New bounds on the strength of some restrictions of Hindman's Theorem

0.22 MB·

by  Lorenzo Carlucci

New bounds on the strength of some restrictions
of Hindman’s Theorem
Lorenzo Carlucci1 [email protected],Leszek Aleksander
7 Kołodziejczyk2 [email protected],Francesco Lepore1 [email protected],
1 and KonradZdanowski3 [email protected]
0
2
1 Department of Computer Science, University of Rome I,

n 2 Instituteof Mathematics, Universityof Warsaw,
a 3 Faculty of Mathematics and NaturalSciences, Cardinal Stefan Wyszyński
J
University in Warsaw

1
2

] Abstract. We prove upper and lower bounds on the effective content
O and logical strength for a variety of natural restrictions of Hindman’s
L Finite Sums Theorem. For example, we show that Hindman’s Theorem
. for sums of length at most 2 and 4 colors implies ACA0. An emerging
h
leitmotiv is that the known lower bounds for Hindman’s Theorem and
t
a for its restriction to sums of at most 2 elements are already valid for
m a number of restricted versions which have simple proofs and better
computability-andproof-theoreticupperboundsthantheknownupper
[
bound for the full version of the theorem. We highlight the role of a

1 sparsity-like condition on the solution set, which we call apartness.
v
5
9 1 Introduction and Motivation
0
6
The Finite Sums Theorem by Neil Hindman [14] (henceforth denoted HT) is
0
. a celebrated result in Ramsey Theory stating that for every finite coloring of
1
the positive integers there exists an infinite set such that all the finite non-
0
7 empty sums of distinct elements from it have the same color. Thirty years ago
1 Blass,HirstandSimpsonprovedin[2]thatallcomputableinstancesofHThave
: some solutions computable in ∅(ω+1) and that for some computable instances
v
i of HT all solutions compute ∅′. In terms of Reverse Mathematics, they showed
X that ACA+ ⊢ HT and that RCA ⊢ HT → ACA . Both bounds hold for the
0 0 0
r particular case of colorings in two colors. Closing the gap between the upper
a
andlowerboundis one ofthe majoropenproblemsinComputable andReverse
Mathematics (see, e.g., [19]).
Blass advocated the study of restrictions of Hindman’s Theorem in which a
bound is put on the length (i.e., number of distinct terms) of sums for which
monochromaticity is guaranteed [1], conjecturing that the complexity of Hind-
man’s Theorem grows as a function of the length of sums. Recently Dzhafarov,
Jockusch,SolomonandWestrick showedin [12]thatthe known ∅′ (ACA ) lower
0
bound on Hindman’s Theorem holds for the restriction to sums of at most 3
terms (with no repetitions, as is the case throughout the paper), and 3 col-
ors (henceforth denoted by HT≤3). They also established that the restrictionto
3
sums of at most 2 terms, and 2 colors (denoted HT≤2), is unprovable in RCA
2 0
and implies SRT2 (the Stable Ramsey’s Theorem for pairs and 2 colors) over
2
RCA +BΣ0.This promptedthe firstauthor (see [5,3]) to look into direct com-
0 2
binatorialreductionsyielding, e.g.,a directimplication fromHT≤2 to IPT2 (the
5 2
Increasing Polarized Ramsey’s Theorem for pairs of Dzhafarov and Hirst [11])
over RCA . Note that IPT2 is strictly stronger than SRT2 (see infra for details).
0 2 2
It should be stressed that no upper bound other than the ∅(ω+1) (ACA+)
0
upper bound on the full Finite Sums Theorem is known to hold for the restric-
tions of the theorem to sums of length (i.e., number of terms) ≤ 2 or ≤ 3. It
is indeed a long-standingopen question in Combinatoricswhether the latter re-
strictions admit a proof that does not establish the full Finite Sums Theorem
(see, e.g., [15], Question 12). On the other hand, Hirst investigated in [17] an
apparentlyslightvariantoftheFiniteSumsTheoremandproveditequivalentto
BΣ0. This prompted the first author to investigate versions of HT for which an
2
upperboundbetterthan∅(ω+1) (ACA+)couldbe established,while retainingas
0
strongalowerboundas possible.In[4](resp.[3])suchrestrictionswereisolated
and provedto attain the known lower bounds for HT (resp. HT≤2), while being
2
provablefromACA (resp.RT2).Alltheseprincipleshaveabuilt-insparsity-like
0 2
condition on the solution set. This condition, called apartness in [4], is crucial
yet was not given a name in earlier work ([14,12]).
We present new results along these lines of research. In Section 3 we prove
an ACA lower bound for HT≤2, and an equivalence with ACA for Hindman’s
0 4 0
theorem restricted to sums of exactly 3 terms (HT=3), with an apartness con-
2
dition, and for some principles from [4]. In Section 4 we establish combinatorial
implications from other restrictions of Hindman’s Theorem to the Increasing
Polarized Ramsey’s Theorem for Pairs. In particular we show that the latter
principle is implied by (and indeed strongly computably reducible to) HT=2
2
with an apartness condition.
2 Restricted Hindman and the Apartness Condition
Letus fixsomenotation.Fortechnicalconvenienceandto avoidtrivialcaseswe
willdealwithcoloringsofthe positiveintegers.We use Nto denotethe positive
integers. If a ∈ N and B is a set we denote by FS≤a(B) (resp. FS=a(B)) the
set of non-empty sums of at most(resp. exactly)a-many distinct elements from
B. More generally, if A and B are sets we denote by FSA(B) the set of all
sums of j-many distinct terms from B, for all j ∈ A. By FS(B) we denote
FSN(B). If X is the set {x ,x ,x ,...} we write {x ,x ,x ,...} to indicate
1 2 3 1 2 3 <
that x <x <x <.... Let us recall the statement of Hindman’s Finite Sums
1 2 3
Theorem [14].
Definition 1 (Hindman’s Finite Sums Theorem). Let k ≥ 2. HT is the
k
following assertion: For every coloring f : N → k there exists an infinite set
H ⊆N such that FS(H) is monochromatic for f. HT denotes ∀kHT .
k
It was proved in [2] that ACA+ ⊢ HT and that RCA ⊢ HT → ACA . We
0 0 2 0
define below two restrictions of Hindman’s Theorem that will feature promi-
nently in the present paper. We then discuss a sparsity-like condition that will
be central to our results.
2.1 Hindman’s Theorem with bounded-length sums
We define two natural types of restrictions of Hindman’s Theorem based on
bounding the length of sums for which homogeneity is guaranteed.
Definition 2 (Hindman’s Theorem with bounded-length sums). Fix
n,k ≥1.
1. HT≤n is the following principle: For every coloring f : N→k there exists
k
an infinite set H ⊆N such that FS≤n(H) is monochromatic for f.
2. HT=n) is the following principle: For every coloring f : N→k there exists
k
an infinite set H ⊆N such that FS=n(H) is monochromatic for f.
The principles HT≤n were discussed in [1] (albeit phrased in terms of finite
k
unions instead of sums) and first studied from the perspective of Computable
and Reverse Mathematics in [12], where the principles HT=n were also defined.
k
The principle HT≤2 is the topic of a long-standing open question in Com-
2
binatorics: Question 12 of [15] asks whether there exists a proof of HT≤2 that
2
does not also provethe full Finite Sums Theorem. On the other hand, the prin-
ciple HT=2 easily follows from Ramsey’s Theorem for pairs: given an instance
2
f :N→2ofHT=2,defineg :[N]2 →2bysettingg(x,y):=f(x+y).Dzhafarov,
2
Jockusch,SolomonandWestrickrecentlyprovedin[12]thatHT≤3 impliesACA
3 0
over RCA and that HT≤2 is unprovable in RCA . They also proved that HT≤2
0 2 0 2
implies SRT2 (the Stable Ramsey’s Theorem for pairs) over RCA +BΣ0. The
2 0 2
first author proved that HT≤2 implies IPT2 (the Increasing Polarized Ramsey’s
5 2
Theorem for pairs) over RCA (see [5]).
0
2.2 The Apartness Condition
Wediscussapropertyofthesolutionset–whichwecalltheapartnesscondition
– that is crucial in Hindman’s original proof and in the proofs of the ∅′ (ACA )
0
lower bounds in [2,12,4]. We use the following notation: Fix a base t ≥ 2. For
n∈Nwedenotebyλ (n)theleastexponentofnwritteninbaset,byµ (n)the
t t
largest exponent of n written in base t, and by i (n) the coefficient of the least
t
termofn written inbaset. We will dropthe subscriptwhen clearfromcontext.
Definition 3 (Apartness Condition). Fix t ≥ 2. We say that a set X ⊆ N
satisfies the t-apartness condition (or is t-apart) if for all x,x′ ∈ X, if x < x′
then µ (x)<λ (x′).
t t
Ourresultsareintermsof2-apartnessexceptinonecase(Lemma1below)where
we have to use 3-apartness for technical reasons. For a Hindman-type principle
P, let “P with t-apartness” denote the corresponding version in which the solu-
tion set is requiredto satisfy the t-apartness condition. Note that the apartness
conditionisinheritedbysubsets.InHindman’soriginalproof2-apartnesscanbe
ensured(Lemma2.2in[14])byasimplecountingargument(Lemma2.2in[13]),
under the assumption that we have a solution to the Finite Sums Theorem, i.e.
an infinite H such that FS(H) is monochromatic. In our terminology, we have
that, for each k ∈N, HT is equivalent to HT with 2-apartness, over RCA .
k k 0
As will be observed below, it is significantly easier to prove lower bounds
on P with t-apartness than on P in all the cases we consider. Moreover, for
all restrictions of Hindman’s Theorem for which a proof is available that does
not also establish the full theorem, the t-apartness condition (for t > 1) can
be guaranteed by construction (see, e.g., [3,4]). This is the case, e.g., for the
principle HT=2: the proof from Ramsey’s Theorem for pairs sketched above
2
yields t-apartness for any t > 1 simply by applying Ramsey’s Theorem relative
toaninfinitet-apartset.Insomecasestheapartnessconditioncanbeensuredat
thecostofincreasingthenumberofcolors.ThisisthecaseofHT≤nasillustrated
k
by the next lemma. The idea of the proof is from the first part of the proof of
Theorem 3.1 in [12], with some needed adjustments.
Lemma 1 (RCA ). For all n ≥ 2, for all d ≥ 1, HT≤n implies HT≤n with
0 2d d
3-apartness.
Proof. Weworkinbase3.Letf :N→dbegiven.Defineg :N→2dasfollows.
f(n) if i(n)=1,
g(n):=
(d+f(n) if i(n)=2.
Let H be an infinite set such that FS≤n(H) is homogeneous for g of color k.
For h,h′ ∈ FS≤n(H) we have i(h) = i(h′). Then we claim that for each k ≥ 0
there is at most one h ∈ H such that λ(h) = k. Suppose otherwise, by way of
contradiction, as witnessed by h,h′ ∈ H. Then i(h) = i(h′) and λ(h) = λ(h′).
Therefore i(h+h′)6=i(h), but h+h′ ∈FS≤n(H). Contradiction.Therefore we
can computably obtain a 3-apart infinite subset of H. ⊓⊔
3 Restricted Hindman and Arithmetical Comprehension
We prove a new ACA lower bound and a new ACA equivalence result for
0 0
restrictions of Hindman’s Theorem. The lower bound proof is in the spirit of
Blass-Hirst-Simpson’s proof that Hindman’s Theorem implies ACA – on which
0
the proof of Theorem 3.1 of [12] is also based – with extra care to work with
sums of length at most two. The upper bound proof is in the spirit of [4].
3.1 HT≤42 implies ACA0
We show that HT≤2 implies ACA over RCA . This is to be compared with
4 0 0
Corollary2.3andCorollary3.4of[12],showing,respectively,thatRCA 0HT≤2
0 2
and that RCA ⊢ HT≤3 → ACA . Blass, towards the end of [1], states without
0 3 0
giving details that inspection of the proof of the ∅′ lower bound for HT in [2]
showsthattheseboundsaretruefortherestrictionoftheFiniteUnionsTheorem
tounionsofatmosttwosets.4 NotethattheFiniteUnionsTheoremhasabuilt-
in apartnesscondition. Blass indicates in Remark 12 of [1] that things might be
differentforrestrictionsoftheFinite Sums Theorem,asthoseconsideredinthis
paper. Also note that the proof of Theorem 3.1 in [12], which stays relatively
close to the argument in [2], requires sums of length 3.
Proposition 1 (RCA ). For any fixed t ≥ 2, HT≤2 with t-apartness implies
0 2
ACA .
0
Proof. Wewritetheprooffort=2.AssumeHT≤2with2-apartnessandconsider
2
an injective function f: N→N. We have to prove that the range of f exists.
For a number n, written as 2n0 +···+2nr in base 2 notation, we call j ∈
{0,...,s}importantin nifsomevalueoff↾[n ,n )isbelown .Heren =0.
j−1 j 0 −1
The coloring c: N→2 is defined by
c(n):=card{i:i is important in n}mod2.
ByHT≤2 with2-apartness,there existsaninfinite setH ⊆N suchthatH is
2
2-apart and FS≤2(H) is monochromatic w.r.t. c. We claim that for each n∈H
and each x < λ(n), x ∈ rg(f) if and only if x ∈ rg(f↾µ(n)). This will give us
a computable definition of rg(f): given x, find the smallest n ∈ H such that
x<λ(n) and check whether x is in rg(f↾µ(n)).
It remains to prove the claim. In order to do this, consider n ∈ H and
assumethatthereissomeelementbelown =λ(n)inrg(f)\rg(f↾µ(n)).Bythe
0
consequenceof Σ0-induction knownas strong Σ0-collection, there is a number ℓ
1 1
suchthat for anyx<λ(n), x∈rg(f) if andonly if x∈rg(f↾ℓ).By 2-apartness,
there is m∈H with λ(m)≥ℓ>µ(n). Write n+m in base 2 notation,
n+m=2n0 +···+2nr +2nr+1 +···+2ns,
where n = λ(n) = λ(n+m), n = µ(n), and n = λ(m). Clearly, j ≤ s
0 r r+1
is important in n + m if and only if either j ≤ r and j is important in n
or j = r +1; hence, c(n) 6= c(n+m). This contradicts the assumption that
FS≤2(H) is monochromatic, thus proving the claim. ⊓⊔
4 TheFiniteUnionsTheoremstatesthateverycoloringofthefinitenon-emptysetsof
N admits an infiniteand pairwise unmeshed set H of finite non-emptysets H such
that every finite non-empty sum of elements of H is of the same color. Two finite
non-emptysubsetsx,yofNareunmeshedifeithermaxx<minyormaxy<minx.
Note that Hindman’s Theorem is equivalent to the Finite Unions Theorem only if
thepairwise unmeshedcondition is present.
Theorem 1 (RCA ). HT≤2 implies ACA .
0 4 0
Proof. By Proposition 1 and Lemma 1. ⊓⊔
3.2 Restrictions of HT equivalent to ACA0
We prove that some restrictions of HT, including HT=3 with 2-apartness, are
2
equivalent to ACA . The first examples of this kind were given in [4], where
0
a family of natural restrictions of Hindman’s Theorem was isolated such that
each of its members admits a simple combinatorialproof, yet each member of a
non-trivial sub-family implies ACA .
0
The weakest principle proved in [4] to be equivalent to ACA is the follow-
0
ing, called the Hindman-Brauer Theorem (with 2-apartness): Whenever N is
2-colored there is an infinite and 2-apart set H ⊆ N and there exist positive
integers a,b such that FS{a,b,a+b,a+2b}(H) is monochromatic.
We improve on [4] by showing that some apparently weak restrictions of
Hindman’s Theorem provable from Ramsey’s Theorem are equivalent to ACA .
0
We first show that the same holds for the following apparently weaker prin-
ciple.
Definition 4. HT∃{a<b} isthefollowingprinciple: Foreverycoloringf :N→2
2
there is an infinite set H ⊆N and positive integers a<b such that FS{a,b}(H)
is monochromatic.
Theorem 2. HT∃{a<b} with 2-apartness is equivalent to ACA over RCA .
2 0 0
Proof. We first prove the upper bound. Given c : N → 2 let g : [N]3 → 8 be
defined as follows:
g(x ,x ,x ):=hc(x ),c(x +x ),c(x +x +x )i.
1 2 3 1 1 2 1 2 3
Fix an infinite and 2-apart set H ⊆ N. By RT3 relativized to H we get an
0 8 0
infinite (and 2-apart) set H ⊆ H monochromatic for g. Let the color be σ =
0
(c ,c ,c ),abinarysequenceoflength3.Then,foreachi∈{1,2,3},g restricted
1 2 3
to FS=i(H) is monochromatic of color c . Obviously for some 3 ≥ b > a > 0 it
i
must be that c =c . Then FS{a,b}(H) is monochromatic of color c .
a b a
ThelowerboundisprovedbyaminoradaptationoftheproofofProposition
1. As the n in that proof take an a-term sum. Then take a (b−a)-term sum as
the m. ⊓⊔
The same proof yields that the following Hindman-Schur Theorem with 2-
apartness from [4] implies ACA : Whenever N is 2-colored there is an infinite
0
2-apart set H and there exist positive integers a,b such that FS{a,b,a+b}(H)
is monochromatic. Indeed, the latter principle implies HT∃{a<b}. Provability in
ACA is shown in [4].
0
WenextadapttheproofofProposition1toshowthatHT=3with2-apartness
2
impliesACA .SinceHT=3 with2-apartnessisalsoeasilydeduciblefromRT3,we
0 2 2
obtain an equivalence. Note that no lower bounds on HT=3 without apartness
2
are known.
Theorem 3 (RCA ). HT=3 with 2-apartness is equivalent to ACA .
0 2 0
Proof. Wefirstprovethelowerbound.WeworkinRCA +HT=3with2-apartness
0 2
and consider an injective function f: N→N. We have to prove that the range
of f exists. The relation j is important in n and the computable c : N → 2 are
defined as in the proof of Proposition 1.
By HT=3 with 2-apartness, there exists an infinite set H such that H is
2
2-apart and FS=3(H) is monochromatic w.r.t. c. Let r < 2 be the color of
FS=3(H) under c. We describe a method of deciding the range of f with H
given as an oracle.
Claim. For each n,k ∈H. If n<k and c(n+k)=r then for each x<λ(n),
x∈rg(f) ⇐⇒ x∈rg(f↾µ(k)).
Proof. Letn,k∈H besuchthatn<kandc(n+k)=r.BystrongΣ0collection,
1
let ℓ be such that for all x<λ(n),
x∈rg(f) ⇐⇒ x∈rg(f↾ℓ)
and take m ∈ H such that λ(m) > ℓ. Now, if for some x < λ(n), x ∈ rg(f)\
rg(f↾µ(k)), then the number of important digits in n+k+m is greater by one
thanthenumberofimportantdigitsinn+k.Then,c(n+k+m)=1−c(n+k)=
1−r which contradicts the fact that r is the color of FS=3(H).
Claim. For each n∈H there exists k ∈H such that n<k and c(n+k)=r.
Proof. Let us fix n and, again, let ℓ be such that for all x<λ(n),
x∈rg(f) ⇐⇒ x∈rg(f↾ℓ).
Then, we define k∈H as any elementsuchthat λ(k)>ℓ.Now,for any m∈H,
if k <m, then c(n+k)=c(n+k+m)=r.
We now describe an algorithm for deciding rg(f). Given x, find n∈H such
that x < λ(n). Then, find k ∈ H such that n < k and c(n + k) = r. By
Lemma 3.2 this part of computation ends successfully. Finally, check whether
x∈rg(f↾µ(k)). By Lemma 3.2 this is equivalent to x∈rg(f).
The upper bound follows by observing that, for n,k ≥ 1 and t ≥2, RCA ⊢
0
RTn → HT=n with t-apartness. Given c : N → k define f : [N]n → k by
k k
f(x ,...,x ) := c(x +···+x ). Let X be an infinite t-apart set. Apply RTn
1 n 1 n k
to f relative to X. ⊓⊔
Let us observe that the above argument works in the case of HT=a with
2
apartness, for any fixed a ≥3. In the proof above take a sum of a−2 elements
inplace ofn.Indeed, a inHT=a couldalsobe nonstandard.This leadsus to the
2
following definition and corollary.
Definition 5. Let HT∃{a≥3} be the following principle: For every coloring f :
2
N→2 there exists an infinite set H ⊆N and there exists a number a≥3 such
that FS{a}(H) is monochromatic for f.
Theorem 4 (RCA ). HT∃{a≥3} with 2-apartness is equivalent to ACA .
0 2 0
3.3 Hindman’s Theorem for Exactly Large Sums
We consider a restriction of Hindman’s Theorem to exactly large sums. A finite
set S ⊆ N is exactly large, or !ω-large, if |S| = min(S)+1. Large sets play a
prominent role in the study of unprovability results for first-order theories of
arithmetic. We denote by [X]!ω the set of exactly large subsets of X and by
FS!ω(X)thesetofpositiveintegersthatcanbeobtainedassumsoftermsofan
exactlylargesubsetofX.Wecallsumsofthistypeexactly large sums(fromX).
Ramsey’s Theorem for exactly large sums (RT!ω) asserts that every 2-coloring
2
f of the exactly large subsets of an infinite set X ⊆ N admits an infinite set
H ⊆X such that f is constant on [H]!ω. It was studied in [6] and there proved
equivalent to ACA+. We introduce an analogue for Hindman’s Theorem.
0
Definition 6 (Hindman’s Theorem for Large Sums). HT!ω denotes the
2
following principle: For every coloring c : N → 2 there exists an infinite set
H ⊆N such that FS!ω(H) is monochromatic under c.
HT!ω (witht-apartness,foranyt>1)followseasilyfromRT!ω.Givenc:N→2
2 2
just set f(S) := c( S), for S an exactly large set (to get t-apartness, restrict
f to an infinite t-apart set). By results from [6] this reduction yields an upper
bound of ∅(ω) on HPT!ω.
2
The argument of Theorem 3 can be easily adapted to show that HT!ω with
2
2-apartness implies ACA . In the proof take, instead of n, an almost exactly
0
large sum n +n +···+n of elements of H. The argument then proceeds
0 1 n0−2
unchanged.
Proposition 2 (RCA ). HT!ω with 2-apartness implies ACA .
0 2 0
Some further results on HT!ω will be proved in Section 4.
2
4 Restricted Hindman and Polarized Ramsey
InthissectionweestablishnewlowerboundsforrestrictedversionsofHindman’s
Theorem by reduction to the Increasing Polarized Ramsey’s Theorem for pairs
[11]. In particular we obtain unprovability in WKL . Most of the restrictions of
0
HT considered in this section do not imply ACA and are therefore provably
0
weakerthanHT.AllproofsofanimplicationtoIPT2 inthepresentsectionyield
2
a strong computable reduction in the sense of [10], not just an implication. P is
strongly computably reducible to Q, written P ≤ Q, if every instance X of P
sc
computes an instance X∗ of Q, such that if Y∗ is any solutionto X∗ then there
is a solution Y to X computable from Y∗.
Definition 7 (Increasing Polarized Ramsey’s Theorem). Fix n,k ≥ 1.
IPTn is the following principle: For every f : [N]n → k there exists a sequence
k
(H ,...,H ) of infinitesets such that all increasing tuples (x ,...,x ) ∈ H ×
1 n 1 n 1
··· × H have the same color under f. The sequence (H ,...,H ) is called
n 1 n
increasing polarized homogeneous (or increasing p-homogeneous) for f.
Note that IPT2 is strictly stronger than SRT2. On the one hand, RCA ⊢
2 2 0
IPT2 →D2 by Proposition 3.5 of [11], and RCA ⊢D2 →SRT2 by Theorem 1.4
2 2 0 2 2
of[8].5 Ontheotherhand,RCA +SRT2 0IPT2:Theorem2.2in[9]showedthat
0 2 2
there is a non-standardmodel of SRT2+BΣ0 having only low sets in the sense
2 2
of the model. Lemma 2.5 in [11] can be formalized in RCA and shows that no
0
model of IPT2 can contain only ∆0 sets.6
2 2
4.1 HT=2 with 2-apartness implies IPT2
2 2
We show that HT=2 with 2-apartness implies IPT2 by a combinatorial proof
2 2
establishing a strong computable reduction. This should be contrastedwith the
fact that no lower bounds on HT=2 without apartness are known.
2
Theorem 5 (RCA ). HT=2 with 2-apartness implies IPT2.
0 2 2
Proof. Let f :[N]2 →2 be given. Define g :N→2 as follows.
0 if n=2m,
g(n):=
(f(λ(n),µ(n)) if n6=2m.
Note that g is well-defined since λ(n) < µ(n) if n is not a power of 2. Let
H ={h ,h ,...} witness HT=2 with 2-apartnessfor g.Letthe colorbe k <2.
1 2 < 2
Let
H :={λ(h ) : i∈N}, H :={µ(h ) : i∈N}.
1 2i−1 2 2i
We claim that (H ,H ) is increasing p-homogeneous for f.
1 2
First observe that we have
H ={λ(h ),λ(h ),λ(h ),...} , H ={µ(h ),µ(h ),µ(h ),...} .
1 1 3 5 < 2 2 4 6 <
This is so because λ(h ) ≤ µ(h ) < λ(h ) ≤ µ(h ) < ... by the 2-apartness
1 1 2 2
condition. We claim that f(x ,x ) = k for every increasing pair (x ,x ) ∈
1 2 1 2
H ×H . Note that (x ,x ) = (λ(h ),µ(h )) for some i < j (the case i = j is
1 2 1 2 i j
impossible by construction of H and H ). We have
1 2
k =g(h +h )=f(λ(h +h ),µ(h +h ))=f(λ(h ),µ(h ))=f(x ,x ),
i j i j i j i j 1 2
since FS=2(H) is monochromatic for g with color k. This shows that (H ,H )
1 2
is increasing p-homogeneous of color k for f. ⊓⊔
With minor adjustments the proof of Theorem 5 yields that IPT2 ≤ HT≤2
2 sc 4
(for a self-contained proof see [5]).
5 Notethatthediagramin[11]doesnottakeintoaccountthelatterresult.D2,defined
2
in [7], is the following assertion: For every 0,1-valued function f(x,s) for which a
lims→∞f(x,s)existsforeach x,thereisaninfiniteset H andak<2suchthatfor
all h∈H we havelims→∞f(h,s)=k.
6 Wethank LudovicPatey for pointing out tous theresults implying strictness.
4.2 IPT2 and the Increasing Polarized Hindman’s Theorem
2
We define a (increasing) polarized version of Hindman’s Theorem. We prove
that its version for pairs and 2 colors with an appropriately defined notion of
2-apartness is equivalent to IPT2.
2
Definition 8 ((Increasing) Polarized Hindman’s Theorem). Fix n ≥ 1.
PHTn (resp. IPHTn) is the following principle: For every f :N→2 there exists
2 2
a sequence (H ,...,H ) of infinite sets such that for some color k < 2, for all
1 n
(resp. increasing) (x ,...,x )∈H ×···×H , f(x +···+x )=k.
1 n 1 n 1 n
We impose a t-apartness condition on a solution (H ,...,H ) of IPHTn by
1 n 2
requiring that the union H ∪···∪H is t-apart. We denote by “IPHTn with
1 n 2
t-apartness” the principle IPHTn with this t-apartnesscondition onthe solution
2
set.
Theorem 6. IPT2 and IPHT2 with 2-apartness are equivalent over RCA .
2 2 0
Proof. We first provethat IPT2 implies IPHT2 with 2-apartness.Given c:N→
2 2
2 define f : [N]2 → 2 in the obvious way setting f(x,y) := c(x + y). Fix
two infinite disjoint sets S ,S such that S ∪S is 2-apart. By Lemma 4.3 of
1 2 1 2
[11], IPT2 implies over RCA its own relativization: there exists an increasing
2 0
p-homogeneous sequence (H ,H ) for f such that H ⊆ S . Therefore H ∪H
1 2 i i 1 2
is 2-apart by construction. Let the color be k < 2. Obviously we have that for
any increasing pair (x ,x ) ∈ H ×H , c(x +x ) = f(x ,x ) = k. Therefore
1 2 1 2 1 2 1 2
(H ,H ) is an increasing p-homogeneous pair for c.
1 2
Next we prove that IPHT2 with 2-apartness implies IPT2. Let f : [N]2 → 2
2 2
be given. Define c : N → 2 by setting c(n) := f(λ(n),µ(n)) if n is not a power
of 2 and c(n)=0 otherwise. Let (H ,H ) be a 2-apart solution to IPHT2 for c,
1 2 2
of color k <2. Then set H+ :={λ(h ):i∈N} and H+ :={µ(h ):i∈N}.
1 2i−1 2 2i
We claim that (H+,H+) is a solution to IPT2 for f. Let (x ,x ) ∈ (H+,H+)
1 2 2 1 2 1 2
be an increasing pair. Then for some h ∈ H and h′ ∈ H such that h < h′ we
1 2
have λ(h)=x and µ(h′)=x . Therefore
1 2
k =c(h+h′)=f(λ(h+h′),µ(h+h′))=f(λ(h),µ(h′))=f(x ,x ).
1 2
⊓⊔
4.3 HT!ω and IPT2
2 2
Proposition 3 (RCA ). IPHT2 with 2-apartness ≤ HT!ω.
0 2 sc 2
Proof. Let f : N → 2 be given, and let H = {h ,h ,h ,...} be an infinite
0 1 2 <
2-apart set such that FS!ω(H) is monochromatic for f of color k < 2. Let
S ,S ,S ,... besuchthateachS isanexactlylargesubsetofH, S =H,
1 2 3 i i∈N i
and maxS < minS , for each i ∈ N. Let s = S . Let H := {s ,s ,...}.
i i+1 i i s 1 2
S
H is 2-apart and consists of the consecutive exactly large sums of elements of
s
P
H.LetH ={t ,t ,...} bethesetconsistingoftheofelementsfromH minus
t 1 2 < s