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If Al and A2 are such units and Al c A2 and (A2 - A11, r, then the corresponding sums will range through not less than 2r, in order that at so much one in every of them might be in J. the outcome now follows from workout 2. nine. We now go back to the case of complicated numbers. we will use j instead of i as a suffix with a view to keep away from confusion with i the place i2 = -1! Theorem eleven. 1. four (Erdos 1945) allow z1, ... , z be complicated numbers with Iz;I % 1 for every j. Then the variety of sums E;=1 E;z; (e; = ±1) 178 ( Combinatorics of finite units mendacity inside of a circle of radius r is at so much Cr2" nin for a few consistent C. evidence If I z I. 1 the place z = x + iy, x and y actual, then both Ix I = or 2 lYI = 2 hence both at the very least half the z; = x; + iy; have Jx;1 2 or at the very least 1/2 them have )y; I. 2. therefore, on changing every one z; by way of iz; if priceless, we will imagine that Jx;J --'l for no less than half the z;. extra, on the grounds that z, could be changed by means of -z; if x; is unfavorable, we will consider that at the least 1/2 the z, have actual half x; - 2 hence consider now that z,, . . . , z, have actual half ,2, t ' In. give some thought to the two' sums E , E ; z ; the place E,+,, ... , E" were given mounted values. We require that the sums E;=, E;z; lie inside of a circle of radius r, and for this reason that the sums E;=, s,x; lie inside of an period of size 2r. by way of Corollary eleven. 1. 2, the variety of such sums is for this reason at such a lot 2r([ J) 2 _r1/2 for a few consistent C; this is why the sums E;(2x;) lie inside of a sector of size 4r, and 2x; ,1 for every j. therefore for every collection of E,+,, ... , E,, we receive at so much Cr2'lt'2 sums mendacity contained in the circle. yet there are 2"-' methods of selecting , E". for that reason the whole variety of sums mendacity contained in the E,+,, circle is at such a lot Cr2'2"-' tin the place C -- C' ' 2"2C considering Zn Cr2" C'r2" = tin = nin t -- n. within the case r = 1 the sure given by means of Theorem eleven. 1. four is a continuing a number of of ([n2]) . in reality, we will be able to receive ([n,2]) itself because the very best top sure. This development, conjectured via Erdos, used to be proved by means of Katona and Kleitman independently. because the facts relies on a generalization of Sperner's theorem, we now flip to such generalizations. eleven. 2 M-part Sperner theorems during this part we current a few generalizations of the elemental theorem of Sperner. We contemplate the set S partitioned into M disjoint The Littlewood-Offord challenge 1 179 non-empty units: S = X, U ... U X. a colorful interpretation is to think about the weather of S being colored via M colors, X being the set of these parts colored through the ith color. A generalization of the antichain is then as follows: a set s1 of subsets of S has the valuables that if A, B E . f and A c B then B - A isn't contained in any of the elements X. by way of shades this estate calls for that there don't exist units A, B E d with A c B and B - A monochromatic. The antichain situation is naturally given by means of M = 1. fairly strangely, even though the situation with M = 2 is much less stringent than the antichain with M = 1, the higher sure ,n/21 n 1 nonetheless holds within the case M = 2.