Discrete subgroups of a real vector space are lattices

Suppose first that $\mathrm{\Gamma }$ is discrete. Then $\{0\}$ is open in the subspace topology on $\mathrm{\Gamma }$, so $\{0\}=\mathrm{\Gamma }\cap U$ for some open $U\subseteq V$; choosing a ball around $0$ inside $U$ gives an $\epsilon >0$ with $\mathrm{\Gamma }\cap B(0,\epsilon )=\{0\}$.

Suppose next that (b) holds. The set $\mathrm{\Gamma }\cap B(0,1)$ is then finite; if it is $\{0\}$ take $\epsilon =1$, and otherwise let $\delta :=\min \{\parallel \gamma \parallel :0\mathrm{\ne }\gamma \in \mathrm{\Gamma }\cap B(0,1)\}>0$, a minimum of a nonempty finite set of positive reals, so that again $\mathrm{\Gamma }\cap B(0,\delta )=\{0\}$. Since balls are translation invariant in the metric of a norm, $B(\gamma ,\delta )=\gamma +B(0,\delta )$ for every $\gamma \in \mathrm{\Gamma }$, so $\mathrm{\Gamma }\cap B(\gamma ,\delta )=\{\gamma \}$: every point of $\mathrm{\Gamma }$ is isolated in $\mathrm{\Gamma }$, that is, $\mathrm{\Gamma }$ is discrete. Hence (b) implies (a).

Assume (b) from here on. By [F1] the lengths of the $\mathbb{R}$-linearly independent finite tuples of elements of $\mathrm{\Gamma }$ form a nonempty subset of $\{0,1,\dots ,n\}$, so a maximum $r$ exists; select an $\mathbb{R}$-linearly independent tuple $v_1,\dots ,v_r\in \mathrm{\Gamma }$, and put $W:={\mathrm{span}}_{\mathbb{R}}(v_1,\dots ,v_r)$, so ${\dim }_{\mathbb{R}}W=r$.

Put ${\mathrm{\Gamma }}_0:=\mathbb{Z}{v}_1\oplus \cdots \oplus \mathbb{Z}{v}_r\subseteq \mathrm{\Gamma }$ and $P:=\{{\sum }_{i=1}^r{t}_i{v}_i:0\le t_i<1\}$. Since $\parallel {\sum }_i{t}_i{v}_i\parallel \le {\sum }_i\mid t_i\mid \parallel v_i\parallel \le {\sum }_i\parallel v_i\parallel$, the set $P$ is bounded, so $F:=\mathrm{\Gamma }\cap P$ is finite by (b).

This proves (a)$\Rightarrow$(b) without selecting a sequence from a bounded set. Fix a basis $e_1,\dots ,e_n$ of $V$ and put $S:={\sum }_{i=1}^n\parallel e_i\parallel >0$. Let $B\subseteq V$ be bounded; if $B=\mathrm{\varnothing }$ the conclusion is immediate. Otherwise choose a ball $B(x_0,R)$ containing it. Let $a_i$ be the coordinates of $x_0$. By [F3] there is $c>0$ with $\parallel v\parallel \ge c\parallel (v_i){\parallel }_{\mathrm{\infty }}$ in these coordinates, so every $x\in B$ satisfies ${\max }_i\mathrm{\mid }{x}_i-a_i\mathrm{\mid }<R\mathrm{/}c$. Set $M:=\max \{1,R\mathrm{/}c\}$; then the coordinate vectors of $B$ lie in the box ${\prod }_i[a_i-M,a_i+M]$. Set $\delta :=\epsilon \mathrm{/}(2S)>0$ and choose an integer $q\ge 1$ with $1\mathrm{/}q<\delta \mathrm{/}(2M)$ by [F6]. Divide each coordinate interval into $q$ equal subintervals and take their finitely many product cells. In one cell, any two coordinate vectors differ by less than $\delta$ in each coordinate, so the corresponding points $x,y$ satisfy $\parallel x-y\parallel \le {\sum }_i\mathrm{\mid }{x}_i-y_i\mathrm{\mid }\parallel e_i\parallel <\delta S=\epsilon \mathrm{/}2<\epsilon$. By step 1.1, each cell therefore contains at most one point of $\mathrm{\Gamma }$. The finite collection of cells covers $B$, so $B\cap \mathrm{\Gamma }$ is finite.

If $\gamma \in \mathrm{\Gamma }\setminus W$, then any relation ${\lambda }_1{v}_1+\cdots +{\lambda }_r{v}_r+\mu \gamma =0$ must have $\mu =0$, since otherwise it would express $\gamma$ as an element of $W$. The independence of $v_1,\dots ,v_r$ then forces every ${\lambda }_i=0$, so adjoining $\gamma$ would give $r+1$ independent elements of $\mathrm{\Gamma }$, contradicting maximality in step 1.3. Thus $\mathrm{\Gamma }\subseteq W$, and since the $v_i$ lie in $\mathrm{\Gamma }$, ${\mathrm{span}}_{\mathbb{R}}\mathrm{\Gamma }=W$.

Every $\gamma \in \mathrm{\Gamma }$ differs from an element of ${\mathrm{\Gamma }}_0$ by an element of $F$: by step 2.2 write $\gamma ={\sum }_i{s}_i{v}_i$ with $s_i\in \mathbb{R}$ and write $s_i=m_i+t_i$ with $m_i\in \mathbb{Z}$ and $0\le t_i<1$; then $\gamma -{\sum }_i{m}_i{v}_i\in \mathrm{\Gamma }\cap P=F$.

The map $F\to \mathrm{\Gamma }\mathrm{/}{\mathrm{\Gamma }}_0$, $f\mapsto f+{\mathrm{\Gamma }}_0$, is surjective by step 3.1. Thus $\mathrm{\Gamma }\mathrm{/}{\mathrm{\Gamma }}_0$ is finite; let its order be $N\ge 1$. By [F5], $N\gamma \in {\mathrm{\Gamma }}_0$ for every $\gamma \in \mathrm{\Gamma }$. The map $T:\mathrm{\Gamma }\to {\mathrm{\Gamma }}_0$, $T(\gamma )=N\gamma$, is an injective homomorphism: if $N\gamma =0$, then $\gamma =0$ because $V$ is a real vector space and $N>0$. Consequently its image is a subgroup of ${\mathrm{\Gamma }}_0\cong {\mathbb{Z}}^r$.

By step 4.1, $T(\mathrm{\Gamma })$ is a subgroup of ${\mathrm{\Gamma }}_0\cong {\mathbb{Z}}^r$. For this use, every subgroup $H\le {\mathbb{Z}}^m$ has a finite $\mathbb{Z}$-basis of length at most $m$, by induction on $m$. For $m=0$ the subgroup is zero and the empty list is a basis. For $m\ge 1$, project $H$ onto its first coordinate. By [F4] the image is $d\mathbb{Z}$ for some $d\ge 0$. If $d=0$, identify $H$ with a subgroup of the last $m-1$ coordinates and apply induction. If $d>0$, choose $h\in H$ with first coordinate $d$; the kernel $H_0$ of that projection is a subgroup of ${\mathbb{Z}}^{m-1}$, so induction gives it a basis of length at most $m-1$. Every $x\in H$ has first coordinate $kd$ for some $k\in \mathbb{Z}$; then $x-kh\in H_0$, so $h$ together with a basis of $H_0$ generates $H$. They are independent because projecting any integer relation to the first coordinate forces the coefficient of $h$ to be zero, after which independence in $H_0$ forces all remaining coefficients to vanish. This proves the claim, including that the basis has at most $m$ elements.

Apply step 5.1 to $T(\mathrm{\Gamma })\le {\mathrm{\Gamma }}_0\cong {\mathbb{Z}}^r$ and pull its $\mathbb{Z}$-basis back through the isomorphism $T:\mathrm{\Gamma }\to T(\mathrm{\Gamma })$. This gives a $\mathbb{Z}$-basis $b_1,\dots ,b_s$ of $\mathrm{\Gamma }$ with $s\le r$. By step 2.2, $v_1,\dots ,v_r$ are linearly independent and span $W$, while ${\mathrm{span}}_{\mathbb{R}}(b_1,\dots ,b_s)=W$ because they generate $\mathrm{\Gamma }$. The finite-dimensional independent-set bound [F1] gives $r\le s$, so $s=r$. If $r>0$ and these $r$ spanning vectors were linearly dependent, one could remove a vector and still span $W$, contradicting [F1] applied to $v_1,\dots ,v_r$; when $r=0$, the empty list is independent. Hence they are $\mathbb{R}$-linearly independent and $\mathrm{\Gamma }=\mathbb{Z}{b}_1\oplus \cdots \oplus \mathbb{Z}{b}_r$, proving (c).

Finally assume (c): $\mathrm{\Gamma }=\mathbb{Z}{v}_1\oplus \cdots \oplus \mathbb{Z}{v}_r$ with $v_1,\dots ,v_r$ $\mathbb{R}$-linearly independent. Extend this tuple to a basis $v_1,\dots ,v_r,w_1,\dots ,w_{n-r}$ of $V$ and let $\phi :{\mathbb{R}}^n\to V$ send the standard basis to it. The function $t\mapsto \parallel \phi (t)\parallel$ is a norm on ${\mathbb{R}}^n$, hence equivalent to the coordinate norm $\mid t{\mid }_{\mathrm{\infty }}={\max }_i\mid t_i\mid$ by [F3], so there is $c>0$ with $\parallel \phi (t)\parallel \ge c\mid t{\mid }_{\mathrm{\infty }}$ for all $t$. A nonzero element of $\mathrm{\Gamma }$ has coordinates $(m_1,\dots ,m_r,0,\dots ,0)$ with some $m_i\in \mathbb{Z}\setminus \{0\}$, so $\mid m{\mid }_{\mathrm{\infty }}\ge 1$ and $\parallel \gamma \parallel \ge c$; thus $\mathrm{\Gamma }\cap B(0,c)=\{0\}$ and $\mathrm{\Gamma }$ is discrete, proving (a). This closes the cycle (a)$\Rightarrow$(b)$\Rightarrow$(c)$\Rightarrow$(a).