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Ramsey Theory 2011-2012 Lecture Notes & Example Sheet
Ramsey Theory
University of Cambridge
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Ramsey Theory
Lectured by I. B. Leader,
Lent Term 2010, Michaelmas Term 2011
Chapter 1 Monochromatic Systems 1
Chapter 2 Partition Regular Equations 10
Chapter 3 Infinite Ramsey Theory 22
Examples Sheets
Prerequisites. None (basic concepts of topology)
There arethreeexamples sheets.
Books.
Bollob ́as,Combinatorics, CUP 1986. (For chapter 3.)
Graham, Rothschild, Spencer,Ramsey Theory, Wiley 1990. (For chapters 1 & 2.)
Last updated: Sat 4thAug, 2012 Please let me know of any corrections: glt1000@cam.ac
Chapter 1 : Monochromatic Systems
Ramsey’s Theorem
LetN={ 1 , 2 , 3 ,.. .}and [n] ={ 1 , 2 ,.. ., n}. For a setX, letX(r)={A⊂X:|A|=r}.
Suppose we have a 2-colouringofN(2)– i. havec:N(2)→ { 1 , 2 }. Can we always find an infinitemonochromatic set– i., an infiniteM⊂Nsuch thatcis constant onM(2)?
Examples. 1. Colourij(that is,{i, j})redifi+jis even, andblueif odd. Yes: can takeM={n:neven}.
Colourijredif max{n: 2n|i+j}is even, andblueotherwise. Yes: can takeM={ 40 , 41 , 42 ,.. .}.
Colourijredifi+jhas an even number of distinct prime factors,blueotherwise. No explicitMis known! However, the answer is yes, by the following.
Theorem 1 (Ramsey’s Theorem). WheneverN(2)is 2-coloured, there exists an infinite monochromatic set.
Proof 1 ∈N. There are infinitely many edges froma 1 , so we can find an infinite set B 1 ⊂N− {a 1 }such that all edges froma 1 toB 1 are the same colourc 1.
Now choosea 2 ∈B 1. There are infinitely many edges froma 2 to points inB 1 −{a 2 }, so we can find an infinite setB 2 ⊂B 1 − {a 2 }such that all edges froma 2 toB 2 are the same colour,c 2.
✤
✣
✜
✢
✦✦
✦
❛❛
❛
✏✏
✏
PP
P
✭❤✭❤✭❤ ✦✦
✦
❛❛
❛
✏P✏✏
PP
✭❤✭❤✭❤
✖✕
✗✔
s s a 1 a 2
c 1 c 2 B 2 B 1
Continue inductively. We obtain a sequencea 1 ,a 2 ,a 3 ,.. .of distinct elements ofN, and a sequencec 1 ,c 2 ,c 3 of colours such that the edgeaiaj(i < j) has colourci. Plainly we must haveci 1 =ci 2 =ci 3 =.. .for some infinite subsequence. Then{ai 1 , ai 2 , ai 3 ,.. .} is an infinite monochromatic set.
Remarks. 1. Called a ‘2-pass’ proof.
The same proof shows that ifN(2)isk-coloured then we get an infinite monochro- matic set. Alternatively, we could view ‘1’ and ‘2 or 3 or... ork’ as a 2-colouring of N(2), and then apply Theorem 1 and use induction onk.
An infinite monochromatic set is much more than having arbitrarily large finite monochromatic sets. For example, consider the colouring inwhich all edges within each of the sets{ 1 , 2 },{ 3 , 4 , 5 },{ 6 , 7 , 8 , 9 },{ 10 , 11 , 12 , 13 , 14 },... are coloured blue and all other edges are coloured red. Here there is no infiniteblue monochromatic set, but there are arbitrarily large finite monochromatic blue sets.
Example sequence (xn)n∈Nin a totally ordered set has a monotone subsequence: colourN(2)by givingij(i < j) the colourupifxi< xjand the colourdownotherwise. The result follows by Theorem 1.
What if we colourN(r)(say forr= 3, 4 ,.. .)? Given a 2-colouring ofN(r), must there be an infinite monochromatic set?
Example(3), colourijk(i < j < k)redifi|j+k,blueotherwise. Yes: can takeM={ 20 , 21 , 22 ,.. .}.
What if we used infinitely many colours? Suppose we have an arbitrary colouring ofN(r), i. we havec:N(r)→Xfor some possibly infiniteX. Of course we cannot guarantee an infinite monochromatic set: just colour each edge a different colour. However, such a colouring is injective, so can we guarantee an infinite set on whichcis either constant or injective? The answer is no – e. give the edgeij(i < j) colouri.
Theorem 4 (The Canonical Ramsey Theorem).Whenever we have a colouring ofN(2) with an arbitrary set of colours, there exists an infinite setMsuch that
(i)cis constant onM(2), or (ii)cis injective onM(2), or (iii)c(ij) =c(kl) iffi=k(for alli,j,k,l∈Mwithi < jandk < l), or (iv)c(ij) =c(kl) iffj=l(for alli,j,k,l∈Mwithi < jandk < l).
Note that this theorem implies Theorem 1: if we have only a finite set of colours then (ii), (iii) and (iv) are impossible.
Proof 2-colourN(4)by givingijkl(by which we mean henceforthi < j < k < l) coloursameifc(ij) =c(kl) and colourdiffif not. By Ramsey for 4-sets, we have an infinite monochromatic setM 1. IfM 1 is colouredsamethenM 1 is monochromatic for c(for given anyijandklinM 1 (2), choose anym < ninM 1 withm > i,j,k,l, then c(ij) =c(mn) =c(kl)). So in this case (i) holds.
Suppose then thatM 1 is coloureddiff. Now 2-colourM 1 (4)by givingijklcoloursame ifc(il) =c(jk) and colourdiffif not. Again by Ramsey, there exists an infinite M 2 ⊂M 1 monochromatic for this colouring. We cannot have coloursame: choose i < j < k < l < m < ninM 2 , thenc(jk) =c(in) =c(lm), which is a contradiction as M 2 ⊂M 1 and soc(jk) 6 =c(lm).
SoM 2 is colourdiff. Now 2-colourM 2 (4)by givingijklcoloursameifc(ik) =c(jl) and colourdiffif not. By Ramsey, we have an infinite monochromatic setM 3 ⊂M 2. We cannot have coloursame: choosei < j < k < l < m < ninM 3 , thenc(ik) = c(jm) =c(ln), which is a contradiction asM 3 ⊂M 1 and soc(ik) 6 =c(ln).
SoM 3 is colourdiff. Now 2-colourM (3) 3 by givingijkcoloursameifc(ij) =c(jk) and colourdiffif not. We get an infinite monochromatic setM 4 ⊂M 3. We cannot have coloursame: choosei < j < k < l, thenc(ij) =c(jk) =c(kl), which is a contradiction asM 4 ⊂M 1 and soc(ij) 6 =c(kl).
Now 2-colourM 4 (3)by givingijkcolourleft-sameifc(ij) =c(ik) and colourleft- diffif not. We get an infinite monochromatic setM 5 ⊂M 4. Finally, 2-colourM 5 (3)by givingijkcolourright-sameifc(ik) =c(jk) and colourright-diffif not. We get an infinite monochromatic setM 6 ⊂M 5.
IfM 6 isleft-diffandright-diffthen (ii) holds.
IfM 6 isleft-sameandright-diffthen (iii) holds.
IfM 6 isleft-diffandright-samethen (iv) holds.
We cannot haveM 6 beingleft-sameandright-same: choosei < j < k, then c(ij) =c(ik) =c(jk), which is a contradiction asM 6 ⊂M 4.
Remarks. 1. We could do it all inonecolouring ofN(4)by colouringijklaccording to the partition of 4induced bycon{i, j, k, l}. The number of colours would be the number of partitions of a set of size
( 4
2
)
.
- In the same way, we can show that if we arbitrarily colourN(r)we get an infinite M⊂Nand a setI⊂[r] such that for anyx 1 ,x 2 ,.. .,xr, andy 1 ,y 2 ,.. .,yrinM(r) we havec(x 1 , x 2 ,... , xr) =c(y 1 , y 2 ,... , yr) ⇐⇒ xi=yifor alli∈I.
So in Theorem 4,I=∅is (i),I={ 1 , 2 }is (ii),I={ 1 }is (iii) andI={ 2 }is (iv). These 2rcolourings are called thecanonical colouringsofN(r).
Van der Waerden’s Theorem
Aim 2-coloured, for allm∈Nthere exists a monochromatic arithmetic progression of lengthm(i.e,a+d,a+ 2d,.. .,a+ (m−1)dall the same colour).
By the familiar compactness argument, this is the same as:
Aim′.For allm∈N, there existsn∈Nsuch that whenever [n] is 2-coloured, there exists a monochromatic arithmetic progression of lengthm.
Indeed, if not, for eachnwe would havecn: [n]→ { 1 , 2 }with no monochromatic arithmetic progression of lengthm. Then infinitely many agree on [1], and of those infinitely many agree on [2], etc. Keep going (as before), and hence obtain a 2-colouring ofNwith no monochromatic arithmetic progression of lengthm.
In our proof (of the second form above), we use the following key idea: we show more generally that for allk,m∈N, there existsnsuch that whenever [n] isk-coloured, there exists a monochromatic arithmetic progression of lengthm. Note that proving a more general result by induction can actually beeasier, because the induction hypothesis is correspondingly stronger.
We writeW(m, k) for the smallestnif it exists – a ‘Van der Waerden number’.
Another idea we use is the following: letA 1.. .,Arbe arithmetic progressions of lengthm
- sayAi={ai, ai+di,... , ai+ (m−1)di}. We say thatA 1 ,A 2 ,.. .,Ararefocusedatf ifai+mdi=ffor alli. For example,{ 1 , 4 }and{ 5 , 6 }are focused at 7.
If in addition eachAiis monochromatic (for a given colouring) and no two are the same colour then we say that they arecolour-focusedatf(for the given colouring).
Thus ifA 1 ,.. ., Akare colour-focused arithmetic progressions of lengthm−1 (in ak-colouring) then we have a monochromatic arithmetic progression of lengthm– by asking ‘what colour is the focus?’
Proposition 5∈N. Then there existsn∈Nsuch that whenever [n] isk-coloured, there exists a monochromatic arithmetic progression of length 3.
Note result will be contained in Theorem 6. It is included here for motivation and understanding.
Claim allr 6 k, there existsnsuch that whenever [n] isk-coloured, we have either - a monochromatic arithmetic progression of lengthm, or - rcolour-focused arithmetic progressions of lengthm−1.
We will then be done: taker=kand look at the focus.
The proof of the claim is by induction onr. Forr= 1 we may taken=W(m− 1 , k). So supposer >1. Ifnis suitable forr−1, we will show thatW(m− 1 , k 2 n)2nis suitable forr.
Given ak-colouring of [W(m− 1 , k 2 n)2n] with no monochromatic arithmetic progression of lengthm, we can break up [W(m− 1 , k 2 n)2n] intoW(m− 1 , k 2 n) blocks of length 2 n, namelyB 1 ,B 2 ,.. .,BW(m− 1 ,k 2 n)whereBi= [2n(i−1) + 1, 2 ni]. Each block can be coloured ink 2 nways, so by definition ofW(m− 1 , k 2 n), we can find blocksBs,Bs+t, .. .,Bs+(m−2)tidentically coloured.
NowBscontainsr−1 colour-focused arithmetic progressions of lengthm−1 (by choice ofn), together with their focus (as the length of each block is 2n). Say we haveA 1 ,A 2 , .. .,Ar− 1 colour-focused atf, whereAi={ai, ai+di,... , ai+ (m−2)di}. Now look at the arithmetic progressionA′i={ai, ai+ (di+ 2nt),... , ai+ (m−2)(di+ 2nt)}for i= 1, 2,.. .,r−1. ThenA′ 1 ,A′ 2 ,.. .,A′r− 1 are colour-focused atf+ (m−1)2nt. But {f, f+ 2nt,... , f+ (m−2)2nt}is monochromatic and a different colour, and thus we havercolour-focused arithmetic progressions of lengthm−1.
This completes the induction, the claim, and hence the proof.
What about bounds onW(m, k)? We define theAckermann(orGrzegorczyk)hierarchy to be the sequence of functionsf 1 , f 2 ,... ,eachN→Ndefined by
f 1 (x) = 2x and forn> 1 , fn+1(x) = fn(x)(1) = fn(fn(.. .fn ︸ ︷︷ ︸ x
(1).. .))
E., f 2 (x) = 2x
f 3 (x) = 2 2 ···
2 }
x
f 4 (1) = 2, f 4 (2) = 2 2 = 4, f 4 (3) = 2 2
22 = 65536, f 4 (4) = 2 2
···
2 }
65536 ,....
Say that a functionf:N→Nisof typenif there existcanddwithf(cx) 6 fn(x) 6 f(dx) for allx. Our bound onW(3, k) was of type 3, and onW(m, k) (as a function ofk) is of typem, and so our bound onW(m) =W(m,2) grows faster thanfnfor alln. This is often a feature of such double inductions, and it was believed for along time that perhapsW(m) really does grow this fast. Shelah (1987) found a proof usingonly induction onm, and gave a boundW(m, k) 6 f 4 (m+k). Graham offered $1000 for a proof thatW(m) 6 f 3 (m).
Gowers (1998) showed thatW(m) 622
222 m+ , ‘almost type 2’. The best lower bound known isW(m)> 2 m/ 8 m.
Corollary 7-coloured, some colour class contains arbitrarily long arith- metic progressions.
Remark cannot guarantee an infinitely long arithmetic progressions. For example:
- 2-colourNby colouring 1 red, then 2 and 3 blue, then 4, 5 and 6 red then 7, 8, 9 and 10 blue, and so on; or
- Enumerate the infinitely long arithmetic progressions asA 1 ,A 2 ,A 3 ,.. .(there are only countably many). Choosex 1 , y 1 ∈A 1 withx 1 < y 1 , then choosex 2 , y 2 ∈A 2 withy 1 < x 2 < y 2 , etc. Now colour eachxired and eachyiblue.
Theorem 8 (Strengthened Van der Waerden).Letm∈N. WheneverNis finitely- coloured, there is an arithmetic progression of lengthmthat, together with its common difference, is monochromatic (i., there exista,a+d,a+ 2d,.. .,a+ (m−1)dandd all the same colour).
Proof proof is by induction onk, the number of colours. The casek= 1 is trivial.
We’ll show that givennsuitable fork−1 (i., such that whenever [n] is (k−1)- coloured there exists a monochromatic arithmetic-progression-with-common-difference of lengthm), thenW
(
n(m−1) + 1, k
)
is suitable fork. Indeed, given ak-colouring of [W
(
n(m−1) + 1, k
)
], we know there exists a monochromatic arithmetic progression of lengthn(m−1) + 1 – saya,a+d,a+ 2d,.. .,a+n(m−1)dis coloured red.
Ifdis red, done – look ata, a+d,.. ., a+(m−1)d. Indeed, ifrdis red for any 1 6 r 6 n, done – look ata, a+rd,.. ., a+ (m−1)rd. So we are done, unless{d, 2 d,.. ., nd}is (k−1)-coloured, but then we are done by induction (by definitionofn).
Remarks. 1. Henceforth we do not care about bounds.
- The casem= 2 is known asSchur’s Theorem: wheneverNisk-coloured, we have monochromaticx, y, zsuch thatx+y=z. We can also prove Schur’s Theorem directly from Ramsey’s Theorem. Given ak-colouringcofN, induce ak-colouringc′of n(n large) byc′(ij) =c(j−i), fori < j. By Ramsey, there exists a monochromatic triangle; i. there existi < j < kwithc′(ij) =c′(ik) =c′(jk). Soc(j−i) =c(k−j) =c(k−i), and since (j−i) + (k−j) = (k−i) we are done.
The Hales-Jewett Theorem
LetXbe a finite set. A subsetLofXn(‘then-dimensional cube on alphabetX’) is called aline(orcombinatorial line) if there exists a non-empty setI⊂[n] andai∈Xfor each i6∈Isuch that
L={(x 1 ,.. ., xn)∈Xn:xi=aifori6∈Iandxi=xj∀i, j∈I}.
We callIthe set ofactive coordinatesforL.
For example, in [3] 2 the lines are as follows.
r
r
r
r
r
r
r
r
r I={ 1 }, with lines{(1,1),(2,1),(3,1)},{(1,2),(2,2),(3,2)},{(1,3),(2,3),(3,3)};
r
r
r
r
r
r
r
r
r I={ 2 }, with lines{(1,1),(1,2),(1,3)},{(2,1),(2,2),(2,3)},{(3,1),(3,2),(3,3)};
r
r
r
r
r
r
r
r
r
I={ 1 , 2 }, with line{(1,1),(2,2),(3,3)}.
lineLin [m]n
′ ︷ ︸︸ ︷
❜
♣
♣
r
♣
❜
♣
r
♣
♣
❜
r
❜♣
❜♣
❜♣
f
❜
♣
♣
r
♣
❜
♣
r
♣
♣
❜
r
❜♣
❜♣
❜♣
f
❜
♣
♣
r
♣
❜
♣
r
♣
♣
❜
r
❜♣
❜♣
❜♣
f
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣♣
♣♣
♣♣
∗
[m]n [m]n [m]n [m]n ︸ ︷︷ ︸ identical
But now letL′ibe the line through the point (L−i, L−) with active coordinate setIi∪I (i= 1, 2,.. .,r−1). ThenL′ 1 ,L′ 2 ,.. .,L′r− 1 are colour-focused at (f, L+). These, together with the line through (f, L−) with active coordinate setI, give usrcolour- focused lines. This completes the induction, the claim, andhence the proof.
Ad-dimensional subspaceord-parameter setSinXnis a set of the following form: there exist disjoint non-empty setsI 1 ,I 2 ,.. .,Id⊂[n] andai∈Xfor eachi∈[n]−(I 1 ∪I 2 ∪· · ·∪Id) such that
S=
{
x∈Xn: xi=aifor alli /∈I 1 ∪I 2 ∪ · · · ∪Id, xi=xjwheneveri, j∈Ikfor somek
}
.
For example inX 3 ,{(x, y,2) :x, y∈X}and{(x, x, y) :x, y∈X}are 2-parameter sets.
Theorem 10 (The Extended Hales-Jewett Theorem). Letm,k,d∈N. Then there existsn∈Nsuch that whenever [m]nisk-coloured, there exists a monochromatic d-parameter set.
‘This should bemuchharder than Hales-Jewett, but... ’
Proof (Xd)n– a cube on alphabetXd. Clearly any line in this (on the alphabetXd) corresponds to ad-parameter set inXdn(on the alphabetX), so we can taken=dHJ(nd, k).
LetS⊂Ndbe a finite set. Ahomothetic copyofSis any set of the forma+λSwhere a∈Ndandλ∈N. For example, inN, a homothetic copy of{ 1 , 2 ,... , m}is precisely an arithmetic progression of lengthm.
Theorem 11 (Gallai’s Theorem). For any finiteS⊂Ndand anyk-colouring ofNd, there exists a monochromatic homothetic copy ofS.
Proof={S(1), S(2),... , S(m)}. Given ak-colouringcofNd, define ak-colouringc′ of [m]n(nlarge) byc′
(
(x 1 ,.. ., xn)
)
=c
(
S(x 1 ) +.. .+S(xn)
)
. By Hales-Jewett, there is a monochromatic line, giving a monochromatic homotheticcopy ofS(withλthe number of active coordinates).
Remarks. 1. Or by a product argument and focusing.
- ForS=
{
(x, y) :x, y∈ { 0 , 1 }
}
, Gallai’s Theorem tells us that whenN 2 is finitely- coloured there exists a monochromatic square. Could we haveused Hales-Jewett for 2- parameter sets (on a 2-point alphabet) instead? No, this would only give us a rectangle.
Chapter 2 : Partition Regular Equations
Rado’s Theorem
Schur says: wheneverNis finitely-coloured, there are monochromaticx, y, zwithx+y=z. Strengthened Van der Waerden says: wheneverNis finitely-coloured, there are monochro- maticx 1 ,x 2 ,y 1 ,.. .,ymsuch thatx 1 +x 2 =y 1 ,x 1 + 2x 2 =y 2 ,... ,x 1 +mx 2 =ym.
LetAbe anm×nmatrix with rational entries. We say thatAispartition regular(PR) (overN) if wheneverNis finitely-coloured, there is a monochromaticx∈NnwithAx= 0.
Examples. 1. Schur states that the matrix (1 1 −1) is PR, i. that there existx, y, zof
the same colour such that (1 1 −1)
x y z
= 0.
- Strengthened Van der Waerden states that
1 1 −1 0... 0
1 2 0 − 1... 0
..
.
..
.
..
.
..
.
..
.
..
.
1 m 0 0... − 1
is PR.
- (2 3 −5) is PR: takex=y=z. But what about (2 3−6) ..?
Notes. 1. Note thatAis PR if and only ifλAis PR for anyλ∈Q− { 0 }, so we could restrict our attention to integer matrices if we wished.
We also talk about the equationAx= 0 being PR.
Not every matrix is PR: for example, (2−1) is not PR. If it were, it would say that we could always findn, 2nof the same colour. Clearly false, e. by 2-colouringx∈N with the parity of max{i: 2idividesx}. Indeed, (λ −1) is PR if and only ifλ= 1.
LetAbe anm×nmatrix with rational entries, say columnsc(1),c(2),... ,c(n)∈Qm, so
A=
↑ ↑ ↑
c(1) c(2)... c(n) ↓ ↓ ↓
.
We say thatAhas thecolumns property(CP) if there is a partitionB 1 ∪ · · · ∪Brof { 1 , 2 ,.. ., n}such that
•
∑
i∈B 1
c(i)= 0 ,
•
∑
i∈Bs
c(i)∈
〈
c(j):j∈B 1 ∪ · · · ∪Bs− 1
〉
fors= 2, 3 ,... , r,
where〈 〉denotes linear span overR. (Or, ‘overQ’ – if a rational vector is areallinear combination of some rational vectors then it is also arationalcombination of them.)
Examples. 1. The matrix (1 1−1) has CP: takeB 1 ={ 1 , 3 }andB 2 ={ 2 }.
- (2 −1) does not have CP. Indeed, (λ −1) has CP if and only ifλ= 1.
For the other direction, we begin with the first non-trivial case, namely (1 λ−1).
Lemma 2. Letλ∈Q. Then wheneverNis finitely-coloured, there exist monochromaticx, yandzwithx+λy=z.
Proofλ= 0 it is trivial, and ifλ <0 we may rewrite our equation asz−λy=x. So we may assume thatλ >0. Writeλ=r/swithr,s∈N.
We need to prove that for allk, there exists annsuch that, whenever [n] isk-coloured, there exist monochromaticx,yandzwithx+ (r/s)y=z. We use induction onk. Fork= 1, taken= max{s, r+ 1}and (x, y, z) = (1, s, r+ 1).
Supposek >1. Givennsuitable fork−1, we shall show thatW(nr+ 1, k) is suitable fork. Given ak-colouring of [W(nr+ 1, k)] we have a monochromatic arithmetic progression of lengthnr+ 1, saya,a+d,.. .,a+nrd, all of colourc.
Look atsd, 2sd,.. .,nsd. If anyisdhas colourcthen we are done, as (a, isd, a+ (r/s)isd) is monochromatic. Otherwise, the set{sd, 2 sd,... , nsd}is (k−1)-coloured, and we are done by induction.
Remarks. 1. Very similar to the proof of Strengthened var der Waerden.
- Lemma 2 seems not to have a proof ‘just by Ramsey’, unlike the caseλ= 1 (Schur).
Theorem 3 (Rado’s Theorem for single equations).Leta 1 ,a 2 ,.. .,an∈Q− { 0 }. Then (a 1 a 2... an) is PR if and only if
∑
i∈Iai= 0 for some non-emptyI⊂[n].
Proof.(⇒) is Proposition 1.
(⇐) Given a finite colouring ofN, fixi 0 ∈I. For suitable monochromaticx,yandz, we shall set
xi=
x ifi=i 0 y ifi6∈I z ifi∈I− {i 0 }
.
We require that
∑
aixi= 0, i. that
ai 0 x+
( ∑
i∈I−{i 0 }
ai
)
z+
(∑
i6∈I
ai
)
y= 0,
i. ai 0 x−ai 0 z+
(∑
i6∈I
ai
)
y= 0, since
∑
i∈I
ai= 0,
i. x+
1
ai 0
(∑
i6∈I
ai
)
y−z= 0,
and suchx,yandzdo indeed exist by Lemma 2.
Rado’s Boundedness Conjecture. LetAbe anm×nmatrix that isnotPR. Then there exists a bad colouring – i. ak-colouring with no monochromatic solution toAx= 0. Iskbounded (for givenm, n)? Equivalently, if there isk=k(m, n) such that anm×n matrix ‘PR forkcolours’ is PR?
This is known for 1×3 matrices (Fox & Kleitman, 2006) – 24 colours suffice.
Proposition 4 any matrix with entries inQ. IfAis PR then it has CP.
Proof may assume that all the entries ofAare integers. Let the columns ofAbec(1), c(2),.. .,c(n). For any primep, colourNby givingxthe colourd(x) (recall = the last non-zero digit in basep). By assumption, there exists a monochromaticx∈Znwith Ax= 0 , i.e 1 c(1)+x 2 c(2)+· · ·+xnc(n)= 0. Say all thexihave colourd.
Partition [n] asB 1 ∪ · · · ∪Braccording to the value ofL(xi), as follows:
- i, j∈Bsfor somes ⇐⇒ L(xi) =L(xj)
- i∈Bs, j∈Btfor somes < t ⇐⇒ L(xi)< L(xj)
For infinitely many primesp, say allp∈P, we get thesameB 1 ,.. .,Br.
Considering
∑
xic(i)= 0 performed in basep(forp∈P), we have
(i)
∑
i∈B 1
dc(i)≡ 0 (modp), and
(ii) for alls>2,
∑
i∈Bs
ptdc(i)+
∑
i∈B 1 ∪···∪Bs− 1
xic(i)≡ 0 (modpt+1) for somet,
where we have writtenu≡v(modp) to meanui≡vi(modp) for alli.
From (i), and asdis invertible, we have
∑
i∈B 1 c (i)≡ 0 (modp) for infinitely manyp, and so
∑
i∈B 1 c (i)= 0.
From (ii), for all 2 6 s 6 rwe have
pt
∑
i∈Bs
c(i)+
∑
i∈B 1 ∪···∪Bs− 1
(d− 1 xi)c(i)≡ 0 (modpt+1).
We claim:
∑
i∈Bsc (i)∈〈c(i):i∈B 1 ∪ · · · ∪Bs− 1 〉.
Suppose not. Then there existsu∈Zmwithu·c(i)= 0 for alli∈B 1 ∪ · · · ∪Bs− 1 but withu·
(∑
i∈Bsc
(i)) 6 = 0.
Dot withu:ptu·
(∑
i∈Bsc
(i))≡0 (modpt+1), i.e·(∑ i∈Bsc
(i))≡0 (modp). But this holds for infinitely manyp, sou·
(∑
i∈Bsc
(i))= 0, contradiction.
Letm,p,c∈N. A setS⊂Nis an (m, p, c)-setwith generatorsx 1 ,x 2 ,.. .,xm∈Nif
S=
{m ∑
i=
λixi:∃jwithλi= 0∀i < j, λj=c, λi∈[−p, p]∀i > j
}
,
where [−p, p] ={−p,−(p−1),.. ., p}. SoSconsists of all numbers in the lists:
cx 1 +λ 2 x 2 +λ 3 x 3 +· · ·+λmxm (λi∈[−p, p]∀i) cx 2 +λ 3 x 3 +· · ·+λmxm (λi∈[−p, p]∀i) .. . cxm− 1 +λmxm (λm∈[−p, p]) cxm
the ‘rows’ ofS
‘An iterated arithmetic-progression-with-common-difference, or progression of progressions.’
The case (m, 1 ,1) of Theorem 5 immediately gives
Corollary 6 (Finite Sums/Folkman’s/Sanders’ Theorem).Forx 1 ,.. .,xm∈N, let
FS(x 1 , x 2 ,... , xm) =
{∑
i∈I
xi:I⊂[m], I 6 =∅
}
.
WheneverNis finitely-coloured, there existx 1 ,x 2 ,.. .,xmwith FS(x 1 , x 2 ,... , xm) monochromatic.
Remarks. 1. This extends Schur (m= 2).
- Similarly, by looking at{ 2 n:n∈N}, we can guarantee a monochromatic
FP(x 1 , x 2 ,... , xm) =
{∏
i∈I
xi:I⊂[m], I 6 =∅
}
.
- How about FS(x 1 ,.. ., xm)∪FP(x 1 ,.. ., xm)? Unknown. Not even known form= 2, where we wantx,y,x+y,xythe same colour. In fact, not even known if there exist xy,x+ythe same colour (except forx=y= 2)!
Proposition 7 CP. Then there existm,p,c∈Nsuch that every (m, p, c)-set contains a solution toAx= 0 , i. we can solveAx= 0 with allxiin the (m, p, c)-set.
Proof columnsc(1),.. .,c(n). AsAhas CP, we have a partitionB 1 ∪ · · · ∪Br of [n] such that ∑
i∈Bs
c(i)∈
〈
c(i):i∈B 1 ∪ · · · ∪Bs− 1
〉
for alls,
say ∑
i∈Bs
c(i) =
∑
i∈B 1 ∪···∪Bs− 1
qisc(i),someqis∈Q.
Then for eachswe have
∑
i∈[n]
disc(i)= 0 , where
dis=
0 ifi6∈B 1 ∪ · · · ∪Bs 1 ifi∈Bs −qis ifi∈B 1 ∪ · · · ∪Bs− 1
.
(Note: ends with a 1.)
Givenx 1 ,x 2 ,.. .,xr∈N, putyi=
∑r s=1disxsfori= 1, 2,.. .,n. Then ∑n
i=
yic(i) =
∑n
i=
∑r
s=
disxsc(i) =
∑r
s=
xs
∑n
i=
disc(i) = 0.
SoAy= 0. Now we are done: takem=r, takecto be the lowest common multiple of the denominators of theqis, and takepto bectimes the maximum of the numerators of theqis.
Thencyis in the (m, p, c)-set generated byx 1 ,x 2 ,.. .,xrandA(cy) = 0.
Theorem 8 (Rado’s Theorem).LetAbe a rational matrix. ThenAis partition regular if and only ifAhas the columns property.
Proof.(⇒) is Proposition 4.
(⇐) follows from Theorem 5 and Proposition 7.
Remarks. 1. Given Rado, things like Schur, Van der Waerden, Finite Sums are trivial CP checks.
- From the proof, we see if a matrix is PR for all of the ‘end in basep’ colourings then it is PR forallcolourings. But no direct proof (i. not via Rado) is known.
Theorem 9 (Consistency Theorem).If A, B are PR then the matrix
(A 0
0 B
)
is PR. In other words, if we can guarantee to solveAx= 0 in some colour class andBy= 0 in some colour class then we can guarantee to solve both in thesamecolour class.
Proof by CP.
Remarks. 1. This is not obvious by considerations of PR alone.
- It can be proved directly – but harder.
Theorem 10. WheneverNis finitely coloured, some colour class contains solutions toall PR equations.
Proof not. Then we haveN=D 1 ∪D 2 ∪ · · · ∪Dk, and, for eachi, a PR matrix Aisuch thatDidoes not contain a solution ofAix= 0. Then the matrix
A 1
..
.
Ak
is PR by Theorem 9, but noDicontains a solution to it, contradiction.
Rado’s Conjecture (1933).Say thatD⊂Nispartition regularif it contains a solution to every PR equation. So Theorem 10 says that ifN=D 1 ∪D 2 ∪ · · · ∪Dkthen someDiis PR. Rado conjectured that ifDis PR andD=D 1 ∪D 2 ∪ · · · ∪Dkthen someDiis PR.
This was proved by Deuber (1973). He introduced (m, p, c)-sets and showed thatDis PR if and only ifDcontains an (m, p, c)-set for allm, p, c(as we know). He showed that for allm, p, c, kthere existn, q, dsuch that whenever an (n, q, d)-set isk-coloured, there exists a monochromatic (m, p, c)-set (like our proof of Theorem 5 but replacing Van der Waerden with Extended Hales-Jewett) – thus proving Rado’s Conjecture.
Ultrafilters
Our next aim is the following:
Hindman’s Theorem 1 , x 2 ,.. .∈N, let FS(x 1 , x 2 ,.. .)=
{∑
i∈I
xi:Ifinite, I 6 =∅
}
.
Then wheneverNis finitely-coloured, there exists a monochromatic FS(x 1 , x 2 ,.. .).
This will be our firstinfinitePR structure of equations.
Remarks. 1. Any ultrafilter extending the cofinite filter is non-principal. Also, ifUis non- principal thenUmust extend the cofinite filter – else ifA∈ Ufor some finiteAthen {x} ∈ Ufor somex∈Aby our remark above aboutB∪C
- The Axiom of Choice is needed in some form to get non-principal ultrafilters.
The set of all ultrafilters onNis denotedβN. We define a topology onβNby taking as a base all sets of the form
CA={U ∈βN:A∈ U},for eachA⊂N.
Thisisa base: it is sufficient to check that
⋃
CA=βNand that the intersection of any two of theCAis another of theCA. Plainly
⋃
CA=βN, andCA∩CB=CA∩BasA,B∈ Uif and only ifA∩B∈ U. Thus open sets are of the form ⋃
i∈I
CAi={U ∈βN:Ai∈ Ufor somei∈I}.
NoteβN−CA=CAc, sinceA /∈ Uif and only ifAc∈ U. So closed sets are of the form ⋂
i∈I
CAi={U ∈βN:Ai∈ Ufor alli∈I}.
We can viewNas a subset ofβNby identifyingn∈Nwith the principal ultrafilter ̃natn. Each point ofNis isolated: for{n ̃}is open as{ ̃n}=C{n}. Also,Nis dense inβN: for every non-empty open set inβNmeetsNas ̃n∈CAwhenevern∈A.
Theorem 13. βNis a compact Hausdorff space.
Proof.Hausdorff 6=V, we have someA∈ UwithA6∈ V. But thenAc∈ Vand so U ∈CAandV ∈CAc, andCA∩CAc=∅.
Compact. Given closed setsFi(i∈I) with the finite intersections property (i. all finite intersections are non-empty), we need to show that
⋂
i∈IFi 6 =∅. Assume without loss of generality that eachFiis basic, i. thatFi=CAifor someAi⊂N.
The setsAi(i∈I), also have the finite intersections property: forCAi 1 ∩ · · · ∩CAin= CAi 1 ∩···∩Ainand soAi 1 ∩ · · · ∩Ain 6 =∅.
So we can define a filterFgenerated by theAi:
F={A⊂N:A⊃Ai 1 ∩ · · · ∩Ainfor somei 1 ,... , in∈I}.
LetUbe an ultrafilter extendingF. ThenAi∈ Ufor alli, soU ∈CAifor alli, and so ∩i∈ICAi 6 =∅as desired.
Remarks. 1. If we view an ultrafilter as a function fromP(N)→ { 0 , 1 }, i. as a point of { 0 , 1 }P(N), then we haveβN⊂ { 0 , 1 }P(N). We can check that the topology onβNis the restriction of the product topology, and also thatβNis a closed subset of{ 0 , 1 }P(N), so is compact by Tychonov.
- Why isβNinteresting? It is the largest compact Hausdorff space in whichNis dense. More precisely, for any compact Hausdorff space and f:N→X, there exists a unique continuousf ̃:βN→Xextendingf. (Not hard to check.) βNis called theStone-Cech compactificationˇ ofN.
N
f → X ↓ րfe βN
LetUbe an ultrafilter andpa statement. We write∀Ux p(x) to mean{x:p(x)} ∈ U, and say thatp(x) holds ‘for mostx’ or ‘forU-mostx’.
Examples. 1. ForUnon-principal,∀Ux x >4.
- ForU= ̃nwe have∀Ux p(x) ⇐⇒p(n).
Ultrafilter quantifiers ‘mesh perfectly’ with logical connectives, as follows.
Proposition 14 an ultrafilter, andpandqstatements. Then
(i)∀Ux
(
p(x)andq(x)
)
⇐⇒
(
∀Ux p(x)
)
and
(
∀Ux q(x)
)
;
(ii)∀Ux
(
p(X)orq(x)
)
⇐⇒
(
∀Ux p(x)
)
or
(
∀Ux q(x)
)
;
(iii)∀Ux p(x) is false ⇐⇒ ∀Ux(notp(x)
)
.
Proof={x:p(x)}and letB={x:q(x)}. Then
(i)A∩B∈ U ⇐⇒ A∈ UandB∈ U; (ii)A∪B∈ U ⇐⇒ A∈ UorB∈ U; (iii)A6∈ U ⇐⇒ Ac∈ U.
Note that∀Ux∀Vy p(x, y) is not in general the same as∀Vy∀Ux p(x, y), even ifU=V. For example, ifUis non-principal then∀Ux∀Uy x < yis true (aseveryxhas∀Uy x < y), but∀Uy∀Ux x < yis false (asnoyhas∀Ux x < y).
ForU,V ∈βN, define
U+V = {A⊂N:∀Ux∀Vy x+y∈A} = {A⊂N:{x∈N:{y:x+y∈A} ∈ V} ∈ U} (‘not to be used’)
Example ̃+ ̃n=m ̃+n.
Note thatU+Visan ultrafilter:
Clearly∅ 6∈ U+V.
IfA∈ U+VandB⊃Athen clearlyB∈ U+V.
Suppose thatA,B∈ U+V, i. (∀Ux∀Vy x+y∈A)and(∀Ux∀Vy x+y∈B). Then by Proposition 14(i) twice, we have:∀Ux∀Vy(x+y∈Aandx+y∈B). And thus:∀Ux∀Vy(x+y∈A∩B), i.e∩B∈ U+V,as required.
Suppose thatA6∈ U+V, i.e
(
∀Ux(∀Vy x+y∈A)
)
.
Then by Proposition 14(iii) twice, we have:∀Ux∀Vy(notx+y∈A). And thus:∀Ux∀Vy(x+y∈Ac), i.e∈ U+V, as required.
Next, note that + :βN×βN→βNis associative. Indeed, for anyU,V,W ∈βN,
U+ (V+W) = {A⊂N:∀Ux∀Vy∀Wz x+y+z∈A} = (U+V) +W.
Ramsey Theory 2011-2012 Lecture Notes & Example Sheet
Module: Ramsey Theory
University: University of Cambridge
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