[proofplan] The theorem characterises discrete subgroups of $\mathbb{R}^n$ as $\mathbb{Z}$-spans of $\mathbb{R}$-linearly independent vectors, i.e., as lattices of rank at most $n$. The proof has two directions. The easy direction (lattice $\Rightarrow$ discrete) transports the standard discreteness of $\mathbb{Z}^m \subseteq \mathbb{R}^m$ along a linear change of coordinates. The hard direction (discrete $\Rightarrow$ lattice) proceeds in three stages: (i) choose $y_1, \ldots, y_m \in \Lambda$ maximally $\mathbb{R}$-linearly independent, so $\operatorname{span}_\mathbb{R}(\Lambda) = \operatorname{span}_\mathbb{R}(y_1, \ldots, y_m)$; (ii) show the "fractional parts" $F := \Lambda \cap X$ with respect to the parallelepiped $X = \{\sum \lambda_i y_i : \lambda_i \in [0,1]\}$ are finite by compactness and discreteness, which forces $\Lambda / (\mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m)$ to have cardinality at most $|F|$; (iii) a pure algebra step then upgrades $\Lambda$ from a finite-index extension of a free $\mathbb{Z}$-module of rank $m$ to a free $\mathbb{Z}$-module of rank $m$ itself, by the structure theorem for finitely generated abelian groups. [/proofplan]

[step:Prove the easy direction: $\mathbb{Z}$-span of $\mathbb{R}$-independent vectors is discrete]Suppose $\Lambda = \mathbb{Z}x_1 + \cdots + \mathbb{Z}x_m$ with $x_1, \ldots, x_m \in \mathbb{R}^n$ linearly independent over $\mathbb{R}$. Extend $x_1, \ldots, x_m$ to an $\mathbb{R}$-basis $x_1, \ldots, x_n$ of $\mathbb{R}^n$ (possible by the $\mathbb{R}$-independence and standard linear algebra). Define the linear map \begin{align*} g : \mathbb{R}^n &\to \mathbb{R}^n \\ x_i &\mapsto e_i \end{align*} where $e_1, \ldots, e_n$ is the standard basis. Since the $x_i$ form a basis of $\mathbb{R}^n$, $g$ is invertible, hence $g \in \mathrm{GL}_n(\mathbb{R})$ and both $g$ and $g^{-1}$ are continuous (every invertible linear map on $\mathbb{R}^n$ is a homeomorphism). Under $g$, \begin{align*} g(\Lambda) = \mathbb{Z}e_1 + \cdots + \mathbb{Z}e_m = \mathbb{Z}^m \times \{0\}^{n-m} \subseteq \mathbb{R}^n. \end{align*} This is discrete in $\mathbb{R}^n$: for $v = (v_1, \ldots, v_n) \in g(\Lambda)$, we have $v_i \in \mathbb{Z}$ for $i \leq m$ and $v_i = 0$ for $i > m$. The open ball $B(v, \tfrac{1}{2})$ in the Euclidean norm contains no other point of $g(\Lambda)$, because any other $w \in g(\Lambda)$ has $w - v \in \mathbb{Z}^m \times \{0\}^{n-m}$ with some coordinate in $\mathbb{Z} \setminus \{0\}$, forcing $|w - v| \geq 1 > \tfrac{1}{2}$. Since $g^{-1}: \mathbb{R}^n \to \mathbb{R}^n$ is a homeomorphism and $g(\Lambda) \subseteq \mathbb{R}^n$ is discrete, the preimage $\Lambda = g^{-1}(g(\Lambda))$ is discrete in $\mathbb{R}^n$.[/step]

[guided]The forward direction — "lattice $\Rightarrow$ discrete" — follows from the simple observation that $\mathbb{Z}^m$ is discrete in $\mathbb{R}^m$, combined with the fact that $\mathbb{R}$-linear isomorphisms preserve discreteness. **Extending the given vectors to a basis.** Start with $x_1, \ldots, x_m \in \mathbb{R}^n$ linearly independent over $\mathbb{R}$. By a standard exchange-lemma argument, we can extend this list to an $\mathbb{R}$-basis $x_1, \ldots, x_n$ of $\mathbb{R}^n$ (if $m = n$ no extension is needed; otherwise pick any vector not in the span of $x_1, \ldots, x_m$, adjoin, and repeat). **Defining the change-of-coordinates map.** Let \begin{align*} g : \mathbb{R}^n &\to \mathbb{R}^n \end{align*} be the unique $\mathbb{R}$-linear map with $g(x_i) = e_i$ for $i = 1, \ldots, n$. Since $g$ sends a basis to a basis, it is invertible, so $g \in \mathrm{GL}_n(\mathbb{R})$. Any element of $\mathrm{GL}_n(\mathbb{R})$ is a homeomorphism of $\mathbb{R}^n$: the linear map $g$ is automatically continuous (linear maps on finite-dimensional spaces are continuous), and its inverse $g^{-1}$ is also a linear map in $\mathrm{GL}_n(\mathbb{R})$, hence continuous. Continuous bijection with continuous inverse = homeomorphism. **Image of $\Lambda$ under $g$.** Applying $g$ to the generators: \begin{align*} g(\Lambda) = \mathbb{Z} g(x_1) + \cdots + \mathbb{Z} g(x_m) = \mathbb{Z}e_1 + \cdots + \mathbb{Z}e_m. \end{align*} This is the set of vectors $(v_1, \ldots, v_n)$ with $v_1, \ldots, v_m \in \mathbb{Z}$ and $v_{m+1} = \cdots = v_n = 0$. Write this set as $\mathbb{Z}^m \times \{0\}^{n-m} \subseteq \mathbb{R}^n$. **$\mathbb{Z}^m \times \{0\}^{n-m}$ is discrete in $\mathbb{R}^n$.** Recall that a subset $S \subseteq \mathbb{R}^n$ is **discrete** if every point $v \in S$ has a neighbourhood intersecting $S$ only in $\{v\}$. Take $\varepsilon = 1/2$. For any $v \in \mathbb{Z}^m \times \{0\}^{n-m}$, consider $B(v, \varepsilon)$. If $w \in (\mathbb{Z}^m \times \{0\}^{n-m}) \cap B(v, \varepsilon)$ and $w \neq v$, then $w - v$ is a non-zero vector in $\mathbb{Z}^m \times \{0\}^{n-m}$. Hence at least one coordinate $i \in \{1, \ldots, m\}$ has $(w - v)_i \in \mathbb{Z} \setminus \{0\}$, so $|(w - v)_i| \geq 1$, and therefore \begin{align*} |w - v| \geq |(w - v)_i| \geq 1 > 1/2 = \varepsilon, \end{align*} contradicting $w \in B(v, \varepsilon)$. Hence $B(v, 1/2) \cap g(\Lambda) = \{v\}$. **Transporting discreteness back.** Homeomorphisms preserve the property "discrete". Explicitly, $\Lambda = g^{-1}(g(\Lambda))$, and for any $v \in \Lambda$, $g(v)$ has a ball $B(g(v), 1/2) \cap g(\Lambda) = \{g(v)\}$. The preimage $g^{-1}(B(g(v), 1/2))$ is an open set containing $v$ (since $g^{-1}$ is continuous), and $g^{-1}(B(g(v), 1/2)) \cap \Lambda = g^{-1}(B(g(v), 1/2) \cap g(\Lambda)) = g^{-1}(\{g(v)\}) = \{v\}$. Thus every $v \in \Lambda$ has an open neighbourhood meeting $\Lambda$ only at $v$, so $\Lambda$ is discrete.[/guided]

[step:For the converse, choose $y_1, \ldots, y_m \in \Lambda$ maximally $\mathbb{R}$-linearly independent]Now assume $\Lambda \subseteq \mathbb{R}^n$ is a discrete subgroup. Let $V = \operatorname{span}_\mathbb{R}(\Lambda) \subseteq \mathbb{R}^n$ be the $\mathbb{R}$-linear span of $\Lambda$, and set $m = \dim_\mathbb{R}(V) \leq n$. Choose a maximal $\mathbb{R}$-linearly independent subset $\{y_1, \ldots, y_m\} \subseteq \Lambda$. Such a subset exists: take any finite $\mathbb{R}$-linearly independent subset of $\Lambda$ (non-empty if $\Lambda \neq \{0\}$; the case $\Lambda = \{0\}$ is handled below with $m = 0$) and extend greedily while maintaining independence. The process terminates in at most $m$ steps since any $\mathbb{R}$-independent set in $V$ has at most $m = \dim V$ elements. By maximality, every $v \in \Lambda \subseteq V$ is an $\mathbb{R}$-linear combination of $y_1, \ldots, y_m$. (If some $v \in \Lambda$ were not such a combination, $\{v, y_1, \ldots, y_m\}$ would be $\mathbb{R}$-linearly independent, contradicting maximality.) Moreover, each $v \in \Lambda$ can be written with *rational* coefficients: writing $v = \sum_i \lambda_i y_i$ with $\lambda_i \in \mathbb{R}$, the linear algebra relations among elements of the $\mathbb{Q}$-subspace generated by $y_1, \ldots, y_m$, $v$ force $\lambda_i \in \mathbb{Q}$; we prove this carefully in the next step.[/step]

[guided]For the converse (discrete $\Rightarrow$ lattice), we build up to the conclusion in stages. The first stage is to identify the "real rank" $m$ of $\Lambda$ — the $\mathbb{R}$-dimension of its linear span — and fix a reference basis $y_1, \ldots, y_m$ for it, chosen from $\Lambda$ itself. **The real span.** Let $V = \operatorname{span}_\mathbb{R}(\Lambda)$. By definition, $V$ is the smallest $\mathbb{R}$-linear subspace of $\mathbb{R}^n$ containing $\Lambda$. Set $m := \dim_\mathbb{R} V \in \{0, 1, \ldots, n\}$. **Picking $\mathbb{R}$-independent elements of $\Lambda$.** Since $\Lambda$ spans $V$ over $\mathbb{R}$, there exist elements of $\Lambda$ that are $\mathbb{R}$-linearly independent in any maximal number, and that number is exactly $m$. We choose $y_1, \ldots, y_m \in \Lambda$ that are $\mathbb{R}$-linearly independent and $m$ in number (if $m = 0$, the list is empty and $\Lambda = \{0\}$, a trivial case). Concretely: if $m \geq 1$, pick any non-zero $y_1 \in \Lambda$; if $m \geq 2$, pick $y_2 \in \Lambda \setminus \mathbb{R}y_1$ (possible because $\dim V \geq 2$, so $\Lambda$ is not contained in the $1$-dimensional subspace $\mathbb{R}y_1$); continue until $m$ vectors are chosen. **Consequence of maximality.** The set $\{y_1, \ldots, y_m\}$ is an $\mathbb{R}$-basis of $V$ (spanning: $V = \operatorname{span}(\Lambda) \supseteq \operatorname{span}(y_1, \ldots, y_m)$ and both have $\mathbb{R}$-dimension $m$; linearly independent by construction). Hence every $v \in V$ — in particular every $v \in \Lambda$ — has a unique expression \begin{align*} v = \sum_{i=1}^m \lambda_i(v) y_i, \qquad \lambda_i(v) \in \mathbb{R}. \end{align*} This expression is the content of "maximally $\mathbb{R}$-linearly independent": any additional vector $v \in \Lambda$ is already in $\operatorname{span}_\mathbb{R}(y_1, \ldots, y_m) = V$. **Why rational coefficients?** The next step will upgrade $\lambda_i \in \mathbb{R}$ to $\lambda_i \in \mathbb{Q}$ by a counting argument: if some $\lambda_i$ were irrational, then the set of integer translates $v - \sum n_i y_i$ would produce infinitely many distinct lattice points inside a compact parallelepiped, contradicting the discreteness of $\Lambda$. The detailed argument is in Step 3.[/guided]

[step:Show that coefficients of elements of $\Lambda$ with respect to $y_1, \ldots, y_m$ are rational, via finiteness of the fundamental domain intersection]Define the half-open parallelepiped \begin{align*} X = \left\{\sum_{i=1}^m \lambda_i y_i : \lambda_i \in [0,1]\right\}. \end{align*} $X$ is the image of the compact set $[0,1]^m$ under the continuous linear map $(\lambda_1, \ldots, \lambda_m) \mapsto \sum_i \lambda_i y_i$, hence $X$ is compact. **Claim.** $X \cap \Lambda$ is finite. *Proof of claim.* The intersection of a compact set with a discrete subset of $\mathbb{R}^n$ is finite. More carefully: every $v \in \Lambda$ has an open neighbourhood $U_v$ with $U_v \cap \Lambda = \{v\}$ (by discreteness). The collection $\{U_v\}_{v \in X \cap \Lambda}$ is an open cover of $X \cap \Lambda$, and $X \cap \Lambda$ is a subset of the compact set $X$. If $X \cap \Lambda$ were infinite, an infinite subset would have a limit point $p \in X$ (Bolzano-Weierstrass in $\mathbb{R}^n$, applied to the bounded sequence; the limit point lies in $X$ since $X$ is closed). Then every neighbourhood of $p$ contains infinitely many elements of $\Lambda$, contradicting discreteness at $p$ (if $p \in \Lambda$) or yielding a non-isolated accumulation of $\Lambda$ away from itself (which also contradicts discreteness by a limit argument). Hence $X \cap \Lambda$ is finite. Let $F = X \cap \Lambda$ and $k = |F| < \infty$. **Claim.** Every $v \in \Lambda$ lies in $F + \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m$, i.e., $v = f + \sum_i n_i y_i$ for some $f \in F$ and $n_i \in \mathbb{Z}$. *Proof of claim.* Write $v = \sum_i \lambda_i y_i$ with $\lambda_i \in \mathbb{R}$ (from Step 2). Decompose each $\lambda_i = n_i + \theta_i$ with $n_i = \lfloor \lambda_i \rfloor \in \mathbb{Z}$ and $\theta_i = \{\lambda_i\} \in [0, 1)$. Then \begin{align*} v - \sum_i n_i y_i = \sum_i \theta_i y_i \in X \end{align*} (since $\theta_i \in [0, 1) \subseteq [0, 1]$), and $v - \sum_i n_i y_i \in \Lambda$ (since $\Lambda$ is a subgroup and each $y_i \in \Lambda$, so $\sum_i n_i y_i \in \Lambda$ and the difference lies in $\Lambda$). Therefore $v - \sum_i n_i y_i \in X \cap \Lambda = F$, as required. **Corollary.** $\Lambda / (\mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m)$ has order at most $k$. *Proof of corollary.* Every coset has a representative in $F$, and $|F| = k$; so the number of cosets is at most $k$.[/step]

[guided]The third step is the heart of the converse argument: we combine discreteness of $\Lambda$ with compactness of the "unit parallelepiped" $X$ to force $\Lambda$ to be a finite extension of a free $\mathbb{Z}$-module. **Defining $X$.** The set \begin{align*} X = \left\{\sum_{i=1}^m \lambda_i y_i : \lambda_i \in [0,1]\right\} \end{align*} is a closed parallelepiped in $V \subseteq \mathbb{R}^n$, with "sides" the vectors $y_i$. **Compactness of $X$.** Define the continuous linear map \begin{align*} \Phi : [0,1]^m &\to \mathbb{R}^n \\ (\lambda_1, \ldots, \lambda_m) &\mapsto \sum_{i=1}^m \lambda_i y_i. \end{align*} $\Phi$ is continuous (linear in finitely many variables). The domain $[0,1]^m$ is compact (Heine-Borel / Tychonoff: product of compact intervals is compact). The continuous image of a compact set is compact, so $X = \Phi([0,1]^m)$ is compact. **Finiteness of $X \cap \Lambda$.** We now use discreteness. By hypothesis, every $v \in \Lambda$ has a radius $\varepsilon_v > 0$ with $B(v, \varepsilon_v) \cap \Lambda = \{v\}$. Suppose for contradiction that $X \cap \Lambda$ is infinite. Then we have an infinite sequence $(v_k)_{k=1}^\infty$ of distinct points in $X \cap \Lambda \subseteq X$. Since $X$ is compact (hence sequentially compact, because $\mathbb{R}^n$ is metric), the sequence has a convergent subsequence, say $v_{k_j} \to p$ for some $p \in X$. Since the $v_{k_j}$ are distinct, the subsequence is not eventually constant, so for every $\delta > 0$ there are infinitely many $j$ with $0 < |v_{k_j} - p| < \delta$. **Case 1: $p \in \Lambda$.** Take $\delta = \varepsilon_p$. Then some $v_{k_j}$ with $v_{k_j} \neq p$ lies in $B(p, \varepsilon_p) \cap \Lambda$, contradicting $B(p, \varepsilon_p) \cap \Lambda = \{p\}$. **Case 2: $p \notin \Lambda$.** Consider the points $w_j = v_{k_j} - v_{k_1} \in \Lambda$. As $v_{k_j} \to p$, we have $w_j \to p - v_{k_1} =: q$. Either $q \in \Lambda$ (then Case 1 applies with $p$ replaced by $q$) or $q \notin \Lambda$; in the latter case, take $\delta$ small enough that $B(q, \delta) \cap \Lambda = \varnothing$ (such $\delta$ exists iff $q$ is not an accumulation point of $\Lambda$). If no such $\delta$ exists, $q$ is an accumulation point of $\Lambda$ not in $\Lambda$, which means infinitely many $w_j \in \Lambda$ converge to $q$; then some $w_j \neq w_{j'}$ have $|w_j - w_{j'}| < \varepsilon_{w_j}$, contradicting $B(w_j, \varepsilon_{w_j}) \cap \Lambda = \{w_j\}$. In all cases we reach a contradiction, so $X \cap \Lambda$ is finite. Write $F := X \cap \Lambda$ and $k := |F| < \infty$. (Cleaner argument: translating the difference $v_{k_{j+1}} - v_{k_j} \in \Lambda$ produces a sequence in $\Lambda$ with $|v_{k_{j+1}} - v_{k_j}| \to 0$, and for any fixed $v_{k_j}$ this gives non-zero elements of $\Lambda$ arbitrarily close to $v_{k_j}$, contradicting discreteness.) **Every $v \in \Lambda$ is $F$ plus a $\mathbb{Z}$-combination.** For any $v \in \Lambda$, write $v = \sum_i \lambda_i y_i$ with $\lambda_i \in \mathbb{R}$ (Step 2). Decompose \begin{align*} \lambda_i = n_i + \theta_i, \qquad n_i := \lfloor \lambda_i \rfloor \in \mathbb{Z}, \quad \theta_i := \{\lambda_i\} \in [0, 1). \end{align*} Define \begin{align*} f := v - \sum_i n_i y_i = \sum_i \theta_i y_i. \end{align*} Then: - $f \in \Lambda$: the element $\sum_i n_i y_i$ is in $\mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m \subseteq \Lambda$ (each $y_i \in \Lambda$, and $\Lambda$ is a subgroup); hence $f = v - \sum_i n_i y_i \in \Lambda$. - $f \in X$: $\theta_i \in [0, 1) \subseteq [0, 1]$, so $f = \sum \theta_i y_i$ is in $X$. - Hence $f \in X \cap \Lambda = F$, and $v = f + \sum_i n_i y_i$. **Consequence for the quotient.** Let $M := \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m \subseteq \Lambda$. The above shows every coset $v + M \in \Lambda / M$ has a representative in $F$. Different elements of $F$ could give the same coset (if $f - f' \in M$), so the map $F \to \Lambda/M$ is surjective, hence \begin{align*} |\Lambda / M| \leq |F| = k < \infty. \end{align*} This is the key finiteness result that powers the final step.[/guided]

[step:Upgrade the finite-index extension to a free $\mathbb{Z}$-module of rank $m$ via the structure theorem]Let $M := \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m \subseteq \Lambda$. Set $d := |\Lambda/M| \in \mathbb{Z}_{\geq 1}$. By Lagrange's theorem applied to the abelian group $\Lambda/M$, $d \cdot v = 0$ in $\Lambda/M$ for every $v \in \Lambda$, i.e., $dv \in M$. Hence \begin{align*} dM \subseteq d\Lambda \subseteq M \subseteq \Lambda. \end{align*} (The first inclusion is trivial since $M \subseteq \Lambda$; the second is the Lagrange consequence; the third is by construction.) **Embedding $\Lambda$ in a free $\mathbb{Z}$-module.** Define \begin{align*} \Lambda' := \tfrac{1}{d} M = \mathbb{Z}\tfrac{y_1}{d} + \cdots + \mathbb{Z}\tfrac{y_m}{d}. \end{align*} Since $d \Lambda \subseteq M$, we have $\Lambda \subseteq \tfrac{1}{d} M = \Lambda'$. The group $\Lambda'$ is free abelian of rank $m$ with basis $\tfrac{y_1}{d}, \ldots, \tfrac{y_m}{d}$ (which are still $\mathbb{R}$-linearly independent because the $y_i$ are). **Applying the structure theorem.** By the structure theorem for finitely generated abelian groups, any subgroup of a free abelian group of finite rank $m$ is itself free abelian of rank at most $m$. Hence $\Lambda \subseteq \Lambda'$ is free abelian of some rank $m' \leq m$. But $\Lambda$ contains the rank-$m$ submodule $M$, so $\operatorname{rank}(\Lambda) \geq \operatorname{rank}(M) = m$. Hence $\operatorname{rank}(\Lambda) = m$. **Finding an $\mathbb{R}$-independent $\mathbb{Z}$-basis.** Let $x_1, \ldots, x_m$ be a $\mathbb{Z}$-basis of $\Lambda$ (which exists by the freeness just established). We must check that the $x_i$ are $\mathbb{R}$-linearly independent. Since $\Lambda \subseteq V = \operatorname{span}_\mathbb{R}(y_1, \ldots, y_m)$ and $\Lambda \supseteq M = \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m$, we have $\operatorname{span}_\mathbb{R}(\Lambda) = V$, so $\operatorname{span}_\mathbb{R}(x_1, \ldots, x_m) = V$. An $m$-element spanning set in an $m$-dimensional $\mathbb{R}$-vector space is a basis, hence $x_1, \ldots, x_m$ are $\mathbb{R}$-linearly independent. **Conclusion.** $\Lambda = \mathbb{Z}x_1 + \cdots + \mathbb{Z}x_m$ with $x_1, \ldots, x_m \in \mathbb{R}^n$ linearly independent over $\mathbb{R}$, as required. This completes the converse direction, and together with Step 1 completes the proof.[/step]

[guided]Step 3 gave us that $\Lambda$ is a finite-index supergroup of the free $\mathbb{Z}$-module $M = \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m$. The final step turns this finite-index condition into the conclusion that $\Lambda$ itself is a free $\mathbb{Z}$-module of rank $m$ with an $\mathbb{R}$-independent basis. **Setting up notation.** Let $M = \mathbb{Z}y_1 + \cdots + \mathbb{Z}y_m \subseteq \Lambda$ and $d = |\Lambda/M| \in \mathbb{Z}_{\geq 1}$ (finite by Step 3). The $y_i$ are $\mathbb{R}$-linearly independent, so $M$ is free abelian of rank $m$. **Lagrange's theorem inside $\Lambda/M$.** $\Lambda/M$ is a finite abelian group of order $d$. By Lagrange's theorem, every element $v + M$ satisfies $d(v + M) = 0$, i.e., $dv \in M$. Equivalently, $d\Lambda \subseteq M$. **Sandwiching $\Lambda$ between free modules of rank $m$.** We now have the chain \begin{align*} dM \subseteq d\Lambda \subseteq M \subseteq \Lambda. \end{align*} Enlarging the outermost inclusion: since $d\Lambda \subseteq M$, every element of $\Lambda$ is $(1/d)$ times an element of $M$, so $\Lambda \subseteq (1/d) M$. Define \begin{align*} \Lambda' := (1/d) M = \mathbb{Z}\tfrac{y_1}{d} + \cdots + \mathbb{Z}\tfrac{y_m}{d}. \end{align*} $\Lambda'$ is free abelian of rank $m$ with basis $\tfrac{y_i}{d}$. (Scaling a free basis by a non-zero scalar gives another free basis of a rescaled free module.) Thus we have the **sandwich** \begin{align*} M \subseteq \Lambda \subseteq \Lambda', \end{align*} where $M$ and $\Lambda'$ are both free abelian of rank $m$. **Applying the structure theorem.** The **structure theorem for finitely generated abelian groups** (equivalently, the structure of submodules of finitely generated free modules over a PID) says: any subgroup of a finitely generated free abelian group is itself free abelian of rank at most the ambient rank. Applied to $\Lambda \subseteq \Lambda'$: \begin{align*} \Lambda \text{ is free abelian of some rank } m' \leq m. \end{align*} **Lower bound on the rank.** Since $M \subseteq \Lambda$ and $M$ has rank $m$, $\Lambda$ also has rank $\geq m$. (Rank is monotone: any $\mathbb{Z}$-basis of $M$ gives $m$ $\mathbb{Z}$-independent elements of $\Lambda$, so $\operatorname{rank}_\mathbb{Z}(\Lambda) \geq m$.) Combined with $m' \leq m$: \begin{align*} \operatorname{rank}_\mathbb{Z}(\Lambda) = m. \end{align*} **Choosing a $\mathbb{Z}$-basis.** Free abelian of rank $m$ means $\Lambda$ admits a $\mathbb{Z}$-basis $x_1, \ldots, x_m$: every element of $\Lambda$ is a unique $\mathbb{Z}$-linear combination of the $x_i$. **Verifying $\mathbb{R}$-independence.** The $x_i$ are $\mathbb{Z}$-independent by construction, but we need $\mathbb{R}$-independence. Consider the $\mathbb{R}$-span: \begin{align*} \operatorname{span}_\mathbb{R}(x_1, \ldots, x_m) \supseteq \operatorname{span}_\mathbb{R}(\mathbb{Z}x_1 + \cdots + \mathbb{Z}x_m) = \operatorname{span}_\mathbb{R}(\Lambda) = V, \end{align*} so the $x_i$ span $V$ over $\mathbb{R}$. A spanning set of $m$ vectors in an $m$-dimensional $\mathbb{R}$-vector space must be a basis (dimension equals both the size of a maximal independent set and the minimal size of a spanning set). Hence $x_1, \ldots, x_m$ are $\mathbb{R}$-linearly independent. **Final assembly.** We have exhibited $x_1, \ldots, x_m \in \mathbb{R}^n$ that are $\mathbb{R}$-linearly independent with \begin{align*} \Lambda = \mathbb{Z}x_1 + \cdots + \mathbb{Z}x_m = \left\{\sum_{i=1}^m n_i x_i : n_i \in \mathbb{Z}\right\}. \end{align*} This is the desired form. Together with Step 1 (the "lattice $\Rightarrow$ discrete" direction), both implications of the theorem are proved. **Remark on the trivial case.** If $\Lambda = \{0\}$, then $m = 0$ and the statement reads $\Lambda = \{\sum_{i=1}^0 n_i x_i\} = \{0\}$ (empty sum), which is true. The argument above works verbatim with $m = 0$: the $\mathbb{R}$-span is $\{0\}$, the parallelepiped $X$ is $\{0\}$, $F = \{0\}$, and $\Lambda = \{0\} = \mathbb{Z} \cdot \emptyset$.[/guided]