An abstract approach in canonizing topological Ramsey spaces

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by  Dimitris Vlitas

AN ABSTRACT APPROACH IN CANONIZING TOPOLOGICAL
RAMSEY SPACES
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DIMITRISVLITAS
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n Abstract. In[To]S.TodorcevicintroducedthenotionofatopologicalRam-
a sey space and a list of axioms required to be satisfied by any such a space.
J HereweshowthatanytopologicalRamseyspacethatsatisfiesastrengthened

6 version of one of the requiredaxioms and a very natural assumption, admits
2 canonization theorem.

]
O
L
.
h 1. Introduction
t
a
In[To]S.TodorcevicintroducedthenotionofatopologicalRamseyspace. Topo-
m
logical Ramsey spaces are structures of the form hR,≤,r,i, satisfying certain con-
[
ditions (see in the next section). Someone should think of R as family of infinite

1
sequence of objects and the function r as finite approximations of these infinite
v
6 sequences. Then ≤ is a quasi-ordering on R. These spaces admit a natural topol-
6 ogy and they are required to satisfy four axioms, A.1−A.4. As a consequence of
5
these axioms one gets that Ramsey subsets of R are exactly those with the Baire
7
0 property and meager sets are Ramsey null. Many of the well known spaces can be
1. seen as instances of topological Ramsey spaces. The most well known example is
0 the Ellentuck space hN[∞],⊆,ri. For A ∈ N[∞], r (A) is the initial segment of A
n
7
formed by taking the first n-elements of A.
1
: Canonical results in Ramsey theory try to describe equivalence relations in a
v
given Ramsey structure, based on the underlying pigeonhole principles. The first
i
X
example of them is the classical Canonization Theorem by P. Erdo˝s and R. Rado
r
a [Er-Ra] which can be presented as follows: Given α≤β ≤ω let
β
:={f(α) : f :α→β is strictly increasing}.
α
(cid:18) (cid:19)
The previous is commonly denoted by [β]α. Then for any n < ω and any finite
coloring of ω there is an isomorphic copy M of ω (i.e. the image of a strictly
n
increasing f : ω → ω) and some I ⊆ n(:= {0,1,...,n−1}) such that any two n-
(cid:0) (cid:1)
element subsets have the same color if and only if they agree on the corresponding
relative positions given by I.
This was extended by P. Pudla´k and V. Ro¨dl in [Pu-Ro] for colorings of a given
uniform family G of finite subsets of ω by showing that given any coloring of G,
thereexistsAaninfinite subsetofω,auniformfamilyT andamappingf :G →T
1
2 DIMITRISVLITAS
such that f(X) ⊆ X for all X ∈ G and such that any two X,Y ∈ G ↾ A have the
same color if and only if f(X)=f(Y).
The P. Erdo˝s-Rado result deals with equivalence relations on the family of k
approximations of elements of members of N[∞]. The Pudla´k-R¨odl result deals
with equivalence relations on uniform families of finite approximations of elements
of members of N[∞]. In this paper we are going to generalize the above results to
anytopologicalRamseyspace. Namelythatgivenafamilyoffiniteapproximations
F ofR,(seeDefinition1)andanequivalencerelationf :F →ω onit,thereexists
an X ∈ R and a map φ, ( see Definition 2 ) so that for any s,t ∈ AX, s,t ∈ F it
holds that f(s)=f(t) if and only if φ(s)=φ(t).
2. Background material
Topological Ramsey spaces are spaces of the form hR,≤,ri, where r :R×ω →
ARis a mapthatgivesus the sequencer(·,n)=r (·) ofapproximationmappings.
n
The basic open sets are of the form:
[s,X]={Y ≤X :(∃n)r (Y)=s},
n
for s ∈ AR and X ∈ R. If r (X) = s we write s ⊑ X. The axioms required to
n
be satisfied by any such a space in order to be topological Ramsey space are the
following.
A.1.
Let X,Y ∈R.
(1) r (X)=∅ for all X ∈R.
0
(2) X 6=Y implies r (X)6=r (Y) for some n∈ω.
n n
(3) r (X)=r (Y) implies n=m and r (X)=r (Y) for all k <n.
n m k k
A.2.
There is a quasi-ordering ≤ on AR such that
fin
(1) For any s∈AR the set
{t∈AR:t≤ s} is finite.
fin
(2) For any X,Y ∈R, X ≤Y if and only if
(∀n)(∃m)r (X)≤ r (Y).
n fin m
(3) For all s,t∈AR
[s⊑t∧t≤ t′ →∃t˜⊑t′ : s≤ t˜].
fin fin
A.3
Let s∈AR, X,Y,Z ∈R.
(1) If [s,X]6=∅ then [s,Y]6=∅ for all Y ∈[s,X].
AN ABSTRACT APPROACH IN CANONIZING TOPOLOGICAL RAMSEY SPACES 3
(2) X ≤Y and [s,X]6=∅ imply that there is Z ∈[s,Y] such that ∅=6 [s,Z]⊆
[s,X].
A.4
LetX ∈R, s∈(AR) ,[s,X]6=∅andO⊆(AR) . ThereexistsY ∈[s,X]such
n n+1
that:
r [s,Y]⊆O or r [s,Y]⊆Oc,
n+1 n+1
where r [s,Y]={t∈(AR) :s⊑t}.
n+1 n+1
WesaythatasubsetHofRisRamsey ifforevery[s,X]6=∅thereisaY ∈[s,X]
such that either [s,Y] ⊂ H or [s,Y] ⊂ Hc, and H is Ramsey null if for every
[s,X]6=∅,thereisY suchthat[s,Y]∩H=∅. In[To]itisshownthatifhR,≤,riis
a topological Ramsey space, then the Ramsey subsets of R are exactly those with
the Baire property. Moreover meager sets are Ramsey null. Then one gets as an
immediate consequence the following two corollaries.
Corollary 1. Let X ∈R, n<ω and c:AX →l be a finitecoloring. There exists
n
an Y ≤X so that c↾AY is constant.
n
and also
Corollary 2. Given c :[s,X]→l a finite Suslin measurable coloring, there exists
Y ∈[s,Y] so that c↾[s,Y] is constant.
Recall that a map f : X →Y between two topological spaces is Suslin measur-
able, if the preimage f−1(U) of every open subset U of Y belong to the minimal
σ−fieldofsubsetsofX thatcontainsitsclosedsetsanditisclosedundertheSuslin
operation [Ke].
For s∈AR and X ∈R we define the depth of s in X as follows:
min{k :s≤ r (X)} if (∃k)s≤ r (X),
fin k fin k
depth (s)=
X
(∞ otherwise.
From now on we will be working with topological Ramsey spaces that admit
a maximal element U. Therefore we are looking on the structures satisfying the
above axioms of the form
hU,≤,ri
where the elements of the space, are the reducts of U, X ≤U. Let
AU ={r (X):X ≤U}
n n
for n∈ω and
AU =∪ AU .
n∈ω n
Similarly for any X ≤ U, n ∈ ω, we define AX = {r (Y) : Y ≤ X} and AX =
n n
∪ AX . Fors∈AU by|s|wedenoteitslength,i.e. theuniquenumbernsothat
n∈ω n
s=r (X), for an X ≤U. For X ≤U by X(n) we denote the sequence of objects
n
4 DIMITRISVLITAS
r (X)\r (X),forn≥1. ThereforeforeachX ≤U wegetacountablesequence
n n−1
(X(n)). Similarly with each t ∈ AU we associate the sequence t(i) of length
n i∈n
n. Let U[k,l) = ∪ U(n). Observe that axiom A.2 implies that if X ≤ U,
n∈[k,l)
s=r (X)ands≤ r (U),thent=r (X)isoftheformt≤ r (U)forsome
n−1 fin k n fin l
l >k. As a consequence to get X(n) we have used the levels U(k),...,U(l−1) of
U. For s∈AU, with depth (s)=k, we define
U
X[s]={t∈AU :s⊑t and t≤ r (Y),for some Y ≤X,j <ω}.
fin j
Observe that X[s] 6= ∅ if and only if [s,X] 6= ∅. In this case we say that X is
compatible with s. Notice also that for every t∈U[s] there exists a set X ⊆[k,l),
where depth (t)=l, so that t\s is made out of ∪ U(n).
U n∈X
Next given s and X, so that [s,X]6=∅, by X/s we denote X\s.
3. Main theorem
We introduce the notion of a Front.
Definition 1. A family F of finite approximations of reducts of U is called a
front, if for every X ≤U, there exists s∈F so that s⊑X and for any two distinct
s,t∈F, is not the case that s⊑t.
We distinguish a specific case of fonts the families of finite approximations of
length n. Namely AU ={s:s=r (X),X ≤U}.
n n
Given a front F on [∅,X], X ≤U we introduce Fˆ defined as follows:
Fˆ ={t∈AU :∃s∈F,t⊑s}
observe that ∅∈Fˆ. For t∈Fˆ\F
F ={s∈F :t⊑s}.
t
Notice that F is a front on U/t. For Y ≤X
t
F ↾Y ={t∈F :t∈AY′,Y′ ≤Y},
Fˆ ↾Y ={t∈Fˆ :t∈AY′,Y′ ≤Y}.
Next we introduce a stronger version of A.4 as follows:
A.4⋆
Let s∈AU , depth (s)=k, X ≤U with [s,X]6=∅ and a coloring
n U
c:[s,X] →ω,
n+1
where [s,X] ={t∈AX :s⊑t}=r [s,X] be given. There exists a map
n+1 n+1 n+1
φ :[s,U] →P((∪[s,U] ) [k,∞))
s n+1 n+1
so that φ (p) ⊆ p(n)∪[k,l), where l = depth (p), [and Y ∈ [s,X] so that for all
s U
p,q ∈ [s,Y] it holds that c(p) = c(q) if and only if φ (p) = φ (q). We will call
n+1 s s
such a mapping φ inner for s.
s
AN ABSTRACT APPROACH IN CANONIZING TOPOLOGICAL RAMSEY SPACES 5
Inotherwords,thereexistsareductY,wherethe coloringcdependents onlyon
a subset of p(n), for any p∈[s,Y] , andthe levels U[k,l) needed to get p(n). In
n+1
some sense φ gives us a subset of the information coded by p(n).
s
A.4⋆ essentially deals with the length one extensions of initial segments. This
necessitatestointroduce the setofalllengthoneextensionsofallmembersofAX.
Let
LX ={w:∃s∈AX,s∪w ∈[s,X] }.
1 |s|+1
Observe that AX ⊆LX , cause r (X)=∅. Let
1 1 0
LX ={w:∃t∈AX,t∪w ∈[t,X] }
n |t|+n
and
LX =∪ LX .
n∈ω n
Next we introduce a partial ordering on LX as follows. Given w,v ∈LX we write
w ≤ v if there exists s ∈ AX so that s∪w∪v ∈ AX. Let w ,...,w elements of
0 n
LX so that either w ≤···≤ w and ∃t ∈AX so that t∪w ∪···∪w ∈AX or
0 n 0 n
w ∈X(n), for all i≤n, and ∃t∈AX such that t∪(w ,...,w )∈AX . Then
i 0 n |t|+1
hw ,...w i ∈LX isdefinedtobethesetofallendextensionofs,s≤ t,made
0 n s 1 fin
outof w ,...,w . When we say that the end extensionin made out of w ,...,w ,
0 n 0 n
we mean all w ,i ≤ n, are needed, not a proper subset is sufficient. Notice that if
i
s∪w∈[s,X] , then hwi =w in this case we say that hi acts trivially on w.
|s|+1 s s
To state the main theorem of this paper we need the following definition.
Definition 2. Let F be a front on [∅,U] and let φ be a function on F. We call φ
is Inner if for every t=t(i) ∈F we have that
i∈n
φ(t)=(φ (w ),...,φ (w )).
s0 0 sm m
Where φ ,...,φ are inner maps, for s = r (t),...,s = r (t) respec-
s0 sm 0 h0 m hm
tively, where h < ··· < h < n. Also w ∈ ht(i ),...,t(i )i ,...,w ∈
0 m 0 0 l t(h0) m
ht(j ),...,t(j )i , {i ,...,i ,...,j ,...,j ,h ,...,h } ⊆ n, and every t ,
0 m t(hm) 0 l0 0 ld 0 m i
i<n appears in at most one combination in {ht(i ),...,t(i )i ,...,ht(j ),...,
0 l t(h0) 0
t(j )i }.
m t(hm)
Now we can state the main theorem of this paper.
Theorem 1. Given a topological Ramsey space hU,≤,ri, that satisfies A.4⋆, a
front F on [∅,U] and a coloring f :F →ω, there exists X ≤U and an Inner map
φ on F ↾ X, so that for every s,t ∈ F ↾ X it holds that f(s) = f(t) if and only if
φ(s)=φ(t).
First we prove the following proposition.
Proposition 1. Suppose the topological Ramsey space hU,≤,ri has the property
that given a property P(·,·), s ∈ AU and X ≤ U, there exists Y ≤ X so that
P(s,Y). Then there exists Z ≤U such that for any s∈AZ it holds that P(s,Z).
6 DIMITRISVLITAS
Similarly for properties of the form P(·,·,·). If given s,t ∈ AU and X ≤ U,
there exists Y ≤ X so that P(s,t,Y). Then there exists Z ≤ U so that P(s,t,Z)
for all s,t∈AZ.
Proof. Let t = r (U) and U. There exists X ≤ U so that P(t ,X ). Set
0 0 0 0 0
t = r (X ). Consider the finite set A = {z ∈ AU : z ≤ t ,i < l}.
1 1 0 0 i i fin 1
For every z ∈ A there exists Y ≤ X so that P(z ,Y). After considering all
i 0 0 i
z ,i ∈ l, we get X ≤ X so that for every z ∈ A , P(z ,X ) holds. Set
i 1 0 i 0 i 1
t = r (X ). Suppose we have constructed t and X . Set t = r (X ).
2 2 1 n n n+1 n+1 n
Consider A = {z ∈ AU : z ≤ t ,i < l′}. For every z ∈ A there exists
n i i fin n+1 i n
Y ≤X so that P(z ,Y). Thereforewe getX ≤X so that for any z ∈A we
n i n+1 n i n
have P(z ,X ). Set t =r (X ). Proceed in that manner. Observe that
i n+1 n+2 n+2 n+1
for all n∈ω t ⊏t . Set Z =∪ t .
n n+1 n∈ω n
Now we provesimilarlythe secondstatement ofourproposition. Lett =r (U)
0 0
andt =r (U)andU. Thereexists X ≤U sothatP(t ,t ,X ). Lett =r (X ).
1 1 1 0 1 1 2 2 1
Consider the finite set A = {z ∈ AU : z ≤ t }. For any (s,t) ∈ [A ]2,
2 fin 2 2
there exists Y ≤ X so that P(s,t,Y). By exhausting all possible such a pairs
1
we get X ≤ X such that for any (s,t) ∈ [A ]2 it holds that P(s,t,X ). Set
2 1 2 2
t = r (X ). Suppose we have constructed t and X . Let t = r (X ) and
3 3 2 n n n+1 n+1 n
A ={z ∈AU :z ≤ t }. Foranypair(s,t)∈[A ]2 thereexistsY ≤X so
n+1 fin n+1 n n
that P(s,t,Y) holds. After considering all possible such a pairs, we get X such
n+1
that for any (s,t) ∈ [A ]2 it holds that P(s,t,X ). Set t = r (X ).
n+1 n+1 n+2 n+2 n+1
Observe that for every n∈ω, t ⊏t . Let Z =∪ t . (cid:3)
n n+1 n∈ω n
Next we make the following definition.
Definition 3. Let f be an equivalence relation on a front F on [∅,U]. Given
X ≤ U and s,t ∈ Fˆ\F we say that X separates s with t if for every Y ≤ X and
s′,t′ ∈F ↾Y where s⊑s′, t⊑t′ it holds that f(s′)6=f(t′). If there is no Z ≤X,
that separates s with t we say that X mixes s with t. We say that X decides for s
and t, if X either mixes or separates them.
Observe that X mixes s with t, if for every Y ≤ X, there exists s′,t′ ∈ F ↾ Y,
s⊑s′,t⊑t′, so that f(s′)=f(t′). The following proposition follows directly from
the definitions.
Proposition 2. The following hold.
(1) If X mixes (separates) s with t, so does any reduct Y ≤X.
(2) For every s,t ∈ Fˆ\F if for any w ∈ [s,X] there exists v ∈ [t,X]
|s|+1 |t|+1
so that X mixes s∪w with t∪v, then X also mixes s with t.
Next we observe the following.
Lemma 1. (Transitivity of mixing) Let s,t,w ∈ Fˆ\F, where the following holds
depth (s)=depth (t)=depth (t′)<ω. If X mixes s with t and t with t′, it also
X X X
mixes s with t′.
AN ABSTRACT APPROACH IN CANONIZING TOPOLOGICAL RAMSEY SPACES 7
Proof. Suppose that X mixes s with t and t with t′, but X separates s with t′.
Consider the two-coloring c :[t′,X] →2 defined by
1 n+1
1 if ∃q ∈[t,X] , and X mixes p with q,
n+1
c (p)=
1
(0 otherwise.
The fact that hU,≤,ri is a topologicalRamsey space,gives us a Y ∈[t′,X]so that
c ↾[t′,Y] =1. Similarlyweconsiderthetwo-coloringc :[t,Y] →2defined
1 n+1 2 n+1
by:
1 if ∃q ∈[s,Y] , and Y mixes p with q,
n+1
c (p)=
2
(0 otherwise.
which gives us a Z ∈ [t,Y] so that c ↾ [t,Z] = 1. But this implies that
2 n+1
Z ≤X mixes s with t′, a contradiction.
(cid:3)
The condition depth (s) = depth (t) = depth (w) < ω is necessary for the
X X X
transitivity of mixing to be valid. In [Vlit] where the topological Ramsey space of
hFIN[∞],≤,riisexamined,weseethattherearecoloringswherethecorresponding
k
notion of mixing is not transitive. An instance of such a coloring is as follows. Let
FIN be the set of all non empty finite subsets of ω. An element X of FIN[∞] is
a sequence X = (x ) so that x ∈ FIN, maxx < minx , for all n ∈ ω.
n n∈ω n n n+1
We write x < x to show that maxx < minx . Let hXi = {x ∪···∪
n n+1 n n+1 n0
x : n < ...n }. For X,Y ∈ FIN[∞], set Y ≤ X if y ∈ hXi for all n ∈ ω.
nk 0 k n
Finallywedefine thatr (X)=(x ) andr =∪ r . ThenhX,≤,ribecomesa
n i i<n n∈ω n
topological Ramsey space, see [To] for a full exposition. Let f : AX →ω defined
2
by c(x ,x )=x ∪x . Consider s=x , t=x ∪x and t′ =x ∪x ∪x . Notice
0 1 0 1 0 0 2 0 1 2
that X mixes s with t and s with t′, but X does not mixes t with t′.
Proposition 3. There exists X ≤U that decides for all s,t∈Fˆ ↾X.
Proof. Given s,t and Y ≤ U it suffices to show that there exists Z ≤ Y which
decides for s and t. Then the statement of the this proposition will follow from
Proposition 1 and the property P(s,t,Z) stating that Y decides for s and t. Con-
sider the two-coloring: c′ :[∅,Y]→2 defined by
1 if ∃p∈[s,Y′] ,q ∈[t,Y′] so that Y′ mixes p with q,
c′(Y′)= |s|+1 |t|+1
(0 otherwise.
The fact that hY,≤,ri is a topological Ramsey space, provides us with Z that
either mixes s with t, in the casethat c′ ↾[∅,Z]=1, or separatesthem, in the case
that c′ ↾[∅,Z]=0.
(cid:3)
Now we prove Theorem 1.
8 DIMITRISVLITAS
Proof. Let f : F → ω be given as in Theorem 1. Assume first that the notion of
mixingintroducedinDefinition 3is transitive. Inthe nextsubsectionwedealwith
the non-transitive case.
Observe that axiom A.4⋆ for any s ∈ Fˆ \F, |s| = n and X with [s,X] 6= ∅,
providesuswithaY ∈[s,X]andφ sothatforp,q ∈[s,Y] , Y mixespwithq if
s n+1
andonlyifφ (p)=φ (q). Thisisdonebyconsideringthecoloringc:[s,U] →ω
s t n+1
defined by c(p) = c(p′) if and only if U mixes p with p′. By Propostiton 1 there
exists X ≤ U so that for every s ∈ Fˆ\F ↾ X there exists such a φ . We assume
s
that U has this property, instead of one of its reducts. Therefore we assume that
U decides any s,t∈ AU and for any s ∈ Fˆ\F, |s| = n, φ defines an equivalence
s
relation on [s,U] .
n+1
Notice that for s ∈ Fˆ \F, if φ = ∅, then U mixes s∪w with s∪v for every
s
w,v ∈[s,U] .
1
Suppose now that our topological Ramsey space hU,≤,ri has the property that
given any s,t∈AU , Y ≤U/(s,t), then Y =s∪(Y[s]) and Y =t∪(Y[t]), where
n s t
Y[s] 6= ∅, Y[t] 6= ∅. In other words Y is compatible with both s and t. Instances
of such a topological Ramsey spaces are the hR ,≤,ri [Do-To] and hFIN[∞],≤,ri
1 k
[To].
Assume that Y mixes s with t and consider the two-coloring c′ : [t,Y ] → 2
t t n+1
defined by
1 if ∃q ∈[s,Y ] ,Y mixes p with q and φ (p)=φ (q),
c′(p)= s n+1 t t s
(0 otherwise.
The fact that hU,≤,ri is a topological Ramsey space, Corollary 1, gives us a
Z ≤ Y where c′ ↾ [t,Z] is constant. If the constant value is equal to one,
t n+1
then on Z we have that for every p ∈[t,Z] there exists a q ∈[s,Z ] , where
n+1 s n+1
Z =s∪(Z\t), Z mixes p with q and also φ (p)=φ (q).
s t s
If the constant value c′ ↾[t,Z] is equal to zero, for every p∈[t,Z] either
n+1 n+1
there exists q ∈ [s,Z] so that Z mixes p with q and φ (p) 6= φ (q), or there is
n+1 t s
not such a q. We require hU,≤,ri to satisfy the following property
(P): if c′ ↾[t,Z] =0, then there exists Z′ ≤Z so that Z′ separates s with t.
n+1
In the case that (P) is not satisfied, i.e. for every Z′ ≤ Z, there exist p ∈
[t,Z′] and q ∈ [s,Z′] , so that Z′ mixes p with q and φ (p) 6= φ (q), no
n+1 n+1 t s
canonization result can be obtained.
Therefore we assume that if Y mixes s with t, then c′ ↾ [t,Z] = 1. Observe
n+1
that if c′ ↾[t,Z] =1, then Z mixes s with t by Proposition 2.
n+1
What we have shown is if hU,≤,ri satisfies (P), given any s,t ∈ AU , there
n
exists Z ≤U so that one of the following two possibilities holds.
(1) Z mixes s with t if and only if for every p ∈ [t,Z] there exists q ∈
n+1
[s,Z] so that Z mixes p with q and φ (q)=φ (p).
n+1 s t
(2) Z separates s with t.
AN ABSTRACT APPROACH IN CANONIZING TOPOLOGICAL RAMSEY SPACES 9
Suppose now that our topological Ramsey space hU,≤,ri has the property that
foranys,t∈AU andX ≤U there existsY ≤X sothat [s,Y]6=∅ and[t,Y]=∅.
n
As we defined above, if X is so that [s,X] 6= ∅ and [t,X] 6= ∅ we say that X is
compatible withs andt. Aninstance ofsuchatopologicalRamseyspaceformsthe
space of strong subtrees hS (U),≤,ri, where U in this context is a tree with fixed
∞
branchingnumberbbutnofinitebranches[To]. Anotheristhespaceofconnections
hF ,≤,ri [Vl].
ω,ω
First we need the following: given s,t ∈ AU , depth (s) = k ≤ depth (t) = l
n U U
we require our space to have the property that: if s has an end extension w,
s∪w ∈[s,U] , made out of U[j,m), then t has also an end extension v, t∪v ∈
n+1
[t,U] ,madeoutofU[j,m),forl≤j ≤m. Supposenot,i.e. thereexisttwofinite
n+1
approximationss,t∈AU ,asabove,sothatforeveryX ≤U,∃Y ≤X andafinite
n
setX ⊂[j,m)suchthatt∪v ∈[t,Y] andv ismadeoutofU =∪ U(n)but
n+1 X n∈X
thereis nos∪w∈[s,Y] wherew ismade outofU . Considerthe two-coloring
n+1 X
c′ :[t,U] →2 defined by
n+1
1 if ∃s∪w ∈[s,U] ,w,v are both made out of U
n+1 X
c′(t∪v)= for some X ⊂[j,m),
0 otherwise.

We get X ≤ U so that c′ ↾ [t,X] = 0 and X is compatible with s as well.
n+1
But then no canonization theorem can be achieved. This is due to the fact that
φ ([s,X] ∩φ ([t,X] ) = ∅ always, even when X mixes s with t. Therefore
s n+1 t n+1
we can assume that our topological Ramsey space satisfies the above property.
We remark here that all the known topological Ramsey spaces satisfy the above
assumption.
We need also to assume that for any X ≤ U, s ∈ AU , m 6= 0, s ∈ Fˆ and
m
v so that s∪ v ∈ [s,X] , there exists Y ∈ [s,X] so that s ∪v ∈/ [s,Y] .
m+1 m+1
Suppose that there exists s∈AU and v so that s∪v ∈[s,Y] , for all Y ≤U.
m m+1
This in general is not possible cause our space will violate axiom A.4. To see that
considerthe coloringc′ :[s,U] →2 definedby c(s∪v)=0 andc(p)=1, for all
m+1
p∈[s,U] , p6=s∪v. ThereisnoY ≤U thatwillsatisfythe conclusionsofA.4.
m+1
The only exception occurs if s∪v is the only element of the set [s,U] . In that
m+1
case observe that φ = ∅. Therefore we can assume that given s ∈ AU , s ∈ Fˆ
s m
andv so that s∪v ∈[s,U] , there exists Y ∈[s,U]so that s∪v ∈/ [s,Y] . In
m+1 m+1
other words we can always go to a reduct that avoids a specific end-extension of s
with length m+1. If this is not the case, then {s∪v}=[s,U] and φ =∅.
m+1 s
Given s,t ∈ AU , recall from above U[t] = {p : t ⊑ p,p = r (Y),Y ≤ U}
n j
and U[s] = {q : s ⊑ q,q = r (X),X ≤ U}. Observe that [t,U] ⊂ U[t]
i n+1
and [s,U] ⊆ U[s] respectively. We have also assumed that depth (s) = k <
n+1 U
depth (t)=l. LetA={q∈[s,U] :q made of U[l,m)},wheremisthesmallest
U n+1
depth of all the length one end extensions of t.
10 DIMITRISVLITAS
At this stage we must assume that our space has the property that there exists
an one-to-one map ι : ∪[t,U] → ∪[s,U[q]] , q ∈ A, satisfying the following
n+1 n+1
properties:
(1) If t⊑p, then ι(t)⊑ι(p),
(2) If depth (p)=m, then depth (ι(p))=m,
U U
(3) If p ∈ [t,U] and p(n) is made out of U[l,m), then ι(p) ∈ [s,U[q]]
n+1 n+1
and ι(p)(n) is made out of U[l,m).
Assuming the existence of such a map and that U mixes s with t, consider the
two-coloringc′ :[s,U[q]] →2 defined by
n+1
1 if ∃p∈[t,U[t]] such that ι(p)=q′,
n+1
c′(q′)= U mixes p with q′ and φ (q′)=ι(φ (p)),
s t
0 otherwise.
Once again, we geta Z ≤U, Z[q]6=∅, Z[t]6=∅ so that either
c′ ↾[s,Z[q]] =1 or c′ ↾[s,Z[q]] =0.
n+1 n+1
In the first case Z mixes s with t and for every q′ ∈ [s,Z[q]] , there exists
n+1
p∈[t,Z] suchthatZ mixespwithq′ andφ (q′)=ι(φ (p)). Werequirehere,as
n+1 s t
above, that hU,≤,ri satisfies property (P), so the alternative c′ ↾ [s,Z[q]] = 0
n+1
is excluded.
Up to this point we have shownthat the topologicalRamsey space hU,≤,ri has
the propertythatgivens,t∈Fˆ, thereexistsZ ≤U thatmixes switht ifandonly
if φ agreewith φ , up to ι, otherwiseZ separatess with t. By Proposition1 there
s t
exists X ≤ Z with the property that given any s,t ∈ AX one has that either X
n
mixes s with t if and only if for every p ∈ [t,X] there exists q ∈ [s,X] so
n+1 n+1
that X mixes p with q and φ (q) = φ (p), up to ι, or X separates s with t. From
s t
now on we will denote F instead of F ↾ X and Fˆ\F instead of Fˆ\F ↾ X since
everything is taking place below X. We will also omit the ι.
We arenowreadytodefinethe innermapφthatisgoingtowitnessthecoloring
being canonical. For s∈Fˆ, |s|=n, we define
φ(s)= φs′(s(|s′|))=(φr0(t)(r1(t)),φr1(t)(t(2)),...,φrn−1(t)(t(n))).
s′⊑s
[
Next we have to show the following four lemmas.
Lemma 2. The following are true for all Y ≤X.
(1) Let s,t ∈ Fˆ\F. If φ 6= ∅ and φ = ∅, there exists w ∈ [s,X] so that
s t |s|+1
X mixes t with s∪w with at most one equivalence class of [s,X] .
|s|+1
(2) If X separates s with t, then its separates s ∪w with t ∪v for all w ∈
[s,X] and v ∈[t,X] .
|s|+1 |t|+1
(3) If s⊏t, s,t∈F and φ(s)=φ(t), then X mixes s with t.
Proof. Suppose that X mixes t with s∪w, and also t with s∪v, and φ (s∪w)6=
s
φ (s∪v). By Lemma 1, we get that X mixes s∪w with s∪v, a contradiction.
s