Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 76 / 3 / ii

Solution

Apply Plünnecke's inequality with the summand set also equal to

A

. There is a nonempty

X \subseteq A

satisfying

∣ X + m A ∣ \leq C^{m} ∣ X ∣

for every

m \geq 0

We need the Ruzsa triangle inequality in a form whose short proof also fixes the signs. For every

z \in R - S

, choose one representation

z = r_{z} - s_{z}

. The map

(z, x) ⟼ (r_{z} + x, s_{z} + x), x \in X,

is injective into

(R + X) \times (S + X)

: its difference recovers

z

, then the fixed representation recovers

x

. Therefore

∣ R - S ∣∣ X ∣ \leq ∣ R + X ∣ ∣ S + X ∣.

Use

R = r A

and

S = s A

and the common set

X

from the previous part:

∣ r A - s A ∣ \leq \frac{∣ r A + X ∣ ∣ s A + X ∣}{∣ X ∣} \leq C^{r + s} ∣ X ∣ \leq C^{r + s} ∣ A ∣.

Thus

∣ r A - s A ∣ \leq C^{r + s} ∣ A ∣.

The shared minimizing subset is important; applying separate growth estimates with unrelated subsets would not justify this bound.

There are no discussions about this article yet.

There are currently no matching articles.