 between a and b and a path p2 between

c and d such that p1 and p2 do not intersect. In [Cam1, Cam2], this is symbolized as

ab|cd. If the four leaves are not distinct, then (a, b, c, d) /∈ R(T ). The class B0 of �nite

structures associated with boron trees in this manner � which we will refer to as boron

tree structures � is a Fraïssé class (see [Cam1, Cam2]). If we mention a path between

two elements of a boron tree structure, we are referring to the corresponding path in the

corresponding graph. The following representation of the aforementioned quadruplet a,

b, c, d is evocative of how we are going to deal with these structures here.

a b c d

53

Chapter 5. Boron Trees 54

Fact 5.1.1. Let T0, T1 be boron trees. Let A0,A1 ∈ B0 be such that Ai = (Ti). If ϕ :

A0 → A1 is an isomorphism of structures, then there is a graph isomorphism λ : T0 → T1

such that λ � A0 = ϕ. (Compare this to Proposition 3.1 in [Cam2].)

Proof. Proceed by induction on |Ai| ≥ 4: Say A0 = {x1, . . . , x|A0|}. Let ϕ : A0 → A1 be

an isomorphism of structures. Let A′0 and A′1 be the substructures of A0 and A1 with

universes A′0 = A0\{x|A0|} and A′1 = A1\{ϕ(x|A0|)}, respectively. Then ϕ � A′0 : A′0 → A′1

is still an isomorphism.

Let v0 ∈ V (T0) be adjacent to x|A0|. Let u0 6= x|A0| be adjacent to v0. Let u1 ∈

V (T0)\{u0, x|A0|} be adjacent to v0. De�ne the boron tree T ′0 as follows: V (T ′0) =

V (T0)\{x|A0|, v0},

E(T ′0) = ([V (T ′0)]2 ∩ E(T0)) ∪ {{u0, u1}}.

In other words, we are deleting x|A0| and v0 and adding an edge between u0 and u1. Then

(T ′0) = A′0.

Let v1 ∈ V (T1) be adjacent to ϕ(x|A0|). Let u′0 6= ϕ(x|A0|) be adjacent to v1. Let

u′1 ∈ V (T1)\{u′0, ϕ(x|A0|)} be adjacent to v1. We can likewise de�ne T ′1 so that (T ′1) = A′1.

By the induction assumption, there is a graph isomorphism λ′ : T ′0 → T ′1 such that

λ′ � A′0 = ϕ � A′0. De�ne λ(v) = λ′(v) for v ∈ V (T ′0) . De�ne λ(v0) = v1 and

λ(x|A0|) = ϕ(x|A0|).

De�ne B(0) to be the structure with B(0) = {0} and RB(0) = ∅. For n > 0, de�ne

B(n) to be the �nite model with universe B(n) = 2n = {f : n → 2} and with RB(n)

de�ned as follows: First, construct a boron tree T with vertex set V (T ) = 2≤n\{∅}.

Suppose n ≥ 1. We place an edge between the two functions in V (T ) of size one. Let

k ≤ n− 1. We also place an edge between any function f : k → 2 and the two functions

g : (k + 1) → 2 with g � k = f . Now, note how 2n is the set of leaves of this tree. For

a, b ∈ B(n), de�ne δ(a, b) = min{j : a(j) 6= b(j)} and a u b = a � δ(a, b) = b � δ(a, b).

Given substructures A and B of B(n), A∪B will refer to the substructure of B(n) with


universe A ∪ B. We will write A tB, if A ∩ B = ∅. Finally, A ⊆ B, means A ⊆ B and

RA = RB ∩ A4.

ba

a u b {

Pick k < n and f0 : k → 2. It is immediately apparent, that the substructure Bf0(n)

of B(n) whose universe is de�ned to be

Bf0(n) = {f : n→ 2 : f � k = f0}

is isomorphic to B(n− k).

Fact 5.1.2. For any A ∈ B0, there is n such that A embeds in B(n).

Proof. Let us proceed by induction on |A| ≥ 4. Say A = {x1, . . . , x|A|}. By induction as-

sumption, there is n0 such that the boron tree substructure with universe {x1, . . . , x|A|−1}

embeds in B(n0), say via map α. Let T be the boron tree corresponding to the structure

A. Let v0 be an element adjacent to x|A|. Let u1, u2 be distinct elements adjacent to v0

and not equal to x|A|. Let T′ be the boron tree T with x|A| and v0 deleted and an edge

added between u0 and u1, similarly to what we did in 5.1.1. Now, de�ne a new graph G

as follows: Set

V (G) = {α(a) u α(b) : a, b ∈ (A\{x|A|})}.

Let {u, v} ∈ E(G) if and only if there is a path between u and v that does not pass

through another element of V (G). Then G is isomorphic to T ′, say via λ. Suppose

the height of λ(u0) is less or equal to λ(u1) . Let xi(k) = λ(u1)(k) for k ∈ |λ(u1)| and

xi(|λ(u1)|) = i. We can send Bλ(u1)(n0) to Bx0(n0 + 1) and map x|A| to the element

x(k) = x1(k) for k ∈ |x1| and x(k) = 0, otherwise.


Proposition 5.1.3 (1-Hrushovski property for boron trees). Let A ∈ B. Then there

is n and an embedding f : A ↪→ B(n) such that for any a0 6= a1 ∈ f ′′A, there is an

automorphism g of B(n) such that g(a1) = a0.

Proof. By Fact 5.1.2, it su�ces to show that if a0 6= a1 ∈ B(n), there there is an

automorphism g of B(n) such that g(a1) = a0.

Note that for any a ∈ B(n), there is an automorphism ϕa of B(n) taking a to the

sequence of all zeroes:

ϕa(b)(k) = (b(k) + a(k))(mod 2).

Set g = ϕ−1a0◦ ϕa1 .

Fact 5.1.4. The class of boron tree structures does not satisfy the full Hrushovski prop-

erty.

Proof. Consider B(2) = {a = 〈0, 0〉 , b = 〈0, 1〉 , c = 〈1, 0〉 , d = 〈1, 1〉}. Suppose we have

κ and an embedding f : B(2) ↪→ B(κ) such that for any x, y, z, w ∈ f ′′B(2) and an

isomorphism g′′{x, y} = {z, w}, there is an automorphism of B(κ) which extends g.

Identify a, b, c, d with their images via f . Set α to be the partial isomorphism de�ned

by α(c) = b, α(b) = a, α(a) = c. Let g1 be the extending automorphism. We then have

au b ⊆ g(d), by de�nition of the boron tree structures. Moreover, g1(d) /∈ {a, b, c, d}. Set

β to be the partial isomorphism de�ned by β(b) = c, β(c) = d, β(d) = b. Let h1 be the

extending automorphism. Then we have c u d ⊆ h1(g1(d)) and h1(g1(d)) /∈ {c, d}.

ba g1(d)

c

h1(g1(d)) d

In the inductive assumption, we have n−1 many distinct di's such that cud ⊆ di and

di /∈ {c, d}. In the n-th inductive step, we extend the isomorphisms α and β to gn and


hn, respectively, to produce distinct d′i = hngn(di) for 1 ≤ i ≤ n − 1 and d′n = hngn(d).

We can then conclude that κ = ω.

5.2 Ramsey Degrees

We �rst establish the existence of monochromatic sets of leaves:

Proposition 5.2.1. Let r, n ∈ ω. Then there exists N = N(r, n) such that if χ : 2N → r

is a colouring of the universe of B(N), then there is a copy B1 of B(n) in B(N) such

that |χ′′B1| = 1.

Proof. Proceed by induction on n: Let K → (2)1r, i.e., K = r + 1. Let k = dlog2Ke. (*)

Let χ : 2k+N(n,r) → r be a colouring of the universe of B(k+N(n, r, )). Given f : k → 2,

by the induction assumption, we have that Bf (k + N(n, r)) contains a monochromatic

copy Bf of B(n). By (*), there are two monochromatic copies � call them Bf1 and Bf2

� such that χ′′Bf1 = χ′′Bf2 . Finally, the substructure of Bf (k + N(n, r)) consisting of

Bf1 tBf2 is isomorphic to B(n+ 1).

Corollary 5.2.2. Let B ∈ B0. Then there is a C ∈ B0, such that for any colouring of

the elements of the universe of C, there is a monochromatic copy of B in C.

Proof. There exists n ∈ ω, such that B embeds in B(n), by Fact 5.1.2. Apply Proposition

5.2.1.

Consider an embedding f : A ↪→ B(N) with A ∈ B0. Set

n = max{j : (∀x, y ∈ f ′′A)(∀i < j)x(i) = y(i)}.

Take any x ∈ f ′′A. The beginning of f ′′A in B(N) is de�ned to equal x � n.

The notion of orientation used in this paper resembles the notion used in [Fo]. We

will show that the Ramsey degree of a boron tree structure is equal to the number of


its possible orientations. Only singletons and boron tree structures with two points have

one possible orientation. We will also show that the Ramsey property is not achieved by

extending these structures with linear orderings and that the class satis�es neither the

ordering property nor the Hrushovski property.

We are about to de�ne the orientation of f ′′A in B(N): If f ′′A is a singleton de�ne

it to be ∅, and de�ne its height to be 1. Suppose, f ′′A is not a singleton. For i ∈ {0, 1},

de�ne xi : (n + 1) → 2 so that xi � n = x � n, while xi(n) = i. In other words, the

function xi is the same as x except with i concatenated at the end. The function f is

called a (C,D)-embedding if and only if Bx0(n) ∩ f ′′A = C and Bx1(n) ∩ f ′′A = D,

for some C,D. The orientation O of f ′′A is an ordered pair 〈O1,O2〉, where O1 is the

orientation of C and O2 is the orientation of D. We de�ne the height of the orientation

O to be 1+ the largest of the heights of the orientations of O1 and O2. Call f−1C the left

side of A and f−1D the right side of A (with respect to f). Observe how, even though

they depend on the embeddings, there are only �nitely many possible orientations. We

will apply these de�nitions to the structures and their underlying sets, interchangeably.

De�ne recursively � the standard orientation � O(n) to be ∅ if n = 0 and

〈O(n− 1),O(n− 1)〉

if n > 0.

Recall from the previous section the de�nition δ(a, b) = min{j : a(j) 6= b(j)}. De�ne

the lexicographic ordering <lex on B(n), as follows: For all a, b ∈ B(n), a <lex b if and

only if a(δ(a, b)) < b(δ(a, b)).

Example 5.2.3. Embed B(2) in B(n) for n ≥ 2. Draw the boron tree corresponding to

B(n) in the plane (i.e., a physical piece of paper) so that the leaf a ∈ B(n) is to the left

of b ∈ B(n) if a <lex b. Let us look at two possible orientations of copies of B(2): A′

below represents a copy of B(2) with the standard orientation O(2), while A′′ is also a


copy of B(2), but has orientation

〈〈∅, 〈∅, ∅〉〉 , ∅〉 .

Note that not all nodes of the boron tree corresponding to B(n) are drawn below.

A′′

A′

Suppose A ⊆ B ⊆ B(N) and α : B ∼= B(m) preserves the lexicographic orderings,

while B has orientation O(m) in B(N). There is only one such α, which saves us from

having to prove that the following de�nition is well-de�ned: The height of the beginning

of A (w.r.t. B) is de�ned to be

max{j : (∀x, y ∈ A′)(∀i < j)α(x)(i) = α(y)(i)}.

Thus, if β is the beginning of A in B(N), we de�ne β's height (w.r.t. B) to be equal to

the height of the beginning of A (w.r.t. B).

Recall thatX ⊆ [Λ](n0

)= Λn is called anm-parameter set corresponding to f ∈ [Λ]

(nm

)if and only if

X =

{f · g : g ∈ [Λ]

(m

0

)}.

Fact 5.2.4. Let n,N ∈ N. An n-parameter subset of {0, 1}N is isomorphic to B(n) and

has orientation O(n).

Proof. Let f ∈ [{0, 1}](Nn

)be the variable word that de�nes the n-parameter subset

X. For a ∈ {0, 1}n = [{0, 1}](n0

), de�ne w(a) to be f · a. De�ne ϕ : X → B(n) by

ϕ(w(a)) = a, for a ∈ {0, 1}n . This preserves the boron tree structure: Say there is a

path between w(a) and w(b) and a path between w(c) and w(d), while these two paths do

not intersect. Without loss of generality, we can assume that w(c) u w(d) 6⊆ w(a), w(b).


Suppose c u d ⊆ a. Let j0 = min{j : cj 6= dj}. In this case, for all j < j0, cj = dj = aj.

Note that j0 determines the �rst spot where w(c) and w(d) di�er. Thus, w(c) u w(d) ⊆

w(a), a contradiction. Likewise, assuming cu d ⊆ b leads to a contradiction. This means

that there is a path between a and b and a path between c and d, while these two do

not intersect. Now, suppose there is a path between a and b and a path between c

and d, while these two paths do not intersect. Without loss of generality, assume that

c u d 6⊆ a, b. Suppose w(c) u w(d) ⊆ w(a). In this case, c u d ⊆ a. Likewise, assuming

w(c) u w(d) ⊆ w(b) leads to a contradiction.

Observe that not every copy of B(n) in {0, 1}N corresponds to an n-parameter set.

In particular, any copy of B(n) in {0, 1}N with orientation other than O(n) is not an

n-parameter set. If O is an orientation of A, de�ne(B(n)A

)O to be the set of substructures

of B(n) isomorphic to A with orientation O. Consequently, the following is � in some

sense � an asymmetric version of the Graham-Rothschild theorem (Theorem 1.2.4) for

the alphabet {0, 1}.

Theorem 5.2.5. Let n, p, r ∈ ω. There exists N = N0(n, p, r) that satis�es the following:

Suppose O is an orientation of A with height p. Then, given a colouring c :(B(N)A

)O → r,

there exists a copy B′ ∈(B(N)B(n)

)O(n)

, such that∣∣∣c′′ ((B(N)

A

)O ∩

(B′

A

))∣∣∣ = 1. We represent

this latter statement by

B(N)→ (B(n))(A,O)r .

Proof. We proceed by induction on p. Corollary 5.2.2 is the base case, i.e., if p = 1 then

|A| = 1. Assume that for any n, r ∈ ω, N0(n, p′, r) is de�ned for p′ < p. Now, �x A. Let

C be the left side of A and D be the right side, as determined from the �xed O. Let

O1,O2 be the orientations corresponding to C and D, respectively. Note that the height

of both O1 and O2 is necessarily less than p.

Claim 5.2.6. Let n, r ∈ ω. There exists N = N1(n, p, r) (distinct from N0) that satis�es

the following: Given a colouring of(B(N)A

)O, there exists a copy of B(n) in B(N) with


orientation O(n) such that every copy of A with orientation O and beginning ∅ has the

same colour.

Proof. We use the inductive hypothesis here twice to de�ne N2 and N3, using the fact

that both O1 and O2 have height less than p. Let N2 = N0(n− 1, p− 1, rX), where X is

set to be the number∣∣∣(B(n−1)

C

)O1

∣∣∣. Let N3 = N0(n − 1, p − 1, rY ), where Y is set to be

the number∣∣∣(B(N2)

D

)O2

∣∣∣. Let N = 1 + max{N2, N3}. Let c :(B(N)A

)O → r be a colouring .

Let bi : 1→ 2 be such that bi(0) = i. Choose C0 to be a copy of B(N3) in Bb0(N) with

orientation O(N3). Choose C1 to be a copy of B(N2) in Bb1(N) with orientation O(N2).

Bb0(N)C0︷︸︸︷ C1︷︸︸︷Bb1(N)

Let c0 be the following induced colouring of(B(N)C

)O1∩(C0

C

):

c0(C′) =

{(c(C′ tD′),D′) : D′ ∈

(B(N)

D

)O2

∩(C1

D

)}

By de�nition of N3, there is a c0-monochromatic copy B1 of B(n− 1) in C0 with orien-

tation O(n− 1). (*)

Let c1 be the following induced colouring of(B(N)D

)O2∩(C1

D

):

c1(D′) =

{(c(D′ tC′),C′) : C′ ∈

(B(N)

C

)O1

∩(B1

C

)}

By de�nition of N2, there is a c1-monochromatic copy B2 of B(n− 1) in C1 with orien-

tation O(n− 1). (**)

Note that B3 = B1tB2∼= B(n) and has orientation O(n). Let A0,A1 be copies of A

in B3, both with orientation O and beginning ∅. There are copies A0j ,A

1j of C,D in Cj


with orientations Oj+1 such thatAi0tAi

1 = Ai. By (*), we have c(A00tA0

1) = c(A10tA0

1).

By (**), we have c(A10 tA0

1) = c(A10 tA1

1).

Claim 5.2.7. Let n, r,m ∈ ω. There exists N = N4(n, p, r,m) that satis�es the following:

Given a colouring of(B(N)A

)O, there exists a copy B′ ∈

(B(n)B

)O(n)

such that every copy

of A with orientation O and the same beginning of height j (w.r.t. B′) is assigned the

same colour, for j ≤ m. In other words, we show that the assigned colour depends only

on the beginning itself.

Proof. We proceed by induction on m. The base case is proved in Claim 5.2.6. Assume

that for any n, p, r, N4(n, p, r,m− 1) is de�ned.

First, observe that if n −m < p and k ≥ n, then for all B′ ∈(

B(k)B(n−m)

)O(n−m)

there

does not exist A′ ∈(B(k)A

)O ∩

(B′

A

)with a beginning of height m (with respect to B′).

Let n, p, r be arbitrary. Set N5 = N4(m + N1(n − m, p, r), p, r,m − 1). Let c :(B(N5)

A

)O → r be a colouring. By de�nition of N5, there is a copy B′′ of B(m + N1(n −

m, p, r)) in B(N5) with orientation O(m + N1(n − m, p, r)) such that every oriented

copy of A with the same beginning of height j (w.r.t. B′′) has the same colour, for

j ≤ m − 1. However, for every beginning β of height m (w.r.t. B′′), there exists Bβ ∈(B(N5)

B(n−m)

)O(n−m)

∩(

B′′

B(n−m)

)itself with beginning β such that every A ∈

(B(N5)

A

)O ∩

(BβA

)with beginning β has same colour, by Claim 5.2.6. This means that we can construct

B′ ∼= B(n) which we seek: Set B′ to be the disjoint union of all Bβ, as β ranges over all

beginnings β of height m (w.r.t. B′′). Therefore, we can set N4(n, p, r,m) = N5.

Now, combine Corollary 1.2.5 and Claim 5.2.7 as follows: Let Λ = {0, 1} = 2. Let

N6 = HJ∗(n, r). Set N7 = N4(N6, p, r, n− p). Suppose c be an r-colouring of(B(N7)

A

)O.

By Claim 5.2.7, we have a copy B′ of B(N6) in B(N7) with orientation O(N6) such

that any two copies of A in B′ with orientation O and the same beginning have the same

colour. (?)


Let α : B′ → B(N6) be the unique isomorphism which preserves the lexicographic

orderings. If A′,A′′ ∈(B(N7)

A

)O∩(B′

A

)have the same beginning, so do α′′A′ and α′′A′′, and

vice-versa. De�ne an r-colouring c′ on the set of beginnings/partial-points Λ≤N6 = 2≤N6

as follows: Let A′ ∈(B(N7)

A

)O ∩

(B′

A

). Let b ∈ 2≤N6 be the beginning of α′′A′. Set

c′(b) = c(A′). By (?), this colouring is well de�ned.

By Corollary 1.2.5, there is an n-parameter subset B′′ of 2N6 = B(N6) such that the

set of all partial points in B′′ is monochromatic. (??)

By Fact 5.2.4, B′′ is isomorphic to B(n) and has orientation O(n) in α′′B′. Moreover,

B′′′ = α−1B′′ ⊆ B′ has orientation O(n) in B(N7), because B′ has orientation O(N6),

while B′′ has orientation O(n) in B′.

We should check that this B′′′ is the one we seek: Let A′,A′′ ∈(B(N7)

A

)O ∩

(B′′′

A

). Let

b′, b′′ be the respective beginnings of α′′A′ and α′′A′′. By (??), c′(b′) = c′(b′′). However,

c′(b′) = c(A′) and c′(b′′) = c(A′′), by de�nition of c′. Therefore, we can �nally set

N0(n, p, r) = N7.

Corollary 5.2.8. Let r ∈ ω,A,B ∈ B0 and O be some orientation of A. Then there is

a N ∈ ω, such that for any colouring c :(B(N)A

)O → r, there exists a copy B′ ∈

(B(N)B

),

such that∣∣∣c′′ ((B(N)

A

)O ∩

(B′

A

))∣∣∣ = 1. We represent this statement by B(N)→ (B)(A,O)r .

Corollary 5.2.9. Let r, p ∈ ω,A,B ∈ B0. Then there is a n ∈ ω, such that for any

r-colouring of the copies of A in B(n), there is a copy B′ of B in B(n), such that the

copies of A in B′ assume at most k(A) colours, where k(A) is the number of all possible

orientations of A.

Proof. Let O1, . . . ,Ok be all the possible orientations of A. Apply Corollary 5.2.8, so

that Bi+1 → (Bi)(A,Oi)r with B1 = B.

It remains to be shown that given any A, there is an N such that for any C and any

r-colouring of the copies of A in C, each copy B′ of B(N) in C is such that the copies


of A in B′ assume at least k(A) colours.

Lemma 5.2.10. Suppose M ≥ n ≥ 2 and f : B(n) ↪→ B(M) is an embedding. Suppose

also that for all copies B′ of B(2) in B(n) with orientation O(2), f ′′B′ has orientation

O(2). Then f ′′B(n) has orientation O(n).

Proof. We prove this by induction on n. If n = 2, then B′ = B(2). Let us proceed to

the inductive step with n ≥ 3: De�ne xi : {0} → {0, 1}, for i ∈ {0, 1} by xi(0) = i.

Claim 5.2.11. For i ∈ {0, 1}, Bxi(n) is either the left or the right side of B(n) (w.r.t. f).

Proof. We shall check this for i = 0. Let {a, b} ⊆ Bx0(n). Now pick any {c, d} ⊆ Bx1(n).

Then C = {a, b, c, d} corresponds to a copy C of B(2) with orientation O(2). Thus, the

orientation of f ′′C is also O(2), by an initial assumption. Suppose {a, b} is not entirely

contained in either side of B(n) (w.r.t. f). This contradicts the fact that the orientation

of f ′′C is O(2).

. . .

a b c d

. . .

f(a)f(b) f(c) f(d)

The conclusion follows from Claim 5.2.11 and the inductive assumption applied to

Bxi(n), for i ∈ {0, 1}.

Let n0 be such that every orientation of A is witnessed in B(n0) and n0 ≥ 2. Set

N = n0 + 1. Let f : B(N) ↪→ B(M) be an embedding. It su�ces to establish the

following:

Lemma 5.2.12. There is a copy B′ of B(n0) in f ′′B(N) such that its orientation is

precisely O(n0), with all terms de�ned as above.


Proof. Let us begin with the following claim:

Claim 5.2.13. We can �nd x : m1 → 2, y : m2 → 2 and a copy B′ of B(n0) in B(M) so

that B′ = By(M) ∩ f ′′Bx(N). In particular, f−1B′ has orientation O(n0).

Proof. Set xi : 1 → 2 by xi(0) = i. Set yi to be the beginning of f ′′Bxi(N). Set y to be

the tallest of yi's and set x to be the corresponding xi.

If y0 u y1 is not equal to the shortest of the two, then we are done. Suppose then,

without loss of generality, that y = y1, x = x1, while y0 u y1 = y0. Suppose (By(M) ∩

B′)\f ′′Bx(N) 6= ∅. Let a ∈ (By(M) ∩ B′)\f ′′Bx(N) ⊆ f ′′Bx0(N). Pick b ∈ f ′′Bx0(N)

such that b u y1 = y0 � such b exists because y1 was assumed to be the tallest of the

beginnings y0 and y1. Pick c, d ∈ f ′′Bx1(N). Then

f−1(a)f−1(b)|f−1(c)f−1(d)

but it is not the case that ab|cd, which contradicts the assumption that f is an embedding.

. . .

y0

y1

d

c a

b

Therefore, we conclude that f ′′Bx(N) = f ′′Bx(N) ∩ By(M). Observe now that

f−1B′ = Bx(N) has orientation O(n0).

Pick a ∈ f ′′B(N)\B′. Let B′′ be a copy of B(2) so that B′′ ⊆ f−1B′ and B′′ has

orientation O(2). Suppose f ′′B′′ does not have orientation O(2). Let f ′′B′′ = {b, c, d, e}

be as in the picture below.


a

bc d ef ′′B′′

Let C be the structure with universe f ′′B′′∪{a}. LetD be the structure with universe

B′′ ∪ {f−1(a)} and let B′′ = {b′, c′, d′, e′} be as in the picture below:

f−1(a)

b′ c′ d′ e′ B′′

Claim 5.2.14. If g : C → D is and isomorphism then g(a) ∈ B′′. In other words,

g(a) 6= f−1(a).

Proof. Suppose g(a) = f−1(a).

Therefore, since bc|de, at most one of the following combinations is possible (*):

1. {g(b), g(c)} = {b′, c′} and {g(d), g(e)} = {d′, e′}.

2. {g(b), g(c)} = {d′, e′} and {g(d), g(e)} = {b′, c′}.

Note that ab|cd, but none of the following is true: g(a)b′|c′d′, g(a)b′|c′e′, g(a)c′|b′d′,

g(a)c′|b′e′. The following are also impossible: g(a)d′|b′e′, g(a)d′|e′c′, g(a)e′|d′b′, g(a)e′|d′c′.

However, by (*), these are the only potential combinations.

This contradicts the assumption that f is an embedding. Thus, by Lemma 5.2.10, B′

has orientation O(n0), as well.

Combining Theorem 5.2.5 and the above discussion, we arrive at the following, using

the notation from the introduction:

Theorem 5.2.15. For any A0 ∈ B0, t(A0,B0) = k(A0).


De�ne the Catalan number Cn by

Cn =1

n+ 1

(2n

n

).

Cn−1 is in fact equal to the number of rooted binary trees with n leaves (see [St]). In

particular, k(B(2)) = C3 = 5 and k(A) ≤ C|A|−1 for all A ∈ B0.

We shall construct a recursive formula for computing k(A), whenever A ∈ B0. Let

G be a binary rooted tree with r as its root. We can de�ne the height h(v) of v ∈ V (G)

to be the length of the shortest path from v to r. De�ne g(G), recursively, as follows:

If V (G) is a singleton, then set g(G) = 1. Let r1, r2 ∈ V (G)\{r} be the two distinct

vertices adjacent to r. The vertex set V (G)\{r} induces two connected components, call

them A1, A2. We can think of each Ai as a rooted binary tree, with the root being the

vertex ri in Ai adjacent to r. If there is a height preserving graph isomorphism between

A1 and A2, then set g(G) = g(A1)g(A2). Otherwise, set g(G) = 2g(A1)g(A2). De�ne

[G] to be the class of rooted binary trees for which there is a height preserving graph

isomorphism to G.

Let T be a boron tree and let A = (T ). Let {u, v} ∈ E(T ). De�ne a graph G{u,v}, as

follows: V (G{u,v}) = V (T ) ∪ {r} where r /∈ V (T ). Set

E(G{u,v}) = {E(T ) ∪ {{r, u}, {r, v}}}\{u, v}

We can think of G{u.v} as a rooted binary tree, with the root being the vertex r. Let

G = {[G{u,v}] : {u, v} ∈ E(T )}.

Suppose G = {[G1], . . . , [Gm]}. Then,

k(A) = Σmi=1g(Gi). (5.1)


To justify our formula, let us de�ne a rooted binary tree associated with an embedding

ϕ : A → B(n). Set V (Gϕ) = {ϕ(a) u ϕ(b) : a, b ∈ A} ⊆ 2≤n. Let x, y ∈ V (Gϕ) and

suppose |x| < |y|. The edge {x, y} ∈ E(Gϕ) if and only if x ⊆ y and there is no

z ∈ V (Gϕ)\{x, y} such that x ⊆ z ⊆ y. Set the root to be the unique element r such

that r ⊆ x for all x ∈ V (Gϕ). Observe that r could potentially be ∅. Now, it su�ces to

prove the following claim.

Claim 5.2.16. Let ϕ0 : A → B(n) and ϕ1 : A → B(n) be embeddings such that ϕ′′0A

and ϕ′′1A have the same orientation O in B(n). Then there is a height preserving graph

isomorphism λ : V (Gϕ0)→ V (Gϕ1).

Proof. Let us proceed by induction on the height of O. Say O = 〈Ol,Or〉, where Ol is

the orientation of the left side Ci of ϕ′′iA, while Or is the orientation of the right side Di

of ϕ′′iA, for i ∈ {0, 1}. Let bi be the beginning of ϕ′′iA, for i ∈ {0, 1}. Then bi is the root

of Gϕi . Set C ′i = ϕ−1i Ci and D′i = ϕ−1

i Di, for i ∈ {0, 1}. Then ϕ0 � C ′0 has the same

orientation as ϕ1 � C ′1. Thus, by the induction hypothesis, there is a height preserving

graph isomorphism λ0 : V (Gϕ0�C′0) → V (Gϕ1�C′1

). Likewise, there is a height preserving

graph isomorphism λ1 : V (Gϕ0�D′0)→ V (Gϕ1�D′1

). Note that

V (Gϕi�C′i) t V (Gϕi�D′i

) = V (Gϕi)\{bi}

for i ∈ {0, 1}. Let ci be the root of Gϕi�C′iand di be the root of Gϕi�D′i

, for i ∈ {0, 1}.

Then

{bi, ci}, {bi, di} ∈ E(Gϕi)

for i ∈ {0, 1}.

De�ne λ : Gϕ0 → Gϕ1 to be such that λ(b0) = b1, while λ � V (Gϕ0�C′0) = λ0 and

λ � V (Gϕ0�D′0) = λ1


Corollary 5.2.17. Let n ≥ 2. Then

k(B(n)) = 1 + Σnk=22k = 2k+1 − 3

Proof. Let T be the boron tree that corresponds to B(n). We can think of 2≤n\{∅} as

T . De�ne the height of v ∈ V (T ) to be the minimum of the distances of v from (0) or

(1). Let {u, v} ∈ E(T ) and {u′, v′} ∈ E(T ). Then the maximum of the heights of u and

v equals the maximum of the heights of u′ and v′ if and only if [G{u,v}] = [G{u′,v′}]. Now,

observe that the �left side� L of G{u,v} is isomorphic (as a rooted tree) to 2≤k\{∅} for

some k. Thus, g(L) = 1. There is also a �left� side L0 and a �right� side R0 of the right

side of G{u,v}. Here, L0 is isomorphic (as a rooted tree) to 2≤k0\{∅} for some k0, while

g(R0) = 2h, where h is the larger of the heights of u and v.

However, we cannot de�ne a formula for k(A) in terms of |A|:

Let T0 be the following boron tree:

Let T1 be the following boron tree:

First, note that, by Fact 5.1.1, since T0 6∼= T1, we also have (T0) 6∼= (T1). However,

|(T0)| = |(T1)| = 6.

For T0, there are exactly two equivalence classes of G{u,v}'s we must consider in

applying the formula 5.1, one corresponding to the edges that contain one of the leaves


and the other class corresponding to the other three edges. We have k((T0)) = 4 + 4 = 8.

For T1, it su�ces to consider the two classes corresponding to the two sets of edges that

contain one of the leaves: k((T1)) = 8 + 16 + . . . > 8.

5.3 Extensions

We can show that the class of (arbitrarily) linearly ordered boron tree structures does not

satisfy the Ramsey property. We do so by showing that linear orderings do not destroy

copies of boron tree structures having di�erent orientations:

De�ne O′0 = O′1 = ∅,

O′0x1...xn−1=⟨O′x1...xn−1

, ∅⟩

O′1x1...xn−1=⟨∅,O′x1...xn−1

⟩for some xi ∈ {0, 1}.

Example 5.3.1. Below is a picture of a copy A′ of B(2) (in B(n), for some n ∈ ω) with

orientation O′010 = O′011.

A′

De�ne

Bx1...xn = {f : 4→ {0, 1} : f(0) = x1, . . . , f(n− 1) = xn} ,

where xi ∈ {0, 1}.

Proposition 5.3.2. Order B(4) with some linear ordering ≺ and linearly order B(2)

lexicographically. There is an order preserving embedding ϕ : B(2) ↪→ B(4) such that the

orientation of the image is O′x1,x2,x3 for some xi ∈ {0, 1}.


Proof. Let n be a positive integer. Note that if B is any subset of B(4), the statement

(∀b ∈ B)a ≺ b is equivalent to the statement a ≺ min B: (⇒) Suppose for all b ∈ B,

a ≺ b. Recall that ≺ is a �nite linear ordering. Therefore, minB ∈ B and consequently

a ≺ minB. (⇐) Suppose a ≺ minB. Let b ∈ B. Then either b = minB or minB ≺ b.

In either case, a ≺ b.

For brevity, we inspect, by induction: Suppose there is b ∈ B0 such that b ≺ min B1.

In this case, set a1 = b and x1 = 1. Otherwise, for all a ∈ B0, we have a � min B1. In

this case, set a1 = minB1 and x1 = 0. Now, assume a1, . . . , ak ∈ B(4) and x1, . . . , xk ∈

{0, 1} have been de�ned with 1 ≤ k < 4. Suppose there is b ∈ Bx1...xk0 such that

b ≺ min Bx1...xk1. In this case, set ak+1 = b and xk+1 = 1. Otherwise, for all a ∈

Bx1...xk0, we have a � min Bx1...xk1. In this case, set ak+1 = minBx1...xk1 and xk+1 = 0.

Therefore, ϕ(〈0, 0〉) = a1, ϕ(〈0, 1〉) = a2, ϕ(〈1, 0〉) = a3, ϕ(〈1, 1〉) = a4 is the required

embedding.

Proposition 5.3.3. Order B(4) with some linear ordering ≺ and linearly order B(2)

lexicographically. There is an order preserving embedding ϕ′ : B(2) ↪→ B(4) such that

the orientation of the image is O(2).

Proof. In fact, we only have to consider B0. There is an order preserving map from B0

to the ordinal 8. Note that if you partition {0, . . . , 7} into two sets of four numbers,

say P0, P1, you can pick two pairs {ai, bi} ⊆ Pi so that either a0 < b0 < a1 < b1 or

a1 < b1 < a0 < b0: Pick j, so that 0 ∈ Pj. Set k = j + 1(mod 2). Set c1 = min(Pj\{0}).

This implies that c1 ≤ 5. Consider the following possibilities:

c1 = 5: Then Pj\{0} = {5, 6, 7} and Pk = {1, 2, 3, 4}. Set ak = 1, bk = 2, aj = 5,

bj = 6.

c1 = 4: Then one of 5, 6, 7 is in Pj and {2, 3} ⊆ Pk. Set ak = 2, bk = 3, aj = 4 and

pick bj ∈ {5, 6, 7} ∩ Pj.

c1 = 3: Then two of 4, 5, 6, 7 are in Pk. Set aj = 0, bj = 3 and pick two distinct

ak < bk from {4, 5, 6, 7} ∩ Pk.


Corollary 5.3.4. Let C be any linearly ordered boron tree structure. Order B(5) with

some linear ordering and linearly order B(2) lexicographically. Call A, B the structures

corresponding to linearly ordered B(2) and B(5), respectively. Then we can de�ne a

colouring c :(CA

)→ r, such that for any B′ ∈

(CB

),∣∣∣c′′(B′A)∣∣∣ ≥ 2, where r is the number

of possible orientations of B(2).

Proof. Without loss of generality, assume C = B(N), while B(N) is ordered with some

linear ordering. List the possible orientations of B(2): O0, . . ., Or−1. If A′ ∈(B(N)A

)Ok,

set c(A′) = k. Let B′ ∈(CB

). By Lemma 5.2.12, there is a copy B′′ of a linearly ordered

B(4) in B′ with orientation being precisely O(4). By the two preceding propositions, B′′

contains copies of A with two di�erent orientations.

Now, we will provide an example to show that the class of linearly ordered boron trees

does not satisfy the ordering property: Consider B(3). Order B(3) with ≺ as follows:

〈0, 0, 0〉, 〈0, 1, 0〉, 〈1, 0, 0〉, 〈1, 1, 0〉, 〈1, 1, 1〉, 〈1, 0, 1〉, 〈0, 1, 1〉, 〈0, 0, 1〉.

1 8 2 7 3 6 4 5

Let n > 3 and order B(n) with the lexicographic ordering. Consider any embedding

ϕ : B(3) ↪→ B(n), not necessarily order preserving. We then have that for some i ∈ {0, 1},

ϕ′′Bxi(3) has the orientation O(2) (refer to the proof of 5.2.12 and the related discussion).

This means that the embedding cannot be order preserving.

Proposition 5.3.5. Let

Let L = (R, S) be a signature such that R is a 4-ary relation symbol and S is a

3-ary relation symbols. Suppose A is a boron tree structure with a boron tree structure

embedding ϕ : A→ B(n), for some n ∈ ω. De�ne o(A, ϕ) to be the (extending) structure

B with signature L such that


1. A = B.

2. RB = RA.

3. For all a, b, c ∈ A, (a, b, c) ∈ SB if and only if

ϕ(a), ϕ(b) <lex ϕ(c)

and

|ϕ(a) u ϕ(b)| > |ϕ(b) u ϕ(c)|.

In particular, A is a reduct of o(A, ϕ). De�ne OB0 to be the class of o(A, ϕ) for all pairs

of boron tree structures A and boron tree structure embeddings ϕ : A→ B(n) (n ∈ ω).

Claim 5.3.6. Suppose the following are true:

1. Ai is a boron tree structure, while ni is such that ϕi : Ai → B(ni) is a boron tree

structure embedding which satis�es property (3) above, for i ∈ {0, 1, 2}.

2. γi : Ai → A2 is a structural embedding, which preserves R and S, for i ∈ {0, 1}.

3. α : A0 → A1 is a structural isomorphism, which preserves R and S.

Then the orientation of (ϕ2 ◦γ0)′′A0 is equal to the orientation of (ϕ2 ◦γ1)′′A1 (in B(n2)).

Proof. Proceed by induction on the height of the orientation.

Let i ∈ {0, 1}. Recall that if Oi = 〈Oil ,Oir〉 is the orientation of (ϕ2 ◦ γi)′′Ai (in

B(n2)), then Oil and Oir are the orientations of the left and right sides of (ϕ2 ◦ γi)′′Ai (in

B(n2)), respectively.

We �rst show that the left side of (ϕ2 ◦γ0)′′A0 is isomorphic (as boron tree structures)

to the left side of (ϕ2 ◦ γ1)′′A1: Let x, y be in the left side of (ϕ2 ◦ γ0)′′A0. Let z be on

the right side of (ϕ2 ◦ γ0)′′A0. Then z >lex x, y and

((γ−10 ϕ−1

2 )(x), (γ−10 ϕ−1

2 )(y), (γ−10 ϕ−1

2 )(z)) ∈ So(A0,ϕ0)


and so

((αγ−10 ϕ−1

2 )(x), (αγ−10 ϕ−1

2 )(y), (αγ−10 ϕ−1

2 )(z)) ∈ So(A1,ϕ1)

Set

x′ = (ϕ2γ1αγ−10 ϕ−1

2 )(x)

y′ = (ϕ2γ1αγ−10 ϕ−1

2 )(y)

z′ = (ϕ2γ1αγ−10 ϕ−1

2 )(z).

Therefore, x′, y′ <lex z′ and

((γ−11 ϕ−1

2 )(x′), (γ−11 ϕ−1

2 )(y′), (γ−11 ϕ−1

2 )(z′)) =

((αγ−10 ϕ−1

2 )(x), (αγ−10 ϕ−1

2 )(y), (αγ−10 ϕ−1

2 )(z))

Suppose x′ and y′ are in opposite sides of (ϕ2 ◦ γ0)′′A0 Suppose x′ is on the left side.

Then y′ is on the right side. Since x′, y′ <lex z′, z′ is on the same side as y′, i.e., the right

side. Thus,

((γ−11 ϕ−1

2 )(x′), (γ−11 ϕ−1

2 )(y′), (γ−11 ϕ−1

2 )(z′)) /∈ So(A1,ϕ1)

which is a contradiction. Suppose x′ is on the right side. Then y′ is on the left side.

Since x′, y′ <lex z′, z′ is in the same side as x′, i.e., the right side. Thus,

((γ−11 ϕ−1

2 )(x′), (γ−11 ϕ−1

2 )(y′), (γ−11 ϕ−1

2 )(z′)) /∈ So(A1,ϕ1)

which is a contradiction. Thus, x′ and y′ must be on the same side. We can show the

same for the right side of (ϕ2 ◦ γ0)′′A0.

De�ne Ali to be the left side of (ϕ2 ◦ γi)′′Ai, for i ∈ {0, 1}. De�ne Ari to be the right

side of (ϕ2 ◦ γi)′′Ai, for i ∈ {0, 1}. Set Bli = (γ−1

i ϕ−12 )′′Ali and Br

i = (γ−1i ϕ−1

2 )′′Ari , for

i ∈ {0, 1}. Then α � Bri : Br

i → Bri and α � Bl

i : Bli → Bl

i are structural isomorphisms,


which preserve R and S.

Now, the induction assumption, applied to Al0 and Al1, implies that O0l = O1

l . Like-

wise, O0r = O1

r .

Theorem 5.3.7. OB0 is a Ramsey class.

Proof. Suppose o(A0, ϕ0) ≤ o(A1, ϕ1) ∈ OB0, where ϕ0 : A0 → B(n0) and ϕ1 : A1 →

B(n1) are boron tree structure embeddings. Let α : A0 → A1 be a structural embedding,

which preserves R and S.

Claim 5.3.8. Let O be the orientation of ϕ′′0A0. Let K ∈ N be such that K > n0. Say

γ : A0 → B(K) is an embedding which preserves R, and S. Then γ′′A0 has orientation

O.

Proof. Let D = ϕ′′0A0. De�ne ϕ′0 : D → B(K) by

(ϕ′0(ϕ0(a)))(j) = 0

for j ∈ (K − n0) and

(ϕ′0(ϕ0(a)))(j +K − n0) = (ϕ0(a))(j)

for j ∈ n0.

De�ne βn : B(n)→ B(n) to be the identity function. Thus,

o(B(n0), βn0), o(B(K), βK) ∈ OB0.

Note that ϕ′0 preserves R and S,. Moreover, (ϕ′0)′′D has orientation O in B(K). By

Claim 5.3.6, γ′′A0 has orientation O.


Let r ∈ N. By Theorem 5.2.5, there is B(N) ∈ B0 such that

B(N)→ (B(n1))(A0,O)r .

Let ϕ : B(N) → B(N) be the identity function. Hence, (B(N), ϕ) ∈ OB0. Let c :(o(B(N),ϕ)o(A0,ϕ0)

)→ r be a colouring. By Claim 5.3.6 and Claim 5.3.8, c induces a colouring

c :(B(N)A0

)O→ r such that c(o(A′, ϕ′)) = c(A′) for A′ ∈

(B(N)A0

)O. There is a copy

B′ ∈(B(N)B(n1)

)O(n1)

such that∣∣∣c′′ ((B′A0

)∩(B(N)A0

)O

)∣∣∣ = 1. Since the orientation of B′ is

O(n1), we can �nd an embedding ϕ2 : A1 → B′ which preserves R and S. Thus,

∣∣∣∣c′′(o(ϕ′′2A1, ϕ � (ϕ′′2A1)

o(A0, ϕ0)

)∣∣∣∣ = 1.

Corollary 5.3.9. Aut(Flim(OB0)) is extremely amenable.

Remark 5.3.10. Let us look closely at amalgamation in this class: Suppose o(A, ϕA),

o(B, ϕB), o(C, ϕC) are in OB0 and there are embeddings λB : A → B, λC : A → C.

Recall our de�nition of a rooted graph Gϕ, when given an embedding ϕ : A → B(n),

from the previous section. We will assign to Gϕ an ordering that coincides with the

lexicographic ordering on B(n). Now, set AB = λ′′BA. De�ne VB to be the set of

u ∈ V (GϕB) such that u = ϕB(x) u ϕB(y) for some x, y ∈ ϕ′′BAB. De�ne AC and VC ,

similarly. Let u, v ∈ VB be such that the height of u is less than the height of v. Set

HB(u,v) to be the elements w ∈ V (GϕB) such that u ( w, but u 6⊆ w. In other words, HB

(u,v)

consists of all the elements along the path between u and v as well as the elements that are

above the elements along this path. The vertex sets VB and VC induce isomorphic ordered

rooted trees GB and GC which are isomorphic via some order preserving isomorphism

α. �Attach along� each edge {u, v} in E(GB), the graphs HB(u,v) and H

C(α(u),α(v)). Embed

the resulting graph in B(k), for some k. Now, replace ϕ′′BAB with a copy of B(k), i.e.,


replace it with Bb(k + |b|) where b is the beginning of ϕ′′BAB. Embed this modi�cation

of ϕ′′BB in B(l). Finally, replace ϕ′′CAC with a copy of B(l). The resulting boron tree

structure gives us the amalgam we seek.

We can embed Flim(OB0) in a structure whose underlying universe consists of se-

quences of 0's and �nitely many 1's and −1's, all indexed by rationals. We can � for

example � imagine each such sequence as tracing a vertical path in R2, while −1 and 1

indicates a left or a right turn, respectively. In particular, on each such path we would

only see �nitely many turns. Set X = {−1, 0, 1}Q. Given f ∈ X, de�ne the support of f

to be

supp(f) = {x ∈ Q : f(x) 6= 0}.

De�ne

Xfin = {−1, 0, 1}Qfin = {f ∈ {−1, 0, 1}Q : |supp(f)| < ℵ0}.

Let f, g ∈ Xfin . Let Qf,g be the longest common initial segment of the increasing

sequences of rational numbers supp(f) and supp(g). We can then view f � Qf,g and

g � Qf,g as sequences of −1's, 0's, and 1's indexed by the increasing sequence of rationals

Qf,g. De�ne f u g to be the longest common initial segment of the sequences f � Qf,g

and g � Qf,g. De�ne an ordering ≺ on Xfin as follows: Suppose f 6= g. Let qf =

min(supp(f)\Qf,g) (assuming the latter set is nonempty). Let qg = min(supp(g)\Qf,g)

(assuming the latter set is nonempty). Suppose qf < qg: If f(qf ) < g(qg), set f ≺ g.

If f(qf ) = −1 and g(qg) = −1 , set f ≺ g . If f(qf ) = 1, set g ≺ f . Now, suppose

min(supp(f)\Qf,g) = ∅. If g(qg) = −1, set g ≺ f . If g(qg) = 1, set f ≺ g. De�ne the

relation S(f, g, h) to be satis�ed if and only if f, g ≺ h and h u g ( g u f .

Now, de�ne R(a, b, c, d) as follows: Let us �rst form a graph. Set V = {x u y : x, y ∈

{a, b, c, d}}. Given x 6= y ∈ V ,{x, y} ∈ E if and only if there is no z ∈ V such that

x ( z ( y. Now, a, b, c, d are the leaves of this graph and we de�ne R(a, b, c, d) to be

satis�ed if and only if there is a path between a and b and a path between c and d, while


these two paths do not intersect.

Claim 5.3.11. We can embed Flim(OB0) into the structure Xfin with R and S de�ned as

above.

Proof. Enumerate Flim(OB0) = {x0, x1, . . .}. Suppose we have embedded {x0, . . . , xn−1}

as {y0, . . . , yn−1}. Let ϕ : {x0, . . . , xn} → B(k) be an embedding that preserves R and

S. We have two corresponding graphs: V1 = {ϕ(xi)uϕ(xj) : i, j ∈ n} and V2 = {yi u yj :

i, j ∈ n} with respective edge sets E1, E2 de�ned as above. We can add a new vertex yn

� corresponding to xn � to an edge (yi0 u yi1 , yi2 u yi3) with

yi0 u yi1 ( yi2 u yi3

as follows: Set q ∈ Q to be such that

maxQyi0 ,yi1< q < min{r : r > maxQyi0 ,yi1

∧ (∃j)r ∈ supp(yj)}

Set yn to be such that supp(yn) = Qyi0 ,yi1∪ {q} and yn � Qyi0 ,yi1

= yi0 u yi1 . Set

yn(q) = −1 or yn(q) = 1 depending on how ϕ(xn) relates to {ϕ(x0), . . . , ϕ(xn−1)} with

respect to the relation <lex: Set A = {xi : ϕ(xi2) u ϕ(xi3) ⊆ ϕ(xi)}. Then either

(∀x ∈ A)ϕ(xn) <lex ϕ(x) or (∀x ∈ A)ϕ(x) <lex ϕ(xn) . In the �rst case, set yn(q) = −1.

Otherwise, set yn(q) = 1.

yi0 u yi1

yi2 u yi3

yn


Below, we show that B0 satis�es an orientation property analogous to the ordering

property:

Lemma 5.3.12. Let o(A1, ϕ1), o(A2, ϕ2) ∈ OB0. Suppose ϕ′′1A1 and ϕ

′′2A2 have the same

orientation. Then there is bijection α : A1 → A2 which preserves R and S.

Proof. We proceed by induction on the height of O. Let Ci be the left side of ϕ′′iAi, with

orientation Ol. Let Di be the right side of ϕ′′iAi, with orientation Or. Set Li = ϕ−1i Ci

and Ri = ϕ−1i Di By the induction assumption, there exist bijections

αl : L1 → L2

and

αr : R1 → R2

which preserve R and S. It su�ces to show that α = αl∪αr preserves R and S: Suppose

(a, b, c) ∈ So(A1,ϕ1) and a, b ∈ L1 and c ∈ R1. Then αl(a), αl(b) ∈ L2 and αr(c) ∈ R2.

Consequently,

(α(a), α(b), α(c)) = (αl(a), αl(b), αr(c)) ∈ So(A2,ϕ2).

Suppose (a, b, c) ∈ So(A2,ϕ2) and a, b ∈ L2 and c ∈ R2. Then α−1l (a), α−1

l (b) ∈ L1 and

α−1r (c) ∈ R1. Consequently,

(α−1(a), α−1(b), α−1(c)) = (α−1l (a), α−1

l (b), α−1r (c)) ∈ So(A2,ϕ2).

Set ϕli = ϕi � Li and ϕri = ϕi � Ri. Recall that given ϕ : A → B(n), V (Gϕ) =

{|ϕ(a) u ϕ(b)| : a, b ∈ A}. Since αl and αr preserve R, we have that Gϕl1∼= Gϕl2

and

Gϕr1∼= Gϕr2

, as rooted binary trees, by Fact5.1.1 and since the orientations of the sides

coincide. Therefore, α also preserves R.

Theorem 5.3.13. Let A ∈ B0. Let ϕ1 : A → B(n1) be an injection that preserves R.


Then there exists B ∈ B0 such that for any injection ϕ2 : A→ B(n2) which preserves R,

we have that o(A, ϕ1) ≤ o(B, ϕ2).

Proof. Let O be the orientation of ϕ′′1A. Let N be as in Lemma 5.2.12. Then for any

embedding ϕ2 : B(N) → B(n2), there is a copy A′ of A in ϕ′′2B(N) such that its

orientation is O. By Lemma 5.3.12, there is a bijection α : A → A′ which preserves R

and S. In other words, we can set B = B(N).

Moreover, the class B0 has a property analogous to being reasonable, in the context

of linear orderings:

Fact 5.3.14. Let o(A1, ϕ1) ∈ OB0. Suppose A1 ≤ A2. Then there exists ϕ2 : A2 →

B(n2) such that o(A1, ϕ1) ≤ o(A2, ϕ2).

Proof. We proceed by induction on |A2| − |A1|. Suppose A2 = {x1, . . . , xn−1, xn}. Set

A′ = {x1, . . . , xn−1}. By the induction assumption, there is ϕ′ : A′ → B(n′) such that

o(A1, ϕ1) ≤ o(A′, ϕ′). Suppose T is a boron tree such that (T ) = A2. Let v ∈ V (T ) be

a vertex adjacent to xn. Let u 6= w be vertices adjacent to v. Let T ′ be the tree with

V (T ′) = V (T )\{xn, v} and E(T ′) = (E(T ) ∩ (V (T ′))2) ∪ {{u,w}}. Let T ′′ be the tree

Gϕ′ with V (Gϕ′′) ⊆ 2≤n′, but with the root removed and an edge added between the two

vertices adjacent to the root. Then there is a graph isomorphism f : T ′ → T ′′. Observe

that f(u) ∈ 2≤n′with dom(f(u)) = k, for some k ≤ n De�ne g : (k + 1) → 2 by setting

g � k = f(u) and g(k) = 0. De�ne h ∈ 2n′+1 by setting h � k = f(u) and h(i) = 1 for

i ∈ n\k. Send Bf(u)(n′) to Bg(n

′) and xn to h. Send the other sequences in B(n′) to

B(n′ + 1) by concatenating them with a 0. Thus, n2 = n′ + 1 will su�ce.

Set F0 = Flim(B0), F = Flim(OB0), G0 = Aut(F0), and S0 = SF. We say a relation

Y ⊆ F 30 is an orientation of F0 if and only if for any �nite substructure B0 ⊆ F0,

its extension B =⟨B0, R

B0 , SB = Y ∩B30

⟩is in the class OB0. Let Or be the set of

orientations of F0. Observe that we can identify Or with a subset of the compact Cantor


space 2ω. Note that G acts on Or via the logic action: Given a, b, c ∈ F0, Y ∈ Or, and

g ∈ G,

(a, b, c) ∈ g · Y if and only if (g−1(a), g−1(b), g−1(c)) ∈ Y.

Thus, the closed and consequently compact closure of theG0-orbit of S0, namelyG0 · S0 ⊆

Or, is the universal minimal �ow of G0. Finally, we can also show that G0 · S0 = Or.

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Hrushovski and Ramsey Properties of Classes of Finite by Jakub … · 2012. 10. 31. · jfgj= kand x(i) = (fg)(i) for i