Given $\{v^i\}_{i \in \mathbb{N}} \subseteq \mathbb{N}^n$, and $\cup_{k=1, \ldots, m} C_j = \mathbb{N}^n$ for some $m$, where each $C_k$ is a cone generated by rational vectors. My question is: does there exist $i,j \in \mathbb{N}$ and $k=1, \ldots, m$, such that $v^i, v^j, v^j - v^i \in C_k?$
For example, when $m=1$, this statement is correct since it reduces to the Dickson's Lemma.
Yes.
The condition that the union of the $C_k$ is $\mathbb N^n$ is not strictly necessary - all you need is that each $v^i$ is in some $C_k$.
Indeed, there must be some $k$ such that infinitely many $v^i$ lie in $C_k$.
By Gordan's lemma, $C_k$ is generated (as a semigroup) by finitely many vectors in $C_k$. Call these $w_1,\dots, w_r$. Write $v^i= \sum_{t=1}^r a_{i,t}w_t$ for all $i$ such that $v^i \in C_k$ - we may have multiple choices of $a_{i,t}$, but just pick one.
Now apply Dickson's lemma to the $a_{i,t}$, obtaining $i$ and $j$ with $v^i\in C_k, v^j \in C_k$, and $a_{i,t} \leq a_{j,t}$ for all $t$. It follows that $v^j - v^i = \sum_{t=1}^r (a_{j,t} - a_{i,t}) w-t \in C_k$, as desired.
