Is the infinite regular tree $T_\infty$ quasi-isometric to the $n$-regular tree $T_n$?

Question

It is known that for $n\geq 3$ all $n$-regular trees $T_n$ are quasi-isometric to each other. This can e.g. be seen by using an edge contraction argument.
Is there also a quasi-isometry between the countably infinite regular tree $T_\infty$ and some $T_n$?

Answer 1

The easiest argument that $T_n$ and $T_\infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:

Every finite valence graph has the "finite packing" property: There exists $\rho\ge 0$ such that for every $R\ge r\ge \rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $\rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.

The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.