The Cubes of the Integer Lattice are Finite, Nested, Exhaust the Lattice and Have (2N+1)^n Points

Let $n\in\mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers, let $[n]$ be the initial segment determined by $n$, let $\mathbb{Z}$ be the set of integers, and let $\mathbb{Z}^{n}$ be the integer lattice, whose points $k$ are read as maps on $[n]$, the component of $k$ at $i\in[n]$ being written $k_{i}$. For $N\in\mathbb{N}$ let $\Gamma_{N}$ be the set of those $k\in\mathbb{Z}^{n}$ with $-N\le k_{i}\le N$ for every $i\in[n]$ (the order of $\mathbb{Z}$ being that of the integers). Then the following hold.

1. (Finite)¶ For every $N\in\mathbb{N}$, $\Gamma_{N}$ is a finite set containing the point all of whose components are $0$; in particular it is nonempty.

2. (Nested)¶ For all $N,M\in\mathbb{N}$ with $N\le M$, $\Gamma_{N}\subseteq\Gamma_{M}$.

3. (Exhausting)¶ For every finite set $E\subseteq\mathbb{Z}^{n}$ there is $N\in\mathbb{N}$ with $E\subseteq\Gamma_{N}$; in particular every $k\in\mathbb{Z}^{n}$ lies in some cube.

4. (Cardinality)¶ For every $N\in\mathbb{N}$, the sum over $\Gamma_{N}$ (nonempty and finite by clause 1) of the map with constant value $1$, a sum over a finite index set of real numbers, is

$$\sum_{k\in\Gamma_{N}}1=(2N+1)^{n},$$

where $N$ is read in $\mathbb{R}$ and the right-hand side is the natural power, as in The Real Numbers: Standing Notation and Background §numbers, and $2=1+1$.