# Triangle inequality for Ruzsa distance

Suppose that $$Z$$ is an additive group, i.e. $$Z$$ is an abelian group under addition. Let $$A$$ be an additive set, i.e. $$A$$ is a non-empty finite subset of $$Z$$.

For any two additive sets $$A$$ and $$B$$ in $$Z$$ define the Ruzsa distance as follows: $$d(A,B):=\log \dfrac{|A-B|}{\sqrt{|A||B|}}$$, where $$\log$$ is a natural logarithm.

My goal is to show that Ruzsa distance obeys triangle inequality: for any additive sets $$A,B,C$$ in $$Z$$ we have $$d(A,C)\leq d(A,B)+d(B,C).$$ It is suffices to show that $$|A-C||B|\leq |A-B||B-C|.$$

Let's construct a function $$f:(A-C)\times B\to (A-B)\times (B-C)$$ defined by rule: $$(a-c,b)\mapsto (a-b,b-c).$$

I have some issues to show that this function is well-defined and injective.

Injectivity: Suppose that $$f((a_1-c_1,b_1))=f((a_2-c_2,b_2)),$$ then $$(a_1-b_1,b_1-c_1)=(a_2-b_2,b_2-c_2),$$ then it implies that $$a_1-a_2=b_1-b_2=c_1-c_2.$$ How to show that $$(a_1-c_1,b_1)=(a_2-c_2,b_2)?$$

Well-definedness: If $$(a_1-c_1,b_1)=(a_2-c_2,b_2),$$ then we need to show that $$(a_1-b_1,b_1-c_1)=(a_2-b_2,b_2-c_2).$$

• @CalvinLin, for well-definedness: if $(a_1-c_1,b_1)=(a_2-c_2,b_2)$, then $b_1=b_2$. How it follows that $a_1=a_2$ and $c_1=c_2$? I do not see that.
– ZFR
Nov 12, 2021 at 19:03
• @CalvinLin, for infectivity I do not think that you are right because this map is actually injective!
– ZFR
Nov 12, 2021 at 19:04

This function, as stated, isn't well-defined: if $$A=B=C=\{0,1\}$$, then you might define $$f(0,0)$$ by $$f(0-0,0)=(0-0,0-0)=(0,0)$$ while you could also select $$f(1-1,0)=(1-0,0-1)=(1,-1).$$ As for coming up with a function that does work, I'd recommend starting by, for each $$x\in A-C$$, selecting a fixed pair $$(a_x,c_x)\in A\times C$$ with $$a_x-c_x=x$$. Then, you won't run into the same issue as above (since you will no longer try to define $$f$$ using $$0=0-0$$ and $$0=1-1$$. Can you show that the function is then well-defined and injective?