Is There any quasi-isomorphism between $\mathbb{R}$ and $\mathbb{R}^2$?

Question

By natural inclusions, $\mathbb{R} \hookrightarrow \mathbb{R}^2$ and $\mathbb{N} \hookrightarrow \mathbb{Z}$, $\mathbb{R}$ and $\mathbb{N}$ are not quasi-isometric with $\mathbb{R}^2$ and $\mathbb{Z}$ respectively but these maps are quasi-isometric embeddings.

Is there any quasi-isometry between $\mathbb{R}$, $\mathbb{R}^2$ and $\mathbb{N}$, $\mathbb{Z}$?

Any net in a metric space $X$ is quasi-isometric to $X$. Is there any other fact about which subsets could be or could not be quasi-isomorphic to it?

Recal from: P.W. Nowak, G. Yu; Large Scale Geometry. European Mathemetical Society (2012).

definition 1.3.1: Let $X$ be a metric space and $C > 0$. A subset $N \subseteq X$ is a $C$-net in $X$ if for every $x \in X$ there exists $y \in N$ such that $d_X(x, y)< C$. We say that $N$ is a net if it is a $C$-net for some $C > 0$.

also

definition 1.3.4: Let $X$,$Y$ be metric spaces. A map $f: X \rightarrow Y$ is called a quasi-isometry if the following conditions are satisfied: (1) there exist constants $L,K > 0$ such that $L^{-1} d_X(x, y)- K \leq d_Y(f(x), f(y)) \leq L d_X(x, y)+K$ for all $x, y \in X$, and (2) the image $f(X)$ is a net in $Y$. We say that metric spaces X and Y are quasi-isometric if there exists a quasi-isometry $f:X \rightarrow Y$ . A map which satisfies (1), but not necessarily (2), is called a quasi-isometric embedding of $X$ into $Y$.

Answer 1

There is a proof in this paper (Proposition 5.7) that $\mathbb{Z}^m$ is quasi-isometric to $\mathbb{Z}^n$ iff $m=n$. This implies $\mathbb{R}^m$ is quasi-isometric to $\mathbb{R}^n$ iff $m=n$, since $\mathbb{R}^n$ is quasi-isometric to $\mathbb{Z}^n$ via the map $$(x_1, \cdots, x_n) \mapsto (\lfloor x_1 \rfloor, \cdots, \lfloor x_n \rfloor)$$

Answer 2

It is easy to see that the natural inclusions $\mathbb N \hookrightarrow \mathbb R_{\geq 0}$ and $\mathbb Z \hookrightarrow \mathbb R$ are quasi-isometries. Both $\mathbb R_{\geq 0}$ and $\mathbb R$ are examples of graphs, but with different numbers of ends. Since the number of ends are quasi-isometric invariants of graphs (See Lemma 2.2 in here, for example), it follows that $\mathbb N$ cannot be quasi-isometric to $\mathbb Z$.

Answer 3

$\mathbb{N}$ and $\mathbb{Z}$ are not quasi-isometric.

Indeed let $f : \mathbb{N}\to \mathbb{Z}$ be a quasi-isometry. We may assume $f(0)=0$. Fix $m\geq 0$. For sufficiently large $n$, $|f(n)| \geq m$. Pick a rank $n_0$ after which this happens. Say for instance $f(n_0)\geq m$.

Clearly any finite set is not a net, so there must be $n\geq n_0$ with $f(n) \leq m$. Pick $n_1$ to be the smallest such natural number, and thus $f(n_1-1) \geq m$, so $|f(n_1) - f(n_1-1)| \geq 2m$. If we had constants $L,K$ such as innthe definition we'd get $2m \leq L+K$, and this for any $m$, a contradiction.

For $\mathbb{R, R^2}$, other answers provided a solution