[proofplan] Part (1) is a pigeonhole argument over a fundamental parallelepiped: since $P$ tiles $\mathbb{R}^n$ under the lattice $\Lambda$, the volume of $S$ decomposes as a sum of volumes of translates $S \cap (P + \gamma)$, which by translation invariance equal the volumes of $(S - \gamma) \cap P$. If these pieces were pairwise disjoint they would fit inside $P$, whose total volume is $\operatorname{covol}(\Lambda)$, contradicting $\operatorname{vol}(S) > \operatorname{covol}(\Lambda)$. Part (2a) applies (1) to the halved set $\tfrac{1}{2}S$, whose volume exceeds $\operatorname{covol}(\Lambda)$, and uses symmetry and convexity of $S$ to convert the output into a non-zero lattice point inside $S$. Part (2b) upgrades the strict volume bound in (2a) to the closed-body equality case by applying (2a) to shrinking dilations and extracting a convergent subsequence using the discreteness of $\Lambda$ and compactness of $2S$. [/proofplan]

[step:Fix a fundamental parallelepiped $P$ and decompose $\operatorname{vol}(S)$ along its $\Lambda$-translates]Choose a $\mathbb{Z}$-basis $b_1, \ldots, b_n$ of $\Lambda$ and let \begin{align*} P &= \Bigl\{\sum_{i=1}^n t_i b_i : t_i \in [0, 1)\Bigr\} \end{align*} be the associated fundamental parallelepiped, so that $\operatorname{vol}(P) = \operatorname{covol}(\Lambda)$ by definition of the covolume. The translates $\{P + \gamma : \gamma \in \Lambda\}$ partition $\mathbb{R}^n$ up to a set of $\mathcal{L}^n$-measure zero. Hence, partitioning $S$ into its intersections with these translates: \begin{align*} \operatorname{vol}(S) &= \sum_{\gamma \in \Lambda} \operatorname{vol}\bigl(S \cap (P + \gamma)\bigr). \end{align*} By translation invariance of $\mathcal{L}^n$ under the shift by $-\gamma$, which maps $S \cap (P + \gamma)$ bijectively onto $(S - \gamma) \cap P$: \begin{align*} \operatorname{vol}\bigl(S \cap (P + \gamma)\bigr) &= \operatorname{vol}\bigl((S - \gamma) \cap P\bigr). \end{align*} Therefore \begin{align*} \operatorname{vol}(S) &= \sum_{\gamma \in \Lambda} \operatorname{vol}\bigl((S - \gamma) \cap P\bigr). \end{align*}[/step]

[guided]The strategy for part (1) is pigeonhole in a measure-theoretic disguise. A finite pigeonhole says: if you place objects into boxes whose total capacity is less than the total size of the objects, two must overlap. We set up this configuration with the fundamental parallelepiped $P$ as the "box" and the pieces $(S - \gamma) \cap P$ as the objects competing for space in $P$. **Fundamental parallelepiped.** Pick any $\mathbb{Z}$-basis $b_1, \ldots, b_n$ of the lattice $\Lambda$ (it has rank $n$ by assumption). The fundamental parallelepiped \begin{align*} P &= \Bigl\{\sum_{i=1}^n t_i b_i : t_i \in [0, 1)\Bigr\} \end{align*} is a half-open $n$-cell whose translates under $\Lambda$ tile $\mathbb{R}^n$ exactly: every point $x \in \mathbb{R}^n$ can be written uniquely as $x = p + \gamma$ with $p \in P$ and $\gamma \in \Lambda$. The volume of $P$ is the covolume: \begin{align*} \operatorname{vol}(P) &= \operatorname{covol}(\Lambda). \end{align*} **Partitioning $S$.** Since $\{P + \gamma\}_{\gamma \in \Lambda}$ is a partition of $\mathbb{R}^n$ (essentially; the boundaries overlap, but boundaries have $\mathcal{L}^n$-measure zero), the measurable set $S$ decomposes as a disjoint union up to measure zero: \begin{align*} S &= \bigsqcup_{\gamma \in \Lambda} \bigl(S \cap (P + \gamma)\bigr). \end{align*} By countable additivity of $\mathcal{L}^n$: \begin{align*} \operatorname{vol}(S) &= \sum_{\gamma \in \Lambda} \operatorname{vol}\bigl(S \cap (P + \gamma)\bigr). \end{align*} **Translation invariance.** The Lebesgue measure $\mathcal{L}^n$ is translation-invariant, so shifting by $-\gamma$ preserves volume. The set $S \cap (P + \gamma)$ shifts to $(S - \gamma) \cap P$: \begin{align*} \operatorname{vol}\bigl(S \cap (P + \gamma)\bigr) &= \operatorname{vol}\bigl((S - \gamma) \cap P\bigr). \end{align*} We have rewritten the total volume of $S$ as a sum of volumes of subsets $(S - \gamma) \cap P \subseteq P$ of a single bounded region.[/guided]

[step:Extract two overlapping translates from the volume inequality]Under the hypothesis $\operatorname{vol}(S) > \operatorname{covol}(\Lambda) = \operatorname{vol}(P)$, suppose for contradiction that the sets $(S - \gamma) \cap P$ are pairwise disjoint for all $\gamma \in \Lambda$. Since each is a subset of $P$, by countable subadditivity: \begin{align*} \sum_{\gamma \in \Lambda} \operatorname{vol}\bigl((S - \gamma) \cap P\bigr) &\leq \operatorname{vol}(P) = \operatorname{covol}(\Lambda). \end{align*} Combined with Step 1, this gives $\operatorname{vol}(S) \leq \operatorname{covol}(\Lambda)$, contradicting the hypothesis. Hence there exist distinct $\gamma, \mu \in \Lambda$ with \begin{align*} \bigl((S - \gamma) \cap P\bigr) \cap \bigl((S - \mu) \cap P\bigr) &\neq \varnothing. \end{align*} Pick a point $p$ in this intersection. Then $p \in P$, $p = x - \gamma$ for some $x \in S$, and $p = y - \mu$ for some $y \in S$. Hence \begin{align*} x - y &= \gamma - \mu \in \Lambda. \end{align*} Since $\gamma \neq \mu$, we have $x - y \neq 0$, so in particular $x \neq y$. This proves part (1).[/step]

[guided]**Pigeonhole via volumes.** We have $\sum_\gamma \operatorname{vol}((S - \gamma) \cap P) = \operatorname{vol}(S) > \operatorname{covol}(\Lambda) = \operatorname{vol}(P)$. All the sets $(S - \gamma) \cap P$ are contained in $P$. If they were pairwise disjoint, their combined volume would satisfy \begin{align*} \sum_\gamma \operatorname{vol}\bigl((S - \gamma) \cap P\bigr) &= \operatorname{vol}\biggl(\bigsqcup_\gamma (S - \gamma) \cap P\biggr) \leq \operatorname{vol}(P), \end{align*} contradicting $\sum_\gamma \operatorname{vol}((S - \gamma) \cap P) > \operatorname{vol}(P)$. So at least two pieces overlap. **Extracting $x, y \in S$.** Two distinct pieces $(S - \gamma) \cap P$ and $(S - \mu) \cap P$ overlap means there is a common point $p \in P$ of the form $p = x - \gamma = y - \mu$ with $x, y \in S$ (the two representations come from the two different pieces). This gives $x - y = \gamma - \mu$, a non-zero lattice vector (since $\gamma \neq \mu$), and forces $x \neq y$ because $\Lambda$ contains no element equal to $0$ except $0$ itself. **Rigor on measure zero.** The boundaries of the translates $P + \gamma$ do overlap, but on a null set. The argument is cleanly valid by working up to $\mathcal{L}^n$-null sets: the equality $\operatorname{vol}(S) = \sum_\gamma \operatorname{vol}((S - \gamma) \cap P)$ holds exactly regardless of the null-set boundary behaviour, and the derived contradiction $\operatorname{vol}(S) \leq \operatorname{vol}(P)$ uses only the volume identity. The existence of a common point follows from the combinatorial fact that two sets cannot cover strictly more volume than their union.[/guided]

[step:Deduce part (2a) by applying part (1) to $\tfrac{1}{2}S$]Assume $S$ is convex, symmetric about $\mathbf{0}$, and $\operatorname{vol}(S) > 2^n \operatorname{covol}(\Lambda)$. Let \begin{align*} S' &= \tfrac{1}{2} S = \{\tfrac{1}{2} s : s \in S\}. \end{align*} Under the dilation $x \mapsto \tfrac{1}{2} x$, the Lebesgue measure scales by $(1/2)^n = 2^{-n}$: \begin{align*} \operatorname{vol}(S') &= 2^{-n} \operatorname{vol}(S) > 2^{-n} \cdot 2^n \operatorname{covol}(\Lambda) = \operatorname{covol}(\Lambda). \end{align*} Part (1) applied to $S'$ produces distinct $y, z \in S'$ with $y - z \in \Lambda$. By definition of $S'$, there exist $a, b \in S$ with $y = a/2$ and $z = b/2$. Then \begin{align*} y - z &= \tfrac{1}{2}(a - b) = \tfrac{1}{2}\bigl(a + (-b)\bigr). \end{align*} By symmetry of $S$ about $\mathbf{0}$, $-b \in S$. By convexity of $S$, the midpoint $\tfrac{1}{2}(a + (-b))$ of $a, -b \in S$ also lies in $S$. Hence \begin{align*} \gamma &:= y - z = \tfrac{1}{2}\bigl(a + (-b)\bigr) \in S, \end{align*} and $\gamma \in \Lambda \setminus \{0\}$ (non-zero because $y \neq z$).[/step]

[guided]**Why halve $S$?** Part (1) gives a difference $x - y \in \Lambda \setminus \{0\}$ with $x, y \in S$. This difference is of the form "something in $S$ minus something in $S$," which is not automatically in $S$. The trick is to first halve $S$: any difference of two points in $\tfrac{1}{2}S$ equals the half of a sum of two points in $S$, and such half-sums lie in $S$ itself by convexity. **Dilation and volume.** The scaling $x \mapsto \tfrac{1}{2}x$ multiplies Lebesgue measure by $(1/2)^n = 2^{-n}$, so \begin{align*} \operatorname{vol}(\tfrac{1}{2}S) &= 2^{-n}\operatorname{vol}(S). \end{align*} The hypothesis $\operatorname{vol}(S) > 2^n \operatorname{covol}(\Lambda)$ converts exactly to $\operatorname{vol}(\tfrac{1}{2}S) > \operatorname{covol}(\Lambda)$, which is the hypothesis of part (1). **Applying part (1) to $\tfrac{1}{2}S$.** Part (1) produces distinct $y, z \in \tfrac{1}{2}S$ with $y - z \in \Lambda$ (automatically non-zero, as $y \neq z$). Writing $y = a/2$ and $z = b/2$ with $a, b \in S$: \begin{align*} y - z &= \frac{a - b}{2} = \frac{a + (-b)}{2}. \end{align*} **Using symmetry and convexity.** The set $S$ is symmetric about $\mathbf{0}$, so $-b \in S$. Now both $a$ and $-b$ lie in $S$, and convexity of $S$ gives the midpoint $(a + (-b))/2 \in S$. Therefore \begin{align*} \gamma := y - z \in S \cap \Lambda \setminus \{0\}, \end{align*} which is exactly what part (2a) asserts. **Why both symmetry and convexity.** Symmetry converts subtraction into a form of addition ($-b \in S$), and convexity averages two points in $S$ back into $S$. Drop either hypothesis and the proof fails: an arbitrary convex set need not contain differences of its points, and an arbitrary symmetric set need not be closed under midpoints.[/guided]

[step:Handle the closed case by taking a subsequence limit of dilations]Now assume $S$ is closed, convex, symmetric about $\mathbf{0}$, and $\operatorname{vol}(S) \geq 2^n \operatorname{covol}(\Lambda)$ (equality is allowed). For each integer $m \geq 1$, let \begin{align*} S_m &= \Bigl(1 + \frac{1}{m}\Bigr) S = \Bigl\{\Bigl(1 + \frac{1}{m}\Bigr) s : s \in S\Bigr\}. \end{align*} Each $S_m$ is convex, symmetric about $\mathbf{0}$, and \begin{align*} \operatorname{vol}(S_m) &= \Bigl(1 + \frac{1}{m}\Bigr)^n \operatorname{vol}(S) \geq \Bigl(1 + \frac{1}{m}\Bigr)^n \cdot 2^n \operatorname{covol}(\Lambda) > 2^n \operatorname{covol}(\Lambda). \end{align*} Part (2a) applies to each $S_m$: there exists a non-zero $\gamma_m \in S_m \cap \Lambda$.[/step]

[guided]**Why the dilation.** Part (2a) requires strict inequality $\operatorname{vol}(S) > 2^n \operatorname{covol}(\Lambda)$. For the closed case we have only $\geq$, so we enlarge $S$ slightly to gain the strict inequality and then shrink back via a limit. **Dilating $S$.** The factor $1 + 1/m > 1$ strictly enlarges $S$, so each $S_m = (1 + 1/m)S$ contains $S$. Dilation by a scalar $\lambda > 0$ scales $\mathcal{L}^n$-measure by $\lambda^n$: \begin{align*} \operatorname{vol}(S_m) &= (1 + 1/m)^n \operatorname{vol}(S). \end{align*} Since $(1 + 1/m)^n > 1$ and $\operatorname{vol}(S) \geq 2^n \operatorname{covol}(\Lambda)$, we have $\operatorname{vol}(S_m) > 2^n \operatorname{covol}(\Lambda)$ strictly. **Properties preserved under dilation.** Dilation by a positive scalar preserves convexity (convex combinations are homogeneous), symmetry about $\mathbf{0}$ (negation commutes with positive scalar multiplication), and — importantly — does not require closedness for applying (2a), since (2a) uses only the strict volume inequality, convexity, and symmetry. **Applying (2a) to each $S_m$.** By part (2a) applied to $S_m$, there exists a non-zero $\gamma_m \in S_m \cap \Lambda$. We now have a sequence $(\gamma_m)_{m \geq 1}$ of non-zero lattice points, each in the respective dilation $S_m$. We must extract a limit that lies in $S$ itself.[/guided]

[step:Extract a constant subsequence using discreteness of $\Lambda$ and compactness of $2S$]For each $m \geq 1$, $\gamma_m \in S_m \subseteq 2S$ (since $1 + 1/m \leq 2$). Hence all $\gamma_m$ lie in the bounded set $2S \cap \Lambda$. [claim:Finiteness of $2S \cap \Lambda$] The intersection $2S \cap \Lambda$ is finite. [/claim] [proof] The set $2S = \{2s : s \in S\}$ is the continuous image of $S$ under the dilation $s \mapsto 2s$, which is a homeomorphism $\mathbb{R}^n \to \mathbb{R}^n$. Since $S$ is closed, $2S$ is closed. Moreover $S$ is bounded: the finite volume $\operatorname{vol}(S) < \infty$ together with convexity implies $S$ is bounded. To see this, note that a convex set of finite volume cannot contain a half-line (a half-line has infinite volume in any convex neighbourhood of it); a bounded convex set is contained in some ball $B(0, R)$, so $2S \subseteq B(0, 2R)$. Therefore $2S$ is closed and bounded in $\mathbb{R}^n$, hence compact. The lattice $\Lambda$ is discrete: every ball $B(0, r)$ meets $\Lambda$ in a finite set. Since $2S \subseteq B(0, 2R)$, the intersection $2S \cap \Lambda \subseteq B(0, 2R) \cap \Lambda$ is a subset of a finite set, hence finite. [/proof] By the claim, the sequence $(\gamma_m)_{m \geq 1}$ takes values in a finite set, so by the pigeonhole principle, some value $\gamma \in 2S \cap \Lambda$ appears infinitely often. Extract a subsequence $(\gamma_{m_k})$ that is constant equal to $\gamma$; since each $\gamma_{m_k} \neq 0$, we have $\gamma \neq 0$.[/step]

[guided]**Boundedness of $S$.** The Minkowski body $S$ has finite volume (at least we assume $\operatorname{vol}(S) \geq 2^n \operatorname{covol}(\Lambda) < \infty$). For a convex set $S$, finite volume implies boundedness: if $S$ contained an unbounded direction, then for every $r > 0$, $S$ would contain a half-line of length $\geq r$ together with a convex neighbourhood, contributing infinite volume. Formally, suppose $S$ is unbounded; then there is a sequence $x_k \in S$ with $|x_k| \to \infty$. By symmetry, $-x_k \in S$, and by convexity, the segments $[-x_k, x_k] \subseteq S$. Any convex set $\operatorname{conv}(B(0, \varepsilon) \cup [-x_k, x_k])$ with $B(0, \varepsilon) \subseteq S$ has volume $\to \infty$ as $k \to \infty$, contradicting finiteness. Hence $S \subseteq B(0, R)$ for some $R > 0$. **Compactness of $2S$.** $S$ closed and bounded in $\mathbb{R}^n$ means $S$ is compact by the Heine-Borel theorem. Dilations $s \mapsto 2s$ are continuous, so $2S$ is the continuous image of a compact set, hence compact. **Discreteness of $\Lambda$.** A lattice $\Lambda \subset \mathbb{R}^n$ of rank $n$ is discrete: there is a minimum distance $\delta > 0$ such that all non-zero $\gamma, \gamma' \in \Lambda$ satisfy $|\gamma - \gamma'| \geq \delta$. Consequently any bounded subset of $\Lambda$ is finite: if $A \subseteq \Lambda$ with $\sup_{\gamma \in A} |\gamma| \leq R$, then $A$ has at most $|A| \leq (\operatorname{vol}(B(0, R+\delta/2))/\operatorname{vol}(B(0, \delta/2))$ elements by a ball-packing argument. **Compactness $\cap$ discreteness.** The intersection $2S \cap \Lambda$ is the intersection of a compact set with a discrete set, hence finite (a standard consequence: any open cover of $2S$ by discs of radius $\delta/2$ centred at lattice points has a finite subcover by compactness, and each disc contains at most one lattice point). **Pigeonhole for the sequence.** The sequence $(\gamma_m)$ takes infinitely many values in a finite set, so by the pigeonhole principle some value $\gamma$ is repeated infinitely often. Pass to that constant subsequence. Because every term of the original sequence is non-zero, the limit value $\gamma$ is also non-zero.[/guided]

[step:Show the limit point $\gamma$ lies in $S$ using closedness]For each $k \geq 1$, $\gamma = \gamma_{m_k} \in S_{m_k} = (1 + 1/m_k) S$, so \begin{align*} \frac{\gamma}{1 + 1/m_k} &\in S. \end{align*} As $k \to \infty$, $1/m_k \to 0$ (since $(m_k)$ is a subsequence of $(m)_{m \geq 1}$, hence $m_k \to \infty$), and $\gamma/(1 + 1/m_k) \to \gamma$ in $\mathbb{R}^n$. Since $S$ is closed, the limit $\gamma$ belongs to $S$. Combined with $\gamma \in \Lambda \setminus \{0\}$ from the previous step, this gives a non-zero lattice point $\gamma \in S \cap \Lambda$, completing the proof of part (2b).[/step]

[guided]**Writing $\gamma$ as a point of $S$.** The relation $\gamma \in S_{m_k} = (1 + 1/m_k)S$ means $\gamma = (1 + 1/m_k) s_k$ for some $s_k \in S$, equivalently \begin{align*} s_k &= \frac{\gamma}{1 + 1/m_k} \in S. \end{align*} **Taking the limit.** As $k \to \infty$, $m_k \to \infty$, so $1/m_k \to 0$ and $(1 + 1/m_k)^{-1} \to 1$. Therefore \begin{align*} s_k = \frac{\gamma}{1 + 1/m_k} \to \gamma \quad \text{in } \mathbb{R}^n. \end{align*} **Closing the sequence in $S$.** The set $S$ is closed in $\mathbb{R}^n$, so it contains all limits of its convergent sequences. The sequence $(s_k) \subseteq S$ converges to $\gamma$, hence $\gamma \in S$. **Why closedness is essential.** Without the closedness hypothesis, the limit $\gamma$ could escape $S$ (for example, if $S$ is open and $\gamma$ lies on its boundary). The closed hypothesis in part (2b) is precisely what allows the equality case $\operatorname{vol}(S) = 2^n \operatorname{covol}(\Lambda)$: we traded strict inequality in (2a) for closedness in (2b), using closedness to recover a lattice point on the boundary via a compactness-and-discreteness limit. **Combining conclusions.** From Step 5, $\gamma \in \Lambda \setminus \{0\}$. From this step, $\gamma \in S$. Hence $\gamma$ is a non-zero element of $S \cap \Lambda$, which is exactly the conclusion of part (2).[/guided]