Proof of fibers of lattice are finite By lattice I mean a subgroup of the additive group $\mathbb{Z}^k$. The lattice $\mathcal{L}$ shall have the property that the only non-negative vector in $\mathcal{L}$ is the origin, i.e.
\begin{align*}
\mathcal{L} \cap \mathbb{N}^k = \{0\}.
\end{align*}
With this prerequisite, I would like to show that the fiber of any point $u\in \mathbb{N}^k$ is a finite set. Here, by the fiber of $u$, I mean the set of all non-negative vectoirs in the same residue class modulo $\mathcal{L}$:
\begin{align*}
\mathcal{F}(u):= (u+\mathcal{L})\cap \mathbb{N}^k = \{ v\in \mathbb{N}^k\,|\, u-v \in \mathcal{L}\}. 
\end{align*}
I took this statement from Section 1.3 in Lectures on Algebraic Statistics and I would be thankful for any hint. For the beginning it would be enough to show the claim for the lattice $\ker_{\mathbb{Z}}A$ for a non-negative integer matrix $A$.  
Unfortunately, I do not know even where to begin. I tried starting with "Suppose the fiber is not finite. Then we have to show that $\mathcal{L}$ contains a non-negative vector that is not the origin." but I don't know how to continue.
 A: SKETCH: You can prove it by induction on $k$. Suppose that $k$ is minimal such that the result fails, and that $u=\langle u_1,\ldots,u_k\rangle\in\Bbb N^k$ has an infinite fibre. Then there are a $j\in\{1,\ldots,k\}$ and an infinite sequence $\left\langle v^{(n)}:n\in\Bbb N\right\rangle$ in $\mathscr{L}$ such that if 
$$w^{(n)}=u+v^{(n)}=\left\langle w_1^{(n)},\ldots,w_k^{(n)}\right\rangle$$ 
for each $n\in\Bbb N$, then each $w^{(n)}\in\Bbb N^k$, and the sequence $\left\langle w_j^{(n)}:n\in\Bbb N\right\rangle$ is strictly increasing. This implies that $v_j^{(n)}>0$ for all sufficiently large $n$, so we might as well assume that $v_j^{(n)}>0$ for all $n\in\Bbb N$. Clearly this implies that $k>1$. 
Now let 
$$\pi_j:\Bbb N^k\to\Bbb N^{k-1}:\langle x_1,\ldots,x_k\rangle\mapsto\langle x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_k\rangle\;,$$
and let $\mathscr{L}'=\pi_j[\mathscr{L}]$. $\mathscr{L}'$ is a lattice in $\Bbb N^{k-1}$ whose only completely non-negative vector is the origin. Let $u'=\pi_j(u)$; by hypothesis $\mathscr{F}(u')$ is finite. It’s not hard to see that $\pi_j\left(w^{(n)}\right)\in\mathscr{F}(u')$ for each $n\in\Bbb N$, so there are a $v'\in\mathscr{L}'$ and a subsequence $\left\langle v^{(n_\ell)}:\ell\in\Bbb N\right\rangle$ such that $\pi_j\left(v^{(n_\ell)}\right)=v'$ for each $\ell\in\Bbb N$. But then
$$\left\{v^{(n_\ell)}-v^{(n_0)}:\ell\in\Bbb N\right\}$$
is an infinite subset of $\mathscr{L}\cap\Bbb N^k$, which is the desired contradition.
A: Assume that some fiber contains infinitely many points. By Dickson's lemma, that fiber only has finitely many minimal elements. Then for a non minimal element $u$ in the fiber there is a minimal element $v$ that divides it and $u-v$ is a strictly positive lattice vector, a contradiction. 
A: Alternatively, since there is a strictly positive vector in the rowspan of the matrix (in algebraic statistics usually the unit vector) you can see that any fiber is contained in the set of integer points in the intersection of a hyperplane perpendicular to that vector with the positive quadrant. That set is finite. 
