PARTITIONS AND SUMS AND PRODUCTS OF INTEGERS

transactions of theamerican mathematical societyVolume 247, January 1979

PARTITIONS AND SUMS AND PRODUCTS OF INTEGERS

BY

NEIL HINDMAN1

Abstract. The principal result of the paper is that, if r < u and (A¡)i<r is a

partition of u, then there exist i < r and infinite subsets B and C of to such

that 2 F e A¡ and IIG G A¡ whenever F and G are finite nonempty subsets

of B and C respectively. Conditions on the partition are obtained which are

sufficient to guarantee that B and C can be chosen equal in the above

statement, and some related finite questions are investigated.

1. Introduction. The finite sum theorem [6, Theorem 3.1] states that if r < w

and {A¡}i<r is a partition of w, then there are some /' < r and some infinite

subset B of w such that 2 F E A¡ whenever F is a finite nonempty subset of

B. It is a trivial consequence of that theorem that the corresponding state-

ment holds with "SF" replaced by "IIF". Indeed, let C, = [x: 2X G A,} for

each i < r. Pick i < r and an infinite subset D of w such that ~2F G C¡

whenever F is a finite nonempty subset of D. Let B = {2X: x G D). (To this

author's knowledge, this consequence was first observed by R. Graham.)

A natural question thus arises. Namely, if r < w and {A¡}i<r is a partition

of w, can one always find i < r and an infinite subset B of w such that

S F E A¡ and JIF G A¡ whenever F is a finite nonempty subset of Bl

Numerous stronger versions also suggest themselves.

One of the strongest versions has as its conclusion that all iterated sums

and products which can be written without repetition of terms lie in A¡. (This

would include expressions like (x, + x6 + x3)(x7x8 + x2(x4 + x9)), but not

like x2 or xxx2 + x,x2x3. For an investigation of sums with repeated terms,

see [7].) As far as this author is aware, no counterexample has been found for

even this strong version.2

A natural candidate for a counterexample to the main question would be a

partition of w with the property that any infinite set with sums in one cell

would have to come from a particular cell of that partition and any infinite

Presented to the Society, January 5, 1978; received by the editors July 11, 1977 and, in revised

form, January 30, 1978.

AMS (MOS) subject classifications (1970). Primary 05A17, 10A45; Secondary 54D35.Key words and phrases. Partitions, ultrafilters, sums, products.

"The author gratefully acknowledges, support received from the National Science Foundationunder grant MCS 76-07995.

1 Added in proof. The author has recently answered the weaker question above in the negative.

© 1979 American Mathematical Society

0002-9947/79 /0000-0009/$ 06.00

227

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

228 NEIL HINDMAN

set with products in one cell would have to come from some other cell. The

principal result of this paper establishes that no such partition exists.

The proof of the main result borrows heavily from Glazer's proof of the

finite sum theorem ([5], see also [3]). (The finite sum theorem has also been

proved by Baumgartner [1]. See [2] for a presentation of Baumgartner's

argument to the effect that this theorem is a theorem of ZF, even though all

three known proofs use choice.)

§2 consists of a proof, using Glazer's methods, of the existence of a

particular ultrafilter on w, a derivation of the main result from the existence

of that ultrafilter, and an investigation of questions related to the existence of

special ultrafilters on w. In §3 we present conditions on a partition of w which

are sufficient to guarantee the existence of one infinite set with all sums and

products in one cell.

There are also several finite versions of the sums and products problem.

The weakest of these has been established by R. Graham (and verified by the

present author) with the aid of a computer. Namely, if {1, 2, 3, ... , 252} =

A0 u Av then there exist i < 2 and distinct x may such that {x,y, x + y, x

■y) G A¡. This result is "sharp"; the statement is not true if 252 is replaced by

251. §4 consists of an investigation of related finite problems.

We write w for the first infinite ordinal. Recall that each ordinal is the set

of its predecessors so that, for example, 3 = {0, 1,2}. Unless otherwise stated

lower case variables range over w. By 9f(A) we mean the set of finite

nonempty subset of A. Thus 9f(A) = [A]<" \ {0}.

2. Multiplicative idempotents in jSw. By ßu we mean the Stone-Cech

compactification of the discrete space w. The points of /?w are the ultrafilters

on w (where the fixed ultrafilters-those with nonempty intersection-are iden-

tified with the points of w). If A G w, then cl^ A = [p G ßu: A G p}. If

p G ßu>, then a basic neighborhood system for/? is (cl^ A : A G p). We will

frequently use the fact that if A G 9(u>) and A has the finite intersection

property then A G p for some p E ßu. For further development of the

Stone-Cech compactification see [4].

In Glazer's proof of the finite sum theorem, he introduced the notion of the

sum of two ultrafilters p and q on w, agreeing that A G p + q if and only if

[x: A — x G p) G q, (where by A — x is meant {y: y + x G A}). He then

showed that there exists an additive idempotent in ßu \ w, thus answering

affirmatively an unpublished question of Galvin.

From the existence of such an ultrafilter, that is an ultrafilter with the

property that [x: A — x G p) G p whenever A G p, the finite sum theorem

follows easily. Previously, the existence of such an ultrafilter had been a

continuum hypothesis [8] or Martin's axiom [Alan Taylor, unpublished]

consequence of the finite sum theorem.


PARTITIONS AND SUMS AND PRODUCTS OF INTEGERS 229

We define here the analogous notion of the product of two ultrafilters, and

show that there exists a multiplicative idempotent with the property that each

member includes an infinite set and all of its finite sums.

2.1 Definition. Let/? and q be members of ßu and let A G w.

(a) For any x, A/x = {y.yx G A};

(b)A G p ■ q if and only if [x: A/x G p) G q.

Lemma 2.2 is well known; its proof may be found in [3].

2.2 Lemma. Let X be a nonempty compact Hausdorff space and let ■ be an

associative operation on X with the property that, for each x in X, the function

Lx defined by the rule Lx(y) = x • y is continuous. Then there is some x in X

such that x ■ x = x.

The following lemma is due to Glazer; its proof is included for com-

pleteness.

2.3 Lemma (Glazer). The operation • is an associative operation on ßu and,

for each p in ßu, the function Z^,, defined by Lp(q) = p ■ q, is continuous.

Proof. The verification that • : ßu X ßu -> ßu, i.e. that p • q is an ultrafil-

ter on w whenever/? and q are, is routine. To see that • is associative, let/;, q,

and r be in ßu and let A G w. Note that, for any x, {z: A/z G p}/x = {y:

A/(xy) G p). Thus

A G (p ■ q) ■ r <r* (x: A / x G p ■ q} Er

~{x: {y:(A/x)/y G p) G q} G r

~{x: {y:A/(xy)Gp} G q) G r

<h>{x: [z: A/z Gp)/x G q] G r

<r+ (z: A/z G p) E q ■ r

<Â Gp- (q-r).

Now let p G ßu. To see that Lp is continuous, let q G ßu and let A Gp- q

(so that cLju A is a basic neighborhood of p • q). Let B — [x: A/x G p). By

the definition of p ■ q, B G q and consequently cl^ B is a basic neighborhood

of q. If r G c\ßu B, then [x: A/x G p) G r so that p • r G cLju A. Thus

Lp[c\ßuB] G clßu A.

One gets immediately from Lemmas 2.2 and 2.3 the existence of a multi-

plicative idempotent in ßoi \ w (using the easily established fact that • : ( ßu \ w)

X ( ßu \ w) -» ßu \ w) and hence a direct proof of the finite product theorem

(without appeal to the finite sum theorem). The result which we now seek is

the existence of a multiplicative idempotent in a particular subspace of

ßu \ u.


230 NEIL HINDMAN

2.4 Definition, (a) IfAGu, then FS (A) = (2F: F E 9f(A)} and FP(A)

= {ÏÏF: F E 9f(A)};(b) r = [A G u: There exists B G [Af such that FS(B) G A };

(c) r = {/,6^/>ç T}.

2.5 Lemma. T is a closed nonempty subset of ßu\u and ■: F X T->T.

Proof. That T =£0 is a consequence of the finite sum theorem and

Theorem 2.5 of [8]. To see that T is closed, let/) G ßu\T and pick A G p\T.

Then clßa A n f =0. That F G ßu\u follows from the fact that every

member of a member of T is infinite.

To see that • : f X T -+ T, let p and q be in T and let ^4 Gp- q. Then [x:

A/x Gp} G q so {x > 0: A/x E p) ¥=0. Pick x ^ 0 such ihdXA/x G p.

Since p G T, there exists ß E [A/xf such that FS'(fi) Ç A/x. Let C =

{xy: y G B). Then FS(C) G A so that y4 E T. Consequently p ■ q E T as

desired. (Notice that we have in fact shown that ■: T X (ßu\ {0})-»I\)

2.6 Theorem. Leí {/!,},<r ¿>e a partition of u. There exist i < r and sets B

and C in [A¡f such that FS(B) G A¡ and FP(C) G Af.

Proof. By Lemmas 2.3 and 2.5, there existsp in T such that/? • p = p. Pick

such p and pick i < r such that A¡ G p. Since p GY, there exists B G [A¡]a

such that FS (B) G A¡.

Since p-p = p, we have that whenever A Gp, [x: A/x Gp) Gp and

hence A n {x: A/x G p) G p. Let x0 E A¡ such that A¡/x0 Gp and let

D0 = A(, Inductively, for n > 0, let D„ = £>„_, n (i>„_,/x„_,) and pick

x„ G Dn n {x: Z)„/x E />} such that x„ > x„_,.

We claim that if min F > n, then UkfEF xk G D„. If \F\ = 1, this is trivial

since the sequence (Dn}n<a is nested. Assume inductively that the result

holds for sets smaller than F and let m = min F. Let G = F \ {m}. Then

min G > m + 1 so HkeG xk G Dm+}. Since Dm+, ç I>m/xm, we have

RkeF*k G A« £ A- Let c = {*«: « < «}■ We thus have that, if F E

9f(C), then HF G A¡.

Note that if p G ßu \ u and/) + p = p, then we have as in Theorem 3.3 of

[8] that p G T. In the event that the multiplicative idempotent obtained in the

proof of Theorem 2.6 is also an additive idempotent, one has for any partition

{Aj)j<r of w an infinite subset of one cell all of whose sums and products are

in that cell. (In fact a considerably stronger conclusion holds-see Theorem

2.13). Corollary 2.11 tells us in fact that each member of T is "close" to an

additive idempotent, that is that T = c\ßa {/> E ßu\u: p + p = p).

The author's original proof of Corollary 2.11 used the continuum hypothe-

sis. We are grateful to K. Prikry and F. Galvin for permission to present their

unpublished results (Theorem 2.8 and Corollary 2.9) which remove the need



for an appeal to the continuum hypothesis for this result. (The proof of

Theorem 2.8 presented here is this author's.)

2.7 Definition. Let <x„>n 0,

x„ > 1,k<n xk. The natural map t for FS({xn: n < w}) is defined by the rule

T(2neJ, x„) = 2ne/r 2" whenever F E 9f(u).

Note that, if x„ > '2k<n xk for n > 0, F and G are in ^(w), and 2„e/- x„

= 2„eG x„, then F = G. Thus t is well defined.

2.8 Theorem (Prikry). Let <xn>„<u be a sequence in u such that for any m

and n, if 2m < x„, then 2m+1|xn + 1. Let t be the natural map for FS({xn:

n < u}). Let p G ßu \ w such that p + p = p and let q = [A G u: There

exists B G p such that t~l[B] G A). Then q G ßu \ u and q + q = q. (Thus,

loosely, inverse images of idempotents are idempotents.)

Proof. (Note that the requirement that/? G ßu\u serves only to eliminate

the event that p = {A G u: 0 G A}.) The verification that q G ßu\u is

routine, and we omit it. (One needs to notice that t is one-to-one and onto

w\{0}.)

To see that q + q = q, let A G q. It suffices, by Lemma 3.1 of [8], to show

that there is some x G A such that A — x G q. Pick B Gp such that

t~1[B] G ,4. Pick .y E B such that y > Oand B - y G p. Letx = r'\y).

Pick m such that y < 2m, and let C = [2m ■ k: k < u). Note that C Gp.

(Otherwise, since w = (J t<y« C + t, there is some t with 0 < / < 2m such

that C + t G p. But if z G C + t, then (C + t) n ((C + t) - z) =0.) Let

D = C n (B - y). Then D G p. We claim that r~l[D] G A - x. Let z G

t"'[£>]. Then

t(z) £ C D (B - y) = C n (B - t(x)).

Since t(z) G B — t(x), t(z) + r(x) G B. Now z = 2nef x„ and x =

2„eG x„ for some F and G in 9f(u). Since t(z) E C and t(z) = 2„ei- 2", we

have min F > m. Since 2m > t(x) = 2n6G 2", we have max G < m. Thus

t(x) + r(z) = t(x + z). Thus t(x + z) E 5, so x + z E A. That is z E A

— x as desired.

2.9 Corollary (Galvtn). Let A G [w]". FAere ex/sta p G ßu\u such that

p + p = p and FS(A) G p.

Proof. By Lemma 2.3 of [8], there is a sequence <x„)n<w such that

FS({xn: n < u})G FS(A) and for any m and n, 2m+l|x„+, whenever 2m <

x„. By Theorem 2.8 there exists q G ßu \ u such that q + q = q and FS({x„:

n < u}) E q. Since FS({x„: n < u}) G FS (A), we have FS(A) G q.

2.10 Corollary. \{p G ßu: p + p = p}\ > c.


232 NEIL HINDMAN

Proof. Choose inductively a sequence <x„>n<<<, in w such that, for each

n > 0, xn > 2yt<„ xk. By Theorem 7 of [9] we may choose a sequence

<â>„<c in [u]a such that, if a < 8 < c, then \Aa n As\ < w. For each a < c,

let B„ = (x„: n G A„). Let a < 8 < c. We claim that \FS(B„) n FS(BS)\ <

u. Let x E FS(BS) n F5(Bä) and pick F E (?/(^(7) and G G 9f(As) such

that

x = 2 -*» an(l * = 2 Xn-

Since, for « > 0, x„ > 2fc<„ x^, we have F = G. Thus

|FS(/?„) n FS(BS)\ =\9f(Aa n ¿a)| < w

as desired.

Pick, via Corollary 2.9, for each a < c some pa G ßu\u such that /?„ + pa

= p„ and FS(Ba) G p„. Let a < 8 < c. Since |FSCB„) n FS(BS)\ < u, we

have pa¥=ps.

2.11 Corollary. F = clßu{p G ßu\u: p + p = p).

Proof. As we have already remarked, if p G ßu \ u and /?+/?=/?, then

p G F. Now let c7 E F and let A G q. Then A G T so pick B G[Af such that

FS(B) G A. Pick, via Corollary 2.9, p G ßu \ u such that p + p — p and

FS(B) G p. Then A Gpsop G clßuA.

We now proceed to show, in Theorem 2.13, that in the event that one has

some p G ßu \ u such that /?+/?=/? and p ■ p = /?, a very strong conclusion

holds.

We introduce now a notion of "finite sums and products", FSPX. Another

notion will be introduced later. U A G u, P is a polynomial in the variables

{y„)„EA, and (xn)n<EA is a sequence in w, we let s(P) = {«: yn occurs in P]

and let, as usual, F«x„>„e/j) denote the number obtained by replacing, for

each n G A, each occurence of yn with xn.

2.12 Definition. Let A G u, let (y„}n(EA be a sequence of variables, and let

<x„>„ey4 be a sequence in w.

(a) Let R0(A) = {yn: n G A). Define inductively, Rn+l(A) = R„(A) u [ym

+ P: P G R„(A) and m < min s(P)} u {ym • P: F E R„(A) and m <

min í(F)}. Let R(A) = Un<„ Ä„(^l).

(b)FS7>,«x„>„6J = {P((x„yneA): P G R(A)}.

An example of a member of FSPi({xn}n<u) is

x2 + x5 + x6(x,0(x,,(x,3 + x,5 + x20))).

2.13 Theorem. If there exists p E ßu \ w such that p + p = /? • p = p, then

whenever {A¡}i<r is a partition of u into finitely many cells, there exist i < r

and an increasing sequence <xn)n<ÙJ such that FSPl({x„)n<u) G A¡.



Proof. Let p G ßu\u such that p+p=p-p=p and let {A¡}¡<r be a

partition of w. Pick /' < r such that A¡ G p. Note that, for each A G p,

A n {x: A — x Gp} n [x: A/x G p) G p.

Let B0 = A¡ and pick x0 E B0 such that B0/x0 G p and B0 — x0 E /?. Define

inductively for n > 0,

Bn = £„-, n (fi„-,/x„_,) n (£„_, - x„_,)

and pick x„ > x„_, such that Bn/x„ Gp and Bn - x„ G p. As in the proof of

Theorem 2.6, one gets that F5,F,«xn>n<u) G A¡.

It is perhaps reasonable to expect that one might be able to show by

invoking the continuum hypothesis, in a fashion similar to the proof of

Theorem 3.3 of [8], that the implication of Theorem 2.13 can be reversed. We

have not in fact been able to accomplish this.

Since T = clßU{p G ßu\u: p +/?=/?} and there exists/? E T such that

P ' P = />> one might hope that any such p would also have the property that

/?+/?=/?. Corollary 2.16. shows that this is not the case. Part (a) of the

following theorem may be of use in finding some simultaneous idempotent.

2.14 Theorem, (a) Ifp G ßu \ u,p + p ^ /?, andp ■ p = /?, then there exists

a sequence (^n)„¥=An+1 G An; and

(3) for each n there exists m such that, for each x G Am, there exists r with

Ar G Ajx.

(b) If (,An}n<u is a sequence in T satisfying conditions (1), (2), and (3) above,

then there exists p G T such that p +/? ¥= p, p -p =/?, and {An: n < w} G p.

Proof, (a) Since/? + /? ¥= p, there exists A Gp such that {x: A — x G/?}

£ p. Let A0 = A n {x: A - x E /?}. Then, for any x E A0, A0 — x & p.

Enumerate A0as {x„: n < u). Let C0 = {y: A0/y G /?} and enumerate C0 as

{yo.n: n < u}- Note that A0 G p and C0 E /?. Inductively let

An+i = knfl (Ak/yk,„-k) nC„\\(A0- xn).

Note that An+l Gp, let C„+, = {y: A„+l/y Gp), and enumerate Cn+1 as

{yn+\.k- k < w}. Note that C„+1 G p.

To verify condition (1), note that (A0 — xn) n An+X =0. Condition (2) is

trivial. To verify condition (3), let n be given and let m = n + 1. Let x E Am

and note that x G C„ so that x = ynt for some /. Let r = t + n + I. Then

Ar G Ajynr_l_n = A„/y„t.

(b) With (A„)n<u as given in the hypothesis, let A = {/? E T: [A„: n < u}

Gp}. The proof that A ¥=0 proceeds exactly like the proof of Theorem 2.5 in


234 NEIL HINDMAN

[8]. If q G T \ A, then for some n, An $ q and hence clßu(u \ An) is a

neighborhood of q missing A. Thus A is closed in ßu.

We claim that • : A X A -» A. To this end, let /? and q be in A and let n be

given. Pick m such that, for all x E Am, there exist r with Ar G A„/x. We

claim that Am G [x: A„/x Gp}. Let x G Am and pick Ar such that Ar G

An/x. Since Ar G p, An/x G p. Since Am G {x: An/x G /?} and Am G q, we

have {x: An/x E /?} E q. Thus, A„ G p ■ q.

We thus have, by Lemmas 2.2 and 2.3, that there exists p in A with

p • p = p. By condition (1), we have (x: A0 — x G /?} n A0 =0 so that

A0 E p + p, and hence/? + p ¥= p.

The following notion of "finite sums and products" will be used in the

proof of Corollary 2.16.

2.15 Definition. Let A Gu and let (,xn)neA be a sequence in w.

(a) FSP2((xn}neA) = {2Geg n,eG xk: § G 9f(9f(A))}.

(b) SP«x„)neA) = {2Ge8n*6G xk: S E 9f(9f(A)) and whenever g =

9 u % with 9 and % nonempty (U 9) n (U %) ¥=0).

As an example, let § = {{1, 2}, (1, 3}, (4, 5}}. Then

¿j 11 Xj. = x,x2 + x,x3 + x4x5.

Thus x,x2 + x,x3 + x4x5 E FSP2((x„)„<a). It is not (on its face) a member

of 5F«xn>n<J since (U {{1, 2}, {1, 3}}) n (U {{4, 5}}) =0. (Of course it

might happen that x,x2 + x,x3 + x4x5 = x,x6.)

The idea of the following proof is to start with a sequence (x„)n<a such

that the expressions in the definition of FSP2((x„)n<u) are unique. Then,

using the notions of FSP2 and SP, we obtain sets (,An)„<a satisfying the

conditions of Theorem 2.14.

2.16 Corollary. There exists p in T such that p-p = p andp + p ¥= p.

Proof. For each n, let x„ = 22", and note that the expressions in the

definition of FSF2«x„>n<u) are unique. That is, if 2GeglUeGx¿ =

~2Fe9JlkeFxk, then § = 9. (To see this, note that 2G6§ ü^c x*. =

2, <= s 2') and the members G of § can be read off the binary expansions of

the members t of S.)Let {■£>„}«<« be a partition of w into infinite sets, and for each n, let

En = Uk>nDk. For each n, let An = (J k>n FSP2((xt)lsDi) u SP((x,\eEi).

We claim that <.4„>„<u is a sequence in T satisfying conditions (1), (2), and

(3) of Theorem 2.14.

That each An G T follows from the fact that

FS({x,: t G Dn}) G FSP2«Xí)l(EDn).

That condition (2) holds is trivial.



To see that condition (3) holds, let n be given, let m = n, and let x G Am.

Now

U FSF2«x,>/eZ)J C FSP2«x,>,e£Jk>n

and

SP{<xt\eEm)QFSP%{{x,\sE)

so x E FSP2((x,yieEn). Pick 9 G 9f(9f(En)) such that x = 2f65f UkeF xk.

Let a = max U 9 and pick r such that a < min Fr.

We claim that Ar G A„/x. To this end let_y E Ar. Then, as above, we may

pick g E 9j(9f(Er)) such that y = 2Geg 11*eG xk. Since min £r > a, we

have that F n G =0 whenever F G 9 and G E g. Thus we have that

x-y= 2 II xk.(f,c)efx§ ke(FuG)

We also have that, if 9i and S are nonempty sets and 9 X § = 9v I) §,

then

( U (FuG)jni U (FuG)]^0.

Thus x-y G SP((xk}keEJ so that.y E A„/x as desired.

Finally we verify that condition (1) of Theorem 2.14 holds. Let x G A0 and

pick 9 G 9f(9f(E0)) such that x = 2Fe$]lk(EF xk. Let a = max U 9 and

pick n such that a < min En. To see that An n (/10 — x) =0> Ie* .V e ^«- We

claim thatj> + x E A0.

Pick g E 9f(9f(E„)) such that .y = 2G6g H.keG xk. Then x + y =

2feg-u8nAe/-xÄ. Since max (J 9 <mm (J §, we have (Uf) n (L)g)

=0. Since, as remarked at the beginning of this proof, expressions of the

form 2Feûg TlkeF xk are unique, we have x + y E SP((txkykeE^. Since

min F„ > max U 5", we have that for no & is U (9 u g ) Ç Z)fc. Thus, again

using the uniqueness of the expression of x + y, we have

x+yZ (J reF2«x,>,eßJ.*>o

Invoking Theorem 2.14 (b), the proof is complete.

We close this section with some results and remarks about the operations

+ and • on ßu. (For other information see [5].) This information is of interest

in the context of this paper primarily because these operations provide what

currently appears to be the most promising attack on the main problem.

Since + and • extend the ordinary operations on w, one might expect them

to be commutative and expect the distributive law to hold. In fact none of

these conclusions is valid, as we see in Theorems 2.17, 2.19, and 2.20.


236 NEIL HINDMAN

The following theorem answers a question of Galvin (in a private

communication). The idea of the second part of the proof is, given p G ßu\

u, to construct a countable subfamily [Bn: n < u} of/? and nearly comple-

mentary sets D and E such that, if r and q are in ßu, [B„: n < u} u {D} G

r, and {Bn: n < u} u {E} G q, then r + q =£ q + r.

2.17 Theorem. The center of the semigroup (ßu, +) is u.

Proof. Recall that we have identified each element n of w with the

ultrafilter [A G u: n G A}. For the purposes of this proof, denote [A G u:

n G A} by n*.

Let n G u and let/? E ßu. We show first that n* + p =/? + n*. Since both

n* + p andp + n* are ultrafilters it suffices to show that n* + p Gp + n*.

To this end, let A G n* + p. Then {x: A — x G n*} G p. But, for any x,

A — x G n* if and only if n + x G A. Thus (x: A — x E «*} = A — n.

Thus A — n G p so n G {x: A — x G /?} and hence A G p + n*.

Now let /? E ßu \ u and, for each r, pick ir with 0 < ir < 2r such that

[2r • k + ir: k < u} G p. (We can do this since, for each r, u = (J ,<2'{2r ■ k

+ i: k < u}.) For each r, let Br = {2r • k + ir: k < u) and note that Br+l G

Br (since otherwise Br+l n Br =0). Thus, for each r, either i'r+, = ir or

Wl-'r + y-We claim that, if 2rQ.k + 1) + 2/r = 2*(2r + 1) + 2is> then s = r and

A = /. We assume that r < s. If f, = is, then 2r(2A: + 1) = 2J(2/ + 1) and the

claim is established. Otherwise r < s and there is some F G {n: r < n < s}

such that is = ir + 2„ef 2". Then

2r(2k + 1) + 2ir = 2*(2/ + 1) + 2Íir + 2 2")V nef /

so that 2r(2A: + 1) = 2r+1(2i-'-r(2/ + 1) + 2„e/- 2"~r) so that 2r+1 divides

2r(2k + 1), a contradiction.

Let A = (22r(2A: + 1) + 2/2r: {r, A:} G u} and note that, if x = 22r+1(2A: +

1) + 2/2r+1 for some r and k, then x & A. For each r, let Cr = Br\ Br+l, let

-D = Ur<u C2r and let E = Ur<u C2/.+ i. In the event that there is some i

such that, for all sufficiently large r, i = ir one has w\(D U E) = {/'}.

Otherwise, w = D u E. In any event, since p G ßu \ u, either D G p or

F E/?.

In the event that D Gp, pick q G ßu such that (5r: r < w} u {F} Ç q.

In the event that E G /?, pick q G ßu such that (fir: r < «} u {0} Ç ?•

We show now that, if/? and q are in ßw, {Br: r < u} \j {D} G p, and {£,.:

r < w} U {E} G q, then A G (q + p) \ (p + q) so that p + q ¥= q + p. Let

x G D and pick r such that x E C2f. We claim that B2r+, G A — x. To this

end let >> E B2r+l, so that 7 = 22r+1 • k + i2r+i for some k. Now x E B2r \



B2r+l. Thus, if i2r+1 = i2r, then

x = 22r+1-i + /2r + 22r

for some s and y = 22r+1 • k + i2r. If /2r+, = i2r + 22r, then x = 22r+1 • j +

i2r for some j and/ = 22r+1 • A + /2r + 22r. In either event we have

x+y = 22r(2(k + s) + 1) + 2/2r

so that x + y G A. Thus we have, since B2r +, E 17, that A — x G q whenever

x E D. So that D G {x: A — x G q} and hence A G q + p.

By a similar argument we can show that, if x E C2r+1, then B2r+2 n (A -

x) =0. Thus, E fi {x: ^4 - x E /?} =0 and heneen E /? + q, as desired.

We abuse terminology slightly and present the following, which is a

corollary to the proof of Theorem 2.17, rather than a corollary to the theorem

itself.

2.18 Corollary. There exist sets A„ and A2 such that F = A, u A2 and

p + q ¥= q + p whenever /? E A, a/u/ q G A2.

Proof. If /? E F, then (2r • A: A: < w} 6 /? for each r. Let .4 = {22'(2A: +

1): {r, A} Ç w} and let A, = {/? E F: A G /?} and A2 = F \ A,. Let /? E A,.Then in the proof of Theorem 2.17, we have each ir = 0 so that A = {22r(2A

+ 1) + 2/"2r: {r,k} G u} as in that proof. The set D chosen in that proof is

then equal to A. Thus A2 = {q G F: {Br: r < w} u {E} G q}.

2.19 Theorem. There is an isomorphism y from (ßu, + ) into (ßu, ■ ).

(Consequently (ßu, ■ ) is not commutative.)

Proof. Define y by agreeing that for/? E ßu, A G y(p) if and only if (x:

2X G A} G p. (The verification that y(p) is an ultrafilter is routine, as is the

verification that y is one-to-one.)

Let p and q be in ßu. Since y(p + q) and y(p) ■ y(q) are ultrafilters, it

suffices to show that y(p + q) G y(p) • y(q). Let A G y(p + q) and let

B = {x: 2X G A). Let C = {x: B - x Gp}. Since B G p + q, C G q. Let

D = [2X: x G C}. Then D G y(q). We claim that D G {x: A/x G y(p)}

and hence that A G y(p) ■ y(q). Let x G D and pick/ E C such that x = 2y.

Then B - y G p. Let E = {2Z: z G B - y). Then E G y(p). It thus suffices

to show that E G A/x. Let w G E and pick z G B — y such that w = 2Z.

Then z + y G B so 2z+y G A. But 2z+y = w ■ x so w G A/x.

2.20 Theorem. There exist p and q in ßu such that p-(q + q)¥=p-q+p-

q and (q + q)-p¥= q-p + q-p.

Proof. Let A = {22n(2A + 1): [n, k} G u}. Since FS({22n: n < u}) G A,

we may by Corollary 2.9, pick /? G ßu\u such that A G p and /? + p = p.


238 NEIL HINDMAN

Let q = [B G u: \ G B}. Then/? - q = q-p = p so

p-q+p-q = q-p + q-p=p+p=p.

But t7 + t7={¿?C.w:2E B}, and hence A £ p • (q + q). Since also/? • (q +

q) = (q + q)-p, we are done.

3. Infinite combinatorial results. We present here, in Theorems 3.1 and 3.4,

two conditions on a partition {A¡}i<r of w, each of which guarantees the

existence of some infinite set with all of its finite sums and finite products in

the same cell. Loosely speaking, the condition presented in Corollary 3.2 is

that the partition include arbitrarily long blocks and that the gaps between

blocks of a given length be bounded. The condition of Theorem 3.1 is the

same except that reference is made to blocks of multiples of some fixed

number c. The basic idea of the proof consists of obtaining a sequence, each

of whose finite products is a long distance from the end of the block into

which it falls. Then sums of those products must lie in the same block.

3.1 Theorem. Let {A¡}¡<r be a partition of u. Assume that there are some

c > 0 and some function f from u to u such that, for any y and n, there exist

i < r andz, withy < z < y + f(n), such that {(z + m)c: m < n) G A¡. Then

there exist i < r and B G [uf such that FS(B) G A¡ and FP(B) G A,.

Proof. We assume that, for each / < r, {me: m < w} n A¡ is infinite. (The

set B which we will obtain will be contained in {me: m < u} so that cells

including only finitely many multiples of c may be merged with other cells,

still leaving an infinite subset of B with all finite sums and products in one of

the orginal cells.)

For each / < r, let ai0 = min{w: me G A¡} and let bi0 = max{/w: ai0 < m

and [tc: ai0 < t < m} G A¡}. Inductively, let a,„+1 = min{«i: bin < m and

me G A¡} and let bin+1 = max{m: ain+l < m and {tc: ain+l < t < m) G

A¡). Thus, for each i,

A¡ n {me: m < w} = U {mc: ain < m < bin).fl<to

Let y0 = c and assume we have chosen (yk}k<m satisfying the following

inductive hypotheses.

0)yk e {*: n < w);

(2) if k > 0, then y k > Ut<ky,; and

(3) if F E 9f(k + 1) and j = max F,

then there exist / and n such that cain < Wt<iFyt < cbin — s.

(In hypothesis (3), recall that A + 1 = [t: t < k + 1}.) Let d = 2m and let

{/?,: t<d} = FP({yt: t < m}) u {1}, with /?0 = Ut<myr Letuo = 0 and

for 0 < / < d, let u, = m + Pd_t(f(u,_x) + «,_, + 1).

We now do a subsidiary induction, choosing (h,}l<d and <e,>,<d sucri mat.



for each / < d:

(a) if / > 0, then/?0 < h,_x < h, and et < e,_,;

(b) e, = h, + /(«,/_,_,) + «,,_,_,; and

(c) there exist i < r and n such that a, „ < hj>, < ey?, < ¿>, „ — m.

To ground this induction, pick / < r and n such that a, „ > pi and

*.> - ai,n > ud> let Ao = min{A: ain < A/?0}, and let e0 = /i0 + f(ud_,) +

"</-1- Note that h0 > /?„. Condition (a) is satisfied vacuously and condition (b)

and the first two inequalities of condition (c) are satisfied directly. Now

hn - e0P0 - Kn - Po(hO + f(»d-l) + «rf-I )

- ¿>,,„ - Po(K - 1) - />o(/(«rf-l) + "rf-1 + 0

> Kn - ai,n - (Po(f(ud-i) + ud-\ + 1) + m) + m > ud - ud + m,

so that condition (c) holds.

Now let 0 < k < d and assume that we have chosen i.h¡)l<k and (e,y,<k

satisfying conditions (a), (b), and (c). Pick / < r and z such that hk_lpk < z

< hk_lpk + f(ud_k) and {(z + f)c: f < ud_k} G A¡. Pick « such that ain <

z < bin and note that bin — z > ud_k. Let hk = min{/i: z < hpk) and let

ek = K + f(ud-k-\) + "tf-*:-1- Condition (b) and the first two inequalities of

condition (c) are satisfied directly. The verification of the third inequality of

condition (c) is essentially the same as it was with A: = 0 (except that ain is

replaced by z).

Since h0 > p0, h,j)k > z > hk_xpk, and pk > 0, we have /?„ < hk_x < hk.

Also, pkek = Pkhk + pj(ud_k_x) + pkud_k_x < z + pk + pj(ud_k_x) +

Pkud-k-\ < K-\Pk + f("d-k) + Pk(f("d-k-i) + ud-k-\ + O = K-iPk +

f(ud-k) + "d-k - m < pk(hk_x + f(ud_k) + ud_k) = pkek_x. Thus,

condition (a) is satisfied.

The subsidiary induction being complete, let >»m = chd_x. Since hd_x > hQ

> Po = II,<m.y„ we have that hypotheses (1) and (2) are satisfied. To verify

hypothesis (3), it suffices to consider F E 9f(m + 1) such that m = max F.

Let /? = 1 if F = {m} and otherwise let /? = n/eiA{m}.y,. Then /? = pk for

some k < d. Pick / < r and n such that ain < A*/?* < ekpk < Z?, n — m. Since

K < ''¿-i < ed-\ < e*: we nave that a,„ < nd_xpk < 6in — m. Thus

«*,.,„ < chd_xPk < cbin - cm < cZ?,.„ - am.

Since lJ,eFy, = chd_xpk, hypothesis (3) is satisfied, and the induction is

complete.

For each / < r, let C, = {F E 9f(u): TLneFy„ G A,}. Pick, by Corollary

3.3 of [6], / < r and a sequence <F„>„<(<) of disjoint members of ^(w) such

that U„eGF„ E C, whenever G G 9f(u). We may presume, by suitably

thinning the sequence <F„>„<U, that if d = max F„, then minFn+1>


240 NEIL HINDMAN

For each n, let x„ = U.keF yk, and let B = (x„: n < u}. Since, for any

G G 9f(u),UneG x„ = ïï-keunecF„yk>we have immediately that FP(B) G A¡.

To see that FS(B) G A„ let" G Ê 9f(u) with \G\ > 2. Let s = max G, let

m = max Fs, and let ¿/ = max Fs_,. Picky and « such that ca, „ < RkeF, 7* <

c67„ - m. Since n*e/r .y* E 4,, we havey = i.

Now 2,6G x, = 2,eGV{i) EkeFi yk + HkeF¡ yk- Thus 2,eG x, > cajn. Also

2 II yk < IT yk < min F, < m./ëC\(j) tef, *<¿

Thus 2,6G x, < cZ?,„. Since each x, is a muliple of c, 2,eG x, E .4,-, as

desired.

3.2 Corollary. Let {A¡}i<r be a partition of u. Assume that there is some

function f from u to u such that, for any y and n, there exist i < r and z with

y < z < y + f(n) and (z + m: m < «} G A¡. Then there exist i < r and

B E [wf such that FS(B) G A¡ and FP(B) G A,,

It is quite possible that, given a partition [A¡}i<r of w, there is only one

/' < r such that FS(B) G A¡ for some B G [uf. (For example let A¡ = [nr +

i: n < u}.) Theorem 3.4 tells us that, in such a situation, we can get /' < r and

B G [u]u such that FS(B) G A¡ and FP(B) G A¡.

In the following proofs, when we write a = b (mod {A¡}i<r) for a partition

(A¡)i<r of w, we mean a is congruent to b modulo the equivalence relation on

w which is induced by {A¡}¡<r.

3.3 Lemma. Let {A¡)i<r be a partition of u. There is an increasing sequence

<*„>«<« '" wN {°} such ihat> 'f {F, G) G 9f(u), k = min(F u G), and

t < Wt<k x„ then t ■ 2„6f xn = t- 2„eG x„ (mod{^,}/<r).

Proof. For each n, let H0n = [n + 1}. Let q > 1 and assume that for each

p < q we have chosen a sequence (Hpn)n<a in ^(w) satisfying the following

inductive hypotheses.

(1) For each n, max Hpn < min Hpn+X;

(2) iip > 0 and n< /?, then Hpn = Hp_u„ and

(3) if m < /?, {F, G} Ç ^(w), m < min (F u G), and / < m then

'2 2<-'s>'2 2^(mod{^.}/<r).nef nGG

Hypothesis (1) is trivially satisfied for/? = 0 and hypotheses (2) and (3) are

vacuous there.

For F and G in "^(w), agree that F« G if and only if, for all t < q,

/2neF2i/?_,,n = /2„eG27/?_,,n (mod K},<r). There are only finitely

many equivalence classes mod « (at most rq '). Pick, by Corollary 3.3 of [6],

a sequence <Fn>n<l0 of pairwise disjoint members of ^(w) such that, if



{H, K) G 9f(u), then U„eW F„ « q let Hqn = (JkeF„Hq_ u.

Hypotheses (1) and (2) can be easily verified. To verify hypothesis (3), let

m < q, let (F, G} G 9f(u) with m < min(F u G), and let / < m. Let

AT = U {F„: n G F and n > q} u [n G F: n < q} and let L = U {F„: n E

G and n > q} u {n G G: n < q}. Then

2 2tf,.„-2 2 a,-.., and 2 2^=2 2»,-m-nef neK n£G n&L

Assume first that m < q. Then m < q — \ and m < min(Ä^ u ¿) so the

conclusion follows from the fact that the hypothesis (3) holds at q — \.

(Notice that, since t < m, this case can hold only û q > 2.) Now assume

m = q, and note that in this case K = (J „eF Fn and L = (J „6G F„. Since K

and L are congruent mod ?», the hypothesis is satisfied.

Let/(0) = 2 and let x0 = 'ZH22. Inductively, given (xk}k<n, let/(«) = 1 +

llk<n xk and let x„ = 2#/(n)/(„). Now let {F, G} G 9f(u), let A = min(F u

G), and let t < IIi<4 x¡. Let K = {/(«): ai E F}, let L = (f(n): n G G} and

let <7 = max(.TY u L). Note that, by hypothesis (2), for each n G F u G,

xn = 2#<7/(n).Thus

2 *„ = 2 2 #,,„ and 2 ^ = 2 2 ^,n-«ef »eï «eC nez.

Since t < Wi<k x¡, t < f(k) and hence / < min(K u L). Thus, by hypothesis

(3),

' 2 x„ = t 2 ^n (mod {^,},<r).nef b£C

3.4 Theorem. Let {A¡}j<r be a partition of u, let i < r, and assume that,

whenever j < r and B G [uf with FS (B) G A¡, one has j = i. Then there is

some B G [uf such that FS(B) G Ai and FP(B) G A,.

Proof. Let <x„>n 0}.

Now, let F E 9j(u) with min F > 0. Then 1 = min(F U {1}) and 1 <

n* 0 and picky < r such that UneF x„ G

Ay We show thaty = /. If \F\ = 1, this follows from the fact that FS(B) G

A¡, so assume \F\ > 1. Let 17 = max F, let G = F \ {q}, and let t = IIn6G x„.

Then t ■ xq = IInS/- x„ so t ■ xq G A¡.

We claim that FS ({t ■ x„: n > q}) G Aj and hence that j = i. To this end,

let H G 9f(u) with min H > q. Then q = min(FT u {<?}) and / < II*<9 xk


242 NEIL HINDMAN

SO

tx„ m / 2 xn (mod{As}s<r).nEH

Thus 2„e// tx„ G Aj as desired.

4. Finite results. We note first that the finite version of Theorem 2.6 holds.

The proof, which we omit, is by a standard compactness argument, invoking

Theorem 2.6.

4.1 Theorem. For each k and r, there is some n such that, whenever {A¡}i<r

is a partition of n, there exist i < r and B and C in [n]k such that FS(B) G A¡

and FP(C)GA¡.

The natural finite version of the main sums and products problem seeks to

have B = C in Theorem 4.1. The simplest of the special cases (which has any

substance at all) takes A = 2 and r = 2. In this case the result holds, letting

n = 4. (Either the cell with 0 has some other member x, in which case

B = (0, x} will do, or one can take B = {1, 2}.) Wishing to be able to

exclude such trivialities, we make the following definition.

4.2 Definition. p(a, r, k) is the supremum of the set of all n such that {x:

a < x < n} may be partitioned into r cells so that no A>element subset A of

{x: a < x < n} has FS(A) u FP(A) contained in one cell.

The definition of p(a, r, k) was phrased in this negative fashion since, with

few exceptions, it is not known that/?(a, r, k) < u. (We shall present here all

of the nontrivial exceptions which we know.)

R. Graham first showed, with computer assistance (unpublished), that

p(\, 2, 2) = 251. He noted that the presence of 1 was important since if x and

1 lie in the same cell, one cannot have either x + 1 or x — 1 (for x > 2) in

that cell.

We proceeded to investigate/?(2, 2, 2) and independently verified (also with

computer assistance) Graham's result. The partition presented in the proof of

Theorem 4.3 is the one we obtained. It is essentially unique, there being a

total of exactly 147,456 such partitions. (The partition is completely

determined, except on 18 numbers. If one of them (namely 12) is assigned to

the same cell as 1, then three other numbers from the set of 18 are forced into

the other cell. Otherwise they may be assigned to either cell at will, giving a

total of 217 + 214 different partitions with the desired property.)

We have been in something of a dilemma with respect to the proof that

each two celled partition of {x: 1 < x < 252} has some distinct x and_y with

{x, y, x + y, x -y) contained in one cell. The proof only required approxi-

mately two seconds of computer time, and involves only 27 cases. However,

these cases required a minimum of 4 and maximum of 47 steps to settle (with



an average around 20). We have thus tried to compromise between saying

"Trust me and my computer" and presenting all details of the proof. We shall

present the cases, and check one of them. The case we will check is

unfortunately the longest, but it is the only case which involves the number

252. (In fact any 2-celled partition of {x: / < x < 247} which has no distinct

x and y with (x, y, x + y, x ■ y} in the same cell must fall under this case.)

4.3 Theorem (Graham). p(\, 2, 2) = 251.

Proof. Let A0 = {x: x is odd and 1 < x < 5} u ({x: x is even and

8 < x < 250} \ {10, 96, 112, 128, 144, 160, 168, 176, 192, 208, 216, 224, 240})and let

Ax = (x: 1 < x < 251} \A0.

(Thus Ax = {x: x is even and 2 < x < 6} U (x: x is odd and 7 < x < 251}

U {10, 96, 112, 128, 144, 160, 168, 176, 192, 208, 216, 224, 240}.) It is aroutine matter to verify that neither A0 nor Ax contains {x,y, x + y, x • y}

for any distinct x and>\

Suppose now that {A0,AX} is a partition of {x: 1 < x < 252} such that,

whenever x ¥ y and i < 2, {x,y, x + y, x ■ y} ¡Z A¡. Without loss of

generality, 1 E A0. Some one of the following 27 cases must hold: Case 1,

{2, 4, 6, 9, 11} Ç A0; case 2, {2, 4, 6, 9} Ç A0 and 11 E Ax; case 3,

{2, 4, 6, 10} G A0 and 9 E Ax; case 4, {2, 4, 6} G A0 and {9, 10} Ç Ax; case

5, {2, 4, 7, 9, 11} ç A0 and 6 £ Ax; case 6, {2, 4, 7, 9} G A0 and {6, 11} G

Ax; case 7, {2, 4, 7} G A0 and {6, 9} GAX; case 8, {2, 4, 8, 10} G A0 and

{6,1} G Ax; case 9, {2, 4, 8} G A0 and {6, 7, 10} G Ax; case 10, {2, 4, 9} Ç

A0 and {6, 7, 8} G Ax, case 11, {2, 4} G A0 and {6, 7, 8, 9} Ç Ax; case 12,

{2, 5, 7, 9, 11} ç A0 and 4 E 4,; case 13, {2, 5, 7, 9} ç /10 and {4, 11} G

Ax\ case 14, {2, 5, 7} G A0 and {4, 9} £ Ax\ case 15, {2, 5, 8} Ç A0 and

{4, 7} G Ax; case 16, {2, 5} G A0 and {4, 7, 8} ç ¿,; case 17, {2, 6} G AQ

and {4,5} G Ax; case 18, {2, 7, 9} Ç /10 and {4, 5, 6} ç Ax; case 19, {2, 7}

G A0 and {4, 5, 6, 9} G Ax; case 20, {2, 8} G A0 and {4, 5, 6, 7} G Ax; case

21, 2 G A0 and {4, 5, 6, 7, 8} G Ax; case 22, {3, 5} G A0 and 2 E /!,; case

23, {3, 6} G A0 and {2, 5} G Ax; case 24, 3 E A0 and {2, 5, 6} G Ax; case

25, 4 G A0 and {2, 3} Ç /!,; case 26, 5e^0 and {2, 3, 4} £ Ax; case 27,

{2, 3, 4, 5} G Ax. We show here that case 22 cannot hold.

Assume, accordingly, that {3, 5} G A0 and 2 G Ax. Since {1, 3} G A&

4 G Ax. Since {1, 5} G A0, 6 E Ax. Since {2, 4, 6} G Ax, 8 E ^0. Since {1, 8}

Ç ^0, 9 E /I, and 7 £ Ax. Since {3, 5, 8} Ç /*0> 15 e A\- Since {2, 7, 9} G

Ax, 14 E ,40. Since {6, 9, 15} Ç ,4,, 54 E A0. Since {1, 14} G A0, 13 E ,4,.

Since {1, 54} G A0, 53 £ ,4, and 55 E Ax. Since {4, 9, 13} G Ax, 36 E /i0.

Since (6, 7, 13} Ç Ax, 42 E ,40. Since {2, 53, 55} G Ax, 106 E A0. Since


244 NEIL HINDMAN

{1, 36} Ç A0, 35 E Ax and 37 E Ax. Since {3, 14, 42} G A0, 17 E Ax. Since

{1, 42} G A0, 41 E Ax. Since {1, 106} ç A0, 105 E Ax. Since {2, 35, 37} G

AX,10G A0. Since {4, 37, 41} G Ax, 148 E A0. Since {2, 15, 17} ç Ax, 30 E

/10. Since {7, 15, 105} G Ax, 22 G A0. Since {5, 14, 70} G A0, 19 E Ax. Since

{1, 148} G A0, 147 E Ax. Since {1, 30} G A0, 31 E /!,. Since {1, 22} E A&

21 E Ax. Since {2, 17, 19} G Ax, 34 E A0. Since {4, 15, 19} G Ax, 60 E A0.

Since {2, 19, 21} G Ax, 38 E yl0. Since {7, 21, 147} G Ax, 28 E A0. Since

{4, 17, 21} G Ax, 68 E ,40> Since U> 34} Ç ô> 33 e A\- Since {L 60) £ ô>61 E Ax and 59 E .4,. Since {1, 38} G A0, 39 £ Ax. Since {8, 28, 36} G A0,

224 E /4,. Since {1, 68} Ç A0, 67 £ Ax. Since {2, 31, 33} G Ax, 62 E ^0.

Since {2, 59, 61} Ç Ax, 118 E ^0. Since {7, 39, 224} G Ax, 32 E /10. Since

{1, 62} ç A0, 63 E Ax. Since {1, 118} G A0, 119 E Ax. Since {4, 63, 67} ç

Ax, 252 £ /10. Since {7, 17, 119} G Ax, 24 £ A0. Since {14, 32, 252} Ç A0,

18 E Ax. Since {3, 8, 24} ç A0, 11 E .4,. But this means that {2, 9, 11, 18} Ç

Ax, a contradiction.

We have no choice in the following result, except to just present it. It took

about 4 minutes of computer time to show that each two celled partition of

{x: 2 < x < 990} has some distinct x and y with {x,y, x + y, x-y}

contained in one cell.

4.4 Theorem./?(2, 2, 2) = 989.

Proof. As remarked above, we shall simply present a partition {A0, Ax} of

{x: 2 < x < 989} such that if x ¥= y and i < 2, then {x,y, x + y, x-y} is

not contained in A¡.

Let A0 = ({2, 4, 5, 7, 8, 11, 13, 14, 16, 17, 18, 19, 25, 26} u {3A:: 27 < 3A

< 483} u {3A + 1: 253 < 3k + 1 < 484} u {x: 487 < x < 989} u

{166, 193, 218, 220, 245, 299, 326, 353, 380, 407, 434, 461, 764, 818, 860,968}) \ ({6A: + 2: 506 < 6A + 2 < 986} u {126, 144, 382, 409, 432, 436, 491,494, 540, 594, 598, 648, 652, 702, 706, 756, 810, 864, 891, 918, 922, 965, 972,

976}). Let Ax = {x: 2 < x < 989} \ A0. (The curious reader will presumably

test this partition on his own computer.)

It is conceivable that/?(3, 2, 2) is computer accessible. It is unlikely, in fact,

that the time consumption would increase as dramatically as it did between

p(\, 2, 2) and/?(2, 2, 2). It is clear that/?(l, 2, 3) is huge-even if it is finite. (It

is a trivial matter to write down a partition of {x: 1 < x < 10"} into 2 cells

so that no 3 element subset has all sums and products in one cell.) We present

the following lower bound for/?(l, 3, 2) only to show that it too is very large

and almost certainly not computer accessible. In our opinion, the only

reasonable hope for a proof that p(n, r, k) is always finite is that the infinite

version be proved.

4.5 Theorem. p(\, 3, 2) > 11, 706, 659.



Proof. Let A0 = ({3k + 1: 3A + 1 < 16} u {9A:: 18 < 9A < 45} u {3A:

54 < 3A < 3417} u {3A + 2: 3422 < 3k + 2 < 11, 706, 659}) \ {27A:: 486

< 27A < 3402}, let Ax = ({3A: 3 < 3A < 51} u {3A: +1: 19 < 3A: + 1 <11, 706, 658} u {27A: 486 < 27A < 3402}) \ {9A: 18 < 9A < 45}, and let

A2 = {3A + 2: 3A + 2 < 3419} u {3A: 3420 < 3A < 11, 706, 657}. It is a

routine matter to verify that if x ¥^ y and / < 3 then {x,y, x + y, x ■ y) is

not contained in A,.

References

1. J. Baumgartner, A short proof of Hindman's theorem, J. Combinatorial Theory Ser. A 17

(1974), 384-386.2. W. Comfort, Some recent applications of ultrafilters to topology, (Proc. Fourth 1976 Prague

Topological Symposium), Lecture Notes in Math., Springer-Verlag, Berlin and New York, 1978.

3. _, Ultrafilters: Some old and some new results, Bull. Amer. Math. Soc. 83 (1977),

417-455.4. L. Gillman and M. Jerison, Rings of continuous functions, Van Nostrand, Princeton, N. J.,

1960.5. S. Glazer, Ultrafilters and semigroup combinatorics, J. Combinatorial Theory Ser. A (to

appear).

6. N. Hindman, Finite sums from sequences within cells of a partition of N, J. Combinatorial

Theory Ser. A 17 (1974), 1-11.7. _, Partitions and sums of integers with repetition, J. Combinatorial Theory Ser. A (to

appear).

8. _, The existence of certain ultrafilters on N and a conjecture of Graham and Rothschild,

Proc. Amer. Math. Soc. 36 (1972), 341-346.9. A. Tarski, Sur la décomposition des ensembles en sous-ensembles presque disjoints, Fund.

Math. 12 (1928), 188-205.

Department of Mathematics, California State University, Los Angeles, California

90032

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