Finding an isomorphism between an infinite tree and a subgraph of $\mathbb{Z}^3$

Question

I was wondering if there exists a construction of an infinite tree, with some properties, that is isomorphic to subgraph of $\mathbb{Z}^3$.

Notation

Let $\Gamma_n$ denote the tree's vertices at distance $n$ from the root. Let $\partial B(n)$ denote the vertices of $\mathbb{Z}^3$ at distance $n$ from the origin, according to $\ell_1$ metric.

I first describe a family of trees $\mathcal{F}$, where a tree is in this family if and only if it satisfies the following properties.

1. For all $i\in\mathbb{N}$, all vertex at distance $i$ from the root, have the same number of children at distance $i+1$ from the root.
2. Every vertex has either one or two children.
3. There exists an isomorphism $\varphi$ from the tree vertices to $\mathbb{Z}^3$, where $$ \forall i\in\mathbb{N},\quad \varphi(\Gamma_i)\subseteq \partial B(i) $$
4. For all $i\in\mathbb{N}$, $|\Gamma_i|\geq i^2/ c$ for some positive constant $c$ (the constant $c$ is absolute constant belonging to $\mathcal{F}$).

Question

Is $\mathcal{F} = \emptyset$? Or perhaps it is not empty, and someone can give me a hint on showing it is not empty?

I would also be happy with seeing proof of the existence of a tree satisfying all properties, where $4$ is weakened to $|\Gamma_i|\geq i^{1+\epsilon}/ c$ for some positive absolute constant $\epsilon$.

Thanks for reading the long explanation.

Answer 1

Let show this in the simpler case where we take $\mathbb{N}^2$ instead of $\mathbb{Z}^3$. We will show that we can find $\mathcal{F}$ with $|\Gamma_i| \geq \frac{i}{c}$. Let $\partial_pB(2^n-2)$ be the vertices of $\mathbb{N}^2$ where both coordinates are even ($\partial B(2^n-2)$ is a diagonal, we take one vertex over two). We have $|\partial_pB(2^n-2)| = 2^{n-1}$ We see easily that no two elements of $\partial_pB(n)$ share a common neighbor. Suppose we found a tree for which $\varphi(\Gamma_{2^n-2}) = \partial_pB(2^n-2)$. At next step, we give each vertex two children (so $|\Gamma_{2^n-1}| = 2^n$), then we find a $2^n$ disjoint paths from the vertices of $\varphi(\Gamma_{2^n-1})$ to the vertices of $\varphi(\Gamma_{2^{n+1}-2})$ (quite easy to find).

$|\Gamma_{2^n-2}| = 2^{n-1} \geq \frac{2^n-2}{2}$, so $|\Gamma_i| \geq \frac{i}{2}\; \forall i$ .

Because $\partial B(n)$ is only 4 times bigger in $\mathbb{Z}^2$ than in $\mathbb{N}^2$, we have $|\Gamma_i| \geq \frac{i}{8}\; \forall i$ when considering $\mathbb{Z}^2$. (but we could get a better constant by growing the tree in all directions).

The same reasoning works with $\mathbb{Z}^3$, it is always possible to find a subset $\partial_p B(n)$ of $\partial B(n)$ with $|\partial_p B(n)| \geq \frac{|\partial B(n)|}{2}$ such that we can split each vertices in two at the next step (Take the subset where the first coordinate is even). It should be even possible to reach a constant of $c = \frac{3}{2}$ by selecting a better subset.