SETS PARTITION REGULAR FOR n EQUATIONS
NEED NOT SOLVE n + 1
VITALY BERGELSON, NEIL HINDMAN, and IMRE LEADER
[Received 18 December 1994—Revised 7 November 1995]
ABSTRACT
An n x m rational matrix A is said to be partition regular if for every finite colouring oi \ there is a
monochromatic vector x s Nm with Ax = 0. A set D c f^J is said to be partition regular for A (or for
the system of equations Ax = 0) if for every finite colouring of D there is a monochromatic x e D'n
with Ax = 6.
In this paper we show that for every n there is a set that is partition regular for every partition
regular system of n equations but not for every system of n + 1 equations. We give several related
results and we also prove a 'uniform" extension of this result: for each n we give a set D which is
uniformly partition regular for n equations in the sense that given any finite colouring of D some one
class solves all partition regular systems of n equations, but D is not partition regular for (and in fact
contains no solution to) a particular partition regular system of n t 1 equations, namely that system
describing a length n •+• 2 arithmetic progression with its increment. We give applications to the
algebraic structure of /3\, the Stone-Cech compactification of the discrete set ^J.
1. Introduction
Theorems in Ramsey Theory often state that whenever a sufficiently large
structure of some kind is partitioned into k cells (or '/c-coloured') a smaller
structure of the same kind must be contained in one cell (or be 'monochrome').
For example, the simplest non-trivial version of Ramsey's Theorem itself says that
whenever the edges of a complete graph on six vertices (a K^) are 2-coloured,
there must be a monochrome triangle. One can naturally ask (so Erdos did)
whether any graph with the property that whenever it is 2-coloured there will
necessarily be a monochrome triangle must contain a K6. In fact it turned out [3]
that there is a graph with this property which contains no K4.
In [14], Spencer gave a similar result for van der Waerden's Theorem. He
provided a simple elegant proof that given any n and k there is a set A such that
whenever A is k-coloured there is a monochrome length n arithmetic progression,
but A contains no length n +1 arithmetic progression. See [12] for some
extensions of Spencer's result.
In a similar vein, it was shown in [11] that given any n and /c in f^J there is a set
5 of finite sets such that, whenever S is /c-coloured, there is a sequence
C\, C2,..., Cn of pairwise disjoint sets with F(7«C,)"=1) monochrome, but 5
contains no FUdQ)^). (Here Ff/((C,)^,) = {{J,.FQ: 0 ^ F g { l , 2,..., n}}.) As
a consequence, one obtains (see [10,Corollary 3.8]) the fact that for any n eN
there is a subset A of N such that whenever A is coloured with any finite number
of colours, there is a sequence (xt)?=] with FS((x,)"=l) monochrome, but A
contains no FS((x,)^i)- (Where FS«x,)?-i) = {2teFx,: 0 * F s { 1 , 2,...,«}}.)
This first two authors acknowledge support received from the National Science Foundation via
grants DMS 9103056 and DMS 9025025 respectively.
1991 Mathematics Subject Classification: 04A20.
Proc. London Math. Soc. (3) 73 (1996) 481-500.
482
VITALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
We describe this situation (where the number of colours is unrestricted) by
saying that A is 'partition regular' for FS((xt)"=\)1.1. DEFINITION. Let ^ be any set of sets. A set B is partition regular for ^ if
whenever B is partitioned into finitely many cells, one cell contains some member
of <g.
We are concerned in this paper with solving systems of equations of the form
Ax = 0 where A is a rational matrix. (It would be equally general, but less
convenient, to assume that all entries of A are integers.) Such a system is said to
be partition regular if, in the sense of Definition 1.1, N is partition regular for the
set <€ of all solution sets for the given system. That is, the given system is partition
regular if and only if whenever f^J is finitely coloured there is a monochrome
solution to the given system.
The problem of which systems are partition regular was completely solved by
Rado [13]. The solution involves a notion known as the 'columns condition'.
and let A be an nXv matrix with rational
1.2. DEFINITION. Let n,veN
entries. Then A satisfies the columns condition if the columns cu c2, •••, cv of A
l^kx<
can be ordered so that there exist m e N and ku k2, •••, km in N with
k2 < ... < km = v such that
(1) 2 f l , c , = 0 a n d
(2) if m > 1 and t G {2, 3,..., m), there exist a, „ a2,,,..., a*,.,., in Q such that
1.3. THEOREM (Rado [13]). Let u,veN
and let A be a uXv matrix with
rational entries. Let x = (xu x2,..., x v ) r . The system of equations Ax = 0 is
partition regular (over N) if and only if A satisfies the columns
condition.
Our aim in this paper is to investigate how 'independent' are the notions of
partition regularity for various systems of equations. We will show in § 3 that for
each n GN there exist a subset D oi N and a partition regular system of n + 1
linear equations Bx = 0 such that D contains no solutions to Bx = 0 but given any
finite colouring of D and any partition regular system of n linear equations (with
rational coefficients) Ax = 0, D will contain a monochrome solution to Ax = 6.
(We remark to the interested reader that the partition regular system of n +1
equations that we use is the system describing a length n + 2 arithmetic
progression together with its increment.)
Our constructions in § 3 are somewhat complicated. We will introduce much of
the machinery in a simpler setting in § 2. There we will establish the following
fact. Given any s e Q+ = {q G Q: q >0} and any finite set F c Q + \ { 1 , s} there is
a set D that is partition regular for the equation x = y + s • z but contains no
solution to x = y +1 • z for any t G F. (We do now know whether one can allow
1 G F.) In particular, given t & {1, s} there is a set partition regular for x = y + s • z
containing no solution to x = y +1 • z. As a consequence, we obtain a set E that
PARTITION REGULAR SETS
483
is partition regular for each single partition regular equation but contains no
solution to the system
x2=x} +2x2,
x4 = X\ + 3x2.
One might ask for the stronger conclusion that whenever E is finitely coloured
some one cell will contain solutions to all single partition regular equations. The
sets we construct in §2 specifically fail to satisfy this requirement—indeed, the
reader will see that this failure is absolutely built in to the way we construct these
sets.
However, it is possible to insist on the far stronger requirement, and in § 4 we
give a proof of this result. This result has consequences for the algebraic structure
of (3N, the Stone-tech compactification of N. For a survey of numerous
applications of the structure of (3N to combinatorial number theory see [8].
All of our equations will have rational coefficients and all of the numbers with
which we will deal will be rational—we will sometimes not mention this fact in
order to improve the flow of the prose.
2. Solving single equations
As a simple consequence of Rado's Theorem (Theorem 1.3), a single equation
is partition regular if and only if some subset of its non-zero coefficients sums to 0.
(This special case is much simpler than the general result. See [5, § 3.2]. We
remark also that it can be derived using recurrence of sets of positive density in a
fashion similar to that used in [1].) The following well-known lemma says that the
only 'interesting' single partition regular equations are x = y + s • z for positive
rationals s.
2.1. LEMMA. Let cx • xx + c2 • x2 + ... + cv • xv = 0 be a partition regular equation.
Then there exists a positive rational s such that any set containing a solution to
x = y + s • z also contains a solution to Cj • x} + c2 • x2 + ... + cv • xv = 0.
Proof. If 2,v=iC, = 0 then any xx=x2 =... =xv solves the equation, so we
may assume we have 2/V=i c, ¥> 0. Since cx • xx + c2 • x2 + ••• + cv • xv = 0 is partition regular, we may presume we have k 2* 2 such that c, is non-zero and
cx + c2 + ... + Q = 0. Let s = \(Z?=lci)/c1\. Note that also s = KSJU+i c,)/c,|.
Assume we have x, y, and z such that x=y + s-z. If (E,u=i c,)/^ <0, let
*! = x, x2 = x3 =... = xk = y, and xk+l = xk+2 =... = xv = z. If (ZU\ ct)lcx > 0, let
*\ = y, x2 = x3 = ... = xk = x, and x*+1 = xk+2 = ... =xv = z.
As a consequence of Lemma 2.1, we are interested in solutions to equations of
the form x = z + s • w. Note that x, z, and w are the three entries of the matrix
product I 1 0 )(
I. The matrix
1 0
is of a special kind which we will use
throughout this paper.
2.2.
DEFINITION.
Let u, v e IU A u X v matrix C ia a monk first entries matrix
484
V1TALY BERGELSON, NEIL HINDMAN, AND 1MRE LEADER
if all entries of C are rational, no row of C is 0 and the first (leftmost) non-zero
entry in each row is 1.
Now we start to describe some sets that will be useful as building blocks for us.
2.3. DEFINITION. Let C be a « X u monic first entries matrix. A sequence
< & X = i is C-useful if
(1) for each m e N, Qm is a finite set of positive rationals,
(2) for each m e M, Qm <= Qm+U and
(3) for each m &N, each q e Qm, and each / e {1, 2,..., v} there exists
x E (Qm + ] U{0})u such that x,, = q and *,-+, =JC I - +2 = ...=*„ = 0 and all
entries of Cx are in Qm+X U {0}.
Given any equation x = z + s • w we will construct a special C-useful sequence
/i
,\
where C = I 1 0 1. The set D defined below will assist in this construction.
\0 1/
2.4. DEFINITION. Let C be a w x v monic first entries matrix, let {Qm)7n=\ be a
C-useful sequence, and let (.ym)m=i be a sequence in N such that for each m <=N
and each q e Qm, q - ym EN. Then
Z = u (ym)l=\) =
2 ^ • ym' F is a finite non-empty subset of N
•-me/7
and qm e Qm for each m e F\.
Our next result is the most difficult result in the paper. It is the 'workhorse' of
this paper. In a sense, having proved it, a lot of the rest of the work is some
(admittedly rather delicate) fine tuning. To be more precise, it is Theorem 2.5
that we shall use when proving the 'positive' parts of statements—that certain sets
are partition regular for certain equations.
Because the proof is rather complicated, we postpone it until the end of this
section. After stating Theorem 2.5 we shall immediately give an example of how
to apply it.
The statement deals with the notion of 'image partition regularity'. (What we
have been calling 'partition regular', namely finding monochrome x with Ax = 6,
is 'kernel partition regular'.) A matrix is image partition regular if whenever N is
finitely coloured there is some x with the entries of Ax monochrome.
2.5. THEOREM. Let C be a uXv monic first entries matrix, let (Qm)°^=i be a
C-useful sequence, and let (ym)Z=\ be a sequence in N such that for each m GN
and each q e Qm, q • ym e N. Then D = D((Qm)^=u
< y , X = i ) is image partition
regular for C. That is, whenever D is finitely coloured, there exists x sNv such that
the entries of Cx belong to D and are monochrome.
As an application of Theorem 2.5, let us find one of the sets claimed in the
PARTITION REGULAR SETS
485
Introduction. Given s e <Q>+ and a finite F g Q + \ {l, s} we wish to produce a set
that is partition regular for the equation x = y + s • z but contains no solution to
x = z + t • w for any / e F. Given such s and F our strategy is simple. We let
1s
I \
C = I 1 0 1 and produce a C-useful sequence ((?m)™=i and a sequence (>'w)^=i
\0 1/
such that D{(Qm)Zi=\, (ym)m=i) contains no solution to x = z + / • w for any r e F.
We see now that by choosing the sequence (ym)m=i to be thin enough, we are
able to concentrate entirely on the individual sets Qm.
2.6. LEMMA. Let C be a monic first entries matrix and let (Qm)m = i be a C-useful
sequence. Let F be a finite set of positive rationals and let c = max F. For each
m e N, let dm = max Qm and let
bm = min{|jc - (z + t - w)\: t e F and x, z, w e Qm U {0} and x # z + t • w}.
Assume that for each m e N we have ym • q e.N for all q e Qm and if m > 1 then
ym ' bm > (c + 1) • SUT^i1 yi< ' dk. Let {r,, t2,..., t,} c F and consider the system of
equations
*3 = X\ + t\ • X2,
^4 = -*1 + h ' X2>
l + 2 ~ X\ + h ' X2-
X
If there is a solution to this system in D((Qm)^=u
is a solution in Qm U {0} with x2 ^ 0.
(ym)m = \), then for some m there
Proof. For each X £ / ) = D ( ( Q j ; = 1 ) (yJZ-i),
pick ^ E X ^ ^
such that x = 2 ^ = i q(x)m ' ym- Assume we have a solution to the specified system
in D. It suffices to show that <7(*/+2)w = q(x\)m + U ' cj{xi)m f ° r e a c h rn e N and
/ e {1,2,...,/}; for then one simply chooses any meN
with q(x2)m^:0.
So
x
let x=xi+2,
z= \,
w=x2,
and t = f, and suppose one has some
msN
such that q(x)m ^q(z)m +1 • q(w)m, and pick the largest such m. Note that
\q(x)m -q(z)m-tq(w)m\ ^ bm.
Case 1: q(x)m > q{z)m +1 • q{w)m. Then we have
0 = x-z-1
•w
m-\
JJ (dk + t - d k ) - y k
k=\
(l + c)' 2
a contradiction.
dk-yk
486
VITALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
Case!: q(x)m <q(z)m + t • q(w)m. Then
0 = z + t • w ~x
= (q(z)m + t • q(w)m - q(x)m) • ym + 2 (q(z)k + ' ' q(*>)k ~ q(x)k) " Yk
k=\
^ bm • ym - 2
dk-yk
again a contradiction.
We are now ready to construct a set which was promised in the Introduction.
2.7. THEOREM. Let s be a positive rational and let F be a finite subset of
Q+\{\,s). Then there is a set D ^N that is partition regular for the equation
x = z + s • w but contains no solutions to x = z + t • w for any t e F.
1 s
Proof Let C be the matrix | 1 0 J. By Lemma 2.6 it suffices to produce a
,0 1
C-useful sequence {Qm)Zi=\ s u c r i t n a t f° r all r e F and all m &N, Qm U{0} does
not contain a solution to x = z +1 • w with tv^O. Indeed, assume we have done
this and choose a sequence (ym)Z=i a s required by Lemma 2.6 and let
D = D({Qm)Zt=\, {ym)Zi=\)- By Theorem 2.5, D is partition regular for the
equation x = z + s • w. By Lemma 2.6, D contains no solution to x = z + t • w for
any t e F.
So we set out to construct (Qm)Z=\- Let
c = max(F U {1/r: t e F} U {2/(1 - r): t e F} U {r/s: r e F} U {s/t: t e F})
and let
rf = min({r e F: f > 1} U {1/r: t e F and r < 1} U {r/5: r e F and r > 5}
U {5/r: r e F and r < 5} U {2/(1 + /): /
G
F and t < 1}).
Note that d > 1.
We show next that it suffices to produce the sequence (Qm)Z=\ so that
(*) for each m e N and each fl and b in Qm with a < b, one has that b/a<d or
b/a>l + c • d, and moreover if b/a <d then there is some q E Qm such
that b - a = q - s.
Indeed, assume that we have constructed (0 w )«=i satisfying (*), and suppose
that we have some m EN and some x, z, w in Qm U {0} and some t e F with
PARTITION REGULAR SETS
487
x = z +1 • w and w i^ 0. Note that z ^ O since if it were zero we would have
x/w = t. This is forbidden by (*) because if / > 1, one has d ^t^c, and if / < 1,
one has d^ \/t *£c.
Next we claim that w/z^d. Suppose instead that w/z>d. Then w/z>
\ + c d > 2 .
If t>\,
w e have x/w = z/w + t^d
a n d x/w = z/w + t < \ + t <
1 + c • d, a contradiction. If / < 1, we have w/z > c ss 2/(1 - t) so z/w < \(l - t)
and so x/w = z/w + t < 5(1 + t), and hence w/x >2/(l + t) s*d; but w/x =
\/t - zl(x • t) < \/t s£ c, a contradiction.
Thus we have w/z =£ d. Consequently x/z = 1 + t • (w/z) ^ 1 + c • d. Hence we
may pick q G Qm such that x - z = s • q, that is t • w = s • q. But then q^w since
t ¥=s, while the ratios t/s and s/t are forbidden for two members of Qm.
It thus suffices to produce a C-useful sequence (Qm)m=\ satisfying (*). But this
is essentially a triviality. We let Qx = {1}. (Or insert any other positive rational.)
Given Qm, which is finite, write Qm = {qu q2, •••, qi}- We need to choose for each
qj some r;, and add r; and r ; + j • q, to Qm + \- (In the definition of C-useful, the
/ = 1 requirement has no content so we are making sure all entries of
I 1 0 If ) are in Qm+\U{0}.) To start out we may assume q, = max Qm\ we
\
I \Qi'
\0 1 /
choose rx such that r, >qt • (1 + c • d) and (r, +s • q\)/rx <d. Given that we
have chosen r,_!, we pick r- such that rt > (r;_, +s • qj-\) • (1 +c • d) and
(rj + s-qj)/rj<d.
Observe that the proof does not work if one allows 1 e F. There is a good
reason for this. Sets of the form D = D((Qm)^,=l, (ym)m = i) are always partition
regular for the equation x =y + z. One way to see this is to pick any qm G Qm and
note that FS((qm • ym)m = \) ^D, where FS((xn)*=l) = {J,neFxn: F is a finite
non-empty subset of N}. Consequently, any finite partition of D will have one cell
containing FS((xn)^=i) for some sequence (xn)^=v (See, for example, [7, Lemma
3.8].)
However, we believe that this fault is due merely to our inability to construct
'better' sets than D.
2.8. CONJECTURE. For each positive rational s ¥^ 1 there is a set D that is partition
regular for x = y + s • z but contains no solution to x = y + z.
In fact, we believe that a much stronger statement is true. Let A and B be
partition regular matrices. Say that A Rado-dominates B if every set which is
partition regular for A is partition regular for B. Say that A solution-dominates B
if every solution to Ax = 0 contains a solution to By = 6. To be more precise, if A
is an n x m matrix and B is an r X k matrix, A solution-dominates B if there is a
function / : {1, 2,..., k}-* {1, 2,..., m) such that whenever Ax = 0 we have
7o>\
B "V.(2) = 6 .
\xml
488
VITALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
Thus we trivially have that if A solution-dominates B then A Rado-dominates B.
2.9. CONJECTURE. Let A and B be partition regular matrices. Then A Radodominates B if and only if A solution-dominates B.
Let us remark that to ask, as we did in Conjecture 2.8, for a set that is partition
regular for one matrix and contains no solution to the second is the same as
asking for a set that is partition regular for one but not for the other. (To see this,
assume D is partition regular for A but not for B. Pick a finite partition 3* of D,
no cell of which contains a solution to B. Some one cell must be partition regular
for A.)
We now turn our attention to constructing a set that is partition regular for
every single partition regular equation but not for a certain partition regular
system of two equations. Before we start, however, let us point out that the
situation just described is not so clear when one deals with a set which is partition
regular for several, perhaps infinitely many matrices but is not partition regular
for some other specified matrix.
However, the specified matrix which we will deal with subsequently will in all
cases be the matrix for the terms of an arithmetic progression together with its
increment. One can derive in this case (as a consequence of Lemma 2.10) using a
construction like that in the proof of Theorem 2.11, that for such a matrix it is
equivalent to ask for a set not partition regular for the matrix or to ask for a set
containing no solution for the matrix. However, we will not spell out this
argument, because the sets we produce always satisfy the stronger conclusion
anyway.
We now start to work towards a set, as mentioned above, that is partition
regular for every single partition regular equation but not for a certain partition
regular system of two equations. The next lemma will be used again in § 3. It will
provide us with a way to piece together partition regular sets in a useful fashion.
2.10. LEMMA. Let 1 EN, let a e {1, 2}, and let (A,)*=, be a sequence of finite
subsets ofN. Assume that no Dn contains a solution to the system
x
a+2 = X\ + 0 • X2,
Then there is a sequence (rn)'=1 in N such that U*=i Tn ' Dn contains no solution to
the same system.
Proof. Let r, = 1 and inductively given n > 1, let kn = max Uy^i1 rj' ty and let
rn = kn • I + 1. Now we claim that if t e {1, 2 , . . . , / - 1} and {x, z, w} £ U"=i rn ' A.
with x = z +1 • w, then there exists n with {JC, z, w) 9 rn • Dn. To see this pick the
largest n such that {x, z, w} n rn • Dn ¥* 0 and suppose {x, z, w} \ rn • Dn ^ 0 . Since
x > z and x > w, there are three possibilities for {x, z, w} n rn • Dn, namely [x],
{x, w}, and {x, z}- Suppose, for example, that {x, z, w} D rn • Dn = {x}. Then
0 < x = z + / • w s£ (1 + /) • kn s% / • kn < rn while rn divides x, a contradiction. The
other two cases are handled similarly.
PARTITION REGULAR SETS
489
Now suppose we have {*,, x2, * 3 ,..., xt+l} c: (J" =1 rn • Dn satisfying the system
For / G {a, a + 1,..., / - 1}, pick n(t) e N such that {x,+2, x t , x 2 }^ rn(l) • Dn(l). Since
x, G rn(() • Dn(() for all t, we have n(t) = n(a). Let n = n(a). Then we have
{xi,x2, ...,*/+,} £r n • £>r so {xjrn,x2/rn,..., xl+Jrn}<^Dn, a contradiction.
Note that by Rado's Theorem the system of Lemma 2.10 with a = 1 is partition
regular. (This is the strengthening of van der Waerden's Theorem which requires
that the increment belong to the chosen colour class as well.)
2.11. THEOREM. There is a subset E of N which is partition regular for every
single partition regular homogeneous linear equation with rational coefficients but
contains no solution to the system
x3 = *, + 2 • x2,
x4 = JCJ + 3 • x2.
Proof. By Lemma 2.1, it suffices to produce a set E which is partition regular
for every equation of the form x = z + s • w for s e Q + but contans no solution to
x 3 =xx
XA
=
+2 • x2,
* I + 3 • x2.
Given s e Q + \{2}, pick by Theorem 2.7 a set HS^N which is partition regular
for x = z + s • w but contains no solution to x = z + 2 • w. Also pick a set H2^N
which is partition regular for x = z + 2 • w but contains no solution to x =
z+3-w.
Given J G Q + and k eN, pick by compactness (see [5, § 1.5]) a finite subset Hsk
of Hs such that whenever Hsk is /c-coloured there is a monochrome solution to
x = z + s • w. (Strictly speaking, we do not need to appeal to compactness, since
we have built up our set in Theorem 2.5 using finite sets.)
Let (AX= i enumerate {Hsk: s e Q + and k E f^J} and choose a sequence (O" = 1
as guaranteed by Lemma 2.10 with a = 2 and / = 4. Let E = U ^ i ^ • Dn. Then by
Lemma 2.10, E contains no solution to
X3 = X]
+2-X2,
It is clear that E is partition regular for every single partition regular equation.
490
Indeed, let E
such that Dn
monochrome
monochrome
V1TALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
be finitely coloured, say with k colours, and let s E Q + and pick n
= Hsk. Colour x in Hsjl by the colour given to rn • x in E. Pick
x, z, w in Hsk with x = z + s • w. Then rn • x, rn • z, and rn • w are
in E and rn • x = rn • z + s • rn • w.
We complete this section by providing the postponed proof of Theorem 2.5. For
those who are familiar with such arguments, we point out that in the special case
where, in the definition of C-useful, we actually always have x, = ... =*,_, = 0,
one could view what we are doing as in some sense reproving Ramsey's Theorem,
but replacing the appeals to the pigeon-hole principle by appeals to the
Hales-Jewett Theorem. We first state the form of the Hales-Jewett Theorem
which we will be using.
By an n-variable word over an alphabet A we mean a word over the alphabet
A U {vx, v2,..., vn} in which each u,- actually occurs, where vu v2,..., vn are
symbols not in A. Given an ^-variable word w(vu v2,..., vn) and (au a2,..., an) E
A", w(ax, a2,..., an) has its obvious meaning—it is the word obtained from
w(vu v2,..., vn) by replacing each occurrence of u, by at.
2.12. LEMMA (Hales-Jewett). Let A be a finite alphabet and let k, n e fU Then
there exists p e N such that whenever the length p words over A are k-coloured,
there is an n-variable word w(vu v2,..., vn) of length p, all of whose instances
w{au a2,..., an) for (au a2,..., an) e An have the same colour.
Proof. The usually stated version of the Hales-Jewett Theorem ([6], or see [5])
is for a single variable. Appy this version to the alphabet A".
2.13. DEFINITION. Given finite subsets Q and SUS2,..., Sp oiN and a sequence
{ym)^=\ in N, the Q-span of (Su S2,..., Sp) with respect to {ymYm = x is
- /,e«2U{0})foreachre{l,2,...,p}and/#6|.
m<=S,
2.14. LEMMA. Let C, (QmTm=\ and (ym)m=\ be as in Theorem 2.5 and let a, k,
n EN. Then there exists p EN such that, given any pairwise disjoint non-empty
subsets Su S2,..., Sp of N with min Uf=1 Sp^ a and any k-colouring <p of the set
D = D((Qm)^=],
(ym)^=\), there exist pairwise disjoint non-empty subsets Ho,
Hu ..., Hn of{\, 2,..., p} and pairwise disjoint non-empty subsets Tu T2,..., Tn ofN
and x E D such that
(1) for allJE {1,2, ...,n),
Tr{J^HiSt,
(2) there exists le X f e W o ( 2 a such that x = ^teHol,
Qa-span of(Su S2,...,SP) over (ym)^ = : ) , and
-^meSiym
(3) for any z in the Qo-span of (Tu T2,..., Tn) over (ym)^=u
(so x is in the
<p(x) = (p(x + z).
Proof. Let A = Qa U{0}. Pick p as guaranteed by Lemma 2.11 for A, n, and
k + l. Let W be the set of length p words over A. Define i//: W \ {0}-» D by
-2 y
t=\
me.S,
PARTITION REGULAR SETS
491
where w = /,/ 2 ... lp and each /, G A. Colour W \{0} using A: colours by (p°ip, and
give 0 its own colour. Pick w(vu v2, ••-, vn) such that all occurrences are
monochrome. Note that not all occurrences are 0 so w(0, 0,..., 0) 5^6. Let
w = /,/ 2 ... lp, where each I, e AU{vu v2, ..., vn}. Let Ho= {t e {1, 2,...,/?}:
/, e Qa). By the observation above, / / o ^ 0 . For j e {1,2,..., n), let / / , =
{/ G { 1 , 2 , ...,p}\ l, = vj} and let 7} = U,e/v,Sr Let x = 2 , e / / o /, • 2 W es,>V Since
min Uf =] 5, s* a, we have A: G D. Conclusions (1) and (2) hold directly.
To verify (3) let q = sxs2 ••• sn be a length n word over A with g 7^6 and let
z =Sr=i^f -S^e^ym- Then x = i//(w(0, 0,..., 0)) and x + z = il/(w(su s2>..., sn)),
SO <p(*) = <p(*+z).
Finally we prove Theorem 2.5.
Proof of Theorem 2.5. Let <p be a finite colouring of Z), say with & colours. Let
y = k • (y — 1) + 1 (so y is the pigeon-hole number to obtain v objects of one
colour). Define numbers n(l), n(2),..., n(y) by reverse induction as follows.
Given n . a e ^ J let p(n, a) be the number guaranteed by Lemma 2.14 for a, k, and
n. L e t « ( y ) = l and for i e {1, 2,..., y - 1}, let n(y - i) = p(n(y - i + 1), i + 1).
For / G {1, 2,..., n(\)} let 50,( = {y +/}. Pick pairwise disjoint non-empty
subsets Hh0, HhU...,Hhn(2)
of {1,2, ...,«(1)} and 1(1) e X , e H ) 0 Q y as guaranteed
by Lemma 2.14. Define jr(l) = 2, e W l i 0 /(I), • 2me5b., ^ and for / e {1, 2,..., n(2)},
let 5,^r = Ufef/, -^o.r Then for any z in the Qy-span of (5 1(1 , S lj2 , ••-, StiW(2)) over
<ym)m=i, we have that <p(x(l)) = <p(x(l) + z).
Inductively, given r e {2, 3,..., y — 1} and pairwise disjoint 5 r _i,i, Sr_1>2, ...,
Sr_!,„(,), recall that n(r) = p(n(r + 1), y - r + 1) and so we may pick pairwise
disjoint non-empty subsets Hr<0, HrX, ..., Hrn(r+X) and /(r) G X , e W o Q y _ r + 1
as guaranteed by Lemma 2.14. Let x(r) = ^,^HrOl(r), •^Zm^Sr_hiym and for
j G {1,2,..., n(r +1)} let £,.,,• = L U ^ ^ r - i , / - Then for any z in the
Q 7 _ r+1 -span of (SrU Sr2,..., 5 r , n(r+1) ) over (ym)^=l one has <p(x(r)) = <p(x(r) + z).
Finally we have S y _ lpl . Let / / y 0 = {l}- p i ck /(y) e g l f and let
x(y) = l(y)-
2
ym=
1
/(y)-
2
>w
Pick by the pigeon-hole principle some 5(1) < 5(2) <... <8(v) such that
<p{x{8(l))) = 9(x{6{2))) =... = <P(JC(8(V))) and let c = V ( J C ( 6 ( 1 ) ) ) .
For r G { l , 2 , . . . , W , let a ( r ) = y - fi(r) + 1, so that ! ( 5 ( r ) ) e X , i % Q B W
Then if r < v and z is in the Q a(r) -span of (5 6(r)>1 , S 6(r)>2 ,..., 5 6(r) ^ (fi(r)+1) ) over
<yj««i, one has <p(jc(5(r))) = <p(x(8(r)) + z) = c.
Now given r G {2, 3,..., u} and t G HS{r)f0, we have l(8(r))t G j2«(r)- Since
(<2w)m=i is a C-useful sequence, we may pick some wrt E (Qa(r)+\ U{0})v such
that wr>f(r) = l(8(r))t and wr>l(r + 1) = wr>t(r + 2) =... = wrp,(v) = 0 and all entries
of Cwr>l are in £> a(r)+ i U{0}.
Now let zv = ^(5(u)), and for j G {1, 2,..., u - 1} let
r=; + l
To complete the proof we show that for each entry /x of Cz one has <p(tx) = c.
To this end let (ah a2,..., flj be a row of C and let /A = E/=ifly*fly-Let / be the
492
VITALY BERGELSON, NEIL H1NDMAN, AND 1MRE LEADER
coordinate of the first non-zero entry of (au a2,..., av), so that a,> = 1. Thus
fx = Zi + 2 / = , + i «> ' Zj.
If / = v, then ix = zv = x(8(v)),
i < v. Then
ti = x(5(i))+ 2
so 8(fx) = c as required. Thus we assume that
2
r-i + l
vv,,,(/)-
2
m
IEWJ(,)0
y™+ 2
^%)-i,i
orZj.
; - /+ l
So, putting
V
K
we find that it suffices to show that either r belongs to the (2O(/)-span of
(5 6 ( / ) 1 , 5S(,)i2. • ••, ^5(,),^(S(/)+i)) over (y w )^ = i or else r = 0. Now, we have
2 fly 2y = flw-x(«(v))+2 fly • ( * ( « ( / ) ) +
2
= r _ 2 i ^ ^ ( 5 ( 0 ) + _ 2 ] 2+x t2
= 2+i f 2
flr-/(«(0)y
, v
V
+ _Z Z
= 2
2
r= i +l
ar-wrt(r)-
me
Wr,U)-
aj-Wr.,u)'m2
2
3
ym
^
V
/\
Z fl/-wr>,(y)-
(eWj(,)o
v
2
2
V
Z >m
2 yw
^S(t)-i,i
r-1
+ 2
2
2
aj'wrAj)-
T — i + L ;-/-+-1 f E n ^ o
u
2
^
m ei^).^,
r
+ 2
2
2fly*„(;)• 2 y^-
Therefore
T =
>
>
r = i' + l
ferij(r)o
VVlZI*
v
= 2
2,
2
w
V+
/
fl',1
• W ' . i (IT I )
t eH^i + \)o
1
/
meSg^+\)-i
r
+ _Z
+
/
m e5j( r )_i,
2
Z
fly*wr.,0)-
2fly• WMO") '
2
Z >>m
>«.
K+u(0+fl1+i-w/+1>,(/ + i))-
2
)»«•
Now observe that if j > r then wr,(/) = 0 and if / < / then a} = 0, so
V
w,-+i.,(0 +fli+ i • w,-+1>,(/ + 1) = 2 fly ' w,- + i.,(/),
t
PARTITION REGULAR SETS
493
and for r ^ /' + 2 we have
r
v
a
W
2 ) " r.t(j) = 2 °J:' Wr,lU)Consequently, we have
T=
2
2 (2«y^.,(/)j- 2 ym.
r = / + l lE//(, r | i0 7=1
'
meSw,Hi(
Now we are given that for r s* / + 1 we have
V
2 fl, • HV.,(/) £ <2«(r)+, U {0} C &,<„ U {0},
7=1
so we have r = 0 or T is in the Q Q(/) -span of (5 6(/) ,,, S8Uh2,..., 5 6 ( ( M ( 6 ( , ) + 1 ) ),
provided we can show that given (ru t^) and (r2, t2) with rj, r2 e {/ + 1, i + 2,..., v}
and r e // a(ri) . o and r2 e //« (ri) . o , if (r,, r,) * (r2, r2) then 5 8(ri) _ lifl n 5 5(r2) _,., 2 = 0 . If
r, = r2, this is immediate, so assume r, < r2. Then 5 5(r2) _ lr/2 is a union of sets of the
form S 6 ( r i ) _ u with 5 e Hr^q for some q > 0 while r, e //r]>0.
3. Solving systems of n equations for n 2= 2
We show in this section that given n s= 2, there is a set £" c f^| which is partition
regular for every partition regular system of n equations, but fails to have any
solution to a specified system of n + 1 equations, namely the equations describing
a length n + 2 arithmetic progression and its increment.
We show first that we can reduce the problem to one involving matrices with
monic first entries. The proof is a minor variation of the standard method of
converting kernel partition regular matrices to image partition regular ones.
(Recall that a matrix is image partition regular if whenever IM is finitely coloured
there is some x with the entries of Ax monochrome.)
3.1. LEMMA. Let n eN. Let a partition regular system of n homogeneous linear
equations with rational coefficients be given. Then there is a monic first entries
matrix C with at most n rows hving more than one non-zero entry, such that, for
any y e Nm, the entries of Cy contain a solution to the given system.
Proof. Let A be the n x u coefficient matrix of the given system. Then A
satisfies the columns condition, so assume the columns cx,c2,...,cu of A have
been ordered as required by the columns condition. We assume that no c, = 5.
Pick v EN and ku k2,..., kv with 1< kx <... < kv = u, so that Sfii c, = 0 and for
t e {2, 3,..., v}, Zfcik.^+i Cj is a linear combination of {cu c2,..., 4,_,} over Q.
Let / = {1} U {/ e {2, 3,..., u}: c, £ Spanjcj, c 2 ,..., ci-J}, and note that since
rank A ^n, we have | / | ^ n . Given t e {2, 3, ..., v}, pick au, ot2>t,..., akiu, in Q
such that S/i*,.,*! ct = Y!t=\ otiit • c,. By elementary linear algebra we may assume
494
VITALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
that if / e /, then au = 0. Let A:o = 0. Now define the uXv matrix C by, for
i e {1,2,..., u} and / e {1, 2,..., v},
Then C is a monic first entries matrix whose only rows with more than one
non-zero entry are labelled by members of /.
Further, given any y\,y2,..., yv in M and / e {1, 2,..., u}, let JC, = 2"=i cu • v,. (So
x = Cy.) Then one can routinely verify that Ax = 0. (See [5, pp. 79-80].)
Now we use a strategy similar to that used in § 2. In outline, here is what we
will do. Given any monic first entries matrix C with at most n rows with more
than one non-zero entry, we produce a C-useful sequence (Qm)m=] such that for
no m does Qm U {0} contain a solution to the system
XT,
X4
=
X\ + X2,
=
X\
•+• 2 •
x2,
with x2^0. Then using Lemma 2.6 we produce a sequence (ym)m=i s o that the
existence of a solution to this system in D = D((Qm)^^u (ym)Z = \) would imply
the existence of a solution in some Qm U {0} with x2 ¥^ 0. Consequently D is
partition regular for C (by Theorem 2.5) but has no solution to the specified
system of n +1 equations. Finally we piece the solutions together as in the proof
of Theorem 2.11.
The construction of the sequence (Qm)m = \ is similar to the corresponding
construction in § 2.
3.2. DEFINITION. Let C be a u X v monic first entries matrix.
(a) Given / e {1, 2,..., u), define fx(l) = min{; e {1, 2, ..., v}: cu ^ 0}.
(b) Given n e.N, say that C is n-sparse provided that for each / e {1, 2, ..., v}
we have
row / of C has more than one non-zero entry}| *£ n.
The reader might wonder why we define n-sparse the way we do, since the
matrices we are considering in this section all have the stronger property that at
most n rows have more than one non-zero entry. It is true that, for this section,
we do not need the generality of n-sparse matrices as defined above. However,
this generality will yield an enormous payoff when we get to § 4.
3.3. LEMMA. Let n EN and let C be a u X v n-sparse monic first entries matrix.
Then there is a C-useful sequence (Qm)m=i such that, for each m eN and each
r e Qm, we have \{s e Qm: 1 ^s/r =s In + 2}| ^ n + 1, and, given r, s e Qm, if
l^s/r^ln
+2, then s/r<2.
495
PARTITION REGULAR SETS
Proof. We construct (Qm)^=\ analogously to the proof of Theorem 2.7 adding
new elements as required by the notion of C-useful, at each stage adding things
much bigger than we have added already.
We show that given any finite set Q of positive rationals and any q e Q there is
a finite set F = F(q, Q) of positive rationals such that
(a) given s e F and r e Q, sir > In + 2;
(b) given r, s e F with r < s, either sir > In + 2 or sir < 2;
+ 2}|^w + 1;
(c) given r e F, \{s e F: l^s/r^2n
(d) for all / e {1, 2,..., v}, there exists x e (Q U F U {0})v such that xt = q and
xi+l = xi+2 = ... = xv = 0 and each entry of Cx is in Q U F U {0}.
(When we have shown we can do this we will proceed to the construction of the
sequence (Qm)~m-X.)
W e construct an upper triangular u X v
0
02,
0
0
matrix
02,i
... aViJ
where for each j , ajj = q. The off-diagonal assignments proceed from left to right
and, within columns, from bottom to top: ah2, 02,3> 0i,3. a3,4> a2*> ••• • When we
have chosen aUj, we let
Rij = C,v U {a,,,} U J Z c,fl • atJ: I e {1, 2,..., u) and /LL(/) = / ]
c
/,r ' 0»,;: / e {1, 2,..., u} and /Lt(/) = / ,
= Gu U K ; } U J au + 2
1
r=/+i
J
where G /; is the set of values which have been chosen before step (/, /). (So
Gi 2 = Q, Gij = /?! y_! if 1< i = ; - 1, and Gu = /?,+, y otherwise.) We then let
Note that for any (/, /) one has \RitJ \ Gu\ =s n + 1. Indeed by assumption there
are at most n rows / of C with fi(l) = / having more than one non-zero entry. If
there is a row / of C with fi(l) = i having exactly one non-zero entry,
V
,
,j
,j
Note also that the manner of the construction guarantees that (d) holds. To see
this, let / E {1, 2,..., v) and let
x =
496
V1TALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
Then let a row / be given. If /u ( / ) > / , then Sfu= 1 cu • x, = 0. If /x(/)=/, then
2,"=i cu • x, = djj = q E Q. If /x(/) = / < ; , then we have
f=l
It thus suffices to show that at step (i, j) we can choose a,, £ QT so that:
(1) given s e /?,, \ C,,; and r G Gtj, s/r>2n + 2; and
(2) given r, s e /?,y \ G,,; with r < s one has s/r <2.
To this end let b = max G,-; and let
d = maxi
2) cu • a,j : / E {1, 2,..., u) and fx(l) = i j .
Pick fl,y 3= max{/? • (2/7 + 2) + d, 4d}. Then given s e RLj \ Gij and r e GtJ one has
n^b and 5 ^a ifj - d^b • (2n +2), so (1) holds. Given r, s e /?,-, \ G,,y with r<s
one has r ^ o,, - rf and 5 =£ aLj + rf, so (2) holds.
We are now ready to choose our C-useful sequence. Let Q, ={1}. Inductively
given Qm = {q\,q2,...,qk),
let Qm + \.o=Qm and let Qm + lA = QmUF(qu
Qm + ]fi)
and Qm+lil+i = Qm + \,, U F(g, + ,, Q m + 1,,). Let Q m + 1 = Qm + hk.
and
By condition (d) the sequence {Qm)Z,~\ is C-useful. Also, given meN
r, 5 £ Q m + 1 with r ^ s , one has some t, pe{\,2
k} with r e g f f l + l i , and
5 G Q m + l p . Immediately t^p.
If r</7 then by (a), s/r>2n+2.
Thus s e
F{q,, Qm + u-\\ So (c) guarantees that \{s E Qm + ,: 1 *s s/r ^ 2n + 2}| ^ « + 1.
The conditions on ratios given in Lemma 3.3 ensure that the sets Qm U {0} do
not contain arithmetic progressions of n + 2 terms with their common difference.
3.4. LEMMA. Let n E N and let C be an n-sparse monic first entries matrix. Then
there is a C-useful sequence (Qm)^=\ such that for each m, <2wU{0} does not
contain a solution to the system
=
X\ "I" 2
with x2 ^ 0.
Proof. Let (Qm)°^=\ be as guaranteed by Lemma 3.3. Let m be given and
suppose Qm U {0} contains a solution to the specified system with x2 ¥=• 0. If we had
*i = 0, we would have x4 = 2 • x2 so xjx2 - 2 contradicting Lemma 3.3. Thus
*! > 0. Now suppose x\ > (n + 1) • x2- Then given / e {3, 4,..., n + 3} we have
PARTITION REGULAR SETS
497
SO
{xux3,X4,...,xn+3}c{s
e Qm: l^s/x]^2n
+2},
a contradiction to Lemma 3.3. Thus we must have 0 < x, =s (n + 1) • x2- But then
given ; E {3, 4,..., n + 3} we have
Xj = x] + (j - 2) • x2 =£ 2 • (n + 1) • x2,
so
{x2,x3,xA,...,xn+3}c{s
e Q m : 1 ^ 5 / x 2 ^ 2 « +2},
again a contradiction to Lemma 3.3.
Analogously to the case of kernel partition regular matrices, we say that a set D
is image partition regular for a u Xv image partition regular matrix C provided
that for any finite colouring of D there is some x eNv with the entries of Cx
belonging to D and monochrome.
3.5. THEOREM. Let n e f\l and let C be an n-sparse first entries matrix. Then
there is a set D which is image partition regular for C but contains no solution to
the system
x4 -
JCJ
+ 2 • x2,
Proof. Pick (Qm)m = \ as guaranteed by Lemma 3.4. Choose a sequence {ym)^n = \
as in L e m m a 2.6 ( w h e r e F = {1, 2, . . . , n + 1}). L e t D = D{(QmTm=x,
(ym)xm = x).
Then by Theorem 2.5, D is partition regular for C. By Lemmas 2.6 and 3.4, D
contains no solution to the specified system.
We now mimic the proof of Theorem 2.11.
3.6. THEOREM. Let n e N. There is a set E which is image partition regular for
every n-sparse monic first entries matrix but contains no solution to the system
==
X\
T
L ' X2f
Proof For each n-sparse monic first entries matrix C there is, by Theorem 3.5,
a set which is partition regular for C and contains no solution to the specified
system, so by compactness there is, for each k e f\l, a finite set HCk which
contains no solution to the specified system and is such that whenever HCk is
/c-coloured, there is some x with the entries of Cx monochrome. (See the
discussion of the use of compactness in the proof of Theorem 2.11.)
Let (A)r=i enumerate {Hc,k' k EN and C is an n-sparse monic first entries
matrix}. Choose a sequence (r5)*=1 as guaranteed by Lemma 2.10 with a = 1 and
498
VITALY BERGELSON, NEIL HINDMAN, AND IMRE LEADER
/ = n + 2. Let E = U.r=i rs' Ds. Then as in the proof of Theorem 2.11, we see that
E is as required.
3.7. COROLLARY. Let n e N. There is a set E which is partition regular for every
partition regular system of n linear homogeneous equations but contains no
solution to the system
Af3 = X,
X4
=
+X2,
X\ 1 Z " X2,
Proof. By Lemma 3.1, the set E produced in Theorem 3.6 is as required.
We shall see in Theorem 4.2 that in fact Theorem 3.6 implies a rather stronger
conclusion.
4. Uniform partition regularity and some subsemigroups o
The Stone-tech compactification pN of the discrete set N supports operations
+ and • extending ordinary addition and multiplication and making (PN, + ) and
(PN, •) compact left topological semigroups. These structures, and the interactions of various ideals and subsemigroups of (PN, + ) and (PN, •), have had
several combinatorial consequences. We take the points of pN to be the
ultrafilters on N. (For general background about fiN and its applications, see the
survey [8].)
We will see here that our results, specifically Theorem 3.6, have consequences
for some special subsemigroups of pN.
4.1. DEFINITION. For n EN, let Gn = {p e f)N: for every B ep and for every
m EN and every n Xm kernel partition regular matrix A, there exists x e Bm
with Ax = 6}.
Thus Gn is the set of ultrafilters, every member of which contains solutions to
every partition regular system of n equations. Each Gn is a subsemigroup of
(fiN, +) and a two-sided ideal of (j3f^l, •), and each Gn + ] 9 Gn. Our aim in this
section is to show that for every n, this inclusion is strict. (We remark that it is not
at all obvious that for each n there exists a n m > « with Gm ¥^ Gn. Indeed, it is not
even obvious that we do not have Gn = Gi for all n.)
We need to strengthen Corollary 3.7 in a 'uniform' way, by insisting that, in any
finite colouring of E, some class should actually contain solutions to all partition
regular systems of n equations. Now it would seem that our methods so far,
involving 'piecing together' finite sets, are absolutely useless for this, since by
their very nature we may need different colour classes for different partition
regular matrices. However, it is now that our rather general definition of
k
n-sparse' pays off.
PARTITION REGULAR SETS
499
4.2. THEOREM. Let n eN. There is a set E which contains no solution to the
system
X4 — X\ 1 Z ' X2>
and whenever E is partitioned into finitely many cells, one of these cells contains a
solution to every partition regular system of n homogeneous linear equations.
Proof. Pick E as guaranteed by Theorem 3.6. Suppose the conclusion fails and
let & = {FU f2> f fk} be a partition of E no cell of which contains solutions to
every partition regular system of n equations. For each i e {1, 2,..., k] pick Ah an
n X m{ kernel partition regular matrix such that for no x E F™' is Atx = 0. Pick by
Lemma 3.1 an n-sparse monic w, X t>, first entries matrix C, such that for any
y e NVi, the entries of Qy contain a solution to Atx - 6. Let C be the diagonal
sum of Cu C2,..., Ck. That is,
°\
0
0
...
Cj
Let w=Sf=iW, and v = Sf=i vt. Then C is a u X i> monic first entries matrix.
Further, a brief consideration shows that C is n -sparse. Pick some / and some
z e f^Ul such that all entries of Cz are in Ft. Then there is some y e NVi such that
all entries of Cty are in Ft and hence /*• contains a solution to Atx = 6, a
contradiction.
We wish to remark that it is very fortunate that the diagonal sums constructed
above are indeed n -sparse.
4.3. COROLLARY. For each n sN, the set Gn+1 is strictly contained in Gn.
Proof. Let E be as guaranteed by Theorem 4.2, and let C = {A^E: A contains
a solution to every partition regular system of n homogeneous linear equations}.
Now, by Theorem 4.2, whenever E is partitioned into finitely many cells, one of
these contains a member of C. So by [5, Theorem 6.2.3], there is an ultrafilter
q on E, every member of which contains a member of C. Let p ={A^N:
A n E e q}. Then p is an ultrafilter on N, and E e p. Thus p E Gn\ Gn+i.
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500
PARTITION REGULAR SETS
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Department of Mathematics
Ohio State University
Columbus
Ohio 43210
U.S.A.
Department of Mathematics
Howard University
Washington
D.C. 20059
U.S.A.
Department of Pure Mathematics and Mathematical Statistics
16 Mill Lane
Cambridge CB2 ISB