In arXiv:1212.5056 [math.CO] “On growth in an abstract plane” by Nick Gill, H. A. Helfgott, Misha Rudnev , in lemma 4.1 is given a proof of the Ruzsa triangle inequality which intrigued me. Later on, at the end of the article the authors give a geometric Ruzsa inequality in a Desarguesian projective plane, based on similar ideas as the ones used in the proof of the Ruzsa triangle inequality.

All this made me write the following.

____________

Let be a non-empty set and be an operation on which has the following two properties:

- for any we have ,
- for any the function is injective.

We may use weaker hypotheses for , namely:

- (weaker) there is a function such that for any ,
- (weaker) there is a function such that is an injective function for any .

* Prop. 1.* Let be a non empty set endowed with an operation which satisfies 1. and 2. (or the weaker version of those). Then for any non empty sets there is an injection

,

where we denote by

In particular, if are finite sets, we have the Rusza triangle inequality

,

where denotes the cardinality of the finite set .

I shall give the proof for hypotheses 1., 2., because the proof is the same for the weaker hypotheses. Also, this is basically the same proof as the one of the mentioned lemma 4.1. The proof of the Ruzsa inequality corresponds to the choice , where is a group (no need to be abelian). The proof of the geometric Ruzsa inequality corresponds to the choice , with the notations from the article, with the observation that this function satisfies weaker 1. and 2.

* Proof.* We can choose functions and such that for any we have . With the help of these functions let

.

We want to prove that is injective. Let . Then, by 1. we have . This gives an unique . Now we know that . By 2. we get that qed.

____________

In a metric space with dilations we have the function approximate difference based at and applied to a pair of closed points . This function has the property that converges uniformly to as goes to . Moreover, there is a local group operation with as neutral element such that , therefore the function satisfies 1. and 2.

As concerns the function , it satisfies the following approximate version of 1.:

- (approximate) for any which are sufficiently close and for any we have, with the notation , the relation

.

We say that a set is separated if for any , the inequality implies . Further I am going to write about sets which are closed to a fixed, but arbitrary otherwise point .

* Prop2.* In a metric space with dilations, let and let be finite sets of points included in a compact neighbourhood of , which are closed to , such that for any the sets and are separated. Then for any there is an injective function

.

* Proof.* As previously, we choose the functions and . Notice that these functions depend on but this will not matter further. I shall use the notation liberally, for example means . Let’s define the function by the same formula as previously:

.

Let and be pairs such that . From 1. (approximate) and from the uniform convergence mentioned previously we get that

.

There is a function such that implies (the last from the previous relation) . For such a , by the separation of we get .

Let . From the hypothesis we have . This implies, via the structure of the function and via the uniform convergence, that (by compactness, this last does not depend on ). By the same reasoning as previously, we may choose such that if . This implies qed.

## 3 thoughts on “Geometric Ruzsa triangle inequalities and metric spaces with dilations”