Maximum subsets of (0, 1] with no solutions to x + y = kz

Maximum subsets of 0, 1] with no solutions to x + y = z Fan R. K. Chung Department of Mathematics University of Pennsylvania Philadelphia, PA 19104 John L. Goldwasser West Virginia University Morgantown, WV 6506 Abstract If is a positive integer we say that a set S of real numbers is -sum-free if there do not exist x, y, x is S such that x + y = z. For greater than or equal to 4 we find the essentially unique measurable -sum-free subset of 0, 1] of maximum size. 1 Introduction We say that a set S of real numbers is sum-free if there do not exist x,y,z is S such that x + y = z. If is a positive integer we say that a set S of real numbers is -sum-free if there do not exist x,y,z in S such that x + y = z we require that not all x,y, and z be equal to each other to avoid a meaningless problem when = ). Let fn, ) denote the maximum size of a -sum-free subset of {1,,...,n}. It is easy to show that fn,1) = n. For n odd there are precisely two such maximum sets: the odd integers and the top half. For n even and greater than 9 there are precisely three such sets: the two maximum sets for the odd number n 1, and the top half. The problem of determining fn,) is unsolved. Roth [R] proved that a subset of the positive integers with positive upper density contains threeterm arithmetic progressions. Szeméredi improved Roth s result to fourterm [S1] and later to -term [S] arithmetic progressions. The current best 1

bounds for fn, ) were established by Salem and Spencer [SS] and Heath- Brown and Szeméredi [H]. Chung and Goldwasser [CG] proved a conjecture of Erdös that fn,3) is roughly n n. They showed that fn,3) = for n 4 and that for n 3 the set of odd integers less than or equal to n is the unique maximum set. Loosely speaing, the set of odd numbers less than or equal to n qualifies as a -sum-free set for odd because of parity considerations while the top half maximum sum-free set qualifies because of magnitude considerations: the sum of two numbers in the top half is too big. There is an obvious way to tae a magnitude -sum-free subset of {1,,...,n} and get an analogue -sum-free subset of the interval 0, ] 1]. The top half maximum sum-free 1 subset of {1,,... n} becomes,1 and the size seems to be preserved. On the other hand it is not so obvious how to get the analogue on 0,1] for the odd numbers maximum sum-free subset of {1,,... n}. One could try to fatten up each odd integral point on [0,n] by as much as possible while eeping it sum-free and then normalize. It turns out one can fatten each odd integer j to j 1 3,j + 1 ) and, after normalization, one ends up with 3 a subset of 0,1] of size roughly 1 3. Chung and Goldwasser have conjectured that if 4, n is sufficiently large, and S is a -sum-free subset of {1,,..., n} of size fn, ), then S is the union of three strings of consecutive integers. Such a set has an analogue -sum-free subset of 0,1] of the same size, so we can learn someting about -sum-free subsets of {1,,...,n} by studying -sum-free subsets of 0,1]. We say that a Lebesgue) measurable subset S of 0,1] is a maximum -sum-free-set if S is -sum-free, has maximum size among all measurable -sum-free subsets of 0,1], and is not a proper subset of any -sum-free subset of 0,1]. So S is a maximum -sum-free set if both S and µs) are maximal where µs) denotes the measure of S. In this paper, for each integer greater than or equal to 3 we will construct a family of -sum-free subsets 0,1], each of which is the union of finitely many intervals Lemma 1). We will find which set in the family has maximum size Theorem 1). Then we will show that for 4 any maximum -sum-free subset of 0,1] must be in the family Section 3). This also gives us a lower bound for lim n fn,), and we conjecture that the bound is the actual value. n

A family of -sum-free sets. Let be a positive integer greater than or equal to 3, m be a positive integer, and a 1 and c be real numbers such that We define sequences {a i } and {b i } by 0 < c < a 1..1) b i = a i i = 1,,...,m a i+1 = b i c i = 1,,...,m 1..) We normalize to get sequences {e 1 } and {f i } defined by e 1 = 1 b m max{a 1,c} e i = 1 b m a i i =,3,...,m f i = 1 b m b i i = 1,,...,m.3) The set S = m i=1 e i,f i ) is the union of m disjoint intervals and is a subset of 0,1]. Furthermore, S is -sum-free because if x e i,f i ),y e j,f j ), z e r,f r ) and if r = max{i,j,r} then x + y < z, while if r < max{i,j,r} then x + y > z. If P is a set of real numbers consisting of f 1 and either e i or f i for i =,3,...,m then it is not hard to see that S P is a maximal -sum-free set; we will show that if c = a 1 and m = 3 then it is in fact a maximum -sum-free set. Now we calculate µs). From equations.) we get a i+1 a i ) = a i a i 1 ) i =,3,...,m 1 a a 1 = a 1 c 3

which has solution ) i a i = c 1 + c i = 1,,...,m where c 1 = a 1 4 ) c.4) If d = max{0,c a 1 } then c = c. µs) = 1 b m = b m = m b i a i ) d b i=1 m ) m 1 c 1 + c m d 1 b m ) 1 + m ) c 1 ) d c 1 c 1 c 1 ) m + c c where we have summed the geometric series and simplified. Now we let y = c a 1 so that c 1 0 < y < by equation.1). Then from equations.4) we get c 1 = c = a 1 ) [ 4 y 1)] a 1 y 4

and c c 1 = So now we substitute and simplify to get µs) = ) y y. { } 1 + ym 1) max 0, )y 1) ) m 1 ) ) y m 1 1 With fixed we note that because of the normalization, µs) is a function of m and y alone. So we have the following result. Lemma 1 Let m be a positive integer, a positive integer greater than or equal to 3, a 1 and c real numbers such that 0 < c < a 1,y = c a 1, and let S m,y) = m i=1 e i,f i ) where {e i } and {f i } are defined by.) and.3). Then S m,y) is a -sum-free set. If c a 1, then 0 < y 1 and.5) µs m,y)) = ) + [ym 1) 1] ) ) m 1 ) ) y m 1 1.6) while if c a 1 then 1 y < and µs m,y)) = ) + 1) y[ + 4) m )] ) m 1 ) ) y m 1. 1.7) For any positive integer greater than we define the set S ) by ) i 1 ) ) i 1 S ) = i=1,. 5

If P ) is formed from S ) by including one end-point of each interval then it is easy to see that P ) is a maximal -sum-free set and µp )) = µs )) = i=1 ) i 1 ) =. We remar that S ) = lim S m,y) for any y 0, ) and that m µs )) = µs,1)). If m is fixed, the expression in.6) is clearly an increasing function [ of y on 0,1], so to maximize µs m,y)) we need only consider y 1, ) and use.7). For fixed we define the functions fm,y) = 1) y [ + 4) m )] ) gm,y) = m 1 ) ) y m 1 1 hm,y) = fm,y) gm, y) [ where m is a positive integer and y 1, ). With y fixed, the function F y m) = fm,y) is an increasing linear function of m with root my) given by my) = + 3 1) y. y ) So the root my) of F y m) is an increasing function of y for y hence = m1) my) < m = + 1 ) [ 1, ) and [ This means that for each y 1, ) there is a positive integer m c y) {,3,..., + 1} such that fm,y) < 0 for m < m c y) and fm,y) 0 for m m c y). It is easy to show that if y is fixed then hm,y) > hm + 1,y) for all m greater than m c y) because gm,y) is positive and exponential 6

in m). So for fixed y the maximum value of hm,y) occurs when m {m c y),m c y) + 1}, which means for some m satisfying m +..8) Now with m fixed and satisfying.8) we let H m y) = hm,y). We differentiate to get where H my) = A [gm,y)] ) A = 1) m 1 1 ) m 1 ) [ + 4) m )] ) m 1 ) 1) [ by.8). So H m y) is strictly decreasing on 1, ) for any m satisfying.8). And hence µ S m,y)) is a maximum if and only if y = 1 and Rm) = hm,1) is a maximum over {,3,..., + }. We have Rm) = 1) + 4) + m ) ) m 1 4) + 1 = + m ) m 1 + 4 which clearly is maximum only at m = 3. In fact it is easy to see that Rm) is decreasing on [3, ) and that lim Rm) = R) = 0. We have proved m the following result. 7

Theorem 1 Let m be a positive integer, a positive integer greater than or equal to 3, a 1 and c real numbers such that 0 < c < a 1,y = c a 1, and let S m,y) = m i=1 e i,f i ) where {e i } and {f i } are defined by.) and.3). Then µs m,y)) is a maximum only when m = 3 and y = 1. The size of this maximum set is ) µ S 3,1)) = + 8 ) ) 4 4) Furthermore, if m is greater than, then µ S m,1)) > µ S m + 1,1)) ) and µ S m,1)) > µ S,1)) = µ S )) =. 3 Maximum -sum-free sets are in the family. Lemma If S is a maximum -sum-free subset of {0,1] where is an integer greater than or equal to 4 and if u = 1.u.b. x S x < } then < u <. Proof: If 1 < u 1 then there exists a real number x in S, Then 0 < x 1 < 1, and for each y [x 1,1], not both y and x y are in S. Because of these forbidden pairs, ). If we now let µ S [x 1,1]) 1 [1 x 1)]. 3.1) S = then S is -sum-free and { } 1 w w S 0,x 1] x 1 ) µs 1 ) = µs 0,x 1] x 1 1 = µs) µs [x 1,1])) x 1 1 x 1)µS) + [1 x 1)]µS) 1 ) x 1 [1 x 1)] > µs) 8

where the first inequality follows from 3.1) and the second follows because µs) > > 1 for 4. We have contradicted the assumption that S is a maximum set, so u 1 [. Now suppose u, 1 ]. For each ǫ > 0 there exists a real number x in S such that 0 u x < ǫ. If u < u then ǫ can be chosen such that x < x, and for each y 0,x], not both y and x y can be in S. Because of this forbidden pairing of 0,x] with [x x,x), and since < x x < x < 1, [ µs) = µ S x, ]) ])) + µ S 0,x],1 u x) + 1 < 1 + ǫ and µs) is not a maximum. If u u then since x x and due to the forbidden pairing of 0,x ] [ ) with,x, µs) 1 ) + u x ) = 1 1)u + u x) 1 1) + u x) = + ǫ. But is the size of S,1), so µs) is not a maximum. Hence u <. If u then the set 9

has size S = { x x S 0,u ] } µs ) = µs 0,u ]) µs) + = µs) + > µs) µs) 1 ) ) ) µs) which again is a contradiction, so µ > completing the proof. Lemma 3 If S is a maximum -sum-free subset of {0,1] where is an integer greater than or equal to 4 and if u = l.u.b. x S x < }, then u,1) S and µ S 0,u ]) [ ]) + µ S,1 =. Proof: First we will show that if S is a maximum -sum-free subset of 0,1] then S u,1) is also -sum-free. If x and y are in S and z u,1) then x+y < z, since u > by Lemma. If x u,1) and z then x + y > z, while if x u,1) and z < then x + y > z, since z u. Thus S u,1) is -sum-free and hence u,1) S. ] As in the proof of Lemma, for each x in S,u there is a forbidden pairing of 0,x ] [ ) and,x, so µ S 0,u ] [ ])),u u. 3.) If the inequality in 3.) is strict, then the set S = S u ]),u ],1 is also -sum-free and µs) < µs ). Thus equality holds in 3.). 10

Lemma 4 If S is a maximum -sum-free subset of {0,1] where is an integer greater than or equal to 4 and if u = l.u.b. x S x < } then µs) > 1 u + 1 ). Proof: If not then since u < µs) < 1 + Lemma ) we have 1 ) 1 ) + = which is a contradiction since this is the size of S,1). Lemma 5 If S is a maximum -sum-free subset { of 0,1] where is an integer greater than or equal to 4 and if u = l.u.b. x S x < } then a) µ S 0, ]) u > 0. { b) If u 3 = l.u.b. x S x < } u then u 3 1 u. c) There exists a positive number c such that u 3 u = u = c. d) [ ) )] u,u,1 S and S 0,c) =. Proof: a) If µ S 0, ]) u = 0 then, since u < 1 ) + u 1 ) < Lemma ), µs) 1 b) if x S u, ) u then there are forbidden pairs in [x u,u ] : If y S [x [ u,u ] then x y S. If we let S [x u,u ] = r and µ S u ]),u = p then 11

because of the forbidden pairing and r < 1 [u x u )] 3.3) p > 1 u 3.4) by Lemma 3 and Lemma 4. From equations 3.3) and 3.4) we get { u p r > 1 x u ). 3.5) [ Now let S = w w S u ]} ] x u,x u,1. It is easy to chec that S is -sum-free and µs u ) = p r) + 1 ) x u = p r) + u x u ) p r) + 1 ) x u r > p r) + 1 p r) p r) + ) = µs) where the inequality follows from equations 3.3) and 3.5) and the 1 last equality follows from Lemma 3. Hence S u, ) u = and u 3 1 u. c) By Lemma, u is equal to some positive number c. Let u 3 u = b and assume b < c. Let S = A B C where c + A = u x x S c,u 3 ),B = u 3 ] ] u,u, and c =,1. 1

If z B C or if {x,y,z} A it is clear that S has no solution to x + y = z. If z A and y A then x + y > c + u > z, so S is -sum-free. And c + µs ) = u µ S c,u 3 )) + 1 ) u + 1 u 3 > S c,u 3 )) + 1 ) u + 1 ) ) since µ S c,u 3 )) > 0 otherwise µs) µ S,1)) and u 3 = b + u < c + u. On the other [ hand, ] if b > c then b is positive and for each x S u,u 3 there is a forbidden pairing of [,x) u. Since x can be arbitrarily close to u 3, 0,x u ] and [ ])) µ S 0,b] u,u 3 b. 3.6) It is not hard to chec that the set ) ] S 0 = S b,u 3 ]) u,u,1 3.7) is -sum-free and that µs) µs 0 ) by 3.6) and Lemma 5b)). We define the set S by where u is chosen so that ] ] S = S b,u 3 ]) u,u,1 u 3 u = u. Since b > c we have u 3 + > u + u, so 13

u = u 3 + + > u + u + = u. Hence µs ) > µs 0 ) µs). It remains to show S is -sum-free. If x S,y S\b,u 3 ] and z S b,u 3 ] then while if x S,y Hence b = c. z < u 3 + u u ) = b + u < x + y ] ],1, and z u,u, then z = < u ) + u 3 ) u + u 3 ) u + < x + y d) If µs 0,c])[ = δ > 0 ) then, because [ ) of the forbidden pairing of 0,c] with each of u,u 3 and,u we have [ µs) µs 0,u 3 ]) + 1 ) ] u δ + 1 ) δ < µs 0 ) where S 0 is given by 3.7) with b = c. And if δ = 0 but ] ])) µ S u,u,1 < u 1 ) + 1 ) 14

we would still have µs) < µs 0 ). So S 0,c) and [ ) )] u,u,1 \S are both sets of measure 0; we will now show each is the empty set. If y S 0,c) choose and r and t in S such that 1 y + < r < t u such [ r and ] t exist because S is missing at most a set of measure zero in,u ). Then < r y < t y u and for each q S)) [r,t],q y S. Hence µs [r y,t y]) = 0, so µ S,u < u and µs) < µs 0). Thus we have shown S 0,c) = ). It is each ) to see that any such maximum S contains u,u and,1. Lemma 6 Let S be a maximum -sum-free subset of 0,1] where is an integer greater { than or equal to 4 with the sequence {u i } defined by u 1 = 1, u i = l.u.b. x S x < } u i 1 for i =,3,.... Then a) There exists a positive integer m 3 such that u m exists but u m+1 does not. [ ] b) There exists a positive number c 1) u m,u m such that [0,c) c) S = and u i+1 u i = c i = 1,,...,m 1. u i,u i ) max { u m,c}. S i = 1,,...,m 1 and t,u m ) S where t = 15

Proof: By Lemma 5 there exists a positive number c such that and u i+1 u i = c 3.8) ) u i,u i S 3.9) for i = 1 and and where S [0,c) =. Since c is a fixed positive number it is clear that the statement in a) is true. We will show by induction that equations 3.8) and 3.9) hold for all positive integers integers less than m. Assume 3.8) and 3.9) hold for all i less than j where j {3,4,...,m 1}; we will show they hold for i = j as well. First we will show that µ S 0,u j )) > 1 u j. 3.10) Since u j+1 exists and S 0,c) = we must have c < u j. Hence the set ) S = j i=1 u i,u i is -sum-free and µ S 0,u j )) = µs) µs ) + 1 ) u j 3.11) 1 u j + 4 u j since µs) µs ). This verifies 3.10) for any greater than 4. But for 4 the value of µs ) is given by Lemma 1 with c < a 1 = u j. Hence y < 1 and we use formula.6). As remared earlier, this is an increasing function of y, so µs ) is not a maximum and µs) µs ) > 0. This maes the inequality in 3.11) strict and verifies 3.10). Next we will show u j+1 1 u j. 3.1) 16

1 If there exists x S u j, ) u j then µ S [x u j,u j ]) 1 [u j x u j )] 3.13) by a forbidden pair argument. If we now let { } [ ]) S uj = w w S c,x u j ) S x u j u j 1,1 then it is easy to show S is -sum-free and µs u j ) µs) = µ S 0,u j ]) µ S [x u j,u j ])) x u j µ S 0,u j ]) u j x u j ) µ S 0,u j ])) x u j u j 1 x u j [u j x u j )] = u j x u j ) µ S 0,u j ]) 1 ) x u j u j > 0 where the first inequality follows by 3.13) and the second by 3.10). Thus we have verified equation 3.1). Now let b = u j+1 u j. We wish to show b = c. If b < c then we let 1 S u j + c) [ ]) = w w S 0,u j+1 ) S u j+1 u j,1. It is easy to chec that S is -sum-free and we have 1 µs ) µs) = u j + c) 1 u µ S 0,u j+1)). j+1 3.14) 17

If µs 0,u j+1 )) = 0 then as in the discussion following inequality 3.11) µs) is given by formula.6) with y < 1, so it cannot be a maximum. Since u j+1 = 1 ) u j + b, each factor in 3.14) is positive, which is a contradiction. If b > c then u j < u j+1 u j and as in the proof of Lemma 5c)) from a forbidden pairing we get [ ))) µ S 0,b] u j,u j+1 b. 3.15) Now we let S = S b,u j+1 )) j i=3 where u = 1 So S is obtained from S by replacing S [ ] u,u u j,u j+1), replacing S ) ) ) u i,u i u,u,1 b + ) > 1 c + ) = u. 3.16) [ ]) 0,b] u j,u j+1 by ) by u,u and possibly omitting finitely many points certain end-points). It is easy to chec that S is - sum-free and µs ) µs) 1 ) u 1 ) u > 0 by 3.15) and 3.16), so again µs) is not a maximum. Therefore ) b = c which verifies 3.8) for i = j. And clearly S u j,u j is -sum-free, which verifies 3.9) for i = j. Thus we have shown by induction that 3.8) and 3.9) hold for i = 1,,...,m 1. { } Since u m+1 does not exist, if we let t = max u m,c then S t,u m ) is -sum-free, so t,u m ) S verifying c). Finally, if c < c + u m 3.17) 18

then we can choose a real number y greater than c such that y < y + u m. But then S {y} is -sum-free which[ violates the maximality ] of S. Hence 3.17) must be false which shows c 1) u m,u m. Theorem If is an integer greater than or equal to 4 and S is a maximum -sum-free subset of 0,1] then S is the union of the set S 3,1) = 3 i=1 e i,f i ) of Lemma 1) and three points, one end-point of each interval. Any of the eight possible ways of choosing the end-points is all right except {e 1,f,e 3 }. Proof: By Lemma 6, if we ignore end-points, S has the form of S m,y) for some m 3. By Theorem 1, S 3,1) is the largest of these. To get a maximal set we need to put in one end-point of each interval, but since e 1 + e 3 = f we cannot choose {e 1,f,e 3 }. The end-points turn out to be f 1 = 4 4 4, f = ) 4 4, f 3 = 1, with e i = f i i = 1,,3. For = 4 one gets S 4 3,1) = ) 1 110, 7 110 110, 14 ) ) 1 110,1. Corollary 1 Let fn,) denote the maximum size of a -sum-free subset fn,) of {1,,... n}. If 4 then lim µ S 3,1)). n n 4 Remars One can get a maximum -sum-free subset of 0,1] by a sort of greedy 8 algorithm if one nows where to start. Put e 1 = 4 in S, 4) and then moving left to right put in anything that goes and eeps it - sum-free. So f 1 = l.u.b.{x [e 1,1] [e 1,x]is -sum-free}. But f 1 cannot be in S, so we have [e 1,f 1 ) so far. Then let e = g.l.b.{x [f 1,1] [e 1,f 1 ) {x} is -sum-free}, and so on. Alternatively, let a 1 = 1 and perform this greedy procedure, stop after three intervals and normalize. 19

] Moving right to left the greedy procedure does not wor. One gets,1 for the first interval and then 1) is the largest number that can be added. However the set is now maximal. In fact is the only real ] 1) number x such that {x},1 is a maximal -sum-free subset of 0,1]. 8 If you first put 4 in S and then wor right to left from 1 4) following the greedy procedure, you do get a maximum -sum-free set. The proof of Theorem is valid for any real number greater than + since then > 1 and the forbidden pairs arguments wor). The proof does not wor for = 3, but µs 3 m,y)) is still maximized when m = 3 and y = 1 µs 3 3,1)) = 77 ). 177 Conjecture 3 Theorem holds for = 3 as well. As mentioned in the Introduction, maximum -sum-free subsets of 0, 1] and of {1,,...,n} have very different structures for = 3. There is no maximum 3-sum-free analogue on 0, 1] of the all odd number maximum 3-sum-free subset of {1,,...,n}). However, we thin they have the same structures for 4. Conjecture 4 Equality holds in Corollary 1. We believe that if n is sufficiently large, to get a maximum -sum-free subset of {1,,...,n} one taes the integers within the three intervals obtained by multiplying each real number in S 3 3,1) by n with slight modification of the end-points due to integer round-off). To prove this integral version one can probably use the general outline of the above proof for 0,1]. There are some technical difficulties due to the fact that if one multiplies each member of a set of integers by a real number greater than 1, the result may not be a set of integers, and even if it is, the size of the set is the same as the size of the original set as opposed to what happens with the measure of a set of real numbers). 0

References [CG] F. R. K. Chung and J. L. Goldwasser, Integer sets containing no solutions to x + y = 3, The Mathematics of Paul Erdős, R. L. Graham and J. Nesetril eds., Springer Verlag, Heidelberg, 1996. [H] D. R. Heath-Brown, Integer sets containing no arithmetic progressions, 1986) preprint. [R] K. Roth, On certain sets of integers, J. London Math. Society, 8 1953), 104-109. [SS] [S1] [S] R. Salem and D. C. Spencer, On sets of integers which contain no three terms in arithmetical progressions, Proc. Nat. Acad. Sci. USA 8 194), 561-563. E. Szemerédi, On sets of integers containing no four elements in arithmetic progression, Acta. Math. Acad. Sci. Hungar. 0 1969), 89-104. E. Szemerédi, On sets of integers containing no elements in arithmetic progression, Acta. Arith. 7 1975), 199-45. 1