Let $\Lambda$ be a lattice in $\mathbb{R}^n$. For $\bar{x} \in \mathbb{R}^n$, let 
$\| \bar{x} \| = max\_{1 \leq i \leq n} \{ |x\_i| \}$, i.e. the sup norm. Let $\lambda\_1, ..., \lambda\_n$
be a successive minima of $\Lambda$ with respect to the sup norm. 
Let $\lambda\_{j} \leq U < \lambda\_{j+1}$.


If we consider the set $S = \{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$, then by the definition of successive minima it follows that every point in $S$ must be a linear combination of $v\_1, ..., v\_j$. Since points achieving successive minima aren't necessarily a basis for the lattice, there can be points in $S$ where the coefficients in front of $v\_i$'s are not integers. I am sure there is an example of a lattice $\Gamma$, where this happens, but I could not come up with one and I am interested in seeing an actual example. I would appreciate any assistance/comments! Thank you! 


PS This question is related to my previous question, which was resolved [Counting number of points in a lattice with bounded sup norm](https://mathoverflow.net/questions/201806/counting-number-of-points-in-a-lattice-with-bounded-sup-norm) 



On successive minima and basis of a lattice
You can read about these things in Siegel-Chandrasekharan: Lectures on the geometry of numbers (Springer, 1988). In particular, Section X.5 contains an example for $n=3$ where the minimizing vectors do not form a basis of the lattice. The same section and the next one also explain that for practical purposes the successive minima can be replaced by slightly larger quantities such that the corresponding vectors do form a lattice basis.


