Given a star body $S \subset \mathbb{R}^n$ with the origin as interior point, the *critical determinant of* $S$---usually denoted as $\Delta(S)$---is the infimum of the determinants of all lattices that intersect $S$ only
at the origin. 


The quantity ${\rm vol}(S)/\Delta(S)$ is a linear invariant of compact, star bodies. If we restrict its domain to the space $\mathcal{K}\_0^n$ of convex bodies in $\mathbb{R}^n$ that contain the origin in their interior, then it is a continuous linear invariant. 


**The question:** Is there a nice little proof or a reference for the statement that the the sublevel sets
$$
\{ K \in \mathcal{K}\_0^n : {\rm vol}(K)/\Delta(K) \leq C \}
$$
are compact and hence the functional must attain a global minimum?


I think this is true (because of Macbeath's compactness theorem and the fact that $\Delta(K)^{-1}$ blows up if the origin is near the boundary of $K$), but have not really checked thoroughly thinking that is must be widely known although I cannot find the explicit statement in Cassel or Lekkerkerker.


**Addendum.** Sergei has given an excellent reason why this is not to be found in Cassel or Lekkerkerker ... However, much work has gone into giving lower bounds of the functional 
$K \mapsto {\rm vol}(K)/\Delta(K)$. For example, the Minkowski-Hlawka theorem says it is bounded below by $1$ (or $\zeta(n)$ to be more accurate). *Is it known whether this functional attains a (global) minimum on* $\mathcal{K}\_0^n$ ?



A question of compactness in the geometry of numbers
It is *not* compact unless you allow the origin at the boundary (or maybe impose some kind of uniform strict convexity).


Consider a rectangle $K=[-1,1]\times[-\delta,1]$ in the plane, where $\delta$ is positive and goes to 0. If $K$ intersects some lattice only at the origin then so does $-K$, by symmetry. But $K\cup -K$ is a convex symmetric set of area 4, hence $\Delta(K)=\Delta(K\cup -K) \ge 1$ by Minkowski's theorem.


Therefore the ratio $vol/\Delta$ is bounded for all such $K$, but they have no limit in $\mathcal K^2\_0$ as $\delta\to 0$.


