Integers and perfect cubes are isometric

Question

Is there a quick way to check if the integers and $\{n^{3}: n \in \mathbb{Z}\}$ are quasi-isometric? It seems that they are since they look the "same" from far away.

Answer 1

They are not quasi-isometric. Let $C=\{n^3 \mid n \in \mathbb{Z} \}$ and suppose that $f: \mathbb{Z} \to C$ is a quasi-isometry. So for some fixed $\lambda\geq 1,$ and $k, l \geq 0$ we have that $$ \frac{1}{\lambda}d(x,y) -k \leq d(fx,fy) \leq \lambda d(x,y)+k, \quad \forall x,y \in \mathbb{Z}$$ and $$\forall z \in C, \exists x \in \mathbb{Z} \quad d(f(x),z) \leq l$$ Consider $x$ and $x+1$, if we put those in the first property we get $1/ \lambda - k \leq d(f(x),f(x+1) ) \leq \lambda +k$, elements in $C$ get arbitrarily far apart, so at some point $f(x)=f(x+1)$, as otherwise we could get their distance to be greater than $\lambda +k$, so $f(x)=f(x+1)=...=f(x+n)=....$ Take $n$ large enough and we get $$0<n/\lambda - k \leq d(f(x),f(x+n))=0$$ a contradiction.

$\mathbb{Z}$ and $C$ do not "look" alike from afar because the elements in $C$ keep getting further and further apart.