Fundamental parallelotope and finite bounded intersections

By [F1] the vectors $b_1,\dots ,b_n$ form a real basis of ${\mathbb{R}}^n$; therefore every $x\in {\mathbb{R}}^n$ has a unique coefficient vector $y=(y_1,\dots ,y_n)\in {\mathbb{R}}^n$ with $x={\sum }_i{y}_i{b}_i$.

For every real $y$ there is exactly one pair $(m,t)\in \mathbb{Z}\times (0,1]$ with $y=m+t$: apply [F5] to $-y$ to get the unique integer $k$ with $k\le -y<k+1$, and put $m:=-k-1$, $t:=y-m$; then $m<y\le m+1$ and hence $t\in (0,1]$. Conversely if $m+t=m^{\mathrm{\prime }}+t^{\mathrm{\prime }}$ with $t,t^{\mathrm{\prime }}\in (0,1]$, then $m-m^{\mathrm{\prime }}=t^{\mathrm{\prime }}-t$ has absolute value $<1$ and is an integer, hence $m=m^{\mathrm{\prime }}$ and $t=t^{\mathrm{\prime }}$.

Let $S\subseteq {\mathbb{R}}^n$ be bounded and suppose first $S\mathrm{\ne }\mathrm{\varnothing }$. By [F7] there are $x_0\in {\mathbb{R}}^n$ and $r>0$ with $S\subseteq B(x_0,r)$, so every $s\in S$ satisfies $\mathrm{\parallel }s\mathrm{\parallel }\le \mathrm{\parallel }s-x_0\mathrm{\parallel }+\mathrm{\parallel }{x}_0\mathrm{\parallel }\le r+\mathrm{\parallel }{x}_0\mathrm{\parallel }=:M$.

(Tiling.) Let $x\in {\mathbb{R}}^n$ have coefficient vector $y$ as in step 1.1 and write $y_i=m_i+t_i$ as in step 1.2. Put $\lambda ={\sum }_i{m}_i{b}_i\in \mathrm{\Lambda }$ and $p={\sum }_i{t}_i{b}_i\in P$; then $x=\lambda +p$. For uniqueness, suppose $\lambda +p={\lambda }^{\mathrm{\prime }}+p^{\mathrm{\prime }}$ with $\lambda ={\sum }_i{m}_i{b}_i$, ${\lambda }^{\mathrm{\prime }}={\sum }_i{m}_i^{\mathrm{\prime }}{b}_i\in \mathrm{\Lambda }$ and $p={\sum }_i{t}_i{b}_i$, $p^{\mathrm{\prime }}={\sum }_i{t}_i^{\mathrm{\prime }}{b}_i\in P$. Then ${\sum }_i((m_i-m_i^{\mathrm{\prime }})+(t_i-t_i^{\mathrm{\prime }}))b_i=0$, and linear independence of the $b_i$ forces $(m_i-m_i^{\mathrm{\prime }})+(t_i-t_i^{\mathrm{\prime }})=0$ for every $i$. Here $m_i-m_i^{\mathrm{\prime }}$ is an integer and $t_i^{\mathrm{\prime }}-t_i\in (-1,1)$, so $m_i-m_i^{\mathrm{\prime }}=t_i^{\mathrm{\prime }}-t_i\in \mathbb{Z}\cap (-1,1)=\{0\}$; thus $m_i=m_i^{\mathrm{\prime }}$ and $t_i=t_i^{\mathrm{\prime }}$ for all $i$, that is $\lambda ={\lambda }^{\mathrm{\prime }}$ and $p=p^{\mathrm{\prime }}$.

Let $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ be the linear map $T(t)={\sum }_i{t}_i{b}_i$, with matrix $B$. By step 1.1 the map $T$ is a bijection, so the square matrix $B$ is invertible and [F2] gives $\det B\mathrm{\ne }0$. Since $T[(0,1{]}^n]=P$ and $(0,1{]}^n$ is Lebesgue measurable of measure $1$ by [F4], [F3] and [F1] give ${\lambda }_n(P)=\mathrm{\mid }\det B\mathrm{\mid }{\lambda }_n((0,1{]}^n)=\mathrm{covol}(\mathrm{\Lambda })$, with the Countable Choice hypotheses of [F3] and [F4] supplied by [A1].

For each $i$ the $i$-th coordinate functional $f_i(z):=(B^{-1}z{)}_i$ is linear, so by [F6] there is $K_i\ge 0$ with $\mathrm{\mid }{f}_i(z)\mathrm{\mid }\le K_i\mathrm{\parallel }z{\mathrm{\parallel }}_2$ for every $z$. If $\lambda ={\sum }_i{m}_i{b}_i\in S$, then $B^{-1}\lambda =(m_1,\dots ,m_n)$, so $f_i(\lambda )=m_i$ and step 1.3 gives $\mathrm{\mid }{m}_i\mathrm{\mid }\le K_{i}M=:C_i$.

Every integer $m_i$ with $\mathrm{\mid }{m}_i\mathrm{\mid }\le C_i$ satisfies $-\lceil C_i\rceil \le m_i\le \lceil C_i\rceil$; the set $\{m\in \mathbb{Z}:\mathrm{\mid }m\mathrm{\mid }\le C_i\}$ is therefore a subset of the finite set $\{-\lceil C_i\rceil ,\dots ,\lceil C_i\rceil \}$ and is finite. Hence $S\cap \mathrm{\Lambda }$ is contained in the image under $(m_1,\dots ,m_n)\mapsto {\sum }_i{m}_i{b}_i$ of the finite set ${\prod }_{i=1}^n\{m\in \mathbb{Z}:\mathrm{\mid }m\mathrm{\mid }\le C_i\}$, so $S\cap \mathrm{\Lambda }$ is finite; for $S=\mathrm{\varnothing }$ it is empty.

Step 2.1 proves the unique tiling, step 2.2 the volume ${\lambda }_n(P)=\mathrm{covol}(\mathrm{\Lambda })$, and step 3.1 the finiteness of $S\cap \mathrm{\Lambda }$ for bounded $S$; these are the three claims of the statement.