What does "closed" mean with respect to lattice properties?

Question

If a lattice $L$ is defined as an additive discrete subgroup of $\mathbb{R}^n$, then Nguyen on page 24 from "The LLL Algorithm - Survey and Applications", with respect to important properties of lattices, says $L$ is closed.

Lemma. Let $L$ be a lattice in $\mathbb{R}^n$

1. There exists $\rho > 0$ such that for all $\mathbf{x} \in L$: $L \cap B(\mathbf{x},\rho) = \left\{ \mathbf{x} \right\}.$
2. L is closed.

What does "closed" mean here in concrete terms? Does this refer to closure with respect to addition, or does this refer to some other kind of closedness? When I first read this, I was thinking of "closed under addition", now I think there might be something more to it, since there are also "closed sets". So I am a bit confused what "closed" means here exactly.

The closed under addition is clear (if that's what was meant). But assuming that closed is meant with respect to a closed set, how exactly can one imagine or justify this?

Reference: Phong Q. Nguyen, Brigitte Vallée, The LLL Algorithm - Survey and Applications, 2010

Answer 1

The answer to the question in the title is: closed with respect to the topology on $\mathbb{R}^n$. But let me also give some more explanations, including a proof of the statements above.

There are many equivalent definitions of being closed; in this case, we are in a metric space, so one can characterize closed sets as the sets containing all their limit points. Recall that a limit point is the limit of a sequence in the subset, which is convergent in the metric space.

We say that a set $S\subset\mathbb{R}^n$ is discrete if every point is isolated, i.e. for every $s\in S$, there exists some $r>0$ such that no point of $S\setminus\lbrace s\rbrace$ has distance less than $r$ to $s$. More mathematically, the ball around $s$ with radius $r$ does not intersect $S$ in other points, $B(s,r)\cap S=\lbrace s\rbrace$. Note that $r$ may depend on the choice of $s$.

Now, uniformly discrete means that there exists a "universal" distance $r>0$, such that for every point $s\in S$, $B(s,r)\cap S=\lbrace s\rbrace$. This is a very important distinction, allow me to give an example to show why.

Example. Take the set $N=\lbrace \frac{1}{n}\mid n\in\mathbb{N}\rbrace$. This lives on the real line $\mathbb{R}$, and is clearly discrete, as for every point $\frac{1}{n}$, the closest points are $\frac{1}{n-1}$ and $\frac{1}{n+1}$, but they both have distance more than $\frac{1}{(n+1)^2}$. Now, $N$ has a sequence $\lbrace \frac{1}{n}\rbrace_{n\in \mathbb{N}}$ (this is in fact the entire set $N$), and this sequence is convergent to $0$ in $\mathbb{R}$. However, $0\notin N$, so $N$ is not closed, even though it is discrete.

Let $L$ be a lattice, i.e. a discrete additive subgroup of $\mathbb{R}^n$. Then a priori, it is not necessarily closed, as we just saw in the example that discreteness does not ensure closedness. But the group structure does in fact ensure uniform discreteness (this is statement 1), and that implies closedness (statement 2). Let us prove these.

1. We need a further fact, which is that addition in $\mathbb{R}^n$ is an isometry, i.e. it preserves the distance. Mathematically, $d(x,y) = d(z+x,z+y)$ for any $x,y,z\in\mathbb{R}^n$. This is easily proven by recalling $d(x,y)=||x-y||$. Now, as $L$ is an additive subgroup, $0\in L$. By discreteness, we have some $r>0$ such that $B(0,r)\cap L=\lbrace 0\rbrace$. Let $s\in L$, and take any point $t\in L\setminus\lbrace s\rbrace$. We see that $$d(s,t)=d(s-s,t-s)=d(0,t-s) \geq r.$$ Here we are critically using that $L$ is an additive subgroup, such that $t-s\in L$, so that the discreteness bound can be applied. This means no point in $L$ has distance less than $r$ to $s$, so $B(s,r)\cap L=\lbrace s\rbrace$. Thus, the $r$ we got from $0$ actually works universally, and so $L$ is uniformly discrete.

2. Let us do this in general, i.e. let $S\subset \mathbb{R}^n$ be any uniformly discrete space. Take any sequence $\lbrace a_n\rbrace_n\subset S$, which is convergent in $\mathbb{R}^n$. Then by definition, for any $r>0$ there exists $K\in\mathbb{N}$ such that for all $n,m>K$, $d(a_n,a_m)<r$. Picking the $r$ from the uniform discreteness of $S$, we get some $K$ such that the sequence satisfies the above property. But as $B(a_n,r)\cap S=\lbrace a_n\rbrace$, we see that all the following elements of the sequence must be $a_n$, i.e. the sequence is constant from some point on. Then its limit point is the constant point $a_n$, which lies in $S$. Thus $S$ contains all its limit points, and so it is closed.