Show that $\mathbb{R}^2$ is not quasi isometric to the $3-$regular tree $T_3$

Question

To do this I would try by contradiction, assume there is a q.i $\phi:\mathbb{R}^2\to T_3$ and map some sufficiently large set and show that the image cannot actually be cobounded. I don't really have any idea how to actually do this in practice or evem if this is the best way to do this.

Answer 1

The best way to do this is to pick the correct quasi-isometry invariant. In this case, use ends. In brief, for proper geodesic metric spaces, the set of ends is a quasi-isometry invariant, in the following sense:

Theorem: If $X,Y$ are proper geodesic metric spaces and if $f:X \to Y$ is a quasi-isometry then $f$ induces a bijection from the ends of $X$ to the ends of $Y$.

Since $\mathbb{R}^2$ has one end and $T_3$ has infinitely many ends, there is no bijection between their ends, so there is no quasi-isometry between them.