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

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 Arithmetic, Order, Discreteness and Intervals of the Integers §ring), so are $c$ and $q(m)$, by the closure of $\mathbb{Z}$ under sums and differences in Arithmetic, Order, Discreteness and Intervals of the Integers §ring. 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 Arithmetic, Order, Discreteness and Intervals of the Integers §ring. 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 Arithmetic, Order, Discreteness and Intervals of the Integers §sign 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 $0\in\mathbb{Z}$ (Arithmetic, Order, Discreteness and Intervals of the Integers §ring), $o\in\mathbb{Z}^{n}$ 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 $k\in\Gamma_{N}$.

Clause 4 (Cardinality). Fix $N\in\mathbb{N}$, and let $P$, $c$, $I_{N}$ and $\Phi$ be as in Steps 1 and 2. All sums below are sums over finite index sets. Write $X=I_{N}$.

Step 4: $\sum_{j\in X}1=\iota(P)$. The numbers $-\iota(N)$ and $\iota(N)$ are integers, and $-\iota(N)\le0\le\iota(N)$ by Step 3, so $-\iota(N)\le\iota(N)$ by transitivity. Apply Arithmetic, Order, Discreteness and Intervals of the Integers §intervals with $a=-\iota(N)$ and $b=\iota(N)$. Here $b-a+1=\iota(N)+\iota(N)+1=\iota(P)$ by Step 1, so the natural number of that clause is $P$, by its uniqueness. Hence $X$, which is nonempty and finite by Step 1, has $P$ elements, and $\sum_{j\in X}1=\iota(P)$ by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §count.

Step 5: $\sum_{t\in X^{m}}1=\iota(P)^{m}$ for every $m\in\mathbb{N}$. By claim 3 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets, every $X^{m}$ is nonempty and finite. Let $Q$ be the set of $m\in\mathbb{N}$ for which the identity holds; we apply Principle of Induction for the Natural Numbers to $Q$.

Base. If $k\in[1]$ then $1\le k\le1$, by claim 4 of Properties of the Order on the Natural Numbers, so $k=1$ by its claim 2. Hence $[1]=\{1\}$, and the map $\rho:X^{1}\to X$, $\rho(t)=t_{1}$, is a bijection: a $1$-tuple is a map on $\{1\}$, so it is determined by its value at $1$, and every value occurs. Claim 2 of Properties of a Sum over a Finite Index Set, applied with $\theta=\rho$ to the map on $X$ with constant value $1$, gives $\sum_{t\in X^{1}}1=\sum_{j\in X}1$. By Step 4 this is $\iota(P)$, which equals $\iota(P)^{1}$ by claim 1 of Properties of Natural Number Powers in a Field. So $1\in Q$.

Step. Let $m\in B$, let $S$ be the successor map, and let $\psi:X^{m}\times X\to X^{S(m)}$ be the bijection of claim 2 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets. The set $X^{m}\times X$ is finite by claim 1 of that lemma, and nonempty. Claim 2 of Properties of a Sum over a Finite Index Set, applied with $\theta=\psi$ to the map with constant value $1$, gives

$$\sum_{t\in X^{S(m)}}1=\sum_{s\in X^{m}\times X}1.$$

The Cartesian product $X^{m}\times X$ is the set of ordered pairs $(t,x)$ with $t\in X^{m}$ and $x\in X$. So Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs, with $B(t)=X$ for every $t$, together with Step 4 and claim 4 (homogeneity) of Properties of a Sum over a Finite Index Set, gives

$$\sum_{s\in X^{m}\times X}1=\sum_{t\in X^{m}}\Bigl(\sum_{x\in X}1\Bigr)=\sum_{t\in X^{m}}\iota(P)=\iota(P)\sum_{t\in X^{m}}1=\iota(P)\,\iota(P)^{m}.$$

By commutativity of multiplication and claim 1 of Properties of Natural Number Powers in a Field, the right-hand side is $\iota(P)^{S(m)}$. So $S(m)\in Q$, and therefore $Q=\mathbb{N}$.

Step 6: conclusion. The surjection $\Phi:X^{n}\to\Gamma_{N}$ of Step 2 is also injective. If $\Phi(t)=\Phi(t')$, then $t_{i}=\Phi(t)_{i}=\Phi(t')_{i}=t'_{i}$ for every $i\in[n]$, so $t=t'$ as maps on $[n]$. Since $\Gamma_{N}$ is nonempty and finite by clause 1, claim 2 of Properties of a Sum over a Finite Index Set, applied with $\theta=\Phi$, and Step 5 with $m=n$ give

$$\sum_{k\in\Gamma_{N}}1=\sum_{t\in X^{n}}1=\iota(P)^{n}.$$

Finally $\iota(P)=\iota(N)+\iota(N)+1=2\iota(N)+1$, where $2=1+1$, by distributivity. This is $2N+1$ with $N$ read in $\mathbb{R}$, as in The Real Numbers: Standing Notation and Background §numbers.