An Improvement to “A Note on Euclidean Ramsey Theory”
Discrete Comput Geom
Consider the finitedimensional Euclidean space Rn. If X, Y ⊆ Rn, then we write
James H. Schmerl 0
Rn −→ (X 0
0 J.H. Schmerl ( ) Department of Mathematics, University of Connecticut , Storrs, CT 062693009 , USA
We characterize those subsets Y ⊆ Rn such that for every infinite X ⊆ Rn, there is a red/blue coloring of Rn having no monochromatic red set similar to X and no monochramtic blue set similar to Y . if for any A ⊆ Rn, there is X ⊆ A similar to X or there is Y ⊆ Rn\A similar to Y . If Rn → (X, Y ) is false, then we write Rn→ (X, Y ). An affine subspace of Rn is a translate of a subspace of Rn. For example, a line is just a 1dimensional affine subspace. If Y ⊆ Rn, then let aff(Y ) be the affine subspace generated by Y . In general, Y  ≥ dim(aff(Y )) + 1. The purpose of this paper is to prove the following theorem, which improves the principal result of [7]. Theorem Suppose that 2 ≤ n < ω and Y ⊆ Rn. The following are equivalent: (1) For every infinite X ⊆ Rn, Rn→ (X, Y ). (2) Either aff(Y ) = Rn or Y  ≥ dim(aff(Y )) + 2.
Euclidean Ramsey theory

The implication (1) ⇒ (2) is Theorem E of [
7
]. That half of (2) ⇒ (1) that
assumes aff(Y ) = Rn is the principal result (Theorem D) of [
7
]. It suffices to prove the
other half of (2) ⇒ (1) in which it is assumed that Y  ≥ dim(aff(Y )) + 2. However,
the proof we give is a unified proof proving both parts of (2) ⇒ (1), using a
modification of the method of [
7
]. The proof in [
7
] of the implication (2) ⇒ (1) of the
Theorem in the case that aff(Y ) = Rn did not apply for n = 2. For just this case, there
was a separate and unpublished proof due to Komjáth. His proof was based on the
proof given by Baumgartner [
1
] of the special case of (2) ⇒ (1) of the Theorem in
which Y is an arithmetic progression of length 3 and X is an infinite arithmetic
progression. The proof given here includes the n = 2 case and, therefore, has both the
Baumgartner and Komjáth results as special cases.
For n = 1, there is also an equivalence as in the Theorem, but in this case, (2) is
simply Y  ≥ 3. The aforementioned proof of Baumgartner [
1
] can easily be modified
to a proof of this n = 1 case. However, this equivalence is also an immediate
consequence of the Theorem. (To be precise, although Baumgartner’s initial motivation in
[
1
] was to answer the question of Erdo˝s, which was just this n = 1 case, he ultimately
proved a stronger result: If V is an arbitrary vector space over the rationals Q, then
there is A ⊆ V such that A contains no infinite arithmetic progressions and V \A
contains no arithmetic progressions of length 3.)
Generally, the method of proof of the Theorem follows the lines of the proof in [
7
].
There, we utilized a countable coloring χ : Rn → ω with the property that there are
no distinct points a, b, c ∈ Rn that are the vertices of an isosceles triangle such that
χ (a) = χ (b) = χ (c). The existence of such a χ was first proved in [
6
]. Earlier, it had
been proved that if Y ⊆ Rn is such that Y  = 3, then there is a coloring χ : Rn → ω
such that whenever Y ⊆ Rn is similar to Y , then {χ (a) : a ∈ Y } ≥ 2. Here we will
make use of the following improvement (Corollary 2.4): if Y ⊆ Rn is a nonempty
finite set, then there is a coloring χ : Rn → ω such that whenever Y ⊆ Rn is similar
to Y , then {χ (a) : a ∈ Y } ≥ Y  − 1.
We begin with Sect. 1, which is pretty much a review of the basic definitions
from [
7
], culminating in the definition of an acceptable coloring. A couple of
fundamental properties of acceptable colorings of Rn are discussed in Sect. 2. In Sect. 3, we
formally define what similarities are, and then some of the connections between them
and acceptable colorings. Finally, in Sect. 4, we complete the proof of the Theorem.
1 Preliminaries
We always think of R as the realclosed ordered field R = (R, +, ·, 0, 1, <). If X,
Y ⊆ R, then X < Y iff x < y for all x ∈ X and y ∈ Y .
In this section, we consider a fixed, but arbitrary, countable, realclosed subfield
F ⊆ R and a fixed, but arbitrary, transcendence basis T ⊆ R for R over F, and we
arrange that T is a subset of the closed unit interval [
0, 1
]. We also consider a fixed n,
where 2 ≤ n < ω.
If D ⊆ R and R ⊆ Rk , then R is Ddefinable if it is definable by a firstorder
formula in the language of R using parameters from F ∪ D, and R is definable if it is
∅definable. A fundamental fact that will be repeatedly used is:
⎧ If U ⊆ Rk is a connected open set, {t0, t1, . . . , tk−1}< ⊆ T ,
⎨ and F : U → R is a definable analytic function such that
⎩ F (t0, t1, . . . , tk−1) = 0, then F is identically 0 on U .
If R ⊆ Rk is Rdefinable, then supp(R) is the unique smallest T ⊆ T such that
R is T definable. If a = (a0, a1, . . . , ak−1) ∈ Rk , then we write supp(a) instead of
supp({a}). We will mainly be interested in supp(a) for a ∈ Rn. When we write T =
{t0, t1, . . . , tm−1}<, it is to be understood that t0 < t1 < · · · < tm−1.
If I ⊆ R and k < ω, then I is an open dyadic interval with norm ν(I ) = k if
there is an integer i such that I is the open interval (i/2k, (i + 1)/2k). A subset
B ⊆ Rm is an mbox of norm ν(B) = k if there are open dyadic intervals I0 < I1 <
· · · < Im−1, each of norm k, such that B = I0 × I1 × · · · × Im−1. We let C be the
set of definable analytic functions f : B → Rn, where B is a box. If a ∈ Rn, then
f ∈ C is an acceptable color for a if supp(a) = {t0, t1, . . . , tm−1}<, B = dom(f ) is
an mbox such that t0, t1, . . . , tm−1 ∈ B, and f (t0, t1, . . . , tm−1) = a. If f ∈ C, then
let ν(f ) = ν(dom(f )). If A ⊆ Rn, then an acceptable coloring of A is a function
χ : A → C such that χ (a) is an acceptable color for every a ∈ A.
2 Acceptable Colorings
Throughout this section we assume that χ : Rn → C is an acceptable coloring.
Lemma 2.1 If distinct a, b, c ∈ Rn and r ∈ F are such that a − b = r b − c and
χ (b) = χ (c), then ν(χ (b)) < ν(χ (a)).
Proof For a contradiction, suppose that a − b = r b − c , f = χ (a), g = χ (b) =
χ (c), and ν(g) ≥ ν(f ). Let
supp(a) = {t0, t1, . . . , tm−1}< ,
supp(b) = {u0, u1, . . . , uk−1}< ,
supp(c) = {v0, v1, . . . , vk−1}< .
Let B1 = dom(f ) = I0 × I1 × · · · × Im−1 and B2 = dom(g) = J0 × J1 × · · · × Jk−1.
Choose this counterexample so as to minimize k + m (where we are allowing χ to
be replaced by another acceptable coloring). By taking the restriction of f to an
mbox B ⊆ B1 such that t0, t1, . . . , tm−1 ∈ B and ν(B) = ν(g), we can assume that
ν(f ) = ν(g).
Let F : B1 × B2 × B2 → R be such that
F (x, y, z) = f (x) − g(y) 2 − r2 g(y) − g(z) 2.
Clearly, F is a definable analytic function such that F (t, u, v) = 0.
Since b = c and χ (b) = χ (c), it must be that supp(b) = supp(c), so we can let
< k be such that u = v .
For each i < m such that Ii = J , replace ti ∈ Ii with the midpoint of Ii thereby
obtaining t from t . Also, replace each ui , vi such that = i < k with the midpoint
of Ji thereby obtaining u , v from u, v, respectively. Then, F (t , u , v ) = 0 by .
Letting a = f (t ), b = g(u ), and c = g(v ), we have a counterexample such that
 supp(a ) ≤  supp(b ) =  supp(c ) = 1. Thus by the minimality of m + k, we can
assume that k ≤ m = 1 and then u0 = v0.
iStufoplploowsestthhaattsuapp−(ag)(u=0)∅.2T−hern2 ga(−u0g)(−u0g) (u20−) r22 =g0(u.0T)h−usg, (gv(0u)0)2 ==a0,, ssoo bfryom
again, g is constantly a. But then supp(g) = ∅, which is a contradiction.
So suppose that k = 1. Then, t0 ∈ {u0, v0}, as otherwise we could replace t0 by the
midpoint of I0 and then we would have the previous case. Without loss of generality,
assume that t0 = u0. Thus, f (u0) − g(u0) 2 − r2 g(u0) − g(v0) 2 = 0, so from
it follows that f (u0) − g(u0) 2 − r2 g(u0) − g(u0) 2 = 0, so that a = f (u0) =
g(u0) = b, which is a contradiction.
The following corollary is Lemma 2.1 of [
7
] and the main result of [
6
].
Corollary 2.2 If distinct a, b, c ∈ Rn and r ∈ F are such that a − b = r b − c ,
then χ (a) = χ (b) or χ (b) = χ (c).
The next corollary of Lemma 2.1 will not be needed but is included here because
of its independent interest.
Corollary 2.3 Let a, b, c, d ∈ Rn be distinct points such that χ (a) = χ (b) and there
are r, s ∈ F such that a − b = r b − c and c − d is one of s d − a , s a − c ,
or s b − c . Then, χ (c) = χ (d).
Proof Let f = χ (a) = χ (b) and suppose that g = χ (c) = χ (d). Then Lemma 2.1
implies that ν(f ) < ν(g) and ν(g) < ν(f ).
A precursor to Corollary 2.2 is the result from [
5
] that if Y ⊆ Rn and Y  = 3,
then Rn has countable coloring ψ such that if Y ⊆ Rn is similar to Y , then {ψ (x) :
x ∈ Y } ≥ 2. This result had improved earlier results in [
2–5
]. The next corollary is a
further strengthening of this result.
Corollary 2.4 Let Y ⊆ Fn be nonempty and finite. If Y ⊆ Rn is similar to Y , then
{χ (x) : x ∈ Y } ≥ Y  − 1.
3 Affine Transformations
In this section we investigate the connection between similarities, and more generally
affine maps, and acceptable colorings.
Let e0, e1, . . . , en−1 ∈ Rn be the standard basis for Rn; thus, ei = (0, 0, . . . , 0, 1, 0,
. . . , 0) where the unique 1 has exactly i preceding 0’s. Also, let en = (0, 0, . . . , 0).
A function α : Rn → Rn is an affine map if whenever x = (x0, x1, . . . , xn−1) ∈ Rn,
then
α(x) = x0α(e0) + x1α(e1) + · · · + xn−1α(en−1) + α(en).
If α is a bijective affine map, then it is an affine transformation. If there is a positive
c ∈ R such that α(x) − α(y) = c x − y for every x, y ∈ Rn, then α is a similarity.
If Y, Y ⊆ Rn, then Y and Y are similar iff there is a similarity α such that Y = α(Y ).
Acceptable colorings were defined only for Rn, but they could easily have been
defined for any Rk . We will need to consider acceptable colorings of Rn(n+1). By
identifying Rn(n+1) with (Rn)n+1, we can consider acceptable colors for elements of
(Rn)n+1. We will refer to acceptable colors for elements of (Rn)n+1 as types. Each
type can be identified with a certain element of Cn+1; thus, the set of types is a subset
of Cn+1. If α is an affine map and τ is a type, then we say that τ is a type of α (or α
has type τ ) if τ is an acceptable color for α(e0), α(e1), . . . , α(en) ∈ (Rn)n+1. Thus,
it makes sense to refer to ν(τ ). Notice that a consequence of is that if τ is a type
of α and β, then α is an affine transformation iff β is, and α is a similarity iff β is.
Lemma 3.1 Let τ be a type and a ∈ Fn. Then there is f ∈ C such that ν(f ) = ν(τ )
and whenever α is an affine map that has type τ , then f is an acceptable color
for α(a).
Proof Let dom(τ ) = B = I0 × I1 × · · · × Im−1 and a = (a0, a1, . . . , an−1) ∈ Fn. Then
define F : (Rn)n+1 → Rn such that
F (x0, x1, . . . , xn) = a0x0 + a1x1 + · · · + an−1xn−1 + xn.
Hence, whenever α : Rn → Rn is an affine map, then
F α(e0), α(e1), . . . , α(en) = α(a).
Clearly, F is a definable analytic function.
Define G : B → Rn by
G(x0, x1, . . . , xm−1) = F τ (x0, x1, . . . , xm−1) ,
so that G is a definable analytic function.
Consider any affine map α that has type τ , and let supp(α) = {t0, t1, . . . , tm−1}<.
Then G(t0, t1, . . . , tm−1) = α(a), so supp(α(a)) ⊆ supp(α). Let J (α) = {j0, j1, . . . ,
js−1}< = {j < m : tj ∈ supp(γ (a))}, and then let B = Ij0 × Ij1 × · · · × Ijs−1 and let
g : B → Rn be defined by g(x0, x1, . . . , xs−1) = G(z0, z1, . . . , zm−1), where zji = xi
if i < s and zj is the midpoint of Ij if j ∈/ {j0, j1, . . . , js−1}<. Therefore, g is an
acceptable color for α(a). Thus, G(t0, t1, . . . , tm−1) = g(tj0 , tj1 , . . . , ts−1).
Now let β be another affine map having type τ , and let supp(β) = {u0, u1, . . . ,
um−1}<. Then β(a) = G(u0, u1, . . . , um−1). By , G(u0, u1, . . . , um−1) = g(uj0 ,
uj1 , . . . , us−1) so that β(a) = g(uj0 , uj1 , . . . , us−1). Thus, J (β) ⊆ J (α), so, by
symmetry, J (β) = J (α). Therefore, g is also an acceptable type for β(a).
If τ ∈ Cn+1 is a type, then, in the light of Lemma 3.1, we let τ : Fn → C be
the function such that whenever a ∈ Fn and α has type τ , then τ (a) is the unique
acceptable color f for α(a) such that ν(f ) = ν(τ ).
Lemma 3.2 Let τ ∈ Cn+1 be a type of an affine transformation, and let f ∈ C. Then,
there are only finitely many a ∈ Fn such that τ (a) = f .
Proof Let dom(τ ) = B = I0 × I1 × · · · × Im−1. As we saw in the proof of Lemma 3.1,
if a ∈ Fn and α is an affine transformation, then supp(α(a)) ⊆ supp(α).
We claim: If j0 < j1 < · · · < js−1 < m, then there is at most one a ∈ Fn for which
there is α having type τ such that f is an acceptable type for α(a), and supp(α(a)) =
{v0, v1, . . . , vs−1}<, where vi ∈ Iji for i < s.
To prove this claim, let both a, b have this property. Thus, there are α, β having
type τ such that f is an acceptable color for α(a), β(b), and
supp(α) = {t0, t1, . . . , tm−1}<,
supp(β) = {u0, u1, . . . , um−1}<,
supp(α(a)) = {tj0 , tj1 , . . . , tjs−1 }<,
supp(α(b)) = {uj0 , uj1 , . . . , ujs−1 }<.
Let a = (a0, a1, . . . , an−1) ∈ Fn, and define F : (Rn)n+1 → Rn such that
F (x0, x1, . . . , xn) = a0x0 + a1x1 + · · · + an−1xn−1 + xn.
Hence, whenever γ : Rn → Rn is an affine map (especially letting γ = α or γ = β),
then
F γ (e0), γ (e1), . . . , γ (en) = γ (a).
Clearly, F is a definable analytic function.
Define G : B → Rn by G(x0, x1, . . . , xm−1) = F (τ (x0, x1, . . . , xm−1)), so that G
is a definable analytic function. Then
G(t0, t1, . . . , tm−1) = α(a) = f (tj0 , tj1 , . . . , tjs−1 ).
Therefore,
β(a) = F β(e0), β(e1), . . . , β(en)
= F τ (u0, u1, . . . , um−1)
= G(u0, u1, . . . , um−1)
= f (uj0 , uj1 , . . . , ujs−1 ) (by )
= β(b).
Therefore, a = b, proving the claim.
We have just seen that each such a is determined by j0 < j1 < · · · < js−1 < m.
Therefore, there are at most ms such that a ∈ Fn.
4 Completing the Proof
In this final section, we complete the proof of the Theorem. Throughout this section,
we consider fixed X, Y ⊆ Rn, where X is infinite and either aff(Y ) = Rn or Y  ≥
dim(aff(Y )) + 2. Clearly, we can assume that X is countable and that Y is minimal
with the required property. We let Y  = m + 2, so m also is fixed throughout this
section. Thus, either m = n − 1 and aff(Y ) = Rn or else 2 ≤ m = dim(aff(Y )) ≤ n − 1.
We let F ⊆ R be a countable realclosed field such that X, Y ⊆ Fn and then
continue with the all the previous assumptions.
In both cases (namely, aff(Y ) = Rn and dim(aff(Y )) < n), there is a common
feature that Y has. Let Y = {y0, y1, . . . , ym, ym+1}. First, suppose that dim(aff(Y )) < n.
Then there are unique c0, c1, . . . , cm ∈ R such that c0 + c1 + · · · + cm = 1 and ym+1 =
c0y0 + c1y1 + · · · + cmym. Then 0 = ci ∈ F for each i ≤ m. Let F : (Rn)m+1 → Rn
be the definable analytic function such that
F (x0, x1, . . . , xm) = c0x0 + c1x1 + · · · + cmxm.
Then
α(ym+1) = F α(y0), α(y1), . . . , α(ym)
for every similarity (even any affine map) α.
Next, suppose that dim(aff(Y )) = n. Then there are two definable analytic
functions F, G : (Rn)m+1 → Rn such that
α(ym+1) ∈ F α(y0), α(y1), . . . , α(ym) , G α(y0), α(y1), . . . , α(ym)
for every similarity α. (See [7, Lemma 2.5].)
It was noted in the previous section that implies that if α, β are affine maps
having type σ , then α is a similarity iff β is. Here we note a generalization of this. An
affine map α is a similarity on Y if there is a positive c ∈ R such that α(x) − α(y) =
c x − y for every x, y ∈ Y . Then, if σ is a type of both α, β, then α is a similarity
on Y iff β is. Hence, we can say unambiguously that σ is a similarity type on Y .
Lemma 4.1 Let k < ω, ≤ m + 1, Y = {y0, y1, . . . , ym, ym+1}, and let f0, f1, . . . ,
f −1 ∈ C be such that ν(fi ) ≤ k. Then there are only finitely many f ∈ C for which
there is a type σ that is a similarity type on Y such that ν(σ ) = k, σ (yi ) ⊆ fi for
i < and σ (yi ) = f for < i ≤ m + 1 .
Proof Fix k < ω, ≤ m + 1, Y = {y0, y1, . . . , ym, ym+1}, and f0, f1, . . . , f −1 ∈ C.
Without loss of generality, suppose that ν(f0) = ν(f1) = · · · = ν(f −1) = k.
< m. By Corollary 2.4 there is no such type σ , for if α were an affine map of
type σ , then α would be a similarity on Y and f would be an acceptable color for
each of α(ym−1), α(ym) and α(ym+1).
= m. Again, there is no such type σ , for if α were an affine map of type σ , then
α would be a similarity on Y , f would be an acceptable color for each of α(ym) and
α(ym+1) and ν(f ) = k = ν(f0), contradicting Lemma 2.1.
= m + 1. Let σ be a type of α such that α is a similarity on Y , ν(σ ) = k,
and σ (yi ) = fi for i ≤ m. Let f = σ (ym+1). We know that there is a
definable analytic F : (Rn)m+1 → Rn such that F (α(y0), α(y1), . . . , α(ym)) = α(ym+1).
For i ≤ m + 1, let ti = (ti0, ti1, . . . , tiri−1) ∈ Rri be such that supp(α(yi )) =
{ti0, ti1, . . . , tiri−1}<. Let Bi = dom(fi ), and let G : B0 × B1 × · · · × Bm → Rn be
such that G(x0, x1, . . . , xm) = F (f (x0), f (x1) . . . , f (xm)). Thus, G is a definable
analytic function such that G(t0, t1, . . . , tm) = f (tm+1) = α(ym+1). Let dom(G) =
I0 × I1 × · · · × Ir−1, where r = r0 + r1 + · · · + rm. Each Ii is such that ν(Ii ) = k. For
each s < r , let ts = tm+1,j if tm+1,j ∈ Is , and let ts be the midpoint of Is otherwise.
(Since ν(f ) = k, this is well defined.) Then f (tm+1,0, tm+1,1, . . . tm+1,rm+1−1) =
G(t0, t1, . . . , tr1 ). Thus, there are only finitely many possible such f .
We now make a crucial definition.
Definition 4.2 We say that ρ is good if ρ : C0 → {0, 1} is a function such that the
following hold:
• C0 is a finite subset of C.
• Whenever f, g ∈ C are distinct and f ⊆ g ∈ C0, then f ∈/ C0.
• Whenever α : Rn → Rn is a similarity and ϕ : α(Y ) → C0 is acceptable, then there
is y ∈ Y such that ρ(ϕ(α(y))) = 0.
Lemma 4.3 Suppose that ρ is good and that τ ∈ Cn+1 is the type of a similarity
such that ν(τ ) ≥ ν(f ) for all f ∈ dom(ρ). Then there exists x ∈ X such that ρ =
ρ ∪ { τ (x), 1 } is good.
Proof Let C0 = dom(ρ), which is finite since ρ is good. Let C1 be the set of those
f ∈ C such that ν(f ) ≤ ν(τ ) and for some g ∈ C0, there is h ∈ C such that f ∪ g ⊆ h.
Clearly, C0 ⊆ C1, and C1 is finite. Let X1 = {a ∈ X : τ (a) ∈ C1}, which is finite by
Lemma 3.2.
Let C2 be the set of all f ∈ C for which there is a type σ of a similarity such that
ν(σ ) = ν(τ ) and σ (y) ∈ C1 ∪ {f } whenever y ∈ Y . Since ρ is good, Lemma 4.1
implies that C2 is finite. Let X2 = {a ∈ X : τ (a) ∈ C2}, which is finite by Lemma 3.2.
Since X is infinite and X1, X2 ⊆ X are finite, we pick some x ∈ X\(X1 ∪ X2). Let
f = τ (x), ρ = ρ ∪ { f, 1 } and C0 = C0 ∪ {f }.
Clearly, f ∈/ C0 = dom(ρ), so ρ : C0 → {0, 1} is a welldefined function. We now
prove that ρ is good.
For a contradiction, suppose that ρ is not good. Obviously, C0 is finite. Also, it is
clear that neither f ⊆ g nor g ⊆ f for g ∈ C0. Thus, there is α : Rn → Rn that is a
similarity on Y (or even a similarity) and there is an acceptable ϕ : α(Y ) → C0 such
that ρ (ϕ (α(y))) = 1 for all y ∈ Y .
Now, if α were such that it had a type σ such that ν(σ ) = ν(τ ), then we would
arrive at a contradiction as follows. We have that ϕ (α(y)) = σ (y) for all y ∈ Y .
But then f ∈ C2 and x ∈ X2, contradicting the choice of f .
We will next show that, even though α may not have such a type σ , another choice
of an affine map will.
Let fj = θ (ej ) for j ≤ n. Let I0 < I1 < · · · < Ir−1 be those intervals occurring
as factors of f0, f1, . . . , fn, and let B = I0 × I1 × · · · × Ir−1. Let (t0, t1, . . . , tr−1) ∈
B ∩ T r , and then let aj = (θ (aj ))(ti0 , ti1 , . . . , tikj −1 ), where the subsequence 0 ≤
i0 < i1 < · · · < ikj −1 is the unique one such that (ti0 , ti1 , . . . , tikj −1 ) ∈ dom(fj ). Now
let β : Rn → Rn be the affine map such that β(ej ) = aj for j ≤ n. By , either β
is a similarity on Y or else β(y0) = β(y1) = · · · = β(ym+1). However, if β(y0) =
β(y1) = · · · = β(ym+1), then implies that f0 = f1 = · · · = fn = f , a contradiction
to Corollary 2.2. Thus, β is a similarity on Y such that β(y) = α(y) for all y ∈ Y , and
β has a type σ such that ν(β) = ν(τ ).
We now complete the proof of the Theorem. Let g0, g1, g2, . . . be a list of all the
elements of C, and let τ0, τ1, τ2, . . . be a list of all types of similarities. We recursively
construct an increasing sequence ρ0 ⊆ ρ1 ⊆ ρ2 ⊆ · · · of good functions. Let ρ0 = ∅,
which clearly is good.
Stage 2k + 1: Let 2k < ω be the least such that ν(gk) ≤ and also ν(g) ≤
whenever g ∈ dom(ρ2k). Let D be the set of those f ⊆ gk such that ν(f ) = and for
no g ∈ dom(ρ2k) is f ⊆ g. Then let ρ2k+1 = ρ2k ∪ { f, 0 : f ∈ D}. Clearly, ρ2k+1 is
good.
Stage 2k + 2: Let 2k+1 < ω be the least such that ν(τk) ≤ and also ν(g) ≤
whenever g ∈ dom(ρ2k+1). Apply Lemma 4.3 successively several times, once for
each τ ⊆ τk such that ν(τ ) = . This results in a good ρ2k+2 ⊇ ρ2k+1 such that
whenever τ ⊆ τk and ν(τ ) = , then there is x ∈ X such that τ (x) ∈ dom(ρ2k+2) and
ρ2k+2( τ (x)) = 1.
Let ρ = k<ω ρk . Then ρ : C0 → {0, 1} for some C0 ⊆ C. The odd stages of the
construction guarantee that for each x ∈ Rn, there is f ∈ C0 that is acceptable for x.
Since each ρk is good, this f is unique. Thus, there is a unique acceptable coloring
χ : Rn → C0. Now define A = {x ∈ Rn : ρ(χ (x)) = 0}. We show that this A verifies
Rn → (X, Y ). We must prove: (1) there is no X ⊆ A that is similar to X; and (2)
there is no Y ⊆ Rn\A similar to Y . Let α : Rn → Rn be a similarity.
(1) Let X = α(X). Let τk be a type of α, and notice that 2k+1 ≥ ν(τk). Let τ ⊆ τk
be such that ν(τ ) = 2k+1 and τ is a type of α. Then, there is x ∈ X such that
τ (x) ∈ dom(ρ2k+2) and ρ2k+2( τ (x)) = 1. Then α(x) ∈ X and ρ(α(x)) =
ρ2k+2(α(x)) = ρ2k+2( τ (x)) = 1, so α(x) ∈/ A.
(2) Let Y = α(Y ) and ϕ = χ Y . Then, for large enough k < ω, ϕ(A ) ⊆ dom(ρk).
Since ρk , is good, there is y ∈ Y such that ρ(χ (y )) = ρk(ϕ(y )) = 0. Therefore,
y ∈ A.
This completes the proof of the Theorem.
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