# Lattice definition as finite density infinite set of vectors closed under addition?

This question came up in the definition of lattices (in the crystallography/group theory sense, not the ordered set sense) in our condensed matter lectures, but I believe it's more appropriate here than on the Physics StackExchange; please move it if I'm wrong.

Suppose we have an infinite set of vectors $V$ in real space $\mathbb{R}^n$ which we interpret as the position vectors of a set of points $S$. If we may write $V$ in the form $\{ \sum_{i=1}^n c_i \mathbf{e}_i : c_i \in\Bbb{Z}\}$ where the $\{\mathbf{e_i}\}$ are a set of $b \le n$ linearly independent vectors then we call $S$ a lattice. Is there a set $V$ which is closed under addition such that $S$ is not a lattice but has finite density of points (i.e. a finite number of points in any given finite volume)?

An example of what I mean in case I am still not formulating the problem correctly: The set of integers $\mathbb{Z}$ in $\mathbb{R}^1$ is a lattice in that we may find a basis vector $(1)$ for which all the vectors in $V$ may be written as integer multiples and further all integer sums of the basis $\{(1)\}$ appear in $V$. On the other hand the set $B=\{m+n\sqrt2 : m,n \in\mathbb{Z}\}$ is not a lattice because one needs two linearly dependent vectors, e.g $(1)$ and $(\sqrt 2)$ in order to write all the vectors in $B$ as an integer sum of basis vectors. $B$ is also closed under addition. However there are an infinite number of points in $B$ in the interval $(0,1)$ say, so it has infinite density. I would like to find a set which is like $B$ in that it satisfies the first two properties listed but has finite density.

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The relation between $S$ and $V$ seems a bit unclear. Are $V$ all pairwise differences between points in $S$? If so, then $v\in V$ and $s\in S$ does not imply $s+v\in S$. –  MvG Feb 12 '14 at 7:12
Yes, the integer multiples of a single nonzero vector in $\mathbb R^2$ is closed under addition and has "finite density", but is not an $2$-dimensional lattice. –  Samuel Feb 13 '14 at 15:38
Oh, yes, you're quite right @MvG ! It probably makes more sense to start from $V$ to construct $S$; I've tried to edit the question appropriately. Hopefully it now makes more sense; I'm sorry if it's still not right. –  maxwelldecoherence Feb 13 '14 at 15:38
@Samuel, that's certainly true! However in some not-very-mathematical sense it's basically a 1-dimensional lattice embedded in $\mathbb{R}^2$, which isn't quite what I'm looking for. I'll add the caveat that the set can also have less than $n$ basis vectors into the question, sorry for not being precise and keep having to change the question. Also sorry about "finite density"; I know it's not right but don't know the correct terminology! –  maxwelldecoherence Feb 13 '14 at 15:48

Another example is the set $\{n: n\geq 0\}$, which is closed under addition but is not a lattice. I believe that is not a satisfactory answer either, so I think what you are looking for is Theorem 17 from Lectures on the Geometry of Numbers by Carl Ludwig Siegel:
Theorem: A non-empty set which is closed under subtraction and which does not contain arbitrarily short vectors, can be written as $V=\{\sum_1^k c_i e_i, c_i\in\mathbb Z\}$ for some finite set of vectors $e_1,\ldots,e_k$. (That we can choose $k\leq n$ is guaranteed by Theorem 18 in the same book.)
Such a set will have "finite density", for if there were an infinite number of points inside a small ball, you could for any $\epsilon>0$ find two vectors, say $x$ and $y$, at a distance at most $\epsilon>0$ from each other, and then $x-y$ would be a vector in your set with length $\epsilon$.
You might want to dd an argument as to why $k\le n$, since by the edited definition, $k>n$ would not be a lattice in that definition, but still satisfy the above theorem. –  MvG Feb 13 '14 at 16:31