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On Pg 116 of Lang's

Let $L$ be a lattice of dimension $N$ in $\mathbb{R}^N$. Let $C$ be a closed, convex, symmetric subset of $\mathbb{R}^N$. If $\mu(C) \ge 2^N \mu(F)$, where $F$ is a fundamental domain, then there exists a lattice point in $C$.

Lang's first proves the case when $C$ is not closed and $\mu(C)>2^N\mu(F)$. But I couldn't follow the proof:

For all $\varepsilon >0 $, $\mu(1+ \varepsilon)C) > \mu(C)$ hence there is a lattice point in $(1+\varepsilon)C$. Let $\varepsilon \rightarrow 0$, shows one of these lattice must be in $C$.

Why does the last line follow? I needed bounded $C$. What am I missing?

Bryan Shih
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The volume of an unbounded convex set $C$ is either $0$ or $\infty$. Indeed, a convex set is either contained in a hyperplane (and then has zero volume) or contains $(N+1)$ points in generic position (and then has nonempty interior). In the latter case, let $B$ be some ball contained in $C$. Since $C$ is unbounded, we have a point $p\in C$ arbitrarily far from $C$. The convex hull of $C\cup \{p\}$ contains a circular cone with fixed radius of the base (the radius of $B$) and atbitrarily large height. Hence $\operatorname{vol}(C) = \infty$.

So, every closed convex set of volume equal to $2^N\mu(F)$ is bounded.