Section 9.1

Problems

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Problem 9.1.1

The sum A+BA+B of subsets A,B⊂RA, B \subset \mathbb {R} or A,B⊂ZmA, B \subset \mathbb {Z}_m is the set A+B:={a+b:a∈A,b∈B}A+B := \left\{ a+b : a \in A, b \in B\right\}.

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(1)

If A,B⊂Z4A, B \subset \mathbb {Z}_4, ∣A+B∣=∣A∣≠0\left|A+B\right| = \left|A\right| \neq 0 and ∣B∣=2\left|B\right| = 2, then ∣A∣∈{2,4}\left|A\right| \in \left\{ 2,4\right\}.

(2)

If A,B⊂Z6A, B \subset \mathbb {Z}_6, ∣A+B∣=∣A∣≠0\left|A+B\right| = \left|A\right| \neq 0 and ∣B∣=2\left|B\right| = 2, then ∣A∣∈{3,6}\left|A\right| \in \left\{ 3,6\right\}.

Problem 9.1.2

The sum A+BA+B of subsets A,B⊂RA, B \subset \mathbb {R} is the set A+B:={a+b:a∈A,b∈B}A+B := \left\{ a+b : a \in A, b \in B\right\}. For any finite subsets A,B⊂RA, B \subset \mathbb {R}

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(1)

max⁡{∣A∣,∣B∣}⩽∣A+B∣⩽∣A∣⋅∣B∣\max \left\{ \left|A\right|, \left|B\right|\right\} \leqslant \left|A+B\right| \leqslant \left|A\right| \cdot \left|B\right|;

(2)

∣A∣⩽∣A+A∣⩽∣A∣(∣A∣+1)2\left|A\right| \leqslant \left|A+A\right| \leqslant \dfrac {\left|A\right|(\left|A\right|+1)}{2}.

Problem 9.1.3
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(1)

If A={2j:j=0,1,…,n−1}A = \left\{ 2^j : j = 0, 1, \ldots , n-1\right\}, then ∣A+A∣=n(n+1)2\left|A+A\right| = \dfrac {n(n+1)}{2}.

(2)

For any finite arithmetic progression P⊂RP \subset \mathbb {R} the inequality ∣P+P∣=2∣P∣−1\left|P+P\right| = 2\left|P\right|-1 holds.

Remark.
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Problem 9.1.3(1) gives an example of a subset for which the upper bound in Problem 9.1.2(2) is attained. Problem 9.1.3(2) gives an example of a subset for which the lower bound in Problem 9.1.2(2) is ``almost attained''. An example of a subset for which this bound is attained is an arbitrary one-element subset.

Problem 9.1.4

Kneser's theorem for R\mathbb {R}. For any finite subsets A,B⊂RA, B \subset \mathbb {R} the inequality ∣A+B∣⩾∣A∣+∣B∣−1\left|A+B\right| \geqslant \left|A\right|+\left|B\right|-1 holds.

The following Zp\mathbb {Z}_p-analogue may be used below without proof.

Cauchy--Davenport theorem. For any prime pp and any two subsets A,B⊂ZpA, B \subset \mathbb {Z}_p the following inequality holds:

∣A+B∣⩾min⁡{∣A∣+∣B∣−1,p}. \left|A+B\right| \geqslant \min \left\{ \left|A\right|+\left|B\right|-1, p\right\} .
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Problem 9.1.5
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(1)

If A,B⊂RA, B \subset \mathbb {R} and ∣A+B∣=∣A∣\left|A+B\right| = \left|A\right|, then ∣B∣=1\left|B\right| = 1.

(2)

If pp is prime, A,B⊂ZpA, B \subset \mathbb {Z}_p and ∣A+B∣=∣A∣\left|A+B\right| = \left|A\right|, then either ∣B∣⩽1\left|B\right| \leqslant 1 or A=ZpA = \mathbb {Z}_p.

Problem 9.1.6

Analogously to the sum of sets, one can define their difference: A−B:={a−b:a∈A,b∈B}A-B := \left\{ a-b : a \in A, b \in B\right\}.

Ruzsa's inequality. For any finite subsets A,B,C⊂RA, B, C \subset \mathbb {R}

∣C∣⋅∣A−B∣⩽∣A−C∣⋅∣B−C∣. \left|C\right| \cdot \left|A-B\right| \leqslant \left|A-C\right| \cdot \left|B-C\right|.
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Problem 9.1.7
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(1)

For any nonempty finite subset X⊂RX \subset \mathbb {R} the inequality ∣X−X∣⩽∣X+X∣2∣X∣\left|X-X\right| \leqslant \dfrac {\left|X+X\right|^2}{\left|X\right|} holds.

(2)

For any nonempty finite subsets X,Y⊂RX, Y \subset \mathbb {R} the inequality ∣X+X∣⩽∣X+Y∣2∣Y∣\left|X+X\right| \leqslant \dfrac {\left|X+Y\right|^2}{\left|Y\right|} holds.

(3)

For any finite subsets X,Y⊂RX, Y \subset \mathbb {R} such that X∩Y≠∅X \cap Y \neq \emptyset, the inequality ∣X+Y∣⩽∣X+X∣⋅∣Y+Y∣∣X∩Y∣\left|X+Y\right| \leqslant \dfrac {\left|X+X\right| \cdot \left|Y+Y\right|}{\left|X \cap Y\right|} holds.

Problem 9.1.8

For any nonempty subsets A,B⊂RA, B \subset \mathbb {R} the following statements are equivalent:

  1. ∣A+B∣=∣A∣⋅∣B∣\left|A+B\right| = \left|A\right| \cdot \left|B\right|;

  2. ∣A−B∣=∣A∣⋅∣B∣\left|A-B\right| = \left|A\right| \cdot \left|B\right|;

  3. ∣{(a1,a2,b1,b2)∈A×A×B×B:a1+b1=a2+b2}∣=∣A∣⋅∣B∣\left|\left\{ (a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1+b_1 = a_2+b_2\right\} \right| = \left|A\right| \cdot \left|B\right|;

  4. ∣{(a1,a2,b1,b2)∈A×A×B×B:a1−b1=a2−b2}∣=∣A∣⋅∣B∣\left|\left\{ (a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1-b_1 = a_2-b_2\right\} \right| = \left|A\right| \cdot \left|B\right|;

  5. ∣A∩({x}−B)∣=1\left|A \cap (\left\{ x\right\} -B)\right| = 1 for any x∈A+Bx \in A+B;

  6. ∣A∩(B+{y})∣=1\left|A \cap (B+\left\{ y\right\} )\right| = 1 for any y∈A−By \in A-B;

  7. (A−A)∩(B−B)={0}(A-A) \cap (B-B) = \left\{ 0\right\}.

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Problem 9.1.9

The sum of arbitrary subsets X1,X2,…,Xn⊂RX_1, X_2, \ldots , X_n \subset \mathbb {R} is defined as

X1+X2+…+Xn:={x1+x2+…+xn:xi∈Xi for any 1⩽i⩽n}. X_1+X_2+\ldots +X_n := \left\{ x_1+x_2+\ldots +x_n : x_i \in X_i \text{ for any } 1 \leqslant i \leqslant n\right\} .

For an arbitrary subset X⊂RX \subset \mathbb {R} and any integer k>0k>0, set kX:=X+X+…+X⏟k timeskX := \underbrace{X+X+\ldots +X}_{k \text{ times}}.

Ruzsa's inequality (Problem 9.1.6) can be generalized to the case of a sum of nn sets.

Plünnecke--Ruzsa inequality. For any nonempty finite subsets A1,A2,…,An,B⊂RA_1, A_2, \ldots , A_n, B \subset \mathbb {R}

∣A1+A2+…+An∣⩽∣A1+B∣⋅∣A2+B∣⋅…⋅∣An+B∣∣B∣n−1. \left|A_1+A_2+\ldots +A_n\right| \leqslant \frac{\left|A_1+B\right| \cdot \left|A_2+B\right| \cdot \ldots \cdot \left|A_n+B\right|}{\left|B\right|^{n-1}}.

For any finite subset A⊂RA \subset \mathbb {R} and arbitrary n∈Nn \in \mathbb {N} the inequalities ∣A∣⩽∣nA∣⩽(∣A∣+n−1n)\left|A\right| \leqslant \left|nA\right| \leqslant \dbinom {\left|A\right|+n-1}{n} hold.

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Problem 9.1.10

Analogously to the sum of sets, one can define their product: AB:={ab:a∈A,b∈B}AB := \left\{ ab : a \in A, b \in B\right\}.

If A,B⊂Z5∖{0}A, B \subset \mathbb {Z}_5 \setminus \left\{ 0\right\}, ∣AB∣=∣A∣≠0\left|AB\right| = \left|A\right| \neq 0 and ∣B∣=2\left|B\right|=2, then ∣A∣∈{2,4}\left|A\right| \in \left\{ 2,4\right\}.

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Problem 9.1.11
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(1)

There exists c>0c>0 such that for any finite set A⊂RA \subset \mathbb {R} the inequality max⁡{∣A+A∣,∣A⋅A∣}⩾c∣A∣5/4\max \left\{ \left|A+A\right|, \left|A \cdot A\right|\right\} \geqslant c\left|A\right|^{5/4} holds.

(2)

There exists c>0c>0 such that for any arithmetic progression P⊂RP \subset \mathbb {R} of length greater than 3 the inequality ∣P⋅P∣>c∣P∣5/4\left|P \cdot P\right| > c\left|P\right|^{5/4} holds.

Remark.
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In fact, if PP is an arithmetic progression of length nn in the set of integers, then the bound from Problem 9.1.11(2) can be improved. Namely, for any ε∈(0,2)\varepsilon \in (0,2) there exists c>0c>0 such that ∣P⋅P∣⩾c∣P∣ε\left|P \cdot P\right| \geqslant c\left|P\right|^{\varepsilon }. The proof of this fact uses nontrivial theorems of number theory and therefore lies beyond the scope of this book.

Problem 9.1.12

Let PP be a set of points in the plane, and LL a set of lines in the plane. The number of incidences I(P,L):=∣{(p,l)∈P×L:p∈l}∣I(P,L) := \left|\left\{ (p,l) \in P \times L : p \in l\right\} \right| is the number of pairs of the form (point on the line, line), where the line and the point are taken from the corresponding sets.

An arbitrary finite nonempty subset A⊆RA \subseteq \mathbb {R} is given. Denote by L\mathscr {L} the family of lines in the plane given by the equation y=a(x−b)y=a(x-b), where a,b∈Aa,b \in A. Let P:=(A+A)×(A⋅A)\mathscr {P} := (A+A) \times (A \cdot A). Prove that the total number of incidences between the lines of L\mathscr {L} and the points of P\mathscr {P} is at least ∣A∣3\left|A\right|^3.

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