Canonical equivalence relations on fronts on $FIN_k$

0.24 MB·

by  Dimitris Vlitas

CANONICAL EQUIVALENCE RELATIONS ON FRONTS ON
FIN
k
7
1
DIMITRISVLITAS
0
2
Abstract. Weprovethat foreveryequivalence relateiononabarrieronthe
n
a spacehFINk[∞],≤,ri,foranyk,thereexistsY ∈FINk[∞] sothat therestric-
J tionofthecoloringonhYiiscanonical.

6
2

]
O
1. Introduction
L
. Canonical results in Ramsey theory try to describe equivalence relations in a
h
t given Ramsey structure, based on the underlying pigeonhole principles. The first
a
m example of them is the classical Canonization Theorem by P. Erdo˝s and R. Rado
[Er-Ra] which can be presented as follows: Given α≤β ≤ω let
[

1 β
:={f(α) : f :α→β is strictly increasing}.
v α
5 (cid:18) (cid:19)
6 The previous is commonly denoted by [β]α. Then for any n < ω and any finite
5
coloring of ω there is an isomorphic copy M of ω (i.e. the image of an strictly
7 n
0 increasing f : ω → ω) and some I ⊆ n(:= {0,1,...,n−1}) such that any two n-
(cid:0) (cid:1)
.
1 element subsets have the same color if and only if they agree on the corresponding
0 relative positions given by I.
7
This was extended by P. Pudla´k and V. Ro¨dl in [Pu-Ro] for colorings of a given
1
: uniform family G of finite subsets of ω by showing that given any coloring of G of
v
i finite subsets of ω, there exists A an infinite subset of ω, a uniform family T and
X
a mapping f : G → T such that f(X) ⊆ X for all X ∈ G and such that any two
r
a X,Y ∈ G ↾ A have the same color if and only if f(X) = f(Y). Since then, many
results of similar nature have been obtained (see [Mil], [Vli]).
In 1992 G.T.Gowers [Gow] obtained a stability result for real valued Lipschitz
functions defined in the unit sphere c . This result is actually a consequence of a
0
deep infinite dimensional Ramsey type result, which gives rise to a Ramsey space.
In this paper we canonize equivalence relations on that space. To state our result
we need to introduce some notions.
Givenapositiveintegerk,letFIN bethesetofmappingsx:ω →{0,1,...,k},
k
called k-vectors, whose support suppx = {n : x(n) 6= 0} is finite and with k
in their range. An element X of FIN[∞] is a sequence X = (x ) so that
k n n∈ω
maxsuppx < minsuppx , for all n ∈ ω. We write x < x to show that
n n+1 n n+1
maxsuppx <minsuppx .
n n+1
1
2 DIMITRISVLITAS
Let T : FIN → FIN be the map defined by T(x)(n) = max{x(n)−1,0}.
k k−1
The k-combinatorial subspace hXi is the set of combinations of the form:
Ti0xn0 +···+Timxnm
with the condition that at least one i = 0, j < m+1. By Ti we mean Tix(n) =
j
max(x(n)−i,0), for i > 0 and T0 = id. Given X = (x ) , we define the length
n n∈l
of X, denoted by |X|, to be equal to l. For X = (x ) , Y = (y ) ∈
n n∈|X| n n∈|Y|
FIN[≤∞], set X ≤ Y if x ∈ hYi for all n < |X|. In this case we say that X is a
k n
block −subsequence of Y. Then ≤ is a partial ordering on FIN[∞].
k
For X ∈ FIN[∞] we define r (X) = (x ) . We set AX = {r (Y) : Y ≤ X}
k n i i∈n n n
and AX = ∪ AX . Next we introduce a topology on FIN[∞] with basic open
n∈ω n k
sets as follows. For s∈AX , by |s|=n we denote is length. We define
n
[s,X]={Y :Y ≤X,r (Y)=s}.
n
These set form the basic sets for a topology on the space FIN[∞].
k
In [To] it is shown that hFIN[∞],≤,ri satisfies axioms A.1 − A.4, so it is a
k
topologicalRamsey space. Two corollaries of being such a space are the following.
Corollary 1. Let c:AX →l be a finite coloring. There exists an Y ≤X so that
n
c↾AY is constant.
n
and
Corollary 2. Let s∈AX and c:[s,X]→l be a finite Suslin measurable coloring.
There exists Y ≤[s,X] 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 a family of finite approximation of elements of FIN[∞] called fronts (see
k
Definition 1), we prove the following.
Theorem. Let f : F → ω be a coloring of a front on [∅,X], for X ∈ FIN[∞].
k
There exists Y ≤X so that f ↾F ↾hY i is canonical.
0 0
Herebycanonical,wemeanthatthereexistsamapφsothatforevery(t ,...,t )∈
0 d−1
F ↾hY i
0
φ(t0,...,td−1)⊆(Ti0ti0 +···+Tiltil,...,Tj0tj0 +···+Tjmtjm),
where {i ,...,i ,...,j ,...,j } ⊆ d and every t , i < d appears in at most one
0 l 0 m i
combination in {Ti0ti0 +···+Tiltil,...,Tj0tj0 +···+Tjmtjm}⊂hXi. Then φ is
so that for any s,t ∈ F it holds f(s) = f(t) if and only if φ(s) = φ(t). We view
elements on FIN , the k-vectors,in the set theorytic way,as sets of orderedpairs.
k
The subsetis takeninthis sense. The proofofthe abovetheoremis divided intwo
parts. In the next section we add the necessary definitions and new concepts and
CANONICAL EQUIVALENCE RELATIONS ON FRONTS ON FINk 3
we present the first part of the proof. In the final section we present the second
part.
2. Main theorem
Theabovepartialordering≤onFIN[∞],allowsthe finitization≤ asfollows:
k fin
for X = (x ) , Y = (y ) , we say that X ≤ Y if and only if X ≤ Y and
i i∈h j j∈m fin
(∀l<m),X (cid:2),Y ↾l.
For s∈AX and X ∈FIN[∞] we define the depth of s in X as follows:
k
min{k :s≤ r (X)} if (∃k)s≤ r (X),
fin k fin k
depth (s)=
X
(∞ otherwise.
Given an non empty basic open set [s,X], |s|=n, let
[s,X] ={t∈AX :s⊑t}.
n+1 n+1
Now we introduce the notion of a Front.
Definition1. Afamily F of finiteapproximations ofreductsof X is called afront,
if for every Y ≤ X, there exists s ∈ F so that s = r (Y) and for any two distinct
n
s,t∈F, is not the case that s⊑t.
We distinguish specific instances of fronts on X, the AX . Given a front F on
n
[∅,X], we introduce Fˆ defined as follows:
Fˆ ={t∈AX :∃s∈F,t⊑s}
observe that ∅∈Fˆ. For t∈Fˆ\F
F ={s∈F :t⊑s}.
t
For Y ≤X F ↾Y ={t∈F :t∈AY′,Y′ ≤Y},
Fˆ ↾Y ={t∈Fˆ :t∈AY′,Y′ ≤Y}.
[∞]
Finally for s ∈ AX and X ∈ FIN by X/s we denote X \s. Similarly for
k
s,t∈AX, by X/(s,t) we denote X \s∩X \t.
The following proposition is a fact that holds in any topological Ramsey space.
For the shake of completeness, we give a proof here in the context of our space.
Proposition 1. Suppose the hFIN[∞],≤,ri has the property that given a property
k
P(·,·), s∈AX and Y ≤X, there exists Z′ ≤Y so that P(s,Z′). Then there exists
Z ≤X such that for any s∈AZ it holds that P(s,Z).
Similarly for properties of the form P(·,·,·). If given s,t ∈ AX and Y ≤ X,
there exists Z′ ≤ Y so that P(s,t,Z′). Then there exists Z ≤ X so that P(s,t,Z)
for all s,t∈AZ.
4 DIMITRISVLITAS
Proof. Let t = r (X) and X. There exists X ≤ X so that P(t ,X ). Set
0 0 0 0 0
t = r (X ) and let X ≤ X so that P(t ,X ) holds. Set t = r (X ). Consider
1 1 0 1 0 1 1 2 2 1
the finite set A = {z ∈ AX : z ≤ t }. For every z ∈ A there exists Y ≤ X
2 fin 2 2 0
so that P(z,Y). After considering all z ∈ A we get X ≤ X and t = r (X )
2 2 1 3 3 2
so that P(z,X ) holds, for all z ∈ A . Suppose we have constructed t and X .
2 2 n n
Set t = r (X ). Consider A = {z ∈ AX : z ≤ t }. For every z ∈ A
n+1 n+1 n n fin n+1 n
there exists Y ≤ X so that P(z,Y). Therefore we get X ≤ X so that for
n n+1 n
anyz ∈A we haveP(z,X ). Set t =r (X ). Proceedinthat manner.
n n+1 n+2 n+2 n+1
Observe that for all n∈ω t ⊏t . Set Z =∪ t .
n n+1 n∈ω n
Nowweprovesimilarlythesecondstatementofourproposition. Lett =r (X)
0 0
andt =r (X)andX. ThereexistsX ≤X sothatP(t ,t ,X ). Lett =r (X ).
1 1 1 0 1 1 2 2 1
Consider the finite set A = {z ∈ AX : 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 ∈AX :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
As mentioned above in [To] is shown that hFIN[∞],≤,ri satisfies a pigeon hole
k
property(axiomA.4 in [To]). To showthat this space satisfiesa strengthenpigeon
holeproperty,(Theorem1below),weintroducesomeconceptsanddefinitionsfrom
[Ab]. WerestrictourattentiontothesubsetofFIN thatcontainsallthek-vectors
k
that are system of staircases.
Definition 2. Given an integer i∈[1,k], let min ,max :FIN →ω be the map-
i i k
pings min x = min x−1({i}), max x = maxx−1({i}), if defined and 0 otherwise.
i i i
A k-vector x is a system of staircases (sos in short) if and only if
1 Range x={0,1,...,k},
2 min x<min x<max x, for i<j ≤k,
i j i
3 for every 1≤i≤k,
Range x↾[min x,min x)={0,...,i−1},
i−1 i
Range x↾(max x,max x]={0,...,i−1},
i i−1
Range x↾[min x,max x]={0,...,k}.
k k
An X = (x ) ∈ FIN[∞] is a system of staircases if and only if every k-vector
n k
x is an sos. In [Ab] it is shownthat for every X ∈FIN[∞] there exists Y ≤X so
n k
that X is sos and that T preserves sos. Next we introduce some mappings.
CANONICAL EQUIVALENCE RELATIONS ON FRONTS ON FINk 5
Definition 3. Let X ∈FIN[∞] be a sos. The mapping min , for i∈[1,k], can be
k i
interpreted as
i if n=min (w),
i
min(w)(n)=
i (0 otherwise.
for w ∈hXi. Extending this, define for I ⊆{1,...k}, the mapping min : hAi →
I k
FIN ⊆FIN by min (w)(n) =i if n =min (w), for i∈I and 0 otherwise,
maxI ≤k I i
i.e. min (w)={(min (w),i):i∈I}, and extended by 0. Similarly, let
I i
i if n=max (w),
i
max(w)(n)=
i (0 otherwise.
and let max : FIN → FIN be defined by max (w) = {(max (s),i) : i ∈ I},
I k maxI I i
again extended by 0. Clearly min = min and max = max , where for
I i∈I i I i∈I i
two mappings f,g :hXi →FIN we define (f ∨g)(w)=f(w)∨g(w).
k ≤k W W
For l ≤i, let θ0 :hXi→FIL be the mappings defined by
i,l l
θ0 (w)={(n,l):n∈(min (w),min (w)),w(n) =l} extended by 0,
i,l i i+1
θ1 (w)={(n,l):n∈(max (w),max (w)),w(n) =l} extended by 0,
i,l i+1 i
θ2(w) = {(n,l): n ∈(min (w),max (w)),w(n) = l} extended by 0, where
l k k
l ≤k.
Definition 4. Let G = {min ,...,min }, G = {max ,...,max }, G =
min 1 k max 1 k midǫ
{θǫ : i ∈ {1,...,k −1},l ∈ {1,...,i−1}}, for ǫ = 0,1 and G = {θ2 : l ∈
i,l mid l
{1,...,k}∪{0}. Set G =G ∪G ∪G ∪G ∪G .
min max mid0 mid1 mid
Given a k-block sequence X, we say that f : hXi → FIN is a staircase
≤k
function if it is in the lattice closure of G. An equivalence relation R on hXi is a
staircase relation if the following holds: sRt if and only if f(s) = f(t), for some
staircase mapping f. In [Ab] is shown that if f is a staircase map then there are
I ⊆ {1,...,k},J ⊆ {j ∈ I : j−1 ∈ I }, (l(ǫ)) with l(ǫ) ≤ j−1, for ǫ = 0,1
ǫ ǫ ǫ ǫ j j∈Jǫ j
and l(2) such that
k
f =:min∨ θ0 ∨θ2 ∨max∨ θ1 .
I0 j_∈J0 j−1,l(j0) l(k2) I1 j_∈J1 j−1,l(j1)
As with the k-vectors, given two functions f,g : hAi→FIN , we write f <g
≤k
to denote that f(w) < g(w), for all w ∈ hAi. Therefore any staircase mapping f
has a unique decomposition f =∪ f with f <f <···<f in G.
i∈n i 0 1 n−1
In [Ab] J. Lopez-Abad showed the following.
Theorem 1. For every k andevery equivalence relation on FIN thereis a system
k
of staircases B such that the equivalence relation restricted to hBi is a staircase
equivalence relation.
Observethattheabovetheoremistheonedimentionalcaseofourmaintheorem
mentioned in the introduction. In other words it takes care of the front AX , for
1
[∞]
X ∈FIN . Next we make the following definition.
k
6 DIMITRISVLITAS
Definition 5. Given F a front on [∅,X] and f : F → ω. Fix s,t ∈ Fˆ \F and
X. X separates s and t if and only if for all w ∈ hX/si and v ∈ hX/ti so that
s∪w,t∪v ∈F ↾X, f(s∪w) 6=f(t∪v). X mixes s with t, if there is no Y ≤X
which separates s with t. X decides for s with t if and only if either X mixes s
with t, or X separates s with t.
Therefore X mixes s with t if and only if for each Y ≤ X, there are w ∈ hY/si
and v ∈hY/ti so that s∪w,t∪v ∈F ↾Y and f(s∪w)=f(t∪v).
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.
Proposition 3. Given X and a front F on [∅,X], there exists Z ≤X so that for
all s,t∈Fˆ ↾Z, Z decides s with t.
Proof. Given s,t and Y ≤ X 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
Proposition1andthepropertyP(s,t,Y)statingthatY decidesforsandt. Assume
that depth (s) ≤ depth (t) and consider the two-coloring: c′ : [t,Y] → 2 defined
X X
by
1 if ∃w∈hY′i,v ∈hY′i so that Y′ mixes s∪w with t∪v,
c′(Y′)=
(0 otherwise.
Corollary 2, provides us with Z that either mixes s with t, in the case that
c′ ↾[t,Z]=1, or separates them, in the case that c′ ↾[t,Z]=0.
(cid:3)
The abovenotionofmixinginducedby Definition5 isnotnecessarilytransitive.
An example of such an equivalence relation is the following. Let X = (x ) ∈
n n∈ω
FIN[∞] and c : AX → ω, defined as follows c(s,t) = s+t = s∪t. Let s = x ,
1 2 0
t=x +x and p=x +x +x . Observe that the depth of s,t and p in X is not
0 2 0 1 2
the same. Then X mixes s with t and s with p, but X does not mixes t with p.
The above example generalizes easily for any k.
In the case of the same depth, mixing is transitive as the next lemma shows.
Lemma 1. Let s,t,p ∈ AX , with depth (s) = depth (t) = depth (p). If X
n X X X
mixes s with t and X mixes t with p, then X mixes s with p.
Proof. Suppose that X mixes s with t and X mixes t with p, but X separates s
with p. Consider the two-coloring c :[p,X] →2 defined by
1 n+1
1 if ∃t∪v ∈[t,X] , and X mixes p∪w with t∪v,
n+1
c (p∪w)=
1
(0 otherwise.
CANONICAL EQUIVALENCE RELATIONS ON FRONTS ON FINk 7
Corollary 1, gives us a Y ∈[p,X] so that c ↾[p,Y] =1. Similarly we consider
1 n+1
the two-coloringc :[t,Y] →2 defined by:
2 n+1
1 if ∃s∪v′ ∈[s,Y] , and Y mixes t∪v with s∪v′,
n+1
c (t∪v)=
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 p, a contradiction.
(cid:3)
Now we are ready to state and prove our main theorem.
Theorem 2. Let f :F →ω be acoloring of a front on X, for X ∈FIN[∞]. There
k
exists Y ≤X so that f ↾F ↾Y is canonical.
0 0
Let F be any front on hXi and f : F → ω any coloring. Then Theorem 1
looks after the case of front F = AX . Consider an arbitrary front F. Definition
1
5 gives us the notion of separation and mixing. We are going to divide the proof
in two parts. In the first part we assume that mixing is transitive. In the second
part we deal with the case that mixing is not transitive. In the first case we are
going to obtain a map φ , φ(t ,...,t ) ⊆ (t ,...,t ) so that for any s,t ∈
0 d−1 0 d−1
F it holds that f(s) = f(t) if and only if φ(s) = φ(t). In the second case we
obtain a map φ so that φ(t0,...,td−1) ⊆ (Ti0ti0 +···+Tiltil,...,Tj0tj0 +···+
Tjmtjm), {i0,...,il,...,j0,...,jm}⊆d, and each ti, i<d appears in at most one
combinationin{Ti0ti0+···+Tiltil,...,Tj0tj0+···+Tjmtjm}⊂hXi. Thenφhas
the property that for any s,t∈F it holds f(s)=f(t) if and only if φ(s)=φ(t).
Assume that mixing is transitive. If s = (s ,...,s ) is an n-tuple and w,v ∈
0 n1
FIN arelengthoneextensionsofs,bys∪wwedenotethen+1-tuplethatextends
k
s by w. From now on for notational simplicity when we write s∪w we mean that
s∪{w} = (s ,...,s ,w) and when we write s∪w+v we mean the n+1-tuple
0 n1
(s ,...,s ,w+v). For any s ∈ Fˆ\F, |s| = n and X ∈ FIN[∞] with [s,X] 6= ∅,
0 n1 k
Theorem1 providesus with a Y ∈[s,X]andφ sothat for s∪w,s∪v ∈[s,Y] ,
s n+1
Y mixes s∪w with t∪v if and only if φ (w)=φ (v). This is done by considering
s t
the equivalence relation c : [s,X] → ω, on hXi, defined by c(w) = c(v) if and
n+1
only if X mixes s∪w with s∪v. By Proposition 1, we can assume that for every
t∈Fˆ\F ↾X,thereexistsastaircasemapφ whichinducesanequivalencerelation
t
on [t,X] .
n+1
Assume that X mixes s with t, s,t ∈ AX , and consider the two-coloring c′ :
n
[t,X] →2 defined by
n+1
1 if Y mixes t∪w with s∪w and φ (w)=φ (w),
c′(t∪w)= t s
(0 otherwise.
The fact that hFIN[∞],≤,ri is a topological Ramsey space gives us a Z ≤ Y
k
where c′ ↾ [t,Z] is constant. If the constant value is equal to one, then on
n+1
8 DIMITRISVLITAS
Z we have that for every t∪w ∈ [t,Z] , Z mixes t ∪w with s∪w and also
n+1
φ (w)=φ (w).
t s
Let c′ ↾ [t,Z] = 0. Then for every t ∪ w ∈ [t,Z] either there exists
n+1 n+1
s∪v ∈ [s,Z] where Z mixes t∪w with s∪v and φ (v) 6= φ (w) or there is
n+1 s t
no s∪v ∈ [s,Z] that Z mixes it with t∪w. This allows to consider the two
n+1
coloring: c :[t,Z] →2 defined by:
1 n+1
1 if there exists s∪v ∈[s,Z] ,Z mixes t∪w with s∪v,
n+1
c1(t∪w)= φs(v)6=φt(w),
0 otherwise.

Once more there exists Z ∈ [t,Z] so that c ↾ [t,Z ] is constant. If the constant
1 1 1
valueisequaltozero,thenZ separatesswitht,acontradictiontotheassumption
1
thatZ mixesswitht. Supposethatc ↾[t,Z ]=1. Thenforeveryt∪w∈[t,Z ]
1 1 1 n+1
there exists s∪v ∈[s,Z ] so that Z mixes t∪w with s∪v and φ (v)6=φ (w).
1 n+1 1 s t
Observe that it might be the case that φ = φ and for every t∪w ∈ [t,Z ] ,
s t 1 n+1
there exists s∪v ∈ [s,Z ] , w 6= v, so that Z mixes t∪w with s∪v and even
1 n+1 1
though φ = φ , φ (v) 6= φ (w) cause w 6= v. The assumption that φ = φ rules
s t s t s t
out the possibility of Z mixing t∪w with s∪w, for any w ∈hZ i.
1 1
Letφ =∪ f ,φ =∪ g ,α,β ∈k+1,andX ⊆α,Y ⊂β,|X|=|Y|,sothat
s i∈α i t j∈β j
foreveryi∈X therej ∈Y sothatf =g . Letφ′ =∪ f andφ′ =∪ g .
i j s i∈α\X i t j∈β\Y j
For every w∈hZ i consider the finite sets:
1
A0 = {Tiw,i < k : φ′(Tiw+v) = φ′(Tiw+v′) and φ′(Tiw+v) = φ′(Tiw+
w s s t t
v′) for all v′,v ∈hZ /wi}.
1
Notice that A0 6=∅ if and only if {f ,g :i∈α\X,j ∈β\Y}⊂F ∪F .
w i j min mid0
Observealso that in the abovedefinition v,v′ play not a role in the value of φ′,φ′,
s t
areneededonlyforTiw+v′ ∈FIN . WithA0 weassociateF0 ={i:Tiw∈A0}.
k w w w
Similarly if {f ,g :i∈α\X,j ∈β\Y}⊂F ∪F we define
i j max mid1
A1 = {Tiw,i < k : φ′(w′ +Tiw) = φ′(v′+Tiw) and φ′(w′ +Tiw) = φ′(v′ +
w s s t t
Tiw) for all v′,w′ ∈hZ /wi}.
1
WithA1 weassociateF1 ={i:Tiw ∈A1}. Observethatgivenφ′,φ′,F0 =F0
w w w s t w v
and F1 =F1, for all v. Let
w v
D0 ={(Tjw,Tiw)∈A0 ×A0 :Z mixes s∪Tjw+w′ with t∪Tiw+v′
w w w 1
for all v′,w′ ∈hZ /wi}.
1
With D0 we associate the set C0 ={(j,i):(Tjw,Tiw)∈D0}.
w w w
CANONICAL EQUIVALENCE RELATIONS ON FRONTS ON FINk 9
Similarly we define
D1 ={(Tjw,Tiw)∈A1 ×A1 :Z mixes s∪w′+Tjw with t∪v′+Tiw
w w w 1
for all v′,w′ ∈hZ /wi}.
1
With D1 we associate the set C1 ={(j,i):(Tjw,Tiw)∈D1}.
w w w
Notice that Ci, i<2, may contains pairs of the form (j,i), where j =i. In the
w
case that φ = φ we have that both C0 and C1 do not contain elements of the
s t w w
form (j,j), j <k. The rest is identical with the case that φ 6=φ .
s t
Observealsothatforeveryw ,w ∈hZ i,i<kandf ∈F ∪F ,f(Tiw +
0 1 1 max mid1 0
w )=f(w ). In the case that f ∈F ∪F , then f(w +Tiw )=f(w ).
1 1 min mid0 0 1 0
Claim 1. Let f,g ∈ F ∪ F ∪F , f 6= g and for i < j, it holds that
min mid0 mid
supp(f(w)) ⊂ [min (w),min (w)) and also supp(g(w)) ⊂ [min (w),min (w)).
i i+1 j j+1
Then for any w ∈ hZ i there always exists Tk−iz, z < w, so that f(Tk−iz+w) 6=
1
f(w) and g(Tk−iz+w)=g(w).
Similarly if f,g ∈ F ∪ F ∪ F , f 6= g and for i < j, it holds that
mid max mid1
supp(f(w))⊂(max (w),max (w)]andalsosupp(g(w))⊂(max (w),max (w)].
i+1 i j+1 j
For any w ∈hZ i there always exists Tk−iz, w < z, so that f(w+Tk−iz)6= f(w)
1
and g(w+Tk−iz)=g(w).
Proof. Let f,g ∈ F ∪F ∪F . Assume that i < j, pick any z < w and
min mid0 mid
considerTk−iz ∈FIN . Noticethatf(Tk−iz+w)6=f(w)andg(Tk−iz+w)=g(w).
i
This is due to the fact that min (Tk−iz +w) = min (w), min (Tk−iz +w) =
j j j+1
min (w) and for every θ0 , l≤j, it holds that θ0 (Tk−iz+w)=θ0 (w).
j+1 j,l j,l j,l
Let f,g ∈ F ∪F ∪F as in the claim. Pick any z > w and consider
max mid1 mid
Tk−iz ∈FIN . Noticethatf(w+Tk−iz)6=f(w)andg(w+Tk−iz)=g(w). Thisis
i
duetothefactthatmax (w+Tk−iz)=max (w),max (w+Tk−iz)=max (w)
j j j+1 j+1
and for every θ1 , l≤j, it holds that θ1 (w+Tk−iz)=θ1 (w).
j,l j,l j,l
(cid:3)
ObservethatonhZ ithereisanone-to-onecorrespondencebetweenequivalence
1
classes induced by φ and φ . Let once more f,g ∈ F ∪F ∪F , f 6= g
s t min mid0 mid
and supp(f(w))⊂[min (w),min (w)), supp(g(w))⊂[min (w),min (w)). The
i i+1 j j+1
case that i = j occurs only in the following two possibilities. First if f = min ,
i
g =θ0 and secondif f =θ0 , g =θ0 , h,h′ ≤i. Let f =min , g =θ0 for h<i
i,h i,h′ i,h i i,h
and consider w = Tk−iw +w . Suppose that Z mixes s∪w with t∪w. Pick
0 1 1
w <z <w and consider v =Tk−iw +Tk−hz+w . Then φ (s∪v)=φ (s∪w)
0 1 0 1 s s
and φ (t∪v) 6= φ (t∪w). Therefore Z mixes s∪w with s∪v and t∪w. If Z
t t 1 1
mixes s∪v with t∪v, would imply that Z mixes t∪w with t∪v, but φ (t∪w)6=
1 t
φ (t∪v). As a result Z separates s∪v with t∪v. In the case that h = i then
t 1
v =Tk−iw +Tk−iz+w . Once more Z separates s∪v with t∪v.
0 1 1
Let now f = θ0 , g = θ0 . The assumption that f 6= g implies that h 6= h′.
i,h′ i,h
SupposeoncemorethatZ mixes s∪w witht∪w, forw =Tk−iw +w . Pickz so
1 0 1
10 DIMITRISVLITAS
that w <z <w and consider v =Tk−iw +Tk−hz+w , in the case that h<h′,
0 1 0 1
and v =Tk−iw +Tk−h′z+w , in the case that h′ < h. In the first case we have
0 1
that φ (s∪v) 6= φ (s∪w) and φ (t∪v) = φ (t∪w). In the second case we have
s s t t
thatφ (s∪v)=φ (s∪w)andφ (t∪v)6=φ (t∪w). AsaconsequenceZ separates
s s t t 1
s∪v with t∪v. Therefore given f ∈φ , g ∈φ , f 6=g, f,g ∈F ∪F ∪F
s t min mid0 mid
and w so that Z mixes s∪w with t∪w, there exists v that results from w by
1
addition, so that Z separatess∪v with t∪v. Let f,g ∈F ∪F ∪F and
1 max mid1 mid
f 6=g. By an identical argument we have that the above statement holds as well.
Toavoidunnecessarylength,fromnowonwards,wearegoingtoassumethatfor
every f ,g so that f ∈φ′ and g ∈φ′ it holds that it is not the case that both
n m n s m t
supp(f (w)),supp(g (w)) ⊂[min (w),min (w)) or supp(f (w)),supp(g (w)) ⊂
n m i i+1 n m
(max (w),max (w)], for the same i≤k.
i+1 i
Assumethatforn =minα\X oneofthefollowingholds. Firstcasesupp(f )⊂
0 n0
[min (w),min (w)) and for all j ∈β\Y, supp(g (w)) ⊂ [min (w),min (w)),
i0 i0+1 j i i+1
i < i or supp(g (w)) ⊂ (max (w),max (w)] for any i > i . Second case
0 j i+1 i 0
supp(f ) ⊂ (max (w),max (w)] and for all j ∈ β\Y, it holds that for i > i
n0 i0+1 i0 0
supp(g (w)) ⊂[min (w),min (w)), or supp(g (w)) ⊂(max (w),max (w)]. Let
j i i+1 j i+1 i
g be sothatm =minβ\Y andeithersupp(g (w))⊂[min (w),min (w))
m0 0 m0 im im+1
or supp(g (w)) ⊂ (max (w),max (w)]. Therefore with f we associate i
m0 im+1 im n0 0
and with g we associate i .
m0 m
For w ∈hZ i we introduce two more sets.
1
Bw0 ={Tiw :i<k, either ∃i′ ∈α\X, or j′ ∈β\Y :fi′(Tiw+w′)6=fi′(w′) or
gj′(Tiw+w′)6=gj′(w′),∀w′ ∈hZ1/wi},
and
Bw1 ={Tiw :06=i<k, either ∃i′ ∈α\X, or j′ ∈β\Y :fi′(w′+Tiw)6=fi′(w′) or
gj′(w′+Tiw)6=gj′(w′),∀w′ <w}.
We claim the following.
Claim 2. Let w ∈hZ i and assume that for all f ∈φ′ and g ∈φ′ it holds that
1 n s m t
f ,g ∈F ∪F . As a result A0 6=∅. There exists
n m min mid0 w
w¯ =Tjlzl+···+Tj0z0+w,
j < ··· < j, so that for every j,i ∈ F0, not necessarily distinct, Z separates
0 l w¯ 1
s∪Tiw¯+v with t∪Tjw¯+v′, for any v,v′ ∈hZ /w¯i.
1
Proof. Let w ∈hZ i with A0 6=∅ be givenand let j =maxF0. Assume that φ 6=
1 w w s
φ . Let (j,j) ∈ C0. Notice that supp(f (w)) < max (w). From above by i < k
t w n0 k 0
we denote the supp(f )(w)⊂[min (w),min (w)). We have assumed also that
n0 i0 i0+1
supp(g (w)) ⊂ [min (w),min (w)) and i ≥ i +1. Pick any z ∈ Z so that
m0 im im m 0 0 1
z0 <w and set w0 =Tk−i0−jz0+w, where Tk−i0−jz0 ∈FINi0+j. Notice that w0
is a sos and φ (Tjw +w′)6=φ (Tjw+w′) and φ (Tjw +w′)=φ (Tjw+w′).
s 0 s t 0 t