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In Lang - Algebraic number theory, theorem 3, chapter V, page 116, there is a version of Minkowski theorem:

Let $L$ be a lattice of dimension $N$ in $\mathbb R^N$, and let $C$ be a closed, convex, symmetric subset of $\mathbb R^N$. If $\mu(C)\geq 2^N\mu(F)$ where $F$ is a fundamental domain for $L$, there there exists a lattice point $\neq 0$ in $C$.

Here is not required $C$ to be bounded, so I'm some difficulties in proving this: if $(1+\epsilon)C\cap L\neq \{0\}$ for each $\epsilon>0$ then $C\cap L\neq\{0\}$. How to prove this without assuming $C$ bounded (hence compact)?

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  • $\begingroup$ In practice, applications of Minkowski's theorem use bounded $C$, so I think the omission of that hypothesis in Lang's book is an oversight rather than being intentional. $\endgroup$
    – KConrad
    Mar 4, 2012 at 16:21

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I claim that either $C$ is bounded or $\mu(C)=\infty$. In either case there is a closed ball $B$ centered at the origin such that $\mu(C\cap B)\geq 2^N\mu(F)$, and then one can apply the argument for $C\cap B$ in place of $C$.

To see the claim, observe first that $C$ is not contained in a hyperplane, hence it contains some closed ball $B(0,r)$ of radius $r>0$ centered at the origin. Let $c\in C\setminus B(0,r)$ be arbitrary. Then $C$ contains the convex hull of $c$ and $B(0,r)$, whose volume is $\gg r^{N-1}|c|$. Hence if $C$ is unbounded, then $\mu(C)=\infty$ as claimed.

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