How to obtain the bound $\lvert R_{n}(\omega)\rvert \leq \beta d(2^{n+1})^{d-1}$ in the Ising Model From Chapter 3, page 85 of Friedli and Velenik, Statistical Mechanics of Lattice Systems: A classical mathematical introduction
https://www.unige.ch/math/folks/velenik/smbook/Ising_Model.pdf
The proof of the existence of the free energy in the Ising model is given.
In this proof we initially look at the "cubes" $D_{n}:=\{1,...,2^{n}\}^{d}$ and let $n$ pass through $\mathbb N$.
Note that by the neat construction, we get that $D_{n+1}$ contains $2^{d}$ "cubes" of $D_{n}$
Now let inverse temperature $\beta$ and magnetic field $h$ be given, and consider the following decomposition of the hamiltonian on the cube $D_{n+1}$:
$\mathcal{H}_{D_{n+1}}=\sum\limits_{i=1}^{2^{d}}\mathcal{H}_{D_{n}}^{(i)}+R_{n}$ where $\mathcal{H}_{D_{n}}^{(i)}$ is the Hamiltonian on $D_{n}^{(i)}$, i.e. the i-th cube of the decomposition and further where $R_{n}$ is the interaction between particles from separate cubes.
Note the hamiltonian in the Ising model is defined in the following way:
$\mathcal{H}_{\Lambda}(\omega):=-\beta\sum\limits_{i,j\in \Lambda\; \lvert i-j\rvert = 1}\omega(i)\omega(j)-h\sum\limits_{i\in \Lambda}\omega(i)\;\;$   where $\omega \in \{-1,+1\}^{\Lambda}$ and $\Lambda \subset \mathbb Z^{d}$.
Now my problem is that in the proof, they claim that the bound $\lvert R_{n}(\omega)\rvert \leq \beta d(2^{n+1})^{d-1}$ works.
I assume that the bound comes about in the following way:
$\beta \times (\text{number of pairs that can interact with one another from separate cubes)}$
"$\text{number of pairs that can interact with one another from separate cubes}$" actually means that the pairs that are nearest neighbours but not in the same cube.
I do not see (despite help with 3d drawings) why "$\text{number of pairs that can interact with one another from separate cubes}$"$=d(2^{n+1})^{d-1}$. Is there a mathematical proof behind this or explanation?
 A: Let $D_{n+1}$ be decomposed into $2^d$ copies of $D_n$. Let $A$ be the set of pairs of particles that are in different $D_n$'s but are adjacent in $D_{n+1}$. We wish to show that $|A| = d(2^{n+1})^{d-1}$.
As is mentioned, the number of particles on a face of $D_{n+1}$ is $(2^{n+1})^{d-1}$. There are $2d$ faces of $D_{n+1}$ which can be grouped into $d$ pairs of opposing faces. Let $B$ be the set of pairs of particles that are on a pair of such opposing faces, but aligned (the line between them is parallel to the other faces). Now, $B$ has size exactly $d(2^{n+1})^{d-1}$. We will construct a bijection from $B$ to $A$.
Given any element of $B$, bring the particles closer along the line segment connecting them until they are neighbors (and on opposite sides of the midpoint). This gives an element of $A$. Conversely, given an element of $A$, move the pair of particles further along the line connecting them until they are on opposite faces of $D_{n+1}$. This gives an element of $B$. The above two maps exhibit a bijection between $A$ and $B$, and thus $|A| = |B| = d(2^{n+1})^{d-1}$.
To describe the above maps explicitly requires some notation. Denote the faces of $D_{n+1}$ by $F_{0i}$ and $F_{1i}$ for $i = 1, 2, \ldots d$ where $F_{0i} = \{x\in D_{n+1}|x_i = 0\}$ and $F_{1i} = \{x\in D_{n+1}|x_i = 2^{n+1}\}$. Then we can write $B  = \cup_{i=1}^dB_i$ with $B_i = \{(x, y) | x \in F_{0i}, y\in F_{1i}, x_j = y_j for j\neq i\}$. The map from $B$ to $A$ takes a pair $(x, y)$ from $B_i$ and maps it to $(x', y')$, where $x'_i = 2^n$, $x'_j = x_j$ for $j\neq i$, and $y'_i = 2^n + 1$, $y'_j = y_j$ for $j\neq i$. It is clear that $x'$ and $y'$ belong to faces of different $D_n$'s and are adjacent. Thus $(x', y')\in A$. The map from $A$ to $B$ is the inverse of the above.
