I have been reading about the notion of an approximate group. These are subsets of groups that are still somewhat symmetric:
$A \subseteq G$ is a $K$ - approximate group if
- it's symmetric $g \in A$ implies $g^{-1} \in A$
- There exists $X \subset G$ with $|X| = K$ such that $A \cdot A \subset X\cdot A$
Let's try this definition out on example. Let $G = \mathbb{Z}$ be an infinite group and $A = \{ [n\sqrt{3}]: n \in \mathbb{Z}\} \subseteq \mathbb{Z}$ be a subset. How many translates are needed to cover $A+A$? It could be argued that: $$ |A+A| < 4 |A| $$ (I think the number is 4, please correct me it's 3 or 5 or something). However, these are not translates.
Here $[ \cdot ]$ is the floor function.
My question is to find a constant $K$ such that $[\sqrt{3}\mathbb{Z}]\subseteq \mathbb{Z}$ is a $K$-approximate group.
Taking integer part turns out to be asymmetric, but it's still not that bad; doesn't fail the definition to egregiously. However, let's try rounding to the nearest integer:
$$ [n]_1 = [ n + 0.5 ] $$
This is called "rounding half-up". So we'd have $- [ \sqrt{3}\mathbb{Z}] = [\sqrt{3}\mathbb{Z}]$ is a symmetric set.
- Melvyn Nathanson Every finite set of integers is an asymptotic approximate group