Proof of The Cubes of the Integer Lattice are Finite, Nested and Exhaust the Lattice

Each result cited below is universally quantified over the data in its own statement.

Conventions. Let $\iota:\mathbb{N}\to\mathbb{R}$ be the canonical map of $\mathbb{R}$, as in The Integers as a Subset of the Real Numbers; by that definition $\iota(m)\in\mathbb{Z}$ for every $m\in\mathbb{N}$, and in the inequalities $-N\le k_{i}\le N$ defining $\Gamma_{N}$ the natural number $N$ stands for the integer $\iota(N)$, as in clause 3 of The Real Numbers and Standard Notation. Thus $\Gamma_{N}$ is the set of $k\in\mathbb{Z}^{n}$ with $-\iota(N)\le k_{i}\le\iota(N)$ for every $i\in[n]$. The order $\le$ of $\mathbb{R}$ is a total order, so it is reflexive, antisymmetric and transitive by clauses 1, 2 and 3 of Total Order on a Set; we write $x<y$ for $x\le y$ and $x\ne y$.

We record two facts about $\iota$. (M1) If $m,m'\in\mathbb{N}$ and $m\le m'$, then $\iota(m)\le\iota(m')$: if $m=m'$ this is reflexivity, and if $m<m'$ then $\iota(m)<\iota(m')$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. (M2) If $m,m'\in\mathbb{N}$ and $\iota(m)\le\iota(m')$, then $m\le m'$: otherwise $m'<m$ by the trichotomy of claim 3 of Properties of the Order on the Natural Numbers, so $\iota(m')<\iota(m)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; together with $\iota(m)\le\iota(m')$, antisymmetry gives $\iota(m)=\iota(m')$, contradicting $\iota(m')\ne\iota(m)$.

Clause 1 (Finite). Fix $N\in\mathbb{N}$.

Step 1: the set $I_{N}=\{j\in\mathbb{Z}:-\iota(N)\le j\le\iota(N)\}$ is finite and nonempty. Let $P=(N+N)+1\in\mathbb{N}$ and $c=\iota(N)+1$. By claims 1 and 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field,

$$\iota(P)=\iota(N+N)+1=\iota(N)+\iota(N)+1=\iota(N)+c .$$

Define $q:[P]\to\mathbb{R}$ by $q(m)=\iota(m)-c$.

(a) $q$ takes values in $I_{N}$. Let $m\in[P]$. Since $\iota(m)$, $\iota(N)$ and $1$ are integers (the first two by The Integers as a Subset of the Real Numbers, the third by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers), so are $c$ and $q(m)$, by the closure of $\mathbb{Z}$ under sums and differences in claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers. By claim 2 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $1\le\iota(m)$; adding $-c$ to both sides, which preserves $\le$ by clause 1 of Ordered Field, gives $1-c\le q(m)$, and $1-c=-\iota(N)$ by the field axioms of Field. Since $m\le P$, (M1) gives $\iota(m)\le\iota(P)$, and adding $-c$ (clause 1 of Ordered Field) gives $q(m)\le\iota(P)-c=\iota(N)$. Hence $q(m)\in I_{N}$.

(b) $q$ maps onto $I_{N}$. Let $j\in I_{N}$ and put $r=j+c$, an integer by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers. Adding $c$ to $-\iota(N)\le j$ (clause 1 of Ordered Field) gives $1\le r$; since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) gives $0<r$. By claim 1 of Arithmetic, Order, Discreteness and Intervals of the Integers the positive integers are exactly the numbers $\iota(m)$ with $m\in\mathbb{N}$, so $r=\iota(m)$ for some $m\in\mathbb{N}$. Adding $c$ to $j\le\iota(N)$ (clause 1 of Ordered Field) gives $\iota(m)=r\le\iota(N)+c=\iota(P)$, so $m\le P$ by (M2), that is $m\in[P]$; and $q(m)=r-c=j$.

By claim 1 of Basic Properties of Finite Sets the set $[P]$ has $P$ elements, and $q$ is a surjection from $[P]$ onto $I_{N}$ by (a) and (b); hence by claim 4 of Basic Properties of Finite Sets the set $I_{N}$ has $p'$ elements for some $p'\in\mathbb{N}$, and in particular is finite and nonempty.

Step 2: $\Gamma_{N}$ is finite. Let $I_{N}^{n}$ be the set of $n$-tuples in $I_{N}$, that is, of maps $[n]\to I_{N}$ (Tuples in a Set). By claim 3 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets, applied to the nonempty finite set $I_{N}$ of Step 1, $I_{N}^{n}$ is nonempty and finite, so by Finite Set it has $p$ elements for some $p\in\mathbb{N}$. For $t\in I_{N}^{n}$ each $t_{i}$ is a real number, so by claim 2 of Euclidean Points as Tuples of Real Numbers there is exactly one $\Phi(t)\in\mathbb{R}^{n}$ with $\Phi(t)_{i}=t_{i}$ for every $i\in[n]$. Every component of $\Phi(t)$ is an integer lying between $-\iota(N)$ and $\iota(N)$, so $\Phi(t)\in\mathbb{Z}^{n}$ by Lattice-Periodic Functions and the Periodic Function Classes §lattice and then $\Phi(t)\in\Gamma_{N}$. Thus $\Phi$ is a map $I_{N}^{n}\to\Gamma_{N}$. It is surjective: given $k\in\Gamma_{N}$, each $k_{i}$ is an integer (Lattice-Periodic Functions and the Periodic Function Classes §lattice) with $-\iota(N)\le k_{i}\le\iota(N)$, so the map $t:[n]\to I_{N}$, $t(i)=k_{i}$, belongs to $I_{N}^{n}$, and $\Phi(t)$ and $k$ have the same components, whence $\Phi(t)=k$ by claim 1 of Euclidean Points as Tuples of Real Numbers. By claim 4 of Basic Properties of Finite Sets, $\Gamma_{N}$ has $m$ elements for some $m\le p$, so it is finite and nonempty.

Step 3: the origin lies in $\Gamma_{N}$. By claim 2 of Euclidean Points as Tuples of Real Numbers there is exactly one $o\in\mathbb{R}^{n}$ with $o_{i}=0$ for every $i\in[n]$. Since $0\in\mathbb{Z}$ (claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers), $o\in\mathbb{Z}^{n}$ by Lattice-Periodic Functions and the Periodic Function Classes §lattice. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $0<\iota(N)$, hence $0\le\iota(N)$, and by sign reversal (claim 4 of Elementary Order Arithmetic in an Ordered Field) $-\iota(N)\le-0=0$. So $-\iota(N)\le o_{i}\le\iota(N)$ for every $i$, and $o\in\Gamma_{N}$.

Clause 2 (Nested). Let $N\le M$. By (M1), $\iota(N)\le\iota(M)$, and by sign reversal (claim 4 of Elementary Order Arithmetic in an Ordered Field) $-\iota(M)\le-\iota(N)$. If $k\in\Gamma_{N}$ and $i\in[n]$, then $-\iota(M)\le-\iota(N)\le k_{i}\le\iota(N)\le\iota(M)$, so $-\iota(M)\le k_{i}\le\iota(M)$ by transitivity (clause 3 of Total Order on a Set). Hence $k\in\Gamma_{M}$, and $\Gamma_{N}\subseteq\Gamma_{M}$.

Clause 3 (Exhausting). Let $E\subseteq\mathbb{Z}^{n}$ be finite. If $E=\emptyset$, then $E\subseteq\Gamma_{1}$. Suppose $E\ne\emptyset$; then $E$ is a nonempty finite set. The set $[n]$ is finite, having $n$ elements by claim 1 of Basic Properties of Finite Sets, and nonempty, since $1\le n$ by claim 4 of Properties of the Order on the Natural Numbers. Sums below are those of Sum over a Finite Index Set. For $k\in E$ put

$$\beta(k)=\sum_{i\in[n]}|k_{i}|,\qquad B=\sum_{k\in E}\beta(k).$$

Each $|k_{i}|$ is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, so each $\beta(k)$ is nonnegative by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §nonnegative. Let $k\in E$ and $i\in[n]$. The singleton $\{i\}$ is a nonempty subset of $[n]$ and the sum of $i'\mapsto|k_{i'}|$ over it is $|k_{i}|$ by claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set; hence $|k_{i}|\le\beta(k)$ by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone. In the same way, using the singleton $\{k\}\subseteq E$, $\beta(k)\le B$. By transitivity (clause 3 of Total Order on a Set), $|k_{i}|\le B$ for all $k\in E$ and $i\in[n]$.

By claim 1 of The Archimedean Property of the Real Numbers there is $N\in\mathbb{N}$ with $B<\iota(N)$. For $k\in E$ and $i\in[n]$, mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) gives $|k_{i}|<\iota(N)$, in particular $|k_{i}|\le\iota(N)$, and then $-\iota(N)\le k_{i}\le\iota(N)$ by the two-sided bound of claim 6 of Properties of the Absolute Value in an Ordered Field. As $k\in\mathbb{Z}^{n}$, this says $k\in\Gamma_{N}$. Hence $E\subseteq\Gamma_{N}$.

Finally, for $k\in\mathbb{Z}^{n}$ the set $\{k\}$ has $1$ element by claim 2 of Basic Properties of Finite Sets, so it is a finite subset of $\mathbb{Z}^{n}$ by Finite Set, and what was just proved gives $N$ with $k\in\Gamma_{N}$.